Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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 P A 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 2 2 max allow (200 N/mm )(863.938 mm ) 172,788 172.8 k N N P A Ans.
53
Embed
Solution Manual for Mechanics of Materials 3rd Edition by Philpot
Solution Manual for Mechanics of Materials 3rd Edition by Philpot
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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 mm4 4
A D d
The normal stress in the tube can be expressed as
P
A
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
2 2
max allow (200 N/mm )(863.938 mm ) 172,788 172.8 kN NP A Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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 kips1.500 in.
18 ksi
P PA
A
The cross-sectional area of the aluminum tube is given by
2 2( )4
A D d
Set this expression equal to the minimum area and solve for the maximum inside diameter d
2 2 2
2 2 2
2 2 2
max
[(2.50 in.) ] 1.500 in.4
4(2.50 in.) (1.500 in. )
4(2.50 in.) (1.500 in. )
2.08330 in.
d
d
d
d
The outside diameter D, the inside diameter d, and the wall thickness t are related by
2D d t
Therefore, the minimum wall thickness required for the aluminum tube is
min
2.50 in. 2.08330 in.0.20835 in. 0.208 in.
2 2
D dt
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.3 Two solid cylindrical rods (1) and (2)
are joined together at flange B and loaded, as
shown in Figure P1.3/4. If the normal stress
in each rod must be limited to 40 ksi,
determine the minimum diameter required
for each rod.
FIGURE P1.3/4
Solution
Cut a FBD through rod (1). The FBD should include the free end of the rod at A.
As a matter of course, we will assume that the internal force in rod (1) is tension
(even though it obviously will be in compression). From equilibrium,
1
1
15 kips 0
15 kips 15 kips (C)
yF F
F
Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again,
we will assume that the internal force in rod (2) is tension. Equilibrium of this
FBD reveals the internal force in rod (2):
2
2
30 kips 30 kips 15 kips 0
75 kips 75 kips (C)
yF F
F
Notice that rods (1) and (2) are in compression. In this situation, we are
concerned only with the stress magnitude; therefore, we will use the force
magnitudes to determine the minimum required cross-sectional areas. If
the normal stress in rod (1) must be limited to 40 ksi, then the minimum
cross-sectional area that can be used for rod (1) is
211,min
15 kips0.375 in.
40 ksi
FA
The minimum rod diameter is therefore
2 2
1,min 1 10.375 in. 0.6909 0.691 9 i4
inn. .A d d
Ans.
Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of
222,min
75 kips1.875 in.
40 ksi
FA
The minimum diameter for rod (2) is therefore
2 2
2,min 2 21.875 in. 1.54509 1.545 in.7 in.4
A d d
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.4 Two solid cylindrical rods (1) and (2) are
joined together at flange B and loaded, as shown in
Figure P1.3/4. The diameter of rod (1) is 1.75 in.
and the diameter of rod (2) is 2.50 in. Determine the
normal stresses in rods (1) and (2).
FIGURE P1.3/4
Solution
Cut a FBD through rod (1). The FBD should include the free end of the rod at A. We
will assume that the internal force in rod (1) is tension (even though it obviously will
be in compression). From equilibrium,
1
1
15 kips 0
15 kips 15 kips (C)
yF F
F
Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we
will assume that the internal force in rod (2) is tension. Equilibrium of this FBD
reveals the internal force in rod (2):
2
2
30 kips 30 kips 15 kips 0
75 kips 75 kips (C)
yF F
F
From the given diameter of rod (1), the cross-sectional area of rod (1) is
2 2
1 (1.75 in.) 2.4053 in.4
A
and thus, the normal stress in rod (1) is
11 2
1
15 kips6.23627 ksi
2.4053 in6.24 ksi )
.(C
F
A
Ans.
