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
1
Slope – Deflection Equations
P
i
w
j
k
Cj
settlement = ∆j
Mij
P
i
w
j
Mji
θi
ψ
θj
2
• Degrees of Freedom
M
θΑ
A
B
1 DOF: θΑ
C
2 DOF: θΑ , θΒ
L
θΑ
A
P
B
θΒ
3
• Stiffness Definition
kAA
kBA
1
B
A
L
k AA =
4 EI
L
k BA =
2 EI
L
4
kBB
kAB
A
B
1
L
k BB =
4 EI
L
k AB =
2 EI
L
5
• 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
L
wL2
12
wL
2
6
• General Case
P
i
w
j
k
Cj
settlement = ∆j
Mij
P
i
w
j
Mji
θi
ψ
θj
7
Mij
P
i
j
w
Mji
θi
L
θj
ψ
4 EI
2 EI
θi +
θj = M
ij
L
L
Mji =
2 EI
4 EI
θi +
θj
L
L
θj
θi
(MFij)∆
+
(MFji)∆
settlement = ∆j
+
P
(MFij)Load
M ij = (
settlement = ∆j
w
(MFji)Load
4 EI
2 EI
2 EI
4 EI
)θ i + (
)θ j + ( M F ij ) ∆ + ( M F ij ) Load , M ji = (
)θ i + (
)θ j + ( M F ji ) ∆ + ( M F ji ) Load 8
L
L
L
L
• Equilibrium Equations
i
P
w
j
k
Cj
Cj M
Mji
Mji
jk
Mjk
j
+ ΣM j = 0 : − M ji − M jk + C j = 0
9
• Stiffness Coefficients
Mij
i
j
Mji
L
θj
θi
kii =
4 EI
L
k ji =
2 EI
L
×θ i
k jj =
4 EI
L
×θ j
1
kij =
2 EI
L
+
1
10
• Matrix Formulation
M ij = (
4 EI
2 EI
)θ i + (
)θ j + ( M F ij )
L
L
M ji = (
2 EI
4 EI
)θ i + (
)θ j + ( M F ji )
L
L
 M ij  (4 EI / L) ( 2 EI / L) θ iI   M ij F 
M  = 
 θ  +  M F 
EI
L
EI
L
(
2
/
)
(
4
/
)
  j   ji 
 ji  
 kii
k ji
[k ] = 
kij 
k jj 
Stiffness Matrix
11
P
i
Mij
w
j
Mji
θi
L
[ M ] = [ K ][θ ] + [ FEM ]
θj
ψ
∆j
([ M ] − [ FEM ]) = [ K ][θ ]
[θ ] = [ K ]−1[ M ] − [ FEM ]
Mij
Mji
θj
θi
Fixed-end moment
Stiffness matrix matrix
+
(MFij)∆
(MFji)∆
[D] = [K]-1([Q] - [FEM])
+
(MFij)Load
P
w
(MFji)Load
Displacement
matrix
Force matrix
12
• Stiffness Coefficients Derivation
Mj
θi
Mi
Real beam
i
j
L
Mi + M j
L
Mi + M j
L/3
M jL
2 EI
L
Mj
EI
Conjugate beam
Mi
EI
θι
MiL
2 EI
M j L 2L
MiL L
)( ) + (
)( ) = 0
2 EI 3
2 EI 3
M i = 2 M j − − − (1)
+ ΣM 'i = 0 : − (
+ ↑ ΣFy = 0 : θ i − (
M L
MiL
) + ( j ) = 0 − − − (2)
2 EI
2 EI
From (1) and (2);
4 EI
)θ i
L
2 EI
)θ i
Mj =(
L
Mi = (
13
• Derivation of Fixed-End Moment
Point load
P Real beam
Conjugate beam
A
A
B
B
L
M
EI
M
M
EI
M
P
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
+ ↑ ΣFy = 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
A
L
B
B
M
EI
M
M
M
EI
M
EI
ML
2 EI
ML
2 EI
w
wL3
24 EI
wL2
8 EI
M
EI
wL3
24 EI
wL2
ML ML 2 wL3
+ ↑ ΣFy = 0 : −
−
+
= 0, M =
2 EI 2 EI 24 EI
12 16
Settlements
Mi = Mj
Real beam
Mj
Conjugate beam
L
Mi + M j
M
L
A
M
EI
B
∆
∆
M
EI
Mi + M j
L
M
EI
ML
2 EI
ML
2 EI
M
M
EI
∆
+ ΣM 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
P
PL
8
PL
8
L2
wL2
12
L
0
PL
4 EI
2 EI
M AB =
θA +
θB + 0 + 1 1
8
L1
L1
0 EI
PL
2 EI
4
M BA =
θA +
θB + 0 − 1 1
8
L1
L1
0
2
P2 L2 wL2
4 EI
2 EI
M BC =
θB +
θC + 0 +
+
L2
L2
8
12
0
2
− P2 L2 wL2
2 EI
4 EI
θB +
θC + 0 +
−
M CB =
8
12
L2
L2
w
wL2
12
L
18
CB
w
P1
P2
C
A
B
L1
L2
CB M
BC
MBA
B
M BA =
M BC
PL
2 EI
4 EI
θA +
θB + 0 − 1 1
8
L1
L1
P L wL
4 EI
2 EI
=
θB +
θC + 0 + 2 2 + 2
L2
L2
8
12
2
+ ΣM B = 0 : C B − M BA − M BC = 0 → Solve for θ B
19
P1
MAB
CB
w
MBA
A
L1
B
P2
MBC
L2
C M
CB
Substitute θB in MAB, MBA, MBC, MCB
0
PL
4 EI
2 EI
θA +
θB + 0 + 1 1
M AB =
L1
L1
8
0 EI
PL
2 EI
4
M BA =
θA +
θB + 0 − 1 1
8
L1
L1
0
2
4 EI
2 EI
P2 L2 wL2
θB +
θC + 0 +
+
M BC =
8
12
L2
L2
0
2
− P2 L2 wL2
2 EI
4 EI
θB +
θC + 0 +
−
M CB =
L2
L2
8
12
20
P1
MAB
A
Ay
CB
w
MBA
B
L1
P2
MCB
MBC
L2
Cy
C
By = ByL + ByR
B
P1
MAB
B
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
6
9
5
14
2EI
1
3
2
21
EI
2
8
3
7
22
Beam-Member Stiffness Matrix
3
1
i
j
2
E, I, A, L
6
5
k14
AE/L
k41
AE/L
k11 = AE/L
4
AE/L = k44
d1 = 1
d4 = 1
1
[k] =
2
3
4
1
AE/L
- AE/L
2
0
0
3
0
0
4
-AE/L
AE/L
5
0
0
6
0
0
5
6
23
3
i
1
6EI/L2 = k32
2
[k] =
E, I, A, L
k62 = 6EI/L2
d2 = 1
k22 =
j
4
6
6EI/L2 = k65
5
6EI/L2 = k35
d5 = 1
12EI/L3 = k52
12EI/L3
1
2
1
AE/L
2
3
12EI/L3 = k25
4
5
0
- AE/L
0
0
12EI/L3
0
- 12EI/L3
3
0
6EI/L2
0
- 6EI/L2
4
-AE/L
0
AE/L
0
5
0
-12EI/L3
0
12EI/L3
6
0
6EI/L2
0
- 6EI/L2
12EI/L3 = k55
6
24
3
i
1
j
2
k33 = 4EI/L
d3 = 1
E, I, A, L
2EI/L = k63
4
6
5
4EI/L = k66
2EI/L = k36
d6 = 1
k23 = 6EI/L2
[k] =
6EI/L2 = k53
k26 = 6EI/L2
6EI/L2 = k56
1
2
3
4
5
6
1
AE/L
0
0
- AE/L
0
0
2
0
12EI/L3
6EI/L2
0
- 12EI/L3
6EI/L2
3
0
6EI/L2
4EI/L
0
- 6EI/L2
2EI/L
4
-AE/L
0
0
AE/L
0
0
5
0
-12EI/L3
-6EI/L2
0
12EI/L3
-6EI/L2
6
0
6EI/L2
2EI/L
0
- 6EI/L2
4EI/L
25
• Member Equilibrium Equations
Mi
Fxi
i
j
E, I, A, L
AE/L
AE/L x δ
i
1
Mj
Fyj
AE/L
AE/L
+
6EI/L2
6EI/L2
1
6EI/L2
1
x ∆i
12EI/L3
12EI/L3
+
+
12EI/L3
4EI/L
x δj
1
+
6EI/L2
=
Fyi
Fxj
2EI/L
1
x ∆j
12EI/L3
4EI/L
2EI/L
x θi
x θj
1
6EI/L2
MFi
FFxi
FFyi
+
6EI/L2
6EI/L2
FFyj
FFxj
6EI/L2
MFj
26
Fxi = ( AE / L)δ i +
Fyi =
(0)δ i +
(0)δ i
Mxi =
Fxj = (−AE / L)δ i
Fyj =
(0)δ i
Mj =
(0)δ i
Fxi 
 
