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A 1.5 in diameter solid shaft is made of a steel alloy having an allowable shear stress of �allow = 12 ksi. a) Determine the maximum torque T that can be transmitted. (Left shaft) b) If 1 in. diameter hold is bored through the shaft. Determine the maximum torque T’ and angle of twist. Use L = 1 ft (Right shaft)

Answer :

Answer: a) T= 7.95kip.in b) T'= 6.38kip.in c) a= 1°

Explanation:

a) ta = allowable shear stress= 12ksi

Using torsion formula

tm= ta= Tc/J tm = Maximum shear stress

12= T(0.75)/(π/2)(0.75^4). radius= d/2 = 1.5/2 = 0.75in

T= 12(π/2)(0.75^3)= 7.95kip.in

b) using torsion formula

tm= ta= T'c/J radius 2= d/2= 1.0/2 = 0.50in

12= T'(0.75)/(π/2)(0.75^4 - 0.5^4)

T'= 6.38 kpi.in

c) angle of twist = a = TL/JG L= 1ft= 12in

a = 6.38 × 12/(π/2)(0.75^4 -0.50^4)(11)(10^3)

a = 0.01745 rad = 1°

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