From the given diameter of rod (2), the cross-sectional area of rod (2) is
2 2
2 (2.50 in.) 4.9087 in.4
A
Accordingly, the normal stress in rod (2) is
22 2
2
75 kips15.2789 ksi
2.4053 in.15.28 ksi (C)
F
A
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.5 Axial loads are applied with rigid bearing plates to the
solid cylindrical rods shown in Figure P1.5/6. The diameter
of aluminum rod (1) is 2.00 in., the diameter of brass rod (2)
is 1.50 in., and the diameter of steel rod (3) is 3.00 in.
Determine the axial normal stress in each of the three rods.
FIGURE P1.5/6
Solution
Cut a FBD through rod (1). The FBD should include the free end A. We will assume that the internal
force in rod (1) is tension (even though it obviously will be in compression). From equilibrium,
1 18 kips 4 kips 4 kips 0 16 kips 16 kips (C)yF F F
FBD through rod (1)
FBD through rod (2)
FBD through rod (3)
Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal
force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):
2 28 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)yF F F
Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in
Bearing plate at A: The area of the bearing plate required for support A is
219,160 lb23.950 in.
800 psiAA
Since the plate is to be square, its dimensions must be
2 use 5 in. 23.95 5 in. bearing plate at0 in. 4.894 in . Awidth Ans.
Bearing plate at B: The area of the bearing plate required for support B is
227,440 lb34.300 in.
800 psiBA
Since the plate is to be square, its dimensions must be
2 use 6 in. 34.30 6 in. bearing plate at0 in. 5.857 in . Bwidth Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.26 The d = 15-mm-diameter solid rod shown in
Figure P1.26 passes through a D = 20-mm-diameter
hole in the support plate. When a load P is applied
to the rod, the rod head rests on the support plate.
The support plate has a thickness of b = 12 mm.
The rod head has a diameter of a = 30 mm and the
head has a thickness of t = 10 mm. If the normal
stress produced in the rod by load P is 225 MPa,
determine:
(a) the bearing stress acting between the support
plate and the rod head.
(b) the average shear stress produced in the rod
head.
(c) the punching shear stress produced in the
support plate by the rod head.
FIGURE P1.26
Solution
The cross-sectional area of the rod is:
2 2
rod (15 mm) 176.715 mm4
A
The tensile stress in the rod is 225 MPa; therefore, the tension force in the rod is
2 2
rod rod rod (225 N/mm )(176.715 mm ) 39,760.782 NF A
(a) The contact area between the support plate and the rod head is
2 2 2
contact (30 mm) (20 mm) 392.699 mm4
A
Thus, the bearing stress between the support plate and the rod head is
2
39,760.782 N
392.699 mm101.3 MPab Ans.
(b) In the rod head, the area subjected to shear stress is equal to the perimeter of the rod times the
thickness of the head.
2(15 mm)(10 mm) 471.239 mmVA
and therefore, the average shear stress in the rod head is
2
39,760.782 N
471.84.4 MP
ma
239 m Ans.
(c) In the support plate, the area subjected to shear stress is equal to the product of the rod head
perimeter and the thickness of the plate.
2(30 mm)(12 mm) 1,130.973 mmVA
and therefore, the average punching shear stress in the support plate is
2
39,760.782 N
1,13035.2 MPa
.973 mm Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.27 The rectangular bar is connected to the support bracket with
a circular pin, as shown in Figure P1.27. The bar width is w = 1.75
in. and the bar thickness is 0.375 in. For an applied load of P =
5,600 lb, determine the average bearing stress produced in the bar
by the 0.625-in.-diameter pin.
FIGURE P1.27
Solution
The average bearing stress produced in the bar by the pin is based on the projected area of the pin. The
projected area is equal to the pin diameter times the bar thickness. Therefore, the average bearing stress
in the bar is
5,600 lb
23,893.33 psi(0.625 in.)(0.375 i
23,n.
900)
psib Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.28 The steel pipe column shown in Figure P1.28 has an
outside diameter of 8.625 in. and a wall thickness of 0.25 in.
