# Session I Structural Analysis

#### Short Description

Slides for the PE exam Class...

#### Description

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

40 0 M (kN-m)

0 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

40 0 M (kN-m)

0 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°

-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°

-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°

-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°

-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

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!

65