P.E. Civil Exam Review:Structural AnalysisStructural Analysis

J.P. MohsenEmail: jpm@louisville.edujp @

Structures

• Determinate• Determinate

• Indeterminate

STATICALLY DETERMINATESTATICALLY DETERMINATE

STATICALLY INDETERMINATE

Stability and Determinacy of Trusses

400 lb.300 lb.

C DB C D

7.5 ft

A

10 ft 10 ft 10 ft 10 ftE

FGH

R

A

RLRR

2j = m + r Truss is determinate

2j m + r indeterminate

J = number of jointsm= number of membersr = number of reactions

2j m + r Unstable

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 anglesright angles.

400 lb.300 lb.

B C D

7.5 ft

10 ft 10 ft 10 ft 10 ft EFGH

RL RR

Member BH.

400 lb.300 lb.

B C D

10 ft 10 ft 10 ft 10 ftA EFGH

RL RR

Analysis of Member BH.

400 lb300 lb 400 lb.300 lb.

B C D

10 ft 10 ft 10 ft 10 ftA EFGH

RL RR

FBH

00 bhy FF+

Applying Equation of Equilibrium to Joint H

FHGFAH

HH

Member BC.

400 lb.300 lb.

B C D

10 ft 10 ft 10 ft 10 ftA EFGH

RL RR

Analysis of Member BC.

400 lb300 lb 400 lb.300 lb.

B C D

10 ft 10 ft 10 ft 10 ftA EFGH

RL RR = 275 lb.

)20(275

+ 05.7200 BCRG FRM

FBC

C DB

400 lb.)(733

5.7)20(275 ncompressiolbsFBC

E

G FFHG

FBG

10 ft

12.5 ft7.5 ft

10 ft

RR

G F 10 ft10 ft

Member DG.

400 lb.300 lb.

B C D

10 ft 10 ft 10 ft 10 ftA EFGH

RL RR

Analysis of Member DG.

400 lb300 lb 400 lb.300 lb.

B C D

10 ft 10 ft 10 ft 10 ftA EFGH

RL RR

FCD

C D

E

G FFGF

FDG

10 ft

12.5 ft7.5 ft

RR

G F 10 ft

Analysis of Member DG.

400 lb300 lb 400 lb.300 lb.

B C D

10 ft 10 ft 10 ft 10 ftA EFGH

RL RR

0 YF 0 YR DGR lbsDGY 275

3DGY

+

FCD

C D

53

DGDGY tensionlbsDG 458

E

G FFGF

FDG

10 ft

12.5 ft7.5 ft

RR

G F 10 ft

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

120 kN-m20 kN/m

60 kN

120 kN m

A

B

C D

EE

2 m 2 m 2 m 2 m

Draw the Free Body Diagram (FBD)Draw the Free Body Diagram (FBD).(Note: The horizontal force at point B is equal to zero.)

120 kN-m20 kN/m

60 kN

120 kN m

A

FB

C D

FEB FE

2 m 2 m 2 m 2 m

Solve for the reactions at supports B and E

120 kN-m20 kN/m

60 kN

Solve for the reactions at supports B and E.

A

FB = 100 kN

C D

FE = 40 kN

2 m 2 m 2 m 2 m

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

∑FY = 0 → -60 – 80 + FE + FB = 0 → -100 + FB = 0 → FB = 100 kN+

120 kN-m20 kN/m

60 kN

Draw the Shear Diagram

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)

kNmkNm 40202

0 0 V (kN)

-40

120 kN-m20 kN/m

60 kN

Show the change in Shear

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Show the change in Shear at B.

0 0

60

V (kN)

kN100

0 0

-40

V (kN)

120 kN-m20 kN/m

60 kN

Draw the Shear Diagram

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Draw the Shear Diagram for segment BC.

0 0

60

20V (kN)

kNmkNm 40202

0 0

-40

V (kN)

120 kN-m20 kN/m

60 kN

Show the change in Shear

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Show the change in Shear at C.

0 0

60

20V (kN)

kN60

0 0

-40 -40

V (kN)

120 kN-m20 kN/m

60 kN

Draw the Shear Diagram

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Draw the Shear Diagram for segment CE.

0 0

60

20V (kN)

kNmkNm 004

0 0

-40 -40 -40

V (kN)

120 kN-m20 kN/m

60 kN

Show the change in Shear

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Show the change in Shear at E.

0 0

60

20V (kN)

kN40

0 0

-40 -40 -40

V (kN)

120 kN-m20 kN/m

60 kN

Completed Shear Diagram

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Completed Shear Diagram

0 0

60

20V (kN)0 0

-40 -40 -40

V (kN)

120 kN-m20 kN/m

60 kN

Draw the Moment Diagram

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Draw the Moment Diagram for segment AB.

0 0

60

20V (kN)0 0

-40 -40 -40

V (kN)

mkNkNm

40

21402

40

M (kN-m)0 02°

-40

120 kN-m20 kN/m

60 kN

Draw the Moment Diagram

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Draw the Moment Diagram for segment AB.

0 0

60

20V (kN)0 0

-40 -40 -40

V (kN)

mkNkNm

40

21402

M (kN-m)0 0

-40

120 kN-m20 kN/m

60 kN

Draw the Moment Diagram

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Draw the Moment Diagram for segment BC.

0 0

60

20V (kN)0 0

-40 -40 -40

V (kN)

40

mkNkNmkNm

80202

21402

40

M (kN-m)0 0

40

2° 2°

2°

-40

120 kN-m20 kN/m

60 kN

Draw the Moment Diagram

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Draw the Moment Diagram for segment CD.

