P.E. Civil Exam Review: Structural Analysis
J.P. Mohsen Email:
[email protected]
Structures • Determinate • Indeterminate
2
STATICALLY DETERMINATE
3
STATICALLY INDETERMINATE
4
Stability and Determinacy of Trusses
300 lb.
400 lb. C
B
D
7.5 ft
A 10 ft
H
10 ft
G
10 ft
F
E
10 ft RR
RL
2j = m + r
Truss is determinate
2j
m+r
indeterminate
2j
m+r
Unstable
J = number of joints m= number of members r = number of reactions
5
Determine the force in members BH, BC, and DG of the truss shown. Note that the truss is composed of triangles 7.5 ft : 10.0 ft : 12.5 ft, so that they are 3:4:5 right angles.
300 lb.
400 lb. C
B
D
7.5 ft
10 ft RL
H
10 ft
G
10 ft
F
E
10 ft RR
6
Member BH.
300 lb.
400 lb. C
B
A
RL
10 ft
H
10 ft
D
G
10 ft
F
E
10 ft RR
7
Analysis of Member BH. 300 lb.
C
B
A
10 ft
400 lb.
H
10 ft
D
G
10 ft
F
10 ft
RL
E RR
FBH
Applying Equation of Equilibrium to Joint H +
FAH H
∑F
y
=0
Fbh = 0
FHG
8
Member BC.
300 lb.
400 lb. C
B
A
RL
10 ft
H
10 ft
D
G
10 ft
F
E
10 ft RR
9
Analysis of Member BC. 300 lb.
400 lb. C
B
A
10 ft
H
D
G
10 ft
10 ft
F
E
10 ft
RL
RR = 275 lb. +
∑M
G
=0
FBC = 400 lb. B
FBC
C
− 20 RR − 7.5 FBC = 0
− 275(20) = − 733 lbs ( compression) 7.5
D 12.5 ft 7.5 ft
FBG FHG
E G
10 ft
F
10 ft RR 10
Member DG.
300 lb.
400 lb. C
B
A
RL
10 ft
H
10 ft
D
G
10 ft
F
E
10 ft RR
11
Analysis of Member DG. 300 lb.
400 lb. C
B
A
10 ft
H
D
G
10 ft
10 ft
F
E
10 ft
RL
RR
C
FCD FDG
G
FGF
D 12.5 ft 7.5 ft E F
10 ft RR 12
Analysis of Member DG. 300 lb.
400 lb. C
B
A
10 ft
H
D
G
10 ft
10 ft
F
RL
+
E
10 ft
RR
∑F
Y
= 0
DGY = 275 lbs
RR − DGY = 0
DGY 3 = DG 5
DG = 458 lbs (tension ) C
FCD FDG
G
FGF
D 12.5 ft 7.5 ft E F
10 ft RR 13
Draw the shear and moment diagrams for the beam shown. Indicate the maximum moment.
60 kN
20 kN/m
120 kN-m
A
D
C B 2m
E 2m
2m
2m
14
Draw the Free Body Diagram (FBD). (Note: The horizontal force at point B is equal to zero.)
60 kN
20 kN/m
120 kN-m
A
D
C FB 2m
FE 2m
2m
2m
15
Solve for the reactions at supports B and E. 60 kN
20 kN/m
120 kN-m
A
D
C FB = 100 kN 2m
2m
FE = 40 kN 2m
2m
+
∑MB = 0 → 60(2) + 120 – 6FE = 0 → FE = 40 kN
+
∑FY = 0 → -60 – 80 + FE + FB = 0 → -100 + FB = 0 → FB = 100 kN
16
60 kN
20 kN/m
A
2m
120 kN-m 2m
100 kN
C
2m
D
Draw the Shear Diagram for segment AB.
2m 40 kN
(2 m )(− 20 kN m ) = −40 kN 0 V (kN)
0 -40
17
60 kN
20 kN/m
A
2m
120 kN-m 2m
100 kN
C
2m
D
Show the change in Shear at B.
2m 40 kN
60
100 kN 0 V (kN)
0 -40
18
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
2m
D
Draw the Shear Diagram for segment BC.
