Analisis struktur 2
INTRODUCTION: STRUCTURAL ANALYSIS
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Deflected Shape of Structures
Method of Consistent Deformations
Maxwell’s Theorem of Reciprocal
Displacement
1
Deflection Diagrams and the Elastic Curve
fixed support
P
∆=0
θ=0
-M
P
roller or rocker
support
∆=0
θ
inflection point
+M
-M
2
P
θ
pined support
∆=0
-M
3
inflection point
• Fixed-connected joint
P
fixed-connected
joint
inflection point
Moment diagram
4
• Pined-connected joint
pined-connected
joint
P
Moment diagram
5
P
inflection point
Moment diagram
6
P1
A
B
C
D
P2
M
+M
x
-M
inflection point
7
P1
P2
+M
x
-M
inflection point
8
Method of Consistent Deformations
Beam 1 DOF
P
MA
Ax = 0
A
A
=
Ay
B
C
P
RB
B
C
∆´B
+
fBB x R
B
A
C
∆´B + fBB RB = ∆B = 0
B
1
9
Beam 2 DOF
Compatibility Equations.
w
∆´1 + f11R1 + f12R2 = ∆1 = 0
3
∆´2 + f21R1 + f22R2 = ∆2 = 0
4
1
5
2
=
0
w
∆´1
∆´1
∆´2
∆´2
+
f11 f12 R1
f12 f22 R
2
=
∆1 0
∆2
+
f21
f11
f12
+
1
×R1
f22
×R2
1
10
Beam 3 DOF
Compatibility Equations.
P1
1
w
P2
θ´1 + f11M1 + f12R2 + f13R3 = θ1 = 0
6
=
∆´2 + f21M1 + f22R2 + f23R3 = ∆2 = 0
4
2
3
P1
5
w
∆´3 + f31M1 + f32R2 + f33R3 = ∆3 = 0
P2
0
∆´2
1
f21
+
f11
θ´1
∆´3
f31
×M1
f11 f12 f13 M1
∆´2
f21 f22 f23 R
2
+
f31 f32 f33 R
∆´3
3
θ1 0
=
∆2 0
∆3
+
f12 f22
θ´1
f32
×R2
f13
f23
+
1
f33
×R3
1
11
Compatibility Equation for n span.
Equilibrium Equations
∆´1 + f11R1 + f12R2 + f13R3 = ∆ 1
∆´2 + f21R1 + f22R2 + f23R3 = ∆2
Fixed-end force matrix
∆´n + fn1R1 + fn2R2 + fnnRn = ∆n
[Q] = [K][D] + [Qf]
If ∆1 = ∆2=……= ∆n = 0 ;
∆ ´1
f11 f12
f13 R1
0
∆´2
f21 f22
f23 R
2 =
0
f n1 f n2
fn n R
n
0
∆´ n
+
Stiffness matrix
Solve for displacement [D];
[D] = [K]-1[Q] - [Qf]
or
[∆´] + [f][R] = 0
[R] = - [f]-1[∆´]
kij =
1
f ij
[fij] = Flexibility matrix dependent on
1
,
EI
1
1
,
EA GJ
12
Maxwell’s Theorem of reciprocal displacements
1=i
A
2=j
EI
1
B
f11 = fii
f21 = fji
f ij = ∫ mi M j
dx
EI
= ∫ mi m j
dx
EI
f ji = ∫ m j M i
dx
EI
= ∫ m j mi
dx
EI
Mi = mi
1
A
B
f22 = fjj
f12 = fij
Mj = mj
f ij = f ji
13
Example 1
Determine the reaction at all supports and the displacement at C.
50 kN
B
C
A
6m
6m
14
SOLUTION
• Principle of superposition
50 kN
MA
C
A
6m
6m
RB
=
RA
B
50 kN
∆´B
+
fBB x RB
Compatibility equation : ∆ ' B + f BB RB = 0
-----(1)
1 kN
15
• Use conjugate beam in obtaining ∆´B and fBB
50 kN
A
B
Real beam
300 kN•m
∆´B
50 kN
6m
6m
9000/EI
6 + (2/3)6 = 10 m
Conjugate beam
900/EI
300 /EI
900/EI
∆´B = M´B = -9000/EI ,
fBB
12 kN•m
1 kN
12 /EI
Real beam
1 kN
72/EI
(2/3)12 = 8 m
576/EI
Conjugate beam
fBB = M´´B = 576/EI,
72/EI
16
• Substitute ∆´B and fBB in Eq. (1)
+ ↑: −
9000 576
+(
) RB = 0
EI
EI
RB = +15.63 kN,
(same direction as 1 kN)
50 kN
300 kN•m
∆´B
50 kN
+
fBB x RB = 15.63 kN
1 kN
=
12 kN•m
1 kN
50 kN
34.37 kN•m A
34.37 kN
C
B
15.63 kN
17
Use conjugate beam in obtaining the displacement
50 kN
6m
113 kN•m
6m
C
∆C
A
B
Real Beam
15.6 kN
34.4 kN
93.6
M
(kN•m)
x (m)
3.28 m
6m
12 m
93.6/EI
-113
Conjugate Beam
-113/EI
281
223
776
( 2) −
(6 ) = −
M 'C =
EI
EI
EI
∆ C = M 'C = −
776
,↓
EI
281/(EI)
223/(EI)
M´C
V´C 2 m
4m
223/(EI)
18
Example 2
Determine the reaction at all supports and the displacement at C. Take E = 200
GPa and I = 5(106) mm4
10 kN
3EI
2EI
B
A
4m
C
2m
2m
19
10 kN
3EI
2EI
B
A
4m
C
2m
2m
=
10 kN
∆´B
+
fBB x R
B
1 kN
Compatibility equation:
∆´B + fBBRB = ∆B = 0
20
• Use conjugate beam in obtaining ∆´B
10 kN
3EI
40 kN•m
2EI
B
A
4m
C
2m
Real Beam
2m
10 kN
10
V (kN)
10
+
x (m)
M
(kN•m)
x (m)
40
177.7/EI
Conjugate Beam
40/3EI = 13.33/EI
26.66/EI
∆´B = M´B = 177.7/EI
21
• Use conjugate beam for fBB
3EI
8 kN•m
2EI
B Real Beam
1 kN
A
1 kN
4m
C
2m
2m
V (kN)
x (m)
-1
-1
8
4
M
(kN•m)
2.67
EI
8
=
3EI
+
x (m)
4/(3EI)=1.33EI
4/(2EI)=2EI
60.44/EI
fBB = M´B = 60.44/EI
12/EI
Conjugate Beam
22
10 kN
3EI
2EI
B
A
C
4m
2m
2m
=
10 kN
∆´B
+
fBB x R
B
1 kN
Compatibility. equation:
∆´B + fBBRB = ∆B = 0
− 177.7 60.44
+(
) RB = 0
EI
EI
RB = +2.941 kN,
(same direction as 1 kN)
+ ↑: −
23
• The quantitative shear and moment diagram and the qualitative deflected curve
10 kN
16.48 kN•m
3EI
2EI
B
A
∆C
C
2.94 kN
4m
2m
2m
7.06 kN
7.06
+
V (kN)
x (m)
-2.94
-2.94
11.76
M
(kN•m)
2.33 m
+
x (m)
1.67 m
-16.48
24
• Use the conjugate beam for find ∆C
3EI
16.48 kN•m
10 kN
2EI
C
A
B
∆C
2m
4m
7.059 kN
2.335 m
11.76
3EI
2m
Real beam
2.941 kN
11.76
2EI
Conjugate beam
16.48
3EI
1.665 m
3.263
EI
M´C=
3.263
6.413
−18.85
(0.555) −
(3.222) =
,↓
EI
EI
EI
(1.665)/3=0.555 m
6.413
−18.85
EI
∆C = M 'C =
= −18.85 mm, ↓
(
200
×
5
)
1.665+(2/3)(2.335) = 3.222 m
25
Example 3
Draw the quantitative Shear and moment diagram and the qualitative deflected
curve for the beam shown below.EI is constant. Neglect the effects of axial load.
5 kN/m
A
B
4m
4m
26
SOLUTION
• Principle of superposition
5 kN/m
θA
4m
4m
θB
B
=
A
5 kN/m
θ´B
+
θ´A
1 kN•m
αBA
+
αAA
×MA
αAB
Compatibility equations:
1 kN•m × M
B
αBB
θ A = 0 = θ ' A + f AA M A + f AB M B
− − − (1)
θ B = 0 = θ 'B + f BA M A + f BB M B
− − − ( 2)
27
• Use formula provided in obtaining θ´A, θ´B, αAA, αBA, αBB, αAB
5 kN/m
θ´A
θ´B
4m
4m
3wL3 3(5)(8) 3 60
θ 'A =
=
=
EI
128 EI
128EI
7 wL3
7(5)(8) 3 46.67
θ 'B =
=
=
EI
384 EI
384 EI
1 kN•m
1 kN•m
αAA
αBA
αAB
8m
8m
αBB
α AA =
M o L 1(8) 2.667
=
=
3EI 3EI
EI
α BB =
M o L 1(8) 2.667
=
=
EI
3EI 3EI
α BA =
M o L 1(8) 1.333
=
=
EI
6 EI 6 EI
α AB =
M o L 1(8) 1.333
=
=
EI
6 EI 6 EI
Note: Maxwell’s theorem of reciprocal displacement, αAB = αBA
28
• Use conjugate beam for αAA, αBA, αBB, αAB
1 kN•m
Real Beam
αAA
(1/8)
Real Beam
αAB
αBA
(1/8)
8m
1 kN•m
αBB
(1/8)
(1/8)
8m
4/EI
4/EI
1/EI
1/EI
Conjugate Beam
2.67/EI
Conjugate Beam
1.33/EI
1.33/EI
2.67/EI
α AA = V ' A =
− 2.667
EI
α AB = V ' A =
− 1.333
EI
α BA = V 'B =
1.333
EI
α BB = V ' B =
2.667
EI
29
5 kN/m
θA
B
θB
4m
Compatibility equation
+
=
4m
5 kN/m
+
θ 'B =
46.67
EI
46.67 1.333
2.667
+(
)M A + (
)M B = 0
EI
EI
EI
Solve simultaneous equations,
×MA
α AA
60 2.667
1.333
+(
)M A + (
)M B = 0
EI
EI
EI
+
60
θ 'A =
EI
1 kN•m
2.667
=
EI
+
A
α BA = V 'B =
1.333
EI
MA = -18.33 kN•m,
+
MB = -8.335 kN•m,
+
1 kN•m
×MB
α AB =
1.333
EI
α BB =
2.667
EI
30
MA = -18.33 kN•m,
MB = -8.335 kN•m,
5 kN/m
18.33 kN•m
A
RA
B
4m
4m
8.335 kN•m
RB
+ ΣMA = 0: 18.33 − 20( 2) + RB (8) − 8.355 = 0
3.753
+ ΣFy = 0: R A + RB − 20 = 0
RB = 3.753 kN,
Ra = 16.25 kN,
31
• Quantitative shear and bending diagram and qualitative deflected curve
5 kN/m
18.33 kN•m
A
B
4m
16.25 kN
4m
8.36 kN•m
3.75 kN
16.25
V
diagram
3.25 m
M
diagram
8.08
-3.75
6.67
-8.36
-18.33
Deflected
Curve
32
Example 4
Determine the reactions at the supports for the beam shown and draw the
quantitative shear and moment diagram and the qualitative deflected curve.
EI is constant.
2 kN/m
C
A
4m
B
4m
33
• Principle of superposition
2 kN/m
C
A
4m
B
4m
Compatibility equations:
∆ 'B + f 'BB RB + f 'CB RC = 0 − − − (1)
2 kN/m
C
A
4m
B
∆'B
∆'C
4m
f 'CB
f 'BB
C
× RB
1 kN
A
4m
B
4m
f 'CC
f 'BC
A
∆ 'C + f 'BC RB + f 'CC RC = 0 − − − (2)
4m
B
× RC
4m
C
1 kN
34
• Solve equation
2 kN/m
C
A
4m
B
Compatibility equations:
4m
−
2 kN/m
B
−
C
A
∆'B = −
f 'BB =
A
64
EI
21.33
EI
149.33 53.33
170.67
+
RB +
RC = 0 − −(2)
EI
EI
EI
∆'C = −
RB = 3.71 kN , ↑
× RB
RC = −0.29 kN , ↓
1 kN
53.33
f ' BC =
EI
A
149.33
EI
53.33
f 'CB =
EI
C
64 21.33
53.33
+
RB +
RC = 0 − −(1)
EI
EI
EI
f 'CC =
C
170.67
EI
1 kN
× RC
35
• Diagram
M A = 8(2) + 0.29(8) − 3.71(4)
= 3.48 kN • m
2 kN/m
C
A
4m
B
4m
Ay = 8 + 0.29 − 3.71 = 4.58 kN 3.71 kN
0.29 kN
V (kN)
4.58
0.29
2.29 m
x (m)
-3.42
M (kN•m)
1.76
x (m)
-3.48
-1.16
Deflected shape
x (m)
Point of inflection
36
Example 5
Draw the quantitative Shear and moment diagram and the qualitative deflected
curve for the beam shown below.
(a) The support at B does not settle
(b) The support at B settles 5 mm.
Take E = 200 GPa, I = 60(106) mm4.
16 kN
B
A
2m
2m
C
4m
37
SOLUTION
• Principle of superposition
16 kN
B
A
C
∆B = 5 mm
2m
16 kN
4m
=
2m
+
∆´B
fBB
× RB
1 kN
Compatibility equation :
+
∆ B = 0 = ∆ ' B + f BB RB
+
∆ B = −0.005m = ∆ ' B + f BB RB
-----(1)
-----(2)
: no settlement
: with settlement
38
• Use conjugate beam method in obtaining ∆´B
16 kN
Real A
beam
12 kN
∆´B
2m
2m
24
M´
diagram
24
Conjugate EI
beam
56
EI
Real
C beam
B
4
3
24
EI
2
3
4m
4 kN
∆´B
16
16
EI
72
EI
2m
32
EI
M´´B
4m
40
EI
V´´B
4
3
40
EI
2
4
3
32 4 40
M
(4) = 0
−
'
'
+
( )−
ΣM
=
0:
+
B
B
EI 3 EI
∆'B = M ' 'B = −
117.33
,↓
EI
39
• Use conjugate beam method in obtaining fBB
fBB
Real A
beam
0.5 kN
B
C
1 kN
4m
4m
0.5 kN
m´
diagram
m´´B
v´´B
-2
Conjugate
beam
4
EI
2
4
3
4
3
−
+ ΣMB = 0:
4
−
EI
−
2
EI
4
−
EI
4
EI
− m' ' B −
f BB
4
EI
4
EI
4 4
4
( )+
(4) = 0
EI 3 EI
10.67
= m' ' B =
,↑
EI
fBB
40
• Substitute ∆´B and fBB in Eq. (1)
+↑; 0= −
117.33
,↓
EI
10.67
=
,↑
EI
∆'B = −
117.33 10.67
+
RB ,
EI
EI
f BB
RB = 11.0 kN,
16 kN
+
fBB
∆´B
4 kN
16 kN
A
RA = 6.5 kN
1 kN
0.5 kN
0.5 kN
=
12 kN
xRB = 11.0 kN
B
11.0 kN
C
no settlement
RC = 1.5 kN
41
• Substitute ∆´B and fBB in Eq. (2)
+ ↑ ; − 0.005m = −
117.33
,↓
EI
10.67
=
,↑
EI
∆'B = −
f BB
117.33 10.67
+
RB
EI
EI
(−0.005m) EI = −117.33 + 10.67 RB
(−0.005)(200 × 60) = −117.33 + 10.67 RB
RB = 5.37 kN,
16 kN
+
fBB
∆´B
4 kN
16 kN
A
RA = 9.31 kN
1 kN
0.5 kN
0.5 kN
=
12 kN
xRB = 5.37
B
5.37 kN
C
with 5 mm settlement
RC = 1.32 kN
42
• Quantitative shear and bending diagram and qualitative deflected curve
16 kN
16 kN
B
A
C
B
A
C
∆B = 5 mm
6.5 kN
11 kN
2m
V
diagram
2m
6.5
1.5 kN
9.31 kN
4m
2m
5.37 kN
2m
V
diagram
-1.32
-6.69
13
+
-
M
diagram
-6
Deflected
Curve
4m
9.31
1.5
-9.5
M
diagram
1.32 kN
Deflected
Curve
18.62
+
5.24
∆B = 5 mm
43
Example 6
Calculate supports reactions and draw the bending moment diagrams for (a)
D2 = 0 and (b) D2 = 2 mm.
