P.E. Civil Exam Review: Strength of Materials

J.P. Mohsen

Email:jpm@louisville.edu

Strength of Materials

Stress vs. Strain

Stress = Force / Area

Strain = Change in Length / Original Length

AP

=

LL

=

2

Stress Strain Diagram Typical for Ductile Materials

Stre

ss

Strain

3

Stress Strain Diagram

Stre

ss

Strain

Proportional limit

4

Stress Strain Diagram

Stre

ss

Strain

Proportional limit

Elastic limit

5

Stress Strain Diagram

Stre

ss

Strain

Proportional limit

Yield point

Elastic limit

6

Stress Strain Diagram

Stre

ss

Strain

7

Stress Strain Diagram

Stre

ss

Strain

Proportional limit

Yield point

Elastic limit

Ultimate strength

8

Stress Strain Diagram

Stre

ss

Strain

Proportional limit

Yield point

Elastic limit

Ultimate strength

Rupture

9

Stress- Strain Relationship

Stre

ss

Strain

Proportional limit

Yield point

Elastic limit

Ultimate strength

Rupture

True rupture

10

Stress Strain Relationship Typical for Ductile Materials

Stre

ss

Strain

Hookes Law E=

ExyportionlineartheofSlope =

=

=

11

Areas under stress-strain curves: (a) modulus of resilience, (b) toughness, and (c) high-strength and high-toughness materials.

Stre

ss

(a)

Strain

Modulus of resilience

Stre

ss

(b)

Strain

Stre

ss

(c)

Strain

High strength

Toughness

Fracture

High toughness

12

Methods for estimating yield stress: (a) offset method and (b) extension method

Stre

ss

(a)

Strain, %

Proportional limit

0.2% offset yield strength

Elastic limit

0.2%

Stre

ss

(b)

Strain, %

0.5% extension yield strength

0.5%

13

Problem 1)

For the beam shown, compute the maximum bending stress in compression and maximum bending stress in tension. Also compute the shear stress at the neutral axis as well as at a section 6 in. from the bottom of the cross-section. Calculate the shear stresses at a point along the beam where the maximum shear force occurs.

1500 lb

1 in

8 in

4 in

3.5 in 1 in

INA = 97.0 in4

NA

1500 lb 4500 lb

4 ft 4 ft 6 ft 6 ft

14

Problem 1 (continued)

For the beam shown, compute the shear stress at the NA and at 6 in. from the bottom of the cross section at the point where the maximum shear force occurs.

1500 lb

1 in

8 in

4 in

3.5 in 1 in

INA = 97.0 in4

NA

1500 lb 4500 lb

4 ft 4 ft 6 ft 6 ft

-1500 lb

1500 lb

3750 lb 3750 lb

V (Ib)

2250

- 2250 - 2250

1500

15

Problem 1 (continued) - For the beam shown, compute the max. bending stress in tension and in compression.

1500 lb

1 in

8 in

4 in

3.5 in 1 in

INA = 97.0 in4

NA

1500 lb 4500 lb

4 ft 4 ft 6 ft 6 ft

-6000 -6000

+7500

M (ft-lb)

------ ---i---

16

Bending Stress in beams :

In the negative moment region In the positive moment region

inCc 5.3= inCc 5.5=

inCt 5.5= inCt 5.3=

psit 324797)5.3)(12)(7500(==psit 408297

)5.5)(12)(6000(==

psic 510397)5.5)(12)(7500(==psic 259897

)5.3)(12)(6000(==

ICM c

c =

ICM

=

ICM t

t =

17

CISModulusSection =

ICM

= =

SM

36.175.5

97 inCIS ===

18

Problem 1 (continued)

For the beam shown, compute the shear stress at the NA and at 6 in. from the bottom of the cross section at the point where the maximum shear force occurs.

1500 lb

1 in

8 in

4 in

3.5 in 1 in

INA = 97.0 in4

NA

1500 lb 4500 lb

4 ft 4 ft 6 ft 6 ft

-1500 lb

1500 lb

3750 lb 3750 lb

V (Ib)

2250

- 2250 - 2250

1500

19

Shear stress in bending bIQV

=3

sec6 12]4[)1()3( inQ tioncrossofbottomfromin ==

312.15]75.2[)1()5.5( inQ axisneturalat ==

psitioncrossofbottomfromin 3.278)1)(97()12()2250(

sec6 ==

psiaxisneturalat 7.350)1)(97()12.15()2250(==

20

What is the value of Q if shear stress is being calculated at:

1 in

8 in

4 in

3.5 in 1 in

INA = 97.0 in4

NA NA

3 in 4.0 in

NA 2.75 in 2.5 in

5.5 in

312]4[)1()3( inQ ==

Neutral axis a section 6 in. above the base

312.15]75.2[)1()5.5( inQ ==

21

Problem 2.