The timber beam is 10.75 in wide, and the upper plate has the
same width. The load imposed on the column by the timber
beam is 80 kips. Determine
(a) The average bearing stress at the surfaces between the pipe
column and the upper and lower steel bearing plates.
(b) The length L of the rectangular upper bearing plate if its
width is 10.75 in. and the average bearing stress between the
steel plate and the wood beam is not to exceed 500 psi.
(c) The dimension “a” of the square lower bearing plate if the
average bearing stress between the lower bearing plate and
the concrete slab is not to exceed 900 psi.
Figure P1.28
Solution
(a) The area of contact between the pipe column and one of the bearing plates is simply the cross-
sectional area of the pipe. To calculate the pipe area, we must first calculate the pipe inside diameter d:
2 2 8.625 in. 2(0.25 in.) 8.125 in.D d t d D t
The pipe cross-sectional area is
2 2 2 2 2
pipe (8.625 in.) (8.125 in.) 6.5777 in.4 4
A D d
Therefore, the bearing stress between the pipe and one of the bearing plates is
2
80 kips12.1623 ksi
6.5777 in.12.16 ksib
b
P
A Ans.
(b) The bearing stress between the timber beam and the upper bearing plate must not exceed 500 psi
(i.e., 0.5 ksi). To support a load of 80 kips, the contact area must be at least
280 kips
160 in.0.5 ksi
b
b
PA
If the width of the timber beam is 10.75 in., then the length L of the upper bearing plate must be
2160 in.
14.8837 in.beam width 10.75
14.88 in
.in.
bAL Ans.
(c) The bearing stress between the concrete slab and the lower bearing plate must not exceed 900 psi
(i.e., 0.9 ksi). To support the 80-kip pipe load, the contact area must be at least
280 kips
88.8889 in.0.9 ksi
b
b
PA
Since the lower bearing plate is square, its dimension a must be
288.8889 in. 9.4 9.43 in281 n. .ibA a a a Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.29 A clevis-type pipe hanger supports an 8-in-
diameter pipe as shown in Figure P1.29. The hanger
rod has a diameter of 1/2 in. The bolt connecting the
top yoke and the bottom strap has a diameter of 5/8 in.
The bottom strap is 3/16-in.-thick by 1.75-in.-wide by
36-in.-long. The weight of the pipe is 2,000 lb.
Determine the following:
(a) the normal stress in the hanger rod
(b) the shear stress in the bolt
(c) the bearing stress in the bottom strap
FIGURE P1.29
Solution
(a) The normal stress in the hanger rod is
2 2
rod
rod 2
(0.5 in.) 0.196350 in.4
2,000 lb10,185.917 psi
0.196350 in.10,190 psi
A
Ans.
(b) The cross-sectional area of the bolt is:
2 2
bolt (0.625 in.) 0.306796 in.4
A
The bolt acts in double shear; therefore, its average shear stress is
bolt 2
2,000 lb3,259.493 psi
2(0.306796 in.3,260 i
) ps Ans.
(c) The bearing stress in the bottom strap is based on the projected area of the bolt in contact with the
strap. Also, keep in mind that there are two ends of the strap that contact the bolt. The bearing stress is
thus
2,000 lb
8,533.334 psi2(0.625 in.)(3/16
8,530 psiin.)
b Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.30 Rigid bar ABC shown in Figure P1.30 is
supported by a pin at bracket A and by tie rod (1).
Tie rod (1) has a diameter of 5 mm, and it is
supported by double-shear pin connections at B and
D. The pin at bracket A is a single-shear
connection. All pins are 7 mm in diameter.
Assume a = 600 mm, b = 300 mm, h = 450 mm, P
= 900 N, and = 55°. Determine the following:
(a) the normal stress in rod (1)
(b) the shear stress in pin B
(c) the shear stress in pin A
FIGURE P1.30
Solution
Equilibrium: Using the FBD shown, calculate
the reaction forces that act on rigid bar ABC.