Fyj 
Mi 
 =
Fxj 
Fyj 
 
M j 
(0) ∆i
(0)θ i +
(−AE / L)δ j +
(0) ∆j +
(0)θ j +
FxiF
(12EI / L3 ) ∆i
(6EI / L2 )θ i
(0)δ j
(−12EI / L3 ) ∆j
(6EI / L2 )θ j
FyiF
(6EI / L2 ) ∆i
(0) ∆i
(4EI / L)θ i
(0)θ i
(0)δ j
( AE / L)δ j
(−6EI / L2 ) ∆j
(0) ∆j
(2EI / L)θ j
(0)θ j
MiF
FxiF
(0)δ j
(0) ∆j
(−6EI / L2 )θ j
FyjF
(0)δ j
(−6EI / L2 ) ∆j
(4EI / L)θ j
MjF
(−12EI / L3 ) ∆i (−6EI / L2 )θ i
(6EI / L2 ) ∆i
(2EI / L)θ 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
 δ i   Fxi 

  
6EI/ L2   ∆i   FyiF 
2EI/ L  θ i  MiF 
  +  F 
0
 δ j   Fxj 
− 6EI/ L2   ∆ j   FyiF 
   F 
4 EI/ L  θ j  M j 
Fixed-end force matrix
[q] = [k][d] + [qF]
End-force matrix
Displacement matrix
27
6x6 Stiffness Matrix
δi
[k ]6×6 =
∆i
θi
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
δj
∆j
− AE/ L
0
− 12 EI/ L3
0
0
6EI/ L2
4 EI/ L
0
− 6EI/ L2
0
AE/ L
0
− 6EI/ L2
0
12 EI/ L3
2EI/ L
0
− 6EI/ L2
θj
0

6EI/ L2 
2EI/ L 

0

− 6EI/ L2 

4EI/ L 
4x4 Stiffness Matrix
∆i
[k ]4×4
=
Vi  12 EI/ L3

Mi  6EI/ L2
Vj − 12 EI/ L3

2
Mj  6EI/ L
θi
6EI/ L2
4 EI/ L
− 6EI/ L2
2EI/ L
∆j
− 12 EI/ L3
− 6EI/ L2
12 EI/ L3
− 6EI/ L2
θj
6EI/ L2 

2EI/ L 
− 6EI/ L2 

4 EI/ L 
28
2x2 Stiffness Matrix
θi
[k ]2×2 =
Mi
Mj
θj
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
1
4
1
6
3
2
2
1
3
2´
Local
w
4´
5
2´
1
1´
4´
2
3´
1´
3´
30
P2y
P
M2
M1
2
Global
1
4
1
6
3
2
2
1
3
2
Member
w
4
5
4
1
6
2
1
3
3
5
2´
4´
2´
4´
Local
1
1´
2
3´
1´
3´
31
P2y
P
M2
M1
2
4
1
Global
w
1
3
2
2
1
6
3
2´
4´
Local
5
2´
1
1´
4´
2
3´
1´
3´
P
(q´F
2)1
[FEF]
1
(q´F1 )1
w
(q´F1)2
(q´F4 )2
(q´F4)1
(q´F3)1
2
(q´F2)2
(q´F3 )2
32
2
Global
4
1
1
3
2´
3
2
2
1
6
5
4´
Local
1
1´
3´
[q] = [T]T[q´]
q1 
q 
 2 =
q3 
 
 q4  1
1
2
3
4
1´
1
0

0

0
2´
0
1
0
0
3´
0
0
1
0
4´
0
0
0

1
q1' 
q 
 2' 
q3' 
 
 q4 ' 
[k] = [T]T[q´] [T]
33
2
Global
4
1
1
6
2
1
3
2
3
5
2´
Local
4´
2
1´
3´
[q] = [T]T[q´]
q3 
q 
 4 =
q5 
 
q6  2
3
4
5
6
1´
1
0

0

0
2´
0
1
0
0
3´
0
0
1
0
4´
0
0
0

1
q1' 
q 
 2' 
q3' 
 
 q4 ' 
[k] = [T]T[q´] [T]
34
2
4
1
1
6
2
1
3
2
3
5
Stiffness Matrix:
1
2
3
4
1
[k]1 =
1
2
1
3
2
4
3
3
[k]2 =
4
4
5
6
[K] =
3
4
5
6
2
3
4
5
6
Member 1
Node 2
Member 2
5
6
35
2
1
Joint Load
Qk
1
6
2
2
1
M
Q 2 1z
 -P2y
Q 3 
Q 4 M3z

=
Q
 1 
Q 5 


Q
 6 
Reaction
Qu
4
3
3
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
2
Global:
4
1
1
3
5
(qF2)1
(qF
[FEF]






5 

6 
1
2
3
4
2
3
5
6
Member 1
 Q2 = M 1 
2
Q = − P 

2y  = 3
 3

 Q4 = M 2 
4 
(q´F6 )2
(qF3)2
4
Node 2
Member 2
2
3
4



Node 2 
w
2
(qF3)1
(qF1 )1
Q1 
 M
Q 2 1
Q 3 -P2y
 M =
Q 4 2
Q 5 


Q
 6 
4)2
(qF4)1
1
1
3
2
2
P
1
6









D1 0


D
 2 
D3 

+
D
 4 
D5 0
 0
D6 
D2  Q 2F 
D   F 
 3  + Q 3 
D4  Q 4F 
(q´F5 )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


6
37
2
1
4
1
3
2
2
D2 
 

D
3
=
3
 

D4 
4 
3
3
2
2
1
6
5
4



Node 2 
  Q2 = M 1  Q2F  

 − Q F  
Q
P
=
−
2y 
 3
 3 
  Q = M  Q F  
2 
 4 
 4
38
2
4
1
1
6
2
1
3
2
3
5
P
qF2
qF4
1
qF1
[FEF]
qF3
Member 1:
1
q 1 
 
2
q 2  =
q 3 
3
 
4
q 4 
1






2
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
2
4
1
1
6
2
1
3
2
3
5
w
qF6
qF4
2
qF3
[FEM]
qF5
Member 2:
 q3 
q 
 4
q5 
 
q6 
1
=
2
3
4
1






2
3
k1
4






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
B
A
1.5 m
3m
9m
3
2
Global
1
2
3
2
Members
1
1
2
2
1
10 kN
1 kN/m
1.5 m
[FEF]
wL2/12
= 6.75
9m
1
wL2/12
= 6.75 PL/8 = 3.75
1.5 m
2
PL/8 = 3.75
42
3
2
1
2
9m
3m
Stiffness Matrix:
θi
Mi
Mj
[k ]2×2 =
[k]1= EI
1
4 EI / L 2EI / L 


