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1
! Deflected Shape of Structures! Method of Consistent Deformations! Maxwell�s Theorem of Reciprocal
Displacement
INTRODUCTION: STRUCTURAL ANALYSIS
2
-M
P
+M
-M
Deflection Diagrams and the Elastic Curve
∆ = 0θ = 0
fixed support
∆ = 0
roller or rockersupport
P
θ
inflection point
3
-M
P
∆ = 0
pined supportθ
4
Moment diagram
P
inflection point
inflection point
� Fixed-connected joint
fixed-connected joint
5
P
� Pined-connected jointpined-connected
joint
Moment diagram
6
P
inflection point
Moment diagram
7
-M
+Mx
M
P2
P1
A B C D
inflection point
8
x-M
+M
P1
P2
inflection point
9
+
AB
C
AB
C
P
=
AB
C
P
∆´B
fBB x RB
1∆´B + fBB RB = ∆B = 0
Method of Consistent DeformationsBeam 1 DOF
RBAy
Ax = 0
MA
10
w
w
=+
+
=∆1
∆2
∆´1
∆´2
R1
R2
×R2
Compatibility Equations.
∆´1 + f11R1 + f12R2 = ∆1 = 0
×R1
f12f22
0
0
Beam 2 DOF
∆´2 + f21R1 + f22R2 = ∆2 = 0
1
1
14
3
52
∆´1 ∆´2
f11 f21
+f11 f12
f12 f22
11
wP1 P2
=
wP1 P2
++
+1
1
f32f22
1
Compatibility Equations.
θ´1 + f11M1 + f12R2 + f13R3 = θ1 = 0
×R2
×R3
=
θ1
∆2
∆3
f31
f33f23f13
24
61
53
∆´2 θ´1∆´3
Beam 3 DOF
∆´2 + f21M1 + f22R2 + f23R3 = ∆2 = 0
∆´3 + f31M1 + f32R2 + f33R3 = ∆3 = 0
×M1
θ´1
∆´2
∆´3
+
f11 f12 f13
f21 f22 f23
f31 f32 f33
M1
R2
R3
0
0
0
f12
f21f11
12
If ∆1 = ∆2=��= ∆n = 0 ;
or [∆´] + [f][R] = 0
[R] = - [f]-1[∆´]
Compatibility Equation for n span.
∆´1 + f11R1 + f12R2 + f13R3 = ∆ 1∆´2 + f21R1 + f22R2 + f23R3 = ∆2
∆´n + fn1R1 + fn2R2 + fnnRn = ∆n
∆ ´1
∆´2 +
f11 f12 R1
R2 =
0
0
∆´ n
f13
f21 f22 f23
f n 1 f n 2 fn n R n 0
[fij] = Flexibility matrix dependent on
Equilibrium Equations
[Q] = [K][D] + [Qf]
Stiffness matrix
Fixed-end force matrix
Solve for displacement [D];
[D] = [K]-1[Q] - [Qf]
GJEAEI1,1,1
ijij f
k 1=
13
A B
A B
EI
1=i 2=j
Mj = mj
f12 = fij
f11 = fii
f22 = fjj
Mi = mi
Maxwell�s Theorem of reciprocal displacements
1
1
f21 = fji
∫=EIdxMmf jiij
∫=EIdxmm ji
∫=EIdxMmf ijji
∫=EIdxmm ij
jiij ff =
14
Example 1
Determine the reaction at all supports and the displacement at C.
AB
C
50 kN
6 m 6 m
15
50 kN
=∆´B
+
fBB
1 kN
x RB
AB
C
50 kN
6 m 6 m
SOLUTION
RB
MA
RA
0' =+∆ BBBB Rf -----(1)Compatibility equation :
� Principle of superposition
∆´B
x RBfBB
16
Conjugate beam
Conjugate beam
50 kN
∆´B
Real beam
50 kN
300 kN�m
6 m 6 m
/EI300 900/EI
6 + (2/3)6 = 10 m9000/EI
� Use conjugate beam in obtaining ∆∆∆∆´B and fBB
900/EI
12 /EI72/EI
(2/3)12 = 8 m576/EI
fBB
1 kN
Real beam
1 kN
12 kN�m
72/EI
∆´B = M´B = -9000/EI ,
fBB = M´´B = 576/EI,
BA
17
RB = +15.63 kN, (same direction as 1 kN)
0)576(9000: =+−↑+ BREIEI
50 kN
∆´B
50 kN
300 kN�m
x RB = 15.63 kN=
AB
C
50 kN
15.63 kN34.37 kN34.37 kN�m
fBB
1 kN1 kN
12 kN�m
+
� Substitute ∆´B and fBB in Eq. (1)
18
93.6/EI
-113/EI
6 m 6 m
AB
C
50 kN
Use conjugate beam in obtaining the displacement
113 kN�m
34.4 kN15.6 kN
EIEIEIM C
776)6(223)2(281' −=−=
↓−==∆ ,776'EI
M CC
Real Beam
Conjugate Beam
M (kN�m) x (m)
3.28 m 6 m 12 m
93.6
-113
223/(EI)
4 m2 m
M´C
V´C 223/(EI)
281/(EI)
∆C
19
10 kN
AC
B
4 m
3EI 2EI
2 m 2 m
Example 2
Determine the reaction at all supports and the displacement at C. Take E = 200GPa and I = 5(106) mm4
20
+
10 kN
=
1 kN
Compatibility equation: ∆´B + fBBRB = ∆B = 0
x RB
∆´B
fBB
10 kN
AC
B
4 m
3EI 2EI
2 m 2 m
21
Conjugate Beam
10 kN
AC
B
4 m
3EI 2EI
2 m 2 m
Real Beam
10 kN
40 kN�m
� Use conjugate beam in obtaining ∆´B
26.66/EI
177.7/EI
x (m)
V (kN) 10 10
+
x (m) M (kN�m)
40
-
40/3EI = 13.33/EI∆´B = M´B = 177.7/EI
22Conjugate Beam
AC
B
4 m
3EI 2EI
2 m 2 m
Real Beam1 kN
4/(2EI)=2EI
� Use conjugate beam for fBB
1 kN
8 kN�m
x (m)V (kN)-
-1 -1
x (m) M (kN�m)
84
+
83EI= 2.67
EI4/(3EI)=1.33EI
60.44/EI
12/EIfBB = M´B = 60.44/EI
23
+
10 kN
=
1 kNx RB
∆´B
fBB
10 kN
AC
B
4 m
3EI 2EI
2 m 2 m
Compatibility. equation: ∆´B + fBBRB = ∆B = 0
RB = +2.941 kN, (same direction as 1 kN)
0)44.60(7.177: =+−
−↑+ BREIEI
24
10 kN
AC
B
4 m
3EI 2EI
2 m 2 m 2.94 kN7.06 kN
16.48 kN�m
7.06
-2.94 -2.94
+- x (m)V (kN)
∆C
-16.48
11.76
-
+ x (m) M (kN�m)
2.33 m
1.67 m
� The quantitative shear and moment diagram and the qualitative deflected curve
25
10 kN
AC
B
4 m
3EI 2EI
2 m 2 m
↓−
=−= ,85.18)222.3(413.6)555.0(263.3EIEIEI
2.941 kN7.059 kN
16.48 kN�m
∆C
� Use the conjugate beam for find ∆∆∆∆C
Real beam
↓−=×
−==∆ ,85.18
)5200(85.18' mmM CC
16.483EI
11.763EI
11.762EI
2.335 m
1.665 m
Conjugate beam
6.413EI
3.263EI
(1.665)/3=0.555 m
1.665+(2/3)(2.335) = 3.222 m
M´C
26
Example 3
Draw the quantitative Shear and moment diagram and the qualitative deflectedcurve for the beam shown below.EI is constant. Neglect the effects of axial load.
A B4 m4 m
5 kN/m
27
αBBαAB
αBAαAA
θ´Bθ´Aθ´A
1 kN�m
5 kN/m
1 kN�m
A B4 m4 m
5 kN/m
SOLUTION
� Principle of superposition
θ´B
=+
αAA αBA
AM×
αBB
BM×αAB
+
Compatibility equations:
θA θB
AM×
BM×
AM×
BM×
)1('0 −−−++== BABAAAAA MfMfθθ
)2('0 −−−++== BBBABABB MfMfθθ
28
� Use formula provided in obtaining θ´A, θ´B, αAA, αBA, αBB, αAB
Note: Maxwell�s theorem of reciprocal displacement, αAB = αBA
1 kN�m
αAA αBA8 m
1 kN�m
αBBαAB
8 m
EIEIEILM o
AA667.2
3)8(1
3===α
EIEIEILM o
BA333.1
6)8(1
6===α
EIEIEILM o
BB667.2
3)8(1
3===α
EIEIEIwL
B67.46
384)8)(5(7
3847'
33
===θ
EIEIEIwL
A60
128)8)(5(3
1283'
33
===θ
4 m4 mθ´Bθ´Aθ´A
5 kN/m
θ´B
EIEIEILM o
AB333.1
6)8(1
6===α
29
αBBαAB
Real Beam1 kN�m
αBBαAB
8 m
αBA
Real Beam1 kN�m
αAA αBAαAA
8 m
Conjugate Beam
4/EI1/EI
(1/8)(1/8)
2.67/EI1.33/EI
(1/8) (1/8)
Conjugate Beam
1/EI4/EI
1.33/EI2.67/EI
EIV AAA
667.2' −==α
EIV BBB
667.2' ==α
EIV AAB
333.1' −==α
EIV BBA
333.1' ==α
� Use conjugate beam for αAA, αBA, αBB, αAB
30
1 kN�m
5 kN/m
1 kN�m
A B4 m4 m
5 kN/m
=+
+
θA θB
BM×
EIAA667.2
=α
AM×
EIV BBA
333.1' ==α
EIB67.46' =θ
EIA60' =θ
EIAB333.1
=αEIBB667.2
=α
Solve simultaneous equations,
MA = -18.33 kN�m, +
MB = -8.335 kN�m, +
Compatibility equation
+ 0)333.1()667.2(60=++ BA M
EIM
EIEI
+ 0)667.2()333.1(67.46=++ BA M
EIM
EIEI
31
MA = -18.33 kN�m,
MB = -8.335 kN�m,
A B4 m4 m
5 kN/m
18.33 kN�m 8.335 kN�m
RBRA
RB = 3.753 kN,
Ra = 16.25 kN,
+ ΣMA = 0: 0355.8)8()2(2033.18 =−+− BR
ΣFy = 0:+3.753
020 =−+ BA RR
32
A B4 m4 m
5 kN/m
18.33 kN�m 8.36 kN�m
3.75 kN16.25 kN
Vdiagram
Mdiagram
DeflectedCurve
-3.75
16.25
3.25 m
-18.33
6.67
-8.36
8.08
� Quantitative shear and bending diagram and qualitative deflected curve
33
Example 4
Determine the reactions at the supports for the beam shown and draw thequantitative shear and moment diagram and the qualitative deflected curve.EI is constant.
A
C
4 m4 m
2 kN/m
B
34
1 kN
)2(0''')1(0'''
−−−=++∆−−−=++∆
CCCBBCC
CCBBBBB
RfRfRfRf
Compatibility equations:
� Principle of superposition
A
C
4 m4 m
2 kN/m
B
4 m4 m BA
C
2 kN/m
B'∆C'∆
A
C
4 m4 m B
BBf ' CBf '
1 kNAC4 m4 m B
BR×
CR×BCf ' CCf '
35A
C
A
C
B
A
C
2 kN/m
1 kN
� Solve equation
A
C
4 m4 m
2 kN/m
B
1 kN
EIB64' −=∆ EIC
33.149' −=∆
EIf BB
33.21' = EIf CB
33.53' =
EIf BC
33.53' = EIf CC
67.170' =
)1(033.5333.2164−−=++− CB R
EIR
EIEI
Compatibility equations:
)2(067.17033.5333.149−−=++− CB R
EIR
EIEI
↓−=
↑=
,29.0,71.3
kNRkNR
C
B
BR×
CR×
36
� Diagram
A
C
4 m4 m
2 kN/m
B
3.71 kN 0.29 kNkNAy 58.471.329.08 =−+=
mkNM A
•=−+=
48.3)4(71.3)8(29.0)2(8
x (m)
V (kN)4.58
0.29
-3.422.29 m
x (m)
M (kN�m)
-3.48
1.76
-1.16
x (m)Deflected shape
Point of inflection
37
Example 5
Draw the quantitative Shear and moment diagram and the qualitative deflectedcurve for the beam shown below.(a) The support at B does not settle(b) The support at B settles 5 mm.Take E = 200 GPa, I = 60(106) mm4.
16 kN
2 m 2 m 4 m
AB C
38
16 kN
2 m 2 m 4 m
AB C
16 kN
SOLUTION
� Principle of superposition
∆´B
1 kN
fBB
BR×
∆´B
fBB
BR×=
+
∆B = 5 mm
Compatibility equation : BBBBB Rf+∆==∆ '0 -----(1) : no settlement+
BBBBB Rfm +∆=−=∆ '005.0 -----(2) : with settlement+
39
16 kN
∆´B
Real beam
A B C
� Use conjugate beam method in obtaining ∆´B
2 m 2 m 4 m12 kN
M´diagram
16
Conjugatebeam
EI24
EI24
EI72
34
32
2 m 4 m
24
EI40
EI56
4 kN
0)4(40)34(32'' =−+−
EIEIM B
EI16
EI32
EI40
V´´B
M´´B
34 4
32
+ ΣMB = 0:
↓−==∆ ,33.117'''EI
M BB
∆´B
Real beam
40
Real beam
1 kN4 m 4 m
AB
CfBB
� Use conjugate beam method in obtaining fBB
0.5 kN 0.5 kN
m´diagram
-2Conjugate
beam
EI2
−EI4
−EI4
− EI4
EI4
v´´B
m´´B
EI4
− EI4
34 4
32
0)4(4)34(4'' =+−−
EIEIm B
+ ΣMB = 0:
↑== ,67.10''EI
mf BBB
fBB
41
� Substitute ∆´B and fBB in Eq. (1)
,67.1033.1170; BREIEI
+−=↑+ RB = 11.0 kN,
xRB = 11.0 kN
=16 kN
AB C
11.0 kN= 6.5 kN = 1.5 kNRCRA
no settlement
↓−=∆ ,33.117'EIB
↑= ,67.10EI
fBB
16 kN
12 kN 4 kN∆´B 1 kN
+
0.5 kN 0.5 kN
fBB
42
� Substitute ∆´B and fBB in Eq. (2)
BREIEI
m 67.1033.117005.0; +−=−↑+
BREIm 67.1033.117)005.0( +−=−
RB = 5.37 kN,
BR67.1033.117)60200)(005.0( +−=×−
xRB = 5.37
=16 kN
AB C
5.37 kN= 9.31 kN = 1.32 kNRCRA
with 5 mm settlement
↓−=∆ ,33.117'EIB
↑= ,67.10EI
fBB
16 kN
12 kN 4 kN∆´B 1 kN+
0.5 kN 0.5 kN
fBB
43
� Quantitative shear and bending diagram and qualitative deflected curve
16 kN
AB C
2 m 2 m 4 m11 kN 6.5 kN 1.5 kN
Vdiagram
6.5
-9.5
1.5
Mdiagram
13
-6
+-
DeflectedCurve
16 kN
∆B = 5 mmA
B C
2 m 2 m 4 m5.37 kN 9.31 kN 1.32 kN
Vdiagram
9.31
-6.69-1.32
Mdiagram
18.625.24+
DeflectedCurve ∆B = 5 mm
44
Example 6
Calculate supports reactions and draw the bending moment diagrams for (a)D2 = 0 and (b) D2 = 2 mm.
4 m 6 m
w 2 kN/m1 2 3
45
+
2 kN/m
=
Compatibility equation:
∆´2 + f22R2 = ∆2
∆´2
1
f22
×R2
4 m 6 m
w 2 kN/m1 2 3
↓−=−=
+××−−=
+−−=∆
,2.6)200)(200(
248
)1041024(24
)4)(2(
)2(24
'
:
323
3232
mm
EI
LLxxEI
wxprovidednsformulatioUse
↑+==
−−×
××−=
−−=
,48.0)200)(200(
2.19
)4610(106
461
)(6
222
22222
mm
EI
xbLLEIPbxf
46
4 m 6 m
w 2 kN/m1 2 3
12.92 kN4.83 kN2.25 kN
Compatibility.equation:
-6.2 + 0.48R2 = -2R2 = 8.75 kN
4 m 6 m
w 2 kN/m1 2 3
For ∆2 = 2 mmFor ∆2 = 0
Compatibility equation:
-6.2 + 0.48R2 = 0 R2 = 12.92 kN
x (m) V (kN)
2.25
-5.75
7.17
-4.83
+--
1.125 m
2.415 m
x (m) V (kN)
2.375 m
3.25 m
-++
-
4.75 5.5
-3.25-6.5
x (m) M (kN�m)
1.27
-7
5.85
+-
+
x (m) M (kN�m)
5.64
3
10.56
++
6.5 kN4.75 kN 8.75 kN
47
L
w
Basic Beams: Single span
wL/2
- wL/2
V
wL2/8
M
APPENDIX
wL/2wL/2L/3L/3
1
EIwL
3845 4
48
� Find ∆∆∆∆1 by Castigliano�s
w
L
L/2 L/2
x1 x2
1 2 3P
22PwL
+22PwL
+
22PwL
+x2
w
V2
M2
22PwL
+x1 V1
M1w
+ ΣΜ = 0;
21
21
1
1
21
1
)22
(2
0)22
(2
MxPwLwxM
xPwLwxM
=++−=
=+−+
49
0
∫∫ ∂∂
+∂
∂=∆=∆
2/
0
22
22/
0
11
1max1 )()(
LL
EIdxM
PM
EIdxM
PM
∫ ∂∂
=2/
0
11
1 )(2L
EIdxM
PM
21
21
1 )22
(2
MxPwLwxM =++−=
11
21
2/
0
1 ))22
(2
()2
(2 dxxPwLwxxEI
L
++−
= ∫
EIwL
3845 4
=
50
L
L /2 L /2
P
1
P/2 P/2
V
P/2
- P/2
+
-
PL/4
M
EILfPf
EIPL
4848
3
1111
3
11 =→==δ
Single span
51
L = 2l
l = L/2 l = L/2
1w
1.25 wl0.375 wl 0.375 wl
0.375 wl
-0.375 wl-0.625 wl
0.625 wl
+-
+-
0.070 wl2 0.070 wl2
-0.125 wl2
+ +
-
EIwl
185
4
max =∆
EIwl
185
4
max =∆
Double span
52
l = L/3 l = L/3 l = L/3
1 w 2
1.1wl1.1wl0.4wl 0.4wl
∆max = wl4
145 EI
V0.4wl
- 0.4wl- 0.6wl
0.5wl
-0.5wl
0.6wl
+
-
+-
+
-
0.08 wl2 0.08 wl2
0.025 wl2
-0.1 wl2 -0.1 wl2
+
- -++M
Triple span
1
! Comparison Between Indeterminate andDeterminate
! Influence line for Statically IndeterminateBeams
! Qualitative Influence Lines for Frames
INFLUENCE LINES FOR STATICALLY INDETERMINATE BEAMS
2
AB CED
RA AB CED
RA
Indeterminate Determinate
Comparison between Indeterminate and Determinate
11
3
AB CED
RAA
B CEDRA
AB CED
ME AB CED
ME
AB CED
VD AB CED
VD
1 1
1
Indeterminate Determinate
11
1
4
f1j
fjj
∆´1 = f1j
+
11 2 3j4
Redundant R1 applied
1
1
=
f11
fj1
×R1×R1
f11
Influence Lines for Reaction
Compatibility equation:
011111 =∆=+ Rff j
)1(11
11 ffR j−=
)(11
11 f
fR j=
5
11 2 3j4
1
=+
×R2
Redundant R2 applied
fjjf2j
1
fj2
f22
Compatibility equation.
0' 22222 =∆=+∆ Rf
022222 =∆=+ Rff j
)1(22
22 ffR j−=
)(22
22 f
fR j=
6
1 2 3j4
fj4
1
1
Influence Lines for Shear
444
)1( jE ff
V =
f44
1
1
7
444
)1( jE fMα
=
1 2 3j4
Influence Lines for Bending Moment
1 1
α44
fj4
1 1
8
R3 )(33
33 f
fR j=
R2 )(22
22 f
fR j=
R1)(
11
11 f
fR j=
1 2 3j4
1
1
� Influence line of Reaction
1
Using Equilibrium Condition for Shear and Bending Moment
111
11 =ff
11
41
ff
11
1
ff j
122
22 =ff
22
2
ff j
22
42
ff
133
33 =ff
33
3
ff j
33
43
ff
9
1 2 3j4
4V4
V4 = R1
R1V4
M41
� Unit load to the right of 4
� Influence line of Shear
V4 = R1
V4 = R1 - 1
1
R2 R3R1
1
1
1
R1
1x
V4
M41
� Unit load to the left of 4
V4 = R1 - 1
01;0 41 =−−=Σ↑+ VRFy
1R1
0;0 41 =−=Σ↑+ VRFy
10
1 2 3j4
� Influence line of Bending moment
M4 = - l + x + l R1
M4 - 1 (l-x) - l R1 = 0 + Σ M4 = 0:
R1
1x
V4
M4l1
� Unit load to the left of 4
M4 = l R1
R1V4
M4
� Unit load to right of
l
4
1
1
R2 R3R1
1
l1 l2
l
M4
4
1R1
1 1M4 = - l + x + l R1
M4 - l R1 = 0 + Σ M4 = 0:
M4 = lR1
11
Influence Line of VI
Maximum positive shear Maximum negative shear
Qualitative Influence Lines for Frames
I
1
1
12
Influence Line of MI
Maximum positive moment Maximum negative moment
I
11
13
A D G
15 m15 mAy GyDy
MA
Dy
1.0
Ay
1.0
Gy
1.0
Influence Line for MOF
14
MH
MA
1
1
A H G
15 m15 m
D
15
RA
1.0
RB
1.0
MB
1
MG
1
A E
G
B C D
16
VG1
VF1
VH1
A E
G
B C D
17
Example 1
Draw the influence line for- the vertical reaction at A and B- shear at C- bending moment at A and C
EI is constant . Plot numerical values every 2 m.
A BC D
2 m 2 m 2 m
18
BBf BBf BBf 1=BBf
� Influence line of RB
1
A BC D
2 m 2 m 2 m
ABf CBf DBfBBf
19
Real BeamA BC D
2 m 2 m 2 m
Conjugate Beam
� Find fxB by conjugate beam
11
6 kN�m
x
=−+EI
xEIEI
x 18726
3
EI6
EI18
EI18
EI72
3x
32xV´x
M´x
x
EI72
EI18
EIx
2
2
EIx
20
x (m)
0
2
4
6
Point
B
D
C
A
x
fxB / fBB
1
0.518
0.148
0
A BC D
2 m 2 m 2 m
1
fBB
fxB
0
72/EI37.33/EI10.67/EI
EIx
EIEIxMf xxB
18726
'3
−+==
EI72
EI33.37
EI67.10
0
/72 = 0.518
/72 = 0.148
1
37.33
10.670
Influence line of RB
17272
==BB
BB
ff
fxB
21
1 kN
Influence line of RA
A BC D
2 m 2 m 2 m
AA
CA
ff
AA
DA
ff
1=AA
AA
ff
22
Conjugate Beam
xV´x
M´x
EIB y
18' =
EIx
2
2
EIx
3x
32x
A BC D
2 m 2 m 2 m
Real Beam
1 kN
� Find fxA by conjugate beam
1 kN
6 kN�m
EIM A
72' =
x
EIB y
18' =
=−EIx
EIx
618 3
EI18
EI6
23
Point
B
D
C
A
fxA / fAA
0
0.482
0.852
1.0
EIx
EIxMf xxA 6
18'3
−==
x (m)
0
2
4
6
x
A BC D
2 m 2 m 2 m
1 kNfAA fxA
72/EI 61.33 /EI34.67 /EI
fxA
0
EI67.34
EI33.61
EI72
34.67
61.3372
1 kN
Influence line of RA
/72 = 1.0/72=0.852
/72 = 0.4821=
AA
AA
ff
24
AB
BA
y
RRRRF
−==−+
=Σ↑+
101
;0
A BC D
2 m 2 m 2 m
1x
RA
MA
RB
0.148.518
1RB
1RB
0.4820.852
1.01 kN
RA
Alternate Method: Use equilibrium conditions for the influence line of RA
RA = 1- RB
25
VC
1
RB
0.1480.518
1
A BC D
2 m 2 m 2 mRA
MA
RB
0.8520.482
-0.148
1 x
1
Using equilibrium conditions for the influence line of VC
VC = 1 - RB
� Unit load to the left of C
RB
1 x
VC
MC
01
;0
=+−+
=Σ↑+
BC
y
RVF
RBVC
MC
VC = - RB
0;0
=++
=Σ↑+
CB
y
VRF
� Unit load to the left of C
26
BA
BA
A
RxMRxM
M
6606)6(1
;0
++−==+−−−
=Σ
A BC D
2 m 2 m 2 mRA
MA
RB
1 x
1
MA
1
RB
0.1480.518
1
-1.112 -0.892
Using equilibrium conditions for the influence line of MA
+
27
MC
C
1
RB
0.1480.518
1
A BC D
2 m 2 m 2 m
1 x
RA
MA
RB
0.074
1
Using equilibrium conditions for the influence line of MC
0.592 RB
1 x
VC
MC
� Unit load to the left of C
RBVC
MC
4 m
MC = 4RB
+04
;0=+−
=Σ
BC
C
RMM
� Unit load to the left of C
4 m
MC = -4 + x + 4RB
04)4(1;0
=+−−−=Σ
BC
C
RxMM+
28
Example 2
Draw the influence line and plot numerical values every 2 m for- the vertical reaction at supports A, B and C- Shear at G and E- Bending moment at G and E
EI is constant.
A B CD E F
2@2=4 m 4@2 = 8 m
G
29
Influence line of RA
1
A B CD E F
2@2=4 m 4@2 = 8 m
G
1=AA
AA
ff
AA
DA
ff
AA
EA
ff
AA
FA
ff
30
4 /EI
1
A B CD E F
4 m 6 m2 m
� Find fxA by conjugate beam
Real beam
0.51.5
0
18.67/EI
64/EI
4/EI
4/EI
Conjugate beam
EI33.5
EI67.10
0
EI67.10 EI
Mf AAA64' ==
31
x1x2
=−EI
xEI
x 13
1 33.512
EIEIEIx 67.18646
32 −+=
V´ x1
M´ x1
x1EIx
21
EI33.5
EIx
4
21
32 1x
31x
Conjugate beam
4/EI
4 m 8 m
64/EI
EI33.5
EI67.18
M´ x2
V´ x2
x2
EIx2
EIx2
22
32x
32 2x
EI67.18
EI64
32
CtoBforEI
xEI
xMf xxA ,33.512
'3
1 −==
EI28
EI64
BtoAforEIEI
xEI
xMf xxA ,6467.186
' 23
22 +−==
fxA
0
0
EI16
−
EI10
−
EI14
−
x (m)
0
2
4
6
Point
C
F
E
D
B
A
8
12
G 10
fxA / fAA
0
-0.1562
-0.25
-0.2188
0
1
0.4375
x1x2
1
fAA fxA
A B CD E FG
4 m 6 m2 m
EI10−
EI16−
EI14−
EI28
EI64
1
Influence line of RA
-0.219 -0.25 -0.156
0.438
1=AA
AA
ff
33
A B CD E FG
4 m 6 m2 m
Using equilibrium conditions for the influence line of RB
RB RCRA
x1
RA
1 -0.219 -0.25 -0.156
10.438
RB
1
0.4850.8751.07810.59
RB
0
0.485
0.875
1.078
1
0.5939
0
x (m)
0
2
4
6
Point
C
F
E
D
B
A
8
12
G 10
RA
0
-0.1562
-0.25
-0.2188
0
0.4375
1
AB
AB
C
RxR
RxRM
812
8
0128;0
−=
=−+−=Σ+
34
A B CD E FG
4 m 6 m2 m
RA
1 -0.219 -0.25 -0.156
10.438
RB RCRA
x1
x (m)
0
2
4
6
Point
C
F
E
D
B
A
8
12
G 10
RA
0
-0.1562
-0.25
-0.2188
0
0.4375
1
RC
1
0.6719
0.375
0.1406
0
-0.0312
0 RC
1
10.672
0.3750.141
-0.0312
Using equilibrium conditions for the influence line of RC
18
5.0
08)8(14;0
+−=
=−−−=Σ+
xRR
RxRM
AC
CA
B
35
A B CD E FG
4 m 6 m2 m
� Check ΣFy = 0
RB RCRA
x1
RA
1 -0.219 -0.25 -0.156
10.438
RB
1
0.490.8751.081
0.59
RC1
10.672
0.3750.141
-0.0312
1
;0
=++
=Σ↑+
CBA
y
RRRF
RC
1
0.6719
0.375
0.1406
0
-0.0312
0
Point
C
F
E
D
B
A
G
RA
0
-0.1562
-0.25
-0.2188
0
0.4375
1
RB
0
0.485
0.875
1.078
1
0.5939
0
ΣR
1
1
1
1
1
1
1
36
VG
RA
1 -0.219 -0.25 -0.156
10.438
1x
VG = RA
RA
AVG
MG
� Unit load to the right of G
0.438
-0.219 -0.25 -0.156-0.5621
A B CD E FG
4 m 6 m2 m
Using equilibrium conditions for the influence line of VG
VG = RA - 1
RA
1x
AVG
MG
� Unit load to the left of G
01;0 =−−=Σ↑+ GAy VRF
37
VE
RC1
10.672
0.3750.141
-0.0312
1x
0.6250.328
0.0312
-0.141-0.375
1
A B CD E FG
4 m 6 m2 m
Using equilibrium conditions for the influence line of VE
RCVE
ME
� Unit load to the left of E
VE = - RC
0;0 =+=Σ↑+ ECy VRF
VE = 1 - RC
� Unit load to the right of E
RC
1 x
VE
ME
01;0 =+−=Σ↑+ CEy RVF
38
MG
RA
1 -0.219 -0.25 -0.156
10.438
1x
-0.438 -0.5 -0.312
1
A B CD E FG
4 m 6 m2 m
Using equilibrium conditions for the influence line of MG
0.876
MG = -2 + x + 2RA
� Unit load to the left of G
RA
1x
AVG
MG2 m
02)2(1;0
=−−+=Σ
AG
G
RxMM+
MG = 2RA
� Unit load to the right of G
RA
AVG
MG2 m
02;0
=−=Σ
AG
G
RMM+
39
RC1
10.672
0.3750.141
-0.0312
ME
1x
1.5
0.6881
4 m 6 m2 m
A B CD E FG
Using equilibrium conditions for the influence line of ME
-0.125
0.564
RCVE
ME
4 m
� Unit load to the left of E
ME = 4RC
+ ΣME = 0;
ME = - 4 + x+ 4RC
� Unit load to the right of E
RC
1 x
VE
ME4 m
04)4(1;0
=+−−−=Σ+
CE
E
RxMM
40
Example 3
For the beam shown(a) Draw quantitative influence lines for the reaction at supports A and B, andbending moment at B.(b) Determine all the reactions at supports, and also draw its quantitative shear,bending moment diagrams, and qualitative deflected curve for
- Only 10 kN downward at 6 m from A- Both 10 kN downward at 6 m from A and 20 kN downward at 4 m from A
10 kN
AC
B
4 m
2EI 3EI
2 m 2 m
20 kN
411
fCA
fAAfEA
fDA
1
/fAA /fAA /fAA
/fAA
fCA
fAAfEA
fDA
AC
B
4 m
2EI 3EI
2 m 2 m
Influence line of RA
42Conjugate Beam
AC
B
4 m
2EI 3EI
2 m 2 m1Real Beam
� Find fxA by conjugate beam
1 kN
8 kN�m
x (m)V (kN)1 1
+
x (m) M (kN�m)
84
+
2EI
fAA = M´A = 60.44/EI
60.44/EI
12/EI
1.33EIEI67.2
43
EI67.2
2EI
60.44/EI
Conjugate Beam12/EI
1.33EI
37.11/EI 17.77/EI 4.88/EI60.44/EI
0
RA = fxA/fAA
0.614 0.294 0.081
1
0
fxA
A C B
� Quantitative influence line of RA
44
A B
4 m 2 m 2 m
Using equilibrium conditions for the influence line of RB and MB
RB = 1 - RA
10.386
0.706 0.919
0
MB = 8RA - (8-x)(1)0
-1.352-1.088 -1.648
0
RA RB
MB
RA
0.614 0.294 0.081
1
0
1x
45
A B
4 m 2 m 2 m
Using equilibrium conditions for the influence line of VB
RA
0.614 0.294 0.081
1
0
-1-0.386
-0.706-0.919
0 VB = RA -1
RA
1x
VB = RA - 1
46
A B
4 m 2 m 2 m
C
Using equilibrium conditions for the influence line of VC and MC
RA
RA
0.614 0.294 0.081
1
0
1x
RB
MB
MC = 4RAMC = 4RA - (4-x)(1)
MC
VC = RAVC = RA - 1
VC1
-0.386 -0.706
0.3240.4561.176
0.294 0.081 0
47
A B
4 m 2 m 2 m
10 kN
RA
0.614 0.294 0.081
1
0
10(0.081)=0.81 kN
MA (kN�m)+
-
-13.53
4.86
V (kN)
-9.19
0.81 0.81
-
The quantitative shear and bending moment diagram and qualitative deflected curve
RB=9.19 kN
MB= 13.53 kN�m
48
A B
4 m 2 m 2 m
10 kNThe quantitative shear and bending moment diagram and qualitative deflected curve
20 kN
20(.294) +1(0.081)= 6.69 kN
V (kN)
-23.31
6.69 6.69
--13.31 -23.31
MA (kN�m)+
-
-46.48
26.760.14
RA
0.614 0.294 0.081
1
0
RB=23.31 kN
MB=46.48 kN�m
49
Example 1
Draw the influence line for - the vertical reaction at B
A BC D
2 m 2 m 2 m
APPENDIX�Muller-Breslau for the influence line of reaction, shear and moment�Influence lines for MDOF beams
50
1 kN
Influence line of RA
A BC D
2 m 2 m 2 m
1=AA
AA
ff AA
CA
ff
AA
DA
ff
51
Conjugate Beam
xV´x
M´x
EIB y
18' =
EIx
2
2
EIx
3x
32x
A BC D
2 m 2 m 2 m
Real Beam
1 kN
� Find fxA by conjugate beam
1 kN
6 kN�m
EIM A
72' =
x
EIB y
18' =
6 /EIEI18
=−EIx
EIx
618 3
52
Point
B
D
C
A
fxA / fAA
0
0.482
0.852
1.0
x (m)
0
2
4
6
x
A BC D
2 m 2 m 2 m
1 kNfAA fxA
72/EI 61.33 /EI34.67 /EI
EIx
EIxMf xxA 6
18'3
−==
fxA
0
EI67.34
EI33.61
EI72
34.67
61.3372
1 kN
Influence line of RA
/72 = 1.0/72=0.852
/72 = 0.4821=
AA
AA
ff
53
� Influence line of RB
1
A BC D
2 m 2 m 2 m
1=BB
BB
ff
BB
AB
ff
BB
CB
ff
BB
DB
ff
54
Real BeamA BC D
2 m 2 m 2 m
Conjugate Beam
� Find fxB by conjugate beam
11
6 kN�m
x
=−+EI
xEIEI
x 18726
3
EI6
EI18
EI18
EI72
3x
32xV´x
M´x
x
EI72
EI18
EIx
2
2
EIx
55
x (m)
0
2
4
6
Point
B
D
C
A
x
fxB / fBB
1
0.518
0.148
0
A BC D
2 m 2 m 2 m
1
fBB
fxB
0
72/EI37.33/EI10.67/EI
/72 = 0.518
/72 = 0.148/72 = 1
1
37.33
10.670
fBB
Influence line of RB
fBB= 1
72
EIx
EIEIxMf xxB
18726
'3
−+==
fxB
0
EI72
EI33.37
EI67.10
56
Example 2
For the beam shown(a) Draw the influence line for the shear at D for the beam(b) Draw the influence line for the bending moment at D for the beamEI is constant.Plot numerical values every 2 m.
A B CD E
2 m 2 m 2 m 2 m
57
A B CD ED
2 m 2 m 2 m 2 m
The influence line for the shear at D
1 kN
1 kN
DVD
1 kN
1 kNDD
ED
ff
1=DD
DD
ff
58
A B CD E
2 m 2 m 2 m 2 m
2 kN�m
1 kN
1 kN2 k
1 kN
1 kN2 kN�m
1 kN2 kN1 kN
� Using conjugate beam for find fxD
1 kN
1 kN
59
A B CD E
2 m 2 m 2 m 2 m
1 kN1 kN
Real beam
V( kN)
x (m)1
-1
M (kN �m) x (m)
4
Conjugate beam
4/EI
M´D
1 kN
1 kN2kN
60
2 m 2 m 2 m 2 m
M´D
Conjugate beam
4/EI
AB
CD E
4/EI
0
8/EI
m38
� Determine M´D at D
EI316
EI38
128/3EI
4/EI
8/EI
m38
EI316
EI340
61
=−=− )2)(3
8()32)(2(4
EIEIEI
EIEIEIEI 352)2)(
340(
3128)
32)(2( =−+=
EIEIEI 376)2)(
340()
32)(2( −=−=
128/3EI
2 m 2 m 2 m 2 m
Conjugate beam
4/EI
AB
CD E
EI38
EI340
V´DL
M´DL
2/EI2/EI
32
EI340
EI340 V´DR
M´DR
2/EI
32
128/3EI
2/EI2/EI
V´E
M´E
32
EI38
62
128/3EI = M´D = fDD
V ´x (m) θ
fxD = M ´ x (m) ∆
EI376
−
EI340
−EI334
−
EI32
EI316
−
EI38
EI352
EI4
−
Influence line of VD = fxD/fDD
76
52
4
0.406 =128
/128 = -0.594 /(128/3) = -0.094
Conjugate beam
EI38128/3EI
2 m 2 m 2 m 2 m
4/EI
AB
CD E
EI340
63
A B CD E
2 m 2 m 2 m 2 m
The influence line for the bending moment at D
1 kN �m1 kN �m
αDD
MD
DD
EDfα
DD
DDfα
64
1 kN�m
0.5 kN1 kN�m
0.5 kN1 kN0.5 kN
2 m 2 m 2 m 2 m
0.5 kN 0.5 kN
0.5 kN1 k
A B CD E
1 kN �m1 kN �m
� Using conjugate beam for find fxD
65
2 m 2 m 2 m 2 m
Real beamA B CD E
0.5 kN1 kN0.5 kN
1 kN �m1 kN �m
V (kN)
x (m)0.5
1
M (kN �m) x (m)
2
2/EI
Conjugate beam
66
2/EI
4/EI
m38
EI38
2/EI
0
4/EI
m38
2 m 2 m 2 m 2 m
2/EI
Conjugate beam
EI34
EI38
EI332
EI34
67
EIEIEI 326)2)(4()
32)(1( =+=
=−=− )2)(3
4()32)(1(2
EIEIEI
2 m 2 m 2 m 2 m
2/EI
Conjugate beam
EI332
EI34
EI4
M´D
V´D
1/EI1/EI
32
EI4
1/EI1/EI
V´E
M´E
32
EI34
68/(32/3) = -0.188-2Influence line of MD
813.03226
=
xD
xDfα
αDD = 32/3EI
2 m 2 m 2 m 2 m
2/EI
Conjugate beam
43EI
4EI
323EI
x (m)V ´ θEI4
EI38
−
EI31
EI34
EI317
−
EI5
fxD = M ´x (m) ∆
-2/EI
θD = 0.469 + 0.531 = 1 rad
θDL = 5/(32/3) = 0.469 rad.θDR = -17/32 = -0.531 rad.