1
1
sin(36.87 )(600 mm)
(900 N)sin (55 )(900 mm) 0
1,843.092 N
AM F
F
(1,843.092 N)cos(36.87 ) (900 N)cos(55 ) 0
958.255 N
x x
x
F A
A
(1,843.092 N)sin (36.87 ) (900 N)sin (55 ) 0
368.618 N
y y
y
F A
A
The resultant force at A is
2 2(958.255 N) ( 368.618 N) 1,026.709 NA
(a) Normal stress in rod (1).
2 2
rod
rod 2
(5 mm) 19.635 mm4
1,843.092 N
19.635 mm93.9 MPa
A
Ans.
(b) Shear stress in pin B. The cross-sectional area of a 7-mm-diameter pin is:
2 2
pin (7 mm) 38.485 mm4
A
Pin B is a double shear connection; therefore, its average shear stress is
pin 2
1,843.092 N
2(38.485 m23.9 M a
)P
mB Ans.
(c) Shear stress in pin A.
Pin A is a single shear connection; therefore, its average shear stress is
pin 2
1,026.709 N
38.485 m26.7 MPa
mA Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.31 The bell crank shown in Figure P1.31 is in
equilibrium for the forces acting in rods (1) and (2).
The bell crank is supported by a 10-mm-diameter
pin at B that acts in single shear. The thickness of
the bell crank is 5 mm. Assume a = 65 mm, b =
150 mm, F1 = 1,100 N, and = 50°. Determine the
following:
(a) the shear stress in pin B
(b) the bearing stress in the bell crank at B
FIGURE P1.31
Solution
Equilibrium: Using the FBD shown, calculate the
reaction forces that act on the bell crank.
2
2
(1,100 N)sin(50 )(65 mm)
(150 mm) 0
365.148 N
BM
F
F
(1,100 N)cos(50 )
365.148 N 0
341.919 N
x x
x
F B
B
(1,100 N)sin(50 ) 0
842.649 N
y y
y
F B
B
The resultant force at B is
2 2(341.919 N) ( 842.649 N) 909.376 NB
(a) Shear stress in pin B. The cross-sectional area of the 10-mm-diameter pin is:
2 2
pin (10 mm) 78.540 mm4
A
Pin B is a single shear connection; therefore, its average shear stress is
pin 2
909.376 N
78.540 mm11.58 MPaB Ans.
(b) Bearing stress in the bell crank at B. The average bearing stress produced in the bell crank by the
pin is based on the projected area of the pin. The projected area is equal to the pin diameter times the
bell crank thickness. Therefore, the average bearing stress in the bell crank is
909.376 N
(10 mm)(5 mm)18.19 MPab Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.32 The beam shown in Figure P1.32 is
supported by a pin at C and by a short link
AB. If w = 30 kN/m, determine the average
shear stress in the pins at A and C. Each
pin has a diameter of 25 mm. Assume L =
1.8 m and = 35°.
FIGURE P1.32
Solution
Equilibrium: Using the FBD shown,
calculate the reaction forces that act on the
beam.
1
1
1.8 msin(35 )(1.8 m) (30 kN/m)(1.8 m) 0
2
47.0731 kN
CM F
F
(47.0731 kN)cos(35 ) 0
38.5600 kN
x x
x
F C
C
1.8 m(1.8 m) (30 kN/m)(1.8 m) 0
2
27.0000 kN
B y
y
M C
C
The resultant force at C is
2 2(38.5600 kN) (27.0000 kN) 47.0731 kNC
Shear stress in pin A. The cross-sectional area of a 25-mm-diameter pin is:
2 2
pin (25 mm) 490.8739 mm4
A
Pin A is a single shear connection; therefore, its average shear stress is
pin 2
47,073.1 N
490.8739 m95.9 MPa
mA Ans.
Shear stress in pin C.
