978-0073398167 Chapter 16 Solution Manual Part 6

subject Type Homework Help
subject Pages 17
subject Words 1019
subject Authors David Mazurek, E. Johnston, Ferdinand Beer, John DeWolf

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page-pf1
PROBLEM 16.47
Solve Prob. 16.46, assuming that the effective length of the column is decreased to
20 ft.
PROBLEM 16.46 A square structural tube having the cross section shown is used
as a column of 26-ft effective length to carry a centric load of 65 kips. Knowing
that the tubes available for use are made with wall thicknesses ranging from
1
4
in.
to
3
4
in. in increments of
1
16
in., use allowable stress design to determine the
lightest tube that can be used. Use
36 ksi
Y
s
=
and
6
29 10 psi.E= ×
page-pf2
consent of McGraw-Hill Education.
page-pf3
PROBLEM 16.48
Two
11
22
3 2 -in.×
angles are bolted together as shown for use as a column of 6-
ft effective length to carry a centric load of 54 kips. Knowing that the angles
available have thickness of
1
4
,
3
8,
and
1
2 in.,
use allowable stress design to
determine the lightest angles that can be used. Use
36 ksi
Y
s
=
and
6
29 10 psi.E= ×
page-pf4
SOLUTION Continued
22
22
36/50.058
cr
cr
all
72 75.616 133.68
0.9522
(29000) 50.058 ksi
( / ) (71.616)
[0.658 ](36) 26.64 ksi
(26.64)(2.88) 45.9 kips 54 kips (not allowed)
1.67 1.67
pp
s
s
s
= = <
= = =
= =
= = = <
e
ee
L
r
E
Lr
A
P
Use
1 13
3 2 in.
2 28
L××
consent of McGraw-Hill Education.
page-pf5
PROBLEM 16.49
A column with the cross section shown has a 13.5-ft effective length. Using a factor
of safety equal to 2.8, determine the allowable centric load that can be applied to
the column. Use E = 29 × 106 psi.
SOLUTION
( ) ( )
min 1 2
3
3
4
2
cr 2
26 4
2
all
2
11 1 1
2 in. 6 in. 10 in. in.
12 2 12 4
18.0130 in
(29 10 psi)(18.0130 in )
[(13.5 ft)(12 in./ft)]
196.451kips
196.451 kips
2.8
e
I II
EI
PL
P
= +
 
= +
 
 
=
=
×
=
=
=
p
p
all
70.2 kips
=P
page-pf6
PROBLEM 16.50
The rigid bar AD is attached to two springs of constant k and is in
equilibrium in the position shown. Knowing that the equal and
opposite loads
P
and
P
remain horizontal, determine the
magnitude
cr
P
of the critical load for the system.
page-pf7
PROBLEM 16.51
A rigid block of mass
m
can be
supported in each of the four ways
shown. Each column consists of an
aluminum tube that has a 44-mm outer
diameter and a 4-mm wall thickness.
Using
70 GPaE=
and a factor of safety
of 2.8, determine the allowable load for
each support condition.
page-pf8
SOLUTION Continued
Allowable capacity:
Case (1):
4 m
e
LL= =
2
5108
(4)
m=
319 kgm=
Case (2):
2 8 m
e
LL
= =
2
5108
(8)
m=
79.8 kgm=
Case (3):
4 m
e
LL= =
2
5108
(4)
m=
319 kgm=
Case (4):
0.699 2.796 m
e
LL= =
2
5108
(2.796)
m=
653 kgm=
page-pf9
PROBLEM 16.52
The steel rod BC is attached to the rigid bar AB and to the fixed
support at C. Knowing that
determine the
diameter of rod BC for which the critical load
cr
P
of the system is
80 lb.
page-pfa
PROBLEM 16.53
Supports A and B of the pin-ended column shown are at a fixed distance L from each
other. Knowing that at a temperature T0 the force in the column is zero and that buckling
occurs when the temperature is
10
,TT T= +∆
express T in terms of b, L, and the
coefficient of thermal temperature
α
.
consent of McGraw-Hill Education.
PROBLEM 16.48
Two
11
22
3 2 -in.×
angles are bolted together as shown for use as a column of 6-
ft effective length to carry a centric load of 54 kips. Knowing that the angles
available have thickness of
1
4
,
3
8,
and
1
2 in.,
use allowable stress design to
determine the lightest angles that can be used. Use
36 ksi
Y
s
=
and
6
29 10 psi.E= ×
SOLUTION Continued
22
22
36/50.058
cr
cr
all
72 75.616 133.68
0.9522
(29000) 50.058 ksi
( / ) (71.616)
[0.658 ](36) 26.64 ksi
(26.64)(2.88) 45.9 kips 54 kips (not allowed)
1.67 1.67
pp
s
s
s
= = <
= = =
= =
= = = <
e
ee
L
r
E
Lr
A
P
Use
1 13
3 2 in.
2 28
L××
consent of McGraw-Hill Education.
PROBLEM 16.49
A column with the cross section shown has a 13.5-ft effective length. Using a factor
of safety equal to 2.8, determine the allowable centric load that can be applied to
the column. Use E = 29 × 106 psi.
SOLUTION
( ) ( )
min 1 2
3
3
4
2
cr 2
26 4
2
all
2
11 1 1
2 in. 6 in. 10 in. in.
12 2 12 4
18.0130 in
(29 10 psi)(18.0130 in )
[(13.5 ft)(12 in./ft)]
196.451kips
196.451 kips
2.8
e
I II
EI
PL
P
= +
 
= +
 
 
=
=
×
=
=
=
p
p
all
70.2 kips
=P
PROBLEM 16.50
The rigid bar AD is attached to two springs of constant k and is in
equilibrium in the position shown. Knowing that the equal and
opposite loads
P
and
P
remain horizontal, determine the
magnitude
cr
P
of the critical load for the system.
PROBLEM 16.51
A rigid block of mass
m
can be
supported in each of the four ways
shown. Each column consists of an
aluminum tube that has a 44-mm outer
diameter and a 4-mm wall thickness.
Using
70 GPaE=
and a factor of safety
of 2.8, determine the allowable load for
each support condition.
SOLUTION Continued
Allowable capacity:
Case (1):
4 m
e
LL= =
2
5108
(4)
m=
319 kgm=
Case (2):
2 8 m
e
LL
= =
2
5108
(8)
m=
79.8 kgm=
Case (3):
4 m
e
LL= =
2
5108
(4)
m=
319 kgm=
Case (4):
0.699 2.796 m
e
LL= =
2
5108
(2.796)
m=
653 kgm=
PROBLEM 16.52
The steel rod BC is attached to the rigid bar AB and to the fixed
support at C. Knowing that
determine the
diameter of rod BC for which the critical load
cr
P
of the system is
80 lb.
PROBLEM 16.53
Supports A and B of the pin-ended column shown are at a fixed distance L from each
other. Knowing that at a temperature T0 the force in the column is zero and that buckling
occurs when the temperature is
10
,TT T= +∆
express T in terms of b, L, and the
coefficient of thermal temperature
α
.

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