1
2
4m
w 2 kN/m 3
6m
44
Compatibility equation:
∆´2 + f22R2 = ∆2
1
2
w 2 kN/m 3
Use formulations provided :
wx
( x 3 − 2 Lx 2 + L3 )
24 EI
(2)(4) 3
=−
(4 − 2 ×10 × 4 2 + 103 )
24 EI
248
=−
= −6.2mm, ↓
(200)(200)
∆ '2 = −
4m
6m
=
2 kN/m
∆´2
Pbx 2
(L − b2 − x2 )
6 LEI
1× 6 × 4
=−
(10 2 − 6 2 − 4 2 )
6 ×10 EI
19.2
=
= +0.48 mm, ↑
(200)(200)
f 22 =
+
f22
×R2
1
45
For ∆2 = 0
For ∆2 = 2 mm
Compatibility.equation:
Compatibility equation:
-6.2 + 0.48R2 = 0
R2 = 12.92 kN
1
2
1
w 2 kN/m 3
12.92 kN
6m
2.25 kN 4 m
-6.2 + 0.48R2 = -2
R2 = 8.75 kN
4.83 kN
4.75 kN 4 m
7.17
V
(kN)
4.75
2.415 m
2.25
+
-
-5.75
1.125 m
x (m)
-4.83
2
w 2 kN/m 3
8.75 kN 6 m
5.5
3.25 m
V +
+
(kN)
2.375 m -3.25
-
1.27
+
-7
x (m)
-6.5
5.85
M
+
(kN•m)
6.5 kN
10.56
5.64
x (m)
M
(kN•m)
+
3
+
x (m)
46
APPENDIX
Basic Beams: Single span
1
wL/2
w
L/3
L/3
wL/2
5wL4
384 EI
L
wL/2
V
- wL/2
wL2/8
M
47
• Find ∆1 by Castigliano’s
2
1
P
3
w
wL P
+
2
2
w
x1
wL P
+
2
2
L/2
x1
L/2
L
M1
V1
wL P
+
2
2
x2
w
M2
V2
x2
wL P
+
2
2
+ ΣΜ = 0;
2
wx
wL P
M1 + 1 − (
+ ) x1 = 0
2
2
2
2
wx1
wL P
M1 = −
+(
+ ) x1 = M 2
2
2
2
48
2
wx
wL P
M1 = − 1 + (
+ ) x1 = M 2
2
2
2
L/2
∆1 = ∆ max =
∫
0
∂M
dx
( 1 )M1 1 +
∂P
EI
L/2
=2∫ (
0
=
2
EI
∫
0
(
∂M 2
dx
)M 2 2
∂P
EI
∂M 1
dx
)M1 1
∂P
EI
L/2
∫
0
L/2
0
x − wx1
wL P
( 1 )(
+(
+ ) x1 )dx1
2
2
2
2
2
5wL4
=
384 EI
49
Single span
1
P
P/2
P/2
PL3
L3
δ 11 =
= f11 P → f11 =
48 EI
48 EI
L /2
L /2
L
P/2
V
+
- P/2
PL/4
M
50
Double span
∆ max
wl 4
=
185 EI
1
∆ max
w
0.375 wl
0.375 wl
1.25 wl
l = L/2
wl 4
=
185 EI
l = L/2
L = 2l
0.625 wl
0.375 wl
+
+
-
-0.375 wl
-0.625 wl
0.070 wl2
+
0.070 wl2
+
-0.125 wl2
51
Triple span
1
∆max = wl4
2
w
145 EI
0.4wl
0.4wl
1.1wl
l = L/3
V
1.1wl
l = L/3
l = L/3
0.6wl
0.5wl
0.4wl
+
+
+
-
-
-
- 0.6wl
0.08 wl2
M
- 0.4wl
-0.5wl
0.08 wl2
0.025 wl2
+
+
-0.1 wl2
+
-0.1 wl2
52
APPROXIMATE ANALYSIS OF STATICALLY
INDETERMINATE STRUCTURES
!
!
!
!
Trusses
Vertical Loads on Building Frames
Lateral Loads on Building Frames: Portal
Method
Lateral Loads on Building Frames: Cantilever
Method Problems
1
Trusses
P1
P2
a
a
F1
Fa
V = R1
Fb
F2
R1
a
(a)
R2
R1
a
(b)
Method 1 : If the diagonals are intentionally designed to be long and slender,
it is reasonable to assume that the panel shear is resisted entirely by the tension diagonal,
whereas the compressive diagonal is assumed to be a zero-force member.
Method 2 : If the diagonal members are intended to be constructed from large
rolled sections such as angles or channels, we will assume that the tension and compression
diagonals each carry half the panel shear.
2
Example 1
Determine (approximately) the forces in the members of the truss shown in
Figure. The diagonals are to be designed to support both tensile and compressive
forces, and therefore each is assumed to carry half the panel shear. The support
reactions have been computed.
F
E
D
3m
C
A
B
4m
10 kN
4m
20 kN
3
20 kN
F
E
D
+ ΣMA = 0:
FFE(3) - 8.33cos36.87o(3) = 0
FFE = 6.67 kN (C)
3m
C
A
B
4m
10 kN
4m
10 kN
20 kN
+ ΣMF = 0:
FAB(3) - 8.33cos36.87o(3) = 0
FAB = 6.67 kN (T)
20 kN
F
θ
θ = 36.87o
FFE
FFB= F
F =F
θ AE
FAB
3m
A
FAF
V = 10 kN
A
10 kN
+ ΣFy = 0:
8.33
θ
6.67 kN
10 kN
20 - 10 - 2Fsin(36.87o) = 0
F = 8.33 kN
FFB = 8.33 kN (T)
FAE = 8.33 kN (C)
+ ΣFy = 0:
FAF - 10 - 8.33sin(36.87o) = 0
FAF = 15 kN (T)
4
θ = 36.87o
D
FED
V = 10 kN
F = FDB
F = FEC
θ
θ
FBC
θ = 36.87o
6.67 kN
3m
D
θ
8.33 kN
C
FDC
10 kN
+ ΣFy = 0:
10 - 2Fsin(36.87o) = 0
+ ΣFy = 0:
FDC - 8.33sin(36.87o) = 0
F = 8.33 kN
FDB = 8.33 kN (T)
FEC = 8.33 kN (C)
+ ΣMC = 0:
FED(3) - 8.33cos36.87o(3) = 0
FDC = 5 kN (C)
θ = 36.87o
6.67 kN
E
θ
θ
6.67 kN
8.33 kN
8.33 kN
FEB
FED = 6.67 kN (C)
+ ΣFy = 0:
+ ΣMD = 0:
FBC(3) - 8.33cos36.87o(3) = 0
FBC = 6.67 kN (T)
FEB = 2(8.33sin36.87o) = 10 kN (T)
5
Example 2
Determine (approximately) the forces in the members of the truss shown in
Figure. The diagonals are slender and therefore will not support a compressive
force. The support reactions have been computed.
10 kN
20 kN
J
I
20 kN
H
20 kN
G
10 kN
F
4m
A
40 kN
4m B
4m C 4m D 4m
E
40 kN
6
10 kN
FJA
J
4m
45o
A
FJB
FJI
FAI = 0
FAB
θ
A
V = 30 kN
0
0
40 kN
+ ΣFy = 0:
40 kN
10 kN
FJA = 40 kN (C)
20 kN
FAI = 0
I
J
+ ΣFy = 0:
40 - 10 - FJBcos 45o = 0
FJB = 42.43 kN (T)
4m
A
+ ΣMA = 0:
FJI(4) - 42.43sin 45o(4) = 0
45o
4m B
FAB(4) = 0
FAB = 0
FIC
FBH = 0
FBC
V = 10 kN
40 kN
FJI = 30 kN (C)
+ ΣMJ = 0:
FIH
FBH = 0
+ ΣFy = 0:
40 - 10 - 20 - FICcos 45o = 0
FIC = 14.14 kN (T)
7
10 kN
20 kN
J
FIH
45o
4m
A
42.3 kN
I
4m B
FIC
FBH = 0
FBC
V = 10 kN
0
+ ΣFy = 0:
FBI
45o 0
45o
45o
B
30 kN
FBI = 42.3 sin 45o = 30 kN (T)
40 kN
+ ΣMB = 0:
FIH(4) - 14.14sin
14.14 kN
45o(4)
14.14 kN
+ 10(4) - 40(4) = 0
FJH = 40 kN (C)
+ ΣMI = 0:
FCH
FBC(4) - 40(4) + 10(4) = 0
FBC = 30 kN (T)
30 kN
45o
45o
C
30 kN
+ ΣFy = 0:
FBI = 2(14.1442.3 sin 45o) = 20 kN (C)
8
Vertical Loads on Building Frames
typical building frame
9
• Assumptions for Approximate Analysis
column
w
column
girder
A
B
L
(a)
w
w
A
Point of
zero A
moment
B
point of zero
0.21L
0.21L moment
L
(b)
w
0.1L
approximate case
(d)
B
L
Simply supported
(c)
assumed
points of
zero moment 0.1L
L
Point of
zero
moment
w
0.1L
0.8L
model
0.1L
(e)
10
Example 3
Determine (approximately) the moment at the joints E and C caused by members
EF and CD of the building bent in the figure.
1 kN/m
F
E
1 kN/m
D
C
B
A
6m
11
1 kN/m
4.8 kN
1 kN/m
4.8 m
4.8 m
0.1L=0.6 m
2.4 kN
0.6 m
0.6 kN
2.4 kN
2.4 kN 2.4 kN
0.6(0.3) + 2.4(0.6) = 1.62 kN•m
3 kN 0.6 m
0.6 kN
1.62 kN•m
0.6 m 3 kN
12
Portal Frames and Trusses
∆
∆
• Frames: Pin-Supported
P
P
h
assumed
hinge
h
l
(a)
Ph/2
Ph/2
l
(b)
L/2
Ph/2
L/2
P
P/2
P/2
Ph/2
Ph/l
Ph/l
h
h
P/2
(d)
Ph/l
P/2
(c)
Ph/l
13
∆
• Frames : Fixed-Supported
∆
P
P
assumed
hinges
h
h
l
(a)
l
(b)
L/2
P
P/2
h/2
Ph/4
Ph/4
P/2
Ph/2l
Ph/4
P/2
Ph/2l
Ph/2l
Ph/2l
Ph/2l
P/2
Ph/4
Ph/4
h/2
Ph/2l
P/2
Ph/4
L/2
P/2
h/2
P/2
Ph/4
(d)
Ph/2l
(c)
Ph/2l
P/2
Ph/4
14
• Frames : Partial Fixity
P
P
assumed
hinges
θ
h
h/3
θ
h/3
l
(a)
(b)
• Trusses
P
P
∆
h
h/2
l
(a)
∆
assumed
hinges
P/2
P/2
l
(b)
15
Example 4
Determine by approximate methods the forces acting in the members of the
Warren portal shown in the figure.
2m
40 kN C
4m
D
2m
E
F
2m
B
H
G
4m
7m
A
I
8m
16
2m
4m
D
40 kN C
2m
E
F
2m
+ ΣMJ = 0:
H
B
G
N(8) - 40(5.5) = 0
3.5 m
J
K
40/2 = 20 kN = V
N = 27.5 kN
20 kN = V
N
N
N
N
V = 20 kN
V = 20 kN + ΣM = 0:
A
3.5 m
M - 20(3.5) = 0
A
V = 20 kN
M
N
I
20 kN = V
M
M = 70 kN•m
N
17
2m
2m
40 kN C FCD D
2m
B
FEF F
E
FEG
F
45o BD
FBH
FGH
2m
45o
G
3.5 m
3.5 m
J
20 kN
K
20 kN = V
27.5 kN
+ ΣFy = 0:
-27.5 + FBDcos 45o = 0
27.5 kN
+ ΣFy = 0:
FBD = 38.9 kN (T)
+ ΣMB = 0:
FCD(2) - 40(2) - 20(3.5) = 0
FCD = 75 kN (C)
27.5 - FEGcos 45o = 0
FEG = 38.9 kN (C)
+ ΣMG = 0:
FEF(2) - 20(3.5) = 0
FEF = 35 kN (T)
+ ΣMD = 0: FBH(2) + 27.5(2) - 20(5.5) = 0
+ ΣME = 0: FGH(2) + 27.5(2) - 20(5.5) = 0
FBH = 27.5 kN (T)
FGH = 27.5 kN (C)
18
y
y
38.9 kN
D
75 kN
x
45o
45o
38.9 kN
+ ΣFy = 0:
FDE
45o
45o
27.5 kN
H
x
27.5 kN
FDH
FDHsin 45o - 38.9sin 45o = 0
FDH = 38.9 kN (C)
+ ΣF = 0:
x
FHE
+ ΣFy = 0:
FHEsin 45o - 38.9sin 45o = 0
FHE = 38.9 kN (T)
75 - 2(38.9 cos 45o) - FDE = 0
FDE = 20 kN (C)
19
Lateral Loads on Building Frames: Portal Method
P
= inflection point
(a)
V
V
(b)
V
V
20
Example 5
Determine by approximate methods the forces acting in the members of the
Warren portal shown in the figure.
5 kN B
D
F
G
3m
A
C
4m
E
4m
H
4m
21
5 kN B
3m
D
M
I
J
A
L
E
H
4m
M
G
O
K
C
4m
5 kN B
F
N
4m
N
D
O
F
G
1.5 m
I
V
J
2V
Jy
Iy
+ ΣF = 0:
x
K
2V
Ky
V
L
Ly
5 - 6V = 0
V = 0.833 kN
22
2m
5 kN B
1.5 m
0.625 kN
0.833 kN
2m D
4.167 kN
4.167 kN
M
2.501kN
1.5 m 0.625kN
J
0.625 kN
1.666 kN
I
2m N
Jy = 0
Iy = 0.625kN
2.501 kN N 2 m F
2m
O0.835 kN
O
1.5 m
0.625kN
K
1.666 kN
Ky = 0
0.625 kN 1.5 m
0.833 kN
L
0.625 kN = Ly
0.625 kN
0.625 kN
G
0.835 kN
0.625kN
I
1.5 m
2m
0.625 kN
0.833kN
J
1.5 m
C
A
0.833 kN
1.25 kN•m
1.666 kN
1.666kN
2.50 kN•m
1.666 kN
K
1.5 m
E
L
1.5 m
H
1.666kN
2.5 kN•m 0.625 kN
0.833 kN
0.833 kN
1.25 kN•m
23
Example 6
Determine (approximately) the reactions at the base of the columns of the frame
shown in Fig. 7-14a. Use the portal method of analysis.
20 kN
G
H
I
5m
30 kN
D
E
F
6m
A
B
8m
C
8m
24
20 kN
G
R
O
30 kN
D
H
S
P
M
E
I
Q
N
F
J
K
L
A
B
C
8m
5m
6m
8m
25
G
20 kN
I
2.5 m
V
2V
Oy
+ ΣF = 0:
x
G
20 kN
V
Py
20 - 4V = 0
Py
V = 5 kN
H
I
5m
D
30 kN
E
F
3m
V´
2V´
Jy
+ ΣF = 0:
x
V´
Ky
20 + 30 - 4V´ = 0
Ly
V´ = 12.5 kN
26
Ry = 3.125 kN
20 kN
G
4m R
Rx = 15 kN
R
3.125 kN
4m H
15 kN
2.5 m
10 kN
2.5 m
Sy = 3.125 kN
S
4m
Sx = 5 kN
5 kN
Oy = 3.125
Py = 0 kN
3.125 kN
O
2.5 m
30 kN
3m
12.5 kN
5 kN
My = 12.5 kN
M
4m
P
2.5 m
22.5 kN M
Mx = 22.5 kN
J
4m
4m N
3m
K
12.5 kN
25 kN
Jy = 15.625 kN
12.5 kN
A
Ax = 15.625 kN
3m
Ax = 12.5 kN
MA = 37.5 kN•m
Ny = 12.5 kN
Nx = 7.5 kN
Ky = 0 kN
15.625kN
J
10 kN
K
B
Bx = 0
25 kN
3m
Bx = 25 kN
MB = 75 kN•m
27
Lateral Loads on Building Frames: Cantilever Method
P
beam
(a)
building frame
(b)
In summary, using the cantilever method, the following assumptions apply to
a fixed-supported frame.
1. A hinge is place at the center of each girder, since this is assumed to be point
of zero moment.
2. A hinge is placed at the center of each column, since this is assumed to be
a point of zero moment.
3. The axial stress in a column is proportional to its distance from the centroid
of the cross-sectional areas of the columns at a given floor level. Since stress equals force
per area, then in the special case of the columns having equal cross-sectional areas,
the force in a column is also proportional to its distance from the centroid of the column areas.
28
Example 7
Determine (approximately) the reactions at the base of the columns of the frame
shown. The columns are assumed to have equal crossectional areas. Use the
cantilever method of analysis.
30 kN
C
D
4m
B
E
15 kN
4m
A
F
6m
29
30 kN
C
I
4m
B
D
K
H
J
E
15 kN
4m
x
G
6m
L
A
F
6m
~
x A 0( A) + 6( A)
∑
x=
=
=3
A
A
A
+
∑
30
3m
3m
+ ΣMM = 0:
C
30 kN
2m
I
Hx
M
-30(2) + 3Hy + 3Ky = 0
D
The unknowns can be related by proportional triangles,
that is
Kx
Hy
Hy
Ky
3
=
Ky
3
or
Hy = Ky
H y = K y = 10 kN
C
30 kN
D
I
4m
B
K
H
J
15 kN
2m
3m
3m
Lx
E
Gy
-30(6) - 15(2) + 3Gy + 3Ly = 0
The unknowns can be related by proportional triangles,
that is
Gy
3
N
Gx
+ ΣMN = 0:
=
Gy
3
or
G y = Ly
G y = L y = 35 kN
Ly
31
Iy = 10 kN
C
30 kN
3m
Ix = 15 kN
2m
Hx = 15 kN
10 kN
3m D
I
15 kN
Kx = 15 kN
10 kN
10 kN
10 kN
10 kN
15 kN
H
2m
Jy = 25 kN
J
15 kN
3m
2m
Jx = 7.5 kN
Gx = 22.5 kN
35 kN
Ax = 35 kN
15 kN
2m
3m
2m
L
25 kN
Lx = 22.5 kN
35 kN
35 kN
22.5 kN
G
K
7.5 kN J
G
A
2m
2m
Ax = 22.5 kN
MA = 45 kN•m
35 kN
L
F
Fy = 35 kN
22.5 kN
2m
Fx = 22.5 kN
MF = 45 kN•m
32
Example 8
Show how to determine (approximately) the reactions at the base of the columns
of the frame shown. The columns have the crossectional areas show. Use the
cantilever of analysis.
