Session I Structural Analysis

8/11/2019 Session I Structural Analysis

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P.E. Civil Exam Review:Structural Analysis

J.P. Mohsen

Email: [email protected]


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Structures

Determinate

Indeterminate

2


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STATICALLY DETERMINATE

3


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STATICALLY INDETERMINATE

4


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Stability and Determinacy of Trusses

400 lb.300 lb.

10 ft 10 ft 10 ft 10 ft

B C D

EFGH

RLRR

7.5 ft

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

A

5


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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.

400 lb.300 lb.

10 ft 10 ft 10 ft 10 ft

B C D

EFGH

RL RR

7.5 ft

6


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Member BH.

400 lb.300 lb.

10 ft 10 ft 10 ft 10 ftA

B C D

EFGH

RL RR

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Analysis of Member BH.

400 lb.300 lb.

10 ft 10 ft 10 ft 10 ftA

B C D

EFGH

RL RR

FBH

FHGFAH

H

00 == bhy FF+Applying Equation of Equi libr ium to Joint H

8


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Member BC.

400 lb.300 lb.

10 ft 10 ft 10 ft 10 ftA

B C D

EFGH

RL RR

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Analysis of Member BC.

400 lb.300 lb.

10 ft 10 ft 10 ft 10 ftA

B C D

EFGH

RL RR

FBC

E

RR

G

C D

FFHG

FBG

B

10 ft

12.5 ft

7.5 ft

400 lb.

= 275 lb.

)(7335.7

)20(275ncompressiolbsFBC =

=

+ 05.7200 == BCRG FRM

10 ft

10


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Member DG.

400 lb.300 lb.

10 ft 10 ft 10 ft 10 ftA

B C D

EFGH

RL RR

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Analysis of Member DG.

400 lb.300 lb.

10 ft 10 ft 10 ft 10 ftA

B C D

EFGH

RL RR

FCD

E

RR

G

C D

FFGF

FDG

10 ft

12.5 ft

7.5 ft

12


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Analysis of Member DG.

400 lb.300 lb.

10 ft 10 ft 10 ft 10 ftA

B C D

EFGH

RL RR

FCD

E

RR

G

C D

FFGF

FDG

10 ft

12.5 ft

7.5 ft

0= YF 0= YR DGR lbsDGY 275=

5

3=

DG

DGY ( )tensionlbsDG 458=

+

13


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120 kN-m20 kN/m

60 kN

A

B

C D

E

2 m 2 m 2 m 2 m

Draw the shear and moment diagrams for the beam shown. Indicate the

maximum moment.

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120 kN-m20 kN/m

60 kN

A

FB

C D

FE

2 m 2 m 2 m 2 m

Draw the Free Body Diagram (FBD).

(Note: The horizontal force at point B is equal to zero.)

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120 kN-m20 kN/m

60 kN

A

FB= 100 kN

C D

FE= 40 kN

2 m 2 m 2 m 2 m

Solve for the reactions at supports B and E.

MB= 0 60(2) + 120 6FE= 0 FE= 40 kN

FY= 0 -60 80 + FE+ FB= 0 -100 + FB= 0 FB= 100 kN

+

+

16


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Draw the Shear Diagram

for segment AB.

0 0 V (kN)

-40

( ) kNm

kNm 40202 =

17


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

V (kN)

Show the change in Shear

at B.

kN100

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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

V (kN)


for segment BC.

( ) kNm

kNm 40202 =

19


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

-40

V (kN)


at C.

kN60

20


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

-40 -40

V (kN)


for segment CE.

( ) kNm

kNm 004 =

21


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

-40 -40

V (kN)


at E.

kN40

22


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

-40 -40

V (kN)

Completed Shear Diagram

23


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

-40 -40

-40

M (kN-m)0 0

2

V (kN)

Draw the Moment Diagram

for segment AB.

( )( ) mkNkNm =

40

2

1402

24


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

-40 -40

-40

M (kN-m)0 0

V (kN)


for segment AB.

( )( ) mkNkNm =

40

2

1402

25


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

-40 -40

-40

M (kN-m)0 0

40

2 2

2

V (kN)


for segment BC.

( )( ) ( )( ) mkNkNmkNm =+

80202

2

1402

26


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

-40 -40

-40 -40

M (kN-m)0 0

40

2 2

2

V (kN)


for segment CD.

