Question

bn
Marks
5
Calculate diameter of shaft required to transmit 90 kW at
1200 rpm. The maximum torque is likely to exceed mean by
33.37 % for a maximum permissible shear stress of 55 N/mm².
Calculate angle of twist for same data if length of shaft is 5 m.
take G = 80 kN/mm2.
5
A 50 mm dia. of wooden shaft having length of 3.7 m is used
to transmit 90000 watt. Determine the lowest and highest rpm
at which stress may not exceed 120 MPa and angle of twist
may not exceed 2".​

Answer 1

Answer:

I Have No IDia You Try

Again Later

sorry

Answer 2

Answer:

For Solid circular steel shaft-

p = 90 kW = 90×103watts,

N = 160 rpm , qmax=60N/mm2,

G=8×104N/mm2

θ=ao=1×π180=0.0175rad

Solution:

P=2πNTavg60

90×103=3π×160×Tavg/60

∴Tavg=5.371×103N−m=5.371×106N−mm

Student may assume Tmax=Tavg=5.371×106N−mm

Using the relation,

TiP=qmaxR

∴5.371×106×32π×60×2=60×2d

∴d3=5.371×106×32π×60×2=455.90×103

∴ d = 76.96 mm

Now, Ip=π32×d4=π32×76.964=3.444×106mm4

Using the relation,

TIp=GoL

∴L=Go−IpT=8×104×0.0175×3.444×1065.371×106

L = 897.71 mm

∴ Length of shaft = L = 0.897m say 0.9m.