26/3EI
69
Example 3
Draw the influence line for the reactions at supports for the beam shown in thefigure below. EI is constant.
A DB C
5 m 5 m 5 m 5 m5 m 5 m
GE F
70
Influence line for RD
A DB C GE F
5 m 5 m 5 m 5 m5 m 5 m
1
fBDfCD fDD fED fFD
fXD
1
fXD/fDD = Influence line for RD
DD
BD
ff 1=
DD
DD
ff
DD
CD
ff
DD
ED
ff
DD
FD
ff
71Conjugate beam
15 + (2/3)(15)
EI5.112
EI15
� Use the consistency deformation method
1
+
x RG
- Use conjugate beam for find ∆´G and fGG
∆´G + fGGRG = 0 ------(1)
1
Real beam
15 m 15 m
A G
1
15
1
Real beam
30 mA G
1
30
1
A G3@5 =15 m 3@5 =15 m
fGG
∆´G
1
=
112.5/EI
EIM CC
5.2812'' ==∆
Conjugate beam
20 m
EI15 EI
450
EIMf GGG
9000''' ==
EI450
72
x RG = -0.3125 kN
090005.2812=+ GR
EIEI
↓−= ,3125.0 kNRG
Substitute ∆´G and fGG in (1) :
1
A G5.625
0.6875 0.3125
11
15
=
1
+
1
30
73
8.182 m
6.818 m
EI16.35
EI98.15
EI01.23
)3
(2
3125.0 22
2 xx−
22 '13.28xMx
EI=+
)2
(6875.0625.5 12
11 xEI
xx −=
)3
2(2
6875. 12
1 xEI
x+
1
fBDfCD fDD fED fFD
A G3@5 =15 m 3@5 =15 m
Real beam5.625
0.6875 0.3125
� Use the conjugate beam for find fXD
28.13EIx1
x2
x1 = 5 m -----> fBD = M´1= 56/EI
x1 = 10 m -----> fCD = M´1= 166.7/EI
x1 = 15 m -----> fDD = M´1= 246.1/EI
x2 = 5 m -----> fFD = M´2= 134.1/EI
x2 = 10 m -----> fED = M´2= 229.1/EI
x2 = 15 m -----> fDD = M´2= 246.1/EI
A G Conjugate beamEI625.5
EI688.4−
Ax1
(5.625-0.6875x1)/EI
V´1
M´1
EI625.5 EI
xx 211 6875.0625.5 −EI
x2
6875.0 21
G
x2
0.3125x2
V´2M´2
EI13.28
EIx
23125.0 2
2
74
� Influence Line for RD
Influence Line for RD
0.228 0.677 1.0 0.931 0.545
1
fXD
EI56
EI7.166
EI1.246
EI2.229
EI1.134
1
fXD/fDD
1.24656
1.2467.166
1.2461.246
1.2462.229
1.2461.134
75
Influence line for RG
A DB C GE F
5 m 5 m 5 m 5 m5 m 5 m
fXG
1fBG fCG
fGGfEGfFG
fXG/fGG
1GG
BG
ff
GG
CG
ff
GG
EG
ff
GG
FG
ff 1=
GG
GG
ff
76Conjugate beam
20 m
EI450
EI30
� Use consistency deformations
1
=
∆´D + fDDRD = 0 ------(2)
- Use conjugate beam for find ∆´D and fDD
1
+1
Real beam
30 mA G
1
30
1
Real beam
15 m 15 m
A G
1
15
X RD
fXG
13@5 =15 m 3@5 =15 m
∆´D
fDD
Conjugate beam15 + (2/3)(15)
EI15
EI5.112
450/EI
EI9000
77
EIEIEIEIMD
5.2812)15(4509000)3
15(5.112'' =−+==∆EIEI
MfDD1125)15
32(5.112'' =×==
↓=−==+ ,5.25.2,011255.2812 kNkNRREIEI DDSubstitute ∆´D and fDD in (2) :
x RD = -2.5 kN
=
1
11
30+
11
15
1.5
7.5
2.5
15 mV´
M´EI
5.112
EI5.1
450/EI
EI9000
M´´
V´´
EI15 EI
5.112
78
EIEIfBG
5.62)532(75.18
−=×−==
� Use the conjugate beam for find fXG
Real beam
1fBG fCG
fGGfEGfFG
3@5 =15 m 3@5 =15 m1.5
7.5
2.5
fGG = M´G = 1968.56/EI
168.75/EI
EIEIfCG
06.125)67.6(75.18−=−==
A G Conjugate beam
EI5.7−
EI15
10 m
15 + (10/3) = 18.33 m
25 + (2/3)(5) = 28.33 m
EI5.112
EI75
EI75.18
A5 m
V´1
M´1
EI5.7−
EI75.18
A6.67 m
V´2
M´2
EI75.18
EI5.7−
EI75.18−
79
=−+ 33
23 75.16856.1968)
3(
2x
EIEIxx
x = 5 m -----> fFG = M´= 1145.64/EI
x = 10 m -----> fEG = M´ = 447.73/EI
Influence line for RG-0.064-0.032
0.227 0.582 1.0
M´ Gx
fGG = M´G = 1968.56/EI
168.75/EI
x
V´2
2
2x
1
fXG
EI5.62−
EI125−
EI73.447
EI64.1145
EI56.1968
56.19685.62−
56.1968125−
56.196873.447
56.196864.1145
56.196856.1968
1
fXG/fGG
80
Using equilibrium condition for the influence line for Ay
A DB C GE F
5 m 5 m 5 m 5 m5 m 5 m
1x
MA
Ay RD RG
Unit load
1 1
Influence Line for RD
0.228 0.678 1.0 0.929 0.542
Influence line for RG-0.064-0.032
0.227 0.582 1.0
0.386Influence line for Ay
0.804
-0.156 -0.124
1.0
CDAy RRRF −−==Σ↑+ 1:0
81
Using equilibrium condition for the influence line for MA
A DB C GE F
5 m 5 m 5 m 5 m5 m 5 m
1x
MA
Ay RD RG
x 15
x 30
RD
0.228 0.678 1.0 0.929 0.542
1x5 10 15 20 25 30
RG-0.064-0.032
0.227 0.582 1.0
Influence line for MA
2.541.75
-0.745 -0.59
CDA RRxM 30151:0 −−=Σ+
1
! General Case! Stiffness Coefficients! Stiffness Coefficients Derivation! Fixed-End Moments! Pin-Supported End Span! Typical Problems! Analysis of Beams! Analysis of Frames: No Sidesway! Analysis of Frames: Sidesway
DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS
2
Slope � Deflection Equations
settlement = ∆j
Pi j kw Cj
Mij MjiwP
θj
θi
ψ
i j
3
Degrees of Freedom
L
θΑ
A B
M
1 DOF: θΑ
PθΑ
θΒΒΒΒ
A BC 2 DOF: θΑ , θΒΒΒΒ
4
L
A B
1
Stiffness
kBAkAA
LEIkAA
4=
LEIkBA
2=
5
L
A B1
kBBkAB
LEIkBB
4=
LEIkAB
2=
6
Fixed-End ForcesFixed-End Moments: Loads
P
L/2 L/2
L
w
L
8PL
8PL
2P
2P
12
2wL12
2wL
2wL
2wL
7
General Case
settlement = ∆j
Pi j kw Cj
Mij MjiwP
θj
θi
ψ
i j
8
wP
settlement = ∆j
(MFij)∆ (MF
ji)∆
(MFij)Load (MF
ji)Load
+
Mij Mji
θi
θj
+
i jwPMij
Mji
settlement = ∆jθj
θi
ψ
=+ ji LEI
LEI θθ 24
ji LEI
LEI θθ 42
+=
,)()()2()4( LoadijF
ijF
jiij MMLEI
LEIM +++= ∆θθ Loadji
Fji
Fjiji MM
LEI
LEIM )()()4()2( +++= ∆θθ
L
9
Mji Mjk
Pi j kw Cj
Mji Mjk
Cj
j
Equilibrium Equations
0:0 =+−−=Σ+ jjkjij CMMM
10
+
1
1
i jMij Mji
θi
θj
LEIkii
4=
LEIk ji
2=
LEIkij
2=
LEIk jj
4=
iθ×
jθ×
Stiffness Coefficients
L
11
[ ]
=
jjji
ijii
kkkk
k
Stiffness Matrix
)()2()4( ijF
jiij MLEI
LEIM ++= θθ
)()4()2( jiF
jiji MLEI
LEIM ++= θθ
+
=
F
ji
Fij
j
iI
ji
ij
MM
LEILEILEILEI
MM
θθ
)/4()/2()/2()/4(
Matrix Formulation
12
[D] = [K]-1([Q] - [FEM])
Displacementmatrix
Stiffness matrix
Force matrixwP(MF
ij)Load (MFji)Load
+
+
i jwPMij
Mji
θj
θi
ψ ∆j
Mij Mji
θi
θj
(MFij)∆ (MF
ji)∆
Fixed-end momentmatrix
][]][[][ FEMKM += θ
]][[])[]([ θKFEMM =−
][][][][ 1 FEMMK −= −θ
L
13
L
Real beam
Conjugate beam
Stiffness Coefficients Derivation: Fixed-End Support
MjMi
L/3
θi
LMM ji +
EIM j
EIMi
θι
EILM j
2
EILMi
2
)1(2
0)3
2)(2
()3
)(2
(:0'
−−−=
=+−=Σ+
ji
jii
MM
LEI
LMLEI
LMM
)2(0)2
()2
(:0 −−−=+−=Σ↑+EI
LMEI
LMF jiiy θ ij
ii
LEIM
LEIM
andFrom
θ
θ
)2(
)4(
);2()1(
=
=
LMM ji +
i j
14
Conjugate beam
L
Real beam
Stiffness Coefficients Derivation: Pinned-End Support
Mi θi
θj
LM i
LMi
EIM i
EILMi
2
32L
θi θj
0)3
2)(2
(:0' =−=Σ+ LLEI
LMM ii
j θ
i j
0)2
()3
(:0 =+−=Σ↑+ jii
y EILM
EILMF θ
LEIM
EILM
ii
i3)
3(1 =→==θ
)6
(EI
LM ij
−=θ)
3(
EILM i
i =θ
15
Fixed end moment : Point Load
Real beam
8,0
162
22:0
2 PLMEI
PLEI
MLEI
MLFy ==+−−=Σ↑+
P
M
M EIM
Conjugate beamA
EIM
B
L
P
A B
EIM
EIML2
EIM
EIML2
EIPL
16
2
EIPL4 EI
PL16
2
16
L
P
841616PLPLPLPL
=+−
+−
8PL
8PL
2P
2PP/2
P/2
-PL/8 -PL/8
PL/8
-PL/16-PL/8
-
PL/4+
-PL/16-PL/8-
17
Uniform load
L
w
A B
w
M
M
Real beam Conjugate beam
A
EIM
EIM
B
12,0
242
22:0
23 wLMEI
wLEI
MLEI
MLFy ==+−−=Σ↑+
EIwL8
2
EIwL
24
3
EIwL
24
3
EIM
EIML2
EIM
EIML2
18
Settlements
M
M
L
∆
MjMi = Mj
LMM ji +L
MM ji +
Real beam
2
6LEIM ∆
=
Conjugate beam
EIM
A B
EIM
∆
,0)3
2)(2
()3
)(2
(:0 =+−∆−=Σ+L
EIMLL
EIMLM B
EIM
EIML2
EIMEI
ML2
∆
19
wP
A B
A B+θA θB
(FEM)AB(FEM)BA
Pin-Supported End Span: Simple Case
BA LEI
LEI θθ 24
+ BA LEI
LEI θθ 42
+
)1()()/2()/4(0 −−−++== ABBAAB FEMLEILEIM θθ
)2()()/4()/2(0 −−−++== BABABA FEMLEILEIM θθ
BABABBA FEMFEMLEIM )()(2)/6(2:)1()2(2 −+=− θ
2)()()/3( BA
BABBAFEMFEMLEIM −+= θ
wP
AB
L
20
AB
wP
A B
A BθA θB
(MF AB)load
(MF BA)load
Pin-Supported End Span: With End Couple and Settlement
BA LEI
LEI θθ 24
+ BA LEI
LEI θθ 42
+
L
(MF AB)∆
(MF BA) ∆
)1()()(24−−−+++== ∆
FABload
FABBAAAB MM
LEI
LEIMM θθ
)2()()(42−−−+++= ∆
FBAload
FBABABA MM
LEI
LEIM θθ
2)(
21])(
21)[(3:
2)1()2(2lim AF
BAloadFABload
FBABBAA
MMMMLEIMbyinateE ++−+=
−∆θθ
MA
wP
AB
∆
21
Fixed-End MomentsFixed-End Moments: Loads
P
L/2 L/2
P
L/2 L/2
8PL
8PL
163)]
8()[
21(
8PLPLPL
=−−+
12
2wL12
2wL
8)]
12()[
21(
12
222 wLwLwL=−−+
22
Typical Problem
0
0
0
0
A C
B
P1P2
L1 L2
wCB
P
8PL
8PL w12
2wL
12
2wL
8024 11
11
LPLEI
LEIM BAAB +++= θθ
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
128042 2
222
22
wLLPLEI
LEIM CBCB −
−+++= θθ
L L
23
MBA MBC
A C
B
P1P2
L1 L2
wCB
B
CB
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0
24
A C
B
P1P2
L1 L2
wCB
Substitute θB in MAB, MBA, MBC, MCB
MABMBA
MBC
MCB
0
0
0
0
8024 11
11
LPLEI
LEIM BAAB +++= θθ
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
128042 2
222
22
wLLPLEI
LEIM CBCB −
−+++= θθ
25
A BP1
MAB
MBA
L1
A CB
P1P2
L1 L2
wCB
ByR CyAy ByL
Ay Cy
MABMBA
MBC
MCB
By = ByL + ByR
B CP2
MBCMCB
L2
26
Example of Beams
27
10 kN 6 kN/m
A C
B 6 m4 m4 m
Example 1
Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.
28
PPL8 wwL2
30FEM
PL8
wL2
20
MBA MBC
B
Substitute θB in the moment equations:
MBC = 8.8 kN�m
MCB = -10 kN�m
MAB = 10.6 kN�m,
MBA = - 8.8 kN�m,
[M] = [K][Q] + [FEM]
10 kN 6 kN/m
A C
B 6 m4 m4 m
0
0
0
0
8)8)(10(
82
84
++= BAABEIEIM θθ
8)8)(10(
84
82
−+= BABAEIEIM θθ
30)6)(6(
62
64 2
++= CBBCEIEIM θθ
20)6)(6(
64
62 2
−+= CBCBEIEIM θθ
0:0 =−−=Σ+ BCBAB MMM
EI
EIEI
B
B
4.2
030
)6)(6(10)6
48
4(2
=
=+−+
θ
θ
29
10 kN 6 kN/m
A C
B 6 m4 m4 m
MBC = 8.8 kN�m
MCB= -10 kN�m
MAB = 10.6 kN�m,
MBA = - 8.8 kN�m,
= 5.23 kN = 4.78 kN = 5.8 kN = 12.2 kN
10 kN�m8.8 kN�m
10.6 kN�m8.8 kN�m
10 kN
10.6 kN�m
8.8 kN�m
A B
ByLAy
2 m
6 kN/m8.8 kN�m
10 kN�mB
CyByR
18 kN
30
10 kN 6 kN/m
A C
B 6 m4 m4 m
10 kN�m10.6 kN�m
5.23 kN 12.2 kN
4.78 + 5.8 = 10.58 kN
V (kN)x (m)
5.23
- 4.78
5.8
-12.2
+-
+
-
M (kN�m) x (m)
-10.6
10.3
-8.8 -10- -
+-
EIB4.2
=θ
Deflected shape x (m)
31
10 kN 6 kN/m
A C
B 6 m4 m4 m
Example 2
Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.
32
PPL8 wwL2
30FEM
PL8
wL2
20
[M] = [K][Q] + [FEM]
10 kN 6 kN/m
A C
B 6 m4 m4 m
)1(8
)8)(10(8
28
4−−−++= BAAB
EIEIM θθ
)2(8
)8)(10(8
48
2−−−−+= BABA
EIEIM θθ
)3(30
)6)(6(6
26
4 2
−−−++= CBBCEIEIM θθ
)4(20
)6)(6(6
46
2 2
−−−−+= CBCBEIEIM θθ
10
10
0
0
0
308
62:)1()2(2 −=− BBAEIM θ
)5(158
3−−−−= BBA
EIM θ
33
MBA MBC
B
)5(158
3−−−−= BBA
EIM θ
)3(30
)6)(6(6
4 2
−−−+= BBCEIM θ
0:0 =−−=Σ+ BCBAB MMM
EI
EIEI
B
B
488.7
)6(030
)6)(6(15)6
48
3(2
=
−−−=+−+
θ
θ
EI
EIEIinSubstitute
A
BAB
74.23
108
28
40:)1(
−=
−+=
θ
θθθ
Substitute θA and θB in (5), (3) and (4):
MBC = 12.19 kN�m
MCB = - 8.30 kN�m
MBA = - 12.19 kN�m)4(
20)6)(6(
62 2
−−−−= BCBEIM θ
34
10 kN 6 kN/m
A C
B 6 m4 m4 m
MBA = - 12.19 kN�m, MBC = 12.19 kN�m, MCB = - 8.30 kN�m
= 3.48 kN = 6.52 kN = 6.65 kN = 11.35 kN
12.19 kN�m
12.19 kN�m8.30 kN�m
10 kN
12.19 kN�m
A B
ByLAy
2 m
6 kN/m12.19 kN�m
8.30 kN�mC
CyByR
18 kN
B
35
Deflected shape x (m)EIB49.7
=θ
10 kN 6 kN/m
A C
B 6 m4 m4 m
V (kN)x (m)
3.48
- 6.52
6.65
-11.35
M (kN�m) x (m)
14
-12.2-8.3
11.35 kN3.48 kN
6.52 + 6.65 = 13.17 kN
EIA74.23−
=θ
36
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
Example 3
Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.
37
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI(4)(62)/12 (4)(62)/12 (10)(8)/8(10)(8)/8
15
12
)1(8
)8)(10(8
)2(28
)2(4−−−++= BAAB
EIEIM θθ
)2(8
)8)(10(8
)2(48
)2(2−−−−+= BABA
EIEIM θθ
0 10
10
)2(8
)8)(10)(2/3(8
)2(3:2
)1()2(2 aEIM BBA −−−−=− θ
)3(12
)6)(4(6
)3(4 2
−−−+= BBCEIM θ
38
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI(4)(62)/12 (4)(62)/12(3/2)(10)(8)/8
EIEIMM
B
BBCBA
/091.13151275.2:0
==+−==−−
θθ
15
12
)2(8
)8)(10)(2/3(8
)2(3 aEIM BBA −−−−= θ
)3(12
)6)(4(6
)3(4 2
−−−+= BBCEIM θ
mkNEIM
mkNEI
EIM
mkNEI
EIM
BCB
BC
BA
•−=−=
•=−=
•−=−=
91.10126
)3(2
18.1412)091.1(6
)3(4
18.1415)091.1(8
)2(3
θ
39
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
MBA = - 14.18 kN�m, MBC = 14.18 kN�m, MCB = -10.91 kN�m
14.18
140.18 kN�m
10 kN
A B
ByLAy = 6.73 kN= 3.23 kN
14.18 kN�m
10.91 kN�m
4 kN/m
C
CyByR
24 kN
14.18 10.91
= 11.46 kN= 12.55 kN
40
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
11.46 kN3.23 kN
10.91 kN�m
V (kN)x (m)
3.23
-6.73
12.55
-11.46
+-
+-
2.86
M (kN�m) x (m)
12.91
-14.18
5.53
-10.91
+
-+
-
6.77 + 12.55 = 19.32 kN
θB = 1.091/EIDeflected shape x (m)
41
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
Example 4
Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.
12 kN�m
42
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
12 kN�m
wL2/12 = 12 wL2/12 = 121.5PL/8 = 15
MBA
MBCB
12 kN�mMBA
MBC
EIB273.3
−=θ
EIA21.7
−=θ
)1(158
)2(3−−−−= BBA
EIM θ
)2(126
)3(4−−−+= BBC
EIM θ
)3(126
)3(2−−−−= BCB
EIM θ
8)8)(10(
8)3(2
8)2(4
++= BAABEIEIM θθ
0 -3.273/EI
012:int =−−− BCBA MMBJo
012)122()1575.0( =−+−−− BEIEI θ
mkNEI
EIM BA •−=−−= 45.1715)273.3(75.0
mkNEI
EIM BC •=+−= 45.512)273.3(2
mkNEI
EIM CB •−=−−= 27.1512)273.3(
43
10 kN
A B
4 kN/m
C
24 kN
17.45 kN�m5.45 kN�m
15.27 kN�m
2.82 kN 13.64 kN10.36 kN7.18 kN
12 kN�m10 kN 4 kN/m
A C
B13.64 kN2.82 kN
15.27 kN�m
17.54 kN
mkNEI
EIM BA •−=−−= 45.1715)273.3(75.0
mkNEI
EIM BC •=+−= 45.512)273.3(2
mkNEI
EIM CB •−=−−= 27.1512)273.3(
44
10 kN 4 kN/m
A C
B6 m4 m4 m
2EI 3EI
12 kN�m
13.64 kN2.82 kN
15.27 kN�m
17.54 kN
V (kN)x (m)3.41 m+
-+
-
2.82
-7.18
10.36
-13.64
M (kN�m) x (m)+
- -+
11.28
-17.45
-5.45-15.27
7.98
Deflected shape x (m)
EIB273.3
=θEIA
21.7−=θ
EIB273.3
−=θ
EIA21.7
−=θ
45
Example 5
Draw the quantitative shear, bending moment diagrams, and qualitativedeflected curve for the beam shown. Support B settles 10 mm, and EI isconstant. Take E = 200 GPa, I = 200x106 mm4.
12 kN�m 10 kN 6 kN/m
A CB
6 m4 m4 m
2EI 3EI10 mm
46
12 kN�m 10 kN 6 kN/m
A CB
6 m4 m4 m
2EI 3EI10 mm
[FEM]∆
∆
A
B
2
6LEI∆
2
6LEI∆
∆
BC
2
6LEI∆
2
6LEI∆
P w
[FEM]load
8PL
8PL
30
2wL30
2wL
)1(8
)8)(10(8
)01.0)(2(68
)2(28
)2(42 −−−+++=
EIEIEIM BAAB θθ
)2(8
)8)(10(8
)01.0)(2(68
)2(48
)2(22 −−−−++=
EIEIEIM BABA θθ
)3(30
)6)(6(6
)01.0)(3(66
)3(26
)3(4 2
2 −−−+−+=EIEIEIM CBBC θθ
)4(30
)6)(6(6
)01.0)(3(66
)3(46
)3(2 2
2 −−−−−+=EIEIEIM CBCB θθ
-12
0
0
47
12 kN�m 10 kN 6 kN/m
A CB
6 m4 m4 m
2EI 3EI10 mm
Substitute EI = (200x106 kPa)(200x10-6 m4) = 200x200 kN� m2 :
)1(8
)8)(10(8
)01.0)(2(68
)2(28
)2(42 −−−+++=
EIEIEIM BAAB θθ
)2(8
)8)(10(8
)01.0)(2(68
)2(48
)2(22 −−−−++=
EIEIEIM BABA θθ
)1(10758
)2(28
)2(4−−−+++= BAAB
EIEIM θθ
)2(10758
)2(48
)2(2−−−−++= BABA
EIEIM θθ
)2(2/12)2/10(10)2/75(758
)2(3:2
)1()2(2 aEIM BBA −−−−−−−+=− θ
16.5
48
+ ΣMB = 0: - MBA - MBC = 0 (3/4 + 2)EIθB + 16.5 - 192.8 = 0
θB = 64.109/ EI
Substitute θB in (1): θA = -129.06/EI
Substitute θA and θB in (5), (3), (4):
MBC = -64.58 kN�m
MCB = -146.69 kN�m
MBA = 64.58 kN�m,
MBA MBC
B
12 kN�m 10 kN 6 kN/m
A CB
6 m4 m4 m
2EI 3EI10 mm
MBC = (4/6)(3EI)θB - 192.8
MBA = (3/4)(2EI)θB + 16.5
49
12 kN�m 10 kN 6 kN/m
A CB
6 m4 m4 m
64.58 kN�m 64.58 kN�m
= -1.57 kN= 11.57 kN
146.69 kN�m
= 47.21 kN= -29.21 kN
10 kN
A B
ByLAy
12 kN�m64.58 kN�m
2 m
6 kN/m
C
CyByR
18 kN
B
64.58 kN�m
146.69 kN�m
MBC = -64.58 kN�m
MCB = -146.69 kN�m
MBA = 64.58 kN�m,
50
12 kN�m 10 kN 6 kN/m
A C
B6 m4 m4 m
2EI 3EI
V (kN)x (m)
11.57 1.57
-29.21-47.21
-+
M (kN�m) x (m)
1258.29 64.58
-146.69
+
-
47.21 kN
146.69 kN�m
11.57 kN
1.57 + 29.21 = 30.78 kN
10 mmθA = -129.06/EI
θB = 64.109/ EIθA = -129.06/EI
Deflected shape x (m)
θB = 64.109/ EI
51
Example 6
For the beam shown, support A settles 10 mm downward, use the slope-deflectionmethod to(a)Determine all the slopes at supports(b)Determine all the reactions at supports(c)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape. (3 points)Take E= 200 GPa, I = 50(106) mm4.
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
52
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
)1(3
)2(4−−−= CCB
EIM θ
)2(1005.43
)5.1(23
)5.1(4−−−+++= ACCA
EIEIM θθ
)2(2
122
1002
)5.4(33
)5.1(3:2
)2()2(2 aEIM CCA −−−+++=− θ
)3(1005.43
)5.1(43
)5.1(2−−−+−+= ACAC
EIEIM θθ12
0.01 m
C
AmkN •=
××
1003
)01.0)(502005.1(62
mkN •100MF
∆
10 mm
6 kN/m
A C
5.412
)3(6 2
= 4.5MF
w
53
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
)1(3
)2(4−−−= CCB
EIM θ
)2(2
122
1002
)5.4(33
)5.1(3 aEIM CCA −−−+++= θ
10 mm
MCBMCA
C
0=+ CACB MM
02
122
1002
)5.4(33
)5.48(=+++
+C
EI θ
radEIC 0015.006.15
−=−
=θ
)3(1005.43
)5.1(4)06.15(3
)5.1(212 −−−+−+−
= AEI
EIEI θSubstitute θC in eq.(3)
radEIA 0034.022.34
−=−
=θ
� Equilibrium equation:
54
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
radEIC 0015.006.15
−=−
=θ radEIA 0034.022.34
−=−
=θ
mkNEI
EIEIM CBC •−=−
== 08.20)06.15(3
)2(23
)2(2 θ
mkNEI
EIEIM CCB •−=−
== 16.40)06.15(3
)2(43
)2(4 θ
kN08.203
08.2016.40=
+kN08.20
B C40.16 kN�m20.08 kN�m
6 kN/m
AC
12 kN�m
40.16 kN�m
18 kN
8.39 kN26.39 kN
55
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
6 kN/m
AC
12 kN�m
40.16 kN�m8.39 kN26.39 kN
B C40.16 kN�m20.08 kN�m
20.08 kN20.08 kN
V (kN)
x (m)
26.398.39
-20.08
+-
M (kN�m)
x (m)20.08
-40.16
12
radC 0015.0−=θ
radA 0034.0−=θ
Deflected shapex (m)
radC 0015.0=θ
radA 0034.0=θ
56
Example 7
For the beam shown, support A settles 10 mm downward, use theslope-deflection method to(a)Determine all the slopes at supports(b)Determine all the reactions at supports(c)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4.
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
57
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
5.412
)3(6 2
=
6 kN/m
AC4.5
mkN •=
××
1003
)01.0)(502005.1(620.01 m
C
A
100
34 CEI∆
∆C
CB
34
3)2(6
2CC EIEI ∆
=∆
)2(3
43
)2(4−−−∆−= CCCB
EIEIM θ
)3(1005.43
)5.1(23
)5.1(4−−++∆++= CACCA EIEIEIM θθ
)4(1005.43
)5.1(43
)5.1(2−−+−∆++= CACAC EIEIEIM θθ
12
)1(3
43
)2(2−−−∆−= CCBC
EIEIM θ
)3(2
122
1002
)5.4(323
)5.1(3:2
)4()3(2 aEIEIM CCCA −−−+++∆+=− θ
CC EIEI
∆=∆
23)5.1(6∆C
C
A
CEI∆
58
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
� Equilibrium equation:
)3
()( CBBCCBy
MMC +−=
6 kN/m
AC
12 kN�m
MCA
18 kN
AyB C
MCBMBC
By3
393
)5.1(1812)( +=
++= CACA
CAyMMC
C
MCB MCA
(Cy)CA(Cy)CB
*)1(0:0 −−−=+=Σ CACBC MMM
*)2(0)()(:0 −−−=+=Σ CAyCByy CCC
)5(15.628333.0167.4*)1( −−−−=∆− CC EIEIinSubstitute θ
)6(75.101167.35.2*)2( −−−−=∆+− CC EIEIinSubstitute θ
mmEIradEIandFrom CC 227.5/27.5200255.0/51.25)6()5( −=−=∆−=−=θ
59
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
� Solve equation
radEIC 00255.051.25
−=−
=θ
mmEIC 227.527.52
−=−
=∆
Substitute θC and ∆C in (4)
radEIA 000286.086.2
−=−
=θ
Substitute θC and ∆C in (1), (2) and (3a)
mkNM BC •= 68.35
mkNMCB •= 67.1
mkNMCA •−= 67.1
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
35.68 kN�m
kN
Ay
55.56
68.3512)5.4(18
=
−−=
kNBy
45.12
55.518
=
−=
60
10 mm
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
35.68 kN�m
12.45 kN 5.55 kN
Deflected shape
x (m)
radC 00255.0−=θ
mmC 227.5−=∆
radA 000286.0−=θ
radC 00255.0=θradA 000286.0=θ
mmC 227.5=∆
M (kN�m)
x (m)1214.57
1.67
-35.68
-+
V (kN)
x (m)0.925 m12.45
-5.55
+
61
Example of Frame: No Sidesway
62
Example 6
For the frame shown, use the slope-deflection method to (a) Determine the end moments of each member and reactions at supports(b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame.
10 kN
C
12 kN/m
A
B
6 m
40 kN
3 m
3 m
1 m
2EI
3EI
63
10 kN
C
12 kN/m
A
B
6 m
40 kN
3 m
3 m
1 m
2EI
3EIPL/8 = 30
PL/8 = 30
� Equilibrium equations
10
MBC
MBA
)3(18366
)2(3−−−++= BBC
EIM θ
*)1(010 −−−=−− BCBA MM
05430310 =−+− BEIθ
EIEIB667.4
)3(14 −
=−
=θ)1(30
6)3(2
−−−+= BABEIM θ
mkNMmkNM
mkNM
BC
BA
AB
•=•−=
•=
33.4933.39
33.25
36/2 = 18
(wL2/12 ) =3636
� Slope-Deflection Equations
)2(306
)3(4−−−−= BBA
EIM θ
Substitute (2) and (3) in (1*)
)3()1(667.4 toinEI
Substitute B−
=θ
64
10 kN
C
12 kN/m
A
B
6 m
40 kN
3 m
3 m
1 m
2EI
3EI39.33
25.33
49.33
A
B
40 kN
39.33
25.33
C
12 kN/m
B49.33
17.67 kN
27.78 kN
Bending moment diagram
-25.33
27.7
-39.3
Deflected curve
-49.33
20.58
10
θB = -4.667/EIθB
θB
MAB = 25.33 kN�m
MBA = -39.33 kN�m
MBC = 49.33 kN�m
65
A
B
C
D
E
25 kN
5 m
5 kN/m
60(106) mm4
240(106) mm4 180(106)
120(106) mm4
3 m 4 m3 m
Example 7
Draw the quantitative shear, bending moment diagrams and qualitativedeflected curve for the frame shown. E = 200 GPa.
66
A
B
C
D
E
25 kN
5 m
5 kN/m
60(106) mm4
240(106) mm4 180(106)
120(106) mm4
3 m 4 m3 m0
BAABEIEIM θθ
5)2(2
5)2(4
+=0
BABAEIEIM θθ
5)2(4
5)2(2
+=
75.186
)4(26
)4(4++= CBBC
EIEIM θθ
75.186
)4(46
)4(2−+= CBCB
EIEIM θθ
CCDEIM θ5
)(3=
104
)3(3+= CCE
EIM θ
0=+ BCBA MM
)1(75.18)68()
616
58( −−−−=++ CB EIEI θθ
0=++ CECDCB MMM
)2(75.8)49
53
616()
68( −−−=+++ CB EIEI θθ
EIEIandFrom CB
86.229.5:)2()1( =−
= θθ
PL/8 = 18.75 18.75
6.667 (wL2/12 ) = 6.667+ 3.333
67
MAB = −4.23 kN�m
MBA = −8.46 kN�m
MBC = 8.46 kN�m
MCB = −18.18 kN�m
MCD = 1.72 kN�m
MCE = 16.44 kN�m
Substitute θB = -1.11/EI, θc = -20.59/EI below
0
0BAAB
EIEIM θθ5
)2(25
)2(4+=
BABAEIEIM θθ
5)2(4
5)2(2
+=
75.186
)4(26
)4(4++= CBBC
EIEIM θθ
75.186
)4(46
)4(2−+= CBCB
EIEIM θθ
CCDEIM θ5
)(3=
104
)3(3+= CCE
EIM θ
68
MAB = -4.23 kN�m, MBA = -8.46 kN�m, MBC = 8.46 kN�m, MCB = -18.18 kN�m,MCD = 1.72 kN�m, MCE = 16.44 kN�m
A
B
5 m
C E
20 kN
A
B
5 m
4.23 kN�m
2.54 kN
(8.46 + 4.23)/5 = 2.54 kN
8.46 kN�m
C
25 kN
B 3 m 3 m2.54 kN 2.54 kN
8.46 kN�m(25(3)+8.46-18.18)/6 = 10.88 kN
14.12 kN18.18 kN�m
16.44 kN�m
(20(2)+16.44)/4= 14.11 kN
5.89 kN
0.34 kN
28.23 kN
14.12+14.11=28.23 kN1.72 kN�m
(1.72)/5 = 0.34 kN
10.88 kN
10.88 kN
2.54-0.34=2.2 kN
2.2 kN
69
Shear diagram
-2.54
-2.54
10.88
-14.12
14.11
-5.89
0.34
+
-
-
+-
+
2.82 m
1.18 m
Deflected curve
1.18 m
0.78 m2.33 m1.29 m
1.67m
1.29 m
0.78 m
Moment diagram
1.18 m
3.46
24.18
1.72
-18.18 -16.44
4.23
-8.46-8.46
2.33 m
+
-
+
- -
+
θB = −5.29/EI
θC = 2.86/EI1.67m
70
Example of Frames: Sidesway
71
A
B C
D
3m
10 kN
3 m
1 m
Example 8
Determine the moments at each joint of the frame and draw the quantitativebending moment diagrams and qualitative deflected curve . The joints at A andD are fixed and joint C is assumed pin-connected. EI is constant for each member
72
A
B C
D
3m
10 kN
3 m
1 m
MABMDC
Ax Dx
Ay Dy
� Boundary Conditions
θA = θD = 0
� Equilibrium Conditions
� Unknowns
θB and ∆
- Entire Frame
*)2(010:0 −−−=−−=Σ→+
xxx DAF
*)1(0:0 −−−=+=Σ BCBAB MMM
� Overview
B- Joint B
MBCMBA
73
)3(3
)(3−−−= BBC
EIM θ
)4(1875.0375.021375.0 −−−∆=∆−∆= EIEIEIM DC
)1(4
64
)1)(3(104
)(222
2
−−−∆
++=EIEIM BAB θ
5.625 0.375EI∆
A
B C
D3m
10 kN
3 m
1 m
(5.625)load
(1.875)load
(0.375EI∆)∆
(0.375EI∆)∆
(0.375EI∆)∆
(0.375EI∆)∆
:0=Σ+ CM
MDC
D
C
4 m
Dx
MAB
MBA
A
B
4 m
Ax
10 kN
∆ ∆
(1/2)(0.375EI∆)∆
:0=Σ+ BM
)5(563.11875.0375.0 −−−+∆+= EIEIA Bx θ4
)( BAABx
MMA +=
)6(0468.04
−−−∆== EIMD DCx
)2(4
64
)1)(3(104
)(422
2
−−−∆
+−=EIEIM BBA θ
5.625 0.375EI∆
� Slope-Deflection Equations
74
� Solve equation
*)2(010 −−−=−− xx DA*)1(0 −−−=+ BCBA MM
)3(3
)(3−−−= BBC
EIM θ
)4(1875.0 −−−∆= EIM DC
)1(375.0625.54
)(2−−−∆++= EIEIM BAB θ
)5(563.11875.0375.0 −−−+∆+= EIEIA Bx θ
)6(0468.0 −−−∆= EIDx
)2(375.0625.54
)(4−−−∆+−= EIEIM BBA θ
Equilibrium Conditions:
Slope-Deflection Equations:
Horizontal reaction at supports:
Substitute (2) and (3) in (1*)
2EI θB + 0.375EI ∆ = 5.625 ----(7)
Substitute (5) and (6) in (2*)
)8(437.8235.0375.0 −−−−=∆−− EIEI Bθ
From (7) and (8) can solve;
EIEIB8.446.5
=∆−
=θ
)6()1(8.446.5 toinEI
andEI
Substitute B =∆−
=θ
MAB = 15.88 kN�mMBA = 5.6 kN�mMBC = -5.6 kN�mMDC = 8.42 kN�mAx = 7.9 kNDx = 2.1 kN
75
A
BC
D
Deflected curve
A
BC
D
Bending moment diagram
MAB = 15.88 kN�m, MBA = 5.6 kN�m, MBC = -5.6 kN�m, MDC = 8.42 kN�m, Ax = 7.9 kN, Dx = 2.1 kN,
A
B C
D
3m
10 kN
3 m
1 m
15.88
5.6
8.42
7.9 kN 2.1 kN
15.88
5.6
8.42
∆ = 44.8/EI ∆ = 44.8/EI
θB = -5.6/EI5.6
7.8
76
B C
DA
3m4 m
10 kN4 m
2 m
pin
2 EI
2.5 EI EI
Example 9
From the frame shown use the slope-deflection method to:(a) Determine the end moments of each member and reactions at supports(b) Draw the quantitative bending moment diagram, and also draw thequalitative deflected shape of the entire frame.