2EI / L 4 EI / L
3
2
4/9
2/9
3
2/9
4/9
2
[k]2 = EI
2
[K] = EI
θj
2
1
4/3
2/3
2
2/3
4/3
1
1
(4/9)+(4/3) 2/3
2
2/3
1
4/3
43
3
2
10 kN
1
1 kN/m
A
6.75
6.75
9m
B
3.75
1.5 m 1.5 m
C
Equilibrium equations: MCB = 0
MBA + MBC = 0
Global Equilibrium: [Q] = [K][D] + [QF]
2
2
MBA + MBC = 0
1
MCB = 0
θB
θC
=
= EI
1
2
(4/9)+(4/3) 2/3
θB
1
2/3
θC
4/3
+
-6.75 + 3.75 = -3
-3.75
0.779/EI
2.423/EI
44
3
2
1
1 kN/m
[FEF]
9m
wL2/12 = 6.75
1
wL2/12 = 6.75
Substitute θB and θC in the member matrix,
Member 1
3
MAB
2
MBA
:
= EI
[q]1 = [k]1[d]1 + [qF]1
3
2
0
4/9
2/9
θA
2/9
4/9
θB= 0.779/EI
6.75
+
-6.75
1 kN/m
4.56 kN
=
-6.40
6.40 kN•m
6.92 kN•m
9m
6.92
1
4.44 kN
45
10 kN
1
2
1.5 m
2
PL/8 = 3.75
1.5 m
2
PL/8 = 3.75
[FEF]
Substitute θB and θC in the member matrix,
Member 2 :
2
MBC
1
MCB
= EI
[q]2 = [k]2[d]2 + [qF]2
2
1
4/3
2/3
θB = 0.779/EI
2/3
4/3
θC = 2.423/EI
+
3.75
-3.75
6.40
=
0
10 kN
6.40 kN•m
2
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 kN•m
7.13 kN
θB = +0.779/EI
10 kN
2.87 kN
θC = +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
(kN•m)
-6.92
-2.87
4.32
3.48
x (m)
-6.40
47
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.
20 kN
9kN/m
40 kN•m
EI
2EI
A
B
4m
C
4m
48
2
4
1
3
2EI
1
1
[k ] =
Vi  12 EI/ L3
Mi 
2
6
/
EI
L

Vj 
− 12 EI/ L3
Mj  6EI/ L2

EI
θi
∆j
2
4 EI/ L
− 6EI/ L
2EI/ L
2
− 12 EI/ L
− 6EI/ L2
12 EI/ L3
− 6EI/ L2
6EI/ L 

2EI/ L 
− 6EI/ L2 

4 EI/ L 
1
0.375EI
2
0.75EI
3
4
2
3
4
-0.75EI
EI
-0.375EI -0.75EI 0.375EI -0.75EI
0.75EI
EI
4
3
0.5625
-0.375
[K] = EI
4
-0.375
3
[k]2
0.75EI - 0.375EI 0.75EI
2EI
3
2
[k]1
1
3
θj
3
6EI/ L
5
2
2
Use 4x4 stiffness matrix,
∆i
6
-0.75EI 2EI
3
3
4
5
6
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 kN•m
18 kN
4m
40 kN•m
12 kN•m
18 kN
2
4
1
Global:
4
Q4 = 40
D3
D4
= EI
=
6
3
2EI
1
1
[Q] = [K][D] + [QF]
3 Q3 = -20
4m
2
2
3
4
3
0.5625
-0.375
4
-0.375
3
5
EI
D3
D4
+
3
18
3
-12
4
-61.09/EI
9.697/EI
50
2
4
1
3
2EI
1
1
9 kN/m
12 kN•m
A
B
1
2
[qF]1
18 kN
Member 1:
12 kN•m
18 kN
[q]1 = [k]1[d]1 + [qF]1
1
q1
1
q2
2
q3L
3
q4L
4
2
3
4
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 kN•m
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 kN•m
B
12.18 kN
51
4
6
3
5
EI
2
2
3
Member 2:
[q]2 = [k]2[d]2 + [qF]2
3
q3R
3
q4R
4
q5
5
q6
6
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
7.82 kN
0
-7.818
d4 = 9.697/EI
0
-13.21
d5 = 0
0
7.818
d6 = 0
0
-18.06
18.06 kN•m
13.21 kN•m
B
EI
d3 =-61.09/EI
2
[q]2
C
7.82 kN
52
9 kN/m
A
67.51 kN•m
48.18 kN
53.21 kN•m
B
B
1
12.18 kN
[q]1
18.06 kN•m
13.21 kN•m
C
2
7.82 kN
7.82 kN
[q]2
20 kN
9kN/m
40 kN•m
D3 = ∆B = -61.09/EI
67.51 kN•m
48.18 kN
48.18
4m
D4 = θB = +9.697/EI 4 m
+
V (kN)
18.06 kN•m
7.818 kN
12.18
x (m)
M
(kN•m)
+
-
-67.51
-7.818
-7.818
53.21
13.21
-
x (m)
-18.08
53
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.
20 kN
9kN/m
10 kN
40 kN•m
EI
2EI
A
B
4m
C
2m
2m
54
20 kN
9kN/m
A
1
3
1
∆i
1
Vi  12 EI/ L3
Mi 
2
EI
L
6
/

Vj 
− 12 EI/ L3
Mj  6EI/ L2

5
3
10 kN
12 kN•m
18 kN
6
2
2
9 kN/m
[FEF]
C
EI
2m
2m
4
12 kN•m
=
B
2
1
[k ]4×4
40 kN•m
2EI
4m
Global
10 kN
θi
∆j
6EI/ L2
− 12 EI/ L3
− 6EI/ L2
12 EI/ L3
− 6EI/ L2
− 6EI/ L2
2EI/ L
5 kN•m
2
18 kN
4 EI/ L
5 kN•m
5 kN
θj
6EI/ L2 

2EI/ L 
− 6EI/ L2 

4 EI/ L 
5 kN
55
2
4
1
3
1
1
1
1
[k]1 =
2
3
4
0.75EI
0.75EI
0.1875EI
4
0.375EI
6
3
2m
2m
3
4
2EI
-0.75EI
EI
3
-0.375EI -0.75EI 0.375EI -0.75EI
3
5
2
12(2EI)/43 0.75EI - 0.375EI 0.75EI
3
[k]2 =
5
2
4m
2
6
EI
4
-0.75EI 2EI/L
5
[K] = EI
6
4
6
3
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 kN•m
A
C
6
2
B4
1
3
1
1
5
2
2
10 kN
9 kN/m
12 kN•m
12 kN•m
[FEF]
1
18 kN
MBA+MBC = 40
= EI
MCB = 0
∆B
θB
θC
3
5 kN•m
2
18 kN
Global: [Q] = [K][D] + [QF]
VBL+VBR = -20
5 kN•m
3
4
5 kN
5 kN
6
3
0.5625
-0.375
0.375
∆B
4
6
-0.375
0.375
3
0.5
0.5
1
θB
θC
+
18 + 5 = 23
-12 + 5 = -7
-5
-116.593/EI
=
-7.667/EI
52.556/EI
57
2
4
1
3
1
1
9 kN/m
12 kN•m
2
12 kN•m
18 kN
1
18 kN
[FEF]
[q]1 = [k]1[d]1 + [qF]1
Member 1:
1
2
3
4
VA
1
0.375EI 0.75EI - 0.375EI 0.75EI
0
18
55.97
MAB
2
0.75EI
0
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
θB =-7.667/EI
+
18
-12
=
-19.97
60.11
9kN/m
91.78 kN•m
4m
55.97 kN
1
60.11 kN•m
19.97 kN
58
4
6
3
10 kN
5
3
2
2
2
5 kN
Member 2:
VBR
3
0.1875EI
MBC
4
0.375EI
MCB
=
5
6
[FEM]
5 kN
[q]1 = [k]1[d]1 + [qF]1
3
VC
5 kN•m
5 kN•m
4
5
6
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
5
-0.0278
θB =-7.667/EI
5
-20.11
0
θC =52.556/EI
+
5
-5
=
10.03
0
10 kN
20.11 kN•m
2m
2m
2
0.0278 kN
10.03 kN
59
9kN/m
91.78 kN•m
10 kN
20.11 kN•m
2m
1
4m
19.97 kN
55.97 kN
60.11 kN•m
20 kN
9kN/m
2m
2
0.0278 kN
10.03 kN
10 kN
θC = 52.556/EI
40 kN•m
91.78 kN•m
55.97 kN
θB = -7.667/EI
10.03 kN
∆B = -116.593/EI
4m
55.97
2m
2m
19.97
+
V (kN)
60.11
M
(kN•m)
-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
C
4m
61
1
2
3
2EI
EI
2
1
θi
Use 2x2 stiffness matrix:
[k]1 =
EI
8
[k ]2×2 =
1
2
1
8
4
2
4
8
[K] =
EI
8
Mi
Mj
θj
4 EI / L 2EI / L 