P
35 kN
6000 mm2
4 mL
Q
5000mm2 4000 mm2
M
N
I
45 kN
6 mE
J
F
6000 mm2
A
6m
R
O
K
G
5000mm2 4000 mm2
B
C
4m
6000 mm2
H
6000 mm2
D
8m
33
P
35 kN
Q
4 mL
M
N
I
45 kN
R
O
J
6 mE
F
A
G
B
6m
K
H
C
4m
D
8m
5000mm2 4000 mm2
6000 mm2
6m
4m
6000 mm2
8m
x
~
x A 6000(0) + 5000(6) + 4000(10) + 6000(18)
∑
=
x=
= 8.48 m
A
6000
+
5000
+
4000
+
6000
∑
34
P
35 kN
2m
Q
Mx
Lx
R
Nx
Ox
My
Ly
Ny
Oy
2.48 m 1.52 m
8.48 m
+ ΣMNA = 0:
9.52 m
-35(2) + Ly(8.48) + My(2.48) + Ny(1.52) + Oy(9.52) = 0
-----(1)
Since any column stress σ is proportional to its distance from the neutral axis
σM
2.48
σN
1.52
σO
9.52
=
=
=
σL
8.48
σL
8.48
σL
8.48
;
σM =
2.48
σL ;
8.48
;
σN =
1.52
σL ;
8.48
;
σO =
9.52
σL ;
8.48
Solving Eqs. (1) - (4) yields
My
−6
5000(10 )
Ny
−6
4000(10 )
Oy
6000(10 −6 )
=
Ly
2.48
(
)
−6
8.48 6000(10 )
− − − − − ( 2)
=
Ly
1.52
(
)
−6
8.48 6000(10 )
− − − − − (3)
=
Ly
9.52
(
)
8.48 6000(10 −6 )
− − − − − ( 4)
Ly = 3.508 kN
Ny = 0.419 kN
My = 0.855 kN
Oy = 3.938 kN
35
35 kN
4m L
M
I
45 kN
3m
O
J
Fx
Ex
N
K
Gx
Fy
Hx
Gy
Ey
Hy
8.48 m
+ ΣMNA = 0:
2.48 m 1.52 m
9.52 m
-45(3) - 35(7) + Ey(8.48) + Fy(2.48) + Gy(1.52) + Hy(9.52) = 0 -----(5)
Since any column stress σ is proportional to its distance from the neutral axis ;
Fy
Ey
σF
σ
2.48
2.48
= E ; σF =
σE ;
=
(
)
− − − − − (6)
−6
−6
2.48 8.48
8.48
5000(10 ) 8.48 6000(10 )
σG
1.52
σH
9.52
=
=
σE
8.48
σE
8.48
1.52
σE ;
8.48
;
σG =
;
9.52
σH =
σE ;
8.48
Gy
Ey
1.52
(
)
8.48 6000(10 −6 )
− − − − − (7)
Ey
9.52
=
(
)
6000(10 −6 ) 8.48 6000(10 −6 )
− − − − − (8)
4000(10 −6 )
Hy
=
Solving Eqs. (1) - (4) yields Ey = 19.044 kN Fy = 4.641 kN Gy = 2.276 kN Hy = 21.38 kN
36
3m
35 kN
2m
Lx= 5.262 kN
Py= 3.508 kN
Px= 29.738 kN
3.508 kN
2m
45 kN
3m
Ex= 64.44 kN
3.508 kN
5.262 kN
Iy= 15.536 kN
I
Ix= 114.702 kN
3m
19.044 kN
19.044 kN
E
64.44 kN
3m
Ax = 19.044 kN
Ax = 64.44 kN
MA = 193.32 kN•m
One can continue to analyze the
other segments in sequence, i.e.,
PQM, then MJFI, then FB, and so on.
37
ANALYSIS OF STATICALLY INDETERMINATE
STRUCTURES BY THE FORCE METHOD
!
!
!
!
!
!
Force Method of Analysis: Beams
Maxwell’s Theorem of Reciprocal Displacements;
Betti’s Law
Force Method of Analysis: Frames
Force Method of Analysis: Trusses
Force Method of Analysis: General
Composite Structures
1
Force Method of Analysis : Beams
1 Degree of freedom
• Compatibility of displacement
M1
P
1
2
C
A
M1
B
R1
P
1
2
C
A
R1
R2
L
R2
=
P
C
A
∆´2
B
L
=
P
• Compatibility of slope
B
θ ´1
+
+
1
α11
× M1
f22 × R2
1
∆´2 + f22 R2
= ∆2 = 0
θ ´1 + α11M1
= θ 1= 0
2
2 Degree of freedom
Ax
P
A
1
B
Ay
R1
P
2
C
R2
B
A
D
∆´2
+
f21
f11
f12
Dy
C
∆´1
A
D
D
xR
R11
1
+
f22
A
D
xR
R22
1
∆´1 + f11 R1 + f12 R2
∆´2 + f21 R1 + f22 R2
= ∆1 = 0
= ∆2 = 0
3
Maxwell’s Theorem of Reciprocal Displacements; Betti’s Law
f21
1
1
2
A
B
f11
f21
mm
m2 M 1
dx = ∫ 2 1 dx
EI
EI
L
L
1 • f 21 = ∫
m1
m2 m1
dx
EI
L
f 21 = ∫
1
A
B
f22
f12
m2
4
1
f12
m2 m1
dx
EI
L
2
f 21 = ∫
1
A
B
f22
f12
mm
m1M 2
dx = ∫ 1 2 dx
EI
EI
L
L
1 • f12 = ∫
m1m2
dx
EI
L
f12 = ∫
m2
f 21 = f12
Maxwell’s Theorem:
1
A
B
f ij = f ji
f21
f11
m1
5
f11, f22
1
2
1
A
B
f11
f21
m1M 1
mm
dx = ∫ 1 1 dx
EI
EI
L
L
1 • f11 = ∫
m2 M 2
mm
dx = ∫ 2 2 dx
EI
EI
L
L
1 • f 22 = ∫
m1
In general,
1
A
B
1 • f ij = f = ∫
mi m j
L
f22
f12
1 • f ji = f ji = ∫
L
EI
m j mi
EI
dx
dx
m2
6
1
A
2
P1
D
d11 = f11 P1
d21 = f21 P1
P2
A
D
d12 = f12 P2
d22 = f22 P2
7
Force Method of Analysis: General
1
Compatibility Eq.
2
w
∆´1 + f11R1 + f12R2 = ∆1 = 0
∆´2 + f21R1 + f22R2 = ∆2 = 0
0
=
∆´1
w
∆´1
∆´2
∆´2
f12 f22 R
2
f11 f12 R1
+
f12 f22 R
2
f21
f11
+
f11 f12 R1
xR1
+
f22
1
xR2
=-
0
∆2
∆´1
∆´2
General form:
1
f12
=
∆1
f11 f12 f1n
f21 f22 f2n
..
.
fn1 fn2 fnn
R1
R2
..
.
Rn
∆´1
∆´2
.
= - ..
∆´n
8
Example 9-1
Determine the reaction at all supports and the displacement at C.
50 kN
B
C
A
6m
6m
9
SOLUTION
Use compatibility of displacement for find reaction
• Principle of superposition
50 kN
MA
A
6m
6m
RB
=
RA
B
C
50 kN
∆´B
+
RBB
fBB x R
Compatibility equation : ∆ ' B + f BB RB = 0
-----(1)
1
10
• Use formulation for ∆´B and fBB
50 kN
A
6m
C
B ∆´C
6m
6θ´C
θ´C
∆´B = ∆´C+ (6 m)θ´C
P(6) 3
P ( 6) 2
∆'B =
+ ( 6)
3EI
2 EI
A
50(6)3
(50)(6) 2 9000
=
+ (6)
=
,↓
3EI
2 EI
EI
C
f BB
∆´B
PL3 (1)(12) 3 576
=
=
=
,↑
EI
3EI
3EI
B
fBB
1
11
• Substitute ∆´B and fBB in Eq. (1): ∆ 'B + f BB RB = 0
+ ↑: −
∆'B =
9000 576
+(
) RB = 0
EI
EI
RB = 15.63 kN,
9000
,↓
EI
f BB =
576
,↑
EI
50 kN
6m
MA
A
6m
C
B
15.63 kN
RA
Equilibrium equation :
+ ΣMA = 0:
+ ΣFy = 0:
M A − 50(6) + 15.63(12) = 0,
+ RA − 50 + 15.63 = 0,
MA = 112.4 kN,
+
Ra = 34.37 kN,
12
• Quantitative shear and bending diagram and qualitative deflected curve
50 kN
6m
112.4 kN•m
A
6m
B
C
15.63 kN
34.37 kN
34.37
V
(kN)
x (m)
-15.63
-15.63
93.78
M
(kN•m)
x (m)
3.28
-112.44
6
12
13
Or use compatibility of slope to obtain reaction
50 kN
• Principle of superposition
6m
MA
6m
B
C
A
=
RA
RB
50 kN
A
B
C
θ ´A
1
+
fAA
A
B
C
xM
MAA
Compatibility equation :
θ ' A + f AA M A =θ A= 0
-----(2)
14
• Use the table on the inside front cover for θ´B and fBB
50 kN
A
C
B
C
B
θ ´A
PL2
θ A'=
16 EI
1
fAA
A
f CC =
L
3EI
Substitute the values in equation: θ ' A + f AA M A =θ A= 0
+:
PL2
L
−
+
MA =0
16 EI 3EI
MA =
3PL 3(50)(12)
=
16
16
= 112.5 kN•m, +
15
Or use Castigliano least work method
50 kN
x1
x2
12RB - 300 = MA
B
C
A
6m
50 - RB = RA
M 12RB - 300
diagram
L
∆B = 0 = ∫ (
0
6m
RB
M2 = RBx2
M1 = (12RB - 300) + (50 - RB)x1
x (m)
∂M M
)
dx
∂RB EI
6
6
1
1
0=
(
12
−
x
)(
12
R
−
300
+
50
x
−
R
x
)
dx
+
x2 ( RB x2 ) dx2
1
B
1
B 1
1
EI ∫0
EI ∫0
2
2
3
3
3
6
900 x1 24 x1
50 x1 x1
x
0 = (144 RB x1 − 3600 x1 +
−
RB −
+
RB ) 0 + 2 RB
2
2
3
3
3
RB =15.63 kN,
6
0
16
Use conjugate beam for find the displacement
50 kN
6m
112 kN•m
6m
C
B
A
∆C
34.4 kN
Real Beam
15.6 kN
93.6
M
(kN•m)
x (m)
6
3.28
12
93.6/EI
-112
Conjugate Beam
-112/EI
281
223
776
( 2) −
(6 ) = −
M 'C =
EI
EI
EI
∆ C = M 'C = −
776
,↓
EI
281/(EI)
223/(EI)
M´C
V´C 2 m
4m
223/(EI)
17
Use double integration to obtain the displacement
50 kN
6m
112 kN•m
6m
C
B
A
∆C
34.4 kN
Real Beam
15.6 kN
93.6
M
(kN•m)
x (m)
3.28
-112
6
12
d 2υ
EI 2 = −112 + 34.4 x1
dx
dυ
x12
EI
= −112 x1 + 34.4 + C1
dx
2
2
x1
x13
EIυ = −112.4 + 34.4 + C1 x1 + C2
2
6
1
62
63
778
(−112( ) + 34.4( ) + 0 + 0) = −
∆C =
,↓
2
6
EI
EI
18
Example 9-2
Draw the quantitative Shear and moment diagram and the qualitative deflected
curve for the beam shown below.The support at B settles 5 mm.
Take E = 200 GPa, I = 60(106) mm4.
16 kN
B
A
C
∆B = 5 mm
2m
2m
4m
19
Use compatibility of displacement to obain reaction
• Principle of superposition
16 kN
B
A
C
∆B = 5 mm
2m
2m
16 kN
4m
=
SOLUTION
+
∆´B
fBB
× RB
1 kN
Compatibility equation :
∆ B = −0.005m = ∆ 'B + f BB RB
-----(1)
20
• Use conjugate beam method for ∆´B
16 kN
Real A
beam
12 kN
B
∆´B
2m
2m
24
M´
diagram
24
Conjugate EI
beam
56
EI
C
4
3
24
EI
4m
4 kN
16
16
EI
72
EI
M´´B
2
3
2m
4m
40
EI
V´´B
4
3
32
EI
40
EI
2
4
3
32 4 40
+ ΣMB = 0: − M ' ' B +
(4) = 0
( )−
EI 3 EI
117.33
∆' B = M ' ' B = −
,↓
21
EI
• Use conjugate beam method for fBB
fBB
Real A
beam
0.5 kN
B
C
1 kN
4m
4m
0.5 kN
m´
diagram
m´´B
vB´B
-2
Conjugate
beam
4
EI
2
4
3
4
3
−
+ ΣMB = 0:
4
−
EI
−
2
EI
4
−
EI
4
EI
4
EI
4
EI
4 4
4
( )+
(4) = 0
EI 3 EI
10.67
f BB = mB ' ' =
,↑
EI
− mB ' '−
22
∆ B = −0.005 m = ∆ ' B + f BB RB
• Substitute ∆´B and fBB in Eq. (1):
+ ↑: − 0.005 = −
117.33 10.67
RB
+
EI
EI
117.33
,↓
EI
10.67
= mB ' ' =
,↑
EI
∆'B = M ' 'B = −
(−0.005) EI = −117.33 + 10.67 RB
f BB
(−0.005)(200 × 60) = −117.33 + 10.67 RB
RB = 5.37 kN,
+
16 kN
xRB = 5.37
1 kN
16 kN
A
0.5 kN
0.5 kN
=
4 kN
12 kN
B
RA = 9.31 kN 5.37 kN
C
RC = 1.32 kN
23
• Quantitative shear and bending diagram and qualitative deflected curve
16 kN
B
A
C
∆B = 5 mm
RA = 9.31 kN 5.37 kN
2m
2m
RC = 1.32 kN
4m
9.31
V
diagram
-1.32
-6.69
M
diagram
Deflected
Curve
18.62
5.24
∆B = 5 mm
24
Or use Castigliano least work method
• Principle of superposition
16 kN
RA = 12 - 0.5RB
16 kN
12
0.5RB
RC = 4 - 0.5RB
4m
=
2m
+
2m
RB
RB
4
0.5RB
25
16 kN
x1 x2
x3
RB
RA = 12 - 0.5RB
2m
x1
2m
RC = 4 - 0.5RB
4m
M1 = (12 - 0.5RB)x1
x3
(4 - 0.5RB)x3 = M3
M2 = 0.5x2RB + 16 - 2RB + 4 x2
V1
12 - 0.5RB
V2
L
∆ B = −0.005 = ∫ (
0
∂M i M i
)
dx
∂RB EI
RB
x2
2
V3
4 - 0.5RB
4 - 0.5RB
4m
2
1
1
− 0.005 =
(
−
0
.
5
x
)(
12
x
−
0
.
5
x
R
)
dx
+
(0.5 x2 − 2)(0.5 x2 RB + 16 − 2 RB + 4 x2 )dx2
1
1
1 B
1
EI ∫0
EI ∫0
4
1
+
(−0.5 x3 )(4 x3 − 0.5 x3 RB )dx3
∫
EI 0
− 0.005 EI = −117.34 + 10.66 RB , RB = 5.38 kN,
26
Example 9-3
Draw the quantitative Shear and moment diagram and the qualitative deflected
curve for the beam shown below.EI is constant. Neglect the effects of axial load.
5 kN/m
A
B
4m
4m
27
SOLUTION
Use compatibility of displacement to obtain reaction
• Principle of superposition
5 kN/m
θA=0
4m
4m
θB=0
B
=
A
5 kN/m
θ´B
+
θ´A
1 kN•m
αBA
+
αAA
×MA
αAB
Compatibility equation :
1 kN•m × M
B
αBB
θ A = 0 = θ ' A +α AA M A + α AB M B
-----(1)
θ B = 0 = θ 'B +α BA M A + α BB M B
-----(2)
28
• Use formulation: θ´A, θ´B, αAA, αBA, αBB, αAB,
5 kN/m
θ´A
θ´B
3wL3 3(5)(8) 3 60
θ 'A =
=
=
EI
128 EI
128EI
7 wL3
7(5)(8) 3 46.67
θ 'B =
=
=
EI
384 EI
384 EI
1 kN•m
αAA
αBA
1 kN•m
αAB
αBB
α AA =
M o L 1(8) 2.67
=
=
3EI 3EI
EI
α BB =
M o L 1(8) 2.67
=
=
3EI 3EI
EI
α BA =
M o L 1(8) 1.33
=
=
6 EI 6 EI
EI
α AB =
M o L 1(8) 1.33
=
=
6 EI 6 EI
EI
Note : Maxwell’s theorem of reciprocal displacement is αAB = αBA
29
• Substitute θ´A, θ´B, αAA, αBA, αBB, αAB, in Eq. (1) and (2)
5 kN/m
θA=0
4m
4m
B
θB=0
θ A = 0 = θ ' A +α AA M A + α AB M B
-----(1)
θ B = 0 = θ 'B +α BA M A + α BB M B
-----(2)
+
+
α AA
α AB =
=
A
60
EI
2.67
=
EI
θ 'A =
θ 'B =
1.33
EI
46.67
EI
1.33
EI
2.67
=
EI
α BA =
0=
60 2.67
1.33
+(
)M A + (
)M B
EI
EI
EI
0=
46.67 1.33
2.67
+(
)M A + (
)M B
EI
EI
EI
α BB
Solving these equations simultaneously, we haave
MA = -18.31 kN•m,
+
MB = -8.36 kN•m,
+
30
MA = -18.31 kN•m,
+
MB = -8.36 kN•m,
+
5 kN/m
18.31 kN•m
A
RA
+ ΣMA = 0:
+ ΣFy = 0:
B
4m
4m
8.36 kN•m
RB
18.31 − 20( 2) + RB (8) − 8.36 = 0,
3.76
+ R A − 20 + RB = 0,
RB = 3.76 kN,
Ra = 16.24 kN,
31
• Quantitative shear and bending diagram and qualitative deflected curve
5 kN/m
18.31 kN•m
A
B
4m
16.24 kN
4m
8.36 kN•m
3.76 kN
16.24
V
diagram
3.25 m
M
diagram
8.08
-3.76
6.67
-8.36
-18.31
Deflected
Curve
32
Force Method of Analysis : Frames
• Principle of superposition
fCC × C x
B
∆´CH
C
Cx
1 kN
Cy
w
=
w
+
Ax
A
Ay
Compatibility equation :
∆ CH = 0 = ∆'CH + f CC C x
33
Example 9-4
Draw the quantitative Shear and moment diagram and the qualitative deflected
curve for the Frame shown below.EI is constant.
