8. 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
11. 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 =
+
(2 EI / L) ( 4 EI / L) θ j M ji F
ji
kii
k ji
[k ] =
kij
k jj
Stiffness Matrix
11
12. 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
13. Stiffness Coefficients Derivation: Fixed-End Support
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
14. Stiffness Coefficients Derivation: Pinned-End Support
Mi
θi
i
L
Mi
L
2L
3
Mi
EI
θi
+ ΣM ' j = 0 : (
MiL
2 EI
M i L 2L
)( ) − θ i L = 0
2 EI 3
ML
θi = ( i )
3EI
θi = 1 = (
θj
j
Real beam
Mi
L
Conjugate beam
θj
+ ↑ ΣFy = 0 : (
MiL
ML
) − ( i ) +θ j = 0
3EI
2 EI
θj =(
MiL
3EI
) → Mi =
L
3EI
− MiL
)
6 EI
14
15. Fixed end moment : Point Load
P Real beam
Conjugate beam
A
A
B
B
L
M
EI
M
M
M
EI
M
EI
ML
2 EI
ML
2 EI
P
PL2
16 EI
PL
4 EI
M
EI
PL2
16 EI
PL
ML ML 2 PL2
+ ↑ ΣFy = 0 : −
−
+
= 0, M =
2 EI 2 EI 16 EI
8 15
17. 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 17
18. 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
18
19. Pin-Supported End Span: Simple Case
P
w
B
A
L
2 EI
4 EI
θA +
θB
L
L
4 EI
2 EI
θA +
θB
L
L
θA
A
θB
+
P
B
w
(FEM)BA
(FEM)AB
A
B
= 0 = (4 EI / L)θ A + (2 EI / L)θ B + ( FEM ) AB
− − − (1)
M BA = 0 = (2 EI / L)θ A + (4 EI / L)θ B + ( FEM ) BA
− − − ( 2)
M AB
2(2) − (1) : 2 M BA = (6 EI / L)θ B + 2( FEM ) BA − ( FEM ) BA
M BA = (3EI / L)θ B + ( FEM ) BA −
( FEM ) BA
2
19
20. Pin-Supported End Span: With End Couple and Settlement
P
w
MA
B
A
L
∆
2 EI
4 EI
θA +
θB
L
L
4 EI
2 EI
θA +
θB
L
L
θA
A
P
θB
w
B
(MF BA)load
(MF AB)load
(MF
AB)∆
A
A
B
B
(MF BA) ∆
4 EI
2 EI
F
F
θA +
θ B + ( M AB )load + ( M AB ) ∆ − − − (1)
L
L
2 EI
4 EI
F
F
θA +
θ B + ( M BA )load + ( M BA ) ∆ − − − (2)
M BA =
L
L
2(2) − (1)
3EI
1
1
M
F
F
F
: M BA =
θ B + [( M BA )load − ( M AB )load ] + ( M BA ) ∆ + A
2
2
2
2
L
M AB = M A =
E lim inate θ A by
20
22. 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
22
23. 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
23
24. 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
24
27. Example 1
Draw the quantitative shear , bending moment diagrams and qualitative
deflected curve for the beam shown. EI is constant.
10 kN
6 kN/m
C
A
4m
4m
B
6m
27
28. 10 kN
6 kN/m
C
A
4m
PL
8
4m
P
B
6m
wL2
30
PL
8
wL2
20
w
FEM
MBA
MBC
[M] = [K][Q] + [FEM]
0
4 EI
2 EI
(10)(8)
θA +
θB +
8
8
8
0
2 EI
4 EI
(10)(8)
M BA =
θA +
θB −
8
8
8
0
4 EI
2 EI
(6)(6 2 )
θB +
θC +
M BC =
6
6
30
0
2 EI
4 EI
(6)(6) 2
θB +
θC −
M CB =
6
6
20
M AB =
B
+ ΣM B = 0 : − M BA − M BC = 0
4 EI 4 EI
(6)(6 2 )
+
=0
(
)θ B − 10 +
8
6
30
2.4
θB =
EI
Substitute θB in the moment equations:
MAB = 10.6 kN•m,
MBC = 8.8 kN•m
MBA = - 8.8 kN•m,
MCB = -10 kN•m 28
29. 10 kN
8.8 kN•m
10.6 kN•m
A
6 kN/m
C
8.8 kN•m
4m
B
4m
MAB = 10.6 kN•m,
6m
MBC = 8.8 kN•m
MBA = - 8.8 kN•m,
10 kN•m
MCB= -10 kN•m
2m
18 kN
10 kN
6 kN/m
10.6 kN•m
A
B
8.8 kN•m B
10 kN•m
8.8 kN•m
Ay = 5.23 kN
ByL = 4.78 kN
ByR = 5.8 kN
Cy = 12.2 kN
29
30. 10 kN
10.6 kN•m
6 kN/m
C
A
4m
5.23 kN
B
4m
6m
10 kN•m
12.2 kN
4.78 + 5.8 = 10.58 kN
V (kN)
5.8
5.23
+
+
x (m)
- 4.78
-
10.3
M
(kN•m)
-10.6
Deflected shape
-12.2
+
-8.8
2.4
θB =
EI
x (m)
-10
x (m)
30
31. Example 2
Draw the quantitative shear , bending moment diagrams and qualitative
deflected curve for the beam shown. EI is constant.