77
B
C
DA
3m4 m
10 kN4 m
2 m
2EI
2.5EI EI
Ax Dx
Ay Dy
MDCMAB
B´
∆
∆
∆ CD C´∆BC
� Overview
� Boundary Conditions
θA = θD = 0
� Equilibrium Conditions
� Unknowns
θB and ∆
- Entire Frame
*)2(010:0 −−−=−−=Σ→+
xxx DAF
*)1(0:0 −−−=+=Σ BCBAB MMM
B- Joint B
MBCMBA
78
B
C
DA 3m4 m
10 kN4 m
2 m
2EI
2.5EI EI36.87° = ∆ tan 36.87° = 0.75 ∆
= ∆ / cos 36.87° = 1.25 ∆
B´
∆
∆∆BC
∆ CD C´
∆
∆BC
∆CD
C
C´
36.87°
)1(5375.04
)(2−−−+∆+= EIEIM BAB θ
)2(5375.04
)(4−−−−∆+= EIEIM BBA θ
)3(2813.04
)2(3−−−∆−= EIEIM BBC θ
)4(375.0 −−−∆= EIM DC
C
C´
BB´
∆BC= 0.75 ∆
0.375EI∆
6EI∆/(4) 2 = 0.375EI∆
B
A
∆´ B´
(6)(2EI)(0.75∆)/(4) 2 = 0.5625EI∆0.5625EI∆
D
C
C´∆CD= 1.25 ∆
(1/2) 0.75EI∆
(1/2) 0.5625EI∆
PL/8 = 5
5
(6)(2.5EI)(1.25∆)/(5)2 = 0.75EI∆
0.75EI∆
� Slope-Deflection Equation
79
B
C
DA
3m4 m
10 kN4 m
2 m
pin2 EI
2.5 EI EI
Dx= (MDC-(3/4)MBC)/4 ---(6)
MBC
4
Ax = (MBA+ MAB-20)/4 -----(5)
B CMBC
C
D
MBC/4MDC
MBC
4
A
B
MBA
10 kN
MAB
� Horizontal reactions
80
� Solve equations
*)2(010 −−−=−− xx DA*)1(0 −−−=+ BCBA MM
Equilibrium Conditions:
Slope-Deflection Equation:
Horizontal reactions at supports:
Substitute (2) and (3) in (1*)
Substitute (5) and (6) in (2*)
From (7) and (8) can solve;
EIEIB56.1445.1 −
=∆=θ
)6()1(56.1445.1 toinEI
andEI
Substitute B−
=∆=θ
MAB = 15.88 kN�mMBA = 5.6 kN�mMBC = -5.6 kN�mMDC = 8.42 kN�mAx = 7.9 kNDx = 2.1 kN
)1(4
654
)(22 −−−∆
++=EIEIM BAB θ
)2(4
654
)(42 −−−∆
+−=EIEIM BBA θ
)3(4
)75.0)(2(34
)2(32 −−−
∆−=
EIEIM BBC θ
)4(5
)25.1)(5.2(32 −−−
∆=
EIM DC
)5(4
)20(−−−
−+= ABBA
xMMA
)6(443
−−−−
=BCDC
x
MMD
)7(050938.05.2 −−−=−∆+ EIEI Bθ
)8(05334.00938.0 −−−=−∆+ EIEI Bθ
81
BC
DA
Deflected shape
θB=1.45/EI
θB=1.45/EI
Bending-moment diagram
BC
DA 11.19
1.91
5.35
5.46
1.91
∆ ∆
MAB = 11.19 kN�m MBA = 1.91 kN�m MBC = -1.91 kN�m MDC = 5.46 kN�m
Ax = 8.28 kN�m Dx = 1.72 kN�m
B
C
DA
3m4 m
10 kN4 m
2 m
pin2 EI
2.5 EI EI11.19 kN�m
8.27 kN
5.46
1.91
1.91
0.478 kN1.73
0.478 kN
82
Example 10
From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected curve.
A
BC
D
3 m
3m
4m
20 kN/mpin
2EI
3EI
4EI
83
A
BC
D
3 m
3m
4m
20 k
N/m
2EI
3EI
4EI
[FEM]load
� Overview
∆ ∆
� Boundary Conditions
θA = θD = 0
� Equilibrium Conditions
� Unknowns
θB and ∆
- Entire Frame
*)2(060:0 −−−=−−=Σ→+
xxx DAF
*)1(0:0 −−−=+=Σ BCBAB MMM
B- Joint B
MBCMBA
84
A
BC
D
3 m
3m
4m
20 k
N/m
2EI
3EI
4EI
[FEM]load
wL2/12 = 15
wL2/12 = 15
A
B C
D
∆ ∆
[FEM]∆∆∆∆
6(2EI∆)/(3) 2= 1.333EI∆
6(2EI∆)/(3) 2= 1.333EI∆ 6(4EI∆)/(4) 2
= 1.5EI∆
1.5EI∆
BBCEIM θ3
)3(3=
∆++= EIEI B 333.115333.1 θ ----------(1)
∆+−= EIEI B 333.115667.2 θ ----------(2)
BEIθ3= ----------(3)
∆= EI75.0 ----------(4)
∆+++= EIEIEIM BAAB 333.1153
)2(23
)2(4 θθ0
∆+−+= EIEIEIM BABA 333.1153
)2(43
)2(2 θθ0
∆+= EIEIM DDC 75.04
)4(3 θ0
(1/2)(1.5EI∆)
� Slope-Deflection Equation
85
C
Dx
MDC
4 m
D
MAB
A
B
60 kN
MBA
1.5 m
1.5 m
Ax + ΣMC = 0:
:0=Σ+ BM
)6(188.04
−−−∆== EIMD DCx
)5(30889.0333.13
)5.1(60
−−−+∆+=
++=
EIEIA
MMA
Bx
ABBAx
θ
� Horizontal reactions
86
� Solve equation
*)1(0 −−−=+ BCBA MM
Equilibrium Conditions
Equation of moment
Horizontal reaction at support
Substitute (2) and (3) in (1*)
Substitute (5) and (6) in (2*)
From (7) and (8), solve equations;
EIEIB67.3451.5
=∆−
=θ
)6()1(67.3451.5 toinEI
andEI
Substitute B =∆−
=θ
)7(15333.1667.5 −−−=∆+ EIEI Bθ
)8(30077.1333.1 −−−−=∆−− EIEI Bθ
*)2(060 −−−=−− xx DA
)1(333.115333.1 −−−∆++= EIEIM BAB θ
)2(333.115667.2 −−−∆+−= EIEIM BBA θ
)3(3 −−−= BBC EIM θ
)4(75.0 −−−∆= EIM DC
)6(188.0 −−−∆= EIDx
)5(30889.0333.1 −−−+∆+= EIEIA Bx θ
mkNM AB •= 87.53mkNM BA •= 52.16
mkNM BC •−= 52.16mkNM DC •= 0.26
Ax = 53.48 kNDx = 6.52 kN
87
A
C
DMoment diagram
D
A
B C
Deflected shape
∆ ∆
A
BC
D
3 m
3m
4m20 k
N/m 16.52 kN�m
53.87 kN�m
53.48 kN
6.52 kN26 kN�m
5.55 kN
5.55 kN
26.
16.52
mkNM AB •= 87.53mkNM BA •= 52.16
mkNM BC •−= 52.16mkNM DC •= 0.26
Ax = 53.48 kNDx = 6.52 kN
53.87
B16.52
1
! Member Stiffness Factor (K)! Distribution Factor (DF)! Carry-Over Factor! Distribution of Couple at Node! Moment Distribution for Beams
! General Beams! Symmetric Beams
! Moment Distribution for Frames: No Sidesway! Moment Distribution for Frames: Sidesway
DISPLACEMENT MEDTHOD OF ANALYSIS: MOMENT DISTRIBUTION
2
Internal members and far-end member fixed at end support:
C D
K(BC) = 4EI/L2, K(CD) = 4EI/L3
COF = 0.5
1
Member Stiffness Factor (K) & Carry-Over Factor (COF)
P
A C
w
B EI D
CB
L1 /2 L1/2 L2 L3
LEIK 4
=LEIkCC
4= L
EIkDC2
=
COF = 0.5
B C
3
Far-end member pinned or roller end support:
C D
K(AB) = 3EI/L1, K(BC) = 4EI/L2, K(CD) = 4EI/L3
COF = 00
1LEIK 3
= LEIkAA
3=
P
A C
w
BEI
D
CB
L1 /2 L1/2 L2 L3
4
K(CD) = 4EI/L3
Joint Stiffness Factor (K)
P
A C
w
BEI
D
CB
L1 /2 L1/2 L2 L3
K(AB) = 3EI/L1 K(BC) = 4EI/L2,
Kjoint = KT = ΣΣΣΣKmember
5
01DF DFBC DFCB DFCDDFBA)
Notes:- far-end pined (DF = 1)- far-end fixed (DF = 0)
B C DA
Distribution Factor (DF)
P
A C
w
BEI
D
CB
L1 /2 L1/2 L2 L3
KKDF
Σ=
KAB/(K(AB) + K(BC) ) KBC/(K(AB) + K(BC) )
K(BC)/(K(BC) + K(CD) )
K(CD)/(K(BC) + K(CD) )
6
CB(DFBC) CB( DFBC)
CB(DFBC) CB( DFBC)
CO=0.5CO=0
Distribution of Couple at Node
CB
B
A CB D
CB
L1 /2 L1/2 L2 L3
(EI)3(EI)2(EI)1
B C DA01DF DFBC DFCA DFCDDFBA
7
50(.333)50(.333)
50(.667) 50(.667) CO=0.5CO=0
B C DA
01DF 0.667 0.5 0.50.333
L1= L2 = L3
50 kN�m
B
16.67
A CB3EI2EI
4 m 4 m 8 m 8 m
3EID
50 kN�m
8
P
A C
w
B DL1 /2 L1/2 L2 L3
B C DA01DF DFBC DFCB DFCDDFBA
L1= L2 = 8 m, L3 = 10 m
Distribution of Fixed-End Moments
MF
B
MF
MF(DFBC) MF( DFBC)
MF(DFBC) MF( DFBC)
MF
0
MF
0.5 0
B
(EI)13(EI)2(EI)1
9
P
A C
w
B DL1 /2 L1/2 L2 L3
B
145.6
8.4
5.6
8.4
4.2
B C DA01DF 0.6 0.5 0.50.4
L1= L2 = 8 m, L3 = 10 m
1630
0
30 16
14
0.5 0
wL22/12=161.5PL1/8=30 wL3
2/12=25
EI
B
10
Moment Distribution for Beams
11
20 kN
A CB3EI2EI
4 m 4 m 8 m
3 kN/m
Example 1
The support B of the beam shown (E = 200 GPa, I = 50x106 mm4 ).Use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams,and qualitative deflected shape.
12
20 kN
A CB3EI2EI
3 kN/m
4 m 4 m 8 m
K1 = 3(2EI)/8 K2 = 4(3EI)/8
161630
DF 0.3331 0.667 0
K1/(K1+ K2) K2/(K1+ K2)
[FEM]load 16 -16-30
4.662 9.338Dist. 4.669CO
-25.34 -11.3325.34ΣΣΣΣ
20 kNA B B C
24 kN
13.17 kN 6.83 kN
11.33 kN�m25.34 kN�m
10.25 kN13.75 kN
0.5 0 0.5 0CO
13
20 kN
A CB3EI2EI
3 kN/m
4 m 4 m 8 m
161620+10
+ ΣMB = 0: -MBA - MBC = 0
(0.75 + 1.5)EIθB - 30 + 16 = 0
θB = 6.22/EI
MBC = 25.33 kN�m
MBA = -25.33 kN�m,
MBA MBC
B
Note : Using the Slope Deflection
)1(308
)2(3−−−−= BBA
EIM θ
)2(168
)3(4−−−+= BBC
EIM θ
mkNEIM BCB •−=−= 33.11168
)3(2 θ
14
20 kN
A C
B
3 kN/m
4 m 4 m 8 m
V (kN)x (m)
6.83
-13.17
13.75
-10.25
+-
+-4.58 m
M (kN�m) x (m) M (kN�m) x (m)
-11.33
-25.33
27.326.13+
-+ -
Deflected shape x (m)
11.336.83 kN 13.17 + 13.75 = 26.92 kN 10.25 kN
15
10 kN
A C
5 kN/m
B3EI2EI
4 m 4 m 8 m 8 m
3EID
50 kN�m
Example 2
From the beam shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.
16
10 kN
A C
5 kN/m
B2EI
4 m 4 m 8 m 8 m
3EID
50 kN�m
26.67 26.6715 400.5 0.50.5
Joint couple -16.65 -33.35
K1 = 3(2EI)/8 K2 = 4(3EI)/8 K3 = 3(3EI)/8
0.571DF 0.3331 0.4280.667 1
K1/(K1+ K2) K2/(K1+ K2) K2/(K2+ K3) K3/(K2+ K3)
50(.333)
50(.667)
50(.333)
50(.667) CO=0.5CO=050 kN�m
B
17
10 kN
A C
5 kN/m
B2EI
4 m 4 m 8 m 8 m
3EID
50 kN�m
26.67 26.6715 400.5 0.50.5
-26.667FEM 40-15 26.667 1.905 1.437Dist. -3.885 -7.782
2.218 1.673Dist. -0.317 -0.636
0.181 0.137Dist. -0.369 -0.740
Joint couple -16.65 -33.35CO -16.675
-3.891CO 0.953
-0.318 1.109CO
K1 = 3(2EI)/8 K2 = 4(3EI)/8 K3 = 3(3EI)/8
-43.28 43.25 -36.22 -13.78ΣΣΣΣ
0.571DF 0.3331 0.4290.667 1
K1/(K1+ K2) K2/(K1+ K2) K2/(K2+ K3) K3/(K2+ K3)
18
10 kN
A B
ByLAy
36.22 kN�m43.25 kN�m
C D
DyCyR
40 kN
13.78 kN�m
B C
CyLByR
40 kN43.25 kN�m
10 kN
A C
5 kN/m
B2EI
4 m 4 m 8 m 8 m
3EID
50 kN�m
13.78 43.2536.22 43.25
= 25.41 kN = 14.59 kN= 0.47 kN = 9.53 kN
= 12.87 kN = 27.13 kN
19
10 kN
AC
5 kN/m
B2EI 3EID
50 kN�m
4 m 4 m 8 m 8 m
M(kN�m) x (m)
21.29
1.88
-36.22
13.7830.32
-43.25Deflected shape
x (m)
V (kN) x (m)0.47
-9.53
12.87
-14.59-27.13
25.41
2.57 m
2.92 m
9.53+12.87=22.4 kN0.47 kN 14.59 kN
27.13+25.41=52.54 kN
20
3 m 3 m 9 m 3 m
50 kN�m
B
A
40 kN 40 kN
C
D
10 kN/m2EI
EI
Example 3
From the beam shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.
21
3 m 3 m 9 m 3 m
50 kN�m
B
A
40 kN
C
D
10 kN/m2EI
EI
DF 0.800 0.20 1
K1/(K1+ K2) K2/(K1+ K2)
30 30 101.25
K1 = 4(2EI)/6 K2 = 3(EI)/9
0.5 0.5
Dist. -9 -2.25
Joint couple 40 10
Dist.FEM 30 101.25-30CO 20 -60
-4.5CO
1 45.5 49 -120ΣΣΣΣ
-120
120 kN�m 40 kN
22
3 m 3 m 9 m 3 m
50 kN�m
B
A
40 kN 40 kN
C
D
10 kN/m2EI
EI
= 27.75 kN = 12.25 kN
1
120 1204945.5
= 37.11 kN = 52.89 kN
= 40 kNByLAy
45.5 kN�m
40 kN
A B1 kN�m
CyLByR
49 kN�m
B C
90 kN120 kN�m
120 kN�m
C D
CyR
40 kN
23
50 kN�m
B
A
40 kN 40 kN
C
D
10 kN/m2EI
EI
3 m 3 m 9 m 3 m12.25+37.11 = 49.36 kN
27.75 kN52.89+40 = 92.89 kN
V (kN) x (m)
27.75
-12.25
37.11 40
-52.893.71 m
+ +
--
+
M(kN�m) x (m)
-45.5 -49
-120
19.8437.75
--
+-
+ 1
Deflected shape
x (m)
24
20 kN
A CB3EI2EI
4 m 4 m 8 m
12 kN�m 3 kN/m15 kN�m
Example 4
The support B of the beam shown (E = 200 GPa, I = 50x106 mm4 ) settles 10 mm.Use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.
25
20 kN
A CB
2EI
12 kN�m 3 kN/m15 kN�m
3EI4 m 4 m 8 m10 mm 161630
0.50.5
12 kN�m15 kN�m
[FEM]∆∆
A
B ∆BC
2
)2(6LEI ∆
375.92
)2(6)2(622 =
∆−
∆LEI
LEI
125.28)3(62 =
∆LEI 125.28)3(6
2 =∆
LEI
DF 0.3331 0.667 0
K1/(K1+ K2) K2/(K1+ K2)
Joint couple -12 5 10
5[FEM]load 16 -16-30[FEM]∆ -28.125 -28.125 9.375
-6CO
12.90 25.85Dist. 12.92CO
-8.72 -26.20-12 23.72ΣΣΣΣ
K1 = 3(2EI)/8 K2 = 4(3EI)/8
26
20 kN
A CB
2EI
12 kN�m 3 kN/m15 kN�m
3EI
4 m 4 m 8 m10 mm16(16)load(20+10)load
Note : Using the slope deflection
)2(75.18208
)2(48
)2(2−−−+−+= BABA
EIEIM θθ
)2(2/12375.9308
)2(3:2
)1()2( aEIM BBA −−−−+−=− θ
)1(75.18208
)2(28
)2(4−−−−++= BAAB
EIEIM θθ
-12
(9.375 )∆
28.125
(28.125)∆
18.75-9.375
+ ΣMB = 0: - MBA - MBC + 15 = 0
(0.75 + 1.5)EIθB - 38.75 - 15 = 0
θB = 23.9/EI
MBC = 23.72 kN�m
MBA = -8.7 kN�m,
MBA MBC
B
15 kN�m
mkN
EIM BCB
•−=
−−=
2.26
125.28168
)3(2 θ
)3(125.28168
)3(4−−−−+= BBC
EIM θ
)4(125.28168
)3(2−−−−−= BCB
EIM θ
27
4 m 4 m 8 m
20 kN
A CB2EI
12 kN�m 3 kN/m15 kN�m
3EI
= 11.69 kN = 12.31 kN= 7.41 kN = 12.59 kN
23.725 8.725 26.205
20 kN
A B
ByLAy
12 kN�m8.725 kN�m 26.205 kN�m
23.725 kN�m
B C
CyByR
24 kN
28
Deflectedshape x (m)
20 kN
A CB2EI
12 kN�m 3 kN/m15 kN�m
3EI
4 m 4 m 8 m
26.2057.41 kN 12.59+11.69 = 24.28 kN 12.31 kN
V (kN)x (m)
7.41
-12.59
11.69
-12.31
+-
+-3.9 m
M (kN�m) x (m) M (kN�m) x (m)
12
-0.93
41.64
-8.725
-26.205-23.725
+
--
10 mm θB = 23.9/EI
29
Example 5
For the beam shown, support A settles 10 mm downward, use the momentdistribution method to(a)Determine all the reactions at supports(b)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4.
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
30
6 kN/m
B A
C3 m 3 m2EI 1.5EI
12 kN�m
10 mm
0.01 mC
A mkN •=
××
1003
)01.0)(502005.1(62
MF∆
100-(100/2) = 50
4.5
4.5+(4.5/2) = 6.75
DF 0.640 0.36 1
K1/(K1+ K2) K2/(K1+ K2)
-40.16 12-20.08 40.16ΣΣΣΣ
K1 = 4(2EI)/3 K2 = 3(1.5EI)/3
0.50.5
Joint couple 12
6CO
-20.08CO-22.59Dist. -40.16
[FEM]∆ 50[FEM]load 6.75
31
kN08.203
08.2016.40=
+kN08.20
B C40.16 kN�m20.08 kN�m
6 kN/m
AC
12 kN�m
40.16 kN�m
18 kN
8.39 kN26.39 kN
6 kN/m
B A
C3 m 3 m2EI 1.5EI
12 kN�m
-10.16 12-20.08 40.16ΣΜΣΜΣΜΣΜ
32
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
6 kN/m
AC
12 kN�m
40.16 kN�m8.39 kN26.39 kN
B C40.16 kN�m20.08 kN�m
20.08 kN20.08 kN
M (kN�m)
x (m)20.08
-40.16
12
Deflected shape
x (m)
V (kN)
x (m)
26.398.39
-20.08
+-
33
Example 6
For the beam shown, support A settles 10 mm downward, use the momentdistribution method to(a)Determine all the reactions at supports(b)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4.
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
34
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
10 mm
6 kN/m
B A
C
12 kN�m
10 mmR
� Overview
B A
R´
∆
C×
*)1(0' −−−=+ CRR
35
� Artificial joint applied 6 kN/m
B A
C3 m 3 m2EI 1.5EI
12 kN�m
10 mm
0.01 mC
A mkN •=
××
1003
)01.0)(502005.1(62
MF∆
100-(100/2) = 50
4.5
4.5+(4.5/2) = 6.75
DF 0.640 0.36 1
K1/(K1+ K2) K2/(K1+ K2)
-40.16 12-20.08 40.16ΣΣΣΣ
K1 = 4(2EI)/3 K2 = 3(1.5EI)/3
0.50.5
Joint couple 12
6CO
-20.08CO-22.59Dist. -40.16
[FEM]∆ 50[FEM]load 6.75
36
kN08.203
08.2016.40=
+kN08.20
B C40.16 kN�m20.08 kN�m
6 kN/m
AC
12 kN�m
40.16 kN�m
18 kN
8.39 kN26.39 kN
6 kN/m
B A
C3 m 3 m2EI 1.5EI
12 kN�m
-40.16 12-20.08 40.16ΣΜΣΜΣΜΣΜ
C
40.16 kN�m40.16 kN�m
26.39 kN20.08R
039.2608.20:0 =+−−=Σ↑+ RFy
kNR 47.46=
37
B A C
3 m 3 m2EI 1.5EI 0.50.5
� Artificial joint removed
R´
∆
∆C
B
EI75
=∆→1003
)2(62 =
∆CEI
100 753
)75)(5.1(62 =EI
EI∆C
A75-(75/2)= 37.5
DF 0.640 0.36 1-100 -100[FEM]∆ +37.5
22.5Dist. 4020CO
-60-80 60ΣΣΣΣ
B C60 kN�m80 kN�m
AC60 kN�m
46.67 kN46.67 kN 20 kN20 kN C 20 kN46.67R´
kNRFy 67.66':0 ==Σ↑+
38
� Solve equation
*)1(67.66'47.46 inkNRandkNRSubstitute ==
6970.0067.6647.46
−==+
CC
6 kN/mB A C2EI 1.5EI
12 kN�m
35.68 kN�m
12.45 kN 5.55 kN
6970.0−=×C
6 kN/mB
A
C
12 kN�m
R = 46.47 kN
20.08 kN�m
20.08 kN10 mm
8.39 kN
BA
R´ = 66.67 kN
∆80 kN�m
46.67 kN 20 kN
39
10 mm
6 kN/m
B A C
3 m 3 m2EI 1.5EI
12 kN�m
35.68 kN�m
12.45 kN 5.55 kN
Deflected shape
x (m)
M (kN�m)
x (m)1214.57
1.67
-35.68
-+
V (kN)
x (m)0.925 m12.45
-5.55
+
40
P P
A CB DL´ L L´
real beam
Symmetric Beam
� Symmetric Beam and Loading
θ θ
+ ΣMC´ = 0: 0)2
)(()(' =+−LL
EIMLVB
EIMLVB 2' == θ
θLEIM 2
=
The stiffness factor for the center span is, therefore,
LEIK 2
=
L2
L2
C´
V´C
B´
V´B
MEI
LMEI
conjugate beam
41
+ ΣMC´ = 0: 0)3
2)(2
)((21)(' =+−
LLEIMLVB
EIMLVB 6' == θ
θLEIM 6
=
The stiffness factor for the center span is, therefore,
� Symmetric Beam with Antisymmetric Loading
P
P
ACB
D
L´ L L´
real beam
θ
θ
conjugate beam
V´C
V´BB´
C´
MEI
MEI
)2
)((21 L
EIM
)2
)((21 L
EIM
L32
LEIK 6
=
42
Example 5a
Determine all the reactions at supports for the beam below. EI is constant.
A
B4 m
D
C6 m 4 m
15 kN/m
43
A
B4 m
D
C6 m 4 m
15 kN/m
,4
33)(
EILEIK AB ==
622
)(EI
LEIK BC ==
,1)()(
)( ==AB
ABAB K
KDF ,692.0
)6/2()4/3()4/3()(
)()(
)( =+
=+
=EIEI
EIKK
KDF
BCAB
ABBA
308.0)6/2()4/3(
)6/2()()()(
)( =+
=+
=EIEI
EIKK
KDF
BCAB
BCBC
[FEM]load 0 -16 +45
-20.07 -8.93Dist.
wL2/12 = 45wL2/15 = 16
DF 0.692 0.3081.0
A B4 m
30 kN83
B C
90 kN
3 m 3 mDC
4 m
30 kN8336.07 kN�m 36.07 kN�m
0.98 kN 29.02 kN
-36.07 +36.07ΣΜ
45 kN 45 kN 29.02 kN 0.98 kN
wL2/12 = 45wL2/15 = 16 wL2/15 = 16
44
A B4 m
30 kN83
B C
90 kN
3 m 3 mDC
4 m
30 kN8336.07 kN�m 36.07 kN�m
29.02 kN 45 kN 45 kN 29.02 kN 0.98 kN0.98 kN
A
B4 m
D
C6 m 4 m
15 kN/m
74.02 kN 74.02 kN 0.98 kN0.98 kN
M (kN�m) x (m)
+
--
-36.07 -36.07
31.42
Deflectedshape
V(kN) x (m)
-29.02
45
-45
29.02
-0.98
0.98
45
Example 5b
Determine all the reactions at supports for the beam below. EI is constant.
A
B
DC
15 kN/m
15 kN/m
4 m 3 m 4 m3 m
46
A
B
DC
15 kN/mFixed End Moment
wL2/15 = 16 11wL2/192 = 30.938
5wL2/192 = 14.063
A
B
DC
15 kN/m
wL2/15 = 165wL2/192 = 14.063
11wL2/192 = 30.938
A
B
DC
15 kN/m
15 kN/m16.87516
16.875 16
47
A B4 m
30 kN83
B C
45 kN
45 kN
DC4 m
30 kN
83
A
B
DC
15 kN/m
15 kN/m
4 m 3 m 4 m3 m
[FEM]load 0 -16 16.875
-0.375 -0.50Dist.
DF 0.429 0.5711.0
,75.04
33)( EIEI
LEIK AB === EIEI
LEIK BC ===
666
)(
,1)( =ABDF ,429.0175.0
75.0)( =+
=BADF 571.0175.0
1)( =+
=BCDF
16.87516
16.875 16
-16.37516.375ΣΜ
16.375 kN�m 16.375 kN�m
24.09 kN5.91 kN 27.96 kN27.96 kN 24.09 kN
5.91 kN
16.87516
48
A B4 m
30 kN83
B C
45 kN
45 kN
16.375 kN�m 16.375 kN�m
24.09 kN5.91 kN 27.96 kN27.96 kN
DC4 m
30 kN
83
24.09 kN5.91 kN
A
B
DC
15 kN/m
15 kN/m52.05 kN 52.05 kN 5.91 kN5.91 kN
V(kN) x (m)
5.91 5.91
-24.09 -24.09
27.96 27.96
-16.375
M (kN�m) x (m)
16.375
Deflected shape
49
Moment Distribution Frames: No Sidesway
50
AB C
D
4 m
48 kN 8 kN/m40 kN�m
5 m5 m
3EI
2.5EI 2.5EI
Example 6
From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports(b) Draw its quantitative shear and bending moment diagrams,and qualitative deflected shape.
51
40 kN�m
AB C
D
4 m
48 kN 8 kN/m
5 m5 m
3EI
2.5EI 2.5EI45 25
KAB = KBC = 3(2.5EI)/5 = 1.5 EI
KBD = 4(3EI)/4 = 3EI
0.50.5
0.5
0.25 0.25DF 0.51 0 1
A B D C
BA BCMember BDAB DB CB
5 5Dist. 10
5CO
-10 -10Joint load -20
-10CO-45FEM 25
20 00 -5ΣΣΣΣ -50 -10
40 kN�m
52
AB C
D
48 kN 8 kN/m40 kN�m
3EI
2.5EI 2.5EI2010
50
16 kN24 kN
50 kN�m
48 kNA B 0
14 kN 34 kN
3.75
5820
40 kNB C 3.75
3.75
3.75
40 kN�m
20
10
50243458 kN
14 kN 16 kN
Member BD DB CBBC20 0
AB0 -5ΣΣΣΣ
BA-50 -10
5
3.75 kN
3.75 kN10 kN�m
5
58
58
53
Moment diagram Deflected shape
AB C
D
4 m
48 kN 8 kN/m40 kN�m
5 m5 m
3EI
2.5EI 2.5EI
58 kN
14 kN 16 kN
1635
5020
10
5
5
50
20
10
54
Moment Distribution for Frames: Sidesway
55
Single Frames
A
B C
D
P∆∆
A
BC
D
P
Artificial joint applied(no sidesway)
R´
A
BC
D
Artificial joint removed(sidesway)
0 = R + C1R´
x C1
RR´
x C1
R
56
0 =R2 + C1R2´ + C2R2´´
0 =R1 + C1R1´ + C2R2´´
P2
P1
P3
P4
∆2
∆1
P2
P1
P3
P4
R2
R1
R2
R1
R1´x C1
∆´∆´R2´
R1´x C1
R2´
x C1
R2´´
x C2
R1´´
R2´´
x C2
R1´´
x C2
∆´´ ∆´´
Multistory Frames
57
A
B C
D
5 m
16 kN
1 m 4 m
Example 7
From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and
qualitative deflected shape.EI is constant.
58
A
BC
D
+
artificial joint removed( sidesway)
A
BC
D
16 kN
=
artificial joint applied(no sidesway)
x C1
A
B C
D
5 m
16 kN
1 m 4 m
5 m
R´R
� Overview
R + C1R´ = 0 ---------(1)
59
� Artificial joint applied (no sidesway)
A
BC
D
16 kN
Ra = 4 mb = 1 m
Pb2a/L2 = 2.56
Pa2b/L2 = 10.24
Fixed end moment:
Equilibrium condition :
ΣFx = 0:+ Ax + Dx + R = 0
60
5 m
1 m 4 m
5 m
A
BC
D
16 kN
R
2.5610.24
0.5
0.5
0.5
0.50 0.50DF 0.500 0.50 0FEM -2.5610.24Dist. 1.28 -5.12-5.12 1.28CO -2.56-2.56 0.64 0.64Dist. 1.28 -0.32-0.32 1.28
CO -0.16-0.16 0.64 0.64Dist. 0.08 -0.32-0.32 0.08CO -0.16-0.16 0.04 0.04
Dist. 0.08 -0.02-0.02 0.08
5.78 1.32-2.88 2.72ΣΣΣΣ -5.78 -2.72
A B C D
2.5610.24
Ax = 1.73 kNDx = 0.81 kN
5.78 kN�m
2.88 kN�m
2.72 kN�m
1.32 kN�m
A
B
5 m
D
C
5 m
1.32-2.88 2.72-5.78
ΣFx = 0:+ 1.73 - 0.81 + R = 0
R = - 0.92 kN
Equilibrium condition :
61
� Artificial joint removed ( sidesway)
Fixed end moment:
5 m
5m
5 m
A
B C
D
R´
∆ ∆
100 kN�m
100 kN�m
100 kN�m
100 kN�m
Since both B and C happen to be displaced the same amount ∆, and AB and DChave the same E, I, and L so we will assume fixed-end moment to be 100 kN�m.
Equilibrium condition :
ΣFx = 0:+ Ax + Dx + R´ = 0
62
5 m
5m
5 m
A
B C
D
R´
∆ ∆
100 kN�m
100 kN�m
100 kN�m
100 kN�m0.5
0.5
0.5
0.50 0.50DF 0.500 0.50 0
A B C D
FEM 100100 100 100Dist. -50-50-50 -50CO -25.0-25.0 -25.0 -25.0Dist. 12.512.5 12.5 12.5
CO 6.56.5 6.5 6.5Dist. -3.125-3.125-3.125 -3.125CO -1.56-1.56 -1.56 -1.56
Dist. 0.780.78 0.78 0.78CO 0.390.39 0.39 0.39
Dist. -0.195-0.195-0.195 -0.195
-60 8080 60ΣΣΣΣ 60 -60
Ax = 28 kNDx = 28 kN
60 kN�m
80 kN�m
A
B
5 m
60 kN�m
80 kN�m
D
C
5 m
100 kN�m
100 kN�m
100 kN�m
100 kN�m
ΣFx = 0:+
-28 - 28 + R´ = 0
R´ = 56 kN
Equilibrium condition:
63
-0.92 + C1(56) = 0
R + C1R´ = 0
Substitute R = -0.92 and R´= 56 in (1) :
C1 =0.9256
A
B C
D
16 kN
R
1.73 kN 0.81
2.72
5.785.78
2.88
2.72
1.32
x C1
A
BC
D
R´
+
28 28
6060
60
80
60
80
A
BC
D
5 m
16 kN
1 m 4 m
5 m=
3.714.79
4.79
1.57
3.71
2.63
1.27 kN1.27
64
Bending moment diagram (kN�m)
A
BC
D
5 m
16 kN
1 m 4 m
5 m
3.714.79
4.79
1.57
3.71
2.63
1.27 kN 1.27 kN
8.22
3.71
2.99 kN13.01 kN∆ ∆
Deflected shape
1.57
4.794.79
3.71
2.63
65
Example 8
From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.
A
BC
D
3 m
3m
4m
20 kN/mpin
2EI
3EI
4EI
66
A
BC
D
3 m , 2EI
3m, 3EI
4m , 4EI
20 kN/m
=
A
B C
D
artificial joint applied(no sidesway)
� Overview
R + C1R´ = 0 ----------(1)
x C1+
A
B C
D
artificial joint removed(sidesway)
R R´
67
R
A
B C
D
3 m , 2EI
3m , 3EI
4m, 4EI
20 kN/m
0.471 0.529DF 1.000 1.00 0FEM 15.00 -15.00Dist. 7.9357.065CO 3.533
0.5 0.5
0.5
15
15
KCD = 3(4EI)/4 = 3EI
KBA = 4(2EI)/3 = 2.667EI
KBC = 3(3EI)/3 = 3EI
A B C D
Ax = 33.53 kN
7.9418.53ΣΣΣΣ -7.94
7.94 kN�m
18.53 kN�m
A
B
3 m60
� Artificial joint applied (no sidesway)
Dx = 0
0
D
C
4 m
ΣFx = 0: +
60 - 33.53 - 0 + R = 0
R = - 26.47 kN
68
� Artificial joint removed ( sidesway)
� Fixed end moment
R´
A
B C
D
3 m, 2EI
3m, 3EI
4m, 4EI
6(2EI∆)/(3) 2
6(2EI∆)/(3) 2
3(4EI∆)/(4) 2
∆ ∆
Assign a value of (FEM)AB = (FEM)BA = 100 kN�m
1003
)2(62 =
∆EI
∆AB = 75/EI
100 kN�m
100 kN�m
R´
A
B C
D
3 m, 2EI
3m, 3EI
4m, 4EI
100 kN�m
100 kN�m 3(4EI)(75/EI)/(4) 2 = 56.25 kN�m
∆ ∆
6(4EI)∆/(4) 2
69
R´
A
B C
D
3 m
3m
4m
0.471 0.529DF 1.000 1.00 0
CO -28.55
0.50.5
0.5
A B C D
Ax = 43.12kN
ΣFx = 0:+
-43.12 - 14.06 + R´ = 0
R´ = 57.18 kN
100
100
56.25
FEM 100 56.25 100Dist. -52.9-47.1 0
14.06 kN
∆ ∆
-52.976.45ΣΣΣΣ 52.9 56.25
56.25 kN�m
D
C
4 m
52.9. kN�m
76.45 kN�m
A
B
3 m
70
=
A
BC
D
3 m
3m
4m
20 kN/m
16.55 kN�m
53.92 kN�m
53.49 kN
6.51 kN
26.04 kN�m
R + C1R´ = 0
-26.47 + C1(57.18) = 0
C1 =26.4757.18
x C1+
R´
A
B C
D
52.9
52.9 kN�m
76.45 kN�m
43.12 kN
14.06
56.25
R
A
B C
D
33.53 kN
7.94 kN�m7.94 kN�m
18.53 kN�m
0
0
Substitute R = -26.37 and R´= 57.18 in (1) :
5.52 kN
5.52 kN
71
A
C
DMoment diagram
D
A
B C
Deflected shape
∆ ∆
A
BC
D
3 m
3m
4m
20 kN/m
16.55 kN�m
53.92 kN�m
53.49 kN
6.51 kN
26.04 kN�m
5.52 kN
5.52 kN 26.04
53.92
16.55
B16.55
72
BC
DA
3m4 m
10 kN
4 m
Example 8
From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected shape.
