let \( L \) be the distance between the pillars chnage in temperature \( =25+10=35 \mathrm{deg} \mathrm{C} \) lienear expansion of the beam \( \Delta L=L^{*} 35^{*} \Delta \) The beam is not allowed to expand as the pillars do not move hence strain in the beam \( =\Delta L / L=35 * 1.1 \mathrm{e}-5= \) \( 3.85 \mathrm{e}-4 \) modulus of elasticity \( \mathrm{Y}= \) stress \( / \) strain \( =2.0 \mathrm{e}+11 \) \( \mathrm{Nm}^{-2} \) stress on the beam \( =F / A=2.0 \mathrm{e}+11 * 3.85 \mathrm{e}-4=7.7 \) e+7 compressional force \( \mathrm{F}=7.7 \mathrm{e}+7 * \mathrm{~A}= \) \[ \text { 7.7e7*45.0e-4 = } 3.465 \mathrm{e}+5 \mathrm{~N} \]
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- The change in temperature is given as \( 25 + 10 = 35 \, \text{deg C} \). Show more…
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When a building is constructed at $-10^{\circ} \mathrm{C}$, a steel beam (cross-sectional area $45 \mathrm{~cm}^{2}$ ) is put in place with its ends cemented in pillars. If the sealed ends cannot move, what will be the compressional force on the beam when the temperature is $25^{\circ} \mathrm{C} ?$ For this kind of steel, $\alpha=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$ and $Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$ Proceed much as in Problem 15.11: so $$ \begin{array}{c} \frac{\Delta L}{L_{0}}=\alpha \Delta T=\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(35^{\circ} \mathrm{C}\right)=3.85 \times 10^{-4} \\ F=Y A \frac{\Delta L}{L_{0}}=\left(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\left(45 \times 10^{-4} \mathrm{~m}^{2}\right)\left(3.85 \times 10^{-4}\right)=3.5 \times 10^{5} \mathrm{~N} \end{array} $$
When a building is constructed at $-10{ }^{\circ} \mathrm{C}$, a steel beam (crosssectional area $45 \mathrm{~cm}^{2}$ ) is put in place with its ends cemented in pillars. If the sealed ends cannot move, what will be the compressional force on the beam when the temperature is $25^{\circ} \mathrm{C}$ ? For this kind of steel, $\alpha=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$ and $Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$. Proceed much as in Problem 15.11: $$ \begin{array}{c} \frac{\Delta L}{L_{0}}=\alpha \Delta T=\left(1.1 \times 10^{-5}{ }^{-5} \mathrm{C}^{-1}\right)\left(35^{\circ} \mathrm{C}\right)=3.85 \times 10^{-4} \\ F=Y A \frac{\Delta L}{L_{0}}=\left(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\left(45 \times 10^{-4} \mathrm{~m}^{2}\right)\left(3.85 \times 10^{-4}\right)=3.5 \times 10^{5} \mathrm{~N} \end{array} $$
How much stress is created in a steel beam if its temperature changes from $-15^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ but it cannot expand? For steel, the Young's modulus $Y=210 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}$ from (Ignore the change in area resulting from the expansion.)
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