# Axially Loaded Member and Tensile Stress oh so lost

## Homework Statement

"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).

http://img208.imageshack.us/img208/121/tensilestressby1.th.jpg [Broken]

## Homework Equations

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.

Last edited by a moderator:

FredGarvin
You will use the cross sectional area at both locations to calculate the normal stresses. The 5mm is required to calculate the area. The 30 mm and 10 mm dimensions are not areas. How do you calculate the area of a rectangle?

Area of a rectangle is length * width

So at section 1, the area A would be 30mm*5mm = 150mm = 0.15m ?
and
Section 2, the area A would be 10mm*5mm = 50mm = 0.05m ?

Then I would just plug those areas into the normal stress = P/A equation for both sections?

Those above areas should read m^2 and not just meters.

normal stress of section 1 = 1590N / 0.15m^2 = 10600 N/m^2
normal stress of section 2 = 1590N / 0.05m^2 = 31800 N/m^2

Could it be that easy? Feel like I'm missing something still

FredGarvin