Question

The steel bracket is used to connect the ends of two cables. if the allowable normal stress for the steel is sallow = 30 ksi, determine the largest tensile force p that can be applied to the cables. assume the bracket is a rod having a diameter of 1.5 in.

Answer 1

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

The static equilibrium is given as:

F = P (Normal force)

Formula for moment at section

M = P(4 + 1.5/2)

= 4.75p

Solve for the cross sectional area

Area = $\frac{\pi d^{2} }{4}$

d = 1.5

$\frac{\pi *1.5^{2} }{4}$

= 1.767 inches²

Solve for inertia

$\frac{\pi *0.75^4}{4}$

= 0.2485inches⁴

Solve for the tensile force from here

$\frac{F}{A} +\frac{Mc}{I}$

30x10³ = $\frac{P}{1.767} +\frac{4.75p*0.75}{0.2485}$

30000 = 14.902 p

divide through by 14.902

2013.15 = P

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

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