________________________________________________________________________
PROBLEM (11.52) Use Fig. 11.16a to determine the maximum load P that can be applied
at both ends of an S 310 x 52 steel link (see Table B.9), given an eccentricity e from the center of
the section measured along the web as shown in Fig. P11.52. Check the value found by
applying Eq. (11.21).
Assumption: σy = 250 MPa, E = 200 GPa, L = 5 m
SOLUTION
From Table B.9, for S310x52:
32 64
6.64 10 95.3 10AmmImm=× =×
mm
33
625 10 120Smmr=× =
We have
3
26
0.06 6.64 10 0.64
625 10
ec eA
rS
−
−
××
== =
×, 541.7
0.12
e
L
r==
So,
155
y
PA MPa= (Fig. 11.16a)
or
63
155 10 (6.64 10 ) 1029
y
Pk
−
=× × = N
Check : Equation (11.21) becomes
69
275 10 1 0.64sec 20.85 200 10
yy
PP
AA
⎡⎤
⎛⎞
⎢⎥
×= + ⎜⎟
⎜⎟
×
⎢⎥
⎝⎠
⎣⎦
Solving by trial and error: 155.5
y
PA MPa
=
Then
63
155.5 10 (6.64 10 ) 1032
y
Pk
−
=× ×= N
It differs 3 kN from that of approximate solution.
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