EI is constant.
73
B C
DA
10 kN
=
artificial joint applied(no sidesway)
x C1
R´
BC
DA
3m4 m
10 kN
4 m
R
+
B C
DA
artificial joint removed(sidesway)
R + C1R´ = 0 ---------(1)
� Overview
74
� Artificial joint applied (no sidesway)
B C
DA
10 kNR
00
ΣFx = 0:+
10 + R = 0
R = - 10 kN
Equilibrium condition :
75
� Artificial joint removed (sidesway)
� Fixed end moment
R´B
C
DA
3m4 m
4 m= ∆ tan 36.87° = 0.75 ∆= 0.75(266.667/EI) = 200/EI
∆CD = ∆ / cos 36.87° = 1.25 ∆ = 1.25(266.667/EI) = 333.334/EI
∆AB = ∆
∆BC
∆CD
C
C´
36.87°
B´
∆
∆
C´∆ CD
36.87°
∆BC
R´B
C
DA
3m4 m
4 m
6EI∆AB/(4) 2
6EI∆AB/(4) 2
6EI∆BC/(4) 2
3EI∆CD/(5) 2
6EI∆CD/(5) 2
Assign a value of (FEM)AB = (FEM)BA = 100 kN�m : 1004
62 =∆ ABEI
∆AB = 266.667/EI
100 kN�m 100 kN�m
76
R´B
C
DA
100 kN�m
100 kN�m
6EI∆BC/(4) 2 = 6(200)/42 = 75 kN�m
3EI∆CD/(5) 2 = 3(333.334)/52 = 40 kN�m
∆BC= 200/EI, ∆CD = 333.334/EI
Equilibrium condition :
ΣFx = 0:+ Ax + Dx + R´ = 0
77
R´B
C
DA
3m4 m
4 m
0.50 0.50DF 0.6250 0.375 1
KBA = 4EI/4 = EI, KBC = 4EI/4 = EI, KCD = 3EI/5 = 0.6EI
100
100
40
7575
0.5
0.5
0.5
A B C D
FEM 100 100 40 -75 -75Dist. -12.5-12.5 13.125 21.875
CO -2.735 1.953 -2.735Dist. -0.977-0.977 1.026 1.709
-81.0591.02ΣΣΣΣ 81.05 -56.48 56.48
B C81.05
56.48
43.02 kN
34.38 kN34.38 kN
C
D
34.38 kN
34.38 kN
56.48
39.91 kNΣFx = 0:+
-43.02 - 39.91 + R´ = 0
R´ = 82.93 kN
A
B
91.02
81.0534.38 kN
34.38 kN
Dist. -5.469-5.469 2.344 3.906CO -6.25 10.938 -6.25
78
BC
DA
3m4 m
10 kN
4 m
5.19 kN
9.77
6.81
10.98
4.15 kN
9.77 6.81
4.15 kN
4.81 kN=
B C
DA
10 kN
00
00
0
R
x C1= 10/82.93+
B C
A 43.02 kN
81.05
56.48
91.02
34.38 kN
81.05 56.48
D
34.38 kN
39.91 kN
R´
C1 = 10/82.93
-10 + C1(82.93) = 0Substitute R = -10 kN and R´= 82.93 kN in (1) :
R + C1R´ = 0 ---------(1)
79
BC
DA
Bending moment diagram(kN�m)
BC
DA
3m4 m
10 kN
4 m
5.19 kN
9.77
6.81
10.98
4.15 kN
9.77 6.81
4.81 kN4.15 kN
BC
DA
Deflected shape
10.98
6.81
9.77
9.77
80
BC
DA
4 m3 m
2m2 m 3 m
40 kN
20 kN
4EI
3EI
4EI
Example 9
From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams,and
qualitative deflected shape.EI is constant.
81
BC
DA
4 m3 m
2m2 m 3 m
40 kN
20 kN
4EI
3EI
4EI
=x C1
R´
� Overview
R + C1R´ = 0 ----------(1)
CB
DA
2m2 m 3 m
40 kN
20 kN
artificial joint applied(no sidesway)
R
artificial joint removed(no sidesway)
+
B C
DA
2m2 m 3 m
82
� Artificial joint applied (no sidesway)
B C
DA
2m2 m 3 m
40 kN
20 kN R15+(15/2)= 22.5 kN�m
PL/8 = 15
Equilibrium condition :
ΣFx = 0:+ Ax + Dx + R = 0
Fixed end moments:
83
B C
DA
2m2 m 3 m
40 kN
20 kNR
0.60 0.40DF 1.000 1.00 022.5 15
KBA = 4(4EI)/3.6 = 4.444EI, KBC = 3(3EI)/3 = 3EI,
0.5 0.5
0.5
A B C D
Dist. -9.0-13.5CO -6.75
FEM 22.5
13.5-6.75ΣΣΣΣ -13.5
15.5 kN24.5 kN23.08 kN 7.75 kN
ΣFx = 0:+
23.08 + 20 -7.75 + R´ = 0
R´ = - 35.33 kN
24.5 kN
24.5 kN13.5 kN�m
6.75 kN�m
B
A
B C
40 kN13.5
C
D
15.5 kN
0
15.5 kN
84
B C
DA
R´
� Artificial joint removed (sidesway)
Fixed end moments:
3(3EI)∆BC/(3) 2
6(4EI)∆CD/(4.47) 2
3(4EI)∆CD/(4.472) 2
6(3EI)∆BC/(3) 2
6(4EI)∆AB/(3.61) 2
6(4EI)∆AB/(3.606) 2
Assign a value of (FEM)AB = (FEM)BA = 100 kN�m : 10061.3
)4(62 =∆ ABEI , ∆AB = 54.18/EI
B C
DA
3 m
4EI 4EI
R´
3.606
m 4.472 m
3EI
100 kN�m
100 kN�m
85
B C
DA
R´
33.69
o
26.5
7°
33.69o
C´∆ CD
C
B´
∆AB = 54.3/EI
B
EIEIEICBBC /59.52/05.30/54.22'' =+==∆
3(3EI)∆BC/(3) 2 = 3(3EI)(52.59/EI) /(3) 2 = 52.59 kN�m
3(4EI)∆CD/(4.472) 2 = 3(4EI)(50.4/EI)/(4.472) 2 = 30.24 kN�m
100 kN�m
100 kN�m B C
DA
R´
B´
BB´CC´
C´∆ = ∆ABcos 33.69° = 45.08/EI
26.57°
∆ tan 33.69 = 30.05/EI
∆ tan 26.57 = 22.54/EI∆ tan 26.57 = 22.54/EI
∆ tan 33.69 = 30.05/EI
C´∆ CD
C
∆CD = ∆/cos 26.57°= 50.4/EI
86
B C
DA
4 m3 m
2m2 m 3 m
4EI
3EI
4EI
R´
23.8523.8568.34 kN
0.5 0.5
0.5
0.60 0.40DF 1.000 1.00 0
A B C D
Dist. -18.96 -28.45FEM 100 100 30.24-52.59
-71.5585.78ΣΣΣΣ 71.55 30.24
19.49 kN
ΣFx = 0:+
-68.34 - 19.49 + R´ = 0
R´ = 87.83 kN
CO -14.223
23.85 kN
71.55 kN�mB
85.78kN�m
A
23.85 kN
C
D
23.85 kN
30.24 kN�m
23.85 kN
B C71.55
100
100
52.59
30.24
87
-35.33 + C1(87.83) = 0
x C1
Substitute R = -35.33 and R´= 87.83 in (1) :
C1 = 35.33/87.83B C
DA
40 kN
20 kN35.33 kN
6.75 kN�m23.08 kN
24.5 kN
13.5 kN�m
7.75 kN
0
15.5 kN
+
90.59 kNB C
D
A
71.55 kN�m
85.78 kN�m
68.34 kN23.85 kN
19.485 kN
23.85 kN
30.24 kN�m
=B
C
DA
40 kN20 kN
15.28 kN�m
27.76 kN�m4.41 kN
14.91 kN15.59 kN
25.09 kN
12.16 kN�m
88
Bending moment diagram
BC
DA
BC
DA
40 kN20 kN
15.28 kN�m
27.76 kN�m
4.41 kN
14.91 kN15.59 kN
25.09 kN
12.16 kN�m
37.65
Deflected shape
B
DA
C
12.1627.76
15.28
1
! Fundamentals of the Stiffness Method! Member Local Stiffness Matrix! Displacement and Force Transformation Matrices! Member Global Stiffness Matrix! Application of the Stiffness Method for Truss
Analysis! Trusses Having Inclined Supports, Thermal Changes
and Fabrication Errors! Space-Truss Analysis
TRUSSES ANALYSIS
2
2-Dimension Trusses
3
Fundamentals of the Stiffness Method
� Node and Member Identification
� Global and Member Coordinates
� Degrees of Freedom
12
3 4
2
1
3
(x1, y1)
(x3, y3)
(x2, y2)
(x4, y4)
1
23
4
56
78
34
56
78
x
y
�Known degrees of freedom D3, D4, D5, D6, D7 and D8� Unknown degrees of freedom D1 and D2
4
AE/LAE/L
x djAE/L x d´j
AE/L
x diAE/LAE/L x d´i
Member Local Stiffness Matrix
x´y´
i
j
q´i
q´j
AE/L
AE/L
jii dL
AEdL
AEq ''' −=
AE/LAE/Ld´ i = 1
d´ j = 1
x di
x dj
jij dL
AEdL
AEq ''' +−=
−
−=
j
i
j
i
dd
LAE
''
1111
''
−
−=
1111
]'[L
AEk
[q´] = [k´][d´] ----------(1)
q´j
q´i
x´y´
x´y´
5
x´y´
m
i
j
(xi,yi)
(xj,yj)x
y
Displacement and Force Transformation Matrices
θy
θx
22 )()(cos
ijij
ijijxx
yyxx
xxL
xx
−+−
−=
−== θλ
22 )()(cos
ijij
ijijyy
yyxx
yyL
yy
−+−
−=
−== θλ
6
x
y
Global
m
i
jdjx
djy
dix
diy
djx
djyx´y´
m
i
j
Local
d´i
d´j
dix
diy
d´j
d´i
� Displacement Transformation Matrices
yiyxixi ddd θθ coscos' +=
=
jy
jx
iy
ix
yx
yx
j
i
dddd
dd
λλλλ00
00''
θy
θx
yjyxjxj ddd θθ coscos' +=
=
yx
yxTλλ
λλ00
00][
λx λy
[d´] = [T][d] ----------(2)
7
x
y
θy
θx
m
i
j
x
y
Global
x´y´
m
i
j
Local
q´i
q´j
� Force Transformation Matrices
θy
θx
xiix qq θcos'=
yiiy qq θcos'=
xjjx qq θcos'=
yjjy qq θcos'=
=
j
i
y
x
y
x
jy
jx
iy
ix
qqqq
''
00
00
λλ
λλ
=
y
x
y
x
TT
where
λλ
λλ
00
00
][
λx
λy
[q] = [T]T[q´] ----------(3)
qjx
qjy
qix
qiy
8
Member Global Stiffness Matrix
[ q ] = [ T ]T ([k´][d´] + [q´F] ) = [ T ]T [ k´ ][T][d] + [ T ]T [q´F] = [k][d] + [qF]
[ k ] [ k ] = [ T ]T[ k´ ][T]
[qF] = [ T ]T [q´F]
[q] = [T]T[q´] ----------(3)
Substitute ( [q´] = [k´][d´] + [q´F] ) into Eq. 3, yields the result,
λyλx
λxλx
−λyλx
−λxλx
λyλy
λxλy
−λyλy
−λxλy
λyλx
λxλx
−λyλx
−λxλx
λyλy
λxλy
−λyλy
−λxλy
V U VU
[ k ] = AEL
VUV
U
00λyλx
λyλx00-11
1-1
AEL[ k ] =
00
λy
λx
λy
λx
00
9
[Qa] = [K][D] + [QF]
[Qk] = [K11][Du] + [K12][Dk] + [QF]
Reaction Boundary Condition
Unknown DisplacementJoint Load
Equilibrium Equation:
Partitioned Form:
Application of the Stiffness Method for Truss Analysis
[Du] = (([Qk] - [QF]) - [K12][Dk])[Ku] -1
Qk
Qu
Du
Dk
=K12
K22
K11
K21
+QF
k
QFu
10
+
−
−=
jF
iF
j
i
j
i
dd
LAE
''
''
1111
''
jy
jx
iy
ix
yx
yx
dddd
λλλλ00
00
+
−
−=
jF
iF
jy
jx
iy
ix
yx
yx
j
i
DDDD
LAE
''
0000
1111
''
λλλλ
Member Forces
x
y
θy
θx
x´y´
m
i
j
q´i
q´j
11
+
−−
−−=
jF
iF
jy
jx
iy
ix
yxyx
yxyx
j
i
DDDD
LAE
''
''
λλλλλλλλ
Dyi
Dxi
Dyj
Dxj
q´j = AEL −λx −λy λx λy qj´
F+
x
y
θy
θx
x´y´
m
i
j
Member Forces
q´i
q´j
12
Member Forces
Dyi
Dxi
Dyj
Dxj
qm = AEL −λx −λy λx λy qj´F+
x
y
θy
θx
x´y´
m
i
j
Member Forces
qm
13
3 m
3 m
4 m 4 m
80 kN
50 kN
5 m
5 m5 m
Example 1
For the truss shown, use the stiffness method to:(a) Determine the deflections of the loaded joint.(b) Determine the end forces of each member and reactions at supports.Assume EA to be the same for each member.
14
Ljyy
Lixx ijij
ij
�)(�)(� −+
−=λ
cosθx = λx cosθy = λy
1
2
3
434
56
78
1
2
(0,0)(-4,-3)
(-4,3)
(4,-3)
31
23 m
3 m
4 m 4 m
80 kN
50 kN
5 m
5 m5 m
Member
#1
#2
λx λy
#3
λyλx
λxλx
λyλy
λxλy
λyλx
λxλx
λyλy
λxλy
−λyλx
−λxλx
−λyλy
−λxλy
−λyλx
−λxλx
−λyλy
−λxλy
Vi Uj VjUi
[ k ]m = AEL
Vi
Uj
Vj
Ui
31
2
-4/5 = -0.8
-4/5 = -0.8
4/5 = 0.8
-3/5 = -0.6
3/5 = 0.6
-3/5 = -0.6
15
-0.48
1.08-0.48
1.92-0.48
0.64
0.36
-0.481
2
1 2
[K] = AE5
1
2
3
434
56
78
1
2
31
2
0.48
0.64
0.36
0.48-0.48
-0.64
-0.36
-0.48
-0.48
-0.64
-0.36
-0.48[ k ]1 =
2 3 41
AE5
23
4
1
[ k ]2 =
2 5 61
AE5
25
6
1
-0.48
0.64
0.36
-0.480.48
-0.64
-0.36
0.48
0.48
-0.64
-0.36
0.48
[ k ]3 =
2 7 81
AE5
27
8
1
-0.48
0.64
0.36
-0.480.48
-0.64
-0.36
0.48
0.48
-0.64
-0.36
0.48
31
2
34
56
78
Member
#1
#2
#3
λx
-0.8
- 0.8
0.8
λy
-0.6
0.6
-0.6
λx2 λx λy λy
2
0.64 0.48 0.36
0.64 -0.48 0.36
0.64 -0.48 0.36
0.48
0.64
0.36
0.48
-0.48
0.64
0.36
-0.48
16
34
56
78
1
2
31
2
Global
Q1 = -50
Q2 = -80
D1
D2
D1
D2=
-250.65/AE
-481.77/AE
3 m
3 m
4 m 4 m
80 kN
50 kN
5 m
5 m5 m
1
2
1
-0.48
2
-0.48
1.08
1.92
= AE5
80 kN
50 kN
0
0+1
2
17
1
2
3
434
56
78
1
2
31
2
Local
= -97.9 kN (C)
[q´F]1 = AE5 0.8 0.6 -0.8 -0.6 D2=
D1=
D4=D3=
-481.77/AE-250.65/AE
0.00.0
= +17.7 kN (T)
D2=D1=
D6=D5=
-481.77/AE-250.65/AE
0.00.0
= -17.7 kN (C)
80 kN
50 kN 36.87o
97.9 kN
17.7 kN
17.7 kN
#1 -0.8 -0.6#2 -0.8 0.6#3 0.8 -0.6
[q´F]2 = AE5 0.8 -0.6 -0.8 0.6
[q´F]3 = AE5 -0.8 +0.6 +0.8 -0.6 D2=
D1=
D8=D7=
-481.77/AE-250.65/AE
0.00.0
Member
#1#2#3
λx λy
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
18
80 kN
50 kN 36.89o
97.9 kN
17.7 kN
17.7 kN
80 kN
50 kN
Member
#1#2#3
λx
-0.8-0.80.8
λy
-0.60.6-0.6
1
2
3
434
56
78
1
2
31
2
97.9(0.8)=78.32 kN
97.9(0.6)=58.74 kN
17.7(0.8)=14.16 kN
17.7(0.6)=10.62 kN
17.7(0.8)=14.16 kN
17.7(0.6)=10.62 kN
ΣFx ´ = 0:+ 17.7 + 17.7 +50cos 36.89 - 97.9cos73.78 - 80cos53.11 = 0, O.K
Check :
19
Example 2
For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the deflections of the loaded joint.The support B settles downward 2.5 mm. Temperature in member BDincrease 20 oC. Take α = 12x10-6 /oC, AE = 8(103) kN.
A
BC
D
8 kN
4 kN
4 m
3 m+20o C
∆B = 2.5 mm
20
λyλx
λxλx
λyλy
λxλy
λyλx
λxλx
λyλy
λxλy
−λyλx
−λxλx
−λyλy
−λxλy
−λyλx
−λxλx
−λyλy
−λxλy
∆B = 2.5 mm
A
BC
D
8 kN
4 kN
4 m
3 m+20o C
Vi Uj VjUi
[ k ]m = AEL
Vi
Uj
Vj
Ui
12
3 4
(0,0)
(-4,-3)
(-4,0)
(0,-3)
2
1
3
1
23
4
56
78
Member
#1
#2
λx λy
#3
-4/4 = -1
-4/5 = -0.8
0
0
-3/5 = -0.6
-3/3 = -1
Ljyy
Lixx ijij
ij
�)(�)(� −+
−=λ
cosθx = λx cosθy = λy
21
0.096
0.128
0.072
0.096
0
0
0.333
0
0
0.25
0
0
0
0.25
0
00
-0.25
0
0
0
-0.25
0
0[k]1 = 8x103
2 3 41
234
1
[k]3 = 8x103
2 7 81
278
1
0
0
0.333
00
0
-0.333
0
0
0
-0.333
0
Member
#1
#2
#3
λx
-1
- 0.8
0
λy
0
-0.6
-1
λx2/L λx λy/L λy
2/L
0.25 0 0
0.128 0.096 0.072
0 0 0.333
0.096
0.128
0.072
0.096
-0.096
-0.128
-0.072
-0.096-0.096
-0.128
-0.072
-0.096
[k]2 = 8x103
2 5 61
256
1
12
3 4
(0,0)
(-4,-3)
(-4,0)
(0,-3)
2
1
3
1
23
4
56
78
1
23
4
56
78
[K] = 8x103
21
21
0.096
0.378
0.405
0.096
22
∆B = 2.5 mm
1.536 kN
1.152 kN
2+20oC1.536 kN
1.152 kN
2+20oC
1.92 kΝ =α(∆T1)AE = (12x10-6)(20)(8x103)
1.92 kN
1.536 kN
1.152 kN
1.536 kN
1.152 kN
12
34
(0,0)
(-4,-3)
(-4,0)
(0,-3)
2
1
3
1
23
4
56
78
∆B = 2.5 mm
+20o C
-1.536
1.536-1.152
1.152
+
1
2
6
5
q1
q2
q1
q5
q2
q6
-1.536
-1.152+1
2
0
-2.5x10-3
5
6-0.096
-0.128
-0.072
-0.096+ 8x103
5 6
= 8x103
21
21
0.096
0.128
0.072
0.096 d1
d2
d1
d5 = 0d2
d6 = -2.5x10-3
0.096
0.128
0.072
0.096
= 8x103
2 5 61
256
1
0.096
0.128
0.072
0.096
-0.096
-0.128
-0.072
-0.096-0.096
-0.128
-0.072
-0.096
[q] = [k]m[d] + [qF]Member 2:
q1
q2
-1.536
-1.152+1
2= 8x103
21
0.096
0.128
0.072
0.096 d1
d2
1.92
1.44+
23
[Q] = [K][D] + [QF]
Global:
A
BC
D
8 kN
4 kN
4 m
3 m+20o C
∆B = 2.5 mm
12
3 4
(0,0)
(-4,-3)
(-4,0)
(0,-3)
2
1
3
1
23
4
56
78
Q1 = -4
Q2 = -8= 8x103
21
21
0.096
0.378
0.405
0.096 D1
D2
-1.536
-1.152+1.92
1.44+
D1
D2
-0.8514x10-3 m
-2.356x10-3 m=
24
Member
#1
#2
#3
λx λy
Local12
3 4
2
1
3
1
23
4
56
78
[q´F]1 = 8x103 1.0 0.0 -1.0 0.04
= -1.70 kN (C)
D2=D1=
D4=D3=
0.00.0
-0.8514x10-3
-2.356x10-3
-1.92+[q´F]2 = 8x103 0.8 0.6 -0.8 -0.65
= -2.87 kN (C)
D2=D1=
D6=D5=
-0.00250.0
-0.8514x10-3
-2.356x10-3
= -6.28 kN (C)
D2=D1=
D8=D7=
0.00.0
-0.8514x10-3
-2.356x10-3[q´F]3 = 8x103 0.0 1.0 0.0 -1.0
3
2+20oC
1.92 kN
1.92 kN
-1 0
- 0.8 -0.6
0 -1
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
25
4 kN
8 kN
6.28 kN
1.70 kN
2.87 kN
12
3 4
2
1
3
1
23
4
56
78
Member
#1
#2
#3
cosθx
-1
- 0.8
0
cosθy
0
-0.6
-1
[q´]m
-1.70
-2.87
-6.28
4 kN
8 kN
2
1
3
1.70 kN
6.28 kN
2.87(0.8) = 2.30 kN
2.87(0.6) = 1.72 kN
26
∆ AD = + 3 mm
AB
C
3 m
D
8 kN
4 kN
4 m 4 m∆ = - 4 m
m
Example 3
For the truss shown, use the stiffness method to:(a) Determine the end forces of each member and reactions at supports.(b) Determine the displacement of the loaded joint.Take AE = 8(103) kN.
27
∆ AD = + 3 mm
AB
C
D
8 kN
4 kN
3 m
4 m 4 m
∆ = - 4 mm
21
4
3
5
1
2
34
56
78
1
2 3 4
(0,0)
(-4,-3) (4,-3)(0,-3)
λyλx
λxλx
λyλy
λxλy
λyλx
λxλx
λyλy
λxλy
−λyλx
−λxλx
−λyλy
−λxλy
−λyλx
−λxλx
−λyλy
−λxλy
Vi Uj VjUi
[ k ]m = AEL
Vi
Uj
Vj
Ui
Member
#1
#2
λx λy
#3
#4
#5
-4/5 =-0.8
0
4/5 = 0.8
4/4 = 1
4/4 = 1
-3/5 = -0.6
-3/3 = -1
-3/5 = -0.6
0
0
Ljyy
Lixx ijij
ij
�)(�)(� −+
−=λ
cosθx = λx cosθy = λy
28
[k]1 = 8x103
2 3 41
234
1
0.096
0.128
0.072
0.096
-0.096
-0.128
-0.072
-0.096-0.096
-0.128
-0.072
-0.096
0.096
0.128
0.072
0.096
[k]2 = 8x103
2 5 61
256
1
0
0
0.333
00
0
-0.333
0
0
0
0.333
0
0
0
-0.333
0
[k]3 = 8x103
2 7 81
278
1
-0.096
0.128
0.072
-0.0960.096
-0.128
-0.072
0.096
0.096
-0.128
-0.072
0.096-0.096
0.128
0.072
-0.096
Member
#1
#2
λy
#3
λx
-0.8
0
0.8
-0.6
-1
-0.6
λx2/L λy
2/Lλxλy/L
0.128 0.0720.096
0 0.3330
0.128 0.072-0.096
21
4
3
5
1
2
34
56
78
1
2 3 4
(0,0)
(-4,-3) (4,-3)(0,-3)
5 m
3 m 5 m
4 m 4 m
29
21
4
3
5
1
2
34
56
78
1
2 3 4
(0,0)
(-4,-3) (4,-3)(0,-3)
5 m
3 m 5 m
4 m 4 m
Member
#4
#5
λyλx
1
1
0
0
λx2/L λy
2/Lλxλy/L
0
0.25
0
00
-0.25
0
0
0
0.25
0
0
0
-0.25
0
0
0
0.25
0
00
-0.25
0
0
0
0.25
0
0
0
-0.25
0
0[k]5= 8x103
6 7 85
678
5
0.25 00
0.25 00
Global Stiffness Matrix
[K] = 8x103
2 5 71
257
1
34 6
8
[k]4= 8x103
4 5 63
456
3
30
Global Stiffness Matrix
0.2560.0 0.477
0.0 0.00.0
0.0 0.0-0.128 0.096
-0.1280.096
0.50-0.25
-0.250.378
[K] = 8x103
2 5 71
257
1
[k]1 = 8x103
2 3 41
234
1
0.096
0.128
0.072
0.096
-0.096
-0.128
-0.072
-0.096-0.096
-0.128
-0.072
-0.096
0.096
0.128
0.072
0.096
[k]2 = 8x103
2 5 61
256
1
0
0
0.333
00
0
-0.333
0
0
0
0.333
0
0
0
-0.333
0
[k]3 = 8x103
2 7 81
278
1
-0.096
0.128
0.072
-0.0960.096
-0.128
-0.072
0.096
0.096
-0.128
-0.072
0.096-0.096
0.128
0.072
-0.096
0
0.250
-0.25
0
00
0
0
0.25
0
-0.250
0
0
0[k]4= 8x103
4 5 63
456
3
0
0.250
-0.25
0
00
0
0
0.25
0
-0.250
0
0
0[k]5= 8x103
6 7 85
678
5
31
∆AE/L = 10.67 kN
∆ AD = + 3 mm1
3.84 kN
2.88 kN
3.84 kN
2.88 kN
∆ AD = + 3 mm
AB
C
D
8 kN
4 kN
3 m
4 m 4 m
∆ = - 4 mm
1
2
21
4
3
534
56
78
Global Fixed end forces
0.00.0
257
1∆AE/L = 4.8 kN
4.8 kN
∆AE/L = 10.67 kN
10.67 kN∆ = -4 m
m
2 Fixed End3.84 kN
2.88 kN-3.84-2.88 + 10.67 = 7.79
32
∆ AD = + 3 mm
AB
C
D
8 kN
4 kN
3 m
4 m 4 m
∆ = -4 mm
Global:
1
2
21
4
3
534
56
78
[Q] = [K][D] + [QF]
Q1 = 4
Q5 = 0Q2 = -8
Q7 = 0
= 8x103
2 5 71
257
1
-0.250.500.00.0
0.378-0.250.096
-0.128
-0.1280.00.0
0.256
0.0960.0
0.4770.0 D1
D5
D2
D7
-3.84
0.07.79
0.0
+
D1
D5
D2
D7
6.4426x10-3 m
2.6144x10-3 m-5.1902x10-3 m
5.2288x10-3 m
=
8 kN
4 kN
33
D1
D5
D2
D7
6.4426x10-3 m
2.6144x10-3 m-5.1902x10-3 m
5.2288x10-3 m
=
-4.8+D2
D1
00
= -1.54 kN (C)
1
2
21
4
3
534
56
78
Member forces
= -3.17 kN (C)
10.67+
[q´F]1 = 8x103 0.8 0.6 -0.8 -0.65
[q´F]2 = 8x103 0.0 1.0 0.0 -1.03
D2
D1
0D5
Member
#1
#2
λiyλix
-0.8 -0.6
0 -1
1
4.8 kN
4.8 kN∆ AD
= + 3 mm
10.67 kN
10.67 kN
2
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
34
D1
D5
D2
D7
6.4426x10-3 m
2.6144x10-3 m-5.1902x10-3 m
5.2288x10-3 m
=
1
2
21
4
3
534
56
78
00
0D5
[q´F]4 = 8x103 -1.0 0.0 1.0 0.04
= 5.23 kN (T)
0D5
0D7
[q´F]5 = 8x103 -1.0 0.0 1.0 0.04
= 5.23 kN (T)
D2
D1
0D7
[q´F]3 = 8x103 -0.8 0.6 0.8 -0.65
= -6.54 kN (C)
Member λx λy
#3#4
#5
0.8 -0.61 0
1 0
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
35
1
2
21
4
3
534
56
78
-0.8
0
0.81
1
-0.6
-1
-0.60
0
Member
#1
#2
λx
#3#4
#5
λy [q´]
4 kN8 kN
3.17 kN
3.17 kN
6.54 kN
6.54 kN
1.54 kN
1.54 kN
5.23 kN5.23 kN
4 kN8 kN
0.92 kN
4 kN
3.17 kN 3.92 kN
-1.54
-3.17
-6.545.23
5.23
36
Special Trusses (Inclined roller supports)
37
λix = cos θi
λiy = sin θi
λjx = cos θj
λjy = sin θj
[ q* ] = [ T ]T[ q´ ] 1
2
3*
4*
56
78
3
2
4
5
1
x *
y *
θi
x
y
θj
1
i
j
q´i
q´j
q*3
q1
q*4
q2
q´iq´j
[T]T
00λiyλix
λjyλjx00 [ T ] = [[ T ]T]T =
=00
λiy
λix
λjy
λjx
00
1
i
j1
2
3*
4*
Transformation Matrices
38
[ k ] = [ T ]T[ k´ ][T]
λjyλjx
λjxλjx
−λiyλjx
−λixλjx
λjyλjy
λjxλjy
−λiyλjy
−λixλjy
λiyλix
λixλix
−λjyλix
−λjxλix
λiyλiy
λixλiy
−λjyλiy
−λjxλiy
Vi Uj VjUi
[ k ]m = AEL
Vi
Uj
Vj
Ui
00λiyλix
λjyλjx00-11
1-1
AEL[ k ]m =
00
λiy
λix
λjy
λjx
00
39
Example 5
For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint. AE is constant.
3 m
4 m
45o
30 kN
40
3 m
4 m45o
2
1
3
1
2
5
63*
4*
Member 1:
15
63*
4*
[q*]θi = 0,λix = cos 0 = 1,λiy = sin 0 = 0
θij = -45o = 135o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707
45o
[q*] = [T*]T[q´] + [T*]T[q´F]
q5
q3*
q6
q4*
q´iq´j
[T*]T
=0001
-0.7070.707
00
q´i q´j
i j1
41
15
63*
4*
[q*]θi = 0 ;λix = cos 0 = 1,λiy = sin 0 = 0
θij = -45o = 135o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707
45o
0-0.707
00.707
00
10
AE4
1-1
-11[k*]1 =
0001
-0.7070.707
00
Member 1:
[k*]1 = AE 63*4*
53*
-0.1250.125
0-0.1768
4*
0.125-0.125
00.1768
5
0.1768-0.1768
00.25
6
0000
q´i q´j
i j1
3 m
4 m45o
2
1
3
1
2
5
63*
4*
[ ]TL
AETk T
−
−=
1111
][*][
42
Member 2:q´i
q´j
i
j
22
1
2
3*4*
θi = -90o = 270o,λix = cos(-90o) = 0,λiy = sin(-90o) = -1
90o+45o
=135o
90o
θj = -135o = 215o ,λix = cos (-135o) = -0.707,λiy = sin(- 135o) = -0.707
0-0.707
0-0.707
-10
00
AE3
1-1
-11[k*]2 =
00-10
-0.707-0.707
00
[k*]2 = AE
2 4*1
23*4*
1
0.16670.1667-0.2357
0
0.16670.1667-0.2357
0
0000
-0.2357-0.23570.3333
03*
3 m
4 m45o
2
1
3
1
2
5
63*
4*
[ ]TL
AETk T
−
−=
1111
][*][
43
[ ]TL
AETk T
−
−=
1111
][][
36.87o
Member 3:
θi = θj = 36.87o ;λix = λjx = cos (36.87o) = 0.8,λiy = λjy = sin(36.87o) = 0.6
00.6
00.8
0.60
0.80
AE5
1-1
-11[k]3 =
00
0.60.8
0.60.800
3
1
2
5
63
q´i
q´j
i
j36.87o
[k]3 = AE
6 25
612
5
0.0960.128-0.096-0.128
0.0720.096-0.072-0.096
-0.096-0.1280.0960.128
-0.072-0.0960.0720.096
1
3 m
4 m45o
2
1
3
1
2
5
63*
4*
44
3 m
4 m45o
2
1
3
1
2
5
63*
4*
[k*]1 = AE 63*4*
53*
-0.1250.125
0-0.1768
4*
0.125-0.125
00.1768
5
0.1768-0.1768
00.25
6
0000
5
6 4*
Global Stiffness:
[k*]2 = AE
2 4*1
23*4*
1
0.16670.1667-0.2357
0
0.16670.1667-0.2357
0
0000
-0.2357-0.23570.3333
03*
[k]3 = AE
6 25
612
5
0.0960.128-0.096-0.128
0.0720.096-0.072-0.096
-0.096-0.1280.0960.128
-0.072-0.0960.0720.096
1[K] = AE 2
3*
11
00.0960.128
2
-0.23570.40530.096
0.2917-0.2357
03*
45
Global :
[Q] = [K][D] + [QF]
Q1 = 30
Q3*= 0
Q2 = 0D1
D3*
D2 = AE 23*
11
00.0960.128
2
-0.23570.40530.096
0.2917-0.2357
03*
D1
D3*
D2 =352.5
-127.3-157.5AE
1
3 m
4 m45o
30 kN
3 m
4 m45o
2
1
3
1
2
5
63*
4*
5
6 4*
46
[q´F]1 = AE 00
0D3*
-1 0 0.707 -0.7074
= -22.50 kN, (C)
D2
D1
0D3*
[q´F]2 = AE 0 1 -0.707 -0.7073
= -22.50 kN, (C)
00
D2
D1
[q´F]3 = AE -0.8 -0.6 0.8 0.65
= 37.50 kN, (T)
D1
D3*
D2 =352.5
-127.3-157.5AE
1
Member Forces :
Member
#1
λiyλix λjx λjy
#2
#3
1 0 0.707 -0.707
0 -1 -0.707 -0.707
0.8 0.6 0.8 0.6
3 m
4 m45o
2
1
3
1
2
5
63*
4*
5
6 4*
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
47
36.87o 45o45o
3 m
4 m45o
2
1
3
1
2
5
63*
4*
Member
Member Force (kN)
[q´]2[q´]1 [q´]3
-22.50 -22.50 37.50
22.50 kN
22.50 kN
37.50 kN
22.50 kN
7.50 kN31.82 kN
Reactions :
3 m
4 m45o
30 kN
48
Example 6
For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint. AE is constant.