2 kN/m
B
6m
C
6m
A
34
SOLUTION
Use compatibility of displacement to obtain reaction
• Principle of superposition
fCC × C x
B
6m
∆´CH
C
Cy
=
6m
2 kN/m
2 kN/m
Cx
1 kN
+
Ax
A
Ay
Compatibility equation :
∆ CH = 0 = ∆'CH + f CC C x
-----(1)
35
6m
3m
Cy
M´2 = (6 + P)x2
x2
P
2 kN/m
2 kN/m
• Use Castigliano’s method for ∆´CH
∆´CH
6m
B
C
V´2
6+P
12 kN
3m
-12 - P
V´1
2x1
-6 - P
Ay
x2
6+P
M´1 = (12 + P)x1- x12
x1
Ax
A
P
x1
-12 - P
-6 - P
0
0
6
6
1
1
2
∂M 'i M 'i
(
x
)
(
12
x
+
x
P
−
x
)dx1 +
( x2 )(6 x2 + x2 P) dx2
dx =
= ∫(
)
1
1
1
1
∫
∫
EI 0
EI 0
∂P EI
0
L
∆ 'CH
6
6
1
1
2
3
2
=
(
12
x
−
x
)
dx
+
(
6
x
) dx2
1
1
1
2
∫
∫
EI 0
EI 0
3
4
3
1 12 x1 x1 6 1 6 x2
=
(
− )0+
(
)
EI
EI 3
3
4
6
0
=
972
,→
EI
36
• Use Castigliano’s method for fCC
fCC
6m
B
C
m´2 = x2P
x2
P
1 kN
P
v´2
Cy
P
6m
x2
P
m´1 = x1P
x1
-P
Ax
A
v´1
Ay
x1
-P
-P
1
1
6
6
1
1
∂m' m'
( x1 )( x1 P)dx1 +
( x2 )( x2 P )dx2
= ∫ ( i ) i dx =
∫
∫
EI 0
EI 0
∂P EI
0
-P
L
f CC
3
3
1 x1 6 1 x2
=
( ) +
( )
EI 3 0 EI 3
6
0
=
144
,→
EI
37
• Substitute ∆´CH and fCC in Eq. (1) ∆ CH = 0 = ∆'CH + f CC C x
+:
0=
-----(1)
972
,→
EI
144
f cc =
,→
EI
∆ CH =
972 144
+
Cx
EI
EI
Cx = -6.75 kN,
6 kN
1 kN
+
× C x = −6.75kN
12 kN
=
C 6.75 kN
B
2 kN/m
2 kN/m
1 kN
0.75 kN
5.25 kN
1 kN
A
6 kN
1 kN
0.75 kN
38
Or use Castigliano least work method:
x2
B
6m
M2 = (6-Cx)x2
C
Cx
2 kN/m
Cx
A
x2
V2
6 - Cx
6 - Cx
6m
M1 = (12 - Cx)x1- x12
x1
12 - Cx
V1
2x1
6 - Cx
x1
12 - Cx
L
∂M i M i
∂U i
= ∫(
)
dx = ∆ CH = 0
∂C x 0 ∂C x EI
6
0=
6 - Cx
6
1
1
2
(
−
x
)(
12
x
−
C
x
−
x
)
dx
+
(− x2 )(6 x2 − C x x2 )dx2
1
1
1
1
x 1
EI ∫0
EI ∫0
3
3
4
3
12 x1 C x x1 x1 6
6x
C x
0 = (−
+
+ ) 0 + (− 2 + x 2
3
3
4
3
3
0 = -972 + 144Cx , Cx = 6.75 kN,
3
6
0
39
• Quantitative shear and bending diagram and qualitative deflected curve
B
6m
C
B
2 kN/m
6.75 kN
C
- 0.75
- 0.75
- 6.75
0.75 kN
6m
5.25
5.25 kN
A
V, (kN)
2.63 m
A
0.75 kN
B
B
C
-4.5
C
1.33 m
-4.5
Deflected curve
A
M, (kN•m)
6.90
A
40
Force Method of Analysis : Truss (Externally indeterminate)
E
A
D
Cx
Ax
C
B
∆'CH + f CC C x = ∆ CH = 0
Cy
Ay
P
=
E
E
D
D
+
A
C
A
B
C
P
1
x Cx
B
∆´C
fCC
41
Truss (Internally indeterminate)
P
3
D
C
6
1
2
5
A
B
4
=
∆'6 + f 66 F6 = ∆ 6 = 0
P
D
D
∆´6
A
C
+
B
f66
C
1
A
xF6
B
42
Example 9-5
Determine the reaction at support A, C, E and all the member forces. Take
= 200 GPa and A = 500. mm2 .
E
E
40 kN
D
4m
A
C
B
5m
5m
43
SOLUTION
Use compatibility of displacement to obtain reaction
• Principle of superposition
RE
E
40 kN
D
4 m Ay
Ax
A
C
B
5m
=
5m
RC
40 kN
+
× Cy
fCC
∆´C
1 kN
Compatibility equation :
∆ C = 0 = ∆'C + f CC RC
-----(1)
44
• Use unit load method for ∆´C and fcc
5.39
m
53.85
20
E
50
5.39
m
+53
.85
A
2.5
0
0
B
=- 7.81 mm,
C
∆´C
0
5m
A
Σ ni´n´iLi
fCC =
N ´i (kN)
=
0
+2.5
AiEi
2(-2.69)2(5.385)
(200x106)(500x10-6)
-2.6
9
-2.6
9
fCC
B
n´i (kN)
+2.5
+
2(2.5)2(5)
(200x106)(500x10-6)
0
2.5
(200x106)(500x10-6)
0
D
0
=
D
1
E
AiEi
(53.85)(-2.69)(5.38)
5m
53.85
∆´C =
40 kN
4 m 20 kN
85
-53.
50
Σ n´iN´iLi
C
1 kN
= 1.41 mm,
45
• Substitute ∆´Cv and fCC in Eq. (1): ∆ C = 0 = ∆'C + f CC RC
∆ 'C = 7.81 mm, ↓
+ ↑: −7.81 + 1.41RC = 0
f CC = 1.41 mm, ↑
RC = 5.54 kN , ↑
20
50
+53
.85
A
40 kN
0
0
36.15 kN
xRC = 5.54 kN
0
B
0
14.46 kN
E
+38
.95
2.5
20 kN 53.85
36.15 kN
A
+13.85
-2.6
9
0
C
N ´i (kN)
38.93 kN
-2.6
9
D
0
0
E
2.5
D
20 kN
85
.
3
5
50
53.85
+
E
1
A
+2.5
B
n´i (kN)
=
53.85
C
+2.5
1 kN
40 kN
D -1
4.90
0 21.8o
B +13.85
N i (kN)
C
5.54 kN
46
Or use Castigliano least work method:
5.39
m
-2.7 RC + 53.85 = RE
4m
Ax = -2.5RC +50 = Ax
5.39
E -2.7 R
m
40 kN
C +5
3.85
D
-2.7
5
RC
8
.
-53
0 21.8o
A
2.5RC
B
2.5RC
C
RC
Ay = 20
5m
5m
Castigliano’s Theorem of Least Work :
∆ CV = 0 = ∑ (
0=
∂N i N i Li
)
∂RC AE
1
[(−2.7)(−2.7 RC + 53.85)(5.39) + ( −2.7)(−2.7 RC )(5.39) + 0 + 0 + 2[(2.5)(2.5 RC )(5)]]
AE
0 = 39.3RC − 783.68 + 39.3RC + 62.5 RC
RC = 5.55 kN,
47
Example 9-6
Determine the force in all member of the truss shown :
(a) If the horizontal force P = 6 kN is applied at joint C.
(b) If the turnbuckle on member AC is used to shorten the member by
1 mm.
(c) If (a) and (b) are both accounted.
Each bar has a cross-sectional area of 500 mm2 and E = 200 GPa.
D
C
2m
A
B
3m
48
SOLUTION
Part (a) : If the horizontal force P = 6 kN is applied at joint C.
• Principle of superposition
D
6 kN
1
C
6
3
A
5
4
2m
B
2
3m
D
C 6 kN
=
3m
D
∆´6
1 E´
E E´
A
C
2m
B
Compatibility equation :
Note : AE + E ' C = L
1
+
E f66
A
∆´6 + f66 F6 = 0
×F6
B
----------(1)
49
• Use unit load method for ∆´6 and f66
3m
D
C 6 kN
+6
D
∆´6
+4
6
-7.
2
A
0
B
1
N ´i (kN)
-14.98
0
n´iN ´iLi (kN2•m)
1
-0.832
A
n´i (kN)
0
2.08
6
3.
0
-2
6. 0
3
-14.98
1
2 m -0.555
4
4
-4.44
3
1
E E´
+6
C
-0.832
0
-0.555
3.6
1
2
3
0
Li (m)
0.616
3.6
2
1
n'i N 'i Li
Ai Ei
=
1
[−4.44 − 26.03 − 2(14.98)]
AE
=
− 60.43
AE
2.08
n´i2Li (kN2•m)
1
B
∆ '6 = ∑
1
0.616
6
3.
2
n' L
1
12.61
f 66 = ∑ i i =
[2(0.616) + 2(2.08) + 2(3.61)] =
Ai Ei
AE
AE
50
• Substitute ∆´6 and f66 in Eq. (1)
−
3m
+6
60.43 12.61
+
( F6 ) = 0
AE
AE
F6 = 4.80 kN, (T)
D
6 kN
∆´6
+4
+6
6
C
-0.832
1
E E´
-7.
2
0
2m
+
1
4
4
-0.832
A
n´i (kN)
0
80
.
+4 2.4
1
+2
+1.34
A
4
Ni (kN)
-0.555 x F6 = 4.80 kN
B
0
6 kN
+2
D
6
1
=
0
N ´i (kN)
1
-0.555
C
-2.66
B
4
51
Part (b) : If the turnbuckle on member AC is used to shorten the member by 1 mm.
f 66 =
12.61
12.61
=
= 1.26(10-4) m = 0.126 mm
AE
(500)(200)
F6 =
D
1 mm
(1 kN ) = 7.94 kN
0.126 mm
C
-0.832
-6.61
D
94
.
+7 7
.94
-6.61
1
1
-0.555
0
1
-0.832
A
n´i (kN)
0
-0.555
B
0
x F6 = 7.94 kN
-4.41
=
6
A
4
Ni (kN)
C
-4.41
B
4
52
Part (c) : If the horizontal force P = 6 kN is applied at joint C and the turnbuckle
on member AC is used to shorten the member by 1 mm are both accounted.
4
(Ni)load (kN)
94
.
+7 7
.94
-6.61
-4.41
B
0
A
4
0
=
A
C
-2.66
-6.61
D
+
80
.
+4 2.4
1
+2
+1.34
6
6 kN
+2
D
1
-3.07
74
.
2
A
4
B
0
C
5.5
3
-4.61
6
-4.41
6 kN
-4.61
D
(Ni)short (kN)
C
(Ni)total (kN)
-7.07
B
4
53
Or use compatibility equation :
∆´6 + f66 F6 = ∆´6 = 0.001
−
60.43 12.61
+
( F6 ) = 0.001
AE
AE
F6 =
∆´6
-7
. 21
6
D
6 kN
+6
+4
0.001AE + 60.43 0.001(500)(200) + 60.43
=
= 12.72 kN, (T)
12.61
12.61
0
+
+6
4
-0.832
A
n´i (kN)
0
-4.58
D
12
-3.06
2
.7
C
5.5
1
-4.58
6
A
4
x F6 = 12.72 kN
f66 -0.555
=
4
(Ni)total (kN)
C
1
-0.555
0
N ´i (kN)
-0.832
1 1
B
0
6 kN
-7.06
B
4
54
Composite Structures
Example 9-7
Find all reaction and the tensile force in the steel support cable. Consider both
bending and axial deformation.
Steel cable
Ac = 2(10-4) m2
Ec = 200(103) kN/m2
C
2m
A
B
5 kN
m2
Ab = 0.06
Ib = 5(10-4) m4
Eb = 9.65(103) kN/m2
6m
55
SOLUTION
C
m
6.32
RC = T
2m
18.43o
A
B
x
6m
5 kN
0.316T
T
M = 0.316Tx - 5x
N = -0.949T
0.949T
V
5 kN
Bx
MB
By
By Castigliano’s Theorem of Least Work ;
∂
∆C = 0 =
(U ib + U in )
∂T
L
L
∂M M
∂N N
∆C = 0 = ∫ (
dx + ∫ ( )
)
dx
∂
T
EI
∂
T
AE
0
0
6
6
1
1
1
0=
(
0
.
316
x
)(
0
.
316
xT
−
5
x
)
dx
+
(
−
0
.
949
)(
−
0
.
949
T
)
dx
+
Eb I b ∫0
Ab Eb ∫0
Ac Ec
6
1
0.316 2 x 3
(0.316 × 5) x 3 6
1
1
2
0=
[(
T) −
]0+
(0.949 xT ) 0 +
( xT )
Eb I b
3
3
Ab Eb
Ac Ec
0 = (1.49T - 23.58) + 9.33(10-3)T + 0.158T
6.32
∫ (1)(T )dx
0
6.32
0
; T = 14.23 kN, (tension) #
56
4.5 kN
C
m
6.32
RC = T = 14.23 kN
13.5 kN
2m
18.43o
A
B
x
5 kN
Bx
6m
By
+ ΣF = 0:
x
Bx = Rc cos θ = 13.5 kN,
+ ΣFy = 0:
By = 5 - Rc sin θ = 0.5 kN,
+ ΣMB = 0:
MB
MB = 13.5(2) - 5(6) = -3 kN•m, +
57
DISPLACEMENT METHOD OF ANALYSIS:
SLOPE DEFLECTION EQUATIONS
!
!
!
!
!
!
!
!
!
General Case
Stiffness Coefficients
Stiffness Coefficients Derivation
Fixed-End Moments
Pin-Supported End Span
Typical Problems
Analysis of Beams
Analysis of Frames: No Sidesway
Analysis of Frames: Sidesway
1
Slope – Deflection Equations
i
P
j
w
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
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 Moments: Loads
P
L/2
PL
8
L/2
PL
8
L
P
2
P
2
w
wL2
12
wL2
12
wL
2
L
wL
2
6
General Case
i
P
j
w
k
Cj
settlement = ∆j
Mij
P
i
w
j
Mji
θi
ψ
θj
7
Mij
P
i
j
w
Mji
θi
L
settlement = ∆j
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 = (
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
j
w
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
(
2
/
)
(
4
/
)
EI
L
EI
L
j ji
ji
kii
k ji
[k ] =
kij
k jj
Stiffness Matrix
11
P
i
Mij
w
j
Mji
θi
[ M ] = [ K ][θ ] + [ FEM ]
L
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: Fixed-End Support
θi
Mi
Mj
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
Mj =(
)θ i
L
Mi = (
13
Stiffness Coefficients Derivation: Pinned-End Support
Mi
θi
i
θj
j
Real beam
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 = (
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
Fixed end moment : Point Load
P Real beam
Conjugate beam
A
A
B
B
L
M
EI
M
M
EI
M
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
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
16
Uniform load
w
Real beam
Conjugate beam
A
A
L
B
B
M
EI
M
M
EI
M
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
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
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
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
M BA =
θA +
θ B + ( M BA
)load + ( M BA
) ∆ − − − (2)
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
Fixed-End Moments
Fixed-End Moments: Loads
P
PL
8
L/2
L/2
PL
8
L/2
PL 1
PL
3PL
+ ( )[−(−
)] =
8
2
8
16
P
L/2
wL2
12
wL2
12
wL2 1
wL2
wL2
+ ( )[−( −
)] =
2
12
8
12
21
Typical Problem
CB
w
P1
P2
C
A
B
L1
P
PL
8
L2
PL
8
wL2
12
L
0
PL
4 EI
2 EI
M AB =
θA +
θB + 0 + 1 1
8
L1
L1
0
PL
2 EI
4 EI
θA +
θB + 0 − 1 1
M BA =
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
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
4 EI
2 EI
P L wL
=
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
P1
MAB
CB
w
MBA
A
B
L1
P2
MBC
L2
C M
CB
Substitute θB in MAB, MBA, MBC, MCB
0
4 EI
2 EI
PL
θA +
θB + 0 + 1 1
M AB =
8
L1
L1
04
2 EI
EI
PL
θA +
θB + 0 − 1 1
M BA =
8
L1
L1
0
2
4 EI
2 EI
P2 L2 wL2
θB +
θC + 0 +
+
M BC =
L2
L2
8
12
0
2
− P2 L2 wL2
2 EI
4 EI
M CB =
θB +
θC + 0 +
−
L2
L2
8
12
24
P1
MAB
CB
w
MBA
P2
MCB
A
Ay
B
L1
MBC
L2
C
Cy
By = ByL + ByR
P1
MAB
B
A
Ay
L1
C
B
MBA
ByL
P2
MBC
ByR
L2
MCB
Cy
25
Example of Beams
26
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
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)
θA +
θB −
M BA =
8
8
8
0
4 EI
2 EI
(6)(6 2 )
M BC =
θB +
θC +
6
6
30
0
2 EI
4 EI
(6)(6) 2
M CB =
θB +
θC −
6
6
20
M AB =
B
+ ΣM B = 0 : − M BA − M BC = 0
(6)(6 2 )
4 EI 4 EI
+
=0
)θ B − 10 +
(
30
8
6
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
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,
MBC = 8.8 kN•m
MBA = - 8.8 kN•m,
MCB= -10 kN•m
10 kN•m
6m
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
10 kN
6 kN/m
10.6 kN•m
C
A
4m
5.23 kN
B
4m
6m
10 kN•m
12.2 kN
4.78 + 5.8 = 10.58 kN
5.8
5.23
V (kN)
+
+
x (m)
- 4.78
-
10.3
M
(kN•m)
x (m)
+
-10.6
Deflected shape
-12.2
-
-8.8
2.4
θB =
EI
-10
x (m)
30
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
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
M CB =
θB +
θC −
− − − (4)
6
6
20
6 EI
2(2) − (1) : 2 M BA =
θ B − 30
8
3EI
M BA =
θ B − 15 − − − (5)
8
32
MBA
MBC
B
M BC
4 EI
(6)
!
