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PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it 
without permission. 
 
 
PROBLEM 10.1 
Knowing that the spring at A is of constant k and that the bar AB is rigid, determine the 
critical load cr .P 
 
SOLUTION 


 
Let θ be the angle change of bar AB. 
 sinF kx kL θ= = 
 
2
0: cos 0
sin cos sin 0
BM FL Px
kL PL
θ
θ θ θ
Σ = − =
− =
 
Using 
 2sin and cos 1, 0kL PLθ θ θ θ θ≈ ≈ − = 
 2( ) 0kL PL θ− = crP kL= 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
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PROBLEM 10.2 
Knowing that the torsional spring at B is of constant K and that the bar AB is rigid, 
determine the critical load cr .P 
 
SOLUTION 

 
Let θ be the angle change of bar AB. 
 , sinM K x L Lθ θ θ= = ≈ 
 0: 0 0BM M Px K PLθ θ= − = − = 
 ( ) 0K PL θ− = cr /P K L= 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
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PROBLEM 10.3 
Two rigid bars AC and BC are connected by a pin at C as shown. Knowing that 
the torsional spring at B is of constant K, determine the critical load crP for the 
system. 
 
SOLUTION 



 
 
Let θ be the angle change of each bar. 
 BM Kθ= 
 0: 0B A
A
M K F L
K
F
L
θ
θ
= − =
=
 
Bar AC. cr
1 1
0: 0
2 2C A
M P L LFθΣ = − = 
 cr
AFP
θ
= cr
K
P
L
= 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it 
without permission. 
 
 
PROBLEM 10.4 
Two rigid bars AC and BC are connected as shown to a spring of constant k. 
Knowing that the spring can act in either tension or compression, determine the 
critical load crP for the system. 
 
SOLUTION 
Let δ be the deflection of point C. 
Using free body AC and 
 
1 3
0 : 0
3C A A
P
M LR P R
L
δδ= − + = = 
Using free body BC and 
 
2
0: 0
3C B
M LR PδΣ = − = 
3
2B
P
R
L
δ= 
Using both free bodies together, 
 0: 0
3 3
0
2
9
0
2
x A BF R R k
P P
k
L L
P
k
L
δ
δ δ δ
δ
Σ = + − =
+ − =
 − = 
 
 
 cr
2
9
kL
P =  
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
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without permission. 
 
 
PROBLEM 10.7 
The rigid rod AB is attached to a hinge at A and to two springs, each of 
constant 2 kip/in.,k = that can act in either tension or compression. Knowing 
that 2 ft,h = determine the critical load. 
 
SOLUTION 
Let θ be the small rotation angle. 
 3
4
D
C
B
x h
x h
x h
θ
θ
θ
≈
≈
≈
 
 
3C C
D D
F kx kh
F kx kh
θ
θ
= ≈
= ≈
 
 0: 3 0A D C BM hF hF PxΣ = + − = 
 2 2
5
9 4 0,
2
kh kh hP P khθ θ+ − = = 
Data: 2.0 kip/in. 2ft 24 in.k h= = = 
 
5
(2.0)(24)
2
P = 120.0 kipsP = 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it 
without permission. 
 
 
PROBLEM 10.11 
A compression member of 20-in. effective length consists of a solid 1-in.-
diameter aluminum rod. In order to reduce the weight of the member 
by 25%, the solid rod is replaced by a hollow rod of the cross section shown. 
Determine (a) the percent reduction in the critical load, (b) the value of the 
critical load for the hollow rod. Use 610.6 10 psiE = × . 
 
SOLUTION 
Solid: 
4
2 4
4 4 2 64 o
o
S o s
d
A d I d
π π π = = = 
 
 
Hollow: ( )2 2 23 3
4 4 4 4H o i S o
A d d A d
π π= − = = 
 2 2
1 1
0.5 in.
4 2i o i o
d d d d= = = 
Solid rod: 4 4(1.0) 0.049087 in
64S
I
π= = 
 
2 2 6
3
cr 2 2
(10.6 10 )(0.049087)
12.839 10 lb
(20)
SEIP
L
π π ×= = = × 
Hollow rod: ( )
4
4 4 4 41(1) 0.046019 in
64 64 2
π π   = − = − =  
   
H o iI d d 
 
2 2 6
3
cr 2 2
(10.6 10 )(0.046019)
12.036 10 lb 12.04 kips
(20)
HEIP
L
π π ×= = = × = 
(a) 
3 3
3
12.839 10 12.036 10
0.0625
12.839 10
S H
S
P P
P
− × − ×= =
×
 Percent reduction = 6.25% 
(b) For the hollow rod, cr 12.04 kipsP =  

 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
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without permission. 
 
 
PROBLEM 10.13 
A column of effective length L can be made by gluing together 
identical planks in either of the arrangements shown. Determine 
the ratio of the critical load using the arrangement a to the critical 
load using the arrangement b. 
 
SOLUTION 
Arrangement (a). 
 4
1
12a
I d= 
 
2 2 4
cr, 2 212
a
e e
EI Ed
P
L L
π π= = 
Arrangement (b). 
3
3
min
3 4
1 1
( ) ( )
12 3 12 3
1 19
( )
12 3 324
y
d d
I I d d
d
d d
   = = +   
   
 + = 
 
 
 
2 2 4
cr, 2 2
19
324 
b
e e
EI Ed
P
L L
π π= = 
 cr,
cr,
1 324 27
12 19 19
a
b
P
P
= ⋅ = cr,
cr,
1.421a
b
P
P
= 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it 
without permission. 
 
