(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A solid steel bar of diameterd_{1}= 60 mm has a hole of diameterd_{2}= 32 mm drilled through it. A steel pin of diameterd_{2}passes through the hole and is attached to supports.

Determine the maximum permissible tensile loadP_{allow}in the bar.

-Yield stress for shear in pin is τ_{Y}= 120 MPa.

-Yield stress for tension in bar is σ_{Y}= 250 MPa.

-Factor of safetyn= 2.0.

2. Relevant equations

σ_{allow}=σ_{Y}/n

τ_{allow}=τ_{Y}/n

In direct tension and compression:

P_{allow}=σ_{allow}A

If compression, and a pin is present:

P_{allow}=τ_{allow}A

Area of regular cross section in bar is area of circle.

Area of cross section with pin present requires additional steps. Because I do not have any images, I can only do my best to explain.

The area of a "circle with a core removed" (side view of a pin through bar), is

A = 2r^{2}(α - (ab)/r^{2})

3. The attempt at a solution

It would be very hard to show my steps of determining the area, but I know the area is correct. I calculated the area of the cross-section with a pin to be

A = 0.0010028 m^{2}.

This is after all the correct conversions and formulas to determine, as outlined in the appendix of my book.

I recognize that because this bar is in tension, the pin does not factor in. I use the equation:

P_{allow}=σ_{allow}A

P_{allow}= (250MPa / 2.0)(0.0010028)

P_{allow}= 0.12535 MPa*m^{2}

0.12535 MPa*m^{2}* 10^{6}Pa/MPa = 125350 Pa*m^{2}

I know a pascal(Pa) is equivalent to 1 N/m^{2}. The m^{2}cancels out, leaving

125350 N.

The solutions in my book says the answer should be 95.6 kN. My answer is 125.35 kN. I was wondering if the book was wrong, or if I'd messed up somewhere?

Thanks for any help. This is my first of hopefully not many posts. Any feedback regarding both my work and the format of my question would be appreciated! :)

D

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# Homework Help: Tension, Compression, Shear Problem.

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