10 kN
6 kN/m
C
A
4m
4m
B
6m
31
32. 10 kN
6 kN/m
C
A
4m
PL
8
P
4m
B
6m
wL2
30
PL
8
w
wL2
20
FEM
10
[M] = [K][Q] + [FEM]
0 4 EI
2 EI
(10)(8)
θA +
θB +
M AB =
− − − (1)
8
8
8 10
2 EI
4 EI
(10)(8)
θA +
θB −
− − − (2)
M BA =
8
8
8
4 EI
2 EI 0 (6)(6 2 )
θB +
θC +
− − − (3)
M BC =
6
6
30
2 EI
4 EI 0 (6)(6) 2
θB +
θC −
− − − (4)
M CB =
6
6
20
6 EI
2(2) − (1) : 2 M BA =
θ B − 30
8
3EI
θ B − 15 − − − (5)
M BA =
8
32
33. MBA
MBC
B
M BC
4 EI
(6)(6 2 )
=
θB +
6
30
M CB =
2 EI
(6)(6)
θB −
6
20
2
− − − (3)
Substitute θA and θB in (5), (3) and (4):
MBA = - 12.19 kN•m
− − − (4)
3EI
θ B − 15 − − − (5)
8
+ ΣM B = 0 : − M BA − M BC = 0
M BA =
MBC = 12.19 kN•m
MCB = - 8.30 kN•m
3EI 4 EI
(6)(6 2 )
+
= 0 − − − (6 )
(
)θ B − 15 +
8
6
30
7.488
θB =
EI
4 EI
2 EI
θA +
θ B − 10
8
8
− 23.74
θA =
EI
Substitute θ B in (1) : 0 =
33
34. 10 kN
12.19 kN•m
A
4m
4m
B
6 kN/m
C
12.19 kN•m
6m
8.30 kN•m
MBA = - 12.19 kN•m, MBC = 12.19 kN•m, MCB = - 8.30 kN•m
2m
10 kN
A
18 kN
B
B
12.19 kN•m
6 kN/m
C
8.30 kN•m
12.19 kN•m
Ay = 3.48 kN
ByL = 6.52 kN
ByR = 6.65 kN
Cy = 11.35 kN
34
35. 10 kN
6 kN/m
C
A
3.48 kN
4m
4m
B
6m
11.35 kN
6.52 + 6.65 = 13.17 kN
V (kN)
6.65
3.48
x (m)
- 6.52
-11.35
14
M
(kN•m)
x (m)
-12.2
Deflected shape
− 23.74
θA =
EI
θB =
7.49
EI
-8.3
x (m)
35
36. Example 3
Draw the quantitative shear , bending moment diagrams and qualitative
deflected curve for the beam shown. EI is constant.
10 kN
4 kN/m
C
A
2EI
4m
B
4m
3EI
6m
36
37. 10 kN
4 kN/m
C
A
2EI
(10)(8)/8
(10)(8)/8
4m
4m
0
M AB
3EI
B (4)(62)/12
6m
(4)(62)/12
10
4(2 EI )
2(2 EI )
(10)(8)
=
θA +
θB +
− − − (1)
8
8
8 10
M BA =
2(2 EI )
4(2 EI )
(10)(8)
θA +
θB −
8
8
15 8
− − − (2)
2(2) − (1)
3(2 EI )
(3 / 2)(10)(8)
: M BA =
θB −
− − − ( 2a )
2
8
8
12
4(3EI )
(4)(6 2 )
θB +
− − − (3)
M BC =
6
12
37
38. 10 kN
4 kN/m
C
A
3EI
2EI
(3/2)(10)(8)/8 B (4)(62)/12
4m
6m
4m
(4)(62)/12
15
3(2 EI )
(3 / 2)(10)(8)
M BA =
θB −
− − − ( 2a )
8
8
12
2
4(3EI )
(4)(6 )
θB +
− − − (3)
M BC =
6
12
− M BA − M BC = 0 : 2.75EIθ B = −12 + 15 = 3
θ B = 1.091 / EI
3(2 EI ) 1.091
(
) − 15 = −14.18 kN • m
8
EI
4(3EI ) 1.091
(
) − 12 = 14.18 kN • m
=
6
EI
2(3EI )
=
θ B − 12 = −10.91 kN • m
6
M BA =
M BC
M CB
38
39. 10 kN
A
2EI 14.18
4m
4 kN/m
C 10.91
B
14.18
3EI
6m
4m
MBA = - 14.18 kN•m, MBC = 14.18 kN•m, MCB = -10.91 kN•m
10 kN
A
24 kN
B
4 kN/m
14.18 kN•m
140.18 kN•m
10.91 kN•m
C
Ay = 3.23 kN
ByL = 6.73 kN
ByR = 12.55 kN
Cy = 11.46 kN
39
40. 10 kN
4 kN/m
C 10.91 kN•m
A
2EI
3.23 kN
4m
V (kN)
3.23
B
3EI
11.46 kN
6m
4m
6.77 + 12.55 = 19.32 kN
12.55
2.86
+
-6.73
+
x (m)
-11.46
12.91
M
(kN•m)
5.53
+
+
-
-10.91
-14.18
Deflected shape
-
x (m)
θB = 1.091/EI
x (m)
40
41. Example 4
Draw the quantitative shear , bending moment diagrams and qualitative
deflected curve for the beam shown. EI is constant.