!
!
Deflected Shape of Structures
Method of Consistent Deformations
Maxwell’s Theorem of Reciprocal
Displacement
1
Deflection Diagrams and the Elastic Curve
fixed support
P
∆=0
θ=0
-M
P
roller or rocker
support
∆=0
θ
inflection point
+M
-M
2
P
θ
pined support
∆=0
-M
3
inflection point
• Fixed-connected joint
P
fixed-connected
joint
inflection point
Moment diagram
4
• Pined-connected joint
pined-connected
joint
P
Moment diagram
5
P
inflection point
Moment diagram
6
P1
A
B
C
D
P2
M
+M
x
-M
inflection point
7
P1
P2
+M
x
-M
inflection point
8
Method of Consistent Deformations
Beam 1 DOF
P
MA
Ax = 0
A
A
=
Ay
B
C
P
RB
B
C
∆´B
+
fBB x R
B
A
C
∆´B + fBB RB = ∆B = 0
B
1
9
Beam 2 DOF
Compatibility Equations.
w
∆´1 + f11R1 + f12R2 = ∆1 = 0
3
∆´2 + f21R1 + f22R2 = ∆2 = 0
4
1
5
2
=
0
w
∆´1
∆´1
∆´2
∆´2
+
f11 f12 R1
f12 f22 R
2
=
∆1 0
∆2
+
f21
f11
f12
+
1
×R1
f22
×R2
1
10
Beam 3 DOF
Compatibility Equations.
P1
1
w
P2
θ´1 + f11M1 + f12R2 + f13R3 = θ1 = 0
6
=
∆´2 + f21M1 + f22R2 + f23R3 = ∆2 = 0
4
2
3
P1
5
w
∆´3 + f31M1 + f32R2 + f33R3 = ∆3 = 0
P2
0
∆´2
1
f21
+
f11
θ´1
∆´3
f31
×M1
f11 f12 f13 M1
∆´2
f21 f22 f23 R
2
+
f31 f32 f33 R
∆´3
3
θ1 0
=
∆2 0
∆3
+
f12 f22
θ´1
f32
×R2
f13
f23
+
1
f33
×R3
1
11
Compatibility Equation for n span.
Equilibrium Equations
∆´1 + f11R1 + f12R2 + f13R3 = ∆ 1
∆´2 + f21R1 + f22R2 + f23R3 = ∆2
Fixed-end force matrix
∆´n + fn1R1 + fn2R2 + fnnRn = ∆n
[Q] = [K][D] + [Qf]
If ∆1 = ∆2=……= ∆n = 0 ;
∆ ´1
f11 f12
f13 R1
0
∆´2
f21 f22
f23 R
2 =
0
f n1 f n2
fn n R
n
0
∆´ n
+
Stiffness matrix
Solve for displacement [D];
[D] = [K]-1[Q] - [Qf]
or
[∆´] + [f][R] = 0
[R] = - [f]-1[∆´]
kij =
1
f ij
[fij] = Flexibility matrix dependent on
1
,
EI
1
1
,
EA GJ
12
Maxwell’s Theorem of reciprocal displacements
1=i
A
2=j
EI
1
B
f11 = fii
f21 = fji
f ij = ∫ mi M j
dx
EI
= ∫ mi m j
dx
EI
f ji = ∫ m j M i
dx
EI
= ∫ m j mi
dx
EI
Mi = mi
1
A
B
f22 = fjj
f12 = fij
Mj = mj
f ij = f ji
13
Example 1
Determine the reaction at all supports and the displacement at C.
50 kN
B
C
A
6m
6m
14
SOLUTION
• Principle of superposition
50 kN
MA
C
A
6m
6m
RB
=
RA
B
50 kN
∆´B
+
fBB x RB
Compatibility equation : ∆ ' B + f BB RB = 0
-----(1)
1 kN
15
• Use conjugate beam in obtaining ∆´B and fBB
50 kN
A
B
Real beam
300 kN•m
∆´B
50 kN
6m
6m
9000/EI
6 + (2/3)6 = 10 m
Conjugate beam
900/EI
300 /EI
900/EI
∆´B = M´B = -9000/EI ,
fBB
12 kN•m
1 kN
12 /EI
Real beam
1 kN
72/EI
(2/3)12 = 8 m
576/EI
Conjugate beam
fBB = M´´B = 576/EI,
72/EI
16
• Substitute ∆´B and fBB in Eq. (1)
+ ↑: −
9000 576
+(
) RB = 0
EI
EI
RB = +15.63 kN,
(same direction as 1 kN)
50 kN
300 kN•m
∆´B
50 kN
+
fBB x RB = 15.63 kN
1 kN
=
12 kN•m
1 kN
50 kN
34.37 kN•m A
34.37 kN
C
B
15.63 kN
17
Use conjugate beam in obtaining the displacement
50 kN
6m
113 kN•m
6m
C
∆C
A
B
Real Beam
15.6 kN
34.4 kN
93.6
M
(kN•m)
x (m)
3.28 m
6m
12 m
93.6/EI
-113
Conjugate Beam
-113/EI
281
223
776
( 2) −
(6 ) = −
M 'C =
EI
EI
EI
∆ C = M 'C = −
776
,↓
EI
281/(EI)
223/(EI)
M´C
V´C 2 m
4m
223/(EI)
18
Example 2
Determine the reaction at all supports and the displacement at C. Take E = 200
GPa and I = 5(106) mm4
10 kN
3EI
2EI
B
A
4m
C
2m
2m
19
10 kN
3EI
2EI
B
A
4m
C
2m
2m
=
10 kN
∆´B
+
fBB x R
B
1 kN
Compatibility equation:
∆´B + fBBRB = ∆B = 0
20
• Use conjugate beam in obtaining ∆´B
10 kN
3EI
40 kN•m
2EI
B
A
4m
C
2m
Real Beam
2m
10 kN
10
V (kN)
10
+
x (m)
M
(kN•m)
x (m)
40
177.7/EI
Conjugate Beam
40/3EI = 13.33/EI
26.66/EI
∆´B = M´B = 177.7/EI
21
• Use conjugate beam for fBB
3EI
8 kN•m
2EI
B Real Beam
1 kN
A
1 kN
4m
C
2m
2m
V (kN)
x (m)
-1
-1
8
4
M
(kN•m)
2.67
EI
8
=
3EI
+
x (m)
4/(3EI)=1.33EI
4/(2EI)=2EI
60.44/EI
fBB = M´B = 60.44/EI
12/EI
Conjugate Beam
22
10 kN
3EI
2EI
B
A
C
4m
2m
2m
=
10 kN
∆´B
+
fBB x R
B
1 kN
Compatibility. equation:
∆´B + fBBRB = ∆B = 0
− 177.7 60.44
+(
) RB = 0
EI
EI
RB = +2.941 kN,
(same direction as 1 kN)
+ ↑: −
23
• The quantitative shear and moment diagram and the qualitative deflected curve
10 kN
16.48 kN•m
3EI
2EI
B
A
∆C
C
2.94 kN
4m
2m
2m
7.06 kN
7.06
+
V (kN)
x (m)
-2.94
-2.94
11.76
M
(kN•m)
2.33 m
+
x (m)
1.67 m
-16.48
24
• Use the conjugate beam for find ∆C
3EI
16.48 kN•m
10 kN
2EI
C
A
B
∆C
2m
4m
7.059 kN
2.335 m
11.76
3EI
2m
Real beam
2.941 kN
11.76
2EI
Conjugate beam
16.48
3EI
1.665 m
3.263
EI
M´C=
3.263
6.413
−18.85
(0.555) −
(3.222) =
,↓
EI
EI
EI
(1.665)/3=0.555 m
6.413
−18.85
EI
∆C = M 'C =
= −18.85 mm, ↓
(
200
×
5
)
1.665+(2/3)(2.335) = 3.222 m
25
Example 3
Draw the quantitative Shear and moment diagram and the qualitative deflected
curve for the beam shown below.EI is constant. Neglect the effects of axial load.
5 kN/m
A
B
4m
4m
26
SOLUTION
• Principle of superposition
5 kN/m
θA
4m
4m
θB
B
=
A
5 kN/m
θ´B
+
θ´A
1 kN•m
αBA
+
αAA
×MA
αAB
Compatibility equations:
1 kN•m × M
B
αBB
θ A = 0 = θ ' A + f AA M A + f AB M B
− − − (1)
θ B = 0 = θ 'B + f BA M A + f BB M B
− − − ( 2)
27
• Use formula provided in obtaining θ´A, θ´B, αAA, αBA, αBB, αAB
5 kN/m
θ´A
θ´B
4m
4m
3wL3 3(5)(8) 3 60
θ 'A =
=
=
EI
128 EI
128EI
7 wL3
7(5)(8) 3 46.67
θ 'B =
=
=
EI
384 EI
384 EI
1 kN•m
1 kN•m
αAA
αBA
αAB
8m
8m
αBB
α AA =
M o L 1(8) 2.667
=
=
3EI 3EI
EI
α BB =
M o L 1(8) 2.667
=
=
EI
3EI 3EI
α BA =
M o L 1(8) 1.333
=
=
EI
6 EI 6 EI
α AB =
M o L 1(8) 1.333
=
=
EI
6 EI 6 EI
Note: Maxwell’s theorem of reciprocal displacement, αAB = αBA
28
• Use conjugate beam for αAA, αBA, αBB, αAB
1 kN•m
Real Beam
αAA
(1/8)
Real Beam
αAB
αBA
(1/8)
8m
1 kN•m
αBB
(1/8)
(1/8)
8m
4/EI
4/EI
1/EI
1/EI
Conjugate Beam
2.67/EI
Conjugate Beam
1.33/EI
1.33/EI
2.67/EI
α AA = V ' A =
− 2.667
EI
α AB = V ' A =
− 1.333
EI
α BA = V 'B =
1.333
EI
α BB = V ' B =
2.667
EI
29
5 kN/m
θA
B
θB
4m
Compatibility equation
+
=
4m
5 kN/m
+
θ 'B =
46.67
EI
46.67 1.333
2.667
+(
)M A + (
)M B = 0
EI
EI
EI
Solve simultaneous equations,
×MA
α AA
60 2.667
1.333
+(
)M A + (
)M B = 0
EI
EI
EI
+
60
θ 'A =
EI
1 kN•m
2.667
=
EI
+
A
α BA = V 'B =
1.333
EI
MA = -18.33 kN•m,
+
MB = -8.335 kN•m,
+
1 kN•m
×MB
α AB =
1.333
EI
α BB =
2.667
EI
30
MA = -18.33 kN•m,
MB = -8.335 kN•m,
5 kN/m
18.33 kN•m
A
RA
B
4m
4m
8.335 kN•m
RB
+ ΣMA = 0: 18.33 − 20( 2) + RB (8) − 8.355 = 0
3.753
+ ΣFy = 0: R A + RB − 20 = 0
RB = 3.753 kN,
Ra = 16.25 kN,
31
• Quantitative shear and bending diagram and qualitative deflected curve
5 kN/m
18.33 kN•m
A
B
4m
16.25 kN
4m
8.36 kN•m
3.75 kN
16.25
V
diagram
3.25 m
M
diagram
8.08
-3.75
6.67
-8.36
-18.33
Deflected
Curve
32
Example 4
Determine the reactions at the supports for the beam shown and draw the
quantitative shear and moment diagram and the qualitative deflected curve.
EI is constant.
2 kN/m
C
A
4m
B
4m
33
• Principle of superposition
2 kN/m
C
A
4m
B
4m
Compatibility equations:
∆ 'B + f 'BB RB + f 'CB RC = 0 − − − (1)
2 kN/m
C
A
4m
B
∆'B
∆'C
4m
f 'CB
f 'BB
C
× RB
1 kN
A
4m
B
4m
f 'CC
f 'BC
A
∆ 'C + f 'BC RB + f 'CC RC = 0 − − − (2)
4m
B
× RC
4m
C
1 kN
34
• Solve equation
2 kN/m
C
A
4m
B
Compatibility equations:
4m
−
2 kN/m
B
−
C
A
∆'B = −
f 'BB =
A
64
EI
21.33
EI
149.33 53.33
170.67
+
RB +
RC = 0 − −(2)
EI
EI
EI
∆'C = −
RB = 3.71 kN , ↑
× RB
RC = −0.29 kN , ↓
1 kN
53.33
f ' BC =
EI
A
149.33
EI
53.33
f 'CB =
EI
C
64 21.33
53.33
+
RB +
RC = 0 − −(1)
EI
EI
EI
f 'CC =
C
170.67
EI
1 kN
× RC
35
• Diagram
M A = 8(2) + 0.29(8) − 3.71(4)
= 3.48 kN • m
2 kN/m
C
A
4m
B
4m
Ay = 8 + 0.29 − 3.71 = 4.58 kN 3.71 kN
0.29 kN
V (kN)
4.58
0.29
2.29 m
x (m)
-3.42
M (kN•m)
1.76
x (m)
-3.48
-1.16
Deflected shape
x (m)
Point of inflection
36
Example 5
Draw the quantitative Shear and moment diagram and the qualitative deflected
curve for the beam shown below.
(a) The support at B does not settle
(b) The support at B settles 5 mm.
Take E = 200 GPa, I = 60(106) mm4.
16 kN
B
A
2m
2m
C
4m
37
SOLUTION
• Principle of superposition
16 kN
B
A
C
∆B = 5 mm
2m
16 kN
4m
=
2m
+
∆´B
fBB
× RB
1 kN
Compatibility equation :
+
∆ B = 0 = ∆ ' B + f BB RB
+
∆ B = −0.005m = ∆ ' B + f BB RB
-----(1)
-----(2)
: no settlement
: with settlement
38
• Use conjugate beam method in obtaining ∆´B
16 kN
Real A
beam
12 kN
∆´B
2m
2m
24
M´
diagram
24
Conjugate EI
beam
56
EI
Real
C beam
B
4
3
24
EI
2
3
4m
4 kN
∆´B
16
16
EI
72
EI
2m
32
EI
M´´B
4m
40
EI
V´´B
4
3
40
EI
2
4
3
32 4 40
M
(4) = 0
−
'
'
+
( )−
ΣM
=
0:
+
B
B
EI 3 EI
∆'B = M ' 'B = −
117.33
,↓
EI
39
• Use conjugate beam method in obtaining fBB
fBB
Real A
beam
0.5 kN
B
C
1 kN
4m
4m
0.5 kN
m´
diagram
m´´B
v´´B
-2
Conjugate
beam
4
EI
2
4
3
4
3
−
+ ΣMB = 0:
4
−
EI
−
2
EI
4
−
EI
4
EI
− m' ' B −
f BB
4
EI
4
EI
4 4
4
( )+
(4) = 0
EI 3 EI
10.67
= m' ' B =
,↑
EI
fBB
40
• Substitute ∆´B and fBB in Eq. (1)
+↑; 0= −
117.33
,↓
EI
10.67
=
,↑
EI
∆'B = −
117.33 10.67
+
RB ,
EI
EI
f BB
RB = 11.0 kN,
16 kN
+
fBB
∆´B
4 kN
16 kN
A
RA = 6.5 kN
1 kN
0.5 kN
0.5 kN
=
12 kN
xRB = 11.0 kN
B
11.0 kN
C
no settlement
RC = 1.5 kN
41
• Substitute ∆´B and fBB in Eq. (2)
+ ↑ ; − 0.005m = −
117.33
,↓
EI
10.67
=
,↑
EI
∆'B = −
f BB
117.33 10.67
+
RB
EI
EI
(−0.005m) EI = −117.33 + 10.67 RB
(−0.005)(200 × 60) = −117.33 + 10.67 RB
RB = 5.37 kN,
16 kN
+
fBB
∆´B
4 kN
16 kN
A
RA = 9.31 kN
1 kN
0.5 kN
0.5 kN
=
12 kN
xRB = 5.37
B
5.37 kN
C
with 5 mm settlement
RC = 1.32 kN
42
• Quantitative shear and bending diagram and qualitative deflected curve
16 kN
16 kN
B
A
C
B
A
C
∆B = 5 mm
6.5 kN
11 kN
2m
V
diagram
2m
6.5
1.5 kN
9.31 kN
4m
2m
5.37 kN
2m
V
diagram
-1.32
-6.69
13
+
-
M
diagram
-6
Deflected
Curve
4m
9.31
1.5
-9.5
M
diagram
1.32 kN
Deflected
Curve
18.62
+
5.24
∆B = 5 mm
43
Example 6
Calculate supports reactions and draw the bending moment diagrams for (a)
D2 = 0 and (b) D2 = 2 mm.
