P.E.

Civil Exam Review:

Structural Analysis

J.P. Mohsen
jpm@louisville.edu

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Index of Topics
NCEES Topics 3

Determinate vs. Indeterminate Structures 4

Stability and Determinacy of Trusses / Truss Analysis 7

Shear and Moment Diagrams 20

Deflection 38

References 43

Appendices 44

2
NCEES Topics
Structural Analysis (10%)
1. Dead Loads
2. Live Loads
3. Construction Loads
4. Trusses
5. Bending
6. Shear
7. Shear Diagrams
8. Moment Diagrams

3
Structures

• Determinate
• Indeterminate

4
Statically Determinate

P
0
0 M0
Hinge

L1 L2 L3 L4

P4

P1

P2 P3

5
Statically Indeterminate

P 2 3
1

L1 L2 L3 L4

6
Stability and Determinacy of Trusses

300 lb. 400 lb.

B C D

7.5 ft

E
10 ft H 10 ft G 10 ft F 10 ft

RA RE

2j  m  r Truss is determinate
J = number of joints
2j  m  r indeterminate m = number of members
2j  m  r unstable r = number of reactions

7
Problem 1

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.

B C D

7.5 ft

10 ft H 10 ft G 10 ft F 10 ft E

RA

8
Problem 1 (continued)
Member BH

300 lb. 400 lb.

B C D

A
10 ft H 10 ft G 10 ft F 10 ft E

RL

9
Problem 1 (continued):
Analysis of member BH.

300 lb. 400 lb.

B C D

A H G F E
10 ft 10 ft 10 ft 10 ft
RR
RL

FBH
Applying equation of equilibrium to joint H

   Fy  0 Fbh  0

FAH FHG
H
10
Problem 1 (continued):

Member BC
300 lb. 400 lb.

B C D

A 10 ft H 10 ft G 10 ft F 10 ft E

RA RE

+ ME  0 + MA  0
300(30)  400(20)  40R A  0 300(10)  400(20)  40R E  0
9000  8000  40R A  0 3000  8000  40R E  0
40R A  17000 40R E  11000
Reaction at A = 425 lbs Reaction at E = 275 lbs

11
Problem 1 (continued):

Analysis of member BC

300 lb. 400 lb.

B C D

A 10 ft H 10 ft G 10 ft F 10 ft E

RA R E  275 lbs

+  M G  0  20 RE  7.5FBC  0 400 lb.

B C D
275(20) FBC
FBC   733 lbs (compression) 12.5 ft
7.5 FBG 7.5 ft
E
FHG
G 10 ft F 10 ft
RE

12
Problem 1 (continued):

Member DG

300 lb. 400 lb.

B C D

A 10 ft H 10 ft G 10 ft F 10 ft E

RL RE

13
Problem 1 (continued):

Analysis of member DG

300 lb. 400 lb.

B C D

A 10 ft H 10 ft G 10 ft F 10 ft E

RL RE

   FY  0 RE  DGY  0  DGY  275 lbs

C D
FCD
12.5 ft
FDG 7.5 ft
DGY 3
  DG  458 lbs (tension) E
DG 5 FGF
G F 10 ft
RE

14
Problem 2

Find the force in the truss members shown.

500 N
C
E
45

D
1 meter 90

A B
45

Pin Rollers

1 meter

15
Problem 3

Find all member forces and specify whether they are in tension or compression.

2 kN

3 kN 3 kN

G E

Ax  0
A 30 60 60 60 60 30 D
B C
3m 3m 3m

Ay  4 kN Dy  4 kN

16
Problem 3 (continued):

FAG

30
A
FAB
x

4 kN

   Fy  0; 4  FAG sin 30  0 FAG  8 kN (C) Ans.


   Fx  0; FAB  8cos 30  0 FAB  6.93 kN (T) Ans.

17
Problem 3 (continued):
y 3 kN

x
30
FGF

8 kN
FGB
 F
y  0; FGB  3cos 30  0 FGB  2.60 kN (C) Ans.

 F
x  0; 8  3sin 30  FGF  0 FGF  6.50 kN (C) Ans.

18
Problem 3 (continued):
y

2.60 kN FBF

60 60
6.93 kN
B FBC

   Fy  0; FBF sin 60  2.60sin 60  0

FBF  2.60 kN (T) Ans.


   Fx  0; FBC  2.60 cos 60  2.60 cos 60  6.93  0

FBC  4.33 kN (T) Ans.

19
Shear and Moment Diagrams

Problem 4:
Draw the shear and moment diagrams for the beam shown. Indicate the maximum moment.

60 kN
20 kN/m
120 kN-m

A C D
B E

2m 2m 2m 2m

20
Problem 4 (continued):

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

C D

FB FE

2m 2m 2m 2m

21
Problem 4 (continued):

Solve for the reactions at supports B and E.

