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# Strength of Materials

1. A solid round steel shaft 3 1/2 inches in diameter transmits 100 HP at 200 RPM. Determine:
a. Torsional stress in the outer fibers
b. Angle of twist per foot of shaft
c. Increase in stress at the outer fibers if the shaft is made hollow with a 1 1/2 inch bore.
d. Stress at the inner fibers of the hollow shaft.
e. Angle of twist per foot of hollow shaft.

2. For the beam shown, plot the shear and moment diagrams. Label each diagram with shear and moment values at the points of interest (see attachment).

#### Solution Preview

1.)
a.)
diameter of the shaft D = 3.5 inch
=> D = 3.5*2.54/100 = 0.0889 m
power transmitted P = 100 HP = 100*746 = 74600 W
RPM N = 200 rpm

Tortional stress fs = T/Z
where
T = torque = 60*P/(2*pi*N) = 60*74600/(2*3.1415*200)
=> T = 3562 N.m

Z = polar modulus = I/R = (pi*D^4/32)/(D/2) = pi*D^3/16
=> Z = 3.1415*0.0889^3/16 = 0.00013795 m

Hence,
fs = 3562/0.00013795 = 2.582*10^7 N/m^2 --Answer
=> fs = 2.779*10^4 N/ft^2 --Answer

b.)
angle of twist/foot = l*fs/(R*G)
Here G= modulus of rigidity = 8*10^10 N/m^2
l = length = 1 foot = 0.3048 m
Hence,
twist/length = 0.3048*2.582*10^7/((0.0889/2)*8*10^10)

c.)
For hollow shaft, I' = pi*(Do^4 - Di^4)/32
Do = 3.5 in = 0.0889 m
Di = 1.5 in = 0.0381 m
Hence,
Z' = I'/(Do/2) = 2*I/Do = pi*(Do^4 - Di^4)/(16*Do)
=> Z' = 3.1415*(0.0889^4 - 0.0381^4)/(16*0.0889) = 0.000133296 m
Hence, new value stress,
fs' ...

#### Solution Summary

This complete solution includes 25 steps of calculations and a BM and SF Diagram attached.

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