Pin C is a double shear connection; therefore, its average shear stress is
pin 2
47,073.1 N
2(490.8739 m47.9 M a
)P
mC Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.33 The bell-crank mechanism shown
in Figure P1.33 is in equilibrium for an
applied load of P = 7 kN applied at A.
Assume a = 200 mm, b = 150 mm, and =
65°. Determine the minimum diameter d
required for pin B for each of the following
conditions:
(a) The average shear stress in the pin may
not exceed 40 MPa.
(b) The bearing stress in the bell crank
may not exceed 100 MPa.
(c) The bearing stress in the support
bracket may not exceed 165 MPa.
FIGURE P1.33
Solution
Equilibrium: Using the FBD shown, calculate
the reaction forces that act on the bell crank.
2
2
(7,000 N)sin(65 )(200 mm)
(150 mm) 0
8,458.873 N
BM
F
F
(7,000 N)cos(65 )
8,458.873 N 0
11,417.201 N
x x
x
F B
B
(7,000 N)sin(65 ) 0
6,344.155 N
y y
y
F B
B
The resultant force at B is
2 2( 11,417.201 N) (6,344.155 N) 13,061.423 NB
(a) The average shear stress in the pin may not exceed 40 MPa. The shear area required for the pin
at B is
2
2
13,061.423 N326.536 mm
40 N/mmVA
Since the pin at B is supported in a double shear connection, the required cross-sectional area for the pin
is
2
pin 163.268 mm2
VAA
and therefore, the pin must have a diameter of
24
(163.268 m 14.m 4) 2 mmd
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
(b) The bearing stress in the bell crank may not exceed 100 MPa. The projected area of pin B on the
bell crank must equal or exceed
2
2
13,061.423 N130.614 mm
100 N/mmbA
The bell crank thickness is 8 mm; therefore, the projected area of the pin is Ab = (8 mm)d. Calculate the
required pin diameter d:
2130.614 mm
816.
3
mm3 mmd Ans.
(c) The bearing stress in the support bracket may not exceed 165 MPa. The pin at B bears on two 6-
mm-thick support brackets. Thus, the minimum pin diameter required to satisfy the bearing stress limit
on the support bracket is
2
2
13,061.423 N79.160 mm
165 N/mmbA
279.160 mm
2(6 mm)6.60 mmd Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.34 A structural steel bar with a 25 mm × 75 mm rectangular cross section is subjected to an axial
load of 150 kN. Determine the maximum normal and shear stresses in the bar.
Solution
The maximum normal stress in the steel bar is
max
(150 kN)(1,000 N/kN)
(25 mm)(75 mm)80 MPa
F
A Ans.
The maximum shear stress is one-half of the maximum normal stress
maxmax
240 MPa
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.35 A steel rod of circular cross section will be used to carry an axial load of 92 kips. The
maximum stresses in the rod must be limited to 30 ksi in tension and 12 ksi in shear. Determine the
required diameter for the rod.
Solution
Based on the allowable 30 ksi tension stress limit, the minimum cross-sectional area of the rod is
2
min
max
92 kips3.0667 in.
30 ksi
FA
For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be
2
min
max
92 kips3.8333 in.
2 2(12 ksi)
FA
Therefore, the rod must have a cross-sectional area of at least 3.8333 in.2 in order to satisfy both the
normal and shear stress limits.
The minimum rod diameter D is therefore
2 2
min min3.8333 in. 2.2092 in. 2.21 in.4
d d
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.36 An axial load P is applied to therectangular bar shown in Figure P1.36. The cross-sectional area of the bar is 400 mm2. Determine the normal stress perpendicular toplane AB and the shear stress parallel toplane AB if the bar is subjected to an axialload of P = 70 kN.
