sm9 68

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_______________________________________________________________________
PROBLEM (*9.68) A sign of weight W is supported by a pipe having an outer-diameter D and
inner-diameter d as shown in Fig. P9.68. For a wind load of P on the sign, determine the factor of safety
s
nagainst failure by permanent deformation.
Given: W = 1.5 kN, D = 125 mm, d = 100 mm, y
σ
= 250 MPa, P = 2 kN
*SOLUTION
WMPT y2.12.1
=
=
44
(125 100 )
32
J
π
=−
46
1015.14 mm×=
46
1008.72 mmJI ×==
22
(125 100 )
4
A
π
=−
23
10416.4 mm×=
Point A:
PMz3
=
3
6
1.2(2 10 )(0.0625) 10.6
14.15 10
t
M
Pa τ
−
×
==
×
33
36
1.5 10 3(2 10 )(0.0625)
4.416 10 7.08 10
x
σ
−−
××
=− −
××
6
(0.34 52.97)10 53.31
M
Pa=− =−
22
1,2 53.31 53.31
( ) (10.67)
22
σ =− ± − +
7.287.26 ±−=
MPaMPaMPa 7.284.552 max21
=
−==
τ
σ
σ
Thus,
66
250 10 55.4 10 , 4.5
s
s
n
n
×=× =
Continued on next slide
A
z
x
My
W
A
I
σ =− −
Tc
J
τ =
T
y
z
W
M
y
x
P
C
B
3 m
A
F
igure S9.68

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
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Copyright Act without the permission of the copyright owner is unlawful.
Point B:
For solid semi-circle:
24
23
rr
QAy
π
π
== 3
2
3r=
For half thin-tube:
33
2(62.5 50 )
3
Q=−
33
10427.79 mm×=
36
6
2 10 (79.427 10 ) 0.897
7.08 10 (0.025)
dPQ
M
Pa
Ib
τ
−
−
××
== =
×
MPa5.11897.06.10 =+=
τ
3
6
1.2(1.5 10 )(0.0625)
0.34 16.23
7.08 10
x
M
Pa σ
−
×
=− − =−
×
22
1,2 16.23 16.23
( ) (11.5)
22
σ =− ± − + 08.1412.8
±
−
=
MPaMPa 2.2296.5 21
−
==
σ
σ
So,
66
250 10 22.2 10 , 11.3
s
s
n
n
×=× =
We have
() 4.5
sall
n=
B
y
x
M
z
W
A
I
σ =− −
td
τ
ττ =+