If you had a steel girder and whilst it was hot you secured it between two strong pins, as it cooled what sort of force would it place on those pins?

AwesomeO said:

If you had a steel girder and whilst it was hot you secured it between two strong pins, as it cooled what sort of force would it place on those pins?

Metallic bond forces

dv said:

AwesomeO said:

If you had a steel girder and whilst it was hot you secured it between two strong pins, as it cooled what sort of force would it place on those pins?

Metallic bond forces

Ok, if you measured that force what would the scale read?

Taking the Young’s modulus to be 200 GPa, and the coefficient of thermal expansion to be 12×10^-6 per K…

I get 2.4 MPa per K. Multiply that by the temperature change and the cross sectional area to get the force. Eg if the cross area is .01 m^2 and the temperature shift is 5K, then the force is 120 kN.

This won’t be quite right, should really allow for the distortion in shape, but it will be about that kind of number.

A metric shitload.

dv said:

Taking the Young’s modulus to be 200 GPa, and the coefficient of thermal expansion to be 12×10^-6 per K…

I get 2.4 MPa per K. Multiply that by the temperature change and the cross sectional area to get the force. Eg if the cross area is .01 m^2 and the temperature shift is 5K, then the force is 120 kN.

This won’t be quite right, should really allow for the distortion in shape, but it will be about that kind of number.

Doesn’t Carbon Steel melt at 1.5k C ?
When you say a “temperature shift is 5K,” do you mean 5k degrees ?
If not what is it DV?

Brett

Thomo said:

dv said:

Taking the Young’s modulus to be 200 GPa, and the coefficient of thermal expansion to be 12×10^-6 per K…

I get 2.4 MPa per K. Multiply that by the temperature change and the cross sectional area to get the force. Eg if the cross area is .01 m^2 and the temperature shift is 5K, then the force is 120 kN.

This won’t be quite right, should really allow for the distortion in shape, but it will be about that kind of number.

Doesn’t Carbon Steel melt at 1.5k C ?
When you say a “temperature shift is 5K,” do you mean 5k degrees ?
If not what is it DV?

Brett

5k degrees? or 5 degrees K?

Thomo said:

dv said:

Taking the Young’s modulus to be 200 GPa, and the coefficient of thermal expansion to be 12×10^-6 per K…

I get 2.4 MPa per K. Multiply that by the temperature change and the cross sectional area to get the force. Eg if the cross area is .01 m^2 and the temperature shift is 5K, then the force is 120 kN.

This won’t be quite right, should really allow for the distortion in shape, but it will be about that kind of number.

Doesn’t Carbon Steel melt at 1.5k C ?
When you say a “temperature shift is 5K,” do you mean 5k degrees ?
If not what is it DV?

Brett

No, K is the SI symbol for kelvin, the unit of absolute temperature. A change of 1K is the same as a change of 1°C, and 0°C is 273.15K.

stumpy_seahorse said:

Thomo said:

dv said:

Taking the Young’s modulus to be 200 GPa, and the coefficient of thermal expansion to be 12×10^-6 per K…

I get 2.4 MPa per K. Multiply that by the temperature change and the cross sectional area to get the force. Eg if the cross area is .01 m^2 and the temperature shift is 5K, then the force is 120 kN.

This won’t be quite right, should really allow for the distortion in shape, but it will be about that kind of number.

Doesn’t Carbon Steel melt at 1.5k C ?
When you say a “temperature shift is 5K,” do you mean 5k degrees ?
If not what is it DV?

Brett

5k degrees? or 5 degrees K?

My physics teacher would give us a rap across the knuckles if we refered to Kelvin as degrees. Celcius is measured in degrees, Kelvin is Kelvin, like there are no degrees Newton.

Thanks btm
wow that is a lot force for a small change in temp

Brett

pommiejohn said:

stumpy_seahorse said:

Thomo said:

Doesn’t Carbon Steel melt at 1.5k C ? When you say a “temperature shift is 5K,” do you mean 5k degrees ? If not what is it DV?

Brett

5k degrees? or 5 degrees K?

My physics teacher would give us a rap across the knuckles if we refered to Kelvin as degrees. Celcius is measured in degrees, Kelvin is Kelvin, like there are no degrees Newton.

YEP.

Capital K is for kelvin. The abbreviation for kilo is a lower case k.

if both the joints were pins, wouldn’t most of that energy go into deformation due to gravity?

If one of the joints was a roller that would make the calculation more complex, surely?

How would you start to go about modelling that?

stan101 said:

if both the joints were pins, wouldn’t most of that energy go into deformation due to gravity?

If one of the joints was a roller that would make the calculation more complex, surely?

How would you start to go about modelling that?

Those considerations are not going to matter, within the terms of the questions as stated.

dv said:

stan101 said:

if both the joints were pins, wouldn’t most of that energy go into deformation due to gravity?

If one of the joints was a roller that would make the calculation more complex, surely?

How would you start to go about modelling that?

Those considerations are not going to matter, within the terms of the questions as stated.

True, but as modified, if one of the pins was a roller the thermal strains would not result in any stress change in the girder (ignoring the increase in bending moment due to the increase in span length).

As stated the dv result is the upper bound force in the girder. The actual force would depend on the stiffness of the pins, and the stiffness of whatever the pins were connected to, and the temperature increase in the restraining mass.

Would all be entirely different if the girder was bimetal.

roughbarked said:

Would all be entirely different if the girder was bimetal.

If it was restrained at the centroid of the cross section, and if everything remained linear elastic, then there would be some additional bending stresses, but the average axial stress (and hence the force on the pins) would be exactly the same*.

• Assuming pins of near infinite stiffness of course.

The Rev Dodgson said:

roughbarked said:

Would all be entirely different if the girder was bimetal.

If it was restrained at the centroid of the cross section, and if everything remained linear elastic, then there would be some additional bending stresses, but the average axial stress (and hence the force on the pins) would be exactly the same*.

• Assuming pins of near infinite stiffness of course.
  

Does Youngs modulus vary much with temperature?