10 kN 12 kN•m
4 kN/m
C
A
2EI
4m
B
4m
3EI
6m
41
42. 10 kN 12 kN•m
4 kN/m
C
A
2EI
3EI
2/12 = 12
1.5PL/8 = 15 B wL
wL2/12 = 12
4m
6m
4m
M BA =
3(2 EI )
θ B − 15 − − − (1)
8
M BC =
4(3EI )
θ B + 12 − − − (2)
6
2(3EI )
θ B − 12 − − − (3)
6
12 kN•m
MBA
MBA
M CB =
B MBC MBC
Jo int B : − M BA − M BC − 12 = 0
-3.273/EI
4(2 EI )
2(3EI )
(10)(8)
θA +
θB +
M AB =
8
8
8
7.21
θA = −
EI
3.273
M BA = 0.75 EI ( −
) − 15 = −17.45 kN • m
EI
3.273
M BC = 2 EI (−
) + 12 = 5.45 kN • m
EI
3.273
M CB = EI (−
) − 12 = −15.27 kN • m
EI
0
− (0.75 EI − 15) − (2 EIθ B + 12) − 12 = 0
θB = −
3.273
EI
42
43. 3.273
) − 15 = −17.45 kN • m
EI
3.273
M BC = 2 EI (−
) + 12 = 5.45 kN • m
EI
3.273
M CB = EI ( −
) − 12 = −15.27 kN • m
EI
M BA = 0.75EI ( −
24 kN
10 kN
4 kN/m
A
5.45 kN•m
B
2.82 kN
17.45 kN•m
7.18 kN
C
10.36 kN
10 kN 12 kN•m
2.82 kN
B
17.54 kN
13.64 kN
4 kN/m
C
A
15.27 kN•m
15.27 kN•m
13.64 kN
43
44. 10 kN 12 kN•m
4 kN/m
C 15.27 kN•m
A
2.82 kN
4m
3EI
2EI
7.21
EI
3.273
θB = −
EI
B
17.54 kN
6m
4m
13.64 kN θ A = −
10.36
V (kN)
2.82
+
+
3.41 m
-
-
-7.18
M
(kN•m)
-13.64
11.28
7.98
+
+
-
x (m)
-
-5.45
-15.27
-17.45
Deflected shape
− 7.21
θA =
EI
x (m)
x (m)
θB =
3.273
EI
44
45. Example 5
Draw the quantitative shear, bending moment diagrams, and qualitative
deflected curve for the beam shown. Support B settles 10 mm, and EI is
constant. Take E = 200 GPa, I = 200x106 mm4.
