K-1Stress Analysis of Unloaded Alumininium Bar

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Discussion Overview

The discussion revolves around the stress analysis of an unloaded aluminum bar subjected to an axial tensile load. Participants explore calculations related to tensile stress, strain, lateral strain, changes in dimensions, and the effects of temperature on length. The scope includes theoretical calculations and practical applications of material properties.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Participants calculate tensile stress using the formula stress = load/area, with the area derived from width and depth.
  • Initial calculations by one participant yield values that are questioned for being too small.
  • Corrected calculations for tensile stress result in 96 MPa, with subsequent recalculations for strain and changes in dimensions.
  • Disagreement arises regarding the signs of lateral strain and changes in dimensions, with some participants suggesting they should be negative due to compressive effects.
  • Clarification is sought on the interpretation of the final length after temperature change, leading to further calculations.

Areas of Agreement / Disagreement

Participants generally agree on the method for calculating stress and strain, but there are competing views on the signs of certain strains and the interpretation of the final length after thermal expansion. The discussion remains unresolved on the correctness of some values and interpretations.

Contextual Notes

Some calculations depend on assumptions about material behavior under load and temperature, and there are unresolved questions regarding the accuracy of certain values and the interpretation of compressive versus tensile effects.

Who May Find This Useful

Individuals interested in material science, engineering mechanics, or those studying stress analysis in materials may find this discussion relevant.

jaymar023
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An unloaded alumininium bar is 1.5m long and has a width and depth of 75mm and 25mm respectively, at room temperature. Determine the following when the bar is subjected to an axial tensile load of 180KN.

a) The tensile stress
b) The tensile strain
c) The lateral strain
d) The change in width and depth of the cross-section
e) The change in length of the bar
f) If the temperature of the unloaded bar is now raised by 200 degrees Celsius, determine its new length.

Take Young's modulus, E = 65GPa, Poisson's ratio, v = 0.33,, Coefficient of thermal expansion, α = 23.4 x 10-6
 
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The answers i calculated were,
a) 1.6MPa
b) 2.46 x 10-5
c) 8.118 x 10-6
d) change in width = 1.845x10-6m
Could not calculate depth
e) 1.2177x10-5m
f) 7.02 x 10-3m

But most of the values I have calculated seem too small to be correct.
 
Hi jaymar023, welcome to PF. You're running into problems right away at (a). How do you calculate stress from load and cross-section area?
 
Stress = Load/Area
 
Agreed. And the relevant area is?
 
Width x Depth = 75 x 10-3 x 25 x 10-3
 
So 180 x 103 / Area = 96MPa
 
Looks good.
 
(a) Nice work, jaymar023. Alternately, leaving all units in N and mm (since 1 MPa = 1 N/mm^2), that would be sigma = P/A = (180 000 N)/[(75 mm)(25 mm)] = 96.0 MPa.

By the way, there should always be a space between the numeric value and its following unit symbol. See international standard for writing units[/color]; i.e., ISO 31-0[/color].

Your approach looked correct on items b and c, so post b and c again using your corrected answer for item a. And post d, e, and f again using your new answers. Change in width and depth of the cross section would just be lateral strain times width or depth, right?
 
  • #10
a) 96 MPa
b) 1.48 x 10-3
c) 4.88 x 10-4
d) Change in width = 4.88 x 10-4 x 25 x 10-3 = 1.22 x 10-5 m
Change in depth = 4.88 x 10-4 x 75 x 10-3 = 3.66 x 10-5 m
e) 2.22 x 10-3 m
f) 7.02 x 10-3 m
 
  • #11
I am still not sure that d) and e) are correct
 
  • #12
jaymar023: Items a through e look correct, except c and d should be negative, because they are compressive strain and contraction. I didn't get the same answer as you got on item f. Notice question f is asking for the new length, not the change in length.
 
  • #13
So for f) the answer should be 8.52 m?
 
  • #14
No, that's not right. Keep trying.
 
  • #15
What about 1.50702 m ?
 
  • #16
That's correct. Nice work.
 
  • #17
thank you for you help, much appreciated.
 

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