2EI / L 4 EI / L
[k]2 =
2
3
2
12
2
3
2
4
EI
8
2
3
2
4
2
3
2
4
62
40 kN
6 kN/m
1
B
A
2
C
EI
∆B = -10 mm
4m
4m
2EI
8m
2
1
40 kN
6 kN/m
[FEM]load
wL2/12 = 32 kN•m
1
32 kN•m
10 mm
6(2EI)∆/L2 = 75 kN•m
Global:
Q2 = 0
Q3 = 0
D2
D3
2
2
12
2
D2
3
2
4
D3
=
185.64/EI
37.5 kN•m
6(EI)∆/L2 = 37.5 kN•m
3
-61.27/EI
40 kN•m
10 mm
[Q] = [K][D] +2[QF]
EI
=
8
2
PL/8 = 40 kN•m
75 kN•m
1
[FEM]∆
3
+
-32 + 40 + 75 -37.5 = 45.5
-40 - 37.5 = -77.5
rad
rad
63
1
2
6 kN/m
[FEM]load
2EI
wL2/12 = 32 kN•m
1
1
32 kN•m
75 kN•m
1
[FEM]∆
10 mm
6(2EI)∆/L2 = 75 kN•m
Member 1:
q1
q2
EI
=
8
[q]1 = [k]1[d]1 + [qF]1
1
2
1
8
4
d1 = 0
2
4
8
d2 = -61.27/EI
76.37 kN•m
6 kN/m
32 + 75 = 107
+
-32 + 75 = 43
76.37 kN•m
=
-18.27 kN•m
18.27 kN•m
8m 1
31.26 kN
16.74 kN
64
2
3
40 kN
[FEM]load
2
PL/8 = 40 kN•m
2
40 kN•m
37.5 kN•m
2
[FEM]∆
10 mm
6(EI)D/L2 = 37.5 kN•m
Member 2:
q2
q3
=
EI
8
[q]2 = [k]2[d]2 + [qF]2
2
3
2
4
2
d2 = -61.27/EI
3
2
4
d3 = 185.64/EI
40 - 37.5 = 2.5
+
-40 - 37.5 = -77.5
18.27 kN•m
=
0
kN•m
40 kN
18.27 kN•m
2
22.28 kN
17.72 kN
65
2
6
4
2EI
EI
2
1
1
Alternate method: Use 4x4 stiffness matrix
1
[K]
3
4
1.5
-0.375
1.5
8
-1.5
4
-1.5
0.375
-1.5
4
-1.5
2
1 12(2)/82
EI 2
1.5
=
8 3
-0.375
4
1.5
[k]1
5
3
3
=
3 12/82
4
0.75
EI
8 5
4
0.75
4
-0.1875 -0.75
6
8
3
4
0.75
5
2
1
2
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
6
0
0
0.75
2
-0.75
4
1.5
4
[k]2
5
6
-0.1875
0.75
-0.75
2
0.1875
-0.75
-0.75
4
6
0
0
0
0
-0.75 -0.1875
0.75
66
40 kN
6 kN/m
2
B
A
C
EI
∆B = -10 mm
2EI
8m
4m
1
32 kN•m
40 kN•m
Q6= 0
Q4 = 0
Q6= 0
D4
D6
5
D4
4
6
D6
12
2
D4
2
4
D6
12
EI 4
=
8 6
2
4
20 kN
20 kN
2
EI 4
=
8 6
=
40 kN•m
2
1
24 kN
24 kN
Global: [Q] = [K][D] + [QF]
4
6
Q4 = 0
5
3
40 kN
32 kN•m
[FEF]
2
1
4m
6 kN/m
6
4
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
2
4
1
32 kN•m
∆B = -10 mm
1
6 kN/m
32 kN•m
[FEF]load
3
1
24 kN
24 kN
Member 1:
[q]1 = [k]1[d]1 + [qF]1
1
1 12(2)/82
2
1.5
q1
q2
2
3
4
1.5
-0.375
1.5
d1 = 0
24
8
-1.5
4
d2 = 0
32
(200x200)
3
8
-0.375
-1.5
0.375
-1.5
q4
4
1.5
4
-1.5
8
q1
31.26
q3
q2
q3
q4
=
=
kN
76.37
kN•m
16.74
kN
-18.27
kN•m
76.37 kN•m
+
d3 = -0.01
-32
d4 = -1.532x10-3
6 kN/m
24
18.27 kN•m
8m 1
31.26 kN
16.74 kN
68
40 kN
6
4
40 kN•m
40 kN•m
-10 mm = ∆B
[FEF]
2
3
5
2
20 kN
20 kN
Member 2:
3
3 12/82
4
0.75
q3
q4
q5
=
4
(200x200)
5
8
6
q3
22.28
kN
q4
18.27
kN•m
17.72
kN
0
kN•m
q6
=
4
-0.1875 -0.75
q6
q5
0.75
0.75
2
5
6
-0.1875
-0.75
0.75
2
0.1875 -0.75
-0.75
4
d3 = -0.01
20
d4 = -1.532x10-3
40
d5 = 0
+
d6 = 4.641x10-3
20
-40
40 kN
18.27 kN•m
2
22.28 kN
17.72 kN
69
40 kN
6 kN/m
B
76.37 kN•m
EI
∆B = -10 mm
17.72 kN
16.74 + 22.28 kN
8m
4m
4m
A
2EI
31.26 kN
V (kN)
C
31.26
+
-
5.21 m
22.28
+
-16.74
-17.72
x (m)
70.85
M
(kN•m)
5.06
-
+
-
-18.27
x (m)
-76.37
Deflected
Curve
∆B = -10 mm
d4 = θB = -1.532x10-3 rad
D6 = θC = 4.641x10-3 rad
70
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 diagram
and qualitative deflected shape.
Take I = 200(106) mm4 and E = 200 GPa and support C settlement 10 mm.
4 kN
0.6 kN/m
20 kN•m
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
3
3
4
1.5
-0.375
1.5
8
-1.5
4
-1.5
0.375
-1.5
4
-1.5
2
6
4
3
EI
=
8
6
8
4
3 12/82
4
0.75
5
-10 mm
[k]2
0.75
4
-0.1875 -0.75
0.75
2
5
6
-0.1875
0.75
-0.75
2
0.1875
-0.75
-0.75
4
• Global: [Q] = [K][D] + [QF]
Q3
Q4
Q6
3
EI 4
=
8
6
3
4
6
0.5625
-0.75
-0.75
12
0.75
2
D3
D4
0.75
2
4
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 kN•m
B
A
EI
2EI
8m
4m
10 mm
4m
Q6 = 20
D3
D4
D6
4 kN•m
=
5
4 kN•m
2
1
3
EI 4
=
8
6
EI
3
4 kN
2.4 kN
2.4 kN
Global: [Q] = [K][D] + [QF]
Q3 = 0
Q4 = 0
6
2
1
1
3.2 kN•m
[FEF]
4
2EI
C
0.6 kN/m
3.2 kN•m
2
2 kN
3
4
6
0.5625
-0.75
-0.75
12
0.75
2
0.75
2
4
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
2
4
0.6 kN/m
3.2 kN•m
1
1
[FEF]load
3
3.2 kN•m
8m
1
2.4 kN
2.4 kN
Member 1:
[q]1 = [k]1[d]1 + [qF]1
1
q1
q2
q3
=
q4
4
1.5
-0.375
1.5
d1 = 0
2.4
8
-1.5
4
d2 = 0
3.2
-0.375
-1.5
0.375
-1.5
d3 = -9.433x10-3
4
1.5
4
-1.5
8
d4 = -1.538x10-3
q1
q3
3
200× 200
3
8
q4
q2
1 12(2)/82
2
1.5
2
8.55
=
kN
43.19
kN•m
-3.75
kN
6.03
kN•m
43.19 kN•m
0.6 kN/m
+
2.4
-3.2
6.03 kN•m
8m 1
8.55 kN
3.75 kN
74
4 kN
6
4
4 kN•m
4 kN•m
[FEF]
2
3
5
8m
10 mm
2
2 kN
2 kN
Member 2:
3
q3
q4
q5
200× 200
=
8
q6
4
3 12/82
4
0.75
5
6
0.75
3.75
kN
q4
-6.0
kN•m
0.25
kN
20.0
kN•m
q6
=
4
-0.1875 -0.75
q3
q5
0.75
2
5
6
-0.1875
-0.75
0.75
d3 = -9.433x10-3
2
2
d4 = -1.538x10-3
4
0.1875 -0.75
-0.75
4
d5 = -0.01
+
d6 = 1.863x10-3
2
-4
4 kN
6.0 kN•m
20 kN•m
2
3.75 kN
0.25 kN
75
4 kN
0.6 kN/m
20 kN•m
B
A
EI
2EI
8m
0.6 kN/m
43.19 kN•m
4m
Deflected
Curve
4 kN
20 kN•m
6.0 kN•m
3.75 kN
8.55
V (kN)
M
(kN•m)
10 mm
4m
6.03 kN•m
8m 1
8.55 kN
C
+
2
3.75
6
-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
For the beam shown, use the stiffness method to:
(a) Determine all the reactions at supports.
(b) Draw the quantitative shear and bending moment diagrams
and qualitative deflected shape.
E = 200 GPa, I = 50x10-6 m4.
30 kN
9kN/m
Hinge
EI
2EI
A
4m
B
C
2m
2m
77
3
4
2
2EI
A
1
1
4m
EI
B
2m
2m
θi
Use 2x2 stiffness matrix,
[k]1 =
[k ]2×2 =
3
1
3
2EI
EI
1
EI
2EI
[K] =
C
2
Mi
Mj
θj
4 EI / L 2EI / L 