10 kN
12 kN•m
6 kN/m
B
C
A
3EI
2EI
4m
4m
10 mm
6m
45
46. 10 kN
12 kN•m
6 kN/m
B
C
A
3EI
2EI
6 EI∆
L2
4m
4m
A
P
B
6 EI∆
L2
[FEM]∆
PL
8
6 EI∆
L2
6m
6 EI∆
L2
∆
PL
8
10 mm
2
wL
30
∆
B
w
C
wL2
30
[FEM]load
-12
4(2 EI )
2(2 EI )
6(2 EI )(0.01) (10)(8)
M AB =
θA +
θB +
+
− − − (1)
8
8
82
8
2(2 EI )
4(2 EI )
6(2 EI )(0.01) (10)(8)
M BA =
θA +
θB +
−
− − − ( 2)
8
8
82
8
0
4(3EI )
2(3EI )
6(3EI )(0.01) (6)(6 2 )
M BC =
θB +
θC −
+
− − − (3)
6
6
62
30
0
2(3EI )
4(3EI )
6(3EI )(0.01) (6)(6) 2
M CB =
θB +
θC −
−
− − − (4)
2
6
6
6
30
46
47. 10 kN
12 kN•m
6 kN/m
B
C
A
3EI
2EI
4m
4m
10 mm
6m
4(2 EI )
2(2 EI )
6(2 EI )(0.01) (10)(8)
θA +
θB +
+
− − − (1)
8
8
82
8
2(2 EI )
4(2 EI )
6(2 EI )(0.01) (10)(8)
=
θA +
θB +
−
− − − ( 2)
8
8
82
8
M AB =
M BA
Substitute EI = (200x106 kPa)(200x10-6 m4) = 200x200 kN• m2 :
4(2 EI )
2(2 EI )
θA +
θ B + 75 + 10 − − − (1)
8
8
2(2 EI )
4(2 EI )
=
θA +
θ B + 75 − 10 − − − (2)
8
8
16.5
M AB =
M BA
2(2) − (1)
3(2 EI )
: M BA =
θ B + 75 − (75 / 2) − 10 − (10 / 2) − 12 / 2 − − − (2a )
8
2
47
48. 10 kN
12 kN•m
6 kN/m
B
C
A
3EI
2EI
4m
4m
10 mm
6m
MBA = (3/4)(2EI)θB + 16.5
MBC = (4/6)(3EI)θB - 192.8
+ ΣMB = 0: - MBA - MBC = 0
MBA
MBC
B
(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):
MBA = 64.58 kN•m,
MBC = -64.58 kN•m
MCB = -146.69 kN•m
48
49. 10 kN
12 kN•m
6 kN/m
B
146.69 kN•m
C
A
64.58 kN•m
4m
4m
64.58 kN•m
6m
MBA = 64.58 kN•m,
MBC = -64.58 kN•m
2m
MCB = -146.69 kN•m
12 kN•m
18 kN
10 kN
A
64.58 kN•m
B
146.69 kN•m
B
Ay = 11.57 kN
ByL = -1.57 kN
6 kN/m
64.58 kN•m
ByR = -29.21 kN
C
Cy = 47.21 kN
49
50. 10 kN
12 kN•m
6 kN/m
C
A
2EI
11.57 kN
3EI
B
4m
47.21 kN
6m
4m
θA = -129.06/EI
θB = 64.109/ EI
1.57 + 29.21 = 30.78 kN
11.57
+
V (kN)
1.57
x (m)
-
-29.21
M
(kN•m)
12
58.29
146.69 kN•m
-47.21
64.58
+
x (m)
-146.69
Deflected shape
x (m)
θA = -129.06/EI
10 mm
θB = 64.109/ EI
50
51. Example 6
For the beam shown, support A settles 10 mm downward, use the slope-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 qualitative
deflected shape. (3 points)
Take E= 200 GPa, I = 50(106) mm4.
6 kN/m
B
2EI
3m
C
1.5EI
12 kN•m
A
10 mm
3m
51
52. 6 kN/m
B
C
2EI
3m
6(32 )
= 4.5
12
6(1.5 × 200 × 50)(0.01)
32
= 100 kN • m
12 kN•m
A
1.5EI
10 mm
3m
6 kN/m
4.5
C
C
MFw
A
0.01 m
MF∆
100 kN • m
A
M CB =
4(2 EI )
θC
3
M CA =
4(1.5EI )
2(1.5 EI )
θC +
θ A + 4.5 + 100 − − − (2)
3
3
12
M AC =
− − − (1)
2(1.5EI )
4(1.5 EI )
θC +
θ A − 4.5 + 100 − − − (3)
3
3
2(2) − (2)
3(1.5EI )
3(4.5) 100 12
: M CA =
θC +
+
+
3
2
2
2
2
− − − ( 2a )
52
53. 6 kN/m
B
2EI
3m
C
1.5EI
12 kN•m
A
10 mm
3m
4(2 EI )
θ C − − − (1)
3
3(1.5 EI )
3(4.5) 100 12
=
θC +
+
+
3
2
2
2
M CB =
M CA
MCB
MCA
− − − ( 2a )
• Equilibrium equation:
M CB + M CA = 0
(8 + 4.5) EI
3(4.5) 100 12
θC +
+
+ =0
C
3
2
2
2
− 15.06
θC =
= −0.0015 rad
EI
2(1.5 EI ) − 15.06 4(1.5 EI )
Substitute θC in eq.(3)
12 =
(
)+
θ A − 4.5 + 100 − − − (3)
EI
3
3
− 34.22
θA =
= −0.0034 rad
EI
53
54. 6 kN/m
B
C
2EI
12 kN•m
1.5EI
3m
A
10 mm
3m
− 15.06
− 34.22
= −0.0015 rad θ A =
= −0.0034 rad
EI
EI
2(2 EI )
2( 2 EI ) − 15.06
=
θC =
(
) = −20.08 kN • m
3
3
EI
4(2 EI )
4(2 EI ) − 15.06
=
θC =
(
) = −40.16 kN • m
3
3
EI
θC =
M BC
M CB
20.08 kN•m
B
20.08 kN
C
40.16 kN•m
40.16 + 20.08
= 20.08 kN
3
18 kN
6 kN/m
40.16 kN•m
C
26.39 kN
12 kN•m
A
8.39 kN
54
55. 12 kN•m
6 kN/m
θ C = −0.0015 rad
B
θ A = −0.0034 rad
C
2EI
3m
B
40.16 kN•m
C
20.08 kN
40.16 kN•m
V (kN)
6 kN/m
C
26.39 kN
26.39
12 kN•m
A
-
Deflected shape
-20.08
20.08
12
-40.16
θ C = 0.0015 rad
8.39 kN
8.39
x (m)
+
M (kN•m)
10 mm
3m
20.08 kN•m
20.08 kN
A
1.5EI
x (m)
x (m)
θ A = 0.0034 rad
55
56. Example 7
For the beam shown, support A settles 10 mm downward, use the
slope-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 qualitative
deflected shape.