3 m
4 m
45o
30 kN
4 m
49
15
63*
4*
[q*]θi = 0o,λix = cos 0o = 1,λiy = sin 0o = 0
θij = -45o
λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707
45o
[q*] = [T*]T[q´] + [T*]T[q´F]
q5
q3*
q6
q4*
q´iq´j
[T*]T
=0001
-0.7070.707
00
q´i q´j
i j1
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78Member 1:
50
3 m
4 m 4 m
15
63*
4*
[q*]θi = 0o,λix = cos 0o = 1,λiy = sin 0o = 0
θij = -45o = 315o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707
45o
0-0.707
00.707
00
10
AE4
1-1
-11[k*]1 =
0001
-0.7070.707
00
[k*]1 = AE 63*4*
53*
-0.1250.125
0-0.1768
4*
0.125-0.125
00.1768
5
0.1768-0.1768
00.25
6
0000
q´i q´j
i j1
2
1
3
4
5
12
5
6 3*4*
78Member 1:
[ ]TL
AETk T
−
−=
1111
][*][
51
q´i
q´j
i
j
22
1
2
3*4*
θi = -90o,λix = cos(-90o) = 0,λiy = sin(-90o) = -1
90o
θj = -135o = 215o,λix = cos (-135o) = -0.707,λiy = sin(- 135o) = -0.707
0-0.707
0-0.707
-10
00
AE3
1-1
-11[k*]2 =
00-10
-0.707-0.707
00
[k*]2 = AE
2 4*1
23*4*
1
0.16670.1667-0.2357
0
0.16670.1667-0.2357
0
0000
-0.2357-0.23570.3333
03*
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78Member 2:
[ ]TL
AETk T
−
−=
1111
][*][
90o+45o
=135o
52
36.87o
Member 3:
θi = θj = 36.87o ;λix = λjx = cos (36.87o) = 0.8,λiy = λjy = sin(36.87o) = 0.6
00.6
00.8
0.60
0.80
AE5
1-1
-11[k]3 =
00
0.60.8
0.60.800
3
1
2
5
63
q´i
q´j
i
j36.87o
[k]3 = AE
6 25
612
5
0.0960.128-0.096-0.128
0.0720.096-0.072-0.096
-0.096-0.1280.0960.128
-0.072-0.0960.0720.096
1
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78
[ ]TL
AETk T
−
−=
1111
][*][
53
3 m
4 m 4 m
θi = θij = 0o; λix = λjx = cos 0o = 1, λiy = λjy = sin 0o = 0
00
01
00
10
AE4
1-1
-11[k]1 =
0001
0100
q´i q´j
i j1
2
1
3
4
5
12
5
6 3*4*
78Member 4:
47
8
[q]1
2
[k]4 = AE 278
17
00.25
0-0.25
8
0000
1
0-0.25
00.25
2
0000
[ ]TL
AETk T
−
−=
1111
][*][
54
Member 5:
θi = - 8.13o;λix = cos (- 8.13o) = 0.9899,λiy = sin(- 8.13o) = -0.1414
00.6
00.8
-0.14140
0.98990
AE5
1-1
-11
5
q´i
q´j
i
j36.87o
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78
53*4*
7
8
[k*]5 =00
-0.14140.9899
0.60.800
[k*]5 = AE
4* 83*
4*78
3*
0.0960.128
0.02263-0.1584
0.0720.096
0.01697-0.1188
-0.1188-0.1584-0.0280.196
0.016970.022630.004-0.028
7
θ j = 36.87o ;
λ jx = cos (36.87o ) = 0.8,
λ jy = sin
(36.87o ) = 0.6
[ ]TL
AETk T
−
−=
1111
][*][
8.13o
550 -0.2357-0.2357
0
[k*]1 = AE 63*4*
53*
-0.1250.125
0-0.1768
4*
0.125-0.125
00.1768
5
0.1768-0.1768
00.25
6
0000
[k*]3 = AE
6 25
612
5
0.0960.128-0.096-0.128
0.0720.096-0.072-0.096
-0.096-0.1280.0960.128
-0.072-0.0960.0720.096
1
3 m
4 m 4 m
2
1
3
4
5
Global Stiffness:1
2
5
6 3*4*
78
5
6 4*
78
[k*]2 = AE
2 4*1
23*4*
1
0.16670.1667-0.2357
0
0.16670.1667-0.2357
0
0000
-0.2357-0.23570.3333
03*
[k]4 = AE 278
17
00.25
0-0.25
8
0000
1
0-0.25
00.25
2
0000
[k*]5 = AE
4* 83*
4*78
3*
0.0960.128
0.02263-0.1584
0.0720.096
0.01697-0.1188
-0.1188-0.1584-0.0280.196
0.016970.022630.004-0.028
7
[K] = AE 23*
11 2
0.0960.378
0.40530.096
3*
0.4877
56
Global :30 kN
4 m 4 m
3 m2
1
3
4
5
12
5
6 3*4*
78
5
6 4*
78
[Q] = [K][D] + [QF]
Q1 = 30
Q3*= 0
Q2 = 0D1
D3*
D2
D1
D3*
D2 =86.612
-13.791-28.535AE
1
= AE 23*
11
00.0960.378
2
-0.23570.40530.096
0.4877-0.2357
03*
57
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78
5
6 4*
78
D1
D3*
D2 =86.612
-13.791-28.535AE
1
[q´F]1 = AE 00
0D3*
-1 0 0.707 -0.7074
= -2.44 kN, (C)
D2
D1
0D3*
[q´F]2 = AE 0 1 -0.707 -0.7073
= -6.26 kN, (C)
00
D2
D1
[q´F]3 = AE -0.8 -0.6 0.8 0.65
= 10.43 kN, (T)
Member Forces :
Member
#1
λiyλix λjx λjy
#2
#3
1 0 0.707 -0.707
0 -1 -0.707 -0.707
0.8 0.6 0.8 0.6
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
58
Member
#4
λiyλix λjx λjy
#5
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78
5
6 4*
78
D1
D3*
D2 =86.612
-13.791-28.535AE
1
D2
D1
00
[q´F]4 = AE -1 0 1 04
= -21.65 kN, (C)
0D3*
00
[q´F]5 = AE -0.9899 0.141 0.8 0.65
= 2.73 kN, (T)1 0 1 0
0.9899 -0.141 0.8 0.6
Member Forces :
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
59
36.87o 45o45o
81.87o
36.87o
Member
Member Force (kN)
[q´]1 [q´]2 [q´]3 [q´]4 [q´]5
-2.44 -6.26 10.43 -21.65 2.73
Reactions :30 kN
4 m 4 m
3 m2
1
3
4
5
12
5
6 3*4*
78
5
6 4*
78
6.26 kN10.43 kN
21.65 kN
2.73 kN
2.44 kN 2.44 kN5.90 kN
6.26 kN
19.47 kN
1.64 kN
6.54 kN
60
AB
C
3 m
D
8 kN
4 kN
4 m36.87o
4 m
λjyλjx
λjxλjx
−λiyλjx
−λixλjx
λjyλjy
λjxλjy
−λiyλjy
−λixλjy
λiyλix
λixλix
−λjyλix
−λjxλix
λiyλiy
λixλiy
−λjyλiy
−λjxλiy
Vi Uj VjUi
[ k *]m = AEL
Vi
Uj
Vj
Ui
Example 7
For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint.Take AE = 8(103) kN.
61
Member 1:
λix = cos 73.74o = 0.28,λiy = sin 73.74o = 0.96
[q*] = [T*]T[q´] + [T*]T[q´F]
λjx = cos 36.87o = 0.8,λjy = sin 36.87o = 0.6
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
x *
y *
73.74o
x
y
36.87o
1
i
j1
2
3*
4*1
i
j
q´i
q´j
q3*
q1
q4*
q2
q´iq´j
=00
0.960.28
0.60.8
00
[T*]T
3
2
4
5
1
1
2
56
78
3*
4*
62
[k]1 = 8x103
4* 1 23*
4*
12
3*
0.0960.128
-0.1536-0.0448
0.0720.096
-0.1152-0.0336
-0.0336-0.04480.053760.01568
-0.1152-0.15360.184320.05376
000.960.280.60.800
[k]1 =00
0.960.28
0.60.8
00
8x103
51
-1-11
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
3
2
4
5
1
1
2
56
78
3*
4*
[ ]TL
AETk T
−
−=
1111
][][
63
x *
y*
36.87x
yMember 2:
λix = cos 36.87o = 0.8,λiy = sin 36.87o = 0.6
[q*] = [T*]T[q´] + [T*]T[q´F]
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
λjx = cos 0o = 1,λjy = sin 0o = 0
q3*
q5
q4*
q6
q´iq´j
[T*]T
[k] = [TT] AEL
[T]1-1
-11
q´i q´j
i j2
=00
0.60.8
0100
i j2 5
6
3*
4*
[k]2 = 8x103
4* 5 63*
4*
56
3*
00.25
-0.15-0.2
0000
0-0.20.120.16
0-0.150.090.12
3
2
4
5
1
1
2
56
78
3*
4*
64
x
y
270o 1
2
784
Member 3:
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
λx = cos 323.13o = 0.8,λy = sin 323.13o = -0.6
Member 4:
[k]4 = 8x103
2 7 81
278
1
-0.0960.1280.096
-0.128
0.072-0.096-0.0720.096
0.096-0.128-0.0960.128
-0.0720.0960.072
-0.096
λx = cos 270o = 0, λy = sin 270o = -1
[k]3 = 8x103
0000
0.3330
-0.3330
00
00
2 5 61
256
10.333
0
-0.3330
x
y
323.13o3
1
2
56
3
2
4
5
1
1
2
56
78
3*
4*
65
x
yMember 5:
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
λx = cos 0o = 1,λy = sin 0o = 0
00.25
0-0.25
0000
00.25
0-0.25
00
00
[k]5 = 8x103
6 7 85
678
5
5 78
56
3
2
4
5
1
1
2
56
78
3*
4*
660.17568
-0.2-0.20.5
00
0 0
-0.0336-0.0448
-0.0448 -0.03360.0
0.2560.4740.0
[k]1 = 8x103
4* 1 23*
4*
12
3*
0.0960.128
-0.1536-0.0448
0.0720.096
-0.1152-0.0336
-0.0336-0.04480.053760.01568
-0.1152-0.15360.184320.05376
[k]4 = 8x103
2 7 81
278
1
-0.0960.1280.096
-0.128
0.072-0.096-0.0720.096
0.096-0.128-0.0960.128
-0.0720.0960.072
-0.096
00.25
0-0.25
0000
00.25
0-0.25
00
00
[k]5 = 8x103
6 7 85
678
5
[k]2 = 8x103
4* 5 63*
4*
56
3*
00.25
-0.15-0.2
0000
0-0.20.120.16
0-0.150.090.12
[k]3 = 8x103
0000
0.3330
-0.3330
00
00
2 5 61
256
10.333
0
-0.3330
[K] = 8x103
2 3* 51
23*
5
1
3
2
4
5
1
12
56
78
3*
4* 6
784*
67
3
2
4
5
1
2
3*
4*
56
78A
BC
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
1
Global: [Q] = [K][D] + [QF]
D1
D3*
D2
D5
1.988x10-3 m
1.996x10-4 m-2.0824x10-3 m
7.984x10-5 m
=
Q1 = 4
Q3*= 0
Q2 = -8
Q5 = 0
D1
D3*
D2
D5
0.17568-0.2
-0.20.5
00
0 0
-0.0336-0.0448
-0.0448 -0.03360.0
0.2560.4740.0
= 8x103
2 3* 51
23*
5
1
8 kN
4 kN
68
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
3
2
4
5
1
2
3*
4*
56
78
1
Member forces
D1
D3*
D2
D5
1.988x10-3 m
1.996x10-4 m-2.0824x10-3 m
7.984x10-5 m
=
0D3*
D2
D1
[q´F]1 = 8x103 -0.28 -0.96 0.8 0.65
= 0.46 kN, (T)
0D3*
0D5
[q´F]2 = 8x103 -0.8 -0.6 1 04
= -0.16 kN, (C)
Member
#1
#2
λiyλix λjx λjy
0.28 0.96 0.8 0.6
0.8 0.6 1 0
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
69
D1
D*3
D2
D5
1.988x10-3 m
1.996x10-4 m-2.0824x10-3 m
7.984x10-5 m
=
3
2
4
5
1
2
3*
4*
56
78
1
[q´F]3 = 8x103 D2
D1
0D5
0 1 0 -13
= -5.55 kN
D2
D1
00
[q´F]4 = 8x103 -0.8 0.6 0.8 - 0.65
= -4.54 kN
0D5
00
[q´F]5 = 8x103 -1 0 1 04
= -0.16 kN
Member
#3
λiyλix λjx λjy
#4
#5
0 -1 0 -1
0.8 -0.6 0.8 -0.6
1 0 1 0
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
70
y*
x *
36.87o
36.87o 36.87o
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
3
2
4
5
1
2
3*
4*
56
78
1
0.46 -0.16 -5.55 -4.54 -0.16
0.46 kN
5.55 kN0.36 kN
4.54 kN
3.79 kN
2.72 kN
Member
Member Force (kN)
[q]2[q]1 [q]3 [q]4 [q]5
0.16 kN 0.16 kN
5.55 kN
71
Space-Truss Analysis
72
+
−
−=
Fj
Fi
j
i
j
i
dd
LEA
''
''
1111
''
[ ] [ ]F
jz
jy
jx
iz
iy
ix
q
dddddd
TL
EA '1111
+
−
−=
[ ]
=
zyx
zyxT
where
λλλλλλ000
000,
[q´] = [k´][d´] + [q´F]
= [k´][T][d] + [q´F]
Member Local Stiffness [k´]:
73
Member Global Stiffness [km]:
[km]= [T]T[k´] [T]
[ ]
−
−
=zyx
zyx
z
y
x
z
y
x
m LEAk
λλλλλλ
λλλ
λλλ
000000
1111
000
000
74
Global equilibrium matrix:
[Q] = [K][D] + [QF]
QI
QII
Du
Dk
KI,II
KII,II
KI,I
KII,I
Reaction Support Boundary Condition
Unknown DisplacementJoint Load
QFI
QFII
= +
Fixed End Forces
75
q´j
q´i
i
j
m
i
j
m
=EAL
λxλx
λzλx
λyλx
λxλy
λyλy
λzλy
λxλz
λyλz
λzλz
λxλx
λzλx
λyλx
λxλy
λyλy
λzλy
λxλz
λyλz
λzλz
−λxλx
−λzλx
−λyλx
−λxλy
−λyλy
−λzλy
−λxλz
−λyλz
−λzλz
−λxλx
−λzλx
−λyλx
−λxλy
−λyλy
−λzλy
−λxλz
−λyλz
−λzλz
diy
dix
djx
diz
djy
djz
qiy
qix
qjx
qiz
qjy
qjz
qjy
qjx
qjzqiy
qix
qiz
+
−
−=
Fj
Fi
j
i
j
i
dd
LEA
''
''
1111
''
[ ] [ ] [ ]F
jz
jy
jx
iz
iy
ix
zyxzyxmj q
dddddd
LEAq '' +
−−−= λλλλλλ
76
x
z
y
4 m4 m
3 m3 m
10 m
80 kN60 kN
O
Example 6
For the truss shown, use the stiffness method to:(b) Determine the end forces of each member.(a) Determine the deflections of the loaded joint.Take E = 200 GPa, A = 1000 mm2.
77
λ1 = (-4/11.18)i + (3/11.18)j + (-10/11.18)k
= -0.3578 i + 0.2683 j - 0.8944 k
λ2 = (+4/11.18)i + (3/11.18)j + (-10/11.18)k= +0.3578 i + 0.2683 j - 0.8944 k
λ3 = (+4/11.18)i + (-3/11.18)j + (-10/11.18)k
= +0.3578 i - 0.2683 j - 0.8944 k
= -0.3578 i - 0.2683 j - 0.8944 k
λ4 = (-4/11.18)i + (-3/11.18)j + (-10/11.18)k
Member
#1
#2
#3
λx λy λz
#4
-0.3578 +0.2683 -0.8944
+0.3578 +0.2683 -0.8944
+0.3578 -0.2683 -0.8944
-0.3578 -0.2683 -0.8944
λm = λxi + λyj + λzk
12
34
12
3
(0, 0, 10)
(-4, 3, 0)
(4, 3, 0)
(4, -3, 0)
(-4, -3, 0)4
1
2
53
x
z
y
4 m4 m
3 m3 m
10 m
80 kN60 kN
O
78
Member
#1#2
#3
λx λy λz
#4
-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944
+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944
Member Stiffness Matrix [k]6x6
[k]m =[k12]3x3
[k22]3x3
[k11]3x3
[k21]3x3
2 31
23
1
+0.80-0.240
+0.320
+0.320-0.096
+0.128
-0.240+0.072
-0.096
[k11]1 =AEL
2 31
23
1
+0.80-0.240
-0.320
-0.320+0.096
+0.128
-0.240+0.072
+0.096
[k11]2 =AEL
2 31
23
1
+0.80+0.240
-0.320
-0.320-0.096
+0.128
+0.240+0.072
-0.096
[k11]3 =AEL
2 31
23
1
+0.80+0.240
+0.320
+0.320+0.096
+0.128
+0.240+0.072
+0.096
[k11]4 =AEL
2 31
23
1
3.20.0
0.0
0.00.0
0.512
0.00.288
0.0
[KI,I] =AEL
79
12
34
12
3
(0, 0, 10)
(-4, 3, 0)
(4, 3, 0)
(4, -3, 0)
(-4, -3, 0)4
1
2
53
x
z
y
4 m4 m
3 m3 m
10 m
80 kN60 kN
O
0.0-80
60
D3
D2
D1
0.00.0
0.0+
2 31
23
1
3.20.0
0.0
0.00.0
0.512
0.00.288
0.0
=AEL
[Q] = [K][D] + [QF]
80
Global equilibrium matrix:
[Q] = [K][D] + [QF]
QI
QII
Du
Dk
KI,II
KII,II
KI,I
KII,I
Reaction Support Boundary Condition
Unknown DisplacementJoint Load
QFI
QFII
= +
Fixed End Forces
(AE/L) = (1x10-3)(200x106)/(11.18) = 17.89x103 kN
0.0-80
60
D3
D2
D1
0.00.0
0.0+
2 31
23
1
3.20.0
0.0
0.00.0
0.512
0.00.288
0.0
=AEL
0.0 mm-15.53 mm
6.551 mm=
D3
D2
D1
= LAE
0.0-277.8
+117.2
81
Member forces:
q´F+
Member
#1#2
#3
λx λy λz
#4
-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944
+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944
dyi
dxi
dxj
dzi
dyj
dzj
[q´j]m = AEL −λx −λy −λz λx λy λz
D3
D2
D1
[ 0 ]
= +116.5 kN (T)
+0.3578 -0.2683 +0.8944[q´j]1 =AEL
0.0
117.2
-277.8L
AE
x
z
y
4 m4 m
3 m3 m
10 m
80 kN60 kN
O
82
Member
#1#2
#3
λx λy λz
#4
-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944
+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944
= +32.61 kN (T)-0.3578 -0.2683 +0.8944[q´j]2 =AEL
0.0
117.2
-277.8L
AE
= -116.5 kN (T)-0.3578 +0.2683 +0.8944[q´j]3 =AEL
0.0
117.2
-277.8L
AE
= -32.61 kN (T)+0.3578 +0.2683 +0.8944[q´j]4 =AEL
0.0
117.2
-277.8L
AE
x
z
y
4 m4 m
3 m3 m
10 m
80 kN60 kN
O
83
5
32.6 kN116.5 kN
116.5 kN
32.6 kN
Member
#1#2
#3
λx λy λz
#4
-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944
+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944
[q´j]m
116.5 32.6
-32.6-116.5
12
34
80 kN60 kN
R5x = (-32.6)(-0.3578) = 11.66 kN
R5y = (-32.6)(-0.2683) = 8.75 kN
R5z = (-32.6)(-0.8944) = 29.16 kN
1
! Development: The Slope-DeflectionEquations
! Stiffness Matrix! General Procedures! Internal Hinges! Temperature Effects! Force & Displacement Transformation! Skew Roller Support
BEAM ANALYSIS USING THE STIFFNESSMETHOD
2
Slope � Deflection Equations
settlement = ∆j
Pi j kw Cj
Mij MjiwP
θj
θi
ψ
i j
3
� Degrees of Freedom
L
θΑ
A B
M
1 DOF: θΑ
PθΑ
θΒΒΒΒ
A BC 2 DOF: θΑ , θΒΒΒΒ
4
L
A B
1
� Stiffness Definition
kBAkAA
LEIkAA
4=
LEIkBA
2=
5
L
A B1
kBBkAB
LEIkBB
4=
LEIkAB
2=
6
� Fixed-End ForcesFixed-End Forces: Loads
P
L/2 L/2
L
w
L
8PL
8PL
2P
2P
12
2wL12
2wL
2wL
2wL
7
� General Case
settlement = ∆j
Pi j kw Cj
Mij MjiwP
θj
θi
ψ
i j
8
wP
settlement = ∆j
(MFij)∆ (MF
ji)∆
(MFij)Load (MF
ji)Load
+
Mij Mji
θi
θj
+
i jwPMij
Mji
settlement = ∆jθj
θi
ψ
=+ ji LEI
LEI θθ 24
ji LEI
LEI θθ 42
+=
,)()()2()4( LoadijF
ijF
jiij MMLEI
LEIM +++= ∆θθ Loadji
Fji
Fjiji MM
LEI
LEIM )()()4()2( +++= ∆θθ
L
9
Mji Mjk
Pi j kw Cj
Mji Mjk
Cj
j
� Equilibrium Equations
0:0 =+−−=Σ+ jjkjij CMMM
10
+
1
1
i jMij Mji
θi
θj
LEIkii
4=
LEIk ji
2=
LEIkij
2=
LEIk jj
4=
iθ×
jθ×
� Stiffness Coefficients
L
11
[ ]
=
jjji
ijii
kkkk
k
Stiffness Matrix
)()2()4( ijF
jiij MLEI
LEIM ++= θθ
)()4()2( jiF
jiji MLEI
LEIM ++= θθ
+
=
F
ji
Fij
j
iI
ji
ij
MM
LEILEILEILEI
MM
θθ
)/4()/2()/2()/4(
� Matrix Formulation
12
[D] = [K]-1([Q] - [FEM])
Displacementmatrix
Stiffness matrix
Force matrixwP(MF
ij)Load (MFji)Load
+
+
i jwPMij
Mji
θj
θi
ψ ∆j
Mij Mji
θi
θj
(MFij)∆ (MF
ji)∆
Fixed-end momentmatrix
][]][[][ FEMKM += θ
]][[])[]([ θKFEMM =−
][][][][ 1 FEMMK −= −θ
L
13
L
Real beam
Conjugate beam
� Stiffness Coefficients Derivation
MjMi
L/3
θi
LMM ji +
EIM j
EIMi
θι
EILM j
2
EILMi
2
)1(2
0)3
2)(2
()3
)(2
(:0'
−−−=
=+−=Σ+
ji
jii
MM
LEI
LMLEI
LMM
)2(0)2
()2
(:0 −−−=+−=Σ↑+EI
LMEI
LMF jiiy θ ij
ii
LEIM
LEIM
andFrom
θ
θ
)2(
)4(
);2()1(
=
=
LMM ji +
i j
14
� Derivation of Fixed-End Moment
Real beam
8,0
162
22:0
2 PLMEI
PLEI
MLEI
MLFy ==+−−=Σ↑+
P
M
M EIM
Conjugate beamA
EIM
B
L
P
A B
EIM
EIML2
EIM
EIML2
EIPL
16
2
EIPL4 EI
PL16
2
Point load
15
L
P
841616PLPLPLPL
=+−
+−
8PL
8PL
2P
2PP/2
P/2
-PL/8 -PL/8
PL/8
-PL/16-PL/8
-
PL/4+
-PL/16-PL/8-
16
Uniform load
L
w
A B
w
M
M
Real beam Conjugate beam
A
EIM
EIM
B
12,0
242
22:0
23 wLMEI
wLEI
MLEI
MLFy ==+−−=Σ↑+
EIwL8
2
EIwL
24
3
EIwL
24
3
EIM
EIML2
EIM
EIML2
17
Settlements
M
M
L
∆
MjMi = Mj
LMM ji +L
MM ji +
Real beam
2
6LEIM ∆
=
Conjugate beam
EIM
A B
EIM
∆
,0)3
2)(2
()3
)(2
(:0 =+−∆−=Σ+L
EIMLL
EIMLM B
EIM
EIML2
EIMEI
ML2
∆
18
� Typical Problem
0
0
0
0
A C
B
P1P2
L1 L2
wCB
8024 11
11
LPLEI
LEIM BAAB +++= θθ
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
128042 2
222
22
wLLPLEI
LEIM CBCB −
−+++= θθ
P
8PL
8PL
L
w12
2wL
12
2wL
L
19
MBA MBC
A C
B
P1P2
L1 L2
wCB
B
CB
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0
20
A C
B
P1P2
L1 L2
wCB
Substitute θB in MAB, MBA, MBC, MCB
MABMBA
MBC
MCB
0
0
0
0
8024 11
11
LPLEI
LEIM BAAB +++= θθ
8042 11
11
LPLEI
LEIM BABA −++= θθ
128024 2
222
22
wLLPLEI
LEIM CBBC ++++= θθ
128042 2
222
22
wLLPLEI
LEIM CBCB −
−+++= θθ
21
A BP1
MAB
MBA
L1
A CB
P1P2
L1 L2
wCB
ByR CyAy ByL
Ay Cy
MABMBA
MBC
MCB
By = ByL + ByR
B CP2
MBCMCB
L2
22
2EI EI
Stiffness Matrix
� Node and Member Identification
� Global and Member Coordinates
� Degrees of Freedom
�Known degrees of freedom D4, D5, D6, D7, D8 and D9� Unknown degrees of freedom D1, D2 and D3
1 21 2 31
2 3
7
8
9
4
5
6
1
2 3
7
8
9
4
5
6
23
i j
E, I, A, L
Beam-Member Stiffness Matrix
d4 = 1
AE/LAE/L AE/L AE/L
64
5
1
2
3
[k] =
1 2 3 4 5 6
3
2
1
4
6
5
d1 = 1
k11 = = k44
0
0
0
0
0
0
0
0
- AE/LAE/L
-AE/L AE/L
k41 k14
24
d2 = 1
[k] =
1 2 3 4 5 6
3
2
1
4
6
5
0
0
0
0
0
0
0
0
- AE/LAE/L
-AE/L AE/L
6EI/L2
6EI/L2
d5 = 1
6EI/L2
6EI/L2
12EI/L312EI/L3 12EI/L3
12EI/L3
6EI/L2
12EI/L3
- 6EI/L2
- 12EI/L3
0
0
6EI/L2
-12EI/L3
0
0
- 6EI/L2
12EI/L3
k22 = = k55
= k52
k62 =
= k32 = k65
= k25
= k35
i j
E, I, A, L6
4
5
1
2
3
25
[k] =
1 2 3 4 5 6
3
2
1
4
6
5
0
0
0
0
0
0
0
0
- AE/LAE/L
-AE/L AE/L
6EI/L2
12EI/L3
- 6EI/L2
- 12EI/L3
0
0
6EI/L2
-12EI/L3
0
0
- 6EI/L2
12EI/L3
4EI/L
6EI/L2 6EI/L2
2EI/L
0
0
2EI/L
-6EI/L2
0
0
-6EI/L2
4EI/L
d6 = 1
d3 = 1 2EI/L 4EI/L
6EI/L26EI/L26EI/L2
4EI/L 2EI/L
6EI/L2
k33 =
k23 = = k53
= k63 = k36
= k56 k26 =
= k66
i j
E, I, A, L6
4
5
1
2
3
26
FFxi
FFyi
MFi
FFxj
FFyj
MFj
+
MFi MF
j
FFxjFF
xi
FFyi FF
yj
1
6EI/L2
6EI/L2
12EI/L312EI/L3
+
x ∆i
12EI/L3
6EI/L2
6EI/L2
12EI/L3
1
+
4EI/L 2EI/L
6EI/L26EI/L2
x θi
6EI/L2
4EI/L 2EI/L
6EI/L2
x δjAE/L
1
AE/L
+
AE/L AE/L
6EI/L2
6EI/L2
1
12EI/L312EI/L3
x ∆j
6EI/L2
6EI/L2
12EI/L3
12EI/L3
1
+
2EI/L
6EI/L26EI/L2
4EI/L
x θj
6EI/L2
2EI/L
6EI/L2
4EI/L
Fxi
Fyi
MiFxj
Fyj
Mj
� Member Equilibrium Equations
Mj
i jFxj
Fxi
Fyi Fyj
=
MiE, I, A, L
x δiAE/L AE/L
1
AE/LAE/L
27
+
−−−−
−−−
−
=
Fj
Fyi
Fxj
Fi
Fyi
Fxi
j
j
j
i
i
i
j
yj
xj
i
yj
xi
MFFMFF
θ∆δθ∆δ
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAELAE
MFFMFF
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00/
22
2323
22
2323
Fjjjjiiij
Fyjjjjiiiyj
Fxijjjiiixj
Fijjjiiixi
Fyijjjiiiyi
Fxijjjiiixi
MθLEI∆LEIδθLEI∆LEIδMFθLEI∆δθLEI∆LEIδFFθ∆δLAEθ∆δLAEFMθLEI∆LEIδθLEI∆LEIδMFθLEI∆LEIδθLEI∆LEIδFFθ∆δLAEθ∆δLAEF
)/4()/6()0()/2()/6()0()/6()0()0()/6()/12()0(
)0()0()/()0()0()/()/2()/6()0()/4()/6()0()/6()/12()0()/6()/12()0(
)0()0()/()0()0()/(
22
223
22
2323
−=−−−=
−=−=−+=
+++−++=
[q] = [k][d] + [qF]
Fixed-end force matrix
End-force matrix
Stiffness matrix
Displacement matrix
28
6x6 Stiffness Matrix
[ ]
−−−−
−−−
−
=×
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAELAE
k
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00/
22
2323
22
2323
66
θi ∆j θj∆i δjδi
Mi
Vj
Mj
Vi
Nj
Ni
4x4 Stiffness Matrix
[ ]
−−−−
−−
=×
LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI
k
/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12
22
2323
22
2323
44
θi ∆j θj∆i
Mi
Vj
Mj
Vi
29
Comment: - When use 4x4 stiffness matrix, specify settlement. - When use 2x2 stiffness matrix, fixed-end forces must be included.
2x2 Stiffness Matrix
θi θj
Mi
Mj[ ]
=× LEILEI
LEILEIk
/4/2/2/4
22
30
21 312
1
2
3
4
5
6
General Procedures: Application of the Stiffness Method for Beam Analysis
wP P2y
M2M1
Local
1´
2´
3´
4´
1
1´
2´
3´
4´
2
Global
31
Member
1
2
3
4
1
3
4
5
6
2
1
2
3
4
5
6
3121 2
Global
wP P2y
M2M1
Local
1´
2´
3´
4´
1
1´
2´
3´
4´
2
32
(q´F4 )2
(q´F3 )2
(q´F1)2
(q´F2)2
w
2
P
(q´F4)1
(q´F3)1
(q´F2)1
(q´F1 )1
1[FEF]
1
2
3
4
5
6
3121 2
Global
Local
1´
2´
3´
4´
1
1´
2´
3´
4´
2
wP P2y
M2M1
33
[q] = [T]T[q´]
1
2
3
4
5
6
3121 2
Global
Local
1´
2´
3´
4´
1
[k] = [T]T[q´] [T]
=
'4
'3
'2
'1
4
3
2
1
1000010000100001
qqqq
qqqq
1´ 4´2´ 3´12341
34
[q] = [T]T[q´]
1
2
3
4
5
6
3121 2
Global
Local
1´
2´
3´
4´
2
[k] = [T]T[q´] [T]
=
'4
'3
'2
'1
6
5
4
3
1000010000100001
qqqq
qqqq
1´ 4´2´ 3´34562
35
Stiffness Matrix:
21 312
1
2
3
4
5
6
[k]1 =
1
2
3
4
1 2 3 4
[k]2 =
3
4
5
6
3 4 5 6
Member 1
[K] =
1
23
4
5
6
1 2 3 4 5 6
Member 2
Node 2
1 2
36
+
=
F
F
F
F
F
F
QQQQQQ
DDDDDD
QQQQQQ
6
5
1
4
3
2
6
5
1
4
3
2
6
5
1
4
3
2 234156
5 612 3 4
000
M1z
-P2y
M3z
KBA
KAB
KBB
KAA
21 312
1
2
3
4
5
6
ReactionQu
Joint LoadQk
Du=Dunknown
Dk = DknownQF
B
QFA
[ ] [ ][ ] [ ]FAuAAk QDKQ +=
[ ] [ ] [ ] [ ])(1 FAkAAu QQKD −+= −
[ ] [ ][ ] [ ]FqdkqForceMember +=:
37
+
6
5
4
3
2
1
DDDDDD
Global:
21 312
1
2
3
4
5
6
1 2
2
3
4
51
6
(q´F6 )2
(q´F5 )2
(qF4)2
(qF3)2
w
2
P
(qF4)1
(qF3)1
(qF2)1
(qF1 )1
1[FEF]
=−=
=
24
23
12
MQPQ
MQ
y
4
3
2
DDD
+F
F
F
QQQ
4
3
2
2 3 4
Node 2
=
234
Member 1
123456
1 2 3 4 5 6
Member 2
Node 2=
6
5
4
3
2
1
QQQQQQ
+=+=
F
F
F4
F4
F
F3
F3
F
F
F
qqQqqQ
6
5
214
213
2
1
)()()()(
M1
-P2y
M2
0
00
38
1 2 312
1
2
3
4
5
6
4
3
2
DDD
2 3 4
Node 2
=
234
−
=−=
=
F
F
F
QQQ
MQPQ
MQ
4
3
2
24
2y3
12
39
Member 1:
21 312
1
2
3
4
5
6
P
qF4
qF3
qF2
qF1
1
[FEF]
4
3
2
1
qqqq
====
44
33
22
1 0
DdDdDd
d
+
F
F
F
F
qqqq
4
3
2
11234
1 2 3 4
= 1k
40
21 312
1
2
3
4
5
6
Member 2:
qF6
qF5
qF4
qF3
w
2
[FEM]
6
5
4
3
qqqq
==
==
00
6
5
44
33
dd
DdDd
+
F
F
F
F
qqqq
6
5
4
31234
1 2 3 4
= 1k
41
AC
B
9 m 3 m
10 kN1 kN/m
1.5 m
Example 1
For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.
42
21Global
123
1Members
23
2
12
AC
B
9 m 3 m
10 kN1 kN/m
1.5 m
wL2/12 = 6.75wL2/12 = 6.75 PL/8 = 3.75PL/8 = 3.75
10 kN
1.5 m1.5 m
2
1 kN/m
9 m 1[FEF]
43
4/3
2/3
2/3
21
123
9 m 3 m
[k]1= EI4/9
2/9
3
4/9
2/92
3
2[k]2 = EI
4/3
4/3
2/3
2/3
2 12
1
[K] = EI
2 12
1
(4/9)+(4/3)
21
123
Stiffness Matrix: θi θj
Mi
Mj[ ]
=× LEILEI
LEILEIk
/4/2/2/4
22
44
A CB
10 kN1 kN/m
123
Equilibrium equations: MCB = 0MBA + MBC = 0
Global Equilibrium: [Q] = [K][D] + [QF]
=2.423/EI
0.779/EI
θC
θB
9 m 1.5 m1.5 m6.756.75
+-3.75
-6.75 + 3.75 = -3MBA + MBC = 0
MCB = 01
2
θC
θB
4/3
2/3
2/3 = EI
2 12
1
(4/9)+(4/3)
3.753.756.75
45
= -6.40
6.92
Member 1 : [q]1 = [k]1[d]1 + [qF]1
MBA
MAB
2
3
= 0.779/EIθB
θA
0
+ -6.75
6.75
Substitute θB and θC in the member matrix,
1
23
wL2/12 = 6.75wL2/12 = 6.75
1 kN/m
9 m 1[FEF]
4.56 kN 4.44 kN
6.92 kN�m 6.40 kN�m
wL2/12 = 6.75wL2/12 = 6.75
= EI4/9
2/9
3
4/9
2/92
1 kN/m
19 m
46
Member 2 : [q]2 = [k]2[d]2 + [qF]2
Substitute θB and θC in the member matrix,
MCB
MBC
1
2 = EI
4/3
4/3
2/3
2/3
2 1
= 0
6.40
θC
θB
= 2.423/EI
= 0.779/EI
2
12
[FEF]PL/8 = 3.75PL/8 = 3.75
10 kN
1.5 m1.5 m
2
7.13 kN 2.87 kN
6.40 kN�m
10 kN
2
PL/8 = 3.75PL/8 = 3.75
+-3.75
3.75
47
A CB
10 kN1 kN/m
9 m 1.5 m1.5 m
6.92 kN�m
11.57 kN 2.87 kN4.56 kN
4.56 kN 4.44 kN 7.13 kN 2.87 kN
1 kN/m6.92 6.40
6.40
10 kN
M (kN�m) x (m)
-6.92 -6.40
3.48 4.32
V (kN) x (m)
4.56
-4.44 -2.874.56 m
7.13
θB = +0.779/EIθC = +2.423/EI
48
20 kN
A CBEI2EI
4 m 4 m
9kN/m 40 kN�m
Example 2
For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.
49
1 2 3
0.5625
3
-0.375
-0.375
1
2
3
4
5
6
[K] = EI
3 4
3
4
Use 4x4 stiffness matrix,
[k]1
0.375EI
-0.75EI2EI
0.75EI - 0.375EI
0.375EI
0.75EI
EI
-0.75EI
2EI
0.75EI
0.75EI
-0.375EI
EI
-0.75EI
-0.75EI
1 2 3 4
1
2
34
[k]2
0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
3 4 5 6
3
4
5
6
2EI EI
1 2
1
2
5
6
θi ∆j θj
Mi
∆i
Vj
Mj
Vi
[ ]
−−−−
−−
=
LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI
k
/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12
22
2323
22
2323
50
20 kN9kN/m 40 kN�m
4 m 4 m
12 kN�m18 kN
12 kN�m
18 kN 18 kN
D4
D3
Global:
D4
D3=
9.697/EI
-61.09/EI
Q4 = 40
Q3 = -203
4 + -12
18 3
4
20 kN
40 kN�m
12 kN�m
1 2
1
2
3
4
5
6
1 2 3
2EI EI
[Q] = [K][D] + [QF]
0.5625
3
-0.375
-0.375= EI
3 4
3
4
51
Member 1:
1
9 kN/m
A B12 kN�m
[qF]1
12 kN�m
18 kN 18 kN
1
9 kN/m
A B
[q]1
53.21 kN�m
48.18 kN67.51 kN�m
12.18 kN
1
1
2
3
4
1 2
2EI 12 kN�m 12 kN�m
18 kN 18 kN
q2
q1
q4L
q3L
67.51
48.18
53.21
-12.18
12
18
-12
18
12(2EI)/43
-0.75EI2EI
0.75EI - 0.375EI
0.375EI
0.75EI
EI
-0.75EI
2EI
0.75EI
0.75EI
-0.375EI
EI
-0.75EI
-0.75EI
1 2 3 4
1
2
34
[q]1 = [k]1[d]1 + [qF]1
d3 = -61.09/EI
d4 = 9.697/EI
d2 = 0
d1 = 0
52
2B C
[q]2
Member 2:
[q]2 = [k]2[d]2 + [qF]2
2
3
4
5
6
2 3
EI
q4R
q3R
q6
q5
0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
3 4 5 6
3
4
5
6
0
0
0
0
d3 =-61.09/EI
d4 = 9.697/EI
d6 = 0
d5 = 0
18.06 kN�m
7.82 kN
13.21 kN�m
7.82 kN
-13.21
-7.818
-18.06
7.818
53
V (kN) x (m)
-7.818
48.18
-7.818-
M (kN�m) x (m)
13.21
-18.08-
-67.51
-
1
9 kN/m
A B
[q]1
53.21 kN�m
48.18 kN67.51 kN�m
12.18 kN2
B C
[q]2
18.06 kN�m
7.82 kN
13.21 kN�m
7.82 kN
12.18+
53.21
+
20 kN
4 m 4 m
9kN/m 40 kN�m
7.818 kN
18.06 kN�m67.51 kN�m
48.18 kN
D3 = ∆B = -61.09/EI
D4 = θB = +9.697/EI
54
20 kN
A CBEI2EI
4 m 2 m
9kN/m 40 kN�m10 kN
2 m
Example 3
For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.