0 0

60

20V (kN)0 0

-40 -40 -40

V (kN)

40

mkNkNm 80402

40 40

M (kN-m)0 0

40

2° 2°

2°

-40 -40

120 kN-m20 kN/m

60 kN

Show the change in

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Show the change in bending moment at D.

0 0

60

20V (kN)0 0

-40 -40 -40

V (kN)

40

80

mkn 120

40 40

M (kN-m)0 0

40

2° 2°

2°

-40 -40

120 kN-m20 kN/m

60 kN

Draw the Moment Diagram

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Draw the Moment Diagram for segment DE.

0 0

60

20V (kN)0 0

-40 -40 -40

V (kN)

40

80

mknkNm 80402

40 40

M (kN-m)0 0

40

2° 2°

2°

-40 -40

120 kN-m20 kN/m

60 kN

Completed Moment

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Completed Moment Diagram.

0 0 V (kN)

60

200 0 V (kN)

-40 -40 -40

40

80

40 40

M (kN-m)0 0

40

2° 2°

2°

-40 -40

120 kN-m20 kN/m

60 kN

Find the maximum

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

Find the maximum moment.

0 0 V (kN)

60

200 0 V (kN)

-40 -40 -40

40

80

mknM 80max

40 40

M (kN-m)0 0

40

2° 2°

2°

-40 -40

Find the force in the truss members shown.

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.

LuS

EA

A C ti l f h b

S = member force with proper sign

A= Cross-sectional area of each member

L= Length of each member

E= modulus of elasticity of materials

These are internal member forces due to original loadingoriginal loading

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

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

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

What are the support reactions for the beam shown?

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

Moment Distribution

• 1) calculate the fixed end moments• 1) calculate the fixed end moments

• 2) Calculate distribution of moments at the clamped ends of the b b th t ti f th t j i tmembers 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 ) gfixed ends moments

Fixed End Moments

P

L/2 L/2FEM = PL

8FEM = PL

8

w

L/2

w

FEM= wL2FEM= wL2

12

L

1212

P

L

a b

FEM= Pa2bL2FEM= Pb2a

L2

Lock the joint B

.

FEM -25 + 25 - 50 + 50

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

Di t ib ti F t KDistribution Factor = K_______________

Sum of K for all members at the joint

K

KKDF 1

1

K

KKDF 2

2

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

Di t ib ti F t KDistribution Factor = K_______________

Sum of K for all members at the joint

KDistribution Factor = KBA_

KBA + KBC

K

KDF 11

K

KKDF 2

2

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

11__10____ = 2/3 D.F. at B for BA

1__ + 1_ 10 20

1__20____ = 1/3 D.F. at B for BC1__ + 1_ 10 20

KAB=4EI = I KBC=4EI = IStiffness K

L 10 L 20

2/3D. F. 1/3

FEM -25 + 25 - 50 + 50

Balancing Joint B

Joint B Released

KAB=4EI = I KBC=4EI = IStiffness K

L 10 L 20

2/3D. F. 1/3

FEM -25 + 25 - 50 + 50

Balancing Joint B

+ 16.67 +8.33

KAB=4EI = I KBC=4EI = IStiffness K

L 10 L 20

2/3D. F. 1/3

FEM -25 + 25 - 50 + 50

Balancing Joint B

C.O.M.

+ 16.67 +8.33

+ 8.33

KAB=4EI = I KBC=4EI = IStiffness K

L 10 L 20

2/3D. F. 1/3

FEM -25 + 25 - 50 + 50

Balancing Joint B

C.O.M.

+16.67 +8.33

+ 8.33 +4.17

KAB=4EI = I KBC=4EI = IStiffness K

L 10 L 20

2/3D. F. 1/3

FEM -25 + 25 - 50 + 50

Balancing Joint B

C.O.M.

+ 16.67 +8.33

+ 8.33 +4.17

Final Moments - 16.67

KAB=4EI = I KBC=4EI = IStiffness K

L 10 L 20

2/3D. F. 1/3

FEM -25 + 25 - 50 + 50

Balancing Joint B

C.O.M.

+ 16.67 +8.33

+ 8.33 +4.17

Final Moments - 16.67 + 41.67

KAB=4EI = I KBC=4EI = IStiffness K

L 10 L 20

2/3D. F. 1/3

FEM -25 + 25 - 50 + 50

Balancing Joint B

C.O.M.

+ 16.67 +8.33

+ 8.33 +4.17

Final Moments - 16.67 + 41.67 - 41.67

KAB=4EI = I KBC=4EI = IStiffness K

L 10 L 20

2/3D. F. 1/3

FEM -25 + 25 - 50 + 50

Balancing Joint B

C.O.M.

+ 16.67 +8.33

+ 8.33 +4.17

Final Moments - 16.67 + 41.67 - 41.67 + 54.17

KAB=4EI = I KBC=4EI = IStiffness K

L 10 L 20

2/3D. F. 1/3

FEM -25 + 25 - 50 + 50

Balancing Joint B

C.O.M.

+ 16.67 +8.33

+ 8.33 +4.17

Final Moments

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

- 16.67 + 41.67 - 41.67 + 54.17

20 k 1.5 K/FT

-16.67 FT.K 41.67 -41.67 54.2

20 k 1.5 K/FT

7.5 k 12.5 k 14.4 k 15.6 k

References

• Hibbeler C R Structural Analysis 3rdHibbeler, C. R., Structural Analysis, 3Edition, Prentice Hall, 1995.

• Chajes, Alexander, Structural Analysis, P ti H ll 1982Prentice Hall, 1982.

Thank You!

• Any Questions?

• Good Luck!