1
2
4m
w 2 kN/m 3
6m
44
Compatibility equation:
∆´2 + f22R2 = ∆2
1
2
w 2 kN/m 3
Use formulations provided :
wx
( x 3 − 2 Lx 2 + L3 )
24 EI
(2)(4) 3
=−
(4 − 2 ×10 × 4 2 + 103 )
24 EI
248
=−
= −6.2mm, ↓
(200)(200)
∆ '2 = −
4m
6m
=
2 kN/m
∆´2
Pbx 2
(L − b2 − x2 )
6 LEI
1× 6 × 4
=−
(10 2 − 6 2 − 4 2 )
6 ×10 EI
19.2
=
= +0.48 mm, ↑
(200)(200)
f 22 =
+
f22
×R2
1
45
For ∆2 = 0
For ∆2 = 2 mm
Compatibility.equation:
Compatibility equation:
-6.2 + 0.48R2 = 0
R2 = 12.92 kN
1
2
1
w 2 kN/m 3
12.92 kN
6m
2.25 kN 4 m
-6.2 + 0.48R2 = -2
R2 = 8.75 kN
4.83 kN
4.75 kN 4 m
7.17
V
(kN)
4.75
2.415 m
2.25
+
-
-5.75
1.125 m
x (m)
-4.83
2
w 2 kN/m 3
8.75 kN 6 m
5.5
3.25 m
V +
+
(kN)
2.375 m -3.25
-
1.27
+
-7
x (m)
-6.5
5.85
M
+
(kN•m)
6.5 kN
10.56
5.64
x (m)
M
(kN•m)
+
3
+
x (m)
46
APPENDIX
Basic Beams: Single span
1
wL/2
w
L/3
L/3
wL/2
5wL4
384 EI
L
wL/2
V
- wL/2
wL2/8
M
47
• Find ∆1 by Castigliano’s
2
1
P
3
w
wL P
+
2
2
w
x1
wL P
+
2
2
L/2
x1
L/2
L
M1
V1
wL P
+
2
2
x2
w
M2
V2
x2
wL P
+
2
2
+ ΣΜ = 0;
2
wx
wL P
M1 + 1 − (
+ ) x1 = 0
2
2
2
2
wx1
wL P
M1 = −
+(
+ ) x1 = M 2
2
2
2
48
2
wx
wL P
M1 = − 1 + (
+ ) x1 = M 2
2
2
2
L/2
∆1 = ∆ max =
∫
0
∂M
dx
( 1 )M1 1 +
∂P
EI
L/2
=2∫ (
0
=
2
EI
∫
0
(
∂M 2
dx
)M 2 2
∂P
EI
∂M 1
dx
)M1 1
∂P
EI
L/2
∫
0
L/2
0
x − wx1
wL P
( 1 )(
+(
+ ) x1 )dx1
2
2
2
2
2
5wL4
=
384 EI
49
Single span
1
P
P/2
P/2
PL3
L3
δ 11 =
= f11 P → f11 =
48 EI
48 EI
L /2
L /2
L
P/2
V
+
- P/2
PL/4
M
50
Double span
∆ max
wl 4
=
185 EI
1
∆ max
w
0.375 wl
0.375 wl
1.25 wl
l = L/2
wl 4
=
185 EI
l = L/2
L = 2l
0.625 wl
0.375 wl
+
+
-
-0.375 wl
-0.625 wl
0.070 wl2
+
0.070 wl2
+
-0.125 wl2
51
Triple span
1
∆max = wl4
2
w
145 EI
0.4wl
0.4wl
1.1wl
l = L/3
V
1.1wl
l = L/3
l = L/3
0.6wl
0.5wl
0.4wl
+
+
+
-
-
-
- 0.6wl
0.08 wl2
M
- 0.4wl
-0.5wl
0.08 wl2
0.025 wl2
+
+
-0.1 wl2
+
-0.1 wl2
52
APPROXIMATE ANALYSIS OF STATICALLY
INDETERMINATE STRUCTURES
!
!
!
!
Trusses
Vertical Loads on Building Frames
Lateral Loads on Building Frames: Portal
Method
Lateral Loads on Building Frames: Cantilever
Method Problems
1
Trusses
P1
P2
a
a
F1
Fa
V = R1
Fb
F2
R1
a
(a)
R2
R1
a
(b)
Method 1 : If the diagonals are intentionally designed to be long and slender,
it is reasonable to assume that the panel shear is resisted entirely by the tension diagonal,
whereas the compressive diagonal is assumed to be a zero-force member.
Method 2 : If the diagonal members are intended to be constructed from large
rolled sections such as angles or channels, we will assume that the tension and compression
diagonals each carry half the panel shear.
2
Example 1
Determine (approximately) the forces in the members of the truss shown in
Figure. The diagonals are to be designed to support both tensile and compressive
forces, and therefore each is assumed to carry half the panel shear. The support
reactions have been computed.
F
E
D
3m
C
A
B
4m
10 kN
4m
20 kN
3
20 kN
F
E
D
+ ΣMA = 0:
FFE(3) - 8.33cos36.87o(3) = 0
FFE = 6.67 kN (C)
3m
C
A
B
4m
10 kN
4m
10 kN
20 kN
+ ΣMF = 0:
FAB(3) - 8.33cos36.87o(3) = 0
FAB = 6.67 kN (T)
20 kN
F
θ
θ = 36.87o
FFE
FFB= F
F =F
θ AE
FAB
3m
A
FAF
V = 10 kN
A
10 kN
+ ΣFy = 0:
8.33
θ
6.67 kN
10 kN
20 - 10 - 2Fsin(36.87o) = 0
F = 8.33 kN
FFB = 8.33 kN (T)
FAE = 8.33 kN (C)
+ ΣFy = 0:
FAF - 10 - 8.33sin(36.87o) = 0
FAF = 15 kN (T)
4
θ = 36.87o
D
FED
V = 10 kN
F = FDB
F = FEC
θ
θ
FBC
θ = 36.87o
6.67 kN
3m
D
θ
8.33 kN
C
FDC
10 kN
+ ΣFy = 0:
10 - 2Fsin(36.87o) = 0
+ ΣFy = 0:
FDC - 8.33sin(36.87o) = 0
F = 8.33 kN
FDB = 8.33 kN (T)
FEC = 8.33 kN (C)
+ ΣMC = 0:
FED(3) - 8.33cos36.87o(3) = 0
FDC = 5 kN (C)
θ = 36.87o
6.67 kN
E
θ
θ
6.67 kN
8.33 kN
8.33 kN
FEB
FED = 6.67 kN (C)
+ ΣFy = 0:
+ ΣMD = 0:
FBC(3) - 8.33cos36.87o(3) = 0
FBC = 6.67 kN (T)
FEB = 2(8.33sin36.87o) = 10 kN (T)
5
Example 2
Determine (approximately) the forces in the members of the truss shown in
Figure. The diagonals are slender and therefore will not support a compressive
force. The support reactions have been computed.
10 kN
20 kN
J
I
20 kN
H
20 kN
G
10 kN
F
4m
A
40 kN
4m B
4m C 4m D 4m
E
40 kN
6
10 kN
FJA
J
4m
45o
A
FJB
FJI
FAI = 0
FAB
θ
A
V = 30 kN
0
0
40 kN
+ ΣFy = 0:
40 kN
10 kN
FJA = 40 kN (C)
20 kN
FAI = 0
I
J
+ ΣFy = 0:
40 - 10 - FJBcos 45o = 0
FJB = 42.43 kN (T)
4m
A
+ ΣMA = 0:
FJI(4) - 42.43sin 45o(4) = 0
45o
4m B
FAB(4) = 0
FAB = 0
FIC
FBH = 0
FBC
V = 10 kN
40 kN
FJI = 30 kN (C)
+ ΣMJ = 0:
FIH
FBH = 0
+ ΣFy = 0:
40 - 10 - 20 - FICcos 45o = 0
FIC = 14.14 kN (T)
7
10 kN
20 kN
J
FIH
45o
4m
A
42.3 kN
I
4m B
FIC
FBH = 0
FBC
V = 10 kN
0
+ ΣFy = 0:
FBI
45o 0
45o
45o
B
30 kN
FBI = 42.3 sin 45o = 30 kN (T)
40 kN
+ ΣMB = 0:
FIH(4) - 14.14sin
14.14 kN
45o(4)
14.14 kN
+ 10(4) - 40(4) = 0
FJH = 40 kN (C)
+ ΣMI = 0:
FCH
FBC(4) - 40(4) + 10(4) = 0
FBC = 30 kN (T)
30 kN
45o
45o
C
30 kN
+ ΣFy = 0:
FBI = 2(14.1442.3 sin 45o) = 20 kN (C)
8
Vertical Loads on Building Frames
typical building frame
9
• Assumptions for Approximate Analysis
column
w
column
girder
A
B
L
(a)
w
w
A
Point of
zero A
moment
B
point of zero
0.21L
0.21L moment
L
(b)
w
0.1L
approximate case
(d)
B
L
Simply supported
(c)
assumed
points of
zero moment 0.1L
L
Point of
zero
moment
w
0.1L
0.8L
model
0.1L
(e)
10
Example 3
Determine (approximately) the moment at the joints E and C caused by members
EF and CD of the building bent in the figure.
1 kN/m
F
E
1 kN/m
D
C
B
A
6m
11
1 kN/m
4.8 kN
1 kN/m
4.8 m
4.8 m
0.1L=0.6 m
2.4 kN
0.6 m
0.6 kN
2.4 kN
2.4 kN 2.4 kN
0.6(0.3) + 2.4(0.6) = 1.62 kN•m
3 kN 0.6 m
0.6 kN
1.62 kN•m
0.6 m 3 kN
12
Portal Frames and Trusses
∆
∆
• Frames: Pin-Supported
P
P
h
assumed
hinge
h
l
(a)
Ph/2
Ph/2
l
(b)
L/2
Ph/2
L/2
P
P/2
P/2
Ph/2
Ph/l
Ph/l
h
h
P/2
(d)
Ph/l
P/2
(c)
Ph/l
13
∆
• Frames : Fixed-Supported
∆
P
P
assumed
hinges
h
h
l
(a)
l
(b)
L/2
P
P/2
h/2
Ph/4
Ph/4
P/2
Ph/2l
Ph/4
P/2
Ph/2l
Ph/2l
Ph/2l
Ph/2l
P/2
Ph/4
Ph/4
h/2
Ph/2l
P/2
Ph/4
L/2
P/2
h/2
P/2
Ph/4
(d)
Ph/2l
(c)
Ph/2l
P/2
Ph/4
14
• Frames : Partial Fixity
P
P
assumed
hinges
θ
h
h/3
θ
h/3
l
(a)
(b)
• Trusses
P
P
∆
h
h/2
l
(a)
∆
assumed
hinges
P/2
P/2
l
(b)
15
Example 4
Determine by approximate methods the forces acting in the members of the
Warren portal shown in the figure.
2m
40 kN C
4m
D
2m
E
F
2m
B
H
G
4m
7m
A
I
8m
16
2m
4m
D
40 kN C
2m
E
F
2m
+ ΣMJ = 0:
H
B
G
N(8) - 40(5.5) = 0
3.5 m
J
K
40/2 = 20 kN = V
N = 27.5 kN
20 kN = V
N
N
N
N
V = 20 kN
V = 20 kN + ΣM = 0:
A
3.5 m
M - 20(3.5) = 0
A
V = 20 kN
M
N
I
20 kN = V
M
M = 70 kN•m
N
17
2m
2m
40 kN C FCD D
2m
B
FEF F
E
FEG
F
45o BD
FBH
FGH
2m
45o
G
3.5 m
3.5 m
J
20 kN
K
20 kN = V
27.5 kN
+ ΣFy = 0:
-27.5 + FBDcos 45o = 0
27.5 kN
+ ΣFy = 0:
FBD = 38.9 kN (T)
+ ΣMB = 0:
FCD(2) - 40(2) - 20(3.5) = 0
FCD = 75 kN (C)
27.5 - FEGcos 45o = 0
FEG = 38.9 kN (C)
+ ΣMG = 0:
FEF(2) - 20(3.5) = 0
FEF = 35 kN (T)
+ ΣMD = 0: FBH(2) + 27.5(2) - 20(5.5) = 0
+ ΣME = 0: FGH(2) + 27.5(2) - 20(5.5) = 0
FBH = 27.5 kN (T)
FGH = 27.5 kN (C)
18
y
y
38.9 kN
D
75 kN
x
45o
45o
38.9 kN
+ ΣFy = 0:
FDE
45o
45o
27.5 kN
H
x
27.5 kN
FDH
FDHsin 45o - 38.9sin 45o = 0
FDH = 38.9 kN (C)
+ ΣF = 0:
x
FHE
+ ΣFy = 0:
FHEsin 45o - 38.9sin 45o = 0
FHE = 38.9 kN (T)
75 - 2(38.9 cos 45o) - FDE = 0
FDE = 20 kN (C)
19
Lateral Loads on Building Frames: Portal Method
P
= inflection point
(a)
V
V
(b)
V
V
20
Example 5
Determine by approximate methods the forces acting in the members of the
Warren portal shown in the figure.
5 kN B
D
F
G
3m
A
C
4m
E
4m
H
4m
21
5 kN B
3m
D
M
I
J
A
L
E
H
4m
M
G
O
K
C
4m
5 kN B
F
N
4m
N
D
O
F
G
1.5 m
I
V
J
2V
Jy
Iy
+ ΣF = 0:
x
K
2V
Ky
V
L
Ly
5 - 6V = 0
V = 0.833 kN
22
2m
5 kN B
1.5 m
0.625 kN
0.833 kN
2m D
4.167 kN
4.167 kN
M
2.501kN
1.5 m 0.625kN
J
0.625 kN
1.666 kN
I
2m N
Jy = 0
Iy = 0.625kN
2.501 kN N 2 m F
2m
O0.835 kN
O
1.5 m
0.625kN
K
1.666 kN
Ky = 0
0.625 kN 1.5 m
0.833 kN
L
0.625 kN = Ly
0.625 kN
0.625 kN
G
0.835 kN
0.625kN
I
1.5 m
2m
0.625 kN
0.833kN
J
1.5 m
C
A
0.833 kN
1.25 kN•m
1.666 kN
1.666kN
2.50 kN•m
1.666 kN
K
1.5 m
E
L
1.5 m
H
1.666kN
2.5 kN•m 0.625 kN
0.833 kN
0.833 kN
1.25 kN•m
23
Example 6
Determine (approximately) the reactions at the base of the columns of the frame
shown in Fig. 7-14a. Use the portal method of analysis.
20 kN
G
H
I
5m
30 kN
D
E
F
6m
A
B
8m
C
8m
24
20 kN
G
R
O
30 kN
D
H
S
P
M
E
I
Q
N
F
J
K
L
A
B
C
8m
5m
6m
8m
25
G
20 kN
I
2.5 m
V
2V
Oy
+ ΣF = 0:
x
G
20 kN
V
Py
20 - 4V = 0
Py
V = 5 kN
H
I
5m
D
30 kN
E
F
3m
V´
2V´
Jy
+ ΣF = 0:
x
V´
Ky
20 + 30 - 4V´ = 0
Ly
V´ = 12.5 kN
26
Ry = 3.125 kN
20 kN
G
4m R
Rx = 15 kN
R
3.125 kN
4m H
15 kN
2.5 m
10 kN
2.5 m
Sy = 3.125 kN
S
4m
Sx = 5 kN
5 kN
Oy = 3.125
Py = 0 kN
3.125 kN
O
2.5 m
30 kN
3m
12.5 kN
5 kN
My = 12.5 kN
M
4m
P
2.5 m
22.5 kN M
Mx = 22.5 kN
J
4m
4m N
3m
K
12.5 kN
25 kN
Jy = 15.625 kN
12.5 kN
A
Ax = 15.625 kN
3m
Ax = 12.5 kN
MA = 37.5 kN•m
Ny = 12.5 kN
Nx = 7.5 kN
Ky = 0 kN
15.625kN
J
10 kN
K
B
Bx = 0
25 kN
3m
Bx = 25 kN
MB = 75 kN•m
27
Lateral Loads on Building Frames: Cantilever Method
P
beam
(a)
building frame
(b)
In summary, using the cantilever method, the following assumptions apply to
a fixed-supported frame.
1. A hinge is place at the center of each girder, since this is assumed to be point
of zero moment.
2. A hinge is placed at the center of each column, since this is assumed to be
a point of zero moment.
3. The axial stress in a column is proportional to its distance from the centroid
of the cross-sectional areas of the columns at a given floor level. Since stress equals force
per area, then in the special case of the columns having equal cross-sectional areas,
the force in a column is also proportional to its distance from the centroid of the column areas.
28
Example 7
Determine (approximately) the reactions at the base of the columns of the frame
shown. The columns are assumed to have equal crossectional areas. Use the
cantilever method of analysis.
30 kN
C
D
4m
B
E
15 kN
4m
A
F
6m
29
30 kN
C
I
4m
B
D
K
H
J
E
15 kN
4m
x
G
6m
L
A
F
6m
~
x A 0( A) + 6( A)
∑
x=
=
=3
A
A
A
+
∑
30
3m
3m
+ ΣMM = 0:
C
30 kN
2m
I
Hx
M
-30(2) + 3Hy + 3Ky = 0
D
The unknowns can be related by proportional triangles,
that is
Kx
Hy
Hy
Ky
3
=
Ky
3
or
Hy = Ky
H y = K y = 10 kN
C
30 kN
D
I
4m
B
K
H
J
15 kN
2m
3m
3m
Lx
E
Gy
-30(6) - 15(2) + 3Gy + 3Ly = 0
The unknowns can be related by proportional triangles,
that is
Gy
3
N
Gx
+ ΣMN = 0:
=
Gy
3
or
G y = Ly
G y = L y = 35 kN
Ly
31
Iy = 10 kN
C
30 kN
3m
Ix = 15 kN
2m
Hx = 15 kN
10 kN
3m D
I
15 kN
Kx = 15 kN
10 kN
10 kN
10 kN
10 kN
15 kN
H
2m
Jy = 25 kN
J
15 kN
3m
2m
Jx = 7.5 kN
Gx = 22.5 kN
35 kN
Ax = 35 kN
15 kN
2m
3m
2m
L
25 kN
Lx = 22.5 kN
35 kN
35 kN
22.5 kN
G
K
7.5 kN J
G
A
2m
2m
Ax = 22.5 kN
MA = 45 kN•m
35 kN
L
F
Fy = 35 kN
22.5 kN
2m
Fx = 22.5 kN
MF = 45 kN•m
32
Example 8
Show how to determine (approximately) the reactions at the base of the columns
of the frame shown. The columns have the crossectional areas show. Use the
cantilever of analysis.