55
Global1 2
1
2
3
4
5
6
1 2 3
20 kN
A CB EI2EI
9kN/m 40 kN�m10 kN
4 m 2 m 2 m
10 kN5 kN�m
5 kN
5 kN�m
5 kN
212 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1[FEF]
θi ∆j θj
Mi
∆i
Vj
Mj
Vi
[ ]
−−−−
−−
=×
LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI
k
/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12
22
2323
22
2323
44
56
1 2
1
2
3
4
5
6
1 2 3 4 m 2 m 2 m
1
2
5
1 2
0.375
0.50.375 0.5 1
[K] = EI
3 4 6
3
46
0.5625
3
-0.375
-0.375
12(2EI)/43
[k]1 =-0.75EI2EI
0.75EI - 0.375EI
0.375EI
0.75EI
EI
-0.75EI
2EI/L
0.75EI
0.75EI
-0.375EI
EI
-0.75EI
-0.75EI
1 2 3 4
1
2
34
[k]2 =0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
3 4 5 6
3
4
5
6
57
MBA+MBC = 40
VBL+VBR = -20
MCB = 0θB
∆B
θC
-12 + 5 = -7-5
θB
∆B
θC
= -7.667/EI
-116.593/EI
52.556/EI
Global: [Q] = [K][D] + [QF]
= EI
0.5625
3
-0.375
-0.375
0.375
0.50.375 0.5 1
3 4 6
3
46
20 kN
A CB
9kN/m 40 kN�m10 kN
21
1
2
3
4
5
6
1 2 3 10 kN5 kN�m
5 kN
5 kN�m
5 kN
212 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1[FEF]
20 kN
40 kN�m
5 kN�m5 kN�m
5 kN
12 kN�m
18 kN
+18 + 5 = 23
58
0.375EI
= -0.75EI2EI
0.75EI - 0.375EI
0.375EI
0.75EI
EI
-0.75EI
2EI/L
0.75EI
0.75EI
-0.375EI
EI
-0.75EI
-0.75EI
1 2 3 4
1
2
34
=91.78
55.97
60.11
-19.97
91.78 kN�m
55.97 kN
Member 1: [q]1 = [k]1[d]1 + [qF]1
MAB
VA
MBA
VBL+
12
18
-12
18=-116.593/EI
=-7.667/EI
0
0
θB
∆B
1
1
2
3
4
1 2
12 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1
[FEF]
19.97 kN
60.11 kN�m4 m
9kN/m
1
12 kN�m
18 kN
12 kN�m
18 kN18 kN
59
VC
MBC
VBR
MCB
+ 5
5
5 -5
=-20.11
-0.0278
0
10.03=52.556/EI
=-116.593/EI
=-7.667/EIθB
∆B
θC
0
10 kN5 kN�m
5 kN
5 kN�m
5 kN
2
[FEM]
10.03 kN
2 m
10 kN
2 m2
= 0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
3 4 5 6
3
4
5
6
0.0278 kN
20.11 kN�m
Member 2: [q]1 = [k]1[d]1 + [qF]1
223
5
6
3
45 kN�m
5 kN
5 kN�m
5 kN
60
20 kN
4 m 2 m
9kN/m 40 kN�m10 kN
2 m
10.03 kN
91.78 kN�m
55.97 kN
V (kN) x (m)
55.97
-0.0278
19.97+
-10.03 -10.03-
91.78 kN�m
55.97 kN 19.97 kN60.11 kN�m4 m
9kN/m
110.03 kN0.0278 kN
20.11 kN�m2 m
10 kN2 m
2
91.78 kN�m
60.11 kN�m
20.11 kN�m
M (kN�m) x (m)
-91.78
-
+20.11
60.11
θB = -7.667/EI
θC = 52.556/EI
∆B = -116.593/EI
61
40 kN
8 m 4 m 4 m
2EI EI
6 kN/m
AB
C∆B = -10 mm
Example 4
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative shear diagram, bending moment diagramand qualitative deflected shape.Take I = 200(106) mm4 and E = 200 GPa and support B settlement 10 mm.
62
[k]1 =4
8
8
4
1 2
1
2
EI8
1 2
1 2 3
2EI EI
1
2
42
2 3
2
3[K] = EI
8
12
2
4
4
2
2 3
2
3[k]2 = EI
8
Use 2x2 stiffness matrix:θi θj
Mi
Mj[ ]
=× LEILEI
LEILEIk
/4/2/2/4
22
63
75 kN�m
6(2EI)∆/L2 = 75 kN�m10 mm
1 37.5 kN�m
6(EI)∆/L2 = 37.5 kN�m10 mm
2[FEM]∆∆∆∆
6(EI)∆/L2 = 37.5 kN�m
D2
D3
=-61.27/EI rad
185.64/EI rad
40 kN
2EI EI
6 kN/m
A
B C
∆B = -10 mm 1 2
1 2 3
[FEM]load
32 kN�mwL2/12 = 32 kN�m
6 kN/m
1 40 kN�mPL/8 = 40 kN�m 2
40 kN8 m 4 m 4 m
32 kN�m 40 kN�mPL/8 = 40 kN�m
75 kN�m 37.5 kN�m
+-32 + 40 + 75 -37.5 = 45.5Q2 = 0
Q3 = 0 -40 - 37.5 = -77.5
D2
D3
2
42
2 3
2
3
12
8EI
=
Global: [Q] = [K][D] + [QF]
64
6(2EI)∆/L2 = 75 kN�m
[FEM]load
32 kN�mwL2/12 = 32 kN�m
6 kN/m
1
75 kN�m
6(2EI)∆/L2 = 75 kN�m10 mm
1[FEM]∆∆∆∆
6 kN/m
18 m
1
2
2EI
1
q1
q2
d1 = 0
d2 = -61.27/EI=
76.37 kN�m
-18.27 kN�m
Member 1: [q]1 = [k]1[d]1 + [qF]1
32 kN�mwL2/12 = 32 kN�m
75 kN�m
+32 + 75 = 107
-32 + 75 = 43
31.26 kN
76.37 kN�m
16.74 kN
18.27 kN�m
4
8
8
4
1 2
1
28EI
=
65
Member 2: [q]2 = [k]2[d]2 + [qF]2
2
2 3
40 kN�mPL/8 = 40 kN�m 2
40 kN
37.5 kN�m
6(EI)D/L2 = 37.5 kN�m10 mm
2
[FEM]load
[FEM]∆∆∆∆
40 kN�m
37.5 kN�m
6(EI)D/L2 = 37.5 kN�m
PL/8 = 40 kN�m
=18.27 kN�m
0 kN�m
q2
q3
+40 - 37.5 = 2.5
-40 - 37.5 = -77.5
d2 = -61.27/EI
d3 = 185.64/EI
17.72 kN
18.27 kN�m
22.28 kN
40 kN
2
2
4
4
2
2 3
2
38EI
=
66
Alternate method: Use 4x4 stiffness matrix
2EI EI
1 2
2
5
6
3
4
1
8EI
= 1.5
1.5
-0.375
12(2)/82
4
-1.5
8
1.5
-1.5
-1.5
-0.375
0.375
1.5
4
-1.5
8
1 2 3 41
2
3
4
[k]1
8EI
=0.75
0.75
-0.1875
12/82
2
-0.75
4
0.75
-0.75
-0.75
-0.1875
0.1875
0.75
2
-0.75
4
3 4 5 63
4
5
6
[k]2
2-0.75
-0.1875-0.75
0.1875-0.75
0.752
-0.754
-0.18750.75
1.5 4
1.54
0.375
-0.3751.5
1.58
-1.5
-0.375-1.5
00
00
00
00
-0.7512
0.5625-0.758
EI=[K]
123456
1 4 5 62 3
67
Global: [Q] = [K][D] + [QF]
40 kN
8 m 4 m 4 m
2EI EI
6 kN/m
A
B C
∆B = -10 mm 1 2
2
5
6
3
4
1
40 kN40 kN�m
20 kN
40 kN�m
20 kN
232 kN�m
24 kN
32 kN�m
24 kN
6 kN/m
1[FEF]
53
2
1
D4
D6=
-61.27/EI = -1.532x10-3 rad
185.64/EI = 4.641x10-3 rad
+(200x200/8)(-0.75)(-0.01) = 37.5
(200x200/8)(0.75)(-0.01) = -37.5
D4
D6
Q4 = 0
Q6= 0
2
4
12
2+
8
-40
4 6
4
68EI
=
-0.75-0.75 D5 = -0.01)
8200200( ×
+ 46
5
D4
D6
Q4 = 0
Q6= 0
2
4
12
2
4 6
4
68EI
= +8
-40
68
+32
24
-32
24
q1
q2
q4
q3
1.5
1.5
-0.375
12(2)/82
4
-1.5
8
1.5
-1.5
-1.5
-0.375
0.375
1.5
4
-1.5
8
1 2 3 41
2
3
4
= (200x200)8
q1
q2
q4
q3=
76.37 kN�m
31.26 kN
-18.27 kN�m
16.74 kN
Member 1: [q]1 = [k]1[d]1 + [qF]1
1
2
3
4
1∆B = -10 mm 32 kN�m
24 kN
32 kN�m
24 kN
6 kN/m
1[FEF]load
d2 = 0
d1 = 0
d4 = -1.532x10-3
d3 = -0.01
6 kN/m
18 m
31.26 kN
76.37 kN�m
16.74 kN
18.27 kN�m
69
Member 2:
+40
20
-40
20
q3
q4
q6
q5
d4 = -1.532x10-3
d3 = -0.01
d6 = 4.641x10-3
d5 = 0
0.75
0.75
-0.1875
12/82
2
-0.75
4
0.75
-0.75
-0.75
-0.1875
0.1875
0.75
2
-0.75
4
3 4 5 63
4
5
6
= (200x200)8
q3
q4
q6
q5=
18.27 kN�m
22.28 kN
0 kN�m
17.72 kN17.72 kN
18.27 kN�m
22.28 kN
40 kN
2
2
5
6
3
4
-10 mm = ∆B
40 kN40 kN�m
20 kN
40 kN�m
20 kN
2[FEF]
70
40 kN
8 m 4 m 4 m
2EI EI
6 kN/m
A
B C
∆B = -10 mm31.26 kN
76.37 kN�m
16.74 + 22.28 kN 17.72 kN
V (kN)
x (m)+ +--
31.26
-16.74
22.28
-17.725.21 m
M (kN�m) x (m)
-76.37
5.06
-18.27
70.85
+
--
D6 = θC = 4.641x10-3 rad
d4 = θB = -1.532x10-3 rad
Deflected Curve
∆B = -10 mm
71
Example 5
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative shear diagram, bending moment diagramand qualitative deflected shape.Take I = 200(106) mm4 and E = 200 GPa and support C settlement 10 mm.
4 kN
8 m 4 m 4 m
2EI EI
0.6 kN/m
AB
C∆C = -10 mm
20 kN�m
72
2EI EI
1 2-10 mm
2
5
6
3
4
1
2
51
Q3
Q6
Q4
0.75 2
0.7524
1.5
1.5
-0.375
12(2)/82
4
-1.5
8
1.5
-1.5
-1.5
-0.375
0.375
1.5
4
-1.5
8
1 2 3 41
2
3
4
[k]1
8EI
=
8EI
=0.75
0.75
-0.1875
12/82
2
-0.75
4
0.75
-0.75
-0.75
-0.1875
0.1875
0.75
2
-0.75
4
3 4 5 63
4
5
6
[k]2
D3
D6
D4
-0.1875
-0.75-0.75 D5 = -0.01
� Global: [Q] = [K][D] + [QF]
�Member stiffness matrix [k]4x4
0.5625-0.75
-0.7512
346
3 4 6
8EI
= )8
200200( ×+
346
5QF
3
QF6
QF4+
73
4 kN4 kN�m
2 kN
4 kN�m
2 kN
23.2 kN�m
2.4 kN
3.2 kN�m
2.4 kN
0.6 kN/m
1[FEF]
4 kN
8 m 4 m 4 m2EI EI
0.6 kN/m
A
BC
10 mm
20 kN�m2EI EI
1 2
6
3
42
51
346
0.5625
0.75-0.75
-0.75122
0.752
3 4 6
48EI
=
D3
D6
D4 +9.375
37.537.5
2.4+2 = 4.4
-4.0-3.2+4 = 0.8+
D3
D6
D4
-377.30/EI = -9.433x10-3 m
+74.50/EI = +1.863x10-3 rad-61.53/EI = -1.538x10-3 rad=
Global: [Q] = [K][D] + [QF]
Q3 = 0
Q6 = 20Q4 = 0
74
+3.2
2.4
-3.2
2.4
q1
q2
q4
q3
q1
q2
q4
q3=
43.19 kN�m
8.55 kN
6.03 kN�m
-3.75 kN
Member 1: [q]1 = [k]1[d]1 + [qF]1
d2 = 0
d1 = 0
d4 = -1.538x10-3
d3 = -9.433x10-3
0.6 kN/m
18 m
8.55 kN
43.19 kN�m
3.75 kN
6.03 kN�m
1
2
3
4
13.2 kN�m
2.4 kN
3.2 kN�m
2.4 kN
0.6 kN/m
1[FEF]load 8 m
1.5
1.5
-0.375
12(2)/82
4
-1.5
8
1.5
-1.5
-1.5
-0.375
0.375
1.5
4
-1.5
8
1 2 3 41
2
3
4
8200200×
=
75
Member 2:
+4
2
-4
2
q3
q4
q6
q5
d4 = -1.538x10-3
d3 = -9.433x10-3
d6 = 1.863x10-3
d5 = -0.01
q3
q4
q6
q5=
-6.0 kN�m
3.75 kN
20.0 kN�m
0.25 kN0.25 kN
6.0 kN�m
3.75 kN
4 kN
2
10 mm2
5
6
3
4 4 kN4 kN�m
2 kN
4 kN�m
2 kN
2[FEF]
8 m
0.75
0.75
-0.1875
12/82
2
-0.75
4
0.75
-0.75
-0.75
-0.1875
0.1875
0.75
2
-0.75
4
3 4 5 63
4
5
6
8200200×
=
20 kN�m
76
4 kN
8 m 4 m 4 m2EI EI
0.6 kN/m
A
BC
10 mm
20 kN�m
Deflected Curve
0.6 kN/m
18 m
8.55 kN
43.19 kN�m
3.75 kN
6.03 kN�m
V (kN)x (m)
+
8.553.75
-0.25
0.25 kN
6.0 kN�m
3.75 kN
4 kN
2
20 kN�m
M (kN�m) x (m)
-43.19
621+
-
20
D3 = -9.433 mm
D6 = +1.863x10-3 radD4 = -1.538x10-3 rad
D5 = -10 mm
77
30 kN
A CB
EI2EI
4 m 2 m
9kN/mHinge
2 m
Example 6
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.E = 200 GPa, I = 50x10-6 m4.
Internal Hinges
78
[K] = EI 1
2
1 2
0.0
0.0
1.0
2.0
A C
B
2EI
4 m 2 m 2 m
EI1 2
43
1
2
[k]1 =EI
2EI
2EI
EI
3 1
3
1[k]2 =
0.5EI
1EI
1EI
0.5EI
2 4
2
4
3 4
Use 2x2 stiffness matrix,θi θj
Mi
Mj[ ]
=× LEILEI
LEILEIk
/4/2/2/4
22
79
0.0
0.0
D1
D2
15 kN�mB C
15 kN�m30 kN
2
12 kN�m
A B
12 kN�m 9kN/m
1[FEF]
= EI 1
2
1 2
0.0
0.0
1.0
2.0
D1
D2=
0.0006 rad
-0.0015 rad
-12
15+
Global matrix:
30 kN9kN/m
Hinge
A C
B1 2
43
12 kN�m 15 kN�m1
2
80
Member 1:
q3
q1
12
-12+=
EI
2EI
2EI
EI
3 1
3
1
18
0.0=
d3
d1
= 0.0
= 0.0006
12 kN�m
A B
12 kN�m 9kN/m
1[FEF]A
B1
31
18 kN�m22.5 kN 13.5 kN
A B
9 kN/m
1
81
B C30 kN
CB2
4
2
15 kN�mB C
15 kN�m 30 kN
2
q2
q4
= -0.0015
= 0.0
d2
d4
15
-15+=
0.5EI
1EI
1EI
0.5EI
2 4
2
4
0.0
-22.5=
Member 2:
22.5 kN�m
20.63 kN9.37 kN
82
30 kN
A CB
EI2EI
4 m 2 m
9kN/mHinge
2 m
-20.63
V (kN) x (m)
22.59.37
-13.5
+
2.5 m
M (kN�m) x (m)
-18
10.13
-22.5
18.75
+-
+--
θBL= 0.0006 rad θBR= -0.0015 rad
13.5 kN22.5 kN
22.5 kN�mB C
30 kN
20.63 kN9.37 kN
A B18 kN�m
9 kN/m
9.37 kN13.5 kN
22.87 kN
83
20 kN
A CBEI2EI
4 m 4 m
9kN/m
Hinge
Example 7
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.E = 200 GPa, I = 50x10-6 m4.
84
A CB1 2
2EI EI 6
7
4
5
4
5
6
7
0.375
0.375
1.0
-0.75
-0.75 2.0
0.0
0[K] = EI 1
2
3
1 2 3
0.5625
[k]1 =0.75EI
EI-0.75EI
2EI/L
0.375EI0.75EI
0.75EI-0.375EI
2EI0.75EI
EI-0.75EI
-0.75EI- 0.375EI
0.375EI-0.75EI
4 5 1 24
51
2
0.375EI
0.375EI-0.1875EI
0.5EI -0.375EI
-0.375EI
[k]2 =
0.1875EI
-0.375EIEI0.375EI - 0.1875EI
0.1875EI
0.375EI0.5EI
-0.375EI EI
1 3 6 71
36
7
1
2
3
85
Q1 = -20
Q3 = 0.0
Q2 = 0.0
D1
D3
D2
18
0.0
-12+
D1
D3
D2 =-0.02382 m
0.008933 rad
-0.008333 rad
[FEM]
12 kN�mA B
12 kN�m
18 kN18 kN1
9kN/m
20 kN
A CB
9kN/m
A CB1 2
6
7
4
5
2EI EI
Global:
= EI 1
2
3
1
0.5625
0.375
-0.75
2
-0.75
2.0
0.0
3
0.375
0
1.0
12 kN�m
18 kN
20 kN
1
2
3
86
A B1
2EI1
24
5
[FEF]
12 kN�mA B
12 kN�m
18 kN18 kN1
9kN/m
+12
18
-12
18
Member 1:
q4
q5
q1
q2
=
0.75EI
EI
-0.75EI
2EI/L
0.375EI
0.75EI
0.75EI
-0.375EI
2EI
0.75EI
EI
-0.75EI
-0.75EI
- 0.375EI
0.375EI
-0.75EI
4 5 1 2
4
5
1
2
= 0.0
= 0.0
= -0.02382
= -0.00833
d5
d4
d2
d1
= 107.32
44.83
0.0
-8.83
A B
107.32 kN�m
8.83 kN44.83 kN
9kN/m
1
87
Member 2:
CB2
EI1
3
6
7
44.66 kN�mB C
11.16 kN11.16 kN
2
+0
0
0
0
= -0.02382
= 0.008933
= 0.0
= 0.0
d3
d1
d7
d6
q1
q3
q7
q6
0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
=
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
1 3 6 71
3
6
7
= 0.0
-11.16
-44.66
11.16
88
20 kN
A CB4 m 4 m
9kN/m
V (kN) x (m)
44.83
8.83
-11.16 -11.16
+
-
M (kN�m) x (m)
-107.32
-44.66-
-
A B
107.32 kN�m
8.83 kN44.83 kN
9kN/m
144.66 kN�m
B C
11.16 kN11.16 kN
2
θBL= -0.008333 rad θBR= 0.008933 rad
89
30 kN
A C
BEI2EI
4 m 2 m
9kN/mHinge
40 kN�m
2 m
Example 8
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.40 kN�m at the end of member AB. E = 200 GPa, I = 50x10-6 m4.
90
[K] = EI 1
2
1 2
0.0
0.0
1.0
2.0
A C
B
2EI
4 m 2 m 2 m
EI1 2
43
[k]1 =EI
2EI
2EI
EI
3 1
3
1[k]2 =
0.5EI
1EI
1EI
0.5EI
2 4
2
4
3 4
1
2
Use 2x2 stiffness matrix: θi θj
Mi
Mj[ ]
=× LEILEI
LEILEIk
/4/2/2/4
22
91
15 kN�mB C
15 kN�m30 kN
2
12 kN�m
A B
12 kN�m 9kN/m
1[FEM]
Global matrix:
A C
B1 2
43
1
2
12 kN�m 15 kN�m
30 kN
A C
BEI2EI
9kN/mHinge
40 kN�m
Q1 = 40
Q2 = 0.0
D1
D2
-12
15+
D1
D2=
0.0026 rad
-0.0015 rad
= EI 1
2
1 2
0.0
0.0
1.0
2.0
40 kN�m
92
12 kN�m
A B
12 kN�m 9kN/m
1[FEF]A
B1
31
40 kN�mA B38 kN�m
9 kN/m
1.5 kN37.5 kN
Member 1:
38
40=
q3
q1
= 0.0
= 0.0026
d3
d1
12
-12+=
EI
2EI
2EI
EI
3 1
3
1
93
CB2
4
2
15 kN�mB C
15 kN�m 30 kN
2
Member 2:
q2
q4
= -0.0015
= 0.0
d2
d4
15
-15+=
0.5EI
1EI
1EI
0.5EI
2 4
2
4
0.0
-22.5=
22.5 kN�mB C
30 kN
20.63 kN9.37 kN
94
9.37 kN
1.5 kN
30 kN
A C
B
4 m 2 m
9kN/mHinge
40 kN�m
2 m
-20.63
V (kN) x (m)
37.5
9.371.5+
M (kN�m) x (m)
-38
40
-22.5
+-
18.75+
--
θBL= 0.0026 radθBR=- 0.0015 rad
7.87 kN40 kN�mA B
38 kN�m
9 kN/m
1.5 kN37.5 kN
22.5 kN�mB C
30 kN
20.63 kN9.37 kN
95
20 kN
A CBEI2EI
4 m 4 m
9kN/m
Hinge40 kN�m
Example 9
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.40 kN�m at the end of member AB. E = 200 GPa, I = 50x10-6 m4
96
A CB1 2
6
7
4
51
2
3
20 kN
A CBEI2EI
4 m 4 m
9kN/m
Hinge40 kN�m
12 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1[FEM]θi ∆j θj
Mi
∆i
Vj
Mj
Vi
[ ]
−−−−
−−
=×
LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI
k
/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12
22
2323
22
2323
44
97
1 22A CB
6
7
4
5
[K] = EI 1
2
3
1 2 3
6
7
4
5
0.5625
0.0
0
0.375
0.375
1.0
-0.75
-0.75
2.0
[k]1 = 0.75EI
EI-0.75EI
2EI
0.375EI0.75EI
0.75EI-0.375EI
2EI0.75EI
EI-0.75EI
-0.75EI- 0.375EI
0.375EI-0.75EI
4 5 1 24
51
2
1
0.375EI
0.375EI-0.1875EI
0.5EI -0.375EI
-0.375EI
[k]2 =
0.1875EI-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI0.5EI
-0.375EI EI
1 3 6 71
36
7
2EI EI1
2
3
98
Q1 = -20
Q3 = 0.0
Q2 = 40
D1
D3
D2
18
0.0
-12+
D1
D3
D2 =-0.01316 m
0.0049333 rad
-0.002333 rad
A CB1 2
6
7
4
5
2EI EI
20 kN
EI
9kN/m
40 kN�m
20 kN
40 kN�m
12 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1[FEF]
= EI 1
2
3
1 2 3
0.5625
0.0
0
0.375
0.375
1.0
-0.75
-0.75
2.0
12 kN�m
18 kNGlobal:
1
2
3
99
1
2EI
[q]1
1
12
1
2EI4
5
+12
18
-12
18
q4
q5
q1
q2
=
0.75EI
EI
-0.75EI
2EI
0.375EI
0.75EI
0.75EI
-0.375EI
2EI
0.75EI
EI
-0.75EI
-0.75EI
- 0.375EI
0.375EI
-0.75EI
4 5 1 2
4
5
1
2
= 87.37
49.85
40
-13.85
12 kN�m
18 kN
12 kN�m
18 kN
9 kN/m
1[FEF]
Member 1: [q]1 = [k]1[d]1 + [qF]1
13.85 kN
40 kN�m87.37 kN�m
49.85 kN
= 0.0
= 0.0
= -0.01316
= -0.002333
d5
d4
d2
d1
12 kN�m
18 kN
12 kN�m
18 kN1
100
22 CB
1
3
EI 6
7
Member 2: [q]2 = [k]2[d]2 + [qF]2
+0
0
0
0
q1
q3
q7
q6
0.375EI
0.375EI
-0.1875EI
0.5EI
-0.375EI
-0.375EI
=
0.1875EI
-0.375EIEI
0.375EI - 0.1875EI
0.1875EI
0.375EI
0.5EI
-0.375EI
EI
1 3 6 71
3
6
7
= 0.0
-6.18
-24.69
6.18
24.69 kN�m
6.18 kN6.18 kN
= -0.01316
= 0.004933
= 0.0
= 0.0
d3
d1
d7
d6
22
[q]2
101
20 kN
A CB
EI2EI
4 m 4 m
9kN/m 40 kN�m
24.69 kN�mB C
6.18 kN6.18 kN
A B
87.37 kN�m 13.85 kN49.85 kN
40 kN�m
M (kN�m) x (m)
-
-87.37
+40
--24.69
θBL= -0.002333 rad θBR = 0.004933 rad
49.85
-6.18
V (kN) x (m)-6.18
-13.85+
102
� Fixed-End Forces (FEF)- Axial- Bending
� Curvature
Temperature Effects
103
Tm> TR
� Thermal Fixed-End Forces (FEF)
Tl
Tt
2bt
mTTT +
=
Tt
Tb > Tt
βF
AFF
BF
BFBMF
AM
Rmaxial TTT −=∆ )(
tbbending TTT −=∆ )(
)()(dTyTy
∆=∆
y β
)(tandT
dTT tb ∆
=−
=β
A
A B
σaxialσaxial
A B
σy
y
TR Tm
Tt
Tb
Tb - Tt
Room temp = TR
ct
cbd
TR
NA
104
- Axial
A B
σaxialσaxial
FAF F
BF
dAFA
axialF
A ∫= σ
dAE axial∫= ε
dATE axial∫ ∆= )(α
axialTEA )(∆= α
AETTF RmAF
axial )()( −= α
dATE axial ∫∆= )(α
Tt
Tb > Tt
Tm> TR
Rmaxial TTT −=∆ )(
TR Tm
105
- Bending
A B
σy
y
dAyMA
yFA ∫= σ
dATyE y∫ ∆= )(α
dAdTyyE∫
∆= )(α
∫∆
= dAydTE 2)(α
EId
TTF ulA
Fbending )()( −
= α
FBMF
AM
dAyE∫= ε
tbbending TTT −=∆ )(
)()(dTyTy
∆=∆
y β Tt
Tb
106
O´
ds´
dx
Afterdeformation
dθ
dθ
Before deformation
y dx
ds = dx y
ρ)()( θρ ddx =
)( θdy )(1dxdθ
ρ=
� Elastic Curve: Bending
107
dxdTyyd )()( ∆
= αθ
dxdTd )()( ∆
= αθ
Tb
Tt
yy
dT∆
� Bending Temperature
dx
Tt
Tb
Tl > Tu
y
Tl > Tu
Tu
O
dθ
2bt
mTTT +
= dθ
MM
Tb
cb
ct
Tt
∆T = Tb - Tt
dT∆
=β
EIM
dT
dxd
=∆
== )(1)( αρ
θ
108
A CB
EI2EI
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm
Example 10
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.Room temp = 32.5oC, α = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4.
109
12 3
1 2
Mean temperature = (40+25)/2 = 32.5Room temp = 32.5oC
FEM+7.5 oC-7.5 oC
a(∆T /d)EI19.78 kN�m = 19.78 kN�m
A CB
EI2EI
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm
mkNEIdTF F
bending •=××−
×=∆
= − 78.19)502002)(182.0
2540)(1012()2)(( 6α
110
[M1] = 3EIθ1 + (1.978x10-3)EI0
Element 1:
M1
M2
q1
q2
[q] = [k][d] + [qF]
+ 1.978
-1.978(10-3) EI= 1
2
2
1
2 1
2
1EI
Element 2:
M3
M1
q3
q1 + 0
0= 0.5
1
1
0.5
1 3
1
3EI
θ1 = -0.659x10-3 rad
12 3
A C
BEI2EI
4 m 4 m
1 2
19.78 kN�m19.78 kN�m182 mm
111
Element 2:
M3
M1
q3
q1= + 0
0
0.5
1
1
0.5
1 3
1
3EI
= -0.659x10-3
= 0= 6.59 kN�m
-26.37 kN�m
= -0.659x10-3
= 0= -3.30 kN�m
-6.59 kN�m
Element 1:
M1
M2 = q1
q2 + 19.78
-19.78
1
2
2
1
2 1
2
1EI
4.95 kN4.95 kN
26.37 kN�m 6.59 kN�m
A B
6.59 kN�m 3.3 kN�m
B C
2.47 kN2.47 kN
12 3
A C
BEI2EI
4 m 4 m
1 2
19.78 kN�m19.78 kN�m182 mm
112
A CB
4 m 4 m
+7.5 oC
-7.5 oC 3.3 kN�m
26.37 kN�m
4.95 kN2.47 kN
2.47 kN
M (kN�m) x (m)
26.37
6.59
-3.30
+-
V (kN) x (m)
-4.95-2.47
- -
θB = -0.659x10-3 radDeflected curve x (m)
113
A CB
EI, AE2EI, 2AE
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm
Example 11
For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.Room temp = 28 oC, a = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4,A = 20(10-3) m2
114
A
CB
EI2EI
4 m 4 m
T2 = 40oC
T1 = 25oC1
2 3
7
8
9
1 24
5
6
Element 1:
mkNm
mkNmLAE /)10(2
)4()/10200)(1020(2)2( 6
2623
=××
=−
mkNm
mmkNLEI
•=×××
=−
)10(20)4(
)1050)(/10200(24)2(4 34626
mkNm
mmkNLEI
•=×××
=−
)10(10)4(
)1050)(/10200(22)2(2 34626
kNm
mmkNLEI )10(5.7
)4()1050)(/10200(26)2(6 3
2
4626
2 =×××
=−
mkNm
mmkNL
EI /)10(75.3)4(
)1050)(/10200(212)2(12 33
4626
3 =×××
=−
115
Fixed-end forces due to temperatures
Mean temperature(Tm) = (40+25)/2 = 32.5 oC ,TR = 28 oC
A CB
EI, AE2EI, 2AE
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm
mkNEIdTF F
bending •=××−
×=∆
= − 78.19)502002)(182.0
2540)(1012()2)(( 6α
19.78 kN�m19.78 kN�m
432 kN432 kNA B
+12 oC
-3 oC
kNmkNmAETF Faxial 432)/10200)(310202)(285.32)(1012()( 2626 =×−××−×=∆= −α
116
FEM
+12 oC
-3 oC
19.78 kN�m
A B
19.78 kN�m
432 kN432 kN
Element 1:
[q] = [k][d] + [qF]
0
0
0
d1
d2
d3
q4
q6
q5
q1
q2q3
=
4
5
6
1
2
3
4 1
- 2x106
0.00
0.00
2x106
0.00
0.00
2
0.00
-3750
-7500
0.00
3750
-7500
3
0.00
7500
10x103
0.00
-7500
20x103
2x106
0.00
0.00
-2x106
0.00
0.00
5
0.00
3750
7500
0.00
-3750
7500
6
0.00
7500
20x103
0.00
-7500
10x103
432
-19.78
0.00
-432
0.00
19.78
+
A CB
EI, AE2EI, 2AE
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm1
2 3
7
8
9
1 24
5
6
117
Element 2:
[q] = [k][d] + [qF]
d1
d3
d2
0
0
0
0
0
0
0
0
0
+
q1
q3
q2
q7
q8q9
- 1x106
0.00
0.00
1x106
0.00
0.00
0.00
-1875
-3750
0.00
1875
-3750
0.00
3750
5x103
0.00
-3750
10x103
1x106
0.00
0.00
-1x106
0.00
0.00
0.00
1875
3750
0.00
-1875
3750
0.00
3750
10x103
0.00
-3750
5x103
7 8 91 2 3
=
1
2
3
7
8
9
A CB
EI, AE2EI, 2AE
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm1
2 3
7
8
9
1 24
5
6
118
D1
D3
D2 =0.000144 m
-719.3x10-6 rad
-0.0004795 m
Q1 = 0.0
Q3 = 0.0
Q2 = 0.0
D1
D3
D2
-432
19.78
0.0+= 1
2
3
1
3x106
0.0
0.0
2
0.0
5625
-3750
3
0.0
-3750
30x103
Global:
A CB
EI, AE2EI, 2AE
4 m 4 m
T2 = 40oC
T1 = 25oC
182 mm1
2 3
7
8
9
1 24
5
6
119
= 144x10-6
= -479.5x10-6
= -719.3x10-6
Element 1:
0
0
0
d1
d2
d3
q4
q6
q5
q1
q2q3
=
4
5
6
1
2
3
4 1
- 2x106
0.00
0.00
2x106
0.00
0.00
2
0.00
-3750
-7500
0.00
3750
-7500
3
0.00
7500
10x103
0.00
-7500
20x103
2x106
0.00
0.00
-2x106
0.00
0.00
5
0.00
3750
7500
0.00
-3750
7500
6
0.00
7500
20x103
0.00
-7500
10x103
432
-19.78
0.00
-432
0.00
19.78
+
=
q4
q6
q5
q1
q2q3
144.0 kN
-23.38 kN�m
-3.60 kN
-144.0 kN
3.60 kN
9.00 kN�m
A B 1
2 3
4
5
6
A B144 kN
3.60 kN
23.38 kN�m
144 kN
3.60 kN
9 kN�m
120
= 144x10-6
= -479.5x10-6
= -719.3x10-6
144 kN
-9 kN�m
-3.6 kN
-144 kN
3.6 kN
-5.39 kN�m
=
q1
q3
q2
q7
q8q9
B C 7
8 9
1
2
3
144 kNB C
3.60 kN
9 kN�m
144 kN
3.60 kN
5.39 kN�m
Element 2:
d1
d3
d2
0
0
0
0
0
0
0
0
0
+
q1
q3
q2
q7
q8q9
=
- 1x106
0.00
0.00
1x106
0.00
0.00
0.00
-1875
-3750
0.00
1875
-3750
0.00
3750
5x103
0.00
-3750
10x103
1x106
0.00
0.00
-1x106
0.00
0.00
0.00
1875
37500
0.00
-1875
3750
0.00
3750
10x103
0.00
-3750
5x103
7 8 91 2 3
1
2
3
7
8
9
121
A CB
EI2EI
4 m 4 m
T2 = 40oC
T1 = 25oC
Isolate axial part from the system
RCRA
RA RA = RC+
Compatibility equation: dC/A = 0
RA = 144 kN
RC = -144 kN
0)4)(285.32(1012)4(2
)4( 6 =−×++ −
AER
AER AA
122
A B
144 kN
3.60 kN
23.38 kN�m
144 kN
3.60 kN
9 kN�m
144 kN
B C3.60 kN
9 kN�m
144 kN
3.60 kN
5.39 kN�m
V (kN) x (m)
-3.6
-
M (kN�m) x (m)
23.388.98
-5.39
+-
D3 = -719.3x10-6 radDeflected curve x (m)
A CB
EI2EI
4 m 4 m
T2 = 40oC
T1 = 25oC
D2 = -0.0004795 mm
123
Skew Roller Support
- Force Transformation- Displacement Transformation- Stiffness Matrix
124
m
i
j
m
i
j
x´y´
x
y
� Displacement and Force Transformation Matrices
12
3
45
6
θyθx
4´
5´
6´
1´
2´
3´
125
λx λy
i
jθy
θx
y´
x´
4´5´
6´
1´
2´
3´
θy
θx
m
i
j
x
y
12
3
45
6
Force Transformation
Lxx ij
x
−=λ
Lyy ij
y
−=λ
yx qqq θθ coscos 5'4'4 −=
xy qqq θθ coscos 5'4'5 −=
6'6 qq =
6
5
4
qqq
−=
10000
xy
yx
λλλλ
6'
5'
4'
qqq
−
−
=
6'
5'
4'
3'
2'
1'
6
5
4
3
2
1
1000000000000000010000000000
qqqqqq
λλλλ
λλλλ
qqqqqq
xy
yx
xy
yx
[ ] [ ] [ ]'qTq T=
126x
y
θyθx
x
y
θyθx
d4 d 4 cos θ x
d 4 cos θ yθy
θx
d5
θx
θy
d 5 cos θ x
d 5 cos θ y
x´
Displacement Transformation
j4´
5´
6´j4
5
6λx λy
yx θdθdd coscos' 544 +=
xy θdθdd' coscos 545 +−=
66' dd =
6'
5'
4'
ddd
−=
10000
xy
yx
λλλλ
6
5
4
ddd
−
−
=
6
5
4
3
2
1
6'
5'
4'
3'
2'
1'
1000000000000000010000000000
dddddd
λλλλ
λλλλ
dddddd
xy
yx
xy
yx
[ ] [ ][ ]dTd ='
127
[ ] [ ] [ ]'qTq T=
[ ] [ ][ ] [ ])'( FT qd'k'T +=
[ ] [ ][ ] [ ] [ ]FTT qTd'k'T '+=
[ ] [ ] [ ][ ][ ] [ ] [ ] [ ][ ] [ ]FFTT qdkqTdTk'Tq +=+= '
[ ] [ ] [ ][ ]TkTkTherefore T ', =
[ ] [ ] [ ]FTF qTq '=
[ ] [ ] [ ]'qTq T=
[ ] [ ][ ]dTd ='
[ ] [ ] [ ][ ]TkTk T '=
128
i j
θjθi 4 *
5 * 6*
1*
2*3 *
λix = cos θi
λiy = sin θi
λjx = cos θj
λjy = sin θj
[q*] = [T]T[q´]
1
Stiffness matrix
i j
4´
5´ 6´
1´
2´3´
*6
5*
*4
*3
2*
*1
qqqqqq
−
−
=
1000000000000000010000000000
jxjy
jyjx
ixiy
iyix
λλλλ
λλλλ
1 4 5 62 31*2*3*4*5*6*
[ T ]T
6'
5'
4'
3'
2'
1'
qqqqqq
129
[ ]
−
−
=
1000000000000000010000000000
jxjy
jyjx
ixiy
iyix
λλλλ
λλλλ
T
1 4 5 62 3123456
[ ]
−−−−
−−−
−
=
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAELAE
k
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00/
'
22
2323
22
2323
1
34
6
2
5
1 2 3 4 5 6
130
[ k ] = [ T ]T[ k´ ][T] =
Ui
Vi
Mi
Uj
Vj
Mj
Vj Mj
- λiy6EIL2
λix6EIL2
2EIL
λjy6EIL2
- λjx6EIL2
4EIL
Ui Vi Mi
- λiy6EIL2
λix6EIL2
4EIL
λjy6EIL2
λjx6EIL2
-
2EIL
Uj
AEL
- λixλiy)(12EIL3
AEL
λiy2 +
12EIL3
λix2 )(
λix6EIL2
λix6EIL2
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
λixλjx +
12EIL3
λiyλjy)-(
λjy6EIL2
AEL
λjx2 +
12EIL3
λjy2 )(
λjy6EIL2
λjx6EIL2
-
AEL
- λjxλjy)(12EIL3
- λjx6EIL2
AEL
λixλjy -
12EIL3
λiyλjx)-(
AEL
- λixλiy)(12EIL3
- λiy6EIL2
- λiy6EIL2
AEL
λixλjx +
12EIL3
λiy λjy)-(
)(AEL
λix2 +
12EIL3
λiy2
AEL
λiyλjy +
12EIL3
λix λjx )-(
AEL
λiyλjy +
12EIL3
λixλjx)-(
12EIL3
λjx2 )
AEL
- ( λixλjy- λiyλjx )12EIL3
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
- λjxλjy)(12EIL3
AEL λjy
2 + (
131
40 kN
4 m4 m 22.02 o
Example 12
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagramsand qualitative deflected shape.Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.
