Axial Deformation due to Temperature Change and Load

[Alexrey]

Homework Statement

A 100 kN load is applied to a flange positioned midway (at B) along a 50 mm diameter steel bar (ABC).
The bar is placed between two rigid supports and is thus constrained.

Determine the position of the flange (distance from the left support) if the temperature of the bar is
decreased by 20⁰C while the 100 kN load is maintained.

$E = 200 GPa$, $α = 11.9×10^{-6} /°C$

Homework Equations

Change in length due to temperature: $\delta_{T}=\alpha \Delta T L$
Change in length due to force: $\delta_{F}=\frac{F L}{A E}$

The Attempt at a Solution

I reasoned that, since the steel bar's length is shrunk from the -20⁰C temperature change, it'll lose contact with the right support and no internal force will be present in the bar to the right of the flange. Thus we'll have 100kN of internal force left of the flange, creating an elongation due to this force. Therefore:

$\delta_{T}=\alpha \Delta T L$
$\delta_{T}=(11.9×10^{-6})(-20)(0.4)$
$\delta_{T}=-9.52×10^{-5} m$

And for the applied force at the flange we'll have:

$\delta_{F}=\frac{F L}{A E}$
$\delta_{F}=\frac{(100×10^3)(0.2)}{π(0.025)^2 (200×10^9}$
$\delta_{F}=5.09×10^{-5} m$

Therefore, since we only want to know the position of the flange from the left support, I would have:

$\delta_{tot}=\frac{1}{2}\delta_{T}+\delta_{F}$
$\delta_{tot}=\frac{1}{2}(-9.52×10^{-5})+5.09×10^{-5}$
$\delta_{tot}=3.3×10^{-6} m$

Thus:

$0.2+3.3×10^{-6} = 0.2000033 m = 200.0033 mm$ from the left support.

Does this approach and final answer seem correct? Any help would be really appreciated. Thank you!

 

[billy_joule]

The bar contracts due to temp. and is compressed due to force.
These effects sum to give the final flange position, you've taken the difference.

 

[William White]

I'm not sure about the logic here

If the bar is constrained, that suggests it is not free to strain in a particular direction.

If a constraint is made in a particular direction, the object is not free to move in that direction. If this bar is constrained axially, it can only strain radially - the radius gets smaller, the length stays the same. Because it is constrained axially, the radius will reduce further than normal (depends on the material, but commonly about 30% more); which means the flange will be free to move (it will become loose)

However,

If the bar is just "balanced" in between the two supports and is not fixed (as you suggest when you say it will detach from the right hand side (why just the right hand side, why not both?) then it will just drop to the floor when it contracts!Not sure about the accuracy of this question...

 

[Chestermiller]

The way I read this question, they don't even specify the level of the constraint imposed by the walls. So, for all we know it could be highly constrained, and will not lose contact with either support when the temperature is decreased. Under these circumstances, the movement of the flange will be the same as if the 100 N force were not even present (since the response to loading and temperature changes is linear and superimposible).

Chet