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

σ

εAP

=σ

LL∆

=ε

2

Stress – Strain Diagram Typical for Ductile Materials

Stre

ss

Strain

∆σ

∆ε

3

Stress – Strain Diagram

Stre

ss

Strain

Proportional limit

∆σ

∆ε

4

Stress – Strain Diagram

Stre

ss

Strain

Proportional limit

Elastic limit

∆σ

∆ε

5

Stress – Strain Diagram

Stre

ss

Strain

Proportional limit

Yield point

Elastic limit

∆σ

∆ε

6

Stress – Strain Diagram

Stre

ss

Strain

∆σ

∆ε

7

Stress – Strain Diagram

Stre

ss

Strain

Proportional limit

Yield point

Elastic limit

Ultimate strength

∆σ

∆ε

8

Stress Strain Diagram

Stre

ss

Strain

Proportional limit

Yield point

Elastic limit

Ultimate strength

Rupture

∆σ

∆ε

9

Stress- Strain Relationship

Stre

ss

Strain

Proportional limit

Yield point

Elastic limit

Ultimate strength

Rupture

True rupture

∆σ

∆ε

10

Stress – Strain Relationship Typical for Ductile Materials

Stre

ss

Strain

∆σ

∆ε Hooke’s Law εσ E=

ExyportionlineartheofSlope =

∆∆

=∆∆

=εσ

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(==σpsit 4082

97)5.5)(12)(6000(==σ

psic 510397

)5.5)(12)(7500(==σpsic 2598

97)5.3)(12)(6000(==σ

ICM c

c =σ

ICM

=σ

ICM t

t =σ

17

CISModulusSection =

ICM

=σ =

SMσ

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

=τ3

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

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

psitioncrossofbottomfromin 3.278)1)(97(

)12()2250(sec6 ==−τ

psiaxisneturalat 7.350)1)(97(

)12.15()2250(==τ

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==τ

bIQV

=τ

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 2580

49.87886.5123200

=×

==σ

3200

-4000

( )( ) psiI

Mctoptop 1279

49.87914.2123200

=×

==σ

---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 3230

49.87886.5124000

=×

==σ

( )( ) psiI

Mctoptop 1599

49.87914.2124000

=×

==σ

------

Tension on top

Compression in bottom 31

Problem 3 results.

Identify maximum tensile and compressive stresses in the beam.

( ) psiT 2580max =σ

( ) psiC 3230max =σ

(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

θxσ

yσ

xyτ

xyτ

xσ

yσ

33

x

y

R

σ

τ

xσ

yσ

xyτ

xyτ

σ

y

x xσ

yσ

xyτConstruction of Mohr’s circle from given stress components.

34

Convention for plotting shear stress on Mohr’s circle.

Shear plotted up Shear plotted down

35

y’

σ

τ

xσ

yσ

yx ′′τ

σ

xyτ

yσ

xσ x

y

y′σ

Θ

yx ′′τy’

x’

x′σ

2σ

Θ1

2 1

1σ

y′σ

1σ

y

a

2

1 2σ

x′σ

xyτ

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

τ (y-axis)

σ (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)

τ (y-axis)

σ (x-axis)

y

x

Problem 4 (continued)

39

R

4 ksi

8 ksi

6 ksi

y

x

(-8, 6)

(4,-6)

τ (y-axis)

σ (x-axis) -2

y

x

Problem 4 (continued)

40

R

2Θ1

-2

4 ksi

8 ksi

6 ksi

y

x

(-8, 6)

(4,-6)

τ (y-axis)

σ (x-axis)

y

x

Problem 4 (continued)

41

τmax

y

1 2

R

2Θ1

-2

4 ksi

8 ksi

6 ksi

y

x

(-8, 6)

x (4,-6)

τ (y-axis)

σ (x-axis)

Problem 4 (continued)

42

(a)

τ

σ

τmax

x

y

1 2

R

4 2Θ1

6 -8 -2

6

(b)

22.5°

10.49

6.49

22.5°

8.49 2

2

ksiR 7266 22 =+=( ) °=Θ⇒°==Θ − 5.22456/6tan2 1

11

ksi

ksi

49.10722

49.6722

2

1

−=−−=

=+−=

σ

σ

ksiR 49.872max ===τ

Problem 4 (continued)

43

Problem 5)

The radius of the 15-in. long bar in Fig. (a) is 3/8 in. Determine the maximum normal stress in the bar at (1) point A; and (2) point B.

z x

y

30 lb

540 lb-in

A B

z x

y

B A

M

T

V

(a) (b) 44

Problem 5 (continued)

The internal force system acting on the cross section at the base of the rod is shown in Fig. (b). It consists of the torque T = 540 lb-in, the bending moment M = 15P = 15(30) = 450 lb-in (acting about the x-axis), and the transverse shear force V = P = 30 lb.