 
PROBLEM 10.15 
A compression member of 7-m effective length is made by welding 
together two L152 102 12.7× × angles as shown. Using 
200 GPa,E = determine the allowable centric load for the 
member if a factor of safety of 2.2 is required. 
 
SOLUTION 
Angle L152 × 102 × 12.7: 2
6 4 6 4
3060 mm
7.20 10 mm 2.59 10 mm
50.3 mm 24.9 mm
x y
A
I I
y x
=
= × = ×
= =
 
Two angles: 6 6 4(2)(7.20 10 ) 14.40 10 mmxI = × = × 
 6 2 6 42[(2.59 10 ) (3060)(24.9) ] 8.975 10 mmyI = × + = × 
 6 4 6 4min 8.975 10 mm 8.975 10 myI I
−= = × = × 
 
2 2 9 6
3
cr 2 2
(200 10 )(8.975 10 )
361.510 N 361.5 kN
(7.0)e
EI
P
L
π π −× ×= = = × = 
 crall
361.5
. . 2.2
P
P
F S
= = all 164.0 kNP = 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it 
without permission. 
 
 
PROBLEM 10.20 
Knowing that 5.2 kN,P = determine the factor of safety for the structure 
shown. Use 200E = GPa and consider only buckling in the plane of the 
structure. 
 
SOLUTION 
Joint B: From force triangle, 
 
5.2
sin 25 sin 20 sin 135
3.1079 kN (Comp)
2.5152 kN (Comp)
BCAB
AB
BC
FF
F
F
= =
° ° °
=
=
 
Member AB: 
4 4
3 4
9 4
18
5.153 10 mm
4 2 4 2
5.153 10 m
AB
d
I
π π
−
   = = = ×   
   
= ×
 
 
2 2 9 9
,cr 2 2
3
,cr
(200 10 )(5.153 10 )
(1.2)
7.0636 10 N 7.0636 kN
7.0636
. . 2.27
3.1079
AB
AB
AB
AB
AB
EI
F
L
F
F S
F
π π −× ×= =
= × =
= = =
 
Member BC: 
4 4
22
4 2 4 2BC
d
I
π π   = =   
   
 
 3 4 9 4
2 2 2 2
2 2 9 9
,cr 2
3
,cr
11.499 10 mm 11.499 10 m
1.2 1.2 2.88 m
(200 10 )(11.499 10 )
2.88
7.8813 10 N 7.8813 kN
7.8813
. . 3.13
2.5152
BC
BC
BC
BC
BC
BC
L
EI
F
L
F
F S
F
π π
−
−
= × = ×
= + =
× ×= =
= × =
= = =
 
Smallest F.S. governs. . . 2.27F S =  
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
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without permission. 
 
 
PROBLEM 10.25 
Column AB carries a centric load P of magnitude 15 kips. Cables BC 
and BD are taut and prevent motion of point B in the xz plane. Using 
Euler’s formula and a factor of safety of 2.2, and neglecting the 
tension in the cables, determine the maximum allowable length L. 
Use 629 10E = × psi. 
 
SOLUTION 
 
4
4
W10 22: 118 in
11.4 in
x
y
I
I
× =
=
 
 
3
3 3
cr
15 10 lb
( . .) (2.2)(15 10 ) 33 10 lb
P
P F S P
= ×
= = × = ×
 
Buckling in xz-plane. 0.7eL L= 
 
2
cr 2
cr0.7(0.7 )
y yEI EIP L
PL
π π= = 
 
6
3
(29 10 )(11.4)
449.2 in.
0.7 33 10
L
π ×= =
×
 
Buckling in yz-plane. 2eL L= 
 
2
cr 2
6
3
cr
(2 )
(29 10 )(118)
505.8 in.
2 2 33 10
x
x
EI
P
L
EI
L
P
π
π π
=
×= = =
×
 
Smaller value for L governs. 449.2 in.L = 37.4 ftL = 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it 
without permission. 
 
 
PROBLEM 10.27 
Column ABC has a uniform rectangular cross section with 12b = mm 
and 22d = mm. The column is braced in the xz plane at its midpoint C 
and carries a centric load P of magnitude 3.8 kN. Knowing that a factor 
of safety of 3.2 is required, determine the largest allowable length L. 
Use 200E = GPa. 
 
SOLUTION 
 
3 3
cr
2
cr 2
cr
( . .) (3.2)(3.8 10 ) 12.16 10 N
e
e
P F S P
EI EI
P L
PL
π π
= = × = ×
= =
 
Buckling in xz-plane. 
cr
e
EI
L L
P
π= = 
 
3 3 3 4
9 4
1 1
(22)(12) 3.168 10 mm
12 12
3.168 10 m
I db
−
= = = ×
= ×
 
 
9 9
3
(200 10 )(3.168 10 )
0.717 m
12.16 10
L π
−× ×= =
×
 
 
Buckling in yz-plane. 
cr
2
2 2
e
e
L EI
L L L
P
π= = = 
 
3 3 3 4
9 4
9 9
3
1 1
(12)(22) 10.648 10 mm
12 12
10.648 10 m
(200 10 )(10.648 10 )
0.657 m
2 12.16 10
I bd
L
π
−
−
= = = ×
= ×
× ×= =
×
 
The smaller length governs. 0.657 mL = 657 mmL = 