132
40 kN40 kN�m
20 kN
40 kN�m
20 kN
[ q´F ][FEF]
1Global
i j
1
i j
Local
40 kN
4 m4 m 22.02o
20sin22.02=7.520cos22.02=18.54
1´
2´
3´
4´
5´
6´
4
5
6
x*
x´1 *
2* 3 *
λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271,
λjy = cos 67.98o = 0.3749
22.02 o
133
x*
x´1
4
5
6
1 *
2* 3 *
� Transformation matrix
Member 1: [ q ] = [ T ]T[q´]
1
i j1´
2´
3´
4´
5´
6´
Global Local
λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271,
λjy = cos 67.98o = 0.3749
*3
2*
*1
6
5
4
qqqqqq
6'
5'
4'
3'
2'
1'
qqqqqq
−=
10000009271.03749.000003749.09271.0000000100000010000001
1´ 4´ 5´ 6´2´ 3´4561*
2*
3*
[ ]
−
−
=
1000000000000000010000000000
jxjy
jyjx
ixiy
iyix
TT
λλλλ
λλλλ
134
[k´]1 = 103
1´2´ 3´4´5´6´
1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
1
i j1´
2´
3´
4´
5´
6´
Local
� Local stiffness matrix
[ ]
−−−−
−−−
−
=×
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAELAE
k
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00/
22
2323
22
2323
66
θi ∆j θj∆i δjδi
Mi
Vj
Mj
Vi
Nj
Ni
135
Stiffness matrix [k´]:
[k´]1 = 103
1´2´ 3´4´5´6´
1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
Stiffness matrix [k*]: [ k* ] = [ T ]T[ k´ ][T]
[k*]1 = 103
4561*
2*
3*
4 1* 2* 3*5 6-139.00.3511.406129.0
1.40651.82
-56.25-0.869-3.75051.8221.90
-3.476
0.0003.75010.001.406
-3.47620.00
150.0
0.0000.000
-139.0-56.250.000
0.0000.93753.7500.351
-0.8693.750
0.0003.75020.001.406
-3.75010.00
x*
x´1
4
5
6
1 *
2* 3 * 1
i j1´
2´
3´
4´
5´
6´
Global Local
4
5
6
2*
136
Global Equilibrium:
Q1 = 0.0
Q3 = 0.0
D1*
D3*=
36.37x10-6 m
0.002 rad
[Q] = [K][D] + [QF]
= 1031*
3*
1*
129
1.406
3*
1.406
20.0
D1*
D3*
-7.5
-40+
x*
x´14
5
6
1 *
2* 3 *
4 m4 m 22.02 o
40 kN
x*
x´4
5
6
2*
40 kN40 kN�m
20 kN
40 kN�m
20 kN
[ q´F ]20sin22.02=7.5
20cos22.02=18.54
137
Member Force : [q] = [k*][D] + [qF]
0
40.020
-7.5418.54-40.0
+
q4
q6
q5
q1*
q2*q3*
= 103
4561*
2*
3*
4 1* 2* 3*5 6-139.00.3511.406129.0
1.40651.82
-56.25-0.869-3.75051.8221.90
-3.476
0.0003.75010.001.406
-3.47620.00
150.0
0.0000.000
-139.0-56.250.000
0.0000.93753.7500.351
-0.8693.750
0.0003.75020.001.406
-3.75010.00
0
00
36.4x10-6
02x10-3
-5.06
6027.5
013.48
0
=
40 kN
13.48 kN
60 kN�m
27.5 kN
5.06 kN
4 m4 m 22.02 o
40 kN
x*
x´1
4
5
6
1 *
2* 3 *
40 kN40 kN�m
7.518.54
40 kN�m
20 kN
4
5
6
2*
138
4 m4 m 22.02 o
40 kN
5.05 kN27.51 kN
60.05 kN�m
13.47 kN
40 kN
+
Bending moment diagram (kN�m)
Deflected shape
50.04
-
-60.05
0.002 rad
139
40 kN
22.02o
8 m 4 m 4 m2EI, 2AE EI, AE
6 kN/m
Example 13
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagramsand qualitative deflected shape.Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.
140
40 kN
22.02 o8 m 4 m 4 m
2EI, 2AE EI, AE
6 kN/m
1 21
2
3
4
5
6
7 *
8 * 9 *
Global
21´
2´3´
4´
5´6´
11´
2´
3´
4´
5´
6´
Local
mkNm
mkNmL
AE
•×=
×=
3
262
10150)8(
)/10200)(006.0(
mkNm
mmkNLEI
•×=
×=
3
426
1020)8(
)0002.0)(/10200(44
mkNLEI
•×= 310102
kNm
mmkNLEI
3
2
426
2
1075.3)8(
)0002.0)(/10200(66
×=
×=
mkNm
mmkNLEI
/109375.0)8(
)0002.0)(/10200(1212
3
3
426
2
×=
×=
141
[k]1 = [k´]1 = 2x103
456123
4 1 2 35 6-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
1 21
2
3
4
5
6
7 *
8 * 9 *
Global Local : Member 1
14
5
6
1
2
3
[ q ]
11´
2´
3´
4´
5´
6´
[ q´]
[q] = [q´] Thus, [k] = [k´]
142
Member 2: Use transformation matrix, [q*] = [T]T[q´]
q1´
q3´
q2´
q4´
q5´
q6´
q1
q3
q2
q7*
q8*q9*
0.92710.3749
-0.37490.9271
000
000
000
0 0
001
10
01
=
1237*
8*
9*
1´ 4´ 5´ 6´2´ 3´
0 0
001
000
000
000
1 21
2
3
4
5
6
7 *
8 * 9 *
λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271,
λjy = cos 67.98o = 0.3749
21
2
3
x*
x´7*
8 * 9* [ q* ]
21´
2´3´
4´
5´6´
[ q´ ]
143
Stiffness matrix [k´]:
[k´]2 = 103
1´2´ 3´4´5´6´
1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
Stiffness matrix [k*]: [k*] = [T]T[k´][T]
[k*]2 = 103
1237*
8*
9*
1 7* 8* 9*2 3-139.00.3511.406129.0
1.40651.82
-56.25-0.869-3.47751.8221.90
-3.476
0.0003.75010.001.406
-3.47620.00
150.0
0.0000.000
-139.0-56.250.000
0.0000.93753.7500.351
-0.8693.750
0.0003.75020.001.406
-3.47710.00
21
23
x*
x´7 *
8 * 9 * [ q* ]
21´
2´3´
4´
5´6´
[ q´ ]
144
1 21
2
3
4
5
6
7 *
8 * 9 *
1
2
3
4
5
6
8 *
[k]1= 2x103
456123
4 1 2 35 6-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
[k*]2 = 103
1237*
8*
9*
1 7* 8* 9*2 3-139.00.3511.406129.0
1.40651.82
-56.25-0.869-3.47751.8221.90
-3.476
0.0003.75010.001.406
-3.47620.00
150.0
0.0000.000
-139.0-56.250.000
0.0000.93753.7500.351
-0.8693.750
0.0003.75020.001.406
-3.47710.00
145
Global:
=-509.84x10-6 rad
18.15x10-6 m
0.00225 rad
58.73x10-6 mD3
D1
D9*
D7*
00
00 +
-32 + 40 = 80
-40-7.5
D3
D1
D9*
D7*
32 kN�m
24 kN
32 kN�m
24 kN
6 kN/m
1
40 kN6 kN/m
1 21
2
3
4
5
6
7*
8 *
9*
1
2
3
4
5
6
8 *
= 1030
0-139
450
101.406
600
1.406
1.406-139
129
010
1.40620
1 3 7* 9*
137*
9*
40 kN40 kN�m
7.518.54
40 kN�m
20 kN
40 kN�m
7.5
40 kN�m32 kN�m
146
Member 1: [ q ] = [k ][d] + [qF]
0
3224
024-32
+
0
00
d1=18.15x10-6
0d3=-509.84 x10-6
q4
q6
q5
q1
q2q3
= 2x103
456123
4 1 2 35 6-150.00.0000.000150.0
0.0000.000
0.000-0.9375-3.7500.000
0.9375-3.750
0.0003.75010.000.000
-3.75020.00
150.0
0.0000.000
-150.00.0000.000
0.0000.93753.7500.000
-0.93753.750
0.0003.75020.000.000
-3.75010.00
-5.45 kN
21.80 kN�m20.18 kN
5.45 kN27.82 kN-52.39 kN�m
=
q4
q6
q5
q1
q2q3
6 kN/m
5.45 kN
20.18 kN
21.80 kN�m
5.45 kN
27.82 kN
52.39 kN�m
32 kN�m
24 kN
32 kN�m
24 kN
6 kN/m
114
5
6
1
2
3
147
q1
q3
q2
q7*
q8*q9*
21
3
x*
x´7*
9 *
= 103
1237*
8*
9*
1 7* 8* 9*2 3-139.00.3511.406129.0
1.40651.82
-56.25-0.869-3.75051.8221.90
-3.476
0.0003.75010.001.406
-3.47620.00
150.0
0.0000.000
-139.0-56.250.000
0.0000.93753.7500.351
-0.8693.750
0.0003.75020.001.406
-3.75010.00
0
4020
-7.518.54-40
+
18.15x10-6
-509.84x10-6
0
58.73x10-6
00.00225
q1
q3
q2
q7*
q8*q9*
-5.45 kN
52.39 kN�m26.55 kN
0 kN14.51 kN0 kN�m
= 5.45 kN
26.55 kN
52.39 kN�m
14.51 kN
40 kN
20sin22.02=7.520cos22.02=18.54
40 kN40 kN�m
20 kN
40 kN�m
20 kN
[ q´F ]
2
2 8 *
Member 2: [ q ] = [k ][d] + [qF]
148
5.45 kN26.55 kN
52.39 kN�m
14.51 kN
40 kN6 kN/m
5.45 kN
27.82 kN
52.39 kN�m
5.45 kN
20.18 kN
21.80 kN�m
40 kN
22.02o
8 m 4 m 4 m
6 kN/m
5.45 kN
20.18 kN
21.80 kN�m
14.51 kN54.37 kN
3.36 m
M (kN�m) x (m)
53.81
-21.8-
V (kN)x (m)
26.55
+20.18
+- -
-27.82
14.51cos 22.02o
=13.45 kN
-52.39-
12.14 +
-13.45
1
! Simple Frames! Frame-Member Stiffness Matrix! Displacement and Force Transformation Matrices! Frame-Member Global Stiffness Matrix
! Special Frames! Frame-Member Global Stiffness Matrix
FRAME ANALYSIS USING THE STIFFNESSMETHOD
2
Simple Frames
3
Frame-Member Stiffness Matrix
0 0 00 - AE/LAE/L
4EI/L - 6EI/L2 2EI/L6EI/L2 00
6EI/L2 - 12EI/L3 6EI/L212EI/L3 00
0
2EI/L
-6EI/L2
0
- 6EI/L2
12EI/L3
0
-6EI/L2
4EI/L
0
6EI/L2
-12EI/L3
AE/L
0
0
-AE/L
0
0
m
x´y´
i
j
3´
5´
6´
2´
4´
1´
3´ 6´2´ 4´1´ 5´
[k´]
4´5´6´
1´2´3´
6EI/L24EI/L
6EI/L24EI/L
AE/L
AE/L
6EI/L2
6EI/L2
12EI/L3
12EI/L3
2EI/L
d 1́ = 1
AE/L
AE/L
d 2́ = 1
12EI/L3
12EI/L3
d 3́ = 1
2EI/L
6EI/L2
6EI/L2
6EI/L26EI/L2AE/L
2EI/L
6EI/L2
6EI/L2d 6́
= 1
6EI/L2
6EI/L22EI/L
6EI/L2
12EI/L3
12EI/L3
4EI/L4EI/L12EI/L3
12EI/L3
6EI/L2
6EI/L2AE/L
6EI/L2d 4́ = 1
d 5́ = 1AE/L
AE/L
4
m
i
j
m
i
j
x´y´
x
y
Displacement and Force Transformation Matrices
12
3
45
6
θyθx
4´
5´
6´
1´
2´
3´
5
λx
q4 = q4´ cos θx - q5´ cos θy
q5 = q4´ cos θy + q5´ cos θx
q6 = q6´
λy
i
jθy
θx
y´
x´
4´5´
6´
1´
2´
3´
θy
θx
m
i
j
x
y
12
3
45
6
Force Transformation
Lxx ij
x
−=λ
Lyy ij
y
−=λ
−=
6'
5'
4'
6
5
4
10000
qqq
λλλλ
qqq
xy
yx
−
−
=
6'
5'
4'
3'
2'
1'
6
5
4
3
2
1
1000000000000000010000000000
qqqqqq
λλλλ
λλλλ
qqqqqq
xy
yx
xy
yx
[ ] [ ] [ ]'qTq T=
6
[q] = [T]T[q´]
= [T]T ( [k´][d´] + [q´F] )
= [T]T [k´][d´] + [T]T [q´F]
[q] = [T]T [k´][T][d] + [T]T [q´F] = [k][d] + [qF]
Therefore, [k] = [T]T [k´][T]
[qF] = [T]T [q´F]
[q] = [T]T[q´]
[d´] = [T][d]
[k] = [T]T [k´][T]
7
[q] = [T]T[q´]= [T]T ( [k´][d´] + [q´F] ) = [T]T[k´][d´] + [T]T[q´F] = [T]T [k´][T][d] + [T]T [q´F]
Frame Member Global Stiffness Matrix
[k] [qF][ k ] = [ T ]T[ k´ ][T] =
Ui
Vi
Mi
Uj
Vj
Mj
Vj Mj
- λiy6EIL2
λix6EIL2
2EIL
λjy6EIL2
- λjx6EIL2
4EIL
Ui Vi Mi
- λiy6EIL2
λix6EIL2
4EIL
λjy6EIL2
λjx6EIL2
-
2EIL
Uj
AEL
- λixλiy)(12EIL3
AEL
λiy2 +
12EIL3
λix2 )(
λix6EIL2
λix6EIL2
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
λixλjx +
12EIL3
λiyλjy)-(
λjy6EIL2
AEL
λjx2 +
12EIL3
λjy2 )(
λjy6EIL2
λjx6EIL2
-
AEL
- λjxλjy)(12EIL3
- λjx6EIL2
AEL
λixλjy -
12EIL3
λiyλjx)-(
AEL
- λixλiy)(12EIL3
- λiy6EIL2
- λiy6EIL2
AEL
λixλjx +
12EIL3
λiy λjy)-(
)(AEL
λix2 +
12EIL3
λiy2
AEL
λiyλjy +
12EIL3
λix λjx )-(
AEL
λiyλjy +
12EIL3
λixλjx)-(
12EIL3
λjx2 )
AEL
- ( λixλjy- λiyλjx )12EIL3
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
- λjxλjy)(12EIL3
AEL λjy
2 + (
8
5 kN
6 m
6 m
AB
C
Example 1
For the frame shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2
9
5 kN
6 m
6 m
AB
C kN/m666.667(6m)
)m10)(60mkN1012(20012
3
462
6
3 =××
=
−
LEI
kN/m200006m
)mkN10)(200m10(600 2
626
=××
=
−
LAE
kN2000(6m)
)m10)(60mkN106(2006
2
462
6
2 =××
=
−
LEI
mkN80006m
)m10)(60mkN104(2004
462
6
•=××
=
−
LEI
mkN40006m
)m10)(60mkN102(2002
462
6
•=××
=
−
LEI
Global :
AB
C
1
2
78 9
4
6 5
12 3
10
Global :
AB
C
1
2
78 9
4
6 5
12 3
A B14´
5´ 6´
1´
2´ 3´
Local :
5´
2´
1´ 3´
4´
6´
2
Using Transformation Matrix:
� Member Stiffness Matrix
[ ]
−−−−
−−−
−
=
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAEAE/L
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00
k'
22
2323
22
2323
Mi
Vj
Mj
Vi
Nj
Ni
θi ∆j θj∆i δjδi
11
A B14´
5´ 6´
1´
2´ 3´
Local :
[q] = [q´]
-> [k]1 = [k´]1
Stiffness Matrix: Member 1
Global:
AB
C
1
2
78 9
4
6 5
12 3
4 6 5 1 2 34
6
5
1
2
3
20000
0
0
-20000
0
0
0
666.667
2000
0
-666.667
2000
0
2000
8000
0
-2000
4000
-20000
0
0
20000
0
0
0
-666.667
-2000
0
666.667
-2000
0
2000
4000
0
-2000
8000
[k]1 =
12
Local:
5´
2´
1´ 3´
4´
6´
2
[q]2 = [ T ]T[ q´]2
q1´
q3´
q2´
q4´q5´
q6´
q1
q3
q2
q7
q8q9
[T]T
Stiffness Matrix: Member 2
=
123789
4´0000
0-1
5´000100
6´000001
1´0
0-1
000
2´100000
3´001000
90o
λjx = cos (-90o) = 0λjy = sin (-90o) = -1
λix = cos (-90o) = 0λiy = sin (-90o) = -1
Global:
AB
C
1
2
78 9
4
6 5
12 3
13
[k]2 = [ T ]T[ k´ ]2[ T ]
1´ 2´ 3´ 4´ 5´ 6´1´
2´
3´
4´
5´
6´
20000
0
0
-20000
0
0
0
666.667
2000
0
-666.667
2000
0
2000
8000
0
-2000
4000
-20000
0
0
20000
0
0
0
-666.667
-2000
0
666.667
-2000
0
2000
4000
0
-2000
8000
[k´]2 =
1 2 3 7 8 9
666.667 20001
2
3
7
8
9
2000
0
0
-666.667
-666.667
0
0
2000
2000
20000 0
0
0
0
-20000
-20000
0
0
8000 -2000
-2000
0
0
4000
4000
666.667 0
0
-2000
-2000
20000 0
0 8000
[k]2 =
14
[k]14 6 5 1 2 3
4
6
5
1
2
3
20000
0
0
-20000
0
0
0666.667
20000
-666.667
2000
02000
80000
-2000
4000
-200000
020000
0
0
0-666.667
-20000
666.667
-2000
02000
40000
-2000
8000
1 2 3 7 8 9666.667 20001
2
3
7
8
9
2000
00
666.667
-666.667
0
0
2000
2000
20000 0
0
0
0
-20000
-20000
0
0
8000 -2000-2000
0
0
4000
4000
666.667 0
0
-2000
2000
20000 0
0 8000
[k]2
Global Stiffness Matrix:
20000
0
-20000
0
0
0
8000
0
-2000
4000
0
-2000
0
4000
-20000
0
0
20666.667
-2000
2000
-2000
16000
20666.667
0
2000
[K]
4
5
1
2
3
4 5 1 2 3
Global:
AB
C
1
2
78 9
4
6 5
12 3
15
AB
C
Q4 = 0Q5 = 0
Q1 = 5Q2 = 0
Q3 = 0
D4
D5
D1
D2
D3
+
0 0
0 0
0
D4
D5
D1
D2
D3
=
0.01316 m
0.01316 m9.199(10-4) rad
-9.355(10-5) m
-1.887(10-3) rad
5 kN
6 m
6 m
AB
C
1
2
Global:
1
2
78 9
4
6 5
12 3
5 kN
=
4
5
1
2
3
4 5 1 2 3-20000 0 0
20666.667 00
2000
2000
20666.667 -2000
-2000 16000
0-2000
4000
08000 0 -2000 4000
-200000
20000
0
0
[Q] = [K][D] + [QF]
16
D4 = 0.01316
D5 = 9.199(10-4)
D6 = 0
D1 = 0.01316
D2 = -9.355(10-5)
D3 = -1.887(10-3)
15 kN
6 m
6 m
AB
C
2
q4
q5
q6
q1
q2
q3
0
0
-1.87
0
1.87
-11.22
Member 1
A B11
2 3
4
6 5
A B1
1.87 kN 11.22 kN�m1.87 kN
4 6 5 1 2 3
4
6
5
1
2
3
20000
0
0
-20000
0
0
0
666.667
2000
0
-666.667
2000
0
2000
8000
0
-2000
4000
-20000
0
0
20000
0
0
0
-666.667
-2000
0
666.667
-2000
0
2000
4000
0
-2000
8000
[q]1 = [k]1[d]1 + [qF]1
17
5 kN
1.87 kN 11.22 kN�m
5 kN1.87 kN
18.77 kN�m
Member 2
q1
q3
q2
q7
q8
q9
D1 = 0.01316
D3 = -1.887(10-3)
D2 = -9.355(10-5)
D7 = 0
D8 = 0
D9 = 0
5
11.22
-1.87
-5
1.87
18.77
1 2 3 7 8 9
666.667 20001
2
3
7
8
9
2000
0
0
-666.667
-666.667
0
0
2000
2000
20000 0
0
0
0
-20000
-20000
0
0
8000 -2000
-2000
0
0
4000
4000
666.667 0
0
-2000
-2000
20000 0
0 8000
[q]2 = [k]2[d]2 + [qF]2
15 kN
6 m
6 m
AB
C
2 1
2 3
78 9
2 2
18
AB
C
Bending moment diagram
A B1
1.87 kN 11.22 kN�m1.87 kN5 kN
6 m
6 m
AB
C
1.87 kN
18.77 kN�m5 kN
1.87 kN
AB
C
Shear diagram
5
5
+
-1.87-
+
18.77
11.22
-11.22
-
5 kN
1.87 kN 11.22 kN�m
5 kN
1.87 kN
18.77 kN�m
2
19
Deflected shape
AB
C
AB
C
Bending moment diagram +
18.77
11.22
-11.22
-
D3=-0.00189 rad
D3=-0.00189 rad
D1=13.16 mmD4=13.16 mm
D4
D5
D1
D2
D3
=
0.01316 m
0.01316 m9.199(10-4) rad
-9.355(10-5) m
-1.887(10-3) rad
Global :
AB
C
1
2
78 9
4
6 5
12 3
D5=0.00092 rad
20
4.5 m
6 m 6 m
3 kN/m
Example 2
For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B(b) Determine all the reactions at supports(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member.
A
B C
21
Global 2
1
1
2 3
4
56
7
8 9
2 7
8 9
1
2 3
[FEM] 9 kN�m
9 kN
9 kN�m
9 kN
3 kN/m
2
Members1
2 3
4
56 1
4.5 m
6 m 6 m
3 kN/m
22
4.5 m
6 m 6 m
3 kN/m
7.5 m
λx = cos θx = 6/7.5 = 0.8
Member 1:
θx
θy 1
2 3
4
56 1
λy = cos θy = 4.5/7.5 = 0.6
mkNm
mkNmL
AE
/160005.7
)/10200)(10600( 2626
=
××=
−
mkNm
mmkNLEI
/33.341)5.7(
)1060)(/10200(12123
4626
3
=
××=
−
kNm
mmkNLEI
1280)5.7(
)1060)(/10200(662
4626
2
=
××=
−
mkNm
mmkNLEI
•=
××=
−
64005.7
)1060)(/10200(44 4626
mkNm
mmkNLEI
•=
××=
−
32005.7
)1060)(/10200(22 4626
23
[ km ] = [ T ]T[ k´ ][T ] =
λx = cos θx
Member m:
θx
θy Uj
VjMj
Ui
Vi
Mim
λy = cos θy
Ui
Vi
Mi
Uj
Vj
Mj
Vj Mj
- λiy6EIL2
λix6EIL2
2EIL
λjy6EIL2
- λjx6EIL2
4EIL
Ui Vi Mi
- λiy6EIL2
λix6EIL2
4EIL
λjy6EIL2
λjx6EIL2
-
2EIL
Uj
AEL
- λixλiy)(12EIL3
AEL
λiy2 +
12EIL3
λix2 )(
λix6EIL2
λix6EIL2
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
λixλjx +
12EIL3
λiyλjy)-(
λjy6EIL2
AEL
λjx2 +
12EIL3
λjy2 )(
λjy6EIL2
λjx6EIL2
-
AEL
- λjxλjy)(12EIL3
- λjx6EIL2
AEL
λixλjy -
12EIL3
λiyλjx)-(
AEL
- λixλiy)(12EIL3
- λiy6EIL2
- λiy6EIL2
AEL
λixλjx +
12EIL3
λiy λjy)-(
)(AEL
λix2 +
12EIL3
λiy2
AEL
λiyλjy +
12EIL3
λix λjx )-(
AEL
λiyλjy +
12EIL3
λixλjx)-(
12EIL3
λjx2 )
AEL
- ( λixλjy- λiyλjx )12EIL3
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
- λjxλjy)(12EIL3
AEL λjy
2 + (
24
λx = cos θx = 6/7.5 = 0.8
Member 1:
θx
θy 1
2 3
4
56 1
λy = cos θy = 4.5/7.5 = 0.6
4.5 m
6 m 6 m
3 kN/m
7.5 m
=
4
5
6
1
2
3
4 1
-10362.879
-7516.162
768
10362.879
768
7516.162
2
-7516.162
-5978.451
-1024
7516.162
5978.451
-1024
3
-768
1024
3200
768
-1024
6400
10362.879
-768
7516.162
-10362.879
-7516.162
-768
5
7516.162
5978.451
1024
-7516.162
-5978.451
1024
6
-768
1024
6400
768
-1024
3200
[k1]
25
λx = cos 0o = 1.0, λy = cos 90o = 0
Member 2:
2 7
8 9
1
2 3
4.5 m
6 m 6 m
3 kN/m
7.5 mmkN
mmkNm
LAE
/200006
)/10200)(10600( 2626
=
××=
−
mkNm
mmkNLEI
/667.666)6(
)1060)(/10200(12123
4626
3
=
××=
−
kNm
mmkNLEI
2000)6(
)1060)(/10200(662
4626
2
=
××=
−
mkNm
mmkNLEI
•=
××=
−
80006
)1060)(/10200(44 4626
mkNm
mmkNLEI
•=
××=
−
40006
)1060)(/10200(22 4626
26
[k2] =0
2EI/L
-6EI/L2
- 6EI/L2
6EI/L2
0
6EI/L2
-12EI/L3
0
0
0
0
-AE/L
0
0
0
6EI/L2
0
- 6EI/L2
- 12EI/L3
0
0
6EI/L2
2EI/L
0
-6EI/L2
0 - AE/L
0
0
4EI/L
12EI/L3
4EI/L
12EI/L3
AE/L
AE/L
3
8
9
2
7
1
3 92 71 8
[k2] =0
4000
-2000
- 2000
2000
0
2000
-666.667
0
0
0
0
-20000
0
0
0
2000
0
- 2000
- 666.667
0
0
2000
4000
0
-2000
0 - 20000
0
0
8000
666.667
8000
666.667
20000
20000
3
8
9
2
7
1
3 92 71 8
2 7
8 9
1
2 3
27
=
4
5
6
1
2
3
4 1
-10362.879
-7516.162
768
10362.879
768
7516.162
2
-7516.162
-5978.451
-1024
7516.162
5978.451
-1024
3
-768
1024
3200
768
-1024
6400
10362.879
-768
7516.162
-10362.879
-7516.162
-768
5
7516.162
5978.451
1024
-7516.162
-5978.451
1024
6
-768
1024
6400
768
-1024
3200
[k1]
[k2] =0
4000
-2000
- 2000
2000
0
2000
-666.667
0
0
0
0
-20000
0
0
0
2000
0
- 2000
- 666.667
0
0
2000
4000
0
-2000
0 - 20000
0
0
8000
666.667
8000
666.667
20000
20000
3
8
9
2
7
1
3 92 71 8
28
D1
D3
D2 =4.575(10-4) m
-5.278(10-4) rad
-1.794(10-3) m
Global:
1
21
2 3
4
5 6
7
8 9
4.5 m
6 m 6 m
3 kN/m
7.5 m9 kN
9 kN�m
9 kN
0
0
0
Q1
Q3
Q2
D1
D3
D2
0
9
9+1
2
3
1
30362.9
768
7516.16
2
7516.16
6645.12
976
3
768
976
14400=
9 kN�m
9 kN
29
λx = cos θx = 6/7.5 = 0.8
Member 1:
θx
θy 1
2 3
4
56 1
λy = cos θy = 4.5/7.5 = 0.6
0
0
0
D1 = 4.575(10-4)
D2 = -1.794(10-3)
D3 = -5.278(10-4)
0
0
0
0
0
0
+ =
q4
q6
q5
q1
q2q3
4
5
6
1
2
3
5 1 2 34 6
k1
1
11.37 kN
11.37 kN
0.09 kN
1.19 kN�m
0.09 kN
0.50 kN�m 1
9.15 kN6.75 kN 1.19 kN�m
9.15 kN
6.75 kN
0.50 kN�m
9.15
0.50
6.75
-9.15
-6.75
-1.19
=
30
Member 2:
2 7
8 9
1
2 3
0
9
9
0
9
-9
+=
q1
q3
q2
q7
q8q9
D1 = 4.575(10-4)
D2 = -1.794(10-3)
D3 = -5.278(10-4)
0
0
0
1
2
3
7
8
9
2 7 8 91 3
k2
9.15
1.19
6.75
-9.15
11.25
-14.70
=
9 kN�m
9 kN
9 kN�m
9 kN
3 kN/m
2[FEM]
3 kN/m
29.15 kN
6.75 kN
1.19 kN�m
9.15 kN11.25 kN
14.70 kN�m
31
3 kN/m
29.15 kN
6.75 kN
1.19 kN�m
9.15 kN11.25 kN
14.70 kN�m
1
9.15 kN6.75 kN 1.19 kN�m
9.15 kN
6.75 kN
0.50 kN�m
3 kN/m
All Reactions
9.15 kN11.25 kN
14.70 kN�m
9.15 kN
6.75 kN
0.50 kN�m
32
-1.19
Shear diagram (kN)
Deflected shape
-0.09
-0.09
6.75
3 kN/m
29.15 kN
6.75 kN
1.19 kN�m
9.15 kN11.25 kN
14.70 kN�m
1
11.37 kN
11.37 kN
0.09 kN
1.19 kN�m
0.09 kN0.50 kN�m
D3 =-5.278(10-4) rad
D2 = -1.79 mm
D1 = 0.46 mm
D1
D3
D2 =4.575(10-4) m
-5.278(10-4) rad
-1.794(10-3) m -11.25
-
+
Bending-moment diagram (kN�m)
-14.70
0.5
33
Example 3
For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member.
4.5 m
6 m 3 m
10 kN
A
BC15 kN
20 kN�m
3 kN/m
3 m
34
[FEM] 13 kN/m 11.25 kN
wL/2 = 11.25 kN
wL2/12 = 14.06 kN�m
14.06 kN�m
Global
21
2 7
8 9
1
2 3
4.5 m
6 m 3 m
10 kN
A
BC15 kN
20 kN�m
3 kN/m
3 m
7.5 kN�m
5 kN
7.5 kN�m
5 kN
2
10 kN
λx = cos θx = 6/7.5 = 0.8
1
Members1
2 3
4
56
1
2 3
4
56
7
8 9
θy = 53.13
θx =36.87
λy = cos θy = 4.5/7.5 = 0.6
6.75 kN
9 kN
11.25(0.6) = 6.75 kN
11.25(0.8) = 9 kN
35
Global:
1
21
2 3
4
5 6
7
8 9
Q1 = 15
Q3 = 20
Q2 = 0
D1
D3
D2
-6.75
-14.06 + 7.5
9 + 5+1
2
3
1
30362.9
768
7516.16
2
7516.16
6645.2
976
3
768
976
14400=
[FEM]13 kN/m 11.25 kN
11.25 kN
14.06 kN�m
14.06 kN�m7.5 kN�m
5 kN
7.5 kN�m
5 kN
2
10 kN
6.75 kN
9 kN
11.25(0.6) = 6.75 kN
9 kN
10 kN
A
BC15 kN
20 kN�m
3 kN/m
14.06 kN�m7.5 kN�m
6.75 kN
5 kN9 kN
15 kN20 kN�m
36
D1
D3
D2 =1.751(10-3) m
2.049(10-3) rad
-4.388(10-3) m
1
21
2 3
4
5 6
7
8 9
37
Member 1:
0
0
0
D1 = 1.751(10-3)
D2 = -4.388(10-3)
D3 = 2.049(10-3)
-6.75
14.06
9
-6.75
9
-14.06
+ =
q4
q6
q5
q1
q2q3
4
5
6
1
2
3
5 1 2 34 6
k1
6.51
26.46
24.17
-20.01
-6.17
4.89
=
[FEM]
13 kN/m 11.25 kN
11.25 kN
14.06 kN�m
14.06 kN�m
6.75 kN
9 kN
11.25(0.6) = 6.75 kN
9 kN
θx
θy 1
2 3
4
56 1
λx = cos 36.87o = 0.8, λy = cos 53.13o = 0.6
14.06 kN�m
14.06 kN�m
6.75 kN
9 kN
11.25(0.6) = 6.75 kN
9 kN
38
3 kN/m
7.5 m1
19.71 kN
19.71 kN
7.07 kN
4.89 kN�m
15.43 kN
26.46 kN�m
=
q4
q6
q5
q1
q2q3
6.51
26.46
24.17
-20.01
-6.17
4.89
θx = 36.87o
θy = 53.13o 1
2 3
4
56 1
3 kN/m
7.5 m
20.01 kN
4.89 kN�m
6.51 kN
24.17 kN
26.46 kN�m
6.17 kN
39
Member 2:
2 7
8 9
1
2 3
0
7.5
5
0
5
-7.5
+=
q1
q3
q2
q7
q8q9
D1 = 1.751(10-3)
D2 = -4.388(10-3)
D3 = 2.049(10-3)
0
0
0
1
2
3
7
8
9
2 7 8 91 3
k2
35.02
15.12
6.17
-35.02
3.83
-8.08
=
35.02 kN
6.17 kN
15.12 kN�m
35.02 kN3.83 kN
8.08 kN�m
[FEM]
7.5 kN�m
5 kN
7.5 kN�m
5 kN
2
10 kN 2
10 kN
7.5 kN�m
5 kN
7.5 kN�m
5 kN
40
3 kN/m
7.5 m
20.01 kN
4.89 kN�m
6.51 kN
24.17 kN
26.46 kN�m
6.17 kN
35.02 kN
6.17 kN
15.12 kN�m
35.02 kN3.83 kN
8.08 kN�m
2
10 kN
10 kN
A
BC15 kN
20 kN�m
3 kN/m
6 m 3 m3 m
6.51 kN
24.17 kN
26.46 kN�m
35.02 kN3.83 kN
8.08 kN�m
41
Shear diagram (kN)
Bending-moment diagram (kN�m)
3 kN/m
7.5 m1
19.71 kN
19.71 kN
7.07 kN
4.89 kN�m
15.43 kN
26.46 kN�m
35.02 kN
6.17 kN
15.12 kN�m
35.02 kN3.83 kN
8.08 kN�m
2
10 kN
15.43D1
D3
D2 =1.751(10-3) m
2.049(10-3) rad
-4.388(10-3) m
Deflected shape
D3 = 2.05(10-3) rad
D2 = -4.39 mm
D1 =1.75 mm
-26.46
4.89
-15.12 -8.08
-7.07
6.17
-3.83
42
Special Frames
43
i j
θjθi 4 *
5 * 6*
1*
2*3 *
λix = cos θi
λiy = sin θi
λjx = cos θj
λjy = sin θj
[ q* ] = [ T ]T[ q´ ]
1
Stiffness matrix
i j
4´
5´ 6´
1´
2´3´
[ T ]T
−
−
=
6'
5'
4'
3'
2'
1'
*6
5*
*4
*3
2*
*1
1000000000000000010000000000
qqqqqq
λλλλ
λλλλ
qqqqqq
jxjy
jyjx
ixiy
iyix1*2*3*4*5*6*
1´ 2´ 3´ 4´ 5´ 6´
44
[ ]
−
−
=
1000000000000000010000000000
jxjy
jyjx
ixiy
iyix
λλλλ
λλλλ
T
1´2´3´4´5´6´
1* 2* 3* 4* 5* 6*
� Member Stiffness Matrix
1´2´3´4´5´6´
1´ 2´ 3´ 4´ 5´ 6´
[ ]
−−−−
−−−
−
=
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAEAE/L
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00
k'
22
2323
22
2323
45
[ k ] = [ T ]T[ k´ ][T] =
Ui
Vi
Mi
Uj
Vj
Mj
Vj Mj
- λiy6EIL2
λix6EIL2
2EIL
λjy6EIL2
- λjx6EIL2
4EIL
Ui Vi Mi
- λiy6EIL2
λix6EIL2
4EIL
λjy6EIL2
λjx6EIL2
-
2EIL
Uj
AEL
- λixλiy)(12EIL3
AEL
λiy2 +
12EIL3
λix2 )(
λix6EIL2
λix6EIL2
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
λixλjx +
12EIL3
λiyλjy)-(
λjy6EIL2
AEL
λjx2 +
12EIL3
λjy2 )(
λjy6EIL2
λjx6EIL2
-
AEL
- λjxλjy)(12EIL3
- λjx6EIL2
AEL
λixλjy -
12EIL3
λiyλjx)-(
AEL
- λixλiy)(12EIL3
- λiy6EIL2
- λiy6EIL2
AEL
λixλjx +
12EIL3
λiy λjy)-(
)(AEL
λix2 +
12EIL3
λiy2
AEL
λiyλjy +
12EIL3
λix λjx )-(
AEL
λiyλjy +
12EIL3
λixλjx)-(
12EIL3
λjx2 )
AEL
- ( λixλjy- λiyλjx )12EIL3
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
- λjxλjy)(12EIL3
AEL λjy
2 + (
46
40 kN
4 m 4 m
3 m
7.416 m
22.02 o
20 kN200 kN�m
6 kN/m
Example 4
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagrams and qualitativedeflected shape.Take I = 200(106) mm4 , A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.
