06 Matrix Beam

BEAM ANALYSIS USING THE STIFFNESS
METHOD
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Development: The Slope-Deflection

Equations
Stiffness Matrix
General Procedures
Internal Hinges
Temperature Effects
Force & Displacement Transformation
Skew Roller Support
Slope Deflection Equations

P
Cj
settlement = j
Mij
Mji
Degrees of Freedom
M
1 DOF:
2 DOF: ,
P
B
Stiffness Definition
kAA
kBA
A
L
k AA =
4 EI
L
k BA =
2 EI
L
kBB
kAB
1
L
k BB =
4 EI
L
k AB =
2 EI
L
Fixed-End Forces
Fixed-End Forces: Loads
P
L/2
PL
8
L/2
PL
8
L
P
2
P
2
w
wL2
12
wL
2
wL2
12
wL
2
General Case
Cj
settlement = j
Mij
Mji
Mij
Mji
i
L
4 EI
2 EI
i +
j = M
ij
L
L
Mji =
2 EI
4 EI
i +
j
L
L
i
(MFij)
+
(MFji)
settlement = j
+
P
(MFij)Load
M ij = (
settlement = j
w
(MFji)Load
4 EI
2 EI
2 EI
4 EI
) i + (
) j + ( M F ij ) + ( M F ij ) Load , M ji = (
) i + (
) j + ( M F ji ) + ( M F ji ) Load 8
L
L
L
L
Equilibrium Equations
i
Cj
Cj M
Mji
Mji
jk
Mjk
j
+ M j = 0 : M ji M jk + C j = 0
Stiffness Coefficients
Mij
Mji
kii =
4 EI
L
k ji =
2 EI
L
k jj =
4 EI
L
kij =
2 EI
L
+
1
10
Matrix Formulation
M ij = (
4 EI
2 EI
) i + (
) j + ( M F ij )
L
L
M ji = (
2 EI
4 EI
) i + (
) j + ( M F ji )
L
L
M ij (4 EI / L) ( 2 EI / L) iI M ij F
M =
+ M F
EI
L
EI
L
(
2
/
)
(
4
/
)
j ji
ji
kii
k ji
[k ] =
kij
k jj
Stiffness Matrix
11
Mij
Mji
[ M ] = [ K ][ ] + [ FEM ]
([ M ] [ FEM ]) = [ K ][ ]
[ ] = [ K ]1[ M ] [ FEM ]
Mij
Mji
Fixed-end moment
Stiffness matrix matrix
+
(MFij)
(MFji)
[D] = [K]-1([Q] - [FEM])

+
(MFij)Load
(MFji)Load
Displacement
matrix
Force matrix
12
Stiffness Coefficients Derivation

Mj
Mi
Real beam
i
Mi + M j
L
Mi + M j
L/3
M jL
2 EI
Mj
EI
Conjugate beam
Mi
EI
MiL
2 EI
M j L 2L
MiL L
)( ) + (
)( ) = 0
2 EI 3
2 EI 3
M i = 2 M j (1)
+ M 'i = 0 : (
+ Fy = 0 : i (
M L
MiL
) + ( j ) = 0 (2)
2 EI
2 EI
From (1) and (2);

4 EI
) i
L
2 EI
) i
Mj =(
L
Mi = (
13
Derivation of Fixed-End Moment

Point load
P Real beam
Conjugate beam
A
M
EI
M
M
EI
M
EI
ML
2 EI
PL2
16 EI
PL
4 EI
ML
2 EI 2
PL
16 EI
M
EI
PL
ML ML 2 PL2
+ Fy = 0 :
+
= 0, M =
2 EI 2 EI 16 EI
8 14
P
PL
8
PL
8
L
P
2
P
2
P/2
PL/8
P/2
-PL/8
-PL/8
-
-PL/8
-PL/16
-
-PL/16
PL/4
+
-PL/8
PL PL PL PL
+
=
+
16
16
4
8
15
Uniform load
w
Real beam
Conjugate beam
A
B
M
EI
M
EI
M
EI
ML
2 EI
ML
2 EI
wL3
24 EI
wL2
8 EI
M
EI
wL3
24 EI
wL2
ML ML 2 wL3
+ Fy = 0 :
+
= 0, M =
2 EI 2 EI 24 EI
12 16
Settlements
Mi = Mj
Real beam
Mj
Conjugate beam
L
Mi + M j
M
EI
M
EI
Mi + M j
L
M
EI
ML
2 EI
ML
2 EI
M
EI
+ M B = 0 : (
ML L
ML 2 L
)( ) + (
)( ) = 0,
2 EI 3
2 EI 3
M=
6 EI
L2
17
Typical Problem
CB
w
P1
P2
C
A
B
L1
PL
8
PL
8
L2
wL2
12
L
0
PL
4 EI
2 EI
M AB =
A +
B + 0 + 1 1
8
L1
L1
0 EI
PL
2 EI
4
M BA =
A +
B + 0 1 1
8
L1
L1
0
2
P2 L2 wL2
4 EI
2 EI
M BC =
B +
C + 0 +
+
L2
L2
8
12
0
2
P2 L2 wL2
2 EI
4 EI
B +
C + 0 +
M CB =
8
12
L2
L2
wL2
12
18
CB
w
P1
P2
C
A
B
L1
L2
CB M
BC
MBA
M BA =
M BC
PL
2 EI
4 EI
A +
B + 0 1 1
8
L1
L1
P L wL
4 EI
2 EI
=
B +
C + 0 + 2 2 + 2
L2
L2
8
12
+ M B = 0 : C B M BA M BC = 0 Solve for B
19
P1
MAB
CB
w
MBA
A
L1
P2
MBC
L2
C M
CB
Substitute B in MAB, MBA, MBC, MCB

0
PL
4 EI
2 EI
A +
B + 0 + 1 1
M AB =
L1
L1
8
0 EI
PL
2 EI
4
M BA =
A +
B + 0 1 1
8
L1
L1
0
2
4 EI
2 EI
P2 L2 wL2
B +
C + 0 +
+
M BC =
8
12
L2
L2
0
2
P2 L2 wL2
2 EI
4 EI
B +
C + 0 +
M CB =
L2
L2
8
12
20
P1
MAB
A
Ay
CB
w
MBA
B
L1
P2
MCB
MBC
L2
Cy
By = ByL + ByR
P1
MAB
A
Ay
L1
MBA
ByL
P2
MBC
ByR
L2
C
MCB
Cy
21
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
5
14
2EI
1
2
21
EI
2
8
3
22
Beam-Member Stiffness Matrix

3
E, I, A, L
k14
AE/L
k41
AE/L
k11 = AE/L
AE/L = k44
d1 = 1
d4 = 1
1
[k] =
AE/L
- AE/L
-AE/L
AE/L
23
6EI/L2 = k32
[k] =
E, I, A, L
k62 = 6EI/L2
d2 = 1
k22 =
6
6EI/L2 = k65
5
6EI/L2 = k35
d5 = 1
12EI/L3 = k52
12EI/L3
1
AE/L
12EI/L3 = k25
4
- AE/L
12EI/L3
- 12EI/L3
6EI/L2
- 6EI/L2
-AE/L
AE/L
-12EI/L3
12EI/L3
6EI/L2
- 6EI/L2
12EI/L3 = k55
24
2
k33 = 4EI/L
d3 = 1
E, I, A, L
2EI/L = k63
5
4EI/L = k66
2EI/L = k36
d6 = 1
k23 = 6EI/L2
[k] =
6EI/L2 = k53
k26 = 6EI/L2
6EI/L2 = k56
AE/L
- AE/L
12EI/L3
6EI/L2
- 12EI/L3
6EI/L2
6EI/L2
4EI/L
- 6EI/L2
2EI/L
-AE/L
AE/L
-12EI/L3
-6EI/L2
12EI/L3
-6EI/L2
6EI/L2
2EI/L
- 6EI/L2
4EI/L
25
Member Equilibrium Equations

Mi
Fxi
E, I, A, L
AE/L
AE/L x
i
Mj
Fyj
AE/L
AE/L
+
6EI/L2
6EI/L2
6EI/L2
1
x i
12EI/L3
12EI/L3
12EI/L3
4EI/L
x j
6EI/L2
Fyi
Fxj
2EI/L
x j
12EI/L3
4EI/L
2EI/L
x i
x j
1
6EI/L2
MFi
FFxi
FFyi
6EI/L2
6EI/L2
FFyj
FFxj
6EI/L2
MFj
26
Fxi = ( AE / L) i +
Fyi =
(0) i +
(0) i
Mxi =
Fxj = (AE / L) i
Fyj =
(0) i
Mj =
(0) i
Fxi

Fyj
Mi
=
Fxj
Fyj

M j
(0) i
(0) i +
(AE / L) j +
(0) j +
(0) j +
FxiF
(12EI / L3 ) i
(6EI / L2 ) i
(0) j
(12EI / L3 ) j
(6EI / L2 ) j
FyiF
(6EI / L2 ) i
(0) i
(4EI / L) i
(0) i
(0) j
( AE / L) j
(6EI / L2 ) j
(0) j
(2EI / L) j
(0) j
MiF
FxiF
(0) j
(0) j
(6EI / L2 ) j
FyjF
(0) j
(6EI / L2 ) j
(4EI / L) j
MjF
(12EI / L3 ) i (6EI / L2 ) i
(6EI / L2 ) i
(2EI / L) i
0
AE/ L
3
0
12
/
EI
L
0
6EI/ L2
0
AE/ L
0
12 EI/ L3
6EI/ L2
0
0
6EI/ L2
4 EI/ L
0
6EI/ L2
2EI/ L
AE/ L
0
12 EI/ L3
0
6EI/ L2
0
0
AE/ L
0
12 EI/ L3
6EI/ L2
0
Stiffness matrix
F
0
i Fxi