2m 40 kN
100 kN
(2 m )(− 20 kN m ) = −40 kN
60 20 0 V (kN)
0 -40
19
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
2m
D
Show the change in Shear at C.
2m 40 kN
100 kN 60
− 60 kN 20 0 V (kN)
0 -40
-40
20
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
2m
D
Draw the Shear Diagram for segment CE.
2m 40 kN
100 kN
(4 m )(0 kN m ) = 0 kN
60 20 0 V (kN)
0 -40
-40
-40
21
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
2m
D
Show the change in Shear at E.
2m 40 kN
100 kN 60
40 kN 20 0 V (kN)
0 -40
-40
-40
22
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
2m
D
Completed Shear Diagram
2m 40 kN
100 kN 60 20
0 V (kN)
0 -40
-40
-40
23
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
2m
D
Draw the Moment Diagram for segment AB.
2m 40 kN
100 kN 60 20
0 V (kN)
0 -40
-40
-40
(2 m )(− 40 kN ) 1 = −40 kN ⋅ m 2
0 M (kN-m)
0 2°
-40
24
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
2m
D
Draw the Moment Diagram for segment AB.
2m 40 kN
100 kN 60 20
0 V (kN)
0 -40
-40
-40
(2 m )(− 40 kN ) 1 = −40 kN ⋅ m 2
0
0 M (kN-m) -40
25
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
2m
D
Draw the Moment Diagram for segment BC.
2m 40 kN
100 kN 60 20
0 V (kN)
0 -40
-40
-40
(2 m )(40 kN ) 1 + (2 m )(20 kN ) = 80 kN ⋅ m 2
2°
40 0 M (kN-m)
0 2°
2°
-40
26
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
2m
D
Draw the Moment Diagram for segment CD.
2m 40 kN
100 kN 60 20
0 V (kN)
0 -40
-40
-40
(2 m )(− 40 kN ) = −80 kN ⋅ m
2°
40 0 M (kN-m)
0 2°
2°
-40
-40
27
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
D
2m
Show the change in bending moment at D.
2m 40 kN
100 kN 60 20
0 V (kN)
0 -40
-40
-40
120 kn ⋅ m 80 2°
40 0 M (kN-m)
0 2°
2°
-40
-40
28
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
D
2m
Draw the Moment Diagram for segment DE.
2m 40 kN
100 kN 60 20
0 V (kN)
0 -40
-40
-40
(2 m )(− 40 kN ) = −80 kn ⋅ m 80 2°
40 0 M (kN-m)
0 2°
2°
-40
-40
29
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
D
2m
Completed Moment Diagram.
2m 40 kN
100 kN 60 20
0 V (kN)
0 -40
-40
-40
80 2°
40 0 M (kN-m)
0 2°
2°
-40
-40
30
60 kN
20 kN/m
A
2m
120 kN-m 2m
C
D
2m
Find the maximum moment.
2m 40 kN
100 kN 60 20
0 V (kN)
0 -40
-40
-40
M max = 80 kn ⋅ m
80 2°
40 0 M (kN-m)
0 2°
2°
-40
-40
31
32
Find the force in the truss members shown. 33
34
35
What are the vertical and horizontal components of deflection at the 30K Load? All members have a cross sectional area of 1 square inch and modulus of elasticity of 29000 ksi.
36
S u L δ = ∑ AE S = member force with proper sign A= Cross-sectional area of each member L= Length of each member E= modulus of elasticity of materials
37
These are internal member forces due to original loading
38
These are internal forces due to a vertical unit load at L2
39
These are internal member forces due to a horizontal unit load at L2
40
41
Please find all member forces and specify whether in tension or compression 42
43
44
45
What are the support reactions for the beam shown?
KAB=4EI = I L 10
KBC=4EI = I L 20
46
Moment Distribution • 1) calculate the fixed end moments • 2) Calculate distribution of moments at the clamped ends of the members by the rotation of that joint • 3) Calculate the magnitude of the moments carried over to the other ends of the members • 4) The addition or subtraction of these latter moments to the original fixed ends moments
47
Fixed End Moments P L/2
L/2
FEM = PL 8
FEM = PL 8 w
L
FEM= wL2 12
FEM= wL2 12
P a
b
L
FEM=
Pb2a L2
FEM= Pa2b L2 48
Lock the joint B
.