47
1Local
12
40 kN
4 m 4 m
3 m
7.416 m
22.02 o
20 kN200 kN�m
6 kN/m
4
5 6
7
8 922.02 o
Global
2
1´
2´ 3´
4´
5´ 6´
1´
2´3´
4´
5´ 6´
1 *
2* 3 *
48
mkNm
mkNmL
AE /101508
)/10200)(006.0( 3262
×=×
=
mkNm
mmkNLEI
•×=×
= 3426
10208
)0002.0)(/10200(44
mkNLEI
•×= 310102
kNm
mmkNLEI 3
2
426
2 1075.3)8(
)0002.0)(/10200(66×=
×=
mkNm
mmkNLEI /109375.0
)8()0002.0)(/10200(1212 3
3
426
3 ×=×
=
[ ]
−−−−
−−−
−
=
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAEAE/L
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00
k'
22
2323
22
2323
Mi
Vj
Mj
Vi
Nj
Ni
θi ∆j θj∆i δjδi
49
2
1´
2´ 3´
4´
5´ 6´
1´
2´3´
1 4´
5´ 6´
Local
8 m
8 m
03750
0
- 3750- 937.5
0
0375010000
0-3750
0 - 150000
00
0
10000
-3750
- 3750
3750
0
3750
-937.5
0
0
00
-150000
0
0
[ k´]1 = [ k´]2 =20000
937.5
20000
937.5
150000
150000
θi ∆j θj∆i δjδi
Mi
Vj
Mj
Vi
Nj
Ni
50
θi =22.02o
Member 1:
1
1 *
2 * 3 *
4
5 6
Global1´
2´3´
1 4´
5´ 6´Local
θj = 0o
λjx = cos (0o) = 1,λjy = sin (0o) = 0
λix = cos (22.02o) = 0.927,λiy = sin (22.02o) = 0.375
40 kN40 kN�m40 kN�m
20 kN20 kN
1
[FEM]
q1´
q3´
q2´
q4´
q5´
q6´
q*1
q*3
q*2
q4
q5q6
[ T ]1T
00
0
00
0
00
0
000
000
000
1 0
1
01
00
0 0
0.9270.375
- 0.375 0.927
10 0
00
[ q* ] = [ T ]T[ q´ ]
=
1´ 4´ 5´ 6´2´ 3´
1*
2*
3*
4*
5*
6*
51
θi =22.02o1
1 *
2 * 3 *
4
5 6
Global1´
2´3´
1 4´
5´ 6´Local
θj = 0o
[ k* ]1 = [ T ]1T[ k´ ]1[ T ]1
[ k* ]1 = 103
1*2*3*456
1*
129.04651.811-1.406
-139.0580.351-1.406
2*
51.811
21.892
3.476-56.240
-0.8693.476
3*
-1.4063.47620.000.00-3.7510.00
4
-139.058-56.240
01500
0
5
0.351-0.869-3.75
00.938-3.75
6
-1.4063.47610.00
0-3.75
20
52
θi =22.02o1
1 *
2 * 3 *
4
5 6
Globalθj = 0o
40 kN40 kN�m40 kN�m
20 kN20 kN
1
[FEM]
[ qF* ] = [ T ]T[ qF´]
[ qF*] = [ T ]1T
0
40
20
0
20
-40
40 kN40 kN�m
20 kN
1
40 kN�m
18.547.5
-7.5018.5440020-40
=
1*
2*
3*
4
5
6
53
Member 2
q1´
q3´
q2´
q4´
q5´
q6´
q4
q6
q5
q7
q8q9
[ T ]2T
0.927-0.375
0.3750.927
1
00
0
00
0
00
0
000
000
000
1
0.927-0.375
0.3750.927
00
0 0
0 0
00
[ q ] = [ T ]T[ q´ ]
=
1´ 4´ 5´ 6´2´ 3´
456
78
9
6 kN/m2
24 kN
24 kN
32 kN�m
32 kN�m
[ q´F]
2
1´
2´ 3´
4´
5´ 6´
[ q´]
λix = λjx = cos (-22.02o) = 0.927,λiy = λjy = sin (-22.02o) = -0.375
2
45 6
[ q ]7
8 922.02o
22.02o
54
2
1´
2´ 3´
4´
5´ 6´
[ q´ ]
[ k ]2 = [ T ]2T[ k´]2[ T ]2
2
45 6
[ q ]7
8 922.02o
22.02o
[ k ]2 = 103
456
78
9
4
129.046-51.811
1.406-129.046
51.8111.4056
5
-51.81121.8923.476
51.811-21.892
3.476
6
1.4063.476
20-1.406-3.476
10
7
-129.04651.811-1.406
129.046-51.811-1.406
8
51.811-21.892-3.476
-51.81121.892-3.476
9
1.4063.476
10-1.406-3.476
20
55
6 kN/m2
24 kN
24 kN
32 kN�m
32 kN�m
[ q´F ]
[ qF* ] = [ T ]T[ qF´ ]
[ qF ] = [ T ]2T
0
32
24
0
24
-32
8.99822.249328.99822.249-32
=
4
5
6
7
8
9
6 kN/m2
32 kN�m
32 kN�m
[ qF ]
22.249 kN
8.998kN
22.249 kN
8.998 kN
2
45 6
[ q ]7
8 922.02o
22.02o
56
[ k* ]1 = 103
1*2*3*456
1*129.04651.811-1.406
-139.0580.351-1.406
2*51.811
21.892
3.476-56.240-0.8693.476
3*-1.4063.47620.000.00-3.7510.00
4-139.058-56.240
01500
0
50.351-0.869-3.75
00.938-3.75
6-1.4063.47610.00
0-3.75
20
Global Stiffness:
12
1 *
2* 3 *
4
5 6
7
8 9
2*
7
8 9
[ k ]2 = 103
456
78
9
4
129.046-51.811
1.406-129.046
51.8111.4056
5
-51.81121.8923.476
51.811-21.892
3.476
6
1.4063.476
20-1.406-3.476
10
7
-129.04651.811-1.406
129.046-51.811-1.406
8
51.811-21.892-3.476
-51.81121.892-3.476
9
1.4063.476
10-1.406-3.476
20
57
Global: 40 kN 20 kN200 kN�m
6 kN/m 12
1 *
2 * 3 *
4
5 6
7
8 9
6 kN/m2
32 kN�m
32 kN�m
[ qF ]
22.249 kN
8.998 kN
22.249 kN
8.998 kN
40 kN40 kN�m
20 kN
1
40 kN�m
18.54 kN7.5
[ q*F ]
2 *
7
8 9
[ Q ] = [ K ][ D ] + [ QF ]
40 kN�m
7.5
40 kN�m 32 kN�m
20 kN 22.249 kN
8.998 kN
20 kN200 kN�m
D1*
D4
D3*
D5
D6
= 0.0
= 0.0
= 0.0
= -20
= -200
Q1*
Q4
Q3*
Q5
Q6
= 103
1*
3*
4
5
6
1*
129.046
-1.406
-139.058
0.352
-1.406
3*
-1.406
20
0
-3.75
10
4
-139.058
0
279.046
-51.811
1.406
5
0.351
-3.75
-51.811
22.829
-0.274
6-1.406
10
1.406
-0.274
40
+ 8.998
-7.5
40
20 +22.249
-40 + 32
58
Global: 40 kN 20 kN200 kN�m
6 kN/m 12
1 *
2 * 3 *
4
5 6
7
8 9
2 *
7
8 9
D1*
D4
D3*
D5
D6
=
-0.0205 m
-0.0112 rad
-0.0191 m
-0.0476 m
-0.0024 rad
59
1
1 *
2 * 3 *
4
5 6
[ q ]
40 kN 40 kN�m
20 kN
1
40 kN�m
18.547.5
[ qF* ]Member 1: [ q*] = [ k*][ d*] + [ qF*]
q1*
q3*
q2*
q4
q5q6
D1*=-0.0205
D3*=-0.0112D2*= 0.0
D4= -0.0191D5= -0.0476D6=-0.0024
-7.50
40
18.54
0
20
-40
+
q1*
q3*
q2*
q4
q5
q6
0
022.63 kN
-8.49 kN19.02 kN7.87 kN�m
=
40 kN
22.63 kN
8.49 kN
19.02 kN
7.87 kN�m
= 103
1*2*3*456
1*
129.04651.811-1.406
-139.0580.351-1.406
2*
51.811
21.892
3.476-56.240
-0.8693.476
3*
-1.4063.47620.000.00-3.7510.00
4
-139.058-56.240
01500
0
5
0.351-0.869-3.75
00.938-3.75
6
-1.4063.47610.00
0-3.75
20
60
Member 2 : [ q ] = [ k ][ d ] + [ qF ]
q4
q6
q5
q7
q8q9
8.998
3222.249
8.99822.249
-32
+
2
4
5 6
[ q ] 7
8 96 kN/m
2
32 kN�m
32 kN�m
[ qF ]
22.249 kN
8.998 kN
22.249 kN
8.998 kN
D4= -0.0191
D6=-0.0024D5= -0.0476
D7 = 0D8 = 0D9 = 0
q4
q6
q5
q7
q8
q9
8.49 kN
-207.87 kN�m-39.02 kN
9.51 kN83.51 kN-248.04 kN�m
= 8.49 kN
39.02 kN
207.87 kN�m
9.51 kN83.51 kN
248.04 kN�m
6 kN/m
=103
456
78
9
4
129.046-51.811
1.406-129.046
51.8111.4056
5
-51.81121.8923.476
51.811-21.892
3.476
6
1.4063.476
20-1.406-3.476
10
7
-129.04651.811-1.406
129.046-51.811-1.406
8
51.811-21.892-3.476
-51.81121.892-3.476
9
1.4063.476
10-1.406-3.476
20
61
40 kN
22.63 kN
8.49 kN
19.02 kN
7.87 kN�m
8.49 kN
39.02 kN
207.87 kN�m
9.51 kN
83.51 kN
248.04 kN�m6 kN/m
22.5 kN81 kN
22.5 kN
33 kN
8.49 kN
20.98 kN
Bending-moment diagram (kN�m)
-
+
207.87
-248.04
83.56
+7.87
Deflected shape
D3*=-0.0112 rad
D1*=- 20.5 mm
D6=-0.0024 rad
D1*
D4
D3*
D5
D6
=
-0.0205 m
-0.0112 rad
-0.0191 m
-0.0476 m
-0.0024 radD4=-19.1 mmD5=-47.6 mm
62
3 m
80 kN�m20 kN/m
20o
50 kN
4 m 2 m
A B
C
Example 5
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagramsand qualitative deflected shape.Take I = 400(106) mm4 , A = 60(103) mm2, and E = 200 GPa for all members.
63
Global
1
2 3
4
5 6
7*9*
8*
1
2 2
1´2´
3´
4´5´
6´
3 m
80 kN�m20 kN/m
20o
50 kN
4 m 2 m
A B
C
Local
1´
2´ 3´
4´
5´ 6´1
76.31o
64
3 m
80 kN�m20 kN/m
20o
50 kN
4 m 2 m
A B
C
1
2 3
4
5 61
Member 1:
kN/m103000m4
)kN/m10)(200m10(60
3
2623
×=
××=
−
LAE
mkN1080m4
)m10)(400kN/m104(2004
3
4626
•×=
××=
−
LEI
mkN1040m4
)m10)(400kN/m102(2002
3
4626
•×=
××=
−
LEI
kN1030m)(4
)m10)(400kN/m106(2006
3
2
4626
2
×=
××=
−
LEI
kN/m1015m)(4
)m10)(400kN/m1012(20012
3
3
4626
3
×=
××=
−
LEI
65
d1
d3
d2
d4
d5
d6
q1
q3
q2
q4
q5q6
0
26.67
40
040
-26.67
+= 103
123
45
6
1 4 5 6
030
40
0-30
80
3000
0
0
-30000
0
2
015
30
0-15
30
3
030
80
0-30
40
-30000
0
3000
0
0
0-15
-30
015
-30
3 m
80 kN�m20 kN/m
20o
50 kN
4 m 2 m
A B
C
26.67 kN�m26.67 kN�m
40 kN40 kN
1
20 kN/m
Member 1: [ q ] = [ k ][ d ] + [ qF ]
1
2 3
4
5 61
56.31o
76.31o
66
Member 2:
2
1´2´
3´
4´5´
6´
45 6
7*9*
8*
2
i
j
kN/m103324m3.61
)kN/m10)(200m10(60
3
2623
×=
××=
−
LAE
mkN1088.64m3.61
)m10)(400kN/m104(2004
3
4626
•×=
××=
−
LEI
kN1036.83m)(3.61
)m10)(400kN/m106(2006
3
2
4626
2
×=
××=
−
LEI
kN/m1020.41m)(3.61
)m10)(400kN/m1012(20012
3
3
4626
3
×=
××=
−
LEI
mkN1044.32m3.61
)m10)(400kN/m102(2002
3
4626
•×=
××=
−
LEI
[ k´ ]2 = 103
1´2´3´4´5´6´
1´ 4´ 5´ 6´3324
00
-332400
2´0
20.4136.83
0-20.4136.83
3´0
36.8388.64
0-36.8344.32
-332400
3324
00
0-20.41-36.83
020.41
-36.83
036.8344.32
0-36.8388.64
67
45 6
7*9*
8*
2
i
j 76.31oλjx = cos (-76.31o) = 0.24,λjy = sin (-76.31o) = -0.97
[ q* ] = [ T ]T[ q´ ]
q1´
q3´
q2´
q4´q5´q6´
q4
q6
q5
q7*q8*q9*
=
4567*8*9*
1´ 4´000
0.24
0-0.97
5´000
0.970.24
0
6´000001
0.55
0-0.83
000
2´0.830.55
0000
3´001000
λix = cos (-56.31o) = 0.55,λiy = sin (-56.31o) = -0.83
56.31o
2
i
j
1´2´
3´
4´5´
6´
[ k* ]2 = [ T ]T[ k´]2[ T ] = 103
4567*8*9*
4 7* 8* 9*5 6-452.884643.585-35.786205.452
-35.786-759.668
1787.474-2689.923
-8.717-759.6683139.053
-8.717
30.64620.43144.321
-35.786-8.71788.643
1036.923
30.646-1524.780
-452.8841787.474
30.646
-1524.7802307.582
20.431643.585
-2689.92320.431
30.64620.43188.643
-35.786-8.71744.321
68
[k]1 = 103
123456
1 4 5 60
30400
-3080
3000
00
-300000
20
1530
0-1530
30
30800
-3040
-300000
3000
00
0-15-300
15-30
1
2 3
4
5 6
7*9*
8*
1
2
[k*]2 = 103
4567*8*9*
4 7* 8* 9*5 6-452.884643.585-35.786205.452
-35.786-759.668
1787.474-2689.923
-8.717-759.6683139.053
-8.717
30.64620.43144.321
-35.786-8.71788.643
1036.923
30.646-1524.780
-452.8841787.474
30.646
-1524.7802307.582
20.431643.585
-2689.92320.431
30.64620.43188.643
-35.786-8.71744.321
69
Global:
= -50= 0= -80= 0= 0
Q4
Q6
Q5
Q7*
Q9*
D4
D6
D5
D7*
D9*
0
-26.6740
00
+
D4
D6
D5
D7*
D9*
-2.199x10-5 m
-2.840x10-4 rad-3.095x10-4 m
0.979x10-3 m6.161x10-4 rad
=
1
2 3
4
5 6
7*9*
8*
1
23 m
80 kN�m20 kN/m
20o
50 kN
4 m 2 m
A B
C 26.67 kN�m
26.67 kN�m
40 kN40 kN
1
20 kN/m
80 kN�m50 kN
= 103
45679
4 7* 9*5 6
205.452
-452.884643.585-35.786
-35.786
30.646-9.56944.321
-35.78688.643
-452.884
4036.923
30.646-1524.780
30.646
-1524.780232.582
-9.569643.58520.431
30.646-9.569
168.643-35.78644.321
70
q1
q3
q2
q4
q5q6
0
26.67
40
040
-26.67
+= 103
123
45
6
1 4 5 6
030
40
0-30
80
3000
0
0
-30000
0
2
015
30
0-15
30
3
030
80
0-30
40
-30000
0
3000
0
0
0-15
-30
015
-30
Member 1:
q1
q3
q2
q4
q5q6
65.97 kN
24.59 kN�m
36.12 kN
-65.97 kN
43.88 kN-40.11 kN�m
=
0
0
0
D4
D5
D6
40.11 kN�m24.59 kN�m
43.88 kN36.12 kN
1
20 kN/m
65.97 kN65.97 kN
1
2 3
4
5 61
71
q4
q6
q5
q7*
q8*q9*
Member 2:
q4
q6
q5
q7*
q8*q9*
15.97 kN
-39.89 kN�m
-43.88 kN
0 kN
46.69 kN0 kN�m
=
= 103
456
7*8*
9*
4 7* 8* 9*5 6-455.21651.27
-35.73
210.67
-35.73
-769.1
1769.34-2678.93
-8.84
-769.13128.82
-8.84
30.5720.26
44.32
-35.73-8.84
88.64
1036.923
30.57-1508.14
-455.211769.34
30.57
-1508.142296.15
20.26
651.27-2678.93
20.26
30.5720.26
88.64
-35.73-8.84
44.32
D4
D6
D5
D*7
0
D*9
15.97 kN43.88 kN 39.89 kN�m
46.69 kN
2
45 6
7*9*
8*
2
i
j
72
Bending-moment diagram (kN�m)
+
40.11 kN�m24.59 kN�m
43.88 kN36.12 kN
1
20 kN/m
65.97 kN65.97 kN
15.97 kN43.88 kN 39.89 kN�m
46.69 kN
2
-24.59-
-40.11
- +
39.89
D4
D6
D5
D7*D9*
-2.199x10-5 m
-2.840x10-4 rad-3.095x10-4 m
0.979x10-3 m6.161x10-4 rad
=
Deflected shape
D4=-2.2x10-5 m
D5=-3.1x10-4 m
D7*=0.979x10-3 mD6 = -2.84x10-4 rad
D9*=6.161x10-4 rad
1
2 3
4
5 6
7*9*
8*
1
2
1
! Trusses! Vertical Loads on Building Frames! Lateral Loads on Building Frames: Portal
Method! Lateral Loads on Building Frames: Cantilever
Method Problems
APPROXIMATE ANALYSIS OF STATICALLYINDETERMINATE STRUCTURES
2
P1 P2
(a)
Trusses
R1 R2
a
a(b)R1
a
a
V = R1
F1
F2
Fb
Fa
Method 1 : If the diagonals are intentionally designed to be long and slender,it is reasonable to assume that the panel shear is resisted entirely by the tension diagonal,whereas the compressive diagonal is assumed to be a zero-force member.
Method 2 : If the diagonal members are intended to be constructed from largerolled sections such as angles or channels, we will assume that the tension and compressiondiagonals each carry half the panel shear.
3
Example 1
Determine (approximately) the forces in the members of the truss shown inFigure. The diagonals are to be designed to support both tensile and compressiveforces, and therefore each is assumed to carry half the panel shear. The supportreactions have been computed.
10 kN 20 kN4 m 4 m
3 m
A BC
DEF
4
+ ΣMA = 0: FFE(3) - 8.33cos36.87o(3) = 0
FFE = 6.67 kN (C)
ΣFy = 0:+ 20 - 10 - 2Fsin(36.87o) = 0
F = 8.33 kN
FFB = 8.33 kN (T)
FAE = 8.33 kN (C)
+ ΣMF = 0: FAB(3) - 8.33cos36.87o(3) = 0
FAB = 6.67 kN (T)
ΣFy = 0:+ FAF - 10 - 8.33sin(36.87o) = 0
FAF = 15 kN (T)
10 kN 20 kN4 m 4 m
3 m
A BC
DEF
10 kN
20 kN
V = 10 kN
FFE
FAB
FAE= FFFB= F
θ = 36.87o
10 kN
3 m
A
F20 kN
θ
θ6.67 kN
8.33FAF
10 kN
θA
5
C
D
10 kN
3 mθ
θ
ΣFy = 0:+ 10 - 2Fsin(36.87o) = 0
F = 8.33 kN
FDB = 8.33 kN (T)
FEC = 8.33 kN (C)
+ ΣMC = 0: FED(3) - 8.33cos36.87o(3) = 0
FED = 6.67 kN (C)
+ ΣMD = 0: FBC(3) - 8.33cos36.87o(3) = 0
FBC = 6.67 kN (T)
θ = 36.87o
V = 10 kN
θ = 36.87oD
θ6.67 kN
8.33 kNFDC
ΣFy = 0:+ FDC - 8.33sin(36.87o) = 0
FDC = 5 kN (C)
ΣFy = 0:+
FEB = 2(8.33sin36.87o) = 10 kN (T)
Eθ θ
6.67 kN6.67 kN
8.33 kN 8.33 kN
θ = 36.87o
FEB
FED
FBC
F = FEC
F = FDB
6
Example 2
Determine (approximately) the forces in the members of the truss shown inFigure. The diagonals are slender and therefore will not support a compressiveforce. The support reactions have been computed.
40 kN 40 kN
4 m
A B C
GHJ
D E
FI
10 kN 20 kN 20 kN 20 kN 10 kN
4 m 4 m 4 m 4 m
7
FBH = 0
FIH
FBC
FIC
40 kN
4 m
A
J
10 kN
45o
V = 10 kN
FBH = 0
ΣFy = 0:+ 40 - 10 - 20 - FICcos 45o = 0
FIC = 14.14 kN (T)
+ ΣMA = 0: FJI(4) - 42.43sin 45o(4) = 0
FJI = 30 kN (C)
+ ΣMJ = 0: FAB(4) = 0
FAB = 0
0
0FJA
40 kN
θA
ΣFy = 0:+ FJA = 40 kN (C)
FAI = 0
FJI
FAB
FJB V = 30 kN
ΣFy = 0:+ 40 - 10 - FJBcos 45o = 0
FJB = 42.43 kN (T)
FAI = 0
40 kN
4 m
A B
J I
10 kN 20 kN
4 m
45o
8
+ ΣMB = 0:
FIH(4) - 14.14sin 45o(4) + 10(4) - 40(4) = 0
FJH = 40 kN (C)
FBC = 30 kN (T)
+ ΣMI = 0: FBC(4) - 40(4) + 10(4) = 0
FBH = 0
FIH
FBC
FIC V = 10 kN
40 kN
4 m
A B
J I
10 kN 20 kN
4 m
45oB
30 kN0
042.3 kNFBI
45o
45o45o
ΣFy = 0:+ FBI = 42.3 sin 45o = 30 kN (T)
ΣFy = 0:+
FBI = 2(14.1442.3 sin 45o) = 20 kN (C)
C30 kN30 kN
14.14 kN14.14 kNFCH
45o45o
9
Vertical Loads on Building Frames
typical building frame
10
w
A B
L(b)
0.21L0.21Lpoint of zero moment
approximate case
w
L
(d)
assumed points of zero moment0.1L 0.1L
model
w
(e)
� Assumptions for Approximate Analysis
0.1L 0.1L0.8L
w
A B
L
Simply supported(c)
column column
girder
w
A BL
(a)
Point ofzeromoment
Point ofzeromoment
11
Example 3
Determine (approximately) the moment at the joints E and C caused by membersEF and CD of the building bent in the figure.
B
D
F
A
C
E
6 m
1 kN/m
1 kN/m
12
1 kN/m
1 kN/m
4.8 m
4.8 kN
0.6 m
0.6 kN 2.4 kN
3 kN
0.6(0.3) + 2.4(0.6) = 1.62 kN�m
0.6 m
0.6 kN2.4 kN
3 kN
1.62 kN�m
0.1L=0.6 m 0.6 m
4.8 m
2.4 kN 2.4 kN
13
P
h
l(a)
P
h
l(b)
Portal Frames and Trusses
� Frames: Pin-Supported∆ ∆
assumedhinge
h
L/2
(c)
P
h
L/2
Ph/l
P/2
Ph/l
P/2
P/2 P/2
Ph/l
Ph/2
Ph/2Ph/2
Ph/2
(d)
Ph/l
14
h/2
� Frames : Fixed-SupportedP
h
l(a)
P/2
Ph/2lP/2
Ph/2l
P
h
l(b)
∆ ∆
assumedhinges
(c)
h/2
L/2P
h/2
L/2
P/2 P/2
Ph/2l
P/2
Ph/2l
P/2
Ph/2l
Ph/2l
Ph/2l
P/2 P/2
Ph/2lPh/4 Ph/4
Ph/4
Ph/4Ph/4
Ph/4
(d)
Ph/4Ph/4
15
� Frames : Partial Fixity
P
(b)
P
h
l(a)
h/3h/3
assumedhinges
θ θ
� Trusses
P
h
l(a)
P
l(b)
P/2 P/2h/2
assumedhinges
∆ ∆
16
Example 4
Determine by approximate methods the forces acting in the members of theWarren portal shown in the figure.
40 kN
7 m
8 m
2 m
4 m
A
B
C
4 m 2 m2 m
D E F
H G
I
17
40/2 = 20 kN = V
N N
20 kN = V
I3.5 m
A
V = 20 kN
NN
V = 20 kN
V = 20 kN
N N
20 kN = VM M
40 kN
3.5 m
2 m
B
C
4 m 2 m2 m
D E F
H G
J K
+ ΣMJ = 0:
N(8) - 40(5.5) = 0
N = 27.5 kN
+ ΣMA = 0:
M - 20(3.5) = 0
M = 70 kN�m
18
20 kN
27.5 kN
40 kN
3.5 m
2 m
B
C
2 m
D
J
FCD
FBDFBH
45o
+ ΣMB = 0: FCD(2) - 40(2) - 20(3.5) = 0
FCD = 75 kN (C)
+ ΣMD = 0: FBH(2) + 27.5(2) - 20(5.5) = 0
FBH = 27.5 kN (T)
ΣFy = 0:+ -27.5 + FBDcos 45o = 0
FBD = 38.9 kN (T)
+ ΣMG = 0: FEF(2) - 20(3.5) = 0
FEF = 35 kN (T)
+ ΣME = 0: FGH(2) + 27.5(2) - 20(5.5) = 0
FGH = 27.5 kN (C)
ΣFy = 0:+ 27.5 - FEGcos 45o = 0
FEG = 38.9 kN (C)
3.5 m
2 m
27.5 kN
20 kN = V
2 m
E F
G
K
FGH
FEG
FEF
45o
19
ΣFy = 0:+ FDHsin 45o - 38.9sin 45o = 0
FDH = 38.9 kN (C)
D75 kN
38.9 kN
FDE
FDH
x
y
45o45o
75 - 2(38.9 cos 45o) - FDE = 0
FDE = 20 kN (C)
ΣFx = 0:+
H 27.5 kN27.5 kN
38.9 kN FHE
x
y
45o45o
ΣFy = 0:+ FHEsin 45o - 38.9sin 45o = 0
FHE = 38.9 kN (T)
20
= inflection point
P
(a)
Lateral Loads on Building Frames: Portal Method
(b)V V V V
21
Example 5
Determine by approximate methods the forces acting in the members of theWarren portal shown in the figure.
4 m 4 m 4 m
5 kN
3 m
A
B
C
D
E
F G
H
22
5 kN
1.5 m
B D F G
I J K L
M N O
4 m 4 m 4 m
5 kN
3 m
A
B
C
D
E
F G
H
I J K L
M N O
V V2V 2V
Iy Jy Ky Ly
ΣFx = 0: 5 - 6V = 0
V = 0.833 kN
+
23
Ly 0.625 kN =
0.625kN
Ky = 0
4.167 kN
0.625 kN
Iy = 0.625kN
2.501kN
0.625kN
Jy = 0
1.5 mA
I
0.625 kN
0.833kN
0.625 kN
0.833 kN
H
L1.5 m
0.833 kN 0.625 kN 1.25 kN�m
C
J1.5 m
1.666 kN
1.666kN 2.50 kN�m
0.835 kN
E
K1.5 m
1.666 kN
1.666kN 2.5 kN�m
0.833 kN0.625 kN 1.25 kN�m
5 kN
1.5 m
B 2 m
0.833 kN
M
I
2.501 kN
0.625kN
F
1.666 kN1.5 m
2 m2 mN O
K
G
0.833 kN
1.5 m
2 mO
L
0.625 kN 0.835 kN
0.625 kN
D
1.666 kN
2 m 2 m
1.5 m
N
J
4.167 kN
24
Example 6
Determine (approximately) the reactions at the base of the columns of the frameshown in Fig. 7-14a. Use the portal method of analysis.
20 kN
30 kN
8 m 8 m
5 m
6 m
G H I
A B C
D E F
25
20 kN
30 kN
8 m 8 m
5 m
6 m
G H I
A B C
D E F
S
N
J K L
O P Q
R
M
26
Py
VOy
VPy
2V
ΣFx = 0: 20 - 4V = 0 V = 5 kN+
Ly
V´Jy
V´ Ky
2V´
ΣFx = 0: 20 + 30 - 4V´ = 0 V´ = 12.5 kN+
20 kN2.5 m
G I
20 kN
30 kN
5 m
3 m
G H I
D E F
27
Oy = 3.125
20 kN2.5 m
G R4 m
5 kN
Rx = 15 kN
Ry = 3.125 kN
Sx = 5 kN
Sy = 3.125 kN
Py = 0 kN
Mx = 22.5 kN
My = 12.5 kN
Jy = 15.625 kN
Nx = 7.5 kN
Ny = 12.5 kN
Ky = 0 kN
B
K3 m
25 kN
A
J3 m
15.625kN12.5 kN
Bx = 25 kNBx = 0 MB = 75 kN�m
Ax = 12.5 kNAx = 15.625 kN MA = 37.5 kN�m
25 kN
2.5 m
N
K
P
M
3 m4 m4 m
10 kN
22.5 kN
12.5 kN
10 kN
H SR15 kN
3.125 kN
2.5 m
4 m4 m
5 kN
3.125 kN
12.5 kN
30 kN
J
O
M2.5 m
3 m 4 m
28
Lateral Loads on Building Frames: Cantilever MethodP
beam(a)
building frame(b)
In summary, using the cantilever method, the following assumptions apply to a fixed-supported frame.
1. A hinge is place at the center of each girder, since this is assumed to be point of zero moment.
2. A hinge is placed at the center of each column, since this is assumed to be a point of zero moment.
3. The axial stress in a column is proportional to its distance from the centroid of the cross-sectional areas of the columns at a given floor level. Since stress equals force per area, then in the special case of the columns having equal cross-sectional areas, the force in a column is also proportional to its distance from the centroid of the column areas.
29
Example 7
Determine (approximately) the reactions at the base of the columns of the frameshown. The columns are assumed to have equal crossectional areas. Use thecantilever method of analysis.
30 kN
15 kN
6 m
CD
BE
A F
4 m
4 m
30
30 kN
15 kN
6 m
CD
BE
A F
4 m
4 m
x
6 m
3)(6)(0~=
++
==∑∑
AAAA
AAx
x
G
H
I
J
K
L
31
I
3 m 3 m
Hy
Hx
Ky
Kx
30 kNC
D2 m
M
+ ΣMM = 0: -30(2) + 3Hy + 3Ky = 0
The unknowns can be related by proportional triangles, that is
yyyy KHor
KH==
33
kNKH yy 10==
Gy
Gx
Ly
LxN
30 kN
15 kN
CD
BE
4 m H
I
J
K
3 m 3 m2 m
+ ΣMN = 0: -30(6) - 15(2) + 3Gy + 3Ly = 0
The unknowns can be related by proportional triangles, that is
yyyy LGor
GG==
33
kNLG yy 35==
32
10 kN
30 kN2 m
C 3 m Ix = 15 kN
Iy = 10 kN
Hx = 15 kN
10 kN
DI15 kN
10 kN
2 m
3 m
Kx = 15 kN
35 kN
15 kN
10 kN
15 kN
G
H
J2 m
2 m 3 m Jx = 7.5 kN
Jy = 25 kN
Gx = 22.5 kN35 kN
2 m
L
K
J
2 m
3 m
15 kN
7.5 kN
25 kN
10 kN
Lx = 22.5 kN
A
G2 m
35 kN22.5 kN
Ax = 22.5 kNAx = 35 kN MA = 45 kN�m
Fx = 22.5 kNFy = 35 kN MF = 45 kN�m
F
L2 m
22.5 kN35 kN
33
Example 8
Show how to determine (approximately) the reactions at the base of the columnsof the frame shown. The columns have the crossectional areas show. Use thecantilever of analysis.
35 kN
45 kN
6 m
6 m
4 m
P Q R
A B C D
L M N O
E F G H
I J K
4 m 8 m
5000mm2
5000mm2
4000 mm2
4000 mm2
6000 mm2
6000 mm2
6000 mm2
6000 mm2
34
35 kN
45 kN
6 m
4 m
P Q R
A B C D
L M N O
E F G H
I J K
6 m 4 m 8 m
6 m 4 m 8 m
6000 mm2 6000 mm24000 mm25000mm2
x
mAAx
x 48.86000400050006000
)18(6000)10(4000)6(5000)0(6000~=
++++++
==∑∑
35
35 kN2 m
P Q R
Ly
Lx My
Mx
Ny
Nx
Oy
Ox
1.52 m2.48 m8.48 m 9.52 m
+ ΣMNA = 0: -35(2) + Ly(8.48) + My(2.48) + Ny(1.52) + Oy(9.52) = 0 -----(1)
)2())10(6000
(48.848.2
)10(5000;
48.848.2;
48.848.2 66 −−−−−=== −−yy
LMLM LM
σσσσ
)3())10(6000
(48.852.1
)10(4000;
48.852.1;
48.852.1 66 −−−−−=== −−yy
LNLN LN
σσσσ
)4())10(6000
(48.852.9
)10(6000;
48.852.9;
48.852.9 66 −−−−−=== −−yy
LOLO LO
σσσσ
Since any column stress σ is proportional to its distance from the neutral axis
Solving Eqs. (1) - (4) yields Ly = 3.508 kN My = 0.855 kN Ny = 0.419 kN Oy = 3.938 kN
36
35 kN
45 kN
3 m
L M N O
I J K4 m
Ey
Ex Fy
Fx
Gy
Gx
Hy
Hx
1.52 m2.48 m8.48 m 9.52 m
+ ΣMNA = 0: -45(3) - 35(7) + Ey(8.48) + Fy(2.48) + Gy(1.52) + Hy(9.52) = 0 -----(5)
)6())10(6000
(48.848.2
)10(5000;
48.848.2;
48.848.2 66 −−−−−=== −−yy
EFEF EF
σσσσ
)7())10(6000
(48.852.1
)10(4000;
48.852.1;
48.852.1 66 −−−−−=== −−yy
EGEG EG
σσσσ
)8())10(6000
(48.852.9
)10(6000;
48.852.9;
48.852.9 66 −−−−−=== −−yy
EHEH EH
σσσσ
Since any column stress σ is proportional to its distance from the neutral axis ;
Solving Eqs. (1) - (4) yields Ey = 19.044 kN Fy = 4.641 kN Gy = 2.276 kN Hy = 21.38 kN
37
One can continue to analyze the other segments in sequence, i.e., PQM, then MJFI, then FB, and so on.
35 kN2 m
3.508 kN
3 m
Ix= 114.702 kN
Iy= 15.536 kN
Lx= 5.262 kN
Ex= 64.44 kN
Px= 29.738 kN
Py= 3.508 kN
45 kN
3 m
I2 m
19.044 kN
5.262 kN3.508 kN
3 m
3 m
E19.044 kN
64.44 kN
Ax = 64.44 kNAx = 19.044 kN MA = 193.32 kN�m
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