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A shaft of circular cross section (diameter = 3 in) is subjected to a bending moment of 3200 lb- ft. Determine the bending stress at the surface of the shaft.

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Answer:


\sigma =145.62 MPa

Step-by-step explanation:

Given that

Diameter of shaft=3 in (1 in=0.0254 m)

Bending moment=3200 lb-ft (1 lb-ft=1.35 N-m)

We know that bending stress in circular shaft given as


\sigma =(32M)/(\pi d^3)

Where M is the bending moment and d is the diameter of shaft.

Given that M=4338.61 N-m

d=0.0672 m

Now by putting the values


\sigma =(32* 4338.61)/(\pi * 0.0672^3)

So
\sigma =145.62 MPa

User Ronald Blaschke
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