6EI/ L2 i FyiF
2EI/ L i MiF
+ F
0
j Fxj
6EI/ L2 j FyiF
F
4 EI/ L j M j
Fixed-end force matrix
[q] = [k][d] + [qF]

End-force matrix
Displacement matrix
27
6x6 Stiffness Matrix
[k ]66 =
Ni AE/ L
0
3
Vi 0
EI
L
12
/
Mi 0
6EI/ L2
Nj AE/ L
0
Vj 0
12 EI/ L3
6EI/ L2
Mj 0
AE/ L
0
12 EI/ L3
0
0
6EI/ L2
4 EI/ L
6EI/ L2
AE/ L
6EI/ L2
12 EI/ L3
2EI/ L
6EI/ L2
j
0
6EI/ L2
2EI/ L
6EI/ L2
4EI/ L
[k ]44
Vi 12 EI/ L3
Mi 6EI/ L2
Vj 12 EI/ L3
2
Mj 6EI/ L
i
6EI/ L2
4 EI/ L
6EI/ L2
2EI/ L
j
12 EI/ L3
6EI/ L2
12 EI/ L3
6EI/ L2
j
6EI/ L2
2EI/ L
6EI/ L2
4 EI/ L
28

i
[k ]22 =
Mi
Mj
4 EI / L 2EI / L
2EI / L 4 EI / L
Comment:
- When use 4x4 stiffness matrix, specify settlement.
- When use 2x2 stiffness matrix, fixed-end forces must be included.
29
General Procedures: Application of the Stiffness Method for Beam Analysis

P2y
P
M2
M1
2
Global
4
1
6
3
2
Local
5
2
4
2
30
P2y
P
M2
M1
2
Global
4
1
6
3
2
Member
5
4
6
2
Local
3
31
P2y
P
M2
M1
Global
Local
5
2
4
2
P
(qF
2)1
[FEF]
(qF1 )1
(qF1)2
(qF4 )2
(qF4)1
(qF3)1
(qF2)2
(qF3 )2
32
2
Global
Local
[q] = [T]T[q]
q1
q
2 =
q3

q4 1
1
2
3
4
1
1
0
2
0
1
0
0
3
0
0
1
0
4
0
0
0
q1'
q
2'
q3'

q4 '
[k] = [T]T[q] [T]

33
2
Global
5
2
Local
4
2
[q] = [T]T[q]
q3
q
4 =
q5

q6 2
3
4
5
6
1
1
0
2
0
1
0
0
3
0
0
1
0
4
0
0
0
q1'
q
2'
q3'

q4 '
[k] = [T]T[q] [T]

34
Stiffness Matrix:
1
[k]1 =
4
3
3
[k]2 =
[K] =
3
4
5
6
Member 1
Node 2
Member 2
5
6
35
2
1
Joint Load
Qk
6
2
M
Q 2 1z
-P2y
Q 3
Q 4 M3z
=
Q
1
Q 5
Q
6
Reaction
Qu
Du=Dunknown 5
2 3 4 1 5 6
2
3
4
1
5
6
KAA
KBA
KAB
KBB
D2 Q 2F
D F
3 Q 3
D 4 Q 4F
0 + F
D1 Q1
D5 0 Q F
0 5F
D6 Q 6
Dk = Dknown
[Q k ] = [K AA ][Du ] + [Q AF ]
QFA
QFB
[Du ] = [K AA ]1 + ([Qk ] [QAF ])
[ ]
Member Force : [q ] = [k ][d ] + q F
36
Global:
(qF2)1
(qF
[FEF]
1
2
3
4
Member 1
Q2 = M 1
2
Q = P
2y = 3
3
Q4 = M 2
4
(qF6 )2
(qF3)2
Node 2
Member 2
2
Node 2
w
2
(qF3)1
(qF1 )1
Q1
M
Q 2 1
Q 3 -P2y
M =
Q 4 2
Q 5
Q
6
4)2
(qF4)1
D1 0
D
2
D3
+
D
4
D5 0
0
D6
D2 Q 2F
D F
3 + Q 3
D4 Q 4F
(qF5 )2
Q1F
F
Q
2
F
F
F
Q 3 = (q 3 ) 1 + (q 3 ) 2
F
F
F
Q
q
q
=
+
(
)
(
)
4
4 1
4 2
Q 5F
F
Q
37
2
1
4
1
2
2
D2

D
3
=
3

D4
4
Node 2
Q2 = M 1 Q2F
Q F
Q
P
=
2y
3
3
Q = M Q F
2
4
4
38
P
qF2
qF4
qF1
[FEF]
qF3
Member 1:
1
q 1

2
q 2 =
q 3
3

4
q 4
3
k1
4
d1 = 0
d
D
=
2
2
+
d 3 = D3
d 4 = D4
q 1F
F
q 2
q 3F
F
q 4
39
w
qF6
qF4
qF3
[FEM]
qF5
Member 2:
q3
q
4
q5

q6
1
=
2
3
4
3
k1
q3F
d 3 = D3
F
d = D
4
4
+ q4
q5F
d5 = 0
F
d
=
0
q6
6
40
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.
10 kN
1 kN/m
B
A
9m
C
1.5 m
3m
41
10 kN
1 kN/m
C
1.5 m
3m
9m
3
Global
Members
2
2
10 kN
1 kN/m
1.5 m
[FEF]
wL2/12
= 6.75
9m
wL2/12
= 6.75 PL/8 = 3.75
1.5 m
2
PL/8 = 3.75
42
2
1
9m
3m
Stiffness Matrix:
i
Mi
Mj
[k ]22 =
[k]1= EI
4 EI / L 2EI / L
2EI / L 4 EI / L
4/9
2/9
2/9
4/9
[k]2 = EI
2
[K] = EI
4/3
2/3
2/3
4/3
(4/9)+(4/3) 2/3
2/3
4/3
43
10 kN
1 kN/m
A
6.75
6.75
9m
3.75
1.5 m 1.5 m
Equilibrium equations: MCB = 0

MBA + MBC = 0
Global Equilibrium: [Q] = [K][D] + [QF]
2
2
MBA + MBC = 0
MCB = 0
B
C
= EI
(4/9)+(4/3) 2/3
2/3
4/3
-6.75 + 3.75 = -3
-3.75
0.779/EI
2.423/EI
44
2
1
1 kN/m
[FEF]
9m
wL2/12 = 6.75
wL2/12 = 6.75
Substitute B and C in the member matrix,

Member 1
3
MAB
MBA
= EI
[q]1 = [k]1[d]1 + [qF]1

3
4/9
2/9
2/9
4/9
B= 0.779/EI
6.75
-6.75
1 kN/m
4.56 kN
-6.40
6.40 kNm
6.92 kNm
9m
6.92
4.44 kN
45
10 kN
1
1.5 m
PL/8 = 3.75
1.5 m
2
PL/8 = 3.75
[FEF]
Substitute B and C in the member matrix,
Member 2 :
2
MBC
MCB
= EI
[q]2 = [k]2[d]2 + [qF]2

2
4/3
2/3
B = 0.779/EI
2/3
4/3
C = 2.423/EI
3.75
-3.75
6.40
10 kN
6.40 kNm
7.13 kN
2.87 kN
46
10 kN
1 kN/m
6.40
6.40
6.92
4.56 kN
4.44 kN
1 kN/m
6.92 kNm
7.13 kN
B = +0.779/EI
10 kN
2.87 kN
C = +2.423/EI
C
A
4.56 kN
B
9m
11.57 kN
2.87 kN
1.5 m 1.5 m
7.13
4.56
V (kN)
x (m)
4.56 m
-4.44
M
(kNm)
-6.92
-2.87
4.32
3.48
x (m)
-6.40
47
Example 2
20 kN
9kN/m
40 kNm
EI
2EI
A
B
4m
C
4m
48
2EI
1
[k ] =
Vi 12 EI/ L3
Mi
2
6
/
EI
L
Vj
12 EI/ L3
Mj 6EI/ L2
EI
j
2
4 EI/ L
6EI/ L
2EI/ L
12 EI/ L
6EI/ L2
12 EI/ L3
6EI/ L2
6EI/ L
2EI/ L
6EI/ L2
4 EI/ L
0.375EI
0.75EI
3
4
-0.75EI
EI
-0.375EI -0.75EI 0.375EI -0.75EI

0.75EI
EI
0.5625
-0.375
[K] = EI
-0.375
[k]2
0.75EI - 0.375EI 0.75EI

2EI
[k]1
1
j
3
6EI/ L
Use 4x4 stiffness matrix,
-0.75EI 2EI
3
3
4
5
6
0.1875EI 0.375EI - 0.1875EI 0.375EI

0.375EI
EI
-0.375EI
0.5EI
-0.1875EI -0.375EI 0.1875EI -0.375EI

0.375EI
0.5EI
-0.375EI
EI
49
20 kN
9kN/m
12 kNm
18 kN
4m
40 kNm
12 kNm
18 kN
Global:
Q4 = 40
D3
D4
= EI
2EI
1
1
[Q] = [K][D] + [QF]
3 Q3 = -20
4m
0.5625
-0.375
-0.375
EI
D3
D4
18
-12
-61.09/EI
9.697/EI
50
2EI
1
9 kN/m
12 kNm
A
[qF]1
18 kN
Member 1:
12 kNm
18 kN
[q]1 = [k]1[d]1 + [qF]1