( )( ) mkNkNm = 80402

27


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

-40 -40

-40 -40

M (kN-m)0 0

40

80

2 2

2

V (kN)

Show the change in

bending moment at D.

mkn 120

28


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0

-40

60

20

-40 -40

-40 -40

M (kN-m)0 0

40

80

2 2

2

V (kN)


for segment DE.

( )( ) mknkNm = 80402

29


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120 kN-m20 kN/m

60 kN

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

0 0 V (kN)

-40

60

20

-40 -40

-40 -40

M (kN-m)0 0

40

80

2 2

2

Completed Moment

Diagram.

30


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32


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Find the force in the truss members shown.

33


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34


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35


36/65

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


37/65

=

EALuS

E= modulus of elasticity of materials

A= Cross-sect ional area of each member

L= Length of each member

S = member force with proper sign

37


38/65

These are internal member forces due tooriginal loading

38


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These are internal forces due to a vertical unit load at L2

39


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These are internal member forces due to a horizontal unit load at L2

40


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41


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Please find all member forces and specify whether in tension or compression

42


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43


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44


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45


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KAB=4EI = I KBC=4EI = IL 10 L 20

What are the support reactions for the beam shown?

46


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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 fixedends moments

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Fixed End Moments

P

FEM = PL

8

FEM = PL

8

w

FEM= wL2

12

FEM= wL2

12

L/2 L/2

L

P

L

a b

FEM= Pa2b

L2FEM= Pb2a

L2

48


49/65

.

FEM -25 + 25 - 50 + 50

Lock the joint B

49


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Distribution Factor = K_______________

Sum of K for all members at the joint

=

K

KDF 11

=

K

KDF 22

50


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Distribution Factor = K_______________

Sum of K for all members at the joint

Distribution Factor = KBA_

KBA + KBC

=K

KDF 11

=

K

KDF 22

51


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1__

10____ = 2/3 D.F. at B for BA

1__ + 1_

10 20

1__

20____ = 1/3 D.F. at B for BC

1__ + 1_

10 20

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Stiffness K

2/3D. F.

FEM

Balancing

Joint B

-25 + 25 - 50 + 50

1/3

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Stiffness K

2/3D. F.

FEM

Balancing

Joint B

-25 + 25 - 50 + 50

+ 16.67 +8.33

1/3

Joint B Released

54


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Stiffness K

2/3D. F.

FEM

Balancing

Joint B

C.O.M.

-25 + 25 - 50 + 50

+ 16.67 +8.33

+ 8.33

1/3

55


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Stiffness K

2/3D. F.

FEM

Balancing

Joint B

C.O.M.

-25 + 25 - 50 + 50

+16.67 +8.33

+ 8.33 +4.17

1/3

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Stiffness K

2/3D. F.

FEM

Balancing

Joint B

C.O.M.

Final

Moments

-25 + 25 - 50 + 50

+ 16.67 +8.33

+ 8.33 +4.17

- 16.67

1/3

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Stiffness K

2/3D. F.

FEM

Balancing

Joint B

C.O.M.

Final

Moments

-25 + 25 - 50 + 50

+ 16.67 +8.33

+ 8.33 +4.17

- 16.67 + 41.67

1/3

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Stiffness K

2/3D. F.

FEM

Balancing

Joint B

C.O.M.

Final

Moments

-25 + 25 - 50 + 50

+ 16.67 +8.33

+ 8.33 +4.17

- 16.67 + 41.67 - 41.67

1/3

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Stiffness K

2/3D. F.

FEM

Balancing

Joint B

C.O.M.

Final

Moments

-25 + 25 - 50 + 50

+ 16.67 +8.33

+ 8.33 +4.17

- 16.67 + 41.67 - 41.67 + 54.17

1/3

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Stiffness K

2/3D. F.

FEM

Balancing

Joint B

C.O.M.

Final

Moments

-25 + 25 - 50 + 50

+ 16.67 +8.33

+ 8.33 +4.17

---------- ------------ ---------- ----------

- 16.67 + 41.67 - 41.67 + 54.17

1/3

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20 k

-16.67 FT.K 41.67 -41.67 54.2

1.5 K/FT

7.5 k 12.5 k14.4 k 15.6 k

20 k 1.5 K/FT

62


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63

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

R f


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References

Hibbeler, C. R., Structural Analysis, 3rdEdition,

Prentice Hall, 1995.

Chajes, Alexander, Structural Analysis, Prentice

Hall, 1982.

64

Th k Y !


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Thank You!

Any Questions?

Good Luck!

Documents

Session I Structural Analysis