FSP02510402381.jpgFSZ

Tyra

Tyra

Answered question

2020-12-25

image

Answer & Explanation

Demi-Leigh Barrera

Demi-Leigh Barrera

Skilled2020-12-26Added 97 answers

a)Torque,
fs=TIp=Tπ32(D4d4)
Where T=1800lb.ft,D=2.4in=0.2ft,d=1.6in=1.33ft
Then find max f_s
b) T=π16fsD
D3=16Tπfs
use T, fsvalues to find D

user_27qwe

user_27qwe

Skilled2021-12-09Added 375 answers

Given that:
Applied torque, T=1800 Ib-ft
Outer diameter =2.4 in
Outer radius, c2=1.2 in
Inner diameter =1.6 in
Inner radius, c1=0.8 in

(a) Maximum shearing stress τmax=Tc2J

Where J= Polar moment of Inertia

J=π2(c24c14) (For hollow circular shaft)

J=π2(1.240.84)

τmax=1800×12×1.2π2(1.240.84)

=9916.5 lb/in2

τmax=9.92 ksi

b) Maximum shearing stress, τmax=9.92 ksi

But we know that, τmax=TcJ

9.92=1800×12×103π2×9.92

c3=1.386

c=1.115

The required diametr of the solid shaft, d=2c

d=2.23 in

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