Properties Of Solids And Liquids Question 555
Question: When a building is constructed at $ -10{}^\circ C $ , a steel beam (cross-sectional area $ 45cm^{2} $ ) is put in place with its ends cemented in pillars. If the sealed ends cannot move, what will be the compressional force in the beam when the temperature is $ 25{}^\circ C $ ? For this kind of steel, $ \alpha =1.1\times {{10}^{-5}}{{}^{o}}{{C}^{-1}} $ and $ Y=2.0\times 10^{11}N/m^{2} $ .
Options:
A) $ 2.5\times 10^{3}N $
B) $ 3.5\times 10^{5}N $
C) $ 2.5\times 10^{5}N $
D) $ 3.5\times 10^{3}N $
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Answer:
Correct Answer: B
Solution:
$ \frac{\Delta L}{L _{0}}=\alpha \Delta T=(1.1\times {{10}^{-5}}{{}^{o}}{{C}^{-1}})(35^{o}C) $
$ [\because \Delta L=L _{0}\alpha \Delta T] $ where, $ \Delta L= $ Change in length and $ L _{0}= $ original length $ =3.85\times {{10}^{-4}} $ So, $ F=YA\frac{\Delta L}{L _{0}} $
$ [ \because \text{Young }’\text{ s}\text{modulus},Y=\frac{F/A}{\Delta L/L _{0}} ] $
$ =(2.0\times 10^{11}N/m^{2}) $
$ (45\times {{10}^{-4}}m^{2})(3.85\times {{10}^{-4}}) $
$ =3.5\times 10^{5}N $
BETA
