Axially Loaded Member and Tensile Stress oh so lost

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Discussion Overview

The discussion revolves around calculating tensile stress in an axially loaded member, specifically focusing on a homework problem involving a test sample with a rectangular cross-section subjected to a tensile load. The participants explore the application of the normal stress equation and the necessary dimensions for calculating cross-sectional area.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a homework problem involving a tensile load of 1590 N and asks how to calculate tensile stress at two sections of a necked bar, noting confusion about the dimensions provided.
  • Another participant suggests using the cross-sectional area at both locations to calculate normal stresses and emphasizes the importance of the thickness in determining the area.
  • A participant correctly identifies the formula for the area of a rectangle as length multiplied by width and calculates the areas for both sections based on the provided dimensions.
  • There is a correction regarding the units, where one participant points out that the areas should be expressed in square meters, not just meters.
  • Normal stress calculations are provided for both sections, with one participant expressing surprise at the simplicity of the calculations involved in a uniaxial tension test.
  • A later reply reassures that the calculations are straightforward in this context, noting the absence of shear stress in this scenario.

Areas of Agreement / Disagreement

Participants generally agree on the approach to calculating tensile stress using the normal stress equation and the importance of using the correct area. However, there is some uncertainty expressed by the original poster regarding the overall process and whether they are missing any steps.

Contextual Notes

Participants discuss the dimensions and units involved in the calculations, highlighting the need for clarity in defining areas and ensuring correct unit conversions. There is an implicit assumption that the tensile load is uniformly distributed across the sections discussed.

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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


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 sort of 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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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?

Thanks for the quick reply
 
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
 
It is in this case. You will never see an easier theoretical state of stress than that of a uniaxial tension test. There is no shear until you start dealing with transformations and Mohr's Circle stuff.
 
Wow, well I expected that to be worse.

Thank you very much
 

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