P
35 kN
6000 mm2
4 mL
Q
5000mm2 4000 mm2
M
N
I
45 kN
6 mE
J
F
6000 mm2
A
6m
R
O
K
G
5000mm2 4000 mm2
B
C
4m
6000 mm2
H
6000 mm2
D
8m
33
P
35 kN
Q
4 mL
M
N
I
45 kN
R
O
J
6 mE
F
A
G
B
6m
K
H
C
4m
D
8m
5000mm2 4000 mm2
6000 mm2
6m
4m
6000 mm2
8m
x
~
x A 6000(0) + 5000(6) + 4000(10) + 6000(18)
∑
=
x=
= 8.48 m
A
6000
+
5000
+
4000
+
6000
∑
34
P
35 kN
2m
Q
Mx
Lx
R
Nx
Ox
My
Ly
Ny
Oy
2.48 m 1.52 m
8.48 m
+ ΣMNA = 0:
9.52 m
-35(2) + Ly(8.48) + My(2.48) + Ny(1.52) + Oy(9.52) = 0
-----(1)
Since any column stress σ is proportional to its distance from the neutral axis
σM
2.48
σN
1.52
σO
9.52
=
=
=
σL
8.48
σL
8.48
σL
8.48
;
σM =
2.48
σL ;
8.48
;
σN =
1.52
σL ;
8.48
;
σO =
9.52
σL ;
8.48
Solving Eqs. (1) - (4) yields
My
−6
5000(10 )
Ny
−6
4000(10 )
Oy
6000(10 −6 )
=
Ly
2.48
(
)
−6
8.48 6000(10 )
− − − − − ( 2)
=
Ly
1.52
(
)
−6
8.48 6000(10 )
− − − − − (3)
=
Ly
9.52
(
)
8.48 6000(10 −6 )
− − − − − ( 4)
Ly = 3.508 kN
Ny = 0.419 kN
My = 0.855 kN
Oy = 3.938 kN
35
35 kN
4m L
M
I
45 kN
3m
O
J
Fx
Ex
N
K
Gx
Fy
Hx
Gy
Ey
Hy
8.48 m
+ ΣMNA = 0:
2.48 m 1.52 m
9.52 m
-45(3) - 35(7) + Ey(8.48) + Fy(2.48) + Gy(1.52) + Hy(9.52) = 0 -----(5)
Since any column stress σ is proportional to its distance from the neutral axis ;
Fy
Ey
σF
σ
2.48
2.48
= E ; σF =
σE ;
=
(
)
− − − − − (6)
−6
−6
2.48 8.48
8.48
5000(10 ) 8.48 6000(10 )
σG
1.52
σH
9.52
=
=
σE
8.48
σE
8.48
1.52
σE ;
8.48
;
σG =
;
9.52
σH =
σE ;
8.48
Gy
Ey
1.52
(
)
8.48 6000(10 −6 )
− − − − − (7)
Ey
9.52
=
(
)
6000(10 −6 ) 8.48 6000(10 −6 )
− − − − − (8)
4000(10 −6 )
Hy
=
Solving Eqs. (1) - (4) yields Ey = 19.044 kN Fy = 4.641 kN Gy = 2.276 kN Hy = 21.38 kN
36
3m
35 kN
2m
Lx= 5.262 kN
Py= 3.508 kN
Px= 29.738 kN
3.508 kN
2m
45 kN
3m
Ex= 64.44 kN
3.508 kN
5.262 kN
Iy= 15.536 kN
I
Ix= 114.702 kN
3m
19.044 kN
19.044 kN
E
64.44 kN
3m
Ax = 19.044 kN
Ax = 64.44 kN
MA = 193.32 kN•m
One can continue to analyze the
other segments in sequence, i.e.,
PQM, then MJFI, then FB, and so on.
37
ANALYSIS OF STATICALLY INDETERMINATE
STRUCTURES BY THE FORCE METHOD
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Force Method of Analysis: Beams
Maxwell’s Theorem of Reciprocal Displacements;
Betti’s Law
Force Method of Analysis: Frames
Force Method of Analysis: Trusses
Force Method of Analysis: General
Composite Structures
1
Force Method of Analysis : Beams
1 Degree of freedom
• Compatibility of displacement
M1
P
1
2
C
A
M1
B
R1
P
1
2
C
A
R1
R2
L
R2
=
P
C
A
∆´2
B
L
=
P
• Compatibility of slope
B
θ ´1
+
+
1
α11
× M1
f22 × R2
1
∆´2 + f22 R2
= ∆2 = 0
θ ´1 + α11M1
= θ 1= 0
2
2 Degree of freedom
Ax
P
A
1
B
Ay
R1
P
2
C
R2
B
A
D
∆´2
+
f21
f11
f12
Dy
C
∆´1
A
D
D
xR
R11
1
+
f22
A
D
xR
R22
1
∆´1 + f11 R1 + f12 R2
∆´2 + f21 R1 + f22 R2
= ∆1 = 0
= ∆2 = 0
3
Maxwell’s Theorem of Reciprocal Displacements; Betti’s Law
f21
1
1
2
A
B
f11
f21
mm
m2 M 1
dx = ∫ 2 1 dx
EI
EI
L
L
1 • f 21 = ∫
m1
m2 m1
dx
EI
L
f 21 = ∫
1
A
B
f22
f12
m2
4
1
f12
m2 m1
dx
EI
L
2
f 21 = ∫
1
A
B
f22
f12
mm
m1M 2
dx = ∫ 1 2 dx
EI
EI
L
L
1 • f12 = ∫
m1m2
dx
EI
L
f12 = ∫
m2
f 21 = f12
Maxwell’s Theorem:
1
A
B
f ij = f ji
f21
f11
m1
5
f11, f22
1
2
1
A
B
f11
f21
m1M 1
mm
dx = ∫ 1 1 dx
EI
EI
L
L
1 • f11 = ∫
m2 M 2
mm
dx = ∫ 2 2 dx
EI
EI
L
L
1 • f 22 = ∫
m1
In general,
1
A
B
1 • f ij = f = ∫
mi m j
L
f22
f12
1 • f ji = f ji = ∫
L
EI
m j mi
EI
dx
dx
m2
6
1
A
2
P1
D
d11 = f11 P1
d21 = f21 P1
P2
A
D
d12 = f12 P2
d22 = f22 P2
7
Force Method of Analysis: General
1
Compatibility Eq.
2
w
∆´1 + f11R1 + f12R2 = ∆1 = 0
∆´2 + f21R1 + f22R2 = ∆2 = 0
0
=
∆´1
w
∆´1
∆´2
∆´2
f12 f22 R
2
f11 f12 R1
+
f12 f22 R
2
f21
f11
+
f11 f12 R1
xR1
+
f22
1
xR2
=-
0
∆2
∆´1
∆´2
General form:
1
f12
=
∆1
f11 f12 f1n
f21 f22 f2n
..
.
fn1 fn2 fnn
R1
R2
..
.
Rn
∆´1
∆´2
.
= - ..
∆´n
8
Example 9-1
Determine the reaction at all supports and the displacement at C.
50 kN
B
C
A
6m
6m
9
SOLUTION
Use compatibility of displacement for find reaction
• Principle of superposition
50 kN
MA
A
6m
6m
RB
=
RA
B
C
50 kN
∆´B
+
RBB
fBB x R
Compatibility equation : ∆ ' B + f BB RB = 0
-----(1)
1
10
• Use formulation for ∆´B and fBB
50 kN
A
6m
C
B ∆´C
6m
6θ´C
θ´C
∆´B = ∆´C+ (6 m)θ´C
P(6) 3
P ( 6) 2
∆'B =
+ ( 6)
3EI
2 EI
A
50(6)3
(50)(6) 2 9000
=
+ (6)
=
,↓
3EI
2 EI
EI
C
f BB
∆´B
PL3 (1)(12) 3 576
=
=
=
,↑
EI
3EI
3EI
B
fBB
1
11
• Substitute ∆´B and fBB in Eq. (1): ∆ 'B + f BB RB = 0
+ ↑: −
∆'B =
9000 576
+(
) RB = 0
EI
EI
RB = 15.63 kN,
9000
,↓
EI
f BB =
576
,↑
EI
50 kN
6m
MA
A
6m
C
B
15.63 kN
RA
Equilibrium equation :
+ ΣMA = 0:
+ ΣFy = 0:
M A − 50(6) + 15.63(12) = 0,
+ RA − 50 + 15.63 = 0,
MA = 112.4 kN,
+
Ra = 34.37 kN,
12
• Quantitative shear and bending diagram and qualitative deflected curve
50 kN
6m
112.4 kN•m
A
6m
B
C
15.63 kN
34.37 kN
34.37
V
(kN)
x (m)
-15.63
-15.63
93.78
M
(kN•m)
x (m)
3.28
-112.44
6
12
13
Or use compatibility of slope to obtain reaction
50 kN
• Principle of superposition
6m
MA
6m
B
C
A
=
RA
RB
50 kN
A
B
C
θ ´A
1
+
fAA
A
B
C
xM
MAA
Compatibility equation :
θ ' A + f AA M A =θ A= 0
-----(2)
14
• Use the table on the inside front cover for θ´B and fBB
50 kN
A
C
B
C
B
θ ´A
PL2
θ A'=
16 EI
1
fAA
A
f CC =
L
3EI
Substitute the values in equation: θ ' A + f AA M A =θ A= 0
+:
PL2
L
−
+
MA =0
16 EI 3EI
MA =
3PL 3(50)(12)
=
16
16
= 112.5 kN•m, +
15
Or use Castigliano least work method
50 kN
x1
x2
12RB - 300 = MA
B
C
A
6m
50 - RB = RA
M 12RB - 300
diagram
L
∆B = 0 = ∫ (
0
6m
RB
M2 = RBx2
M1 = (12RB - 300) + (50 - RB)x1
x (m)
∂M M
)
dx
∂RB EI
6
6
1
1
0=
(
12
−
x
)(
12
R
−
300
+
50
x
−
R
x
)
dx
+
x2 ( RB x2 ) dx2
1
B
1
B 1
1
EI ∫0
EI ∫0
2
2
3
3
3
6
900 x1 24 x1
50 x1 x1
x
0 = (144 RB x1 − 3600 x1 +
−
RB −
+
RB ) 0 + 2 RB
2
2
3
3
3
RB =15.63 kN,
6
0
16
Use conjugate beam for find the displacement
50 kN
6m
112 kN•m
6m
C
B
A
∆C
34.4 kN
Real Beam
15.6 kN
93.6
M
(kN•m)
x (m)
6
3.28
12
93.6/EI
-112
Conjugate Beam
-112/EI
281
223
776
( 2) −
(6 ) = −
M 'C =
EI
EI
EI
∆ C = M 'C = −
776
,↓
EI
281/(EI)
223/(EI)
M´C
V´C 2 m
4m
223/(EI)
17
Use double integration to obtain the displacement
50 kN
6m
112 kN•m
6m
C
B
A
∆C
34.4 kN
Real Beam
15.6 kN
93.6
M
(kN•m)
x (m)
3.28
-112
6
12
d 2υ
EI 2 = −112 + 34.4 x1
dx
dυ
x12
EI
= −112 x1 + 34.4 + C1
dx
2
2
x1
x13
EIυ = −112.4 + 34.4 + C1 x1 + C2
2
6
1
62
63
778
(−112( ) + 34.4( ) + 0 + 0) = −
∆C =
,↓
2
6
EI
EI
18
Example 9-2
Draw the quantitative Shear and moment diagram and the qualitative deflected
curve for the beam shown below.The support at B settles 5 mm.
Take E = 200 GPa, I = 60(106) mm4.
16 kN
B
A
C
∆B = 5 mm
2m
2m
4m
19
Use compatibility of displacement to obain reaction
• Principle of superposition
16 kN
B
A
C
∆B = 5 mm
2m
2m
16 kN
4m
=
SOLUTION
+
∆´B
fBB
× RB
1 kN
Compatibility equation :
∆ B = −0.005m = ∆ 'B + f BB RB
-----(1)
20
• Use conjugate beam method for ∆´B
16 kN
Real A
beam
12 kN
B
∆´B
2m
2m
24
M´
diagram
24
Conjugate EI
beam
56
EI
C
4
3
24
EI
4m
4 kN
16
16
EI
72
EI
M´´B
2
3
2m
4m
40
EI
V´´B
4
3
32
EI
40
EI
2
4
3
32 4 40
+ ΣMB = 0: − M ' ' B +
(4) = 0
( )−
EI 3 EI
117.33
∆' B = M ' ' B = −
,↓
21
EI
• Use conjugate beam method for fBB
fBB
Real A
beam
0.5 kN
B
C
1 kN
4m
4m
0.5 kN
m´
diagram
m´´B
vB´B
-2
Conjugate
beam
4
EI
2
4
3
4
3
−
+ ΣMB = 0:
4
−
EI
−
2
EI
4
−
EI
4
EI
4
EI
4
EI
4 4
4
( )+
(4) = 0
EI 3 EI
10.67
f BB = mB ' ' =
,↑
EI
− mB ' '−
22
∆ B = −0.005 m = ∆ ' B + f BB RB
• Substitute ∆´B and fBB in Eq. (1):
+ ↑: − 0.005 = −
117.33 10.67
RB
+
EI
EI
117.33
,↓
EI
10.67
= mB ' ' =
,↑
EI
∆'B = M ' 'B = −
(−0.005) EI = −117.33 + 10.67 RB
f BB
(−0.005)(200 × 60) = −117.33 + 10.67 RB
RB = 5.37 kN,
+
16 kN
xRB = 5.37
1 kN
16 kN
A
0.5 kN
0.5 kN
=
4 kN
12 kN
B
RA = 9.31 kN 5.37 kN
C
RC = 1.32 kN
23
• Quantitative shear and bending diagram and qualitative deflected curve
16 kN
B
A
C
∆B = 5 mm
RA = 9.31 kN 5.37 kN
2m
2m
RC = 1.32 kN
4m
9.31
V
diagram
-1.32
-6.69
M
diagram
Deflected
Curve
18.62
5.24
∆B = 5 mm
24
Or use Castigliano least work method
• Principle of superposition
16 kN
RA = 12 - 0.5RB
16 kN
12
0.5RB
RC = 4 - 0.5RB
4m
=
2m
+
2m
RB
RB
4
0.5RB
25
16 kN
x1 x2
x3
RB
RA = 12 - 0.5RB
2m
x1
2m
RC = 4 - 0.5RB
4m
M1 = (12 - 0.5RB)x1
x3
(4 - 0.5RB)x3 = M3
M2 = 0.5x2RB + 16 - 2RB + 4 x2
V1
12 - 0.5RB
V2
L
∆ B = −0.005 = ∫ (
0
∂M i M i
)
dx
∂RB EI
RB
x2
2
V3
4 - 0.5RB
4 - 0.5RB
4m
2
1
1
− 0.005 =
(
−
0
.
5
x
)(
12
x
−
0
.
5
x
R
)
dx
+
(0.5 x2 − 2)(0.5 x2 RB + 16 − 2 RB + 4 x2 )dx2
1
1
1 B
1
EI ∫0
EI ∫0
4
1
+
(−0.5 x3 )(4 x3 − 0.5 x3 RB )dx3
∫
EI 0
− 0.005 EI = −117.34 + 10.66 RB , RB = 5.38 kN,
26
Example 9-3
Draw the quantitative Shear and moment diagram and the qualitative deflected
curve for the beam shown below.EI is constant. Neglect the effects of axial load.
5 kN/m
A
B
4m
4m
27
SOLUTION
Use compatibility of displacement to obtain reaction
• Principle of superposition
5 kN/m
θA=0
4m
4m
θB=0
B
=
A
5 kN/m
θ´B
+
θ´A
1 kN•m
αBA
+
αAA
×MA
αAB
Compatibility equation :
1 kN•m × M
B
αBB
θ A = 0 = θ ' A +α AA M A + α AB M B
-----(1)
θ B = 0 = θ 'B +α BA M A + α BB M B
-----(2)
28
• Use formulation: θ´A, θ´B, αAA, αBA, αBB, αAB,
5 kN/m
θ´A
θ´B
3wL3 3(5)(8) 3 60
θ 'A =
=
=
EI
128 EI
128EI
7 wL3
7(5)(8) 3 46.67
θ 'B =
=
=
EI
384 EI
384 EI
1 kN•m
αAA
αBA
1 kN•m
αAB
αBB
α AA =
M o L 1(8) 2.67
=
=
3EI 3EI
EI
α BB =
M o L 1(8) 2.67
=
=
3EI 3EI
EI
α BA =
M o L 1(8) 1.33
=
=
6 EI 6 EI
EI
α AB =
M o L 1(8) 1.33
=
=
6 EI 6 EI
EI
Note : Maxwell’s theorem of reciprocal displacement is αAB = αBA
29
• Substitute θ´A, θ´B, αAA, αBA, αBB, αAB, in Eq. (1) and (2)
5 kN/m
θA=0
4m
4m
B
θB=0
θ A = 0 = θ ' A +α AA M A + α AB M B
-----(1)
θ B = 0 = θ 'B +α BA M A + α BB M B
-----(2)
+
+
α AA
α AB =
=
A
60
EI
2.67
=
EI
θ 'A =
θ 'B =
1.33
EI
46.67
EI
1.33
EI
2.67
=
EI
α BA =
0=
60 2.67
1.33
+(
)M A + (
)M B
EI
EI
EI
0=
46.67 1.33
2.67
+(
)M A + (
)M B
EI
EI
EI
α BB
Solving these equations simultaneously, we haave
MA = -18.31 kN•m,
+
MB = -8.36 kN•m,
+
30
MA = -18.31 kN•m,
+
MB = -8.36 kN•m,
+
5 kN/m
18.31 kN•m
A
RA
+ ΣMA = 0:
+ ΣFy = 0:
B
4m
4m
8.36 kN•m
RB
18.31 − 20( 2) + RB (8) − 8.36 = 0,
3.76
+ R A − 20 + RB = 0,
RB = 3.76 kN,
Ra = 16.24 kN,
31
• Quantitative shear and bending diagram and qualitative deflected curve
5 kN/m
18.31 kN•m
A
B
4m
16.24 kN
4m
8.36 kN•m
3.76 kN
16.24
V
diagram
3.25 m
M
diagram
8.08
-3.76
6.67
-8.36
-18.31
Deflected
Curve
32
Force Method of Analysis : Frames
• Principle of superposition
fCC × C x
B
∆´CH
C
Cx
1 kN
Cy
w
=
w
+
Ax
A
Ay
Compatibility equation :
∆ CH = 0 = ∆'CH + f CC C x
33
Example 9-4
Draw the quantitative Shear and moment diagram and the qualitative deflected
curve for the Frame shown below.EI is constant.