Take E= 200 GPa, I = 50(106) mm4.
6 kN/m
B
2EI
3m
C
1.5EI
12 kN•m
A
10 mm
3m
56
57. 12 kN•m
6 kN/m
B
C
2EI
3m
C
B
∆C
6( 2 EI )∆ C 4 EI∆ C
=
2
3
3
M BC
M CB
M CA =
2(2 EI )
4 EI
=
θC −
∆C
3
3
4(2 EI )
4 EI
=
θC −
∆C
3
3
A
1.5EI
10 mm
3m
EI∆ C
C
∆C
4 EI∆ C
3
A
6 kN/m
4.5
100
− − − (1)
C
A
C
− − − (2)
0.01 m
A
6(1.5 EI ) ∆ C
= EI∆ C
2
3
6(32 )
= 4.5
12
6(1.5 × 200 × 50)(0.01)
32
= 100 kN • m
4(1.5 EI )
2(1.5EI )
θC +
θ A + EI∆ C + 4.5 + 100 − −(3)
3
3
12
2(1.5 EI )
4(1.5 EI )
M AC =
θC +
θ A + EI∆ C − 4.5 + 100 − −(4)
3
3
2(3) − (4)
3(1.5 EI )
EI
3(4.5) 100 12
: M CA =
θ C + ∆C +
+
+
− − − (3a)
2
3
2
2
2
2
57
58. 12 kN•m
6 kN/m
B
C
2EI
3m
A
1.5EI
3m
18 kN
• Equilibrium equation:
6 kN/m
MBC
B
By
C
(C y ) CB = −(
MCB
MCA
C
M BC + M CB
(C y ) CA
)
3
MCB
MCA
(Cy)CB
10 mm
C
12 kN•m
A
Ay
M + 12 + 18(1.5) M CA + 39
= CA
=
3
3
ΣM C = 0 : M CB + M CA = 0 − − − (1*)
ΣC y = 0 : (C y ) CB + (C y ) CA = 0 − − − (2*)
(Cy)CA
Substitute in (1*) 4.167 EIθ C − 0.8333EI∆ C = −62.15 − − − (5)
Substitute in ( 2*) − 2.5 EIθ C + 3.167 EI∆ C = −101.75 − − − (6)
From (5) and (6) θ C = −25.51 / EI = −0.00255 rad
∆ C = −52.27 / EI = −5.227 mm
58
59. 6 kN/m
B
2EI
C
3m
1.5EI
12 kN•m
A
10 mm
3m
• Solve equation
θC =
− 25.51
= −0.00255 rad
EI
Substitute θC and ∆C in (1), (2) and (3a)
M BC = 35.68 kN • m
− 52.27
∆C =
= −5.227 mm
EI
M CB = 1.67 kN • m
Substitute θC and ∆C in (4)
θA =
M CA = −1.67 kN • m
− 2.86
= −0.000286 rad
EI
6 kN/m
35.68 kN•m
B
B y = 18 − 5.55
= 12.45 kN
2EI
3m
C
1.5EI
3m
12 kN•m
A
18(4.5) − 12 − 35.68
6
= 5.55 kN
Ay =
59
60. 6 kN/m
35.68 kN•m
B
C
2EI
12.45 kN
3m
12 kN•m
A
1.5EI
10 mm
5.55 kN
3m
V (kN) 12.45
θ C = −0.00255 rad
+
0.925 m
x (m)
∆ C = −5.227 mm
-5.55
θ A = −0.000286 rad
M (kN•m)
1.67
14.57
+
12
-
x (m)
-35.68
Deflected shape
∆ C = 5.227 mm
θ C = 0.00255 rad
x (m)
θ A = 0.000286 rad
60
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
12 kN/m
C
3m
B
3EI
40 kN
3m
2EI
A
1m
6m
62
63. 10 kN
36/2 = 18
12 kN/m
• Equilibrium equations
C
B
3m
40 kN
3m
A
1m
2EI 2
36
(wL /12 ) =36
PL/8 = 30
3EI
PL/8 = 30
6m
• Slope-Deflection Equations
M AB =
M BA =
M BC
10
MBC
MBA
10 − M BA − M BC = 0 − − − (1*)
Substitute (2) and (3) in (1*)
10 − 3EIθ B + 30 − 54 = 0
− 14 − 4.667
=
(3EI )
EI
2(3EI )
θ B + 30 − − − (1)
6
θB =
4(3EI )
θ B − 30 − − − (2)
6
Substitute θ B =
3( 2 EI )
θ B + 36 + 18 − − − (3)
=
6
− 4.667
in (1) to (3)
EI
M AB = 25.33 kN • m
M BA = −39.33 kN • m
M BC = 49.33 kN • m
63
64. 10 kN
12 kN/m
20.58
C
3m
B
49.33
39.33
3EI
40 kN
3m
25.33
A
1m
2EI
10
-39.3
-49.33
MAB = 25.33 kN•m
27.7
MBA = -39.33 kN•m
-25.33
MBC = 49.33 kN•m
6m
Bending moment diagram
12 kN/m
B
B
C
θB
49.33
39.33
27.78 kN
θB
θB = -4.667/EI
40 kN
A
17.67 kN
25.33
Deflected curve
64
65. Example 7
Draw the quantitative shear, bending moment diagrams and qualitative
deflected curve for the frame shown. E = 200 GPa.
