14 2042012Assignment18Solution

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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 10.10 
Determine the critical load of a steel tube that is 5 m long and has a 100-mm outer 
diameter and a 16-mm wall thickness. Use 200 GPa.E = 
 
SOLUTION 
 1 50 mm 50 16 34 mm
2o o i o
c d c c t= = = ! = ! = 
 ( )4 4 6 4 6 43.859 10 mm 3.859 10 m4 o iI c c
" != ! = # = # 
 
2 2 9 6
3
cr 2 2
(200 10 )(3.859 10 ) 305 10 N
(5.0)
EI
P
L
" " !# #= = = #
 
 cr 305 kNP = W 
�
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
" " "§ ·= = =¨ ¸
© ¹
 
Hollow: ( )2 2 23 34 4 4 4H o i S oA d d A d
" "= ! = = 
 2 21 1 0.5 in.
4 2i o i o
d d d d= = = 
Solid rod: 4 4(1.0) 0.049087 in
64S
I
"= = 
 
2 2 6
3
cr 2 2
(10.6 10 )(0.049087) 12.839 10 lb
(20)
SEIP
L
" " #= = = # 
Hollow rod: ( )
4
4 4 4 41(1) 0.046019 in
64 64 2
" " ª º§ ·= ! = ! =« »¨ ¸
© ¹« »¬ ¼
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
" " #= = = # = 
(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% W�
(b) For the hollow rod, cr 12.04 kipsP = W 
�
 
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 ) 361.5 10 N 361.5 kN
(7.0)e
EI
P
L
" " !# #= = = # = 
 crall
361.5
. . 2.2
P
P
F S
= = all 164.0 kNP = W�
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
dI ! !
"
§ · § ·= = = #¨ ¸ ¨ ¸
© ¹ © ¹
= #
 
 
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
! ! "# #= =
= # =
= = =
 
Member BC: 
4 422
4 2 4 2BC
dI ! !§ · § ·= =¨ ¸ ¨ ¸
© ¹ © ¹
 
 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
! !
"
"
= # = #
= + =
# #= =
= # =
= = =
 
Smallest F.S. governs. . . 2.27F S = W