The cross-sectional properties of the bar are:

( ) 4344

10532.154

8/34

inrI −×===ππ

( ) 433 1006.3110532.1522 inIJ −− ×=×==

2

4rJInertiaofmomentPolar π=

45

z

x

y

ksi520.6=τksi865.10=σ

τ (ksi)

σ (ksi)

x

y

6.520

10.865

5.433

Problem 5 (continued) PART 1

State of stress at point A with the corresponding Mohr’s circle.

46

Problem 5 (continued)

The bending stress is calculated:

z

x

y

ksi520.6=τksi865.10=σ

τ (ksi)

x

y

6.520

10.865

5.433

( )( ) ksipsiI

Mr 865.108651010532.15

8/34503 ==

×== −σ

47

z

x

y

ksi520.6=τksi865.10=σ σ (ksi)

τ (ksi)

x

y

6.520

10.865

5.433

Problem 5 (continued)

And the shear stress is calculated:

( ) ksipsiJ

TrT 520.65206

1006.318/3540

3 ==×

== −τ

The shear stress due to the transverse shear force V is zero at A.

48

z

x

y

ksi520.6=τksi865.10=σ σ (ksi)

τ (ksi)

x

y

6.520

10.865

5.433

Problem 5 (continued)

The maximum normal stress at point A is:

ksi92.13487.8433.5max =+=σ

49

Problem 5 (continued) PART 2

State of stress at point B with the corresponding Mohr’s circle.

z

x

y

ksi611.6=τ

B

τ (ksi)

σ (ksi)

z

y

6.611

σmax

50

Problem 5 (continued)

The shear stress due to torque is τT = 6.520 ksi, as before. But because the point lies on the neutral axis, the bending stress is zero. There is, however, an additional shear stress caused by the transverse shear force V. The magnitude of this shear stress is τV = VQ/(Ib), where b = 2r = ¾ in. and Q is the first moment of half the cross-sectional area about the neutral axis. Referring to the figure below, Q is calculated:

==

ππ

34

2''

2 rrZAQ

x

z

C

π34' rZ =( )

38/32

32 33

=

=

r

331016.35 in−×=

51

Problem 5 (continued)

Therefore:

z

x

y ksi611.6=τ

B

τ (ksi)

σ (ksi)

z

y

6.611

σmax

( )( )( ) ksipsi

IbVQ

V 091.05.904/310532.15

1016.35303

3

==×

×== −

−

τ

52

Problem 5 (continued)

Since τT and τV act on the same planes they can be added. So the total shear

stress is:

z

x

y ksi611.6=τ

B

τ (ksi)

σ (ksi)

z

y

6.611

σmax

ksiVT 611.6091.0520.6 =+=+= τττ

Vτ

53

Problem 5 (continued)

The Mohr’s circle for this state of pure shear yields for the maximum normal stress at B:

z

x

y ksi611.6=τ

B

τ (ksi)

σ (ksi)

z

y

6.611

σmax

ksi61.6max =σ

54

Poisson’s ratio

If a structural member is subjected to axial tension, the material elongates and there is a reduction in its cross-section. When the member is under compression, the opposite happens. The ratio of the lateral strain to axial strain is Poisson’s ratio.

axial

lateral

εεν −

=

Most solids have a Poisson’s ratio between 0.10 and 0.45 55

General State of Stress

z

x

y

1σ

2σ

3σ

τ

σ

1σ

3σ

2σ

−−−=

2,

2,

2max 133221 σσσσσσ

τ abs

56

Problem 6)

For the state of stress shown, determine the maximum in-plane shear stress and the absolute maximum shear stress.

50 ksi

20 ksi

y

x

57

Problem 6 (continued)

The given stresses are:

50 ksi

20 ksi

y

x

501 == xσσ

202 == yσσ

58

Problem 6 (continued)

The maximum in-plane shear stress τmax is equal to the radius of the circle that represents transformation in the xy-plane.

τ (ksi)

σ (ksi)

501 =σ202 =σ

10

25

15

Transformation in xy-plane

Transformation in zx-plane

Transformation in yz-plane

ksi15max =τ

59

Problem 6 (continued)

The absolute maximum shear stress equals the radius of the largest circle, which represents the transformation in the zx-plane.

τ (ksi)

σ (ksi)

501 =σ202 =σ

10

25

15

Transformation in xy-plane

Transformation in zx-plane

Transformation in yz-plane

ksiabs 25=τ

60

References:

Pytel and Kiusalaas, “Mechanics of Materials,” Thomson, 2003. McCormac and Nelson, “Design of Reinforced Concrete,” Wiley, 2003.

61

Thank You

and

GOOD LUCK!!!

62