- #1
Alexrey
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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!