1
q1
q2
q3L
q4L
12(2EI)/430.75EI - 0.375EI 0.75EI

0.75EI
2EI
-0.75EI
EI
-0.375EI -0.75EI 0.375EI -0.75EI

0.75EI
EI
-0.75EI 2EI
9 kN/m
A
67.51 kNm
48.18 kN
1
[q]1
d1 = 0
18
48.18
d2 = 0
12
67.51
d3 = -61.09/EI
18
-12.18
d4 = 9.697/EI
-12
53.21
53.21 kNm
B
12.18 kN
51
EI
2
Member 2:
[q]2 = [k]2[d]2 + [qF]2
3
q3R
q4R
q5
q6
0.1875EI 0.375EI - 0.1875EI 0.375EI

0.375EI
EI
-0.375EI
0.5EI
-0.1875EI -0.375EI 0.1875EI -0.375EI

0.375EI
0.5EI
-0.375EI
7.82 kN
-7.818
d4 = 9.697/EI
-13.21
d5 = 0
7.818
d6 = 0
-18.06
18.06 kNm
13.21 kNm
B
EI
d3 =-61.09/EI
2
[q]2
C
7.82 kN
52
9 kN/m
A
67.51 kNm
48.18 kN
53.21 kNm
12.18 kN
[q]1
18.06 kNm
13.21 kNm
7.82 kN
7.82 kN
[q]2
20 kN
9kN/m
40 kNm
D3 = B = -61.09/EI
67.51 kNm
48.18 kN
48.18
4m
D4 = B = +9.697/EI 4 m
V (kN)
18.06 kNm
7.818 kN
12.18
x (m)
M
(kNm)
+
-
-67.51
-7.818
-7.818
53.21
13.21
-
x (m)
-18.08
53
Example 3
20 kN
9kN/m
10 kN
40 kNm
EI
2EI
A
B
4m
C
2m
2m
54
20 kN
9kN/m
A
3
1
Vi 12 EI/ L3
Mi
2
EI
L
6
/
Vj
12 EI/ L3
Mj 6EI/ L2
5
3
10 kN
12 kNm
18 kN
9 kN/m
[FEF]
EI
2m
2m
4
12 kNm
[k ]44
40 kNm
2EI
4m
Global
10 kN
6EI/ L2
12 EI/ L3
6EI/ L2
12 EI/ L3
6EI/ L2
6EI/ L2
2EI/ L
5 kNm
18 kN
4 EI/ L
5 kNm
5 kN
6EI/ L2
2EI/ L
6EI/ L2
4 EI/ L
5 kN
55
3
1
1
1
[k]1 =
2
3
4
0.75EI
0.75EI
0.1875EI
0.375EI
2m
2m
2EI
-0.75EI
EI
-0.375EI -0.75EI 0.375EI -0.75EI
12(2EI)/43 0.75EI - 0.375EI 0.75EI
[k]2 =
4m
2
EI
4
-0.75EI 2EI/L
5
[K] = EI
6
0.5625
-0.375
0.375
4
6
-0.375
0.375
3
0.5
0.5
1
0.375EI - 0.1875EI 0.375EI

EI
-0.375EI
0.5EI
-0.1875EI -0.375EI 0.1875EI -0.375EI

0.375EI
0.5EI
-0.375EI
EI
56
20 kN
9kN/m
10 kN
40 kNm
C
6
B4
3
1
5
2
10 kN
9 kN/m
12 kNm
12 kNm
[FEF]
18 kN
MBA+MBC = 40
= EI
MCB = 0
B
B
C
5 kNm
18 kN
Global: [Q] = [K][D] + [QF]

VBL+VBR = -20
5 kNm
5 kN
5 kN
6
0.5625
-0.375
0.375
4
6
-0.375
0.375
3
0.5
0.5
1
B
C
18 + 5 = 23
-12 + 5 = -7
-5
-116.593/EI
-7.667/EI
52.556/EI
57
3
1
9 kN/m
12 kNm
12 kNm
18 kN
18 kN
[FEF]
[q]1 = [k]1[d]1 + [qF]1
Member 1:
1
VA
0.375EI 0.75EI - 0.375EI 0.75EI
18
55.97
MAB
0.75EI
12
91.78
VBL
MBA
3
4
2EI
-0.75EI
EI
-0.375EI -0.75EI 0.375EI -0.75EI B =-116.593/EI

0.75EI
EI
-0.75EI 2EI/L
B =-7.667/EI
18
-12
-19.97
60.11
9kN/m
91.78 kNm
4m
55.97 kN
60.11 kNm
19.97 kN
58
10 kN
5
3
5 kN
Member 2:
VBR
0.1875EI
MBC
0.375EI
MCB
5
6
[FEM]
5 kN
[q]1 = [k]1[d]1 + [qF]1

3
VC
5 kNm
5 kNm
0.375EI - 0.1875EI 0.375EI

EI
-0.375EI
0.5EI
-0.1875EI -0.375EI 0.1875EI -0.375EI

0.375EI
0.5EI
-0.375EI
EI
B =-116.593/EI
-0.0278
B =-7.667/EI
-20.11
C =52.556/EI
5
-5
10.03
0
10 kN
20.11 kNm
2m
2m
2
0.0278 kN
10.03 kN
59
9kN/m
91.78 kNm
10 kN
20.11 kNm
2m
1
4m
19.97 kN
55.97 kN
60.11 kNm
20 kN
9kN/m
2m
2
0.0278 kN
10.03 kN
10 kN
C = 52.556/EI
40 kNm
91.78 kNm
55.97 kN
B = -7.667/EI
10.03 kN
B = -116.593/EI
4m
55.97
2m
2m
19.97
+
V (kN)
60.11
M
(kNm)
-91.78
+
-
-0.0278
-10.03
x (m)
-10.03
20.11
x (m)
60
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 diagram
and qualitative deflected shape.
Take I = 200(106) mm4 and E = 200 GPa and support B settlement 10 mm.
40 kN
6 kN/m
B
A
2EI
8m
EI
B = -10 mm
4m
4m
61
2EI
EI
2
i
Use 2x2 stiffness matrix:
[k]1 =
EI
8
[k ]22 =
[K] =
EI
8
Mi
Mj
4 EI / L 2EI / L
2EI / L 4 EI / L
[k]2 =
2
12
EI
8
62
40 kN
6 kN/m
B
A
C
EI
B = -10 mm
4m
4m
2EI
8m
40 kN
6 kN/m
[FEM]load
wL2/12 = 32 kNm
32 kNm
10 mm
6(2EI)/L2 = 75 kNm
Global:
Q2 = 0
Q3 = 0
D2
D3
12
D2
D3
185.64/EI
37.5 kNm
6(EI)/L2 = 37.5 kNm

3
-61.27/EI
40 kNm
10 mm
[Q] = [K][D] +2[QF]

EI
=
8
PL/8 = 40 kNm
75 kNm
[FEM]
-32 + 40 + 75 -37.5 = 45.5

-40 - 37.5 = -77.5
rad
rad
63
6 kN/m
[FEM]load
2EI
wL2/12 = 32 kNm
32 kNm
75 kNm
[FEM]
10 mm
6(2EI)/L2 = 75 kNm
Member 1:
q1
q2
EI
=
8
[q]1 = [k]1[d]1 + [qF]1

1
d1 = 0
d2 = -61.27/EI
76.37 kNm
6 kN/m
32 + 75 = 107
-32 + 75 = 43
76.37 kNm
-18.27 kNm
18.27 kNm
8m 1
31.26 kN
16.74 kN
64
40 kN
[FEM]load
PL/8 = 40 kNm
40 kNm
37.5 kNm
[FEM]
10 mm
6(EI)D/L2 = 37.5 kNm

Member 2:
q2
q3
EI
8
[q]2 = [k]2[d]2 + [qF]2

2
d2 = -61.27/EI
d3 = 185.64/EI
40 - 37.5 = 2.5
-40 - 37.5 = -77.5
18.27 kNm
kNm
40 kN
18.27 kNm
2
22.28 kN
17.72 kN
65
2EI
EI
2
1
Alternate method: Use 4x4 stiffness matrix
1
[K]
1.5
-0.375
1.5
-1.5
-1.5
0.375
-1.5
-1.5
1 12(2)/82
EI 2
1.5
=
8 3
-0.375
4
1.5
[k]1
3 12/82
4
0.75
EI
8 5
0.75
4
-0.1875 -0.75
8
3
0.75
5
1
2
0.375
1.5
1.5
8
-0.375
-1.5
EI 3
=
8 4
5
-0.375
-1.5
0.5625
1.5
0
4
0
-0.75
-0.1875
12
-0.75
-0.75
0.1875
2
-0.75
0.75
-0.75
1.5
4
[k]2
-0.1875
0.75
-0.75
0.1875
-0.75
-0.75
0
0
0
0
-0.75 -0.1875
0.75
66
40 kN
6 kN/m
B
A
C
EI
B = -10 mm
2EI
8m
4m
32 kNm
40 kNm
Q6= 0
Q4 = 0
Q6= 0
D4
D6
5
D4
4
6
D6
12
D4
D6
12
EI 4
=
8 6
2
4
20 kN
20 kN
EI 4
=
8 6
40 kNm
24 kN
24 kN
4
6
Q4 = 0
3
40 kN
32 kNm
[FEF]
4m
6 kN/m
200 200 4
+(
)
8
6
-0.75 D = -0.01
5
-0.75
(200x200/8)(-0.75)(-0.01) = 37.5
(200x200/8)(0.75)(-0.01) = -37.5
-61.27/EI = -1.532x10-3
rad
185.64/EI = 4.641x10-3
rad
+
+
8
-40
8
-40
67
32 kNm
B = -10 mm
6 kN/m
32 kNm
[FEF]load
24 kN
24 kN
Member 1:
[q]1 = [k]1[d]1 + [qF]1