FEM
-25
+ 25
- 50
+ 50
49
KAB=4EI = I L 10
Distribution Factor =
KBC=4EI = I L 20
K_______________ Sum of K for all members at the joint
DF1 =
K1 ∑K
DF2 =
K2 ∑K 50
KAB=4EI = I L 10
KBC=4EI = I L 20
Distribution Factor =
K_______________ Sum of K for all members at the joint
Distribution Factor =
KBA_ KBA + KBC
DF1 =
K1 ∑K
DF2 =
K2 ∑K 51
KAB=4EI = I L 10
1__ 10____ = 2/3 1__ + 1_ 10 20 1__ 20____ = 1/3 1__ + 1_ 10 20
KBC=4EI = I L 20
D.F. at B for BA
D.F. at B for BC
52
Stiffness K
KAB=4EI = I L 10
2/3
D. F.
FEM
KBC=4EI = I L 20
-25
+ 25
1/3
- 50
+ 50
Balancing Joint B
53
Joint B Released
Stiffness K
KAB=4EI = I L 10
2/3
D. F.
FEM
Balancing Joint B
KBC=4EI = I L 20
-25
+ 25
+ 16.67
1/3
- 50
+ 50
+8.33
54
Stiffness K
KAB=4EI = I L 10
D. F.
FEM
-25
2/3
1/3
+ 25
- 50
+ 50
+ 16.67 +8.33
Balancing Joint B
C.O.M.
KBC=4EI = I L 20
+ 8.33
55
Stiffness K
KAB=4EI = I L 10
D. F.
FEM
-25
Balancing Joint B
KBC=4EI = I L 20
2/3
1/3
+ 25
- 50
+ 50
+16.67 +8.33
C.O.M. + 8.33
+4.17
56
Stiffness K
KAB=4EI = I L 10
D. F.
FEM
-25
Final Moments
2/3
1/3
+ 25
- 50
+ 50
+ 16.67 +8.33
Balancing Joint B C.O.M.
KBC=4EI = I L 20
+ 8.33
- 16.67
+4.17
57
Stiffness K
KAB=4EI = I L 10
D. F.
FEM
-25
KBC=4EI = I L 20
2/3
1/3
+ 25
- 50
+ 50
+ 16.67 +8.33
Balancing Joint B C.O.M.
+ 8.33
Final Moments
- 16.67
+4.17
+ 41.67
58
Stiffness K
KAB=4EI = I L 10
D. F.
FEM
-25
KBC=4EI = I L 20
2/3
1/3
+ 25
- 50
+ 50
+ 16.67 +8.33
Balancing Joint B C.O.M.
+ 8.33
Final Moments
- 16.67
+4.17
+ 41.67
- 41.67
59
Stiffness K
KAB=4EI = I L 10
D. F.
FEM
-25
KBC=4EI = I L 20
2/3
1/3
+ 25
- 50
+ 50
+ 16.67 +8.33
Balancing Joint B C.O.M.
+ 8.33
Final Moments
- 16.67
+4.17
+ 41.67
- 41.67
+ 54.17
60
Stiffness K
KAB=4EI = I L 10
2/3
D. F.
FEM
KBC=4EI = I L 20
-25
Balancing Joint B
+ 25
+ 16.67
1/3
- 50
+ 50
+8.33
C.O.M.
+4.17
+ 8.33 ---------Final Moments
- 16.67
------------
+ 41.67
----------
- 41.67
----------
+ 54.17 61
1.5 K/FT
20 k
41.67
-16.67 FT.K
-41.67
20 k
7.5 k
54.2
1.5 K/FT
12.5 k
14.4 k
15.6 k
62
Member forces for the truss on slide 33 -----
AE=Zero member CE=500 N Compression BD= Zero member BC=500 N Compression AC= 707 N Tension AB= Zero member
63
References • Hibbeler, C. R., Structural Analysis, 3rd Edition, Prentice Hall, 1995. • Chajes, Alexander, Structural Analysis, Prentice Hall, 1982.
64
Thank You! • Any Questions?
• Good Luck!
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