2
EI
/
L
4
EI
/
L


2
[k]2 =
EI
2
1EI
4 0.5EI
1
2
1
2.0
0.0
2
0.0
1.0
4
0.5EI
1EI
78
30 kN
9kN/m
Hinge
3
4
2
A
1
[FEF]
1
A
=
0.0
EI
=
30 kN
15 kN•m
B
15 kN•m
C
2
1
2
1
2.0
0.0
D1
2
0.0
1.0
D2
0.0006
D1
D2
B
12 kN•m
B
Global matrix:
0.0
1
9kN/m
12 kN•m
C
2
+
-12
15
rad
-0.0015 rad
79
3
12 kN•m
1
A
[FEF]
1
9kN/m
1
A
B
12 kN•m
B
Member 1:
q3
q1
=
3
1
3
2EI
EI
1
EI
2EI
d3 = 0.0
d1 = 0.0006
+
12
-12
=
18
0.0
9 kN/m
A
18 kN•m
22.5 kN
1
B
13.5 kN
80
4
B
C
2
2
15 kN•m
B
30 kN
15 kN•m
C
2
Member 2:
2
q2
q4
=
2
4
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 kN•m
20.63 kN
81
30 kN
9 kN/m
A
18 kN•m
B
13.5 kN
13.5 kN
22.5 kN
9.37 kN
C
B
22.87 kN
22.5 kN•m
20.63 kN
9.37 kN
30 kN
9kN/m
Hinge
EI
θBR= -0.0015 rad
2EI
θBL= 0.0006 rad
A
4m
B
2m
C
2m
22.5
V (kN)
9.37
+
2.5 m
M
(kN•m)
-13.5
10.13
+
- -18
x (m)
-20.63
18.75
+
-
x (m)
-22.5
82
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 diagrams
and qualitative deflected shape.
E = 200 GPa, I = 50x10-6 m4.
20 kN
9kN/m
2EI
A
B
4m
EI
Hinge
C
4m
83
5
7
4
1 3
2EI
1
A
2
2 B
4
1
5
4
0.375EI
5
[k]1 = 1
0.75EI
2EI
-0.375EI
-0.75EI
2
0.75EI
EI
[k]2 =
1
0.1875EI
3
0.375EI
6
7
C
2
0.75EI - 0.375EI 0.75EI
-0.75EI
EI
0.375EI -0.75EI
-0.75EI
2EI/L
[K] =
1
6
EI
3
6
7
1
2
3
1
0.5625
-0.75
0.375
EI 2
-0.75
2.0
0
3
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 kN•m
Global:
B
1
[FEM]
1
Q1 = -20
Q2 = 0.0
=
C
12 kN•m
18 kN
3
0.5625
-0.75
0.375
D1
EI 2
-0.75
2.0
0
D2
3
0.375
0.0
1.0
D3
D1
D3
2
1
Q3 = 0.0
D2
2
6
EI
2 B
9kN/m
A
18 kN
7
18
+
-12
0.0
-0.02382 m
=
-0.008333 rad
0.008933 rad
85
5
9kN/m
4
2EI
2
1
A
1
12 kN•m
A
18 kN
B
12 kN•m
B
1
[FEF]
18 kN
Member 1:
4
1
5
q4
4
0.375EI
q5
5
0.75EI
2EI
1 -0.375EI
-0.75EI
0.75EI
EI
q1
q2
=
2
2
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 kN•m
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
6
EI
2
3
C
Member 2:
1
q1
1
0.1875EI
q3
3
0.375EI
q6
=6
q7
7
6
3
7
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
C
B
2
11.16 kN
d1 = -0.02382
0
d3 = 0.008933
0
d6= 0.0
d7= 0.0
+
0
0
-11.16
=
0.0
11.16
-44.66
44.66 kN•m
11.16 kN
87
20 kN
9kN/m
θBL= -0.008333 rad B
A
θBR= 0.008933 rad C
4m
4m
9kN/m
107.32 kN•m
A
B
1
44.83 kN
C
B
2
8.83 kN
44.66 kN•m
11.16 kN
11.16 kN
44.83
V (kN)
+
8.83
x (m)
-11.16
-11.16
M
(kN•m)
-107.32
-
x (m)
-44.66
88
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 diagrams
and qualitative deflected shape.
40 kN•m at the end of member AB. E = 200 GPa, I = 50x10-6 m4.
30 kN
9kN/m
A
Hinge
2EI
4m
40 kN•m
EI
B
2m
C
2m
89
3
4
2
2EI
A
1
1
4m
C
2
B
2m
2m
θj
θi
Use 2x2 stiffness matrix:
[k ]2×2 =
[k]1 =
EI
3
1
3
2EI
EI
1
EI
2EI
[K] =
Mi
Mj
4 EI / L 2EI / L 