1
1 12(2)/82
2
1.5
q1
q2
1.5
-0.375
1.5
d1 = 0
24
-1.5
d2 = 0
32
(200x200)
3
8
-0.375
-1.5
0.375
-1.5
q4
1.5
-1.5
q1
31.26
q3
q2
q3
q4
kN
76.37
kNm
16.74
kN
-18.27
kNm
76.37 kNm
d3 = -0.01
-32
d4 = -1.532x10-3
6 kN/m
24
18.27 kNm
8m 1
31.26 kN
16.74 kN
68
40 kN
40 kNm
40 kNm
-10 mm = B
[FEF]
20 kN
20 kN
Member 2:
3
3 12/82
4
0.75
q3
q4
q5
(200x200)
5
8
6
q3
22.28
kN
q4
18.27
kNm
17.72
kN
kNm
q6
-0.1875 -0.75
q6
q5
0.75
0.75
-0.1875
-0.75
0.75
2
0.1875 -0.75
-0.75
d3 = -0.01
20
d4 = -1.532x10-3
40
d5 = 0
d6 = 4.641x10-3
20
-40
40 kN
18.27 kNm
2
22.28 kN
17.72 kN
69
40 kN
6 kN/m
B
76.37 kNm
EI
B = -10 mm
17.72 kN
16.74 + 22.28 kN
8m
4m
4m
2EI
31.26 kN
V (kN)
31.26
+
5.21 m
22.28
+
-16.74
-17.72
x (m)
70.85
M
(kNm)
5.06
-
+
-
-18.27
x (m)
-76.37
Deflected
Curve
B = -10 mm
d4 = B = -1.532x10-3 rad
D6 = C = 4.641x10-3 rad
70
Example 5
For the beam shown:
(c) Draw the quantitative shear diagram, bending moment diagram
Take I = 200(106) mm4 and E = 200 GPa and support C settlement 10 mm.
4 kN
0.6 kN/m
20 kNm
B
A
EI
2EI
8m
4m
C
C = -10 mm
4m
71
2
2EI
1
Member stiffness matrix [k]4x4
1
1 12(2)/82
EI 2
1.5
=
8 3
-0.375
4
1.5
[k]1
EI
2
1.5
-0.375
1.5
-1.5
-1.5
0.375
-1.5
-1.5
EI
=
8
3 12/82
4
0.75
5
-10 mm
[k]2
0.75
4
-0.1875 -0.75
0.75
-0.1875
0.75
-0.75
0.1875
-0.75
-0.75

Q3
Q4
Q6
3
EI 4
=
8
6
0.5625
-0.75
-0.75
12
0.75
2
D3
D4
0.75
D6
5
QF3
3 -0.1875
200 200
+(
) 4 -0.75 D5 = -0.01 + QF4
8
QF6
6 -0.75
72
4 kN
0.6 kN/m
20 kNm
B
A
EI
2EI
8m
4m
10 mm
4m
Q6 = 20
D3
D4
D6
4 kNm
5
4 kNm
3
EI 4
=
8
6
EI
3
4 kN
2.4 kN
2.4 kN
Q3 = 0
Q4 = 0
6
2
3.2 kNm
[FEF]
2EI
0.6 kN/m
3.2 kNm
2 kN
0.5625
-0.75
-0.75
12
0.75
2
0.75
D3
9.375
D4 + 37.5 +
D6
37.5
2 kN
2.4+2 = 4.4
-3.2+4 = 0.8
-4.0
-377.30/EI = -9.433x10-3 m
-61.53/EI = -1.538x10-3 rad
+74.50/EI = +1.863x10-3 rad
73
0.6 kN/m
3.2 kNm
1
[FEF]load
3.2 kNm
8m
2.4 kN
2.4 kN
Member 1:
[q]1 = [k]1[d]1 + [qF]1

1
q1
q2
q3
q4
1.5
-0.375
1.5
d1 = 0
2.4
-1.5
d2 = 0
3.2
-0.375
-1.5
0.375
-1.5
d3 = -9.433x10-3
1.5
-1.5
d4 = -1.538x10-3
q1
q3
200 200
3
8
q4
q2
1 12(2)/82
2
1.5
8.55
kN
43.19
kNm
-3.75
kN
6.03
kNm
43.19 kNm
0.6 kN/m
2.4
-3.2
6.03 kNm
8m 1
8.55 kN
3.75 kN
74
4 kN
4 kNm
4 kNm
[FEF]
8m
10 mm
2 kN
2 kN
Member 2:
3
q3
q4
q5
200 200
=
8
q6
3 12/82
4
0.75
5
6
0.75
3.75
kN
q4
-6.0
kNm
0.25
kN
20.0
kNm
q6
-0.1875 -0.75
q3
q5
0.75
-0.1875
-0.75
0.75
d3 = -9.433x10-3
d4 = -1.538x10-3
0.1875 -0.75
-0.75
d5 = -0.01
d6 = 1.863x10-3
2
-4
4 kN
6.0 kNm
20 kNm
2
3.75 kN
0.25 kN
75
4 kN
0.6 kN/m
20 kNm
B
A
EI
2EI
8m
0.6 kN/m
43.19 kNm
4m
Deflected
Curve
4 kN
20 kNm
6.0 kNm
3.75 kN
8.55
V (kN)
M
(kNm)
10 mm
4m
6.03 kNm
8m 1
8.55 kN
3.75
-0.25
21
+
-43.19
0.25 kN
3.75 kN
D3 = -9.433 mm
D4 = -1.538x10-3 rad
x (m)
20
x (m)
D5 = -10 mm
D6 = +1.863x10-3 rad
76
Internal Hinges
Example 6
(a) Determine all the reactions at supports.
(b) Draw the quantitative shear and bending moment diagrams
E = 200 GPa, I = 50x10-6 m4.
30 kN
9kN/m
Hinge
EI
2EI
A
4m
C
2m
2m
77
4
2
2EI
1
4m
EI
2m
2m
i
Use 2x2 stiffness matrix,
[k]1 =
[k ]22 =
2EI
EI
EI
2EI
[K] =
Mi
Mj
4 EI / L 2EI / L
2
EI
/
L
4
EI
/
L
[k]2 =
EI
1EI
4 0.5EI
2.0
0.0
0.0
1.0
4
0.5EI
1EI
78
30 kN
9kN/m
Hinge
3
4
2
[FEF]
0.0
EI
30 kN
15 kNm
B
15 kNm
C
2
2.0
0.0
D1
0.0
1.0
D2
0.0006
D1
D2
B
12 kNm
B
Global matrix:
0.0
9kN/m
12 kNm
-12
15
rad
-0.0015 rad
79
12 kNm
1
A
[FEF]
9kN/m
1
12 kNm
B
Member 1:
q3
q1
2EI
EI
EI
2EI
d3 = 0.0
d1 = 0.0006
12
-12
18
0.0
9 kN/m
A
18 kNm
22.5 kN
B
13.5 kN
80
4
B
15 kNm
B
30 kN
15 kNm
C
Member 2:
2
q2
q4
1EI
0.5EI
4 0.5EI
1EI
d2 = -0.0015
d4 = 0.0
15
-15
0.0
-22.5
30 kN
C
B
9.37 kN
22.5 kNm
20.63 kN
81
30 kN
9 kN/m
A
18 kNm
13.5 kN
13.5 kN
22.5 kN
9.37 kN
22.87 kN
22.5 kNm
20.63 kN
9.37 kN
30 kN
9kN/m
Hinge
EI
BR= -0.0015 rad
2EI
BL= 0.0006 rad
4m
2m
2m
22.5
V (kN)
9.37
+
2.5 m
M
(kNm)
-13.5
10.13
+
- -18
x (m)
-20.63
18.75
+
-
x (m)
-22.5
82
Example 7
E = 200 GPa, I = 50x10-6 m4.
20 kN
9kN/m
2EI
A
B
4m
EI
Hinge
C
4m
83
1 3
2EI
1
2 B
0.375EI
[k]1 = 1
0.75EI
2EI
-0.375EI
-0.75EI
0.75EI
EI
[k]2 =
0.1875EI
0.375EI
6
7
0.75EI - 0.375EI 0.75EI

-0.75EI
EI
0.375EI -0.75EI
-0.75EI
2EI/L
[K] =
1
EI
0.5625
-0.75
0.375
EI 2
-0.75
2.0
0.375
0.0
1.0
0.375EI - 0.1875EI 0.375EI

EI
-0.375EI
0.5EI
-0.1875EI -0.375EI 0.1875EI -0.375EI

0.375EI
0.5EI
-0.375EI
EI
84
20 kN
B
9kN/m
A
C
5
4
1 3
2EI
1
A
12 kNm
Global:
[FEM]
1
Q1 = -20
Q2 = 0.0
12 kNm
18 kN
3
0.5625
-0.75
0.375
D1
EI 2
-0.75
2.0
D2
0.375
0.0
1.0
D3
D1
D3
Q3 = 0.0
D2
EI
2 B
9kN/m
A
18 kN
18
-12
0.0
-0.02382 m
-0.008333 rad
0.008933 rad
85
9kN/m
2EI
12 kNm
A
18 kN
12 kNm
[FEF]
18 kN
Member 1:
4
q4
0.375EI
q5
0.75EI
2EI
1 -0.375EI
-0.75EI
0.75EI
EI
q1
q2
0.75EI - 0.375EI 0.75EI

