ch01-03 stress & strain & properties

01 Solutions 46060 5/6/10 2:43 PM Page 43
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•1–61. Determine the maximum magnitude P of the load
the beam will support if the average shear stress in each pin
is not to allowed to exceed 60 MPa.All pins are subjected to
double shear as shown, and each has a diameter of 18 mm.
C
Referring to the FBD of member AB,Fig.a,
a+©M A = 0; F BC sin 30°(6) - P(2) - P(4) = 0 F BC = 2P
A
2 m
2 m
30
2 m
B
: + ©F x = 0; A x - 2P cos 30° = 0 A x = 1.732P
+ c ©F y = 0; A y - P - P + 2P sin 30° = 0 A y = P
Thus, the force acting on pin A is
F A = 2 A x 2 + A y 2 = 2 (1.732P) 2 + P 2 = 2P
All pins are subjected to same force and double shear. Referring to the FBD of the
pin, Fig. b,
P
P
V = F 2 = 2P 2
= P
The cross-sectional area of the pin is A = p . Thus,
4 (0.0182 ) = 81.0(10 - 6 )p m 2
t allow = V A ; 60(106 ) =
P
81.0(10 - 6 )p
P = 15 268 N = 15.3 kN
Ans.
43