FIGURE P1.36
Solution The angle θ for the inclined plane is 35°. The normal force N perpendicular to plane AB is found from cos (40 kN)cos35 57.3406 kNN P θ= = ° = and the shear force V parallel to plane AB is sin (70 kN)sin 35 40.1504 kNV P θ= − = − ° = −
The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is
2
2400 mm 488.3098 mmcos cos35n
AAθ
= = =°
The normal stress σn perpendicular to plane AB is
2
(57.3406 kN)(1,000 N/kN) 117.4268 MPa488
117.4 MP.3098 mm
ann
NA
σ = = = = Ans.
The shear stress τnt parallel to plane AB is
2( 40.1504 kN)(1,000 N/kN) 82.2231 MPa
48882.2 M
.3098 mPa
mntn
VA
τ −−= = = − = Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.37 An axial load P is applied to the 1.75 in.
by 0.75 in. rectangular bar shown in Figure
P1.37. Determine the normal stress
perpendicular to plane AB and the shear stress
parallel to plane AB if the bar is subjected to
an axial load of P = 18 kips.
FIGURE P1.37
Solution
The angle for the inclined plane is 60°. The
normal force N perpendicular to plane AB is
found from
cos (18 kips)cos60 9.0 kipsN P
and the shear force V parallel to plane AB is
sin (18 kips)sin60 15.5885 kipsV P
The cross-sectional area of the bar is (1.75 in.)(0.75 in.) = 1.3125 in.2, but the area along inclined plane
AB is
2
21.3125 in./ cos 2.6250 in.
cos60nA A
The normal stress n perpendicular to plane AB is
2
9.0 kips3.4286 ksi
2.6250 in3.43 ks
.in
n
N
A Ans.
The shear stress nt parallel to plane AB is
2
15.5885 kips5.9385 ksi
2.62505.94 ks
ii
n.nt
n
V
A Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.38 A compression load of P = 80 kips is applied to a 4 in. by 4
in. square post, as shown in Figure P1.38/39. Determine the normal
stress perpendicular to plane AB and the shear stress parallel to
plane AB.
FIGURE P1.38/39
Solution
The angle for the inclined plane is 55°. The normal force N
perpendicular to plane AB is found from
cos (80 kips)cos55 45.8861 kipsN P
and the shear force V parallel to plane AB is
sin (80 kips)sin55 65.5322 kipsV P
The cross-sectional area of the post is (4 in.)(4 in.) = 16 in.2, but the area
along inclined plane AB is
2
216 in./ cos 27.8951 in.
cos55nA A
The normal stress n perpendicular to plane AB is
2
45.8861 kips1.6449 ksi
27.8951 1.645 ksi
in.n
n
N
A Ans.
The shear stress nt parallel to plane AB is
2
65.5322 kips2.3492 ksi
27.89512.35 ks
ii
n.nt
n
V
A Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.39 Specifications for the 50 mm × 50 mm square bar
shown in Figure P1.38/39 require that the normal and shear
stresses on plane AB not exceed 120 MPa and 90 MPa,
respectively. Determine the maximum load P that can be
applied without exceeding the specifications.
FIGURE P1.38/39
Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle are
(1 cos2 )2
n
P
A (a)
and
sin 22
nt
P
A (b)
The cross-sectional area of the square bar is A = (50 mm)2 = 2,500 mm
2, and the angle for plane AB is
55°.
The normal stress on plane AB is limited to 120 MPa; therefore, the maximum load P that can be
supported by the square bar is found from Eq. (a):
2 22 2(2,500 mm )(120 N/mm )
911,882 N1 cos2 1 cos2(55 )
nAP
The shear stress on plane AB is limited to 90 MPa. From Eq. (b), the maximum load P based the shear
stress limit is
2 22 2(2,500 mm )(90 N/mm )
478,880 Nsin 2 sin 2(55 )
ntAP
Thus, the maximum load that can be supported by the bar is
max 479 kNP Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.40 Specifications for the 6 in. × 6 in. square post shown in
Figure P1.40 require that the normal and shear stresses on
plane AB not exceed 800 psi and 400 psi, respectively.