25 kN
5 kN/m
B
240(106) mm4
5m
180(106)
C
120(106) mm4
60(106) mm4
A
E
D
3m
3m
4m
65
66. 25 kN
PL/8 = 18.75
18.75
B
240(106) mm4 C
5m
5 kN/m
180(106)
6.667+ 3.333
120(106) mm4
(wL2/12 ) = 6.667
60(106) mm4
A
E
D
3m
3m
0
4(2 EI )
2(2 EI )
M AB =
θA +
θB
5
5
0
2(2 EI )
4(2 EI )
M BA =
θA +
θB
5
5
4( 4 EI )
2(4 EI )
M BC =
θB +
θ C + 18.75
6
6
2(4 EI )
4(4 EI )
M CB =
θB +
θ C − 18.75
6
6
3( EI )
θC
M CD =
5
3(3EI )
M CE =
θ C + 10
4
4m
M BA + M BC = 0
8 16
8
( + ) EIθ B + ( ) EIθ C = −18.75 − − − (1)
5 6
6
M CB + M CD + M CE = 0
16 3 9
8
( ) EIθ B + ( + + ) EIθ C = 8.75 − − − (2)
6 5 4
6
From (1) and (2) : θ B =
− 5.29
EI
θC =
2.86
EI
66
67. Substitute θB = -1.11/EI, θc = -20.59/EI below
4(2 EI ) 0 2(2 EI )
M AB =
θA +
θB
5
5
0
2(2 EI )
4(2 EI )
M BA =
θA +
θB
5
5
4( 4 EI )
2(4 EI )
M BC =
θB +
θ C + 18.75
6
6
2(4 EI )
4(4 EI )
θB +
θ C − 18.75
M CB =
6
6
3( EI )
θC
M CD =
5
3(3EI )
θ C + 10
M CE =
4
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
67
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
20 kN
25 kN
16.44 kN•m
3m
3m C
C
2.54 kN B
2.54 kN 2.54-0.34
=2.2 kN
8.46 kN•m
18.18 kN•m
(20(2)+16.44)/4
(25(3)+8.46-18.18)/6
14.12 kN
= 14.11 kN
= 10.88 kN
10.88 kN
B
8.46 kN•m
(8.46 + 4.23)/5
= 2.54 kN
5m
A
E 2.2 kN
5.89 kN
14.12+14.11=28.23 kN
1.72 kN•m
(1.72)/5 = 0.34 kN
B
5m
2.54 kN
A
0.34 kN
4.23 kN•m
10.88 kN
28.23 kN
68
69. 24.18
1.29 m 2.33 m
14.11
10.88
1.18 m
+
+
-2.54
2.82 m
-
-14.12
-
-2.54
+
0.78 m
-
-8.46
-5.89 -8.46 +
-
-18.18
0.34
4.23
Shear diagram
0.78 m
+
3.46
1.72
+
-
1.18 m
-16.44
1.67m
Moment diagram
1.29 m 2.33 m
θB = −5.29/EI
1.18 m
θC = 2.86/EI
1.67m
Deflected curve
69
71. Example 8
Determine the moments at each joint of the frame and draw the quantitative
bending moment diagrams and qualitative deflected curve . The joints at A and
D are fixed and joint C is assumed pin-connected. EI is constant for each member
3m
1m
B
10 kN
C
3m
A
D
71
72. • Overview
• Unknowns
C
B
1m
10 kN
θB and ∆
• Boundary Conditions
3m
MAB
Ax
3m
A
Ay
θA = θD = 0
MDC
D D
• Equilibrium Conditions
x
- Joint B
Dy
B
MBA
MBC
ΣM B = 0 : M BA + M BC = 0 − − − (1*)
- Entire Frame
+
→ ΣFx = 0 : 10 − Ax − Dx = 0 − − − (2*)
72
74. Equilibrium Conditions:
M BA + M BC = 0 − − − (1*)
10 − Ax − Dx = 0 − − − (2*)
• Solve equation
Substitute (2) and (3) in (1*)
2EI θB + 0.375EI ∆ = 5.625 ----(7)
Slope-Deflection Equations:
Substitute (5) and (6) in (2*)
2( EI )
M AB =
θ B + 5.625 + 0.375EI∆ − − − (1)
− 0.375EIθ B − 0.235EI∆ = −8.437 − − − (8)
4
4( EI )
M BA =
θ B − 5.625 + 0.375EI∆ − − − (2) From (7) and (8) can solve;
4
− 5.6
44.8
3( EI )
θB =
∆=
M BC =
θ B − − − (3)
EI
EI
3
− 5.6
44.8
M DC = 0.1875EI∆ − − − (4)
Substituteθ B =
and ∆ =
in (1)to (6)
EI
EI
