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Stress and strain compound bar question

  • #1
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A compound bar is made using stainless steel tube with an aluminium alloy bar positioned concentrically inside it, and rigidly fixed at each end. The cross sectional area of the tube is 400mm^2 and the cross sectional area of the bar is 300mm^2. The combination is in tension and limiting stress in the aluminium alloy is 60MN/m^2.
Calculate the maximum (1) Stress
(2) Tensile load
σ alumimium =60MN/m^2
E of (s)stainless=180GN/m^2
E of (a)aluminium=120GN/m^2

Area of the tube is 400mm^2
Area of aluminium bar is 300mm^2
Area of stainless tube is 300-400=100mm^2

σa/Ea=σs/Es→σa=σs*Ea/Es

σa=σs*120/180→σa=σs*.666

60=σs*.666→σs=60/.666=90

Answer (1)σs=90MN/m^2

(2)Tensile load
F=Fs+Fa
F=σs*As+σa*Aa
F=60*100+90*300
F=6000+27000
F=33000 or 33KN

Past paper question Is the answers right or wrong?
 
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Answers and Replies

  • #2
haruspex
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I don't understand your equations. You start with
σ aluminium =60MN/m^2
, then calculate
σa=40MN/m^2
. Please define your variables.
 
  • #3
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i don't understand your equations. You start with , then calculate . Please define your variables.
so its wrong.
 
  • #4
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I have amended the original solution
 
  • #5
haruspex
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I have amended the original solution
You still haven't defined your symbols, so I have to guess what they mean.
It looks like σs should mean, by analogy with σa, the limiting stress of the steel. Why should that satisfy the equation σa/Ea=σs/Es? What is the reasoning behind that equation?

The steel tube is extended by some amount, resulting in a certain tension.
The aluminium core is extended by some amount, resulting in a certain tension.
What value is common to those two equations?
 
  • #6
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You still haven't defined your symbols, so I have to guess what they mean.
It looks like σs should mean, by analogy with σa, the limiting stress of the steel. Why should that satisfy the equation σa/Ea=σs/Es? What is the reasoning behind that equation?

The steel tube is extended by some amount, resulting in a certain tension.
The aluminium core is extended by some amount, resulting in a certain tension.
What value is common to those two equations?
I dont understand what you mean by defining symbols, σs=stress of steel, σa=stress of aluminium. Is this what you mean.

As for the equation it comes from total force=force of steel+force of aluminium. F=Fs+Fa

Not sure if this is what you want.
 
  • #7
haruspex
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I dont understand what you mean by defining symbols, σs=stress of steel, σa=stress of aluminium. Is this what you mean.
As for the equation it comes from total force=force of steel+force of aluminium. F=Fs+Fa

Not sure if this is what you want.
Sorry, I wasn't thinking straight. Yes, I agree with your revised solution until here:
F=σs*As+σa*Aa
F=60*100+90*300
To recap:
σa = 60MN/m2
σs = 90MN/m2
Aa = 300mm2
As = 100mm2
 
  • #8
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Sorry, I wasn't thinking straight. Yes, I agree with your revised solution until here:

To recap:
σa = 60MN/m2
σs = 90MN/m2
Aa = 300mm2
As = 100mm2
Ya trying to remember what I was thinking last night, yes your right I need to revise that total force equation did it wrong. Cheers
 
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  • #9
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F=σs*Asa*Aa

F=90*100+60*300

F=9000+18000

F=27000 or 27KN/m2
 
  • #10
haruspex
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F=σs*Asa*Aa

F=90*100+60*300

F=9000+18000

F=27000 or 27KN/m2
Looks good, except that I don't understand where the /m2 comes from. This is a force.
 
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