2
EI
/
L
4
EI
/
L


2
[k]2 =
EI
2
1EI
4 0.5EI
1
2
1
2.0
0.0
2
0.0
1.0
4
0.5EI
1EI
90
30 kN
9kN/m
Hinge
A
2EI
3
40 kN•m
EI
B
C
4
2
A
1
[FEM]
1
A
=
Q2 = 0.0
EI
D1
D2
B
12 kN•m
15 kN•m
B
15 kN•m
C
2
1
2
1
2.0
0.0
D1
2
0.0
1.0
D2
0.0026
=
30 kN
B
Global matrix:
Q1 = 40
1
9kN/m
12 kN•m
C
2
+
-12
15
rad
-0.0015 rad
91
3
12 kN•m
1
A
[FEF]
1
9kN/m
1
A
B
12 kN•m
B
Member 1:
q3
q1
=
3
1
3
2EI
EI
1
EI
2EI
d3 = 0.0
d1 = 0.0026
+
12
-12
=
38
40
9 kN/m
A
38 kN•m
B 40 kN•m
37.5 kN
1.5 kN
92
4
B
C
2
2
15 kN•m
B
30 kN
15 kN•m
C
2
Member 2:
2
q2
q4
=
2
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 kN•m
20.63 kN
93
1.5 kN
30 kN
9 kN/m
A
38 kN•m
B 40 kN•m
9.37 kN
7.87 kN
1.5 kN
37.5 kN
C
B
22.5 kN•m
20.63 kN
9.37 kN
30 kN
9kN/m
40 kN•m
Hinge
A
θBL= 0.0026 B
rad
4m
C
θBR=- 0.0015 rad
2m
2m
37.5
V (kN)
+
9.37
1.5
x (m)
-20.63
M
(kN•m)
40
+
- -38
18.75
+
-
x (m)
-22.5
94
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 diagrams
and qualitative deflected shape.
40 kN•m at the end of member AB. E = 200 GPa, I = 50x10-6 m4
20 kN
9kN/m
2EI
A
4m
40 kN•m
B
EI
Hinge
C
4m
95
20 kN
9kN/m
2EI
A
40 kN•m
B
EI
Hinge
4m
C
4m
5
7
1
4
A
12 kN•m
[FEM]
1
3
6
C
2
2 B
9 kN/m
12 kN•m
1
18 kN
18 kN
[k ]4×4
=
∆i
Vi  12 EI/ L3
Mi 
2
 6EI/ L
Vj 
− 12 EI/ L3
Mj  6EI/ L2

θi
∆j
θj
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
5
7
4
2EI
A
1
4
[k]1 =
1
4
0.375EI
5
0.75EI
1
5
2
0.75EI
EI
-0.75EI
EI
0.375EI -0.75EI
-0.75EI
3
6
0.1875EI
3
0.375EI
EI
-0.375EI
0.5EI
-0.1875EI -0.375EI 0.1875EI -0.375EI
7
0.375EI
2
3
1
0.5625
-0.75
0.375
EI 2
-0.75
2.0
0
3
0.375
0.0
1.0
7
1
6
1
2EI
[K] =
[k]2 =
C
2
0.75EI - 0.375EI 0.75EI
1 -0.375EI
1
6
EI
2 B
2EI
-0.75EI
2
3
0.375EI - 0.1875EI 0.375EI
0.5EI
-0.375EI
EI
97
9kN/m
20 kN
5
4
2EI
A
1
C
2
2 B
1
18 kN
18 kN
Global:
1
Q1 = -20
=
Q3 = 0.0
D1
D3
6
EI
12 kN•m
[FEF]
D2
7
9 kN/m
12 kN•m
Q2 = 40
EI
40 kN•m
1 3
2
3
1
0.5625
-0.75
0.375
D1
EI 2
-0.75
2.0
0
D2
3
0.375
0.0
1.0
D3
18
+
-12
0.0
-0.01316 m
=
-0.002333 rad
0.0049333 rad
98
5
4
2
2EI
1
12 kN•m
1
5
4
0.375EI
q5
5
0.75EI
2EI
1 -0.375EI
-0.75EI
0.75EI
EI
q2
2
18 kN
[q]1 = [k]1[d]1 + [qF]1
q4
q1
1
18 kN
4
=
12 kN•m
[FEF]
1
Member 1:
9 kN/m
2
0.75EI - 0.375EI 0.75EI
87.37 kN•m
-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 kN•m
1
13.85 kN
[q]1
99
7
1
B
Member 2:
q1
1
0.1875EI
q3
3
0.375EI
=6
q7
7
C
2
3
[q]2 = [k]2[d]2 + [qF]2
1
q6
6
EI
6
3
7
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
+
0
-6.18
0
0.0
0
0
=
6.18
-24.69
24.69 kN•m
6.18 kN
2
6.18 kN
[q]2
100
20 kN
9kN/m
40 kN•m
EI
θBR = 0.004933 rad C
4m
2EI
θBL= -0.002333 rad
A
B
4m
B
A
87.37 kN•m 49.85 kN
40 kN•m
13.85 kN
C
B
24.69 kN•m
6.18 kN
6.18 kN
49.85
V (kN)
+
13.85
x (m)
-
-6.18
-6.18
M
(kN•m)
+
-87.37
40
-
x (m)
-24.69
101
Temperature Effects
• Fixed-End Forces (FEF)
- Axial
- Bending
• Curvature
102
tan β =
• Thermal Fixed-End Forces (FEF)
Room temp = TR
T
TR
t
M AF
FBF
Tm> TR
FAF
Tb > Tt
A
σaxial
A
d
ct
Tt β
NA
Tm =
cb
F
M
B
B
σaxial
B
σy
Tb - Tt
TR
B
Tl
(∆T ) axial = Tm − TR
y
A
Tt + Tb
2
Tm
Tt
y
Tb − Tt
∆T
=( )
d
d
( ∆T y ) = y (
∆T
)
d
β
Tb
(∆T ) bending = Tb − Tt
103
- Axial
σaxial
σaxial
Tt
Tm> TR
FAF
A
Tb > Tt
TR
Tm
FBF
B
(∆T ) axial = Tm − TR
FAF = ∫ σ axial dA
A
= ∫ Eε axial dA
= ∫ Eα (∆T ) axial dA
= Eα (∆T ) axial ∫ dA
= EAα (∆T ) axial
F
( Faxial
) A = α (Tm − TR ) AE
104
- Bending
σy
Tt
y
y
M AF
A
B
M BF
(∆Ty ) = y (
∆T
)
d
β
Tb
(∆T ) bending = Tb − Tt
M = ∫ yσ y dA
F
A
A
= ∫ yEε dA
= ∫ yEα (∆Ty ) dA
= ∫ yEαy (
= Eα (
∆T
) dA
d
∆T
) ∫ y 2 dA
d
F
( Fbending
)A = α(
Tl − Tu
) EI
d
105
• Elastic Curve: Bending
O´
dθ
ρ
(dx ) = ρ ( dθ )
y ( dθ )
ds = dx
y
dx
Before
deformation
y
ds´
1
ρ
=(
dθ
)
dx
dθ
dx
After
deformation
106
• Bending Temperature
Tt
O
dx
Tl > Tu
Tb
dθ
Tt + Tb
2
M
Tm =
Tu
∆T = Tb - Tt
Tt
dθ
y
M
∆T
β=
d
ct
cb
Tl > Tu
Tb
Tt
∆T
y
d
y
Tb
∆T
) dx
d
∆T
(dθ ) = α ( )dx
d
dθ
M
1
∆T
( ) = = α( ) =
dx
d
EI
ρ
(dθ ) y = αy (
107
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 diagrams
and qualitative deflected shape.
Room temp = 32.5oC, α = 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
2
B
C
4m
3
1
1
FEM
EI
2EI
2
Mean temperature = (40+25)/2 = 32.5
Room temp = 32.5oC
19.78 kN•m
a(∆T /d)EI = 19.78 kN•m
o
+7.5 C
F
Fbending
=α(
-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 kN•m 19.78 kN•m
182 mm
A
1
2
2EI
1
3
B
4m
M1
=
EI
2
1
2
2
1
q2
1
1
2
q1
1
3
1
1
0.5
q1
3
0.5
1
q3
Element 2:
M1
M3
=
EI
4m
[q] = [k][d] + [qF]
Element 1:
M2
EI
C
+
+
-1.978
1.978
(10-3) EI
0
0
0
[M1] = 3EIθ1 + (1.978x10-3)EI
θ1 = -0.659x10-3
rad
110
2
19.78 kN•m 19.78 kN•m
A
182 mm
2EI
1
3
1
2
EI
B
4m
C
4m
Element 1:
M2
M1
=
EI
2
1
2
2
1
q2 = 0
1
1
2
q1 = -0.659x10-3
Element 2:
M1
M3
=
EI
+
1
3
1
1
0.5
q1 = -0.659x10-3
3
0.5
1
q3 = 0
26.37 kN•m
A
6.59 kN•m
B
4.95 kN
4.95 kN
-19.78
19.78
+
0
0
6.59 kN•m
B
=
=
-26.37 kN•m
6.59 kN•m
-6.59 kN•m
-3.30 kN•m
3.3 kN•m
C
2.47 kN
2.47 kN
111
26.37 kN•m
+7.5 oC
A
4.95 kN
-7.5 oC
B
C
2.47 kN
4m
4m
2.47 kN
V (kN)
3.3 kN•m
x (m)
-2.47
-4.95
26.37
M
(kN•m)
Deflected
curve
+
6.59
-
θB = -0.659x10-3 rad
x (m)
-3.30
x (m)
112
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 diagrams
and qualitative deflected shape.
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
1
3
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
C
B
4m
4m
Fixed-end forces due to temperatures
F
Fbending
=α(
∆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
= α (∆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 kN•m
+12 oC
A
-3
oC
19.78 kN•m
B
432 kN
115
6
9
T2 = 40oC
5
182 mm
1
4
A
T1 =
3
2
2EI, 2AE
25oC
2
1
EI, AE
4m
FEM
Element 1:
432 kN
[q] = [k][d] +
19.78 kN•m
+12 oC
-3
A
[qF]
19.78 kN•m
432 kN
oC
4
5
6
B
1
2
q4
4
2x106
0.00
0.00
- 2x106 0.00
q5
5
0.00
3750
7500
0.00
q6
6
0.00
7500 20x103
q1
q2
q3
=
7
C
B
4m
8
1
0.00
3
0.00
0
432
7500
0
0.00
-7500 10x103
0
-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
2
0.00 -3750
3
0.00
-432
116
6
9
T2 = 40oC
5
182 mm
1
4
A
T1 =
3
2
2EI, 2AE
25oC
8
2
1
7
C
B
4m
EI, AE
4m
Element 2:
[q] = [k][d] + [qF]
1
2
3
7
8
9
0.00
d1
0
-1875
3750
d2
0
0.00
-3750
5x103
d3
0
0.00
1x106
0.00
0.00
0
1875
-3750
0
0
-3750 10x103
0
0
q1
1
1x106
0.00
0.00
- 1x106 0.00
q2
2
0.00
1875
3750
0.00
q3
3
0.00
3750 10x103
q7
q8
q9
=
7 -1x106 0.00
8
0.00 -1875
-3750
0.00
9
0.00
5x103
0.00
3750
+
0
117
6
9
T2 = 40oC
5
182 mm
4
A
1
T1 =
2EI, 2AE
25oC
Q1 = 0.0
Q2 = 0.0
=
Q3 = 0.0
D1
D2
D3
=
8
2
1
EI, AE
7
C
B
4m
Global:
3
2
4m
1
2
3
1
3x106
0.0
0.0
D1
2
0.0
5625
-3750
D2
3
0.0
-3750
30x103
D3
0.000144
m
-0.0004795
m
-719.3x10-6
rad
-432
+
0.0
19.78
118
Element 1:
4
5
6
2
1
3
q4
4
2x106
0.00
0.00
- 2x106 0.00
0.00
0
432
q5
5
0.00
3750
7500
0.00
7500
0
0.00
q6
6
0.00
7500 20x103
-7500 10x103
0
-19.78
q1
=
q2
q3
1
-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
2
0.00 -3750
3
0.00
d1 = 144x10-6
-432
6
q4
144.0
kN
q5
-3.60
kN
q6
-23.38
kN•m
-144.0
kN
3.60
kN
9.00
kN•m
q1
q2
q3
=
5
A 4
23.38 kN•m
144 kN
A
3.60 kN
3
2
B 1
9 kN•m
B
3.60 kN
144 kN
119
Element 2:
1
2
3
7
8
9
0.00
d1 = 144x10-6
0
-1875
3750
d2 = -479.5x10-6
0
0.00
-3750
5x103
d3 = -719.3x10-6
0
0.00
1x106
0.00
0.00
0
1875
-3750
0
0
-3750 10x103
3
0
0
q1
1
1x106
0.00
0.00
- 1x106 0.00
q2
2
0.00
1875
3750
0.00
q3
3
0.00
37500 10x103
q7
q8
q9
=
7 -1x106 0.00
8
0.00 -1875
-3750
0.00
9
0.00
5x103
0.00
3750
q1
144
kN
q2
-3.6
kN
q3
-9
kN•m
-144
kN
3.6
kN
-5.39
kN•m
q7
q8
q9
=
2
B 1
9 kN•m
144 kN
B
3.60 kN
+
0
9
8
C 7
5.39 kN•m
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 kN•m
23.38 kN•m
144 kN
144 kN
A
C
B
144 kN
144 kN
B
3.60 kN
3.60 kN
5.39 kN•m
9 kN•m
B
C
3.60 kN
3.60 kN
V (kN)
x (m)
-3.6
23.38
M
(kN•m)
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´
5´
y´
6´
j
2´
3´
4´
m
1´
i
y
6
5
θy
3
2
i
1
j
θx
4
m
x
124
Force Transformation
6´
x´
θy
5´
y´
θy
2´
3´
λx =
1´
i
λy =
y
q6 = q6'
x j − xi
L
y j − yi
L
5
4
m
3
2
i
1
q 4 
q 
 5
q 6 
x
− λy
λx
0
0 
1 
q4' 
q 
 5' 
q6' 
λx
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
λx
λy
0
0
− λy
λx
0
λx
=  λy
 0
q 1   λ x
  