-0.75EI
EI
d4 = 0.0
18
d5 = 0.0
12
0.375EI -0.75EI
d1 = -0.02382
-0.75EI
d2 = -0.00833
107.32 kNm
A
2EI/L
18
-12
44.83
107.32
-8.83
0.0
9kN/m
1
B
8.83 kN
44.83 kN
86
7
1
B
EI
2
Member 2:
1
q1
0.1875EI
q3
0.375EI
q6
=6
q7
0.375EI - 0.1875EI 0.375EI

EI
-0.375EI
0.5EI
-0.1875EI -0.375EI 0.1875EI -0.375EI

0.375EI
0.5EI
EI
-0.375EI
B
2
11.16 kN
d1 = -0.02382
d3 = 0.008933
d6= 0.0
d7= 0.0
0
0
-11.16
0.0
11.16
-44.66
44.66 kNm
11.16 kN
87
20 kN
9kN/m
BL= -0.008333 rad B
BR= 0.008933 rad C
4m
4m
9kN/m
107.32 kNm
A
44.83 kN
B
2
8.83 kN
44.66 kNm
11.16 kN
11.16 kN
44.83
V (kN)
8.83
x (m)
-11.16
-11.16
M
(kNm)
-107.32
x (m)
-44.66
88
Example 8
40 kNm at the end of member AB. E = 200 GPa, I = 50x10-6 m4.
30 kN
9kN/m
A
Hinge
2EI
4m
40 kNm
EI
B
2m
2m
89
4
2
2EI
4m
2m
2m
j
Use 2x2 stiffness matrix:
[k ]22 =
[k]1 =
EI
2EI
EI
EI
2EI
[K] =
Mi
Mj
4 EI / L 2EI / L
2
EI
/
L
4
EI
/
L
[k]2 =
EI
1EI
4 0.5EI
2.0
0.0
0.0
1.0
4
0.5EI
1EI
90
30 kN
9kN/m
Hinge
2EI
40 kNm
EI
C
4
2
A
[FEM]
Q2 = 0.0
EI
D1
D2
B
12 kNm
15 kNm
B
15 kNm
C
2
2.0
0.0
D1
0.0
1.0
D2
0.0026
30 kN
Global matrix:
Q1 = 40
9kN/m
12 kNm
-12
15
rad
-0.0015 rad
91
12 kNm
1
A
[FEF]
9kN/m
1
12 kNm
B
Member 1:
q3
q1
2EI
EI
EI
2EI
d3 = 0.0
d1 = 0.0026
12
-12
38
40
9 kN/m
A
38 kNm
B 40 kNm
37.5 kN
1.5 kN
92
4
B
15 kNm
B
30 kN
15 kNm
C
Member 2:
2
q2
q4
1EI
4 0.5EI
4
0.5EI
1EI
d2 = -0.0015
d4 = 0.0
15
-15
0.0
-22.5
30 kN
C
B
9.37 kN
22.5 kNm
20.63 kN
93
1.5 kN
30 kN
9 kN/m
A
38 kNm
B 40 kNm
9.37 kN
7.87 kN
1.5 kN
37.5 kN
22.5 kNm
20.63 kN
9.37 kN
30 kN
9kN/m
40 kNm
Hinge
BL= 0.0026 B
rad
4m
BR=- 0.0015 rad

2m
2m
37.5
V (kN)
9.37
1.5
x (m)
-20.63
M
(kNm)
40
+
- -38
18.75
+
x (m)
-22.5
94
Example 9
40 kNm at the end of member AB. E = 200 GPa, I = 50x10-6 m4
20 kN
9kN/m
2EI
A
4m
40 kNm
EI
Hinge
C
4m
95
20 kN
9kN/m
2EI
A
40 kNm
EI
Hinge
4m
C
4m
7
1
4
A
12 kNm
[FEM]
6
C
2 B
9 kN/m
12 kNm
1
18 kN
18 kN
[k ]44
Vi 12 EI/ L3
Mi
2
6EI/ L
Vj
12 EI/ L3
Mj 6EI/ L2
6EI/ L2
12 EI/ L3
6EI/ L2
12 EI/ L3
6EI/ L2
6EI/ L2
2EI/ L
6EI/ L2
4 EI/ L
4 EI/ L
6EI/ L2
2EI/ L
96
2EI
[k]1 =
0.375EI
0.75EI
0.75EI
EI
-0.75EI
EI
0.375EI -0.75EI
-0.75EI
0.1875EI
0.375EI
EI
-0.375EI
0.5EI
-0.1875EI -0.375EI 0.1875EI -0.375EI
0.375EI
0.5625
-0.75
0.375
EI 2
-0.75
2.0
0.375
0.0
1.0
1
6
2EI
[K] =
[k]2 =
0.75EI - 0.375EI 0.75EI
1 -0.375EI
EI
2 B
2EI
-0.75EI
0.375EI - 0.1875EI 0.375EI
0.5EI
-0.375EI
EI
97
9kN/m
20 kN
5
4
2EI
2 B
18 kN
18 kN
Global:
Q1 = -20
Q3 = 0.0
D1
D3
EI
12 kNm
[FEF]
D2
9 kN/m
12 kNm
Q2 = 40
EI
40 kNm
1 3
0.5625
-0.75
0.375
D1
EI 2
-0.75
2.0
D2
0.375
0.0
1.0
D3
18
-12
0.0
-0.01316 m
-0.002333 rad
0.0049333 rad
98
5
4
2EI
12 kNm
0.375EI
q5
0.75EI
2EI
1 -0.375EI
-0.75EI
0.75EI
EI
q2
18 kN
[q]1 = [k]1[d]1 + [qF]1
q4
q1
18 kN
12 kNm
[FEF]
Member 1:
9 kN/m
0.75EI - 0.375EI 0.75EI
87.37 kNm
-0.75EI
EI
d4 = 0.0
18
d5 = 0.0
12
0.375EI -0.75EI
d1 = -0.01316
-0.75EI
d2 = -0.002333
2EI
2EI
49.85 kN
18
-12
49.85
87.37
-13.85
40
40 kNm
1
13.85 kN
[q]1
99
7
1
B
Member 2:
q1
0.1875EI
q3
0.375EI
=6
q7
[q]2 = [k]2[d]2 + [qF]2

1
q6
EI
0.375EI - 0.1875EI 0.375EI d1 = -0.01316

EI
-0.375EI
0.5EI
-0.1875EI -0.375EI 0.1875EI -0.375EI

0.375EI
0.5EI
-0.375EI
EI
d3 = 0.004933
d6= 0.0
d7= 0.0
-6.18
0.0
0
0
6.18
-24.69
24.69 kNm
6.18 kN
6.18 kN
[q]2
100
20 kN
9kN/m
40 kNm
EI
BR = 0.004933 rad C
4m
2EI
BL= -0.002333 rad
4m
B
A
87.37 kNm 49.85 kN
40 kNm
13.85 kN
24.69 kNm
6.18 kN
6.18 kN
49.85
V (kN)
13.85
x (m)
-6.18
-6.18
M
(kNm)
+
-87.37
40
-
x (m)
-24.69
101
Temperature Effects
Fixed-End Forces (FEF)

- Axial
- Bending
Curvature
102
tan =
Thermal Fixed-End Forces (FEF)

Room temp = TR
T
TR
M AF
FBF
Tm> TR
FAF
Tb > Tt
axial
ct
Tt
NA
Tm =
cb
F
M
B
B
axial
Tb - Tt
TR
Tl
(T ) axial = Tm TR
y
A
Tt + Tb
2
Tm
Tt
Tb Tt
T
=( )
d
d
( T y ) = y (
T
)
d
Tb
(T ) bending = Tb Tt
103
- Axial
axial
axial
Tt
Tm> TR
FAF
Tb > Tt
TR
Tm
FBF
(T ) axial = Tm TR
FAF = axial dA
A
= E axial dA
= E (T ) axial dA
= E (T ) axial dA
= EA (T ) axial
F
( Faxial
) A = (Tm TR ) AE
104
- Bending
Tt
y
y
M AF
M BF
(Ty ) = y (
T
)
d
Tb
(T ) bending = Tb Tt
M = y y dA
F
A
= yE dA
= yE (Ty ) dA
= yEy (
= E (
T
) dA
d
T
) y 2 dA
d
F
( Fbending
)A = (
Tl Tu
) EI
d
105
Elastic Curve: Bending
O
d
(dx ) = ( d )
y ( d )
ds = dx
y
dx
Before
deformation
ds
=(
d
)
dx
dx
After
deformation
106
Bending Temperature
Tt
dx
Tl > Tu
Tb
d
Tt + Tb
2
M
Tm =
Tu
T = Tb - Tt
Tt
d
y
T
=
d
ct
cb
Tl > Tu
Tb
Tt
T
y
d
y
Tb
T
) dx
d
T
(d ) = ( )dx
d
d
M
1
T
( ) = = ( ) =
dx
d
EI
(d ) y = y (
107
Example 10
Room temp = 32.5oC, = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4.
T2 = 40oC
182 mm
A
T1 = 25oC
4m
EI
2EI
B
C
4m
108
T2 = 40oC
182 mm
A
T1 = 25oC
4m
C
4m
3
1
1
FEM
EI
2EI
Mean temperature = (40+25)/2 = 32.5

