Assignment #3 Solution - Spring 2014

8/11/2019 Assignment #3 Solution - Spring 2014

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MAE 311: Machines and Mechanisms I Spring 2014

Assignment #3 Due Date: 3/7/2014

Copyright 2014 – Phillip M. Cormier and Jobaidur Khan

Problem 1 (25 Points)

The steel eyebolt shown in Figure 1 is loaded with a force of 75 pounds. The bolt is formed from round

stock with a diameter of 0.25 inches, and has in inner bend radius of 0.5 inches in the eye and at the end

of the shank. Estimate the stress at the inner and outer surfaces of section B, where = 30°.

Figure 1: Problem 1 Diagram

Sol. d = 0.25 in, r o = 0.75 on, r

i = 0.5 in,

From table 3-4, for circular cross-section, we get, R = 0.125 in, r c = 0.5 + 0.125 = 0.625 in

in618686.0

125.0625.0625.02

125.0

r r r 2

r r

22

2

22

cc

2

n

in006314.0618686.0625.0r r e nc

in118686.05.0618686.0r r c ini

in131314.0618686.075.0r r c ioo

Cross-section area,2

2

in049087.04

dA

The angle of the line of radius centers is,

30

2

1sin

R dR

2dR sin 11

Moment, in.lbf 44.2330Sin225.05.075Sin2dR FM

kpsi7.18 psi716,185.0006314.0049087.0

118686.044.23

049087.0

30Sin75

Aer

Mc

A

FSin

i

ii

psik 5.12 psi478,1275.0006314.0049087.0

131314.044.23

049087.0

30Sin75

Aer

Mc

A

FSin

o

oo






Problem 2 (25 Points)

The latch spring shown in Figure 2 supports a load of 3 pounds. Given the noted dimensions, determine

the following:

a) Determine the stresses at the top and bottom of the beam, just before the bend. b) Using curved beam theory, determine the inner and outer stresses at the bend.

c) Compare the stresses between the curved portion and bent portion to determine an effectivestress concentration factor.


Sol. (a)

kpsi02.8 psi8021

12in1094.0in75.0

in1094.05.0in4lb3

I

Mc3

(b) From table 3-4, for rectangular cross-section,

we get, r i = 0.125 in, r o = r i + h = 0.125 + 0.1094 = 0.2344 in

in174006.0

125.02344.0ln

1094.0

r r ln

hr

io

n

in005694.0174006.01797.0r r e nc

in049006.0125.0174006.0r r c ini

in060394.0174006.02344.0r r c ioo

Cross-section area,2in08205.01094.075.0 bhA

M = – 3 lb

4 in = – 12 lb.in

The negative sign is due to convention, now the stresses,

kpsi1.10 psi070,10125.0005694.008205.0

049006.012

Aer

Mc

i

ii

kpsi62.6 psi618,62344.0005694.008205.0

060394.012

Aer

Mc

o

oo






(c) Stress concentrations,

26.102.8

1.10K i

i

825.0

02.8

62.6K o

o

Problem 3 (50 Points Total)

Compute the factor of safety for points A and B using both the distortion energy theory and max shear

stress theory. The bar is made from AISI 1006 cold-drawn steel, and is loaded with force F = 550 N, P

= 4000 N, and T = 25 N-m. Determine the required diameter change to bring the most restrictive caseup to a factor of safety of 3.


Sol. Using Table A-20, yp = 280 MPaLet’s analyze the point B,

x from bending

xz

y

T

x from P






For point B the bending stress from F will cause a normal stress in the x direction,

MPa Pam

m N

d

M

I

c M F F bend x 16610166

015.0

1.05503232 6

33

For point B the axial stress (tension) from P will cause a normal stress in the x direction,

MPa Pa

m

N

d

P

A

P axial 6.22106.22

015.0

400044 6

22

So, total axial stress at B, x = x,bend + axial = 188.6 MPa

Shear stress for torsion,

MPa7.37m015.0

m. N2516

d

T16

J

Tr 33T

So, total shear stress, MPa7.37Txy

So, according to DE, 4.17.3736.188280

32222 n

nn xy x

yp

So, according to MSST, 38.17.3746.188280

42222 n

nn xy x

yp

Let’s analyze the point A,

For point A the axial stress (tension) from P will cause a normal stress in the x direction,

MPa Pa

m

N

d

P

A

P axial 6.22106.22

015.0

400044 6

22

x from P

xz

y

T

F






Shear stress for torsion,

MPa7.37m015.0

m. N2516

d

T16

J

Tr 33T

Shear stress from F,

MPam

N

d

F

A

F F 15.4

015.03

55016

3

16

3

422

So, total shear stress, MPa F T xz 55.3315.47.37

So, according to DE, 27.455.3536.22280

32222 n

nn xz x

yp

So, according to MSST, 75.355.3546.22280

42222 n

nn xz x

yp

This is obvious from the previous calculation that MSST in point B gives most restrictive case. So, the

design calculation becomes,

333

FF bendx

d

2.560

d

m1.0 N55032

d

M32

I

cM

For point B the axial stress (tension) from P will cause a normal stress in the x direction,

222

5093400044

d d

N

d

P

A

P axial

So, total axial stress at B, 23 50932.560d d

axial bend x x

Shear stress for torsion,333T

d

3.127

d

m. N2516

d

T16

J

Tr

So, total shear stress,3Txy

d

3.127

So, according to MSST,

mmmd

d d

d d d n xy x

yp

2002.0

03.127450932.560910280

3.1274

50932.560

3

10280 4

22236

2

3

2

23

622

The diameter needs to be increased by 33%.

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Assignment #3 Solution - Spring 2014