60 kN
20 kN/m
120 kN-m

C D

FB  100 kN FE  40 kN

2m 2m 2m 2m

+  M B  0  60(2)  120  6 FE  0  FE  40 kN

+  FY  0  60  80  FE  FB  0  100  FB  0  FB  100 kN

22
Problem 4 (continued):
Draw the shear diagram for segment AB.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

kN
(2m )( 20 )  40kN
m
0 0 V (kN)

-40

23
Problem 4 (continued):

Show the change in shear at B.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

100 kN
0 0 V (kN)

-40

24
Problem 4 (continued):

Draw the shear diagram for segment BC.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20
2 m  20 kN m   40 kN
0 0 V (kN)

-40

25
Problem 4 (continued):
Show the change in shear at C.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20
V (kN)
 60 kN
0 0

-40 -40

26
Problem 4 (continued):

Draw the shear diagram for segment CE.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

4 m 0 kN m   0 kN
20

00 0 V (kN)

-40 -40 -40

27
Problem 4 (continued):
Show the change in shear at E.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20 40 kN
00 0 V (kN)

-40 -40 -40

28
Problem 4 (continued):
Completed shear diagram.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20

0 0 V (kN)

-40 -40 -40

29
Problem 4 (continued):
Draw the moment diagram for segment AB.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20
2 m  40 kN  1   40 kN  m
0 0 V (kN) 2
-40 -40 -40

0 M (kN-m)
2⁰
-40

30
Problem 4 (continued):
Draw the moment diagram for segment AB.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20 2 m  40 kN  1   40 kN  m

2
0 0 V (kN)

-40 -40 -40

0 M (kN-m)
2⁰
-40

31
Problem 4 (continued):
Draw the moment diagram for segment BC.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

 2 m  40 kN  
60 1
2   (2m )(20kN )  80kN  m
20  
0 0 V (kN)

-40 -40 -40

2⁰
40

0 M (kN-m)
2⁰ 2⁰
-40

32
Problem 4 (continued):
Draw the moment diagram for segment CD.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20
2 m  40 kN   80 kN  m
0 0 V (kN)

-40 -40 -40

40
2⁰

0 M (kN-m)
2⁰ 2⁰
-40 -40

33
Problem 4 (continued):
Show the change in bending moment at D.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20

0 0 V (kN) 120kn  m
-40 -40 -40

40 80
2⁰

0 M (kN-m)
2⁰ 2⁰
-40 -40

34
Problem 4 (continued):
Draw the moment diagram for segment DE.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20 2 m  40 kN   80 kn  m
0 0 V (kN)

-40 -40 -40

40 80
2⁰

0 0 M (kN-m)
0 2⁰ 2⁰
-40 -40

35
Problem 4 (continued):
Completed moment diagram.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20

0 0 V (kN)

-40 -40 -40

40 80
2⁰

0 0 M (kN-m)
0 2⁰ 2⁰
-40 -40

36
Problem 4 (continued):
Find the maximum moment.

60 kN
20 kN/m
120 kN-m

A 2m 2m C 2m D 2m
100 kN 40 kN

60

20

0 0 V (kN)

-40 -40 -40

40 80
2⁰

0 0 M (kN-m) M max  80 kn  m
0 2⁰ 2⁰
-40 -40

37
Deflection
Problem 5:
The reaction at point “A” is
a. Zero
b. 40 lbs ↑
c. 40 lbs ↓
d. 40 lbs ↑ plus 400 ft-lbs
100 lbs.
e. 40 lbs ↓ plus 400 ft-lbs
2’ 2’
’

2’

A 2’ B

6’ 4’

38
Problem 6:

8 ft
A
2 ft

6 ft 300 lb/ft

B 39
Problem 6 (continued):

8 ft
A

2 ft

3 ft

1800 lbs

3 ft

40
Problem 7:
Please find the reaction at all supports.

6 ft

8 ft

10 ft

41
B
Problem 7 (continued):

A 4
3 3
4
6 ft

8 ft

10 ft

B 42
References
 Hibbeler, C. R., Structural Analysis, 3rd Edition, Prentice Hall, 1995.

 Chajes, Alexander, Structural Analysis, Prentice Hall, 1982.

Any questions? Good luck!

43
Appendix A: Problem 2 Solution

500 N
C
E
45

D
1 meter 90 BC = 500 N (C), AC = 707 N(T)
CE = 500 N (C), AE = 0, BD = 0,
AB = 0

A B
45

Pin Rollers

1 meter

44
Appendix B: 6 Solution

8 ft
A

2 ft

+  M B  0  300(6)(6 / 2)  AX (6  2)  0
3 ft
AX  675 lbs 
1800 lbs
+  FX  0  300(6)  675  BX  0
BX  1125 lbs  3 ft

+ M A  0  300(6)(5)  BX (6  2)  BY (8)  0

BY  0 lbs
B

45
Appendix C: Problem 7 Solution

A 4
3 3
4
6 ft

+ M B  0  1000(5)  BY (8)  0
8 ft
 BY  625 lbs

+  FX  0  AX  1000(3 / 5)  0
10 ft
 AX  600 lbs

+  FY  0  AY  625  1000(4 / 5)  0

 AY  175 lbs

B
46