q 2   λy
q 3   0
 =
q 4   0
q 5   0
  
q 6   0
6
j
q4 = q4' cos θ x − q5' cos θ y
λy
q5 = q4' cos θ y − q5' cos θ x
4´
j θx
θx
λx
0
− λy
[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
6
x´
θx
s
o
c
d4
θx4´ d4
6´
j θy
θy
s
o
c
d4
5´
5
4
j
θy
θx
d5
d
θy
θy
os
c
5
d
θy
d ' 4 = d 4 cos θ x + d 5 cos θ y
d '6 = d 6
θx
s
o
c
5
θx
λy
d' 5 = −d 4 cos θ y + d 5 cos θ x
x
y
λx
d 4' 
d 
 5' 
d 6' 
 λx
= − λy
 0
d 1'   λ x
  
d 2'  − λy
d 3'   0
 =
d 4'   0
d 5'   0
  
d 6'   0
θx
x
0
0 
1 
λy
λx
0
d 4 
d 
 5
d 6 
λx
0
0
0
0
0
0
1
0
0 λx
0 − λy
0
0
λy
0
0
0
0
0
λy
λx
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
T
T
T
Therefore,
F
T
F
F
[k ] = [T ]T [k '][T ]
[q ] = [T ] [q ' ]
F
T
F
[q ] = [T ]T [q ']
[d '] = [T ][d ]
[k ] = [T ]T [k '][T ]
127
Stiffness matrix
2*
5*
3*
1*
3´
6*
2´
1
θi
4*
i
θj
i
λjx = cos θj
λiy = sin θi
λjy = sin θj
4´
1´
j
λix = cos θi
6´
5´
j
[q*] = [T]T[q´]
q 1* 
q 
 2* 
q 3* 
 
q 4 * 
q 5* 
 
q 6 * 
1*
2*
=
3*
4*
5*
6*
1
 λix
λ
 iy
0

0
0

 0
2
− λiy
λix
0
0
0
0
3
0
0
1
0
0
0
4
5
0
0
0
0
0
0
λ jx − λ jy
λ jy
λ jx
0
0
6
0
0 
0

0
0

1 
q 1' 
q 
 2' 
q 3' 
 
q 4' 
q 5' 
 
q 6' 
[ T ]T
128
1
2
[T ]
=
3
4
5
6
1
[k ']
=
1
2
3
4
5
6
1
2
λiy
 λix
− λ
 iy λix
 0
0

0
 0
 0
0

0
 0
2
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 λ jx
0 − λ jy
0
0
3
0
6EI/ L2
4 EI/ L
0
− 6EI/ L2
2EI/ L
5
0
0
0
λ jy
λ jx
0
4
6
0
0 
0

