- #1
markles
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Hello I recently got handed this problem and I had a fair idea on how to go about it or so I thought! My only problem is I do not know how to find d? I would be grateful if someone could show me how to find d and show me the equation used
Thank you
A steel shaft 3m long is transmitting 1MW at 240rev/min. The working conditions to
be satisfied by the shaft are:
(a) that the shaft must not twist more than 0.02radian on a length of 10 diameters;
(b) that the working stress must not exceed 60MN/m2.
If the modulus of rigidity of the steel 80GN/m2 what is
(i). the diameter of the shaft required;
(ii). the actual working stress;
(iii). the angle of twist of the 3m length?
T/J=Gxtheta/L=t/R
T=power/W
rads=degrees x pi/180
(1x106) /(240/2pi/60)= T = 39.793KN
J= pi x d4/32
diameter=d
T= Torque
G =mod of rigidity
J= polar second moment of area?
R= radius
d=?
G=80x109
G x theta (0.02) = 1.6x109/10d
what is the diameter? and how would i go about finding it?
thanks
Thank you
Homework Statement
A steel shaft 3m long is transmitting 1MW at 240rev/min. The working conditions to
be satisfied by the shaft are:
(a) that the shaft must not twist more than 0.02radian on a length of 10 diameters;
(b) that the working stress must not exceed 60MN/m2.
If the modulus of rigidity of the steel 80GN/m2 what is
(i). the diameter of the shaft required;
(ii). the actual working stress;
(iii). the angle of twist of the 3m length?
Homework Equations
T/J=Gxtheta/L=t/R
T=power/W
rads=degrees x pi/180
The Attempt at a Solution
(1x106) /(240/2pi/60)= T = 39.793KN
J= pi x d4/32
diameter=d
T= Torque
G =mod of rigidity
J= polar second moment of area?
R= radius
d=?
G=80x109
G x theta (0.02) = 1.6x109/10d
what is the diameter? and how would i go about finding it?
thanks
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