"The axial load for a given test sample carries 1590 N. Calculate tensile stress at sections (1) and (2) assuming the sample thickness is 5mm. (Rectangular cross section).
normal stress = P/A
Pallow = (normal stress)allow * A
The Attempt at a Solution
This is a new one for me. All the previous homework problems involved a prismatic bar (a bar of uniform cross-section) and not this 'necked' bar.
Since P are pulling forces, the bar is in tension.
I know my first step would be to convert 5mm into meters = 0.005m
But then I'm sorta lost.
Do I just use the normal stress equation above for each section; 1 and 2? Keeping P = 1590N and then changing A for section 1 to 30mm = 0.03m, and for section 2 to 10mm = 0.01m?
But then where does 5mm come into play?
Thanks for all the help.
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