Room temp = 32.5oC
19.78 kNm
a(T /d)EI = 19.78 kNm
o
+7.5 C
F
Fbending
=(
-7.5 oC
T
40 25
)(2 EI ) = (12 10 6 )(
)(2 200 50) = 19.78 kN m
d
0.182
109
2
19.78 kNm 19.78 kNm
182 mm
1
2
2EI
4m
M1
EI
q2
q1
0.5
q1
0.5
q3
Element 2:
M1
M3
EI
4m
[q] = [k][d] + [qF]
Element 1:
M2
EI
-1.978
1.978
(10-3) EI
0
0
0
[M1] = 3EI1 + (1.978x10-3)EI
1 = -0.659x10-3
rad
110
2
19.78 kNm 19.78 kNm
A
182 mm
2EI
1
2
EI
4m
4m
Element 1:
M2
M1
EI
q2 = 0
q1 = -0.659x10-3
Element 2:
M1
M3
EI
0.5
q1 = -0.659x10-3
0.5
q3 = 0
26.37 kNm
A
6.59 kNm
B
4.95 kN
4.95 kN
-19.78
19.78
0
0
6.59 kNm
B
-26.37 kNm
6.59 kNm
-6.59 kNm
-3.30 kNm
3.3 kNm
C
2.47 kN
2.47 kN
111
26.37 kNm
+7.5 oC
A
4.95 kN
-7.5 oC
2.47 kN
4m
4m
2.47 kN
V (kN)
3.3 kNm
x (m)
-2.47
-4.95
26.37
M
(kNm)
Deflected
curve
6.59
-
B = -0.659x10-3 rad
x (m)
-3.30
x (m)
112
Example 11
Room temp = 28 oC, a = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4,
A = 20(10-3) m2
T2 = 40oC
182 mm
A
T1 =
25oC
4m
EI, AE
2EI, 2AE
B
C
4m
113
6
5
9
T2 = 40oC
A
4
T1 = 25oC
2
2EI
4m
1
B
8
2
EI
7
C
4m
Element 1:
(2 AE ) 2(20 10 3 m 2 )(200 106 kN / m 2 )
=
= 2(106 ) kN / m
L
( 4 m)
4(2 EI ) 4 2(200 106 kN / m 2 )(50 10 6 m 4 )
=
= 20(103 ) kN m
L
( 4 m)
2(2 EI ) 2 2(200 106 kN / m 2 )(50 10 6 m 4 )
=
= 10(103 ) kN m
L
( 4 m)
6(2 EI ) 6 2(200 106 kN / m 2 )(50 10 6 m 4 )
3
=
=
7
.
5
(
10
) kN
2
2
L
( 4 m)
12(2 EI ) 12 2(200 106 kN / m 2 )(50 10 6 m 4 )
3
=
=
3
.
75
(
10
) kN / m
3
3
L
( 4 m)
114
T2 = 40oC
182 mm
A
T1 =
25oC
EI, AE
2EI, 2AE
4m
4m
Fixed-end forces due to temperatures

F
Fbending
=(
T
40 25
)(2 EI ) = (12 10 6 )(
)(2 200 50) = 19.78 kN m
d
0.182
Mean temperature(Tm) = (40+25)/2 = 32.5 oC ,

TR = 28 oC
F
Faxial
= (T ) AE = (12 10 6 )(32.5 28)(2 20 10 3 m 2 )(200 10 6 kN / m 2 ) = 432 kN
432 kN
19.78 kNm
+12 oC
A
-3
oC
19.78 kNm
B
432 kN
115
9
T2 = 40oC
5
182 mm
4
A
T1 =
2EI, 2AE
25oC
EI, AE
4m
FEM
Element 1:
432 kN
[q] = [k][d] +
19.78 kNm
+12 oC
-3
[qF]
19.78 kNm
432 kN
oC
B
1
q4
2x106
0.00
0.00
- 2x106 0.00
q5
0.00
3750
7500
0.00
q6
0.00
7500 20x103
q1
q2
q3
7
C
4m
0.00
3
0.00
432
7500
0.00
-7500 10x103
-19.78
-3750
0.00
2x106
0.00
0.00
d1
-7500
0.00
3750
-7500
d2
0.00
7500 10x103
0.00
-7500 20x103
d3
19.78
-2x106 0.00
0.00 -3750
0.00
-432
116
9
T2 = 40oC
5
182 mm
4
A
T1 =
2EI, 2AE
25oC
8
2
7
C
4m
EI, AE
4m
Element 2:
[q] = [k][d] + [qF]

1
9
0.00
d1
-1875
3750
d2
0.00
-3750
5x103
d3
0.00
1x106
0.00
0.00
1875
-3750
-3750 10x103
q1
1x106
0.00
0.00
- 1x106 0.00
q2
0.00
1875
3750
0.00
q3
0.00
3750 10x103
q7
q8
q9
7 -1x106 0.00
8
0.00 -1875
-3750
0.00
0.00
5x103
0.00
3750
117
9
T2 = 40oC
5
182 mm
4
A
T1 =
2EI, 2AE
25oC
Q1 = 0.0
Q2 = 0.0
Q3 = 0.0
D1
D2
D3
8
2
EI, AE
7
C
4m
Global:
4m
3x106
0.0
0.0
D1
0.0
5625
-3750
D2
0.0
-3750
30x103
D3
0.000144
-0.0004795
-719.3x10-6
rad
-432
0.0
19.78
118
Element 1:
q4
2x106
0.00
0.00
- 2x106 0.00
0.00
432
q5
0.00
3750
7500
0.00
7500
0.00
q6
0.00
7500 20x103
-7500 10x103
-19.78
q1
q2
q3
-3750
0.00
0.00
2x106
0.00
0.00
-7500
0.00
3750
-7500
d2 = -479.5x10-6
0.00
7500 10x103
0.00
-7500 20x103
d3 = -719.3x10-6
19.78
-2x106 0.00
0.00 -3750
0.00
d1 = 144x10-6
-432
6
q4
144.0
kN
q5
-3.60
kN
q6
-23.38
kNm
-144.0
kN
3.60
kN
9.00
kNm
q1
q2
q3
5
A 4
23.38 kNm
144 kN
3.60 kN
2
B 1
9 kNm
B
3.60 kN
144 kN
119
Element 2:
9
0.00
d1 = 144x10-6
-1875
3750
d2 = -479.5x10-6
0.00
-3750
5x103
d3 = -719.3x10-6
0.00
1x106
0.00
0.00
1875
-3750
-3750 10x103
3
q1
1x106
0.00
0.00
- 1x106 0.00
q2
0.00
1875
3750
0.00
q3
0.00
37500 10x103
q7
q8
q9
7 -1x106 0.00
8
0.00 -1875
-3750
0.00
0.00
5x103
0.00
3750
q1
144
kN
q2
-3.6
kN
q3
-9
kNm
-144
kN
3.6
kN
-5.39
kNm
q7
q8
q9
2
B 1
9 kNm
144 kN
3.60 kN
8
C 7
5.39 kNm
C
3.60 kN
144 kN
120
Isolate axial part from the system

T2 = 40oC
RA
A
T1 = 25oC
EI
2EI
B
4m
RC
RA
C
4m
RA = RC
+
Compatibility equation:
dC/A = 0
R A (4) RA ( 4)
+
+ 12 10 6 (32.5 28)(4) = 0
AE
2 AE
RA = 144 kN
RC = -144 kN
121
T2 = 40oC
EI
2EI
A
T1 = 25oC
4m
4m
9 kNm
23.38 kNm
144 kN
144 kN
144 kN
144 kN
B
3.60 kN
3.60 kN
5.39 kNm
9 kNm
B
C
3.60 kN
3.60 kN
V (kN)
x (m)
-3.6
23.38
M
(kNm)
Deflected
curve
8.98
-
D2 = -0.0004795 mm
D3 = -719.3x10-6 rad
x (m)
-5.39
x (m)
122
Skew Roller Support
- Force Transformation
- Displacement Transformation
- Stiffness Matrix
123
Displacement and Force Transformation Matrices

x
6
j
2
3
6
5
3
2
i
x
124
Force Transformation
6
2
3
x =
y =
q6 = q6'
x j xi
L
y j yi
L
5
4
3
2
i
q 4
q
5
q 6
y
x
0
0
1
q4'
q
5'
q6'
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
x
y
0
0
y
x
0
x
= y
0
q 1 x

q 2 y
q 3 0
=
q 4 0
q 5 0

q 6 0
q4 = q4' cos x q5' cos y
q5 = q4' cos y q5' cos x
j x
[q ] = [T ]T [q ']
0 q 1'

0 q 2'
0 q 3'

0 q 4'
0 q 5'

1 q 6'
125
Displacement Transformation
y
x
s
o
c
d4
x4 d4
6
j y
y
s
o
c
d4
5
5
4
d5
y
os
c
5
d ' 4 = d 4 cos x + d 5 cos y
d '6 = d 6
x
s
o
c
5
d' 5 = d 4 cos y + d 5 cos x
x
y
d 4'
d
5'
d 6'
x
= y
0
d 1' x

d 2' y
d 3' 0
=
d 4' 0
d 5' 0

d 6' 0
x
x
0
0
1
y
x
d 4
d
5
d 6
0
0
0
0
0
0
1
0
0 x
0 y
0
0
0
0
0
0
0
y
x
0
0 d 1

