Fixed end beam flexural stress under differential temperature 3

[Yolac]

Hi guys, for a both end fixed beam subject to differential temperature, say top part hotter than the bottom part, what sort of bending moment diagram will we be getting? I think it would be constant tension at top (rectangular moment diagram) while my colleague says its constant tension at bottom of beam. Who is correct?

 

[paddingtongreen]

BAretired, I avoided the non linear case because I have a reservation, something I haven't quite worked out, the brain's too slow these days. The linear case causes circular bending but I have to convince myself that the non linear does too.

As I see it at the moment, if it is non linear, there will be shear plane(s) for the length of the beam. That shear will cause complimentary vertical shear, causing bending moment. I have to convince myself that that moment causes circular bending.

I am trying to think my way through a case where all of the upper half of a beam is heated to an even temperature, above that of the uniformly cool lower half.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

paddington,

Say that the upper half is heated to 100^o above starting temperature and the lower half is unchanged. Now, the upper half would like to expand, but it can't because there is a fixed support at each end. So it is stressed to 100*C*E where C is the coefficient of expansion and E is the modulus of elasticity.

The temperature in the lower half has not changed, so the stress in the lower half is zero.

The stress over the beam depth is a step function, having a large value in the top half and zero in the bottom half. But there is no bending because none of the fibers can change in length. There are no horizontal shear stresses developing between the top and bottom half because there is zero strain in both halves.

BA

 

[paddingtongreen]

Yup, I came to that conclusion on my way to bed last night.

Separating and then adding loading effects has helped me considerably over the years, but in this case it was a complication.

Thanks

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

You're welcome.

BA

 

[IDS]

It seems that the old dead horse is in need of further flogging (or if it isn't dead, then it's still alive )

I think the procedure of splitting the beam into statically determinate sections is the right way to approach this problem, but it needs to be done in the right order:

1) Split the member into statically determinate sections.
2) Apply non-load related strains (i.e. temperature and shrinkage)
3) Apply any real external restraints
4) If necessary, apply imaginary forces to each section to restore strain compatibility.
5) Apply equal and opposite forces to the composite section to restore equilibrium

For the example in the original post, consider three different end conditions:

1) Fully fixed both ends
2) Fully fixed left, restrained against rotation only right
3) Simply supported both ends, laterally restrained at bottom left

and two different strain conditions
A) Top half of beam at uniform +10 degrees relative to the bottom half.
B) Left hand end of top half at +10 degrees relative to the bottom half and right hand end

1) Split the member into statically determinate sections.

A) Split the beam along the Neutral Axis (NA) and remove all end restraints
B) Split the top section into two halves

2) Apply non-load related strains (i.e. temperature and shrinkage)

A) Top half expands uniformly
B) Top Left hand section expands uniformly

3) Apply any real external restraints

1A) Apply restraint force to end of top half to return to original position
1B) Slide the top left section so that its left hand end is in the original position. No force required since it is unrestrained at the right
2) and 3) No external restraint to horizontal movement

4) If necessary, apply imaginary forces to each section to restore strain compatibility.

1A) External restraints have already restored strain compatibility, so nothing to do
1B, 2B) Apply force to right hand end of top left section
2A) Apply force to right hand end of top half
3A) Apply forces to both ends of top half
3B) Apply forces to both ends of top left section

5) Apply equal and opposite forces to the composite section to restore equilibrium

1A) Nothing to do
1B, 2B, 3B) Apply restoring force to right hand end of top left section at mid height (i.e. top quarter point at mid span of the composite beam)
3A) Apply restoring force to both ends of top half at mid height (i.e. top quarter point at both ends of the composite beam)

End result:

1A) No rotation or deflection, top half of beam in compression, outward force applied to external restraints at the level of the top quarter point.

1B) No rotation at the ends, but a counter-clockwise moment is applied at the mid point, with resulting reverse curvature, deflections and stresses.

2A) As 1A (no rotation or deflection) except bottom half of beam in tension, and compression in top half reduced. Moment applied to external restraints.

2B) As 1B except stresses, rotations and deflections modified by lack of translational restraint at ends.

3A, 3B) Beam hogs with no nett moment or axial force on any section.


So the main point of all that is: apply the real restraints before messing around with imaginary restraint forces and equal and opposite eqilibrium forces.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

1B) No rotation at the ends, but a counter-clockwise moment is applied at the mid point, with resulting reverse curvature, deflections and stresses.

Which is of course clockwise in the Northern Hemisphere

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[hokie66]

My head spins just reading this without trying to contribute. Anybody got a practical problem?