Horizontal reaction at supports:
Ax = 0.375EIθ B + 0.1875EI∆ + 1.563 − − − (5)
Dx = 0.0468EI∆ − − − (6)
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
74
75. C
B
1m
10 kN
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,
5.6
3m
8.42
D 2.1 kN
15.88
7.9 kN
3m
A
5.6
C
B
B
7.8
A
∆ = 44.8/EI
C
∆ = 44.8/EI
5.6
15.88
D
8.42
Bending moment diagram
A
θB = -5.6/EI
D
Deflected curve
75
76. 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 the
qualitative deflected shape of the entire frame.
B
C
2 EI
10 kN
2m
pin
4m
2.5 EI
EI
A
D
4m
3m
76
77. • Overview
B ∆ 2EI
B´
10 kN
2m
A
EI
MAB
Ax
Ay
4m
∆ CD
C
C´
∆BC
∆
2.5EI
• Unknowns
θB and ∆
MDC
• Boundary Conditions
Dx
D
3m
4m
θA = θD = 0
Dy
• Equilibrium Conditions
- Joint B
B
MBA
MBC
ΣM B = 0 : M BA + M BC = 0 − − − (1*)
- Entire Frame
+
→ ΣFx = 0 : 10 − Ax − Dx = 0 − − − (2*)
77
78. • Slope-Deflection Equation
C´
B ∆
∆ CD
∆BC
5
B´ 2EI C ∆
10 kN
36.87°
4m
2.5EI
EI
2m
PL/8 = 5
D
A
4m
3m
0.375EI∆
B
∆´
B´
= ∆ / cos 36.87° = 1.25 ∆
C´
∆CD
∆BC = ∆ tan 36.87° = 0.75 ∆
36.87°
∆
C
2( EI )
θ B + 0.375EI∆ + 5 − − − (1)
4
(6)(2.5EI)(1.25∆)/(5)2 = 0.75EI∆
4( EI )
M BA =
θ B + 0.375EI∆ − 5 − − − (2)
C´
4
∆CD= 1.25 ∆
3(2 EI )
C
M BC =
θ B − 0.2813EI∆ − − − (3)
4
M AB =
M DC = 0.375EI∆ − − − (4)
A
D
0.75EI∆
6EI∆/(4)
2=
0.375EI∆
0.5625EI∆
B
B´
(1/2) 0.5625EI∆
C´
C
(1/2) 0.75EI∆
(6)(2EI)(0.75∆)/(4) 2 = 0.5625EI∆
∆BC= 0.75 ∆
78
79. • Horizontal reactions
B
2 EI
pin
C
10 kN
2m
4m
2.5 EI
EI
A
D
4m
3m
MBC
MBA
B
B
10 kN
C
MBC
MBC
4
4
C
A
Ax = (MBA+ MAB-20)/4 -----(5)
MAB
D
Dx= (MDC-(3/4)MBC)/4 ---(6)
MDC
MBC/4
79
80. Equilibrium Conditions:
M BA + M BC = 0 − − − (1*)
10 − Ax − Dx = 0 − − − (2*)
Slope-Deflection Equation:
• Solve equations
Substitute (2) and (3) in (1*)
2.5EIθ B + 0.0938 EI∆ − 5 = 0 − − − (7)
Substitute (5) and (6) in (2*)
2( EI )
6 EI∆
0.0938EIθ B + 0.334EI∆ − 5 = 0 − − − (8)
θB + 5 + 2
− − − (1)
4
4
From (7) and (8) can solve;
6 EI∆
4( EI )
M BA =
θB −5 + 2
− − − (2)
4
4
1.45
− 14.56
θB =
∆=
3(2 EI )
3(2 EI )(0.75∆)
EI
EI
M BC =
θB −
− − − (3)
2
4
4
1.45
− 14.56
3(2.5EI )(1.25∆)
Substituteθ B =
and ∆ =
in (1)to (6)
M DC =
− − − (4)
EI
EI
52
Horizontal reactions at supports:
MAB = 15.88 kN•m
MBA = 5.6 kN•m
( M BA + M AB − 20)
− − − (5)
Ax =
MBC = -5.6 kN•m
4
MDC = 8.42 kN•m
3
Ax = 7.9 kN
M DC − M BC
4
Dx =
− − − (6)
Dx = 2.1 kN
4
80
M AB =
81. 1.91
1.91
B
2 EI
A
pin
C
10 kN
2m
MAB = 11.19 kN•m
MBA = 1.91 kN•m
MBC = -1.91 kN•m
2.5 EI
EI
11.19 kN•m
8.27 kN
0.478 kN
4m
5.46
4m
1.73
D
MDC = 5.46 kN•m
Ax = 8.28 kN•m
Dx = 1.72 kN•m
0.478 kN
3m
∆
∆
B
1.91
1.91
B
C
C
5.35
A
θB=1.45/EI
θB=1.45/EI
11.19
5.46
D
Bending-moment diagram
A
D
Deflected shape
81
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.
