A concentrated load of P = 74 kips is applied to beam AB , as shown in the figur

Question

A concentrated load of P = 74 kips is applied to beam AB, as shown in the figure. Rod (1) has a diameter of 1.4 in., and its yield strength is 67 ksi. Pin A is supported in a double shear connection, and the ultimate shear strength of pin A is 77 ksi.
(a) Determine the normal stress in rod (1).
(b) Determine the factor of safety with respect to the yield strength for rod (1).
(c) If a factor of safety of 2.6 with respect to the ultimate strength is specified for pin A, determine the minimum required
pin diameter.

Explanation / Answer

deflection @B due to applied load = deflection due to tension in the rod

(PL3/3EI) +4.7( PL2/2EI) =(T Sin 30.170 * (12.9)3)/3EI

74*(8.2)3/3EI + 4.7*74*(8.2)2/2EI = (T Sin 30.170 * (12.9)3)/3EI

T = 70.34 kips

a) normal stress in rod (1) =T/A

A=Pi*1.4*1.4/4

therefore normal stress in rod (1) = 45.69 ksi

b) factor of safety = yield strength / normal stress

F.O.S = 67/45.69 = 1.466

C) P = 2*fs*pi*(Dmin)2/4

74 = 2*(77/2.6)*pi*(Dmin)2/4

(Dmin)2 = 1.590 in

Dmin = 1.260 in

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