Home Work #3

Home Work #3Due Date: 11 Mar, 2010

(Turn in your assignment at the mail box of S581 outside the ME general

office)

The solutions must be written on single-side A4 papers only.

HW 3-Problem #1A steel rod 110 ft long hangs inside a tall tower and holds a 200-pound weight at its lower end (see figure).

If the diameter of the circular rod is ¼ inch, calculate the maximum normal stress σmax in the rod, taking into account the weight of the rod itself.

Weight density of steel: 490lb/ft3

The maximum force Fmax in the rod occurs at the upper end

3 2 2

maxmax

2

1 1(490 / )( ( ) )(110 )8 12

18.37200

(200 18.37) 4448.671( )8

o

O

W AL lb ft ft ft

lbW lb

F W W lb psiA A

HW 3-Problem #2A loading crane consisting of a steel girder ABC supported by a cable BD is subjected to a load P. The cable has an effective cross-sectional area A=0.471 in2. The dimensions of the crane are H=9 ft, L1=12 ft, and L2=4 ft.

a.If the load P=9000 lb, what is the average tensile stress in the cable?

b.If the cable stretches by 0.382 in, what is the average strain?

Ray

Rax

TBy

TBx

P

A B C

2

3

0

0

( 1 2) 1

12000

15( ) 200009

20000 424628.50.4710.382 2.122 10(15 12)

By

By

B By

Baverage

average

Ma

L L P L T

T lb

T T lb

T lb psiA in

inL

(a)

(b)

HW 3-Problem #3

The data shown in the accompanying table were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 0.505 in and a gage length of 2 in. At fracture, the elongation between the gage marks was 0.12 in and the minimum diameter was 0.42 in.

Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2 in, and percent reduction in area.

HW 3-Problem #3

Data for the conventional stress-strain curve

Ultimate stress112832.9lb/in2

Yield stress69000lb/in2

Proportinal limit65000lb/in2

0.1%

Percentage of reduction in area = %

As shown in the plot,Proportinal limit≈ 65000 lb/in2

Young’s Modulus =

Yield stress of 0.1% offset ≈ 69000 lb/in2

59911.28139 4992.606 296860.00 9̀5 0.0001

ksi

(Any two point in the proportinal line is OK)

Ultimate stress ≈ 112832.9 lb/in2 Percentage of elongation =0.12/2 = 6%

2 2

2

1 (0.525 0.42 )4 30.81 0.5254

HW 3-Problem #4

A bar of monel metal as in the figure (length L=8 in, diameter d=0.25 in) is loaded axially by a tensile force P=1500 lb. Determine the increase in length of the bar and the percent decrease in this cross-sectional area.

Monel metal material property:

Modulus of elasticity E=25000 ksi

Poisson’s Ratio ν=0.32

3 32

4

4 5

2 5 2

2

1500 825000 10 9.7785 10(0.125)

(0.32) 3.9104 10

3.9104 10 0.25 9.776 10

1/ 4 (0.25 (0.25 9.776 10 ) ) 0.0781/ 4 (0.25)

X

XX

d X

PLEA

d

AA

Percentage of reduction in area

HW 3-Problem #5

The connection shown in the figure consists of five steel plates, each 3/16 in. thick, jointed by a single ¼-in. diameter bolt. The total load transferred between the plates is 1200 lb, distributed among the plates as shown.

a.Calculate the largest shear stress in the bolt, disregarding friction between the plates.

b.Calculate the largest bearing stress acting against the bolt.

ABC

D

360

480

360

600

600

A360360

600B

360 A

240

BC

360

480600A

240

D

600

600

ABC

360

480

360

①②

③

④

As shown in the figure above, the shear force of A, D= 360 lbThe shear force of B,C= 240lbThe largest shear force is:

3 2maxmax

2

4 2maxmax

360 7.334 101( )4 4

600 1.28 101 34 16

V lbinA

Flbin

A

The largest bearing force is:

HW 3-Problem #6An elastomeric bearing pad consisting of two steel plates bonded to a chloroprene elastomer (an artificial rubber) is subjected to a shear force V during a static loading test. The pad has dimensions a=150 mm and b=250mm, and the elastomer has thickness t=50 mm. When the force V equals 12 kN, the top plate is found to have displaced laterally 8.0mm with respect to the bottom plate.

What is the shear modulus of elasticity G of the chloroprene.

Vaver ab

e

aver VabGGe

Shear strain :

6 23 3

0.16

120000 2 10150 10 250 10 0.16

Ve ab

dt

G Nm

A steel bar AD has a cross-sectional area of 0.4 in2 and is loaded by forces P1=2700 lb, P2=1800 lb, and P3=1300 lb. The lengths of the segments of the bar are a=60 in, b=24 in, and c=36 in.

a.Assuming that the modulus of elasticity E=30*106 psi, calculate the change in length δ of the bar. Does the bar elongate or shorten?

b.By what amount P should the load P3 be increased so that the bar does not change in length when the three loads are applied?

HW 3-Problem #7

C D B C BA

F3 P3 F2 P2-P3 F1 P1+P2-P3

P3 has to be increased by 2610-1310=1310 lb

336

32 36

1 2 36

3

1300 36 3.9 1030 10 0.4

( ) (1800 1300) 24 1 1030 10 0.4

( ) (1800 1300) 60 0.01630 10 0.4

0.0131 ( )) 036 (180

CDCD

BCBC

ABAB

total

CD BC AB

P Lin

E AP P L inE A

P P P L inE Ain elongate

bP

3 3

3

0 )24 (2700 1800 )60 02610

P PP lb

A bar ABC of length L consists of two parts of equal lengths but different diameters. Segment AB has diameter d1=100 mm, and segment BC has diameter d2=60 mm. Both segments have length L/2=0.6 m. A longitudinal hole of diameter d is drilled through segment AB for one-half of its length (distance L/4=0.3 m). The bar is made of plastic having modulus of elasticity E=4.0 Gpa. Compressive loads P=110 kN act at the ends of the bar.

If the shortening of the bar is limited to 8.0mm, what is the maximum allowable diameter dmax of the hole?

HW 3-Problem #8

A’

' '

3 3 3

9 3 2 9 3 29 3 2 2

3 2 2 3

0.008

110 10 0.6 110 10 0.3 110 10 0.34 10 (3 10 ) 4 10 (50 10 ) 4 10 ((50 10 ) ( ) )

2

(50 10 ) ( ) 2.357 102

23.87

AA A B BC m

d

d

d mm