0
0

1 
5
− AE/ L
0
0
− 12 EI/ L3
0
− 6EI/ L2
0
AE/ L
0
12 EI/ L3
0
− 6EI/ L2
6


6EI/ L2 
2EI/ L 

0

2
− 6EI/ L

4 EI/ L 
0
129
[ k ] = [ T ]T[ k´ ][T] =
Ui
Ui (
AE
Vi
L
(
AE
Vj - (
Mj
-
L
Mi
Uj -(
λix2 +
-
AE
L
AE
L
12EI
L3
12EI
L3
6EI
L2
λixλjx +
λixλjy-
-
Vi
6EI
L2
λiy2 )
) λixλiy
(
(
AE
L
AE
L
L3
12EI
L3
λiy
12EI
λiy2 +
L2
λiy λjy)
-(
λiyλjx )
-(
AE
L
AE
L
) λixλiy
L3
6EI
λiy
12EI
-
λiyλjx -
λiyλjy +
6EI
L2
Uj
Mi
12EI
L3
-
λix2 )
λix
12EI
L3
12EI
L3
λix
λixλjy)
λix λjx ) -
6EI
λiy -(
L2
6EI
L2
λix
-(
AE
L
AE
L
λixλjx +
6EI
L
L2
6EI
L2
λjy
6EI
L2
λjx
(
AE
L
(
AE
L
L3
L3
L3
L3
6EI
L
L2
AE
λixλjy )
AE
-(
L
L
λjy
λixλjy -
λiyλjy +
6EI
-
12EI
12EI
2EI
λiyλjy ) -(
λjy
λjx2 +
-
12EI
12EI
λiyλjx -
4EI
Vj
λjy2 )
) λjxλjy
(
AE
L2
-
L
Mj
12EI
L3
12EI
L3
6EI
L2
λiy
L2
λix
L2
2EI
L
6EI
) λjxλjy
λjx
6EI
6EI
λixλjx )
AE
12EI
λjx2 )
( L λjy2 +
L3
-
-
λjx
12EI
L3
λiyλjx )
λjy
L2
-
6EI
L2
λjx
4EI
L
130
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 diagrams
and 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.
40 kN
4m
4m
22.02 o
131
40 kN
4m
4m
22.02o
6
5
i
3´
Global
2*
1
2´
3*
1*
x´
22.02 o
j
x*
λjx = cos 22.02o = 0.9271,
λjy = cos 67.98o = 0.3749
4
λix = cos 0o = 1,
λiy = cos 90o = 0
40 kN
40 kN•m
[FEF]
20 kN
[ q´F ]
i
6´
Local
5´
1
1´
j
4´
40 kN•m
20sin22.02=7.5
20cos22.02=18.54 20 kN
132
• Transformation matrix
6
5
3´
Global
2*
1
4
λix = cos 0o = 1,
λiy = cos 90o = 0
Member 1: [ q ] = [
q 4 
 
q5 
q 6 
 
q 1* 
q 2* 
 
q 3* 
3*
1*
2´
x´
*
x
λjx = cos 22.02o = 0.9271,
λjy = cos 67.98o = 0.3749
T ]T[q´]
=
4
5
6
1*
2*
3*
1´
1
0

0

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
i
6´
Local
5´
1
1´
[T ]T
λix
λ
 iy
0
=
0
0

 0
− λiy
0
0
0
1
0
0
0
0 λ jx
0 λ jy
0 0
λix
4´
j
0
0
0
0
0
0
− λ jy
λ jx
0
0
0
0

0
0

1
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
3´
• Local stiffness matrix
2´
i
δi
[k ]6×6 =
[k´]1
=
6´
Local
5´
1
1´
∆i
j
θi
δj
4´
∆j
θj
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
0


6EI/ L2 
2EI/ L 

0

2
− 6EI/ L

4EI/ L 
6´
0.000
3.750
10.00
134
6
5
3´
Global
2*
1
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
1´
j
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
6
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
4´
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
5
2*
1
4
4m
4m
22.02 o
20 kN
x´
x*
40 kN
40 kN•m
Global Equilibrium:
3*
1*
[ q´F ]
40 kN•m
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
m
0.002
rad
+
-7.5
-40
136
6
40 kN
5
2*
1
4
4m
4m
Member Force :
22.02
[q] = [k*][D] +
[qF]
o
40 kN•m
q4
q5
4
5
q6
6
0.000 3.750 20.00 1.406 -3.750 10.00
q1*
q2*
q3*
=
x´
x*
40 kN•m
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
0
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 kN•m
5.06 kN
27.5 kN
13.48 kN
137
40 kN
40 kN
4m
4m
60.05 kN•m
22.02 o
5.05 kN
27.51 kN
13.47 kN
50.04
+
-60.05
Bending moment diagram
(kN•m)
0.002 rad
Deflected shape
138
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 diagrams
and 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.
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
6
22.02 o
AE (0.006 m 2 )(200 ×106 kN / m 2 )
=
L
(8 m)
4m
= 150 ×103 kN • m
3
5
2
8*
4
2
1
1
9*
7*
Global
3´
6´
2´
5´
1´
1
5´
2´
Local
= 20 ×103 kN • m
2 EI
= 10 ×103 kN • m
L
3´
4´
4 EI 4(200 ×106 kN / m 2 )(0.0002 m 4 )
=
L
(8 m)
1´
2
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
6
3
5
2
8*
1
1
4
2
9*
7*
Global
Local : Member 1
6
3
5
[ q]
4
3´
2
2´
1
1
[q] = [q´]
Thus, [k] = [k´]
6´
[ q´]
1´
1
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
6
0.000
4´
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
5´
141
6
3
5
[
2
1
λix = cos 0o = 1,
λiy = cos 90o = 0
9*
7*
2
3´
8*
q* ]
8*
1
1
4
3
2
2
9*
7*
[ q´ ]
2´
x´
1´
5´
6´
4´
2
x*
λjx = cos 22.02o = 0.9271,
λjy = cos 67.98o = 0.3749
Member 2: Use transformation matrix, [q*] = [T]T[q´]
1´
2´
3´
4´
5´
6´
q1
q2
1
2
1
0
0
1
0
0
0
0
0
0
0
0
q1´
q2´
q3
3
0
0
1
0
0
0
q3´
7*
8*
0
0
0
0
0
0
0
0
q4´
q5´
9*
0
0
0
1
q6´
q7*
q8*
q9*
=
0.9271 -0.3749
0.3749 0.9271
0
0
142
2
3
[ q* ]
1
3´
8*
9*
7*
2
1´
2
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
3
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
5´
4´
x*
Stiffness matrix [k´]:
[k´]2 =
x´
[ q´ ]
2´
6´
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
6
3
5
2
4
1
1
2
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
6
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
3
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
6
40 kN
6 kN/m
3
5
2
4
8*
2
1
1
40 kN
6 kN/m
32 kN•m
9*
40 kN•m
32 kN•m
40 kN•m
1
24 kN
24 kN
Global:
1
3
= 103 7*
0
0
0
0
9*
1
3
450
0
-139
0
0
60
1.406
10
7*
-139
1.406
129
1.406
D1
18.15x10-6
m
D3
-509.84x10-6
rad
58.73x10-6
m
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
6
3
5
2
4
32 kN•m
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
6
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
kN•m
5.45
27.82
kN
kN
-52.39
kN•m
q1
q2
q3
=
24 kN
24 kN
Member 1: [ q ] = [k ][d] + [qF]
3.750
32 kN•m
1
1
1
6 kN/m
21.80 kN•m
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 kN•m
5.45 kN
5.45 kN
20.18 kN
27.82 kN
146
2
3
1
40 kN
8*
9*
7*
2
40 kN•m
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
kN•m
0
14.51
kN
kN
0
kN•m
q7*
q8*
q9*
=
20sin22.02=7.5
20cos22.02=18.54 20 kN
2
x*
q1
q2
[
40 kN•m
q´F ]
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 kN•m
5.45 kN
26.55 kN
14.51 kN
147
6 kN/m
21.80 kN•m
52.39 kN•m
52.39 kN•m
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 kN•m
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
(kN•m) -21.8
+
-52.39
x (m)
148