0 d 2
0 d 3

0 d 4
0 d 5

1 d 6
[d '] = [T ][d ]
126
[q ] = [T ]T [q ']
[ ]
= [T ] [k' ][d' ] + [T ] [q ' ]
[q ] = [T ] [k' ][T ][d ] + [T ] [q' ] = [k ][d ] + [q ]
= [T ] ([k' ][d' ] + q 'F )
T
Therefore,
[k ] = [T ]T [k '][T ]
[q ] = [T ] [q ' ]
F
[q ] = [T ]T [q ']
[d '] = [T ][d ]
[k ] = [T ]T [k '][T ]
127
Stiffness matrix
2*
5*
3*
1*
6*
4*
j
i
jx = cos j
iy = sin i
jy = sin j
ix = cos i
[q*] = [T]T[q]
q 1*
q
2*
q 3*

q 4 *
q 5*

q 6 *
1*
2*
=
3*
4*
5*
6*
1
ix
iy
0
0
0
2
iy
ix
0
0
0
0
3
0
0
1
0
0
0
4
5
0
0
0
0
0
0
jx jy
jy
jx
0
0
6
0
0
0
0
0
q 1'
q
2'
q 3'

q 4'
q 5'

q 6'
[ T ]T
128
1
2
[T ]
3
4
5
6
[k ']
1
2
3
4
5
6
1
2
iy
ix

iy ix
0
0
0
0
0
0
0
0
0
AE/ L
3
EI
L
0
12
/
0
6EI/ L2
0
AE/ L
0
12 EI/ L3
6EI/ L2
0
3
4
0
0
0
0
1
0
0 jx
0 jy
0
0
3
0
6EI/ L2
4 EI/ L
0
6EI/ L2
2EI/ L
5
0
0
0
jy
jx
0
6
0
0
0
0
0
AE/ L
0
0
12 EI/ L3
0
6EI/ L2
0
AE/ L
0
12 EI/ L3
0
6EI/ L2
6EI/ L2
2EI/ L
2
6EI/ L
4 EI/ L
0
129
[ k ] = [ T ]T[ k ][T] =
Ui
Ui (
AE
Vi
L
(
AE
Vj - (
Mj
Mi
Uj -(
ix2 +
AE
L
AE
L
12EI
L3
12EI
L3
6EI
L2
ixjx +
ixjy-
Vi
6EI
L2
iy2 )
) ixiy
AE
L
AE
L
L3
12EI
L3
iy
12EI
iy2 +
L2
iy jy)
-(
iyjx )
-(
AE
L
AE
L
) ixiy
L3
6EI
iy
12EI
iyjx -
iyjy +
6EI
L2
Uj
Mi
12EI
L3
ix2 )
ix
12EI
L3
12EI
L3
ix
ixjy)
ix jx ) -
6EI
iy -(
L2
6EI
L2
ix
-(
AE
L
AE
L
ixjx +
6EI
L2
6EI
L2
jy
6EI
L2
jx
AE
L
AE
L
L3
L3
L3
L3
6EI
L2
AE
ixjy )
AE
-(
jy
ixjy -
iyjy +
6EI
12EI
12EI
2EI
iyjy ) -(
jy
jx2 +
12EI
12EI
iyjx -
4EI
Vj
jy2 )
) jxjy
AE
L2
Mj
12EI
L3
12EI
L3
6EI
L2
iy
L2
ix
L2
2EI
L
6EI
) jxjy
jx
6EI
6EI
ixjx )
AE
12EI
jx2 )
( L jy2 +
L3
jx
12EI
L3
iyjx )
jy
L2
6EI
L2
jx
4EI
L
130
Example 12
For the beam shown:
(c) Draw the quantitative bending moment diagrams
Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members.
Include axial deformation in the stiffness matrix.
40 kN
4m
4m
22.02 o
131
40 kN
4m
4m
22.02o
6
5
i
3
Global
2*
3*
1*
x
22.02 o
j
x*
jx = cos 22.02o = 0.9271,
jy = cos 67.98o = 0.3749
ix = cos 0o = 1,
iy = cos 90o = 0
40 kN
40 kNm
[FEF]
20 kN
[ qF ]
6
Local
5
1
40 kNm
20sin22.02=7.5
20cos22.02=18.54 20 kN
132
Transformation matrix
6
5
3
Global
2*
4
ix = cos 0o = 1,
iy = cos 90o = 0
Member 1: [ q ] = [
q 4

q5
q 6

q 1*
q 2*

q 3*
3*
1*
2
x
*
x
jx = cos 22.02o = 0.9271,
jy = cos 67.98o = 0.3749
T ]T[q]
4
5
6
1*
2*
3*
1
1
0
0
0
2
0
1
0
0
0
0
3
4
0
0
0
0
1
0
0 0 . 9271
0 0 . 3749
0
0
6
Local
5
1
[T ]T
ix
iy
0
=
0
0
iy
0
0
0
1
0
0
0
0 jx
0 jy
0 0
ix
0
0
0
0
0
0
jy
jx
0
0
0
0
0
0
5
6
0
0 q 1'

0
0 q 2'
0
0 q 3'

0 . 3749 0 q 4'
0 . 9271 0 q 5'

0
1 q 6'
133
Local stiffness matrix

2
i
[k ]66 =
[k]1
6
Local
5
1
Ni AE/ L
0
3
Vi 0
12
/
EI
L
Mi 0
6EI/ L2
Nj AE/ L
0
Vj 0
12 EI/ L3
6EI/ L2
Mj 0
0
6EI/ L2
4 EI/ L
0
6EI/ L2
2EI/ L
1
2
1 150.0 0.000
2 0.000 0.9375
3 0.000 3.750
103 4 -150.0 0.000
5 0.000 -0.9375
3
4
5
0.000 -150.0
0.000
3.750 0.000 -0.9375
20.00 0.000 -3.750
0.000
-3.750
150.0
0.000
0.000
0.9375
0.000
-3.750
3.750
10.00
0.000
-3.750
20.00
6 0.000
0
AE/ L
0
12 EI/ L3
0
6EI/ L2
0
AE/ L
0
12 EI/ L3
0
6EI/ L2
6EI/ L2
2EI/ L
2
6EI/ L
4EI/ L
6
0.000
3.750
10.00
134
6
5
3
Global
2*
3*
1*
4
Stiffness matrix [k]:
1
2
1 150.0 0.000
2 0.000 0.9375
3 0.000 3.750
[k]1 = 103 4 -150.0 0.000
5 0.000 -0.9375
3.750
2
x
i
6
Local
5
1
x*
3
4
5
0.000
0.000 -150.0
3.750 0.000 -0.9375
20.00 0.000 -3.750
6
0.000
3.750
10.00
0.000
-3.750
150.0
0.000
0.000
0.9375
0.000
-3.750
10.00
0.000
-3.750
20.00
4
5
4
5
150.0 0.000
0.000 0.9375
6
1*
0.000 -139.0
3.750 0.351
2*
-56.25
-0.869
3*
0.000
3.750
0.000
20.00
1.406
-3.750
10.00
1* -139.0 0.351 1.406

2* -56.25 -0.869 -3.750
129.0
51.82
3* 0.000
1.406
6 0.000
Stiffness matrix [k*]: [ k* ] = [ T ]T[ k ][T]
[k*]1
= 103
3.750
3.750
10.00
51.82
1.406
21.90 -3.476
-3.476
20.00
135
6
40 kN
2*
4
4m
4m
22.02 o
20 kN
x
x*
40 kN
40 kNm
Global Equilibrium:
3*
1*
[ qF ]
40 kNm
20sin22.02=7.5
20cos22.02=18.54 20 kN
[Q] = [K][D] + [QF]

1*
Q1 = 0.0
Q3 = 0.0
D1*
D3*
1*
103 3*
129
1.406
3*
1.406
D1*
20.0
D3*
36.37x10-6
0.002
rad
-7.5
-40
136
6
40 kN
2*
4
4m
4m
Member Force :
22.02
[q] = [k*][D] +
[qF]
40 kNm
q4
q5
4
5
q6
0.000 3.750 20.00 1.406 -3.750 10.00
q1*
q2*
q3*
x
x*
40 kNm
18.54
20 kN
4
5
6
1*
2*
3*
150.0 0.000 0.000 -139.0 -56.25 0.000
0.000 0.9375 3.750 0.351 -0.869 3.750
103
40 kN
3*
1*
7.5
0
0
0
20
-5.06
27.5
40.0
60
1* -139.0 0.351 1.406 129.0 51.82 1.406 36.4x10-6

0
2* -56.25 -0.869 -3.750 51.82 21.90 -3.476
-3
2x10
*
3 0.000 3.750 10.00 1.406 -3.476 20.00
-7.54
18.54
-40.0
0
13.48
0
40 kN
60 kNm
5.06 kN
27.5 kN
13.48 kN
137
40 kN
40 kN
4m
4m
60.05 kNm
22.02 o
5.05 kN
27.51 kN
13.47 kN
50.04
+
-60.05
Bending moment diagram
(kNm)
0.002 rad
Deflected shape
138
Example 13
For the beam shown:
(c) Draw the quantitative bending moment diagrams
Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members.
Include axial deformation in the stiffness matrix.
40 kN
6 kN/m
2EI, 2AE
8m
EI, AE
4m
22.02o
4m
139
40 kN
6 kN/m
EI, AE
2EI, 2AE
8m
4m
22.02 o
AE (0.006 m 2 )(200 106 kN / m 2 )
=
L
(8 m)
4m
= 150 103 kN m
8*
9*
7*
Global
3
5
1
Local
= 20 103 kN m
2 EI
= 10 103 kN m
L
4 EI 4(200 106 kN / m 2 )(0.0002 m 4 )

=
L
(8 m)
6 6 EI 6(200 106 kN / m 2 )(0.0002 m 4 )

=
2
L
(8 m) 2
4
= 3.75 103 kN
12 EI 12(200 106 kN / m 2 )(0.0002 m 4 )
=
L2
(8 m) 3
= 0.9375 103 kN / m
140
8*
9*
7*
Global
Local : Member 1
6
[ q]
4
2
1
[q] = [q]
Thus, [k] = [k]
6
[ q]
4
5
4
5
6
1
2
0.000
150.0 0.000 0.000 -150.0
0.000 0.9375 3.750 0.000 -0.9375
0.000
3
0.000
3.750
20.00
0.000
-3.750
10.00
-150.0 0.000 0.000