Determine the maximum load P that can be applied without
exceeding the specifications.
FIGURE P1.40
Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle are
(1 cos2 )2
n
P
A (a)
and
sin 22
nt
P
A (b)
The cross-sectional area of the square post is A = (6 in.)2 = 36 in.
2, and the angle for plane AB is 40°.
The normal stress on plane AB is limited to 800 psi; therefore, the maximum load P that can be
supported by the square post is found from Eq. (a):
22 2(36 in. )(800 psi)
49,078 lb1 cos2 1 cos2(40 )
nAP
The shear stress on plane AB is limited to 400 psi. From Eq. (b), the maximum load P based the shear
stress limit is
22 2(36 in. )(400 psi)
29,244 lbsin 2 sin 2(40 )
ntAP
Thus, the maximum load that can be supported by the post is
max 29,200 l 29.2 ipb k sP Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.41 A 90 mm wide bar will be used to carry an axial
tension load of 280 kN as shown in Figure P1.41. The
normal and shear stresses on plane AB must be limited
to 150 MPa and 100 MPa, respectively. Determine the
minimum thickness t required for the bar.
FIGURE P1.41
Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle are
(1 cos2 )2
n
P
A (a)
and
sin 22
nt
P
A (b)
The angle for plane AB is 50°.
The normal stress on plane AB is limited to 150 MPa; therefore, the minimum cross-sectional area A
required to support P = 280 kN can be found from Eq. (a):
2
2
(280 kN)(1,000 N/kN)(1 cos2 ) (1 cos2(50 )) 771.2617 mm
2 2(150 N/mm )n
PA
The shear stress on plane AB is limited to 100 MPa; therefore, the minimum cross-sectional area A
required to support P = 280 kN can be found from Eq. (b):
2
2
(280 kN)(1,000 N/kN)sin 2 sin 2(50 ) 1,378.7309 mm
2 2(100 N/mm )nt
PA
To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin =
1,379.7309 mm2. Since the bar width is 90 mm, the minimum bar thickness t must be
2
min
1,378.7309 mm15.3192 mm
90 m15.3
m2 mmt Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.42 A rectangular bar having width w =
6.00 in. and thickness t = 1.50 in. is subjected
to a tension load P as shown in Figure
P1.42/43. The normal and shear stresses on
plane AB must not exceed 16 ksi and 8 ksi,
respectively. Determine the maximum load P
that can be applied without exceeding either
stress limit.
FIGURE P1.42/43
Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle are
(1 cos2 )2
n
P
A (a)
and
sin 22
nt
P
A (b)
The angle for inclined plane AB is calculated from
3
tan 3 71.56511
The cross-sectional area of the bar is A = w×t = (6.00 in.)(1.50 in.) = 9.0 in.2.
The normal stress on plane AB is limited to 16 ksi; therefore, the maximum load P can be found from
Eq. (a):
22 2(9.0 in. )(16 ksi)
1,440 ksi1 cos2 1 cos2(71.5651 )
nAP
The shear stress on plane AB is limited to 8 ksi. From Eq. (b), the maximum load P based the shear
stress limit is
22 2(9.0 in. )(8 ksi)
240 kipssin 2 sin 2(71.5651 )
ntAP
Thus, the maximum load that can be supported by the bar is
max 240 kipsP Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
P1.43 In Figure P1.42/43, a rectangular bar
having width w = 1.25 in. and thickness t is
subjected to a tension load of P = 30 kips. The
normal and shear stresses on plane AB must not
exceed 12 ksi and 8 ksi, respectively. Determine
the minimum bar thickness t required for the bar.
FIGURE P1.42/43
Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle are
(1 cos2 )2
n
P
A (a)
and
sin 22
nt
P
A (b)
The angle for inclined plane AB is calculated from
3
tan 3 71.56511
The normal stress on plane AB is limited to 12 ksi; therefore, the minimum cross-sectional area A
required to support P = 30 kips can be found from Eq. (a):