How Much Force Does a Jammed Aluminium Ingot Exert When Heated?

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Homework Help Overview

The discussion revolves around calculating the induced force on a jammed aluminium ingot when heated from 20°C to 500°C. The problem involves understanding the relationship between temperature change, material properties, and induced stress in a constrained scenario.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for calculating force based on the modulus of elasticity and dimensions of the ingot. There is a focus on verifying the calculations and the significance of the values used, particularly the original length of the ingot.

Discussion Status

Some participants have confirmed the calculation of the change in length due to temperature, while others have raised questions about the original length used in the calculations and the significance of the units, particularly the conversion from GPa to Pa. There is no explicit consensus on the correctness of the final answer, but there is ongoing verification of the steps involved.

Contextual Notes

Participants note a potential typo in the original length of the ingot and discuss the implications of significant figures in the calculations. The context of the problem is framed within homework constraints, emphasizing the need for careful consideration of the provided data.

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Homework Statement



(c) A standard aluminium ingot is 746 mm in length at 20°C. When the ingots reach a temperature of 500°C. Calculate:(ii) the induced force in an ingot assuming it has become jammed in the furnace and is NOT free to expand
y = Modulus of elasticity for aluminium: 69 GPa
A=0.019m^2
Change in length= 7.95mm
Original Length=740mm

Homework Equations



F=Ay(change in length / original length)

The Attempt at a Solution


[/B]
F=0.019*69x10^9(7.95/740)
F=14084391 Newtons

I've been to solve the above and have been told my original answer was incorrect. I've relooked at my work and come up with the above.

Any helps as to whether this looks correct would be greatly appreciated.

Thanks
 
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AutumnBeds said:

Homework Statement



(c) A standard aluminium ingot is 746 mm in length at 20°C. When the ingots reach a temperature of 500°C. Calculate:(ii) the induced force in an ingot assuming it has become jammed in the furnace and is NOT free to expand
y = Modulus of elasticity for aluminium: 69 GPa
A=0.019m^2
Change in length= 7.95mm
Original Length=740mm

Homework Equations



F=Ay(change in length / original length)

The Attempt at a Solution


[/B]
F=0.019*69x10^9(7.95/740)
F=14084391 Newtons

I've been to solve the above and have been told my original answer was incorrect. I've relooked at my work and come up with the above.

Any helps as to whether this looks correct would be greatly appreciated.

Thanks
I assume you correctly calculated the increase in length due to temperature and got 7.95mm.
How did 746mm turn into 740mm? Is that a typo?
You quote too many significant figures for the given data.
Other than that, your work looks right.
 
haruspex said:
I assume you correctly calculated the increase in length due to temperature and got 7.95mm.
How did 746mm turn into 740mm? Is that a typo?
You quote too many significant figures for the given data.
Other than that, your work looks right.

Yeah that looks like a typo. The increase in length has been confirmed correct as part of another question.

Thanks for the confirmation.
 
AutumnBeds said:

Homework Statement



(c) A standard aluminium ingot is 746 mm in length at 20°C. When the ingots reach a temperature of 500°C. Calculate:(ii) the induced force in an ingot assuming it has become jammed in the furnace and is NOT free to expand
y = Modulus of elasticity for aluminium: 69 GPa
A=0.019m^2
Change in length= 7.95mm
Original Length=740mm

Homework Equations



F=Ay(change in length / original length)

The Attempt at a Solution


[/B]
F=0.019*69x10^9(7.95/740)
F=14084391 Newtons

I've been to solve the above and have been told my original answer was incorrect. I've relooked at my work and come up with the above.

Any helps as to whether this looks correct would be greatly appreciated.

Thanks

AutumnBeds said:

Homework Statement



(c) A standard aluminium ingot is 746 mm in length at 20°C. When the ingots reach a temperature of 500°C. Calculate:(ii) the induced force in an ingot assuming it has become jammed in the furnace and is NOT free to expand
y = Modulus of elasticity for aluminium: 69 GPa
A=0.019m^2
Change in length= 7.95mm
Original Length=740mm

Homework Equations



F=Ay(change in length / original length)

The Attempt at a Solution


[/B]
F=0.019*69x10^9(7.95/740)
F=14084391 Newtons

I've been to solve the above and have been told my original answer was incorrect. I've relooked at my work and come up with the above.

Any helps as to whether this looks correct would be greatly appreciated.

Thanks
Hi Autumn beds,

I'm stuck on the same question about the induced force on the ingot and was wondering where have you got the 10^9 from in your equation
F=0.019*69x10^9(7.95/740)
F=14084391 Newtons
I've looked through my notes and can't find anything else in the given information in relation to this, I know this was over a year ago for you.

Regards Greg.
 
Gregs6799 said:
where have you got the 10^9 from in your equation
Converting GPa to Pa.
 

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