2 0.000 -0.9375 -3.750
3 0.000 3.750 10.00
150.0
0.000
0.000
0.9375
0.000
-3.750
0.000
-3.750
20.00
[k]1 = [k]1 = 2x103 1
3.750
141
2
1
ix = cos 0o = 1,
iy = cos 90o = 0
9*
7*
8*
q* ]
8*
3
2
9*
7*
[ q ]
6
4
x*
jx = cos 22.02o = 0.9271,
jy = cos 67.98o = 0.3749
Member 2: Use transformation matrix, [q*] = [T]T[q]

1
q1
q2
1
2
1
0
0
1
0
0
0
0
0
0
0
0
q1
q2
q3
q3
7*
8*
0
0
0
0
0
0
0
0
q4
q5
9*
q6
q7*
q8*
q9*
0.9271 -0.3749
0.3749 0.9271
0
142
3
[ q* ]
1
8*
9*
7*
1
2
1 150.0 0.000
2 0.000 0.9375
3 0.000 3.750
103 4 -150.0 0.000
5 0.000 -0.9375
3
4
5
0.000
0.000 -150.0
3.750 0.000 -0.9375
20.00 0.000 -3.750
6
0.000
3.750
10.00
0.000
-3.750
150.0
0.000
0.000
0.9375
0.000
-3.750
3.750
10.00
0.000
-3.750
20.00
1
2
1
2
150.0 0.000
0.000 0.9375
3
7*
0.000 -139.0
3.750 0.351
8*
-56.25
-0.869
9*
0.000
3.750
0.000
20.00
1.406
-3.477
10.00
7* -139.0 0.351 1.406

8* -56.25 -0.869 -3.477
129.0
51.82
9* 0.000
1.406
6 0.000
x*
Stiffness matrix [k]:
[k]2 =
[ q ]
Stiffness matrix [k*]: [k*] = [T]T[k][T]
[k*]2 = 103
3.750
3.750
10.00
51.82
1.406
21.90 -3.476
-3.476
20.00
143
2
4
9*
7*
4
5
4
5
6
1
2
0.000
150.0 0.000 0.000 -150.0
0.000 0.9375 3.750 0.000 -0.9375
3
0.000
3.750
0.000
20.00
0.000
-3.750
10.00
-150.0 0.000 0.000

2 0.000 -0.9375 -3.750
3 0.000 3.750 10.00
150.0
0.000
0.000
0.9375
0.000
-3.750
0.000
-3.750
20.00
1
2
1
2
150.0 0.000
0.000 0.9375
3
7*
0.000 -139.0
3.750 0.351
8*
-56.25
-0.869
9*
0.000
3.750
0.000
20.00
1.406
-3.477
10.00
7* -139.0 0.351 1.406

8* -56.25 -0.869 -3.477
9* 0.000 3.750 10.00
129.0
51.82
[k]1= 2x103 1
[k*]2 = 103
8*
3.750
3.750
51.82
1.406
1.406
21.90 -3.476
-3.476
20.00
144
40 kN
6 kN/m
2
4
8*
2
40 kN
6 kN/m
32 kNm
9*
40 kNm
32 kNm
40 kNm
24 kN
24 kN
Global:
1
3
= 103 7*
0
0
0
0
9*
450
0
-139
0
0
60
1.406
10
7*
-139
1.406
129
1.406
D1
18.15x10-6
D3
-509.84x10-6
rad
58.73x10-6
0.00225
rad
D7*
D9*
18.54
20 kN
7*
7.5
9*
0
10
1.406
20
D1
D3
D7*
D9*
0
-32 + 40 = 8
-7.5
-40
145
2
4
32 kNm
q4
q5
4
5
4
5
6
1
2
0.000
150.0 0.000 0.000 -150.0
0.000 0.9375 3.750 0.000 -0.9375
q6
0.000
q1
q2
q3
= 2x103
20.00
0.000
-3.750
1 -150.0 0.000 0.000

2 0.000 -0.9375 -3.750
3 0.000 3.750 10.00
150.0
0.000
0.000
0.9375
0.000
-3.750
q4
q5
-5.45
20.18
kN
kN
q6
21.80
kNm
5.45
27.82
kN
kN
-52.39
kNm
q1
q2
q3
24 kN
24 kN
Member 1: [ q ] = [k ][d] + [qF]
3.750
32 kNm
1
6 kN/m
21.80 kNm
3
0.000 0
3.750 0
10.00 0
0
24
0.000 d1=18.15x10-6
-3.750 0
20.00 d3=-509.84
x10-6
6 kN/m
32
0
24
-32
52.39 kNm
5.45 kN
5.45 kN
20.18 kN
27.82 kN
146
3
1
40 kN
8*
9*
7*
40 kNm
x
20 kN
Member 2: [ q ] = [k ][d] + [qF]

1
2
3
7*
q1
1 150.0 0.000 0.000 -139.0
q2
2 0.000 0.9375 3.750 0.351
q3
3 0.000 3.750 20.00 1.406
q7* = 103 7* -139.0 0.351 1.406 129.0
q8*
8* -56.25 -0.869 -3.750 51.82
q9*
9* 0.000 3.750 10.00 1.406
-5.45
26.55
kN
kN
q3
52.39
kNm
0
14.51
kN
kN
kNm
q7*
q8*
q9*
20sin22.02=7.5
20cos22.02=18.54 20 kN
2
x*
q1
q2
40 kNm
qF ]
8*
-56.25
-0.869
9*
0.000
3.750
18.15x10-6
0
0
20
-3.750
10.00
-509.84x10-6
40
1.406
21.90 -3.476
51.82
-3.476
20.00
58.73x10-6
0
0.00225
-7.5
18.54
-40
40 kN
52.39 kNm
5.45 kN
26.55 kN
14.51 kN
147
6 kN/m
21.80 kNm
52.39 kNm
52.39 kNm
40 kN
5.45 kN
5.45 kN
5.45 kN
26.55 kN
27.82 kN
20.18 kN
40 kN
6 kN/m
21.80 kNm
14.51 kN
14.51cos 22.02o
=13.45 kN
5.45 kN
20.18 kN
V (kN)
54.37 kN
4m
8m
26.55
20.18
+
3.36 m
4 m 14.51 kN
22.02o
+
-
x (m)
-13.45
-27.82
53.81
12.14
M
(kNm) -21.8
+
-52.39
x (m)
148

06 Matrix Beam

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BEAM ANALYSIS USING THE STIFFNESS

Development: The Slope-Deflection

Slope Deflection Equations

[D] = [K]-1([Q] - [FEM])

Stiffness Coefficients Derivation

From (1) and (2);

Derivation of Fixed-End Moment

Substitute B in MAB, MBA, MBC, MCB

Beam-Member Stiffness Matrix

Member Equilibrium Equations

Fixed-end force matrix

[q] = [k][d] + [qF]

6x6 Stiffness Matrix

4x4 Stiffness Matrix

2x2 Stiffness Matrix

General Procedures: Application of the Stiffness Method for Beam Analysis

[k] = [T]T[q] [T]

[k] = [T]T[q] [T]

[Du ] = [K AA ]1 + ([Qk ] [QAF ])

Member Force : [q ] = [k ][d ] + q F

= 6.75 PL/8 = 3.75

Equilibrium equations: MCB = 0

Substitute B and C in the member matrix,

[q]1 = [k]1[d]1 + [qF]1

[q]2 = [k]2[d]2 + [qF]2

-0.375EI -0.75EI 0.375EI -0.75EI

0.75EI - 0.375EI 0.75EI

Use 4x4 stiffness matrix,

0.1875EI 0.375EI - 0.1875EI 0.375EI

-0.1875EI -0.375EI 0.1875EI -0.375EI

[q]1 = [k]1[d]1 + [qF]1

12(2EI)/430.75EI - 0.375EI 0.75EI

-0.375EI -0.75EI 0.375EI -0.75EI

0.1875EI 0.375EI - 0.1875EI 0.375EI

-0.1875EI -0.375EI 0.1875EI -0.375EI

-0.375EI -0.75EI 0.375EI -0.75EI

12(2EI)/43 0.75EI - 0.375EI 0.75EI

0.375EI - 0.1875EI 0.375EI

-0.1875EI -0.375EI 0.1875EI -0.375EI

Global: [Q] = [K][D] + [QF]

[q]1 = [k]1[d]1 + [qF]1

0.375EI 0.75EI - 0.375EI 0.75EI

-0.375EI -0.75EI 0.375EI -0.75EI B =-116.593/EI

[q]1 = [k]1[d]1 + [qF]1

0.375EI - 0.1875EI 0.375EI

-0.1875EI -0.375EI 0.1875EI -0.375EI

6(EI)/L2 = 37.5 kNm

[Q] = [K][D] +2[QF]

-32 + 40 + 75 -37.5 = 45.5

[q]1 = [k]1[d]1 + [qF]1

6(EI)D/L2 = 37.5 kNm

[q]2 = [k]2[d]2 + [qF]2

-40 - 37.5 = -77.5

[q]1 = [k]1[d]1 + [qF]1

Global: [Q] = [K][D] + [QF]

[q]1 = [k]1[d]1 + [qF]1

BR= -0.0015 rad

BL= 0.0006 rad

0.75EI - 0.375EI 0.75EI

0.375EI - 0.1875EI 0.375EI

Stiffness matrix [k]: [ k ] = [ T ]T[ k ][T]

Stiffness matrix [k]: [k] = [T]T[k][T]