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

[AutumnBeds]

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

 

[haruspex]

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.

 

[AutumnBeds]

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.

 

[Gregs6799]

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

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.

 

[haruspex]

where have you got the 10^9 from in your equation

Converting GPa to Pa.