Download Solution Manual for Mechanics of Materials 3rd Edition by Philpot PDF

TitleSolution Manual for Mechanics of Materials 3rd Edition by Philpot
File Size1.7 MB
Total Pages53
Document Text Contents
Page 1

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P1.1 A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as

a compression member. If the axial normal stress in the member must be limited to 200 MPa,

determine the maximum load P that the member can support.





Solution

The cross-sectional area of the stainless steel tube is


2 2 2 2 2

( ) [(60 mm) (50 mm) ] 863.938 mm
4 4

A D d

The normal stress in the tube can be expressed as



The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable

normal stress, rearrange this expression to solve for the maximum load P

Ans.

Page 2

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P1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip

load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness

required for the tube.





Solution

From the definition of normal stress, solve for the minimum area required to support a 27-kip load

without exceeding a stress of 18 ksi

2

min

27 kips
1.500 in.

18 ksi

P P
A

A


The cross-sectional area of the aluminum tube is given by



Set this expression equal to the minimum area and solve for the maximum inside diameter d





The outside diameter D, the inside diameter d, and the wall thickness t are related by



Therefore, the minimum wall thickness required for the aluminum tube is

Ans.

Page 26

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P1.19 The five-bolt connection shown in Figure P1.19 must

support an applied load of P = 265 kN. If the average shear stress

in the bolts must be limited to 120 MPa, determine the minimum

bolt diameter that may be used for this connection.


FIGURE P1.19

Solution

There are five bolts, and it is assumed that each bolt supports an equal portion of the external load P.

Therefore, the shear force carried by each bolt is


265 kN

53 kN 53,000 N
5 bolts

V

Since the average shear stress must be limited to 120 MPa, each bolt must provide a shear area of at

least:



Each bolt in this connection acts in double shear; therefore, two cross-sectional bolt surfaces are

available to transmit shear stress in each bolt.



The minimum bolt diameter must be

Ans.

Page 27

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P1.20 A coupling is used to connect a 2 in. diameter

plastic pipe (1) to a 1.5 in. diameter pipe (2), as

shown in Figure P1.20. If the average shear stress in

the adhesive must be limited to 400 psi, determine

the minimum lengths L1 and L2 required for the joint

if the applied load P is 5,000 lb.


FIGURE P1.24

Solution

To resist a shear force of 5,000 lb, the area of adhesive required on each pipe is


2

adhesive

5,000 lb
12.5 in.

400 psi
V

V
A



Consider the coupling on pipe (1). The adhesive is applied to the circumference of the pipe, and the

circumference C1 of pipe (1) is



The minimum length L1 is therefore

Ans.



Consider the coupling on pipe (2). The circumference C2 of pipe (2) is



The minimum length L2 is therefore

Ans.

Page 52

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P1.44 The rectangular bar has a width of w =

3.00 in. and a thickness of t = 2.00 in. The

normal stress on plane AB of the rectangular

block shown in Figure P1.44/45 is 6 ksi (C)

when the load P is applied. Determine:

(a) the magnitude of load P.

(b) the shear stress on plane AB.

(c) the maximum normal and shear stresses in

the block at any possible orientation.


FIGURE P1.44/45

Solution

The general equation for normal stress on an inclined plane in terms of the angle  is

(1 cos2 )
2

n

P

A
   (a)

and the angle  for inclined plane AB is


3

tan 0.75 36.8699
4

     

The cross-sectional area of the rectangular bar is A = (3.00 in.)(2.00 in.) = 6.00 in.
2
.



(a) Since the normal stress on plane AB is given as 6 ksi, the magnitude of load P can be calculated from

Eq. (a):


2

2 2(6.0 in. )(6 ksi)
56.25 kips

1 cos2 1 co
56.3 kip

s2(36.8699 )
sn

A
P




   

  
Ans.



(b) The general equation for shear stress on an inclined plane in terms of the angle  is

sin 2
2

nt

P

A
 

therefore, the shear stress on plane AB is


2

56.25 kips
sin 2(36.8699 )

2(6.00 in. )
4.50 ksi

nt
    Ans.



(c) The maximum normal stress at any possible orientation is


max 2

56.25 kips
9.3750 ksi

6.00 i
9.38 k

n
si

.

P

A
     Ans.

and the maximum shear stress at any possible orientation in the block is


max 2

56.25 kips
4.6875 ksi

2 2(6.00
4.69 ksi

in. )

P

A
     Ans.

Page 53

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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that

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P1.45 The rectangular bar has a width of w = 100

mm and a thickness of t = 75 mm. The shear stress

on plane AB of the rectangular block shown in

Figure P1.44/45 is 12 MPa when the load P is

applied. Determine:

(a) the magnitude of load P.

(b) the normal stress on plane AB.

(c) the maximum normal and shear stresses in the

block at any possible orientation.


FIGURE P1.44/45

Solution

The general equation for shear stress on an inclined plane in terms of the angle  is

sin 2
2

nt

P

A
  (a)

and the angle  for inclined plane AB is


3

tan 0.75 36.8699
4

     

The cross-sectional area of the rectangular bar is A = (100 mm)(75 mm) = 7,500 mm
2
.



(a) Since the shear stress on plane AB is given as 12 MPa, the magnitude of load P can be calculated

from Eq. (a):


2 2

2 2(7,500 mm )(12 N/mm )
187,500 N

sin 2 sin 2(36.869
187.5 k

9
N

)

nt
A

P



   


Ans.



(b) The general equation for normal stress on an inclined plane in terms of the angle  is

(1 cos2 )
2

n

P

A
  

therefore, the normal stress on plane AB is


2

187,500 N
(1 cos2(36.8699 ))

2(7,500 mm
16.00 MPa

)
n

     Ans.



(c) The maximum normal stress at any possible orientation is


max 2

187,500 N

7,500 mm
25.0 MPa

P

A
    Ans.

and the maximum shear stress at any possible orientation in the block is

Ans.

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