The vertical shear force acting on the I-section shown is 100 kN. Compute the maximum shear stress in bending acting on the this beam section.

20 mm

20 mm

20 mm

160 mm

120 mm

22

20 mm

20 mm

20 mm

160 mm

120 mm

NA

TheoremAxisParallelAdbhI 23

12+=

2

3

1

423

1 000,520,19]90)[20)(120(12)20)(120( mmI NAabout =+=

43

2 667,826,612)160)(20( mmI NAabout ==

423

3 000,520,19]90)[20)(120(12)20)(120( mmI NAabout =+=

4666,866,45 mmI NAabouttotal =20 mm

23

Problem 2 (continued) Maximum shearing stress occurs at the neutral axis.

3000,280]90)[20)(120(]40)[80)(20( mmNAatQ =+=

NAatmmb 20=

KNV 100=

4666,866,45 mmI NAabouttotal =

2/03.0)20()45866666(

)280000()100( mmKN==

bIQV

=

24

Problem 3.

The simply supported beam has the T-shaped cross section shown. Determine the values and locations of the maximum tensile and compressive bending stresses.

400 lb/ft 1000 lb

A

B

D

10 ft 4 ft

y

x

RA = 1600 lb RB = 3400 lb

6 in

8 in

0.8 in

0.8 in

25

Problem 3 (continued) Shear Diagram

400 lb/ft 1000 lb

A

B

D 10 ft 4 ft

y

x

RA = 1600 lb RB = 3400 lb

1600 1000

-2400

4 ft V (lb)

26

Moment Diagram 400 lb/ft

1000 lb

A

B

D 10 ft 4 ft x

RA = 1600 lb RB = 3400 lb

1600 1000

-2400

4 ft V (lb)

M (lb-ft) 4 ft

-4000

3200

2

27

Find y-bar.

21

2211

AAyAyAbary

++

=

( ) ( )8.44.6

4.88.444.6++

=

in886.5=

6 in

0.8 in

8 in

A2

A1 y1

y2

y-bar

N.A.

C

0.8 in

28

Compute Moment of Inertia, I.

6 in

0.8 in

8 in

A2

A1 y1

y2

y-bar

N.A.

C

0.8 in

( ) ( ) ( ) ( )

++

+= 2

32

3

886.54.88.412

8.06886.544.612

88.0I

449.87 inI =29

Stresses at x = 4 ft

8.8 in

y1 y-bar

N.A. C

cbot = 5.886 in.

ctop = 2.914 in.

4 ft

( )( ) psiI

Mcbotbot 258049.87

886.5123200=

==

3200

-4000

( )( ) psiI

Mctoptop 127949.87

914.2123200=

==

---i---

Compression on top

Tension in bottom 30

Stresses at x = 10 ft

8.8 in

y1 y-bar

N.A. C

cbot = 5.886 in.

ctop = 2.914 in.

10 ft

3200

-4000

( )( ) psiI

Mcbotbot 323049.87

886.5124000=

==

( )( ) psiI

Mctoptop 159949.87

914.2124000=

==

------

Tension on top

Compression in bottom 31

Problem 3 results.

Identify maximum tensile and compressive stresses in the beam.

( ) psiT 2580max =

( ) psiC 3230max =

(bottom of the section at x = 4 ft)

(bottom of the section at x = 10 ft)

32

Original State of Stress

y

x

a

a

x

y

xy

xy

x

y

33

x

y

R

x

y

xy

xy

y

x x

y

xyConstruction of Mohrs circle from given stress components.

34

Convention for plotting shear stress on Mohrs circle.

Shear plotted up Shear plotted down

35

y

x

y

yx

xy

y

x x

y

y

yx y

x

x

2

1

2 1

1

y

1

y

a

2

1 2

x

xy

b

x

x

36

Problem 4 )

For the state of stress shown, determine (a) the principal stresses; and (b) the maximum in-plane shear stress. Show the results on properly oriented elements.

4 ksi

8 ksi

6 ksi

y

x

37

(y-axis)

(x-axis)

(4,-6)

(-8, 6)

4 ksi

8 ksi

6 ksi

y

x

y

x

Problem 4 (continued)

38

R

4 ksi

8 ksi

6 ksi

y

x

(-8, 6)

(4,-6)

(y-axis)

(x-axis)