2 kN/m
B
6m
C
6m
A
34
SOLUTION
Use compatibility of displacement to obtain reaction
• Principle of superposition
fCC × C x
B
6m
∆´CH
C
Cy
=
6m
2 kN/m
2 kN/m
Cx
1 kN
+
Ax
A
Ay
Compatibility equation :
∆ CH = 0 = ∆'CH + f CC C x
-----(1)
35
6m
3m
Cy
M´2 = (6 + P)x2
x2
P
2 kN/m
2 kN/m
• Use Castigliano’s method for ∆´CH
∆´CH
6m
B
C
V´2
6+P
12 kN
3m
-12 - P
V´1
2x1
-6 - P
Ay
x2
6+P
M´1 = (12 + P)x1- x12
x1
Ax
A
P
x1
-12 - P
-6 - P
0
0
6
6
1
1
2
∂M 'i M 'i
(
x
)
(
12
x
+
x
P
−
x
)dx1 +
( x2 )(6 x2 + x2 P) dx2
dx =
= ∫(
)
1
1
1
1
∫
∫
EI 0
EI 0
∂P EI
0
L
∆ 'CH
6
6
1
1
2
3
2
=
(
12
x
−
x
)
dx
+
(
6
x
) dx2
1
1
1
2
∫
∫
EI 0
EI 0
3
4
3
1 12 x1 x1 6 1 6 x2
=
(
− )0+
(
)
EI
EI 3
3
4
6
0
=
972
,→
EI
36
• Use Castigliano’s method for fCC
fCC
6m
B
C
m´2 = x2P
x2
P
1 kN
P
v´2
Cy
P
6m
x2
P
m´1 = x1P
x1
-P
Ax
A
v´1
Ay
x1
-P
-P
1
1
6
6
1
1
∂m' m'
( x1 )( x1 P)dx1 +
( x2 )( x2 P )dx2
= ∫ ( i ) i dx =
∫
∫
EI 0
EI 0
∂P EI
0
-P
L
f CC
3
3
1 x1 6 1 x2
=
( ) +
( )
EI 3 0 EI 3
6
0
=
144
,→
EI
37
• Substitute ∆´CH and fCC in Eq. (1) ∆ CH = 0 = ∆'CH + f CC C x
+:
0=
-----(1)
972
,→
EI
144
f cc =
,→
EI
∆ CH =
972 144
+
Cx
EI
EI
Cx = -6.75 kN,
6 kN
1 kN
+
× C x = −6.75kN
12 kN
=
C 6.75 kN
B
2 kN/m
2 kN/m
1 kN
0.75 kN
5.25 kN
1 kN
A
6 kN
1 kN
0.75 kN
38
Or use Castigliano least work method:
x2
B
6m
M2 = (6-Cx)x2
C
Cx
2 kN/m
Cx
A
x2
V2
6 - Cx
6 - Cx
6m
M1 = (12 - Cx)x1- x12
x1
12 - Cx
V1
2x1
6 - Cx
x1
12 - Cx
L
∂M i M i
∂U i
= ∫(
)
dx = ∆ CH = 0
∂C x 0 ∂C x EI
6
0=
6 - Cx
6
1
1
2
(
−
x
)(
12
x
−
C
x
−
x
)
dx
+
(− x2 )(6 x2 − C x x2 )dx2
1
1
1
1
x 1
EI ∫0
EI ∫0
3
3
4
3
12 x1 C x x1 x1 6
6x
C x
0 = (−
+
+ ) 0 + (− 2 + x 2
3
3
4
3
3
0 = -972 + 144Cx , Cx = 6.75 kN,
3
6
0
39
• Quantitative shear and bending diagram and qualitative deflected curve
B
6m
C
B
2 kN/m
6.75 kN
C
- 0.75
- 0.75
- 6.75
0.75 kN
6m
5.25
5.25 kN
A
V, (kN)
2.63 m
A
0.75 kN
B
B
C
-4.5
C
1.33 m
-4.5
Deflected curve
A
M, (kN•m)
6.90
A
40
Force Method of Analysis : Truss (Externally indeterminate)
E
A
D
Cx
Ax
C
B
∆'CH + f CC C x = ∆ CH = 0
Cy
Ay
P
=
E
E
D
D
+
A
C
A
B
C
P
1
x Cx
B
∆´C
fCC
41
Truss (Internally indeterminate)
P
3
D
C
6
1
2
5
A
B
4
=
∆'6 + f 66 F6 = ∆ 6 = 0
P
D
D
∆´6
A
C
+
B
f66
C
1
A
xF6
B
42
Example 9-5
Determine the reaction at support A, C, E and all the member forces. Take
= 200 GPa and A = 500. mm2 .
E
E
40 kN
D
4m
A
C
B
5m
5m
43
SOLUTION
Use compatibility of displacement to obtain reaction
• Principle of superposition
RE
E
40 kN
D
4 m Ay
Ax
A
C
B
5m
=
5m
RC
40 kN
+
× Cy
fCC
∆´C
1 kN
Compatibility equation :
∆ C = 0 = ∆'C + f CC RC
-----(1)
44
• Use unit load method for ∆´C and fcc
5.39
m
53.85
20
E
50
5.39
m
+53
.85
A
2.5
0
0
B
=- 7.81 mm,
C
∆´C
0
5m
A
Σ ni´n´iLi
fCC =
N ´i (kN)
=
0
+2.5
AiEi
2(-2.69)2(5.385)
(200x106)(500x10-6)
-2.6
9
-2.6
9
fCC
B
n´i (kN)
+2.5
+
2(2.5)2(5)
(200x106)(500x10-6)
0
2.5
(200x106)(500x10-6)
0
D
0
=
D
1
E
AiEi
(53.85)(-2.69)(5.38)
5m
53.85
∆´C =
40 kN
4 m 20 kN
85
-53.
50
Σ n´iN´iLi
C
1 kN
= 1.41 mm,
45
• Substitute ∆´Cv and fCC in Eq. (1): ∆ C = 0 = ∆'C + f CC RC
∆ 'C = 7.81 mm, ↓
+ ↑: −7.81 + 1.41RC = 0
f CC = 1.41 mm, ↑
RC = 5.54 kN , ↑
20
50
+53
.85
A
40 kN
0
0
36.15 kN
xRC = 5.54 kN
0
B
0
14.46 kN
E
+38
.95
2.5
20 kN 53.85
36.15 kN
A
+13.85
-2.6
9
0
C
N ´i (kN)
38.93 kN
-2.6
9
D
0
0
E
2.5
D
20 kN
85
.
3
5
50
53.85
+
E
1
A
+2.5
B
n´i (kN)
=
53.85
C
+2.5
1 kN
40 kN
D -1
4.90
0 21.8o
B +13.85
N i (kN)
C
5.54 kN
46
Or use Castigliano least work method:
5.39
m
-2.7 RC + 53.85 = RE
4m
Ax = -2.5RC +50 = Ax
5.39
E -2.7 R
m
40 kN
C +5
3.85
D
-2.7
5
RC
8
.
-53
0 21.8o
A
2.5RC
B
2.5RC
C
RC
Ay = 20
5m
5m
Castigliano’s Theorem of Least Work :
∆ CV = 0 = ∑ (
0=
∂N i N i Li
)
∂RC AE
1
[(−2.7)(−2.7 RC + 53.85)(5.39) + ( −2.7)(−2.7 RC )(5.39) + 0 + 0 + 2[(2.5)(2.5 RC )(5)]]
AE
0 = 39.3RC − 783.68 + 39.3RC + 62.5 RC
RC = 5.55 kN,
47
Example 9-6
Determine the force in all member of the truss shown :
(a) If the horizontal force P = 6 kN is applied at joint C.
(b) If the turnbuckle on member AC is used to shorten the member by
1 mm.
(c) If (a) and (b) are both accounted.
Each bar has a cross-sectional area of 500 mm2 and E = 200 GPa.
D
C
2m
A
B
3m
48
SOLUTION
Part (a) : If the horizontal force P = 6 kN is applied at joint C.
• Principle of superposition
D
6 kN
1
C
6
3
A
5
4
2m
B
2
3m
D
C 6 kN
=
3m
D
∆´6
1 E´
E E´
A
C
2m
B
Compatibility equation :
Note : AE + E ' C = L
1
+
E f66
A
∆´6 + f66 F6 = 0
×F6
B
----------(1)
49
• Use unit load method for ∆´6 and f66
3m
D
C 6 kN
+6
D
∆´6
+4
6
-7.
2
A
0
B
1
N ´i (kN)
-14.98
0
n´iN ´iLi (kN2•m)
1
-0.832
A
n´i (kN)
0
2.08
6
3.
0
-2
6. 0
3
-14.98
1
2 m -0.555
4
4
-4.44
3
1
E E´
+6
C
-0.832
0
-0.555
3.6
1
2
3
0
Li (m)
0.616
3.6
2
1
n'i N 'i Li
Ai Ei
=
1
[−4.44 − 26.03 − 2(14.98)]
AE
=
− 60.43
AE
2.08
n´i2Li (kN2•m)
1
B
∆ '6 = ∑
1
0.616
6
3.
2
n' L
1
12.61
f 66 = ∑ i i =
[2(0.616) + 2(2.08) + 2(3.61)] =
Ai Ei
AE
AE
50
• Substitute ∆´6 and f66 in Eq. (1)
−
3m
+6
60.43 12.61
+
( F6 ) = 0
AE
AE
F6 = 4.80 kN, (T)
D
6 kN
∆´6
+4
+6
6
C
-0.832
1
E E´
-7.
2
0
2m
+
1
4
4
-0.832
A
n´i (kN)
0
80
.
+4 2.4
1
+2
+1.34
A
4
Ni (kN)
-0.555 x F6 = 4.80 kN
B
0
6 kN
+2
D
6
1
=
0
N ´i (kN)
1
-0.555
C
-2.66
B
4
51
Part (b) : If the turnbuckle on member AC is used to shorten the member by 1 mm.
f 66 =
12.61
12.61
=
= 1.26(10-4) m = 0.126 mm
AE
(500)(200)
F6 =
D
1 mm
(1 kN ) = 7.94 kN
0.126 mm
C
-0.832
-6.61
D
94
.
+7 7
.94
-6.61
1
1
-0.555
0
1
-0.832
A
n´i (kN)
0
-0.555
B
0
x F6 = 7.94 kN
-4.41
=
6
A
4
Ni (kN)
C
-4.41
B
4
52
Part (c) : If the horizontal force P = 6 kN is applied at joint C and the turnbuckle
on member AC is used to shorten the member by 1 mm are both accounted.
4
(Ni)load (kN)
94
.
+7 7
.94
-6.61
-4.41
B
0
A
4
0
=
A
C
-2.66
-6.61
D
+
80
.
+4 2.4
1
+2
+1.34
6
6 kN
+2
D
1
-3.07
74
.
2
A
4
B
0
C
5.5
3
-4.61
6
-4.41
6 kN
-4.61
D
(Ni)short (kN)
C
(Ni)total (kN)
-7.07
B
4
53
Or use compatibility equation :
∆´6 + f66 F6 = ∆´6 = 0.001
−
60.43 12.61
+
( F6 ) = 0.001
AE
AE
F6 =
∆´6
-7
. 21
6
D
6 kN
+6
+4
0.001AE + 60.43 0.001(500)(200) + 60.43
=
= 12.72 kN, (T)
12.61
12.61
0
+
+6
4
-0.832
A
n´i (kN)
0
-4.58
D
12
-3.06
2
.7
C
5.5
1
-4.58
6
A
4
x F6 = 12.72 kN
f66 -0.555
=
4
(Ni)total (kN)
C
1
-0.555
0
N ´i (kN)
-0.832
1 1
B
0
6 kN
-7.06
B
4
54
Composite Structures
Example 9-7
Find all reaction and the tensile force in the steel support cable. Consider both
bending and axial deformation.
Steel cable
Ac = 2(10-4) m2
Ec = 200(103) kN/m2
C
2m
A
B
5 kN
m2
Ab = 0.06
Ib = 5(10-4) m4
Eb = 9.65(103) kN/m2
6m
55
SOLUTION
C
m
6.32
RC = T
2m
18.43o
A
B
x
6m
5 kN
0.316T
T
M = 0.316Tx - 5x
N = -0.949T
0.949T
V
5 kN
Bx
MB
By
By Castigliano’s Theorem of Least Work ;
∂
∆C = 0 =
(U ib + U in )
∂T
L
L
∂M M
∂N N
∆C = 0 = ∫ (
dx + ∫ ( )
)
dx
∂
T
EI
∂
T
AE
0
0
6
6
1
1
1
0=
(
0
.
316
x
)(
0
.
316
xT
−
5
x
)
dx
+
(
−
0
.
949
)(
−
0
.
949
T
)
dx
+
Eb I b ∫0
Ab Eb ∫0
Ac Ec
6
1
0.316 2 x 3
(0.316 × 5) x 3 6
1
1
2
0=
[(
T) −
]0+
(0.949 xT ) 0 +
( xT )
Eb I b
3
3
Ab Eb
Ac Ec
0 = (1.49T - 23.58) + 9.33(10-3)T + 0.158T
6.32
∫ (1)(T )dx
0
6.32
0
; T = 14.23 kN, (tension) #
56
4.5 kN
C
m
6.32
RC = T = 14.23 kN
13.5 kN
2m
18.43o
A
B
x
5 kN
Bx
6m
By
+ ΣF = 0:
x
Bx = Rc cos θ = 13.5 kN,
+ ΣFy = 0:
By = 5 - Rc sin θ = 0.5 kN,
+ ΣMB = 0:
MB
MB = 13.5(2) - 5(6) = -3 kN•m, +
57
DISPLACEMENT METHOD OF ANALYSIS:
SLOPE DEFLECTION EQUATIONS
!
!
!
!
!
!
!
!
!
General Case
Stiffness Coefficients
Stiffness Coefficients Derivation
Fixed-End Moments
Pin-Supported End Span
Typical Problems
Analysis of Beams
Analysis of Frames: No Sidesway
Analysis of Frames: Sidesway
1
Slope – Deflection Equations
i
P
j
w
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
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 Moments: Loads
P
L/2
PL
8
L/2
PL
8
L
P
2
P
2
w
wL2
12
wL2
12
wL
2
L
wL
2
6
General Case
i
P
j
w
k
Cj
settlement = ∆j
Mij
P
i
w
j
Mji
θi
ψ
θj
7
Mij
P
i
j
w
Mji
θi
L
settlement = ∆j
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 = (
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
j
w
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
(
2
/
)
(
4
/
)
EI
L
EI
L
j ji
ji
kii
k ji
[k ] =
kij
k jj
Stiffness Matrix
11
P
i
Mij
w
j
Mji
θi
[ M ] = [ K ][θ ] + [ FEM ]
L
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: Fixed-End Support
θi
Mi
Mj
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
Mj =(
)θ i
L
Mi = (
13
Stiffness Coefficients Derivation: Pinned-End Support
Mi
θi
i
θj
j
Real beam
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 = (
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
Fixed end moment : Point Load
P Real beam
Conjugate beam
A
A
B
B
L
M
EI
M
M
EI
M
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
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
16
Uniform load
w
Real beam
Conjugate beam
A
A
L
B
B
M
EI
M
M
EI
M
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
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
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
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
M BA =
θA +
θ B + ( M BA
)load + ( M BA
) ∆ − − − (2)
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
Fixed-End Moments
Fixed-End Moments: Loads
P
PL
8
L/2
L/2
PL
8
L/2
PL 1
PL
3PL
+ ( )[−(−
)] =
8
2
8
16
P
L/2
wL2
12
wL2
12
wL2 1
wL2
wL2
+ ( )[−( −
)] =
2
12
8
12
21
Typical Problem
CB
w
P1
P2
C
A
B
L1
P
PL
8
L2
PL
8
wL2
12
L
0
PL
4 EI
2 EI
M AB =
θA +
θB + 0 + 1 1
8
L1
L1
0
PL
2 EI
4 EI
θA +
θB + 0 − 1 1
M BA =
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
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
4 EI
2 EI
P L wL
=
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
P1
MAB
CB
w
MBA
A
B
L1
P2
MBC
L2
C M
CB
Substitute θB in MAB, MBA, MBC, MCB
0
4 EI
2 EI
PL
θA +
θB + 0 + 1 1
M AB =
8
L1
L1
04
2 EI
EI
PL
θA +
θB + 0 − 1 1
M BA =
8
L1
L1
0
2
4 EI
2 EI
P2 L2 wL2
θB +
θC + 0 +
+
M BC =
L2
L2
8
12
0
2
− P2 L2 wL2
2 EI
4 EI
M CB =
θB +
θC + 0 +
−
L2
L2
8
12
24
P1
MAB
CB
w
MBA
P2
MCB
A
Ay
B
L1
MBC
L2
C
Cy
By = ByL + ByR
P1
MAB
B
A
Ay
L1
C
B
MBA
ByL
P2
MBC
ByR
L2
MCB
Cy
25
Example of Beams
26
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
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)
θA +
θB −
M BA =
8
8
8
0
4 EI
2 EI
(6)(6 2 )
M BC =
θB +
θC +
6
6
30
0
2 EI
4 EI
(6)(6) 2
M CB =
θB +
θC −
6
6
20
M AB =
B
+ ΣM B = 0 : − M BA − M BC = 0
(6)(6 2 )
4 EI 4 EI
+
=0
)θ B − 10 +
(
30
8
6
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
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,
MBC = 8.8 kN•m
MBA = - 8.8 kN•m,
MCB= -10 kN•m
10 kN•m
6m
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
10 kN
6 kN/m
10.6 kN•m
C
A
4m
5.23 kN
B
4m
6m
10 kN•m
12.2 kN
4.78 + 5.8 = 10.58 kN
5.8
5.23
V (kN)
+
+
x (m)
- 4.78
-
10.3
M
(kN•m)
x (m)
+
-10.6
Deflected shape
-12.2
-
-8.8
2.4
θB =
EI
-10
x (m)
30
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
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
M CB =
θB +
θC −
− − − (4)
6
6
20
6 EI
2(2) − (1) : 2 M BA =
θ B − 30
8
3EI
M BA =
θ B − 15 − − − (5)
8
32
MBA
MBC
B
M BC
4 EI
(6)