3m
20 kN/m
B
pin
C
3EI
3m
2EI
4m
4EI
A
D
82
83. • Overview
• Unknowns
∆
∆
3m
20 kN/m
B
3EI
2EI
θB and ∆
C
• Boundary Conditions
4EI
θA = θD = 0
4m
• Equilibrium Conditions
- Joint B
A
[FEM]load
D
B
3m
MBA
MBC
ΣM B = 0 : M BA + M BC = 0 − − − (1*)
- Entire Frame
+
→ ΣFx = 0 : 60 − Ax − Dx = 0 − − − (2*)
83
84. • Slope-Deflection Equation
B
20 kN/m
B
3m
6(2EI∆)/(3)
= 1.333EI∆
2
C
3EI
= 15
wL2/12
2EI
4EI
6(2EI∆)/(3) 2
= 1.333EI∆
A
A
[FEM]load
D
0
M BA
1.5EI∆
4m
wL2/12 = 15
M AB
∆
C
∆
3m
(1/2)(1.5EI∆)
6(4EI∆)/(4) 2
= 1.5EI∆
[FEM]∆
D
4(2 EI )
2(2 EI )
=
θA +
θ B + 15 + 1.333EI∆ = 1.333EIθ B + 15 + 1.333EI∆ ----------(1)
3
3
0
2( 2 EI )
4(2 EI )
=
θA +
θ B − 15 + 1.333EI∆ = 2.667 EIθ B − 15 + 1.333EI∆ ----------(2)
3
3
3(3EI )
θ B = 3EIθ B ----------(3)
3
0
3( 4 EI )
=
θ D + 0.75 EI∆ = 0.75 EI∆ ----------(4)
4
M BC =
M DC
84
85. • Horizontal reactions
MAB
C
B
1.5 m
+ ΣM B = 0 :
60 kN
4m
1.5 m
M BA + M AB + 60(1.5)
3
Ax = 1.333EIθ B + 0.889 EI∆ + 30 − − − (5)
Ax =
A
Ax
MBA
D
Dx
MDC
+ ΣMC = 0:
Dx =
M DC
= 0.188EI∆ − − − (6)
4
85
86. Equilibrium Conditions
M BA + M BC = 0 − − − (1*)
60 − Ax − Dx = 0 − − − (2*)
Equation of moment
M AB = 1.333EIθ B + 15 + 1.333EI∆ − − − (1)
M BA = 2.667 EIθ B − 15 + 1.333EI∆ − − − (2)
M BC = 3EIθ B
− − − (3)
M DC = 0.75 EI∆ − − − ( 4)
Horizontal reaction at support
Ax = 1.333EIθ B + 0.889 EI∆ + 30 − − − (5)
Dx = 0.188EI∆ − − − (6)
• Solve equation
Substitute (2) and (3) in (1*)
5.667 EIθ B + 1.333EI∆ = 15 − − − (7)
Substitute (5) and (6) in (2*)
− 1.333EIθ B − 1.077 EI∆ = −30 − − − (8)
From (7) and (8), solve equations;
θB =
− 5.51
EI
∆=
34.67
EI
34.67
− 5.51
and ∆ =
EI
EI
M AB = 53.87 kN • m
M BA = 16.52 kN • m
M BC = −16.52 kN • m
Substituteθ B =
in (1)to (6)
M DC = 26.0 kN • m
Ax = 53.48 kN
Dx = 6.52 kN
86
87. B
3m
20 kN/m
C
3m
M AB = 53.87 kN • m
M BA = 16.52 kN • m
M BC = −16.52 kN • m
53.87 kN•m 4m
M DC = 26.0 kN • m
16.52 kN•m
Ax = 53.48 kN
53.48 kN
A
D
5.55 kN
26 kN•m
Dx = 6.52 kN
6.52 kN
5.55 kN
∆
16.52
C
B
∆
B
16.52
C
53.87
A
Moment diagram
26.
D
A
Deflected shape
D
87
