sm11 45

________________________________________________________________________
PROBLEMS (*11.45) A structural ASTM-A36 steel tube has length L, outside diameter do ,
thickness t, and both ends pinned. A compressive load P acts with an eccentricity e as shown
in Fig. P11.43. Calculate:
(a) The maximum stress in the tube.
(b) The allowable load based on a safety factor n with respect to material yielding.
Given: d = 70 mm, t = 5 mm, L = 2.5 m, e = 32.5 mm, P = 10 kN, ns = 3,
E = 210 GPa, σy = 250 MPa (from Table B.4)
*SOLUTION
2702(5)60
io
dd t mm=−=− =
22 2 2
( ) (70 60 ) 1021
44
oi
Add m
4
m
π
π
=−= −=
44 4 4 3
( ) (70 60 ) 542.4 10
64 64
oi 4
I
dd mm
π
π
=−= −=×
33
542.4(10) 2.5 10
23.05 108.5
1021 28.05
IL
rmm
Ar
×
== = = =
(a) Use Eq. (11.15b)
max 2
1sec
2
Pec LP
Ar rAE
σ
⎡⎤
⎛⎞
=+
⎢⎥
⎜⎟
⎜⎟
⎢⎥
⎝⎠
⎣⎦
where
3
3
10 10 9.804
1.02 10
P
M
Pa
A−
×
==
×, 22
(32.5)(35) 2.141
(23.05)
ec
r==
Thus
3
max 69
10 10
(9.804) 1 (2.141)sec 54.25 (1021)(10 )(210 10 )
σ
−
⎡⎤
⎛⎞
×
⎢⎥
=+ ⎜⎟
⎜⎟
×
⎢⎥
⎝⎠
⎣⎦
(1)
32.3
M
Pa=
(b) Apply Eq.(11.16)
2
1sec
2
y
y
Pec L P
Ar rAE
σ
⎡⎤
⎛⎞
=+
⎢⎥
⎜⎟
⎜⎟
⎢⎥
⎝⎠
⎣⎦
416 3
250(10 ) 1 (2.141)sec 54.25
1021(10 ) (1021)(210 10 )
yy
PP
−
⎡
⎤
⎛⎞
⎢
⎥
=+
⎜⎟
⎜⎟
×
⎢
⎥
⎝⎠
⎣
⎦
(2)
Solving numerically: Equation (11.22) gives then 57.92 .
y
P=kN
57.92 19.31
3
y
all s
P
Pk
n
== = N
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