Two bars are unstressed and have lengths of 25cm and 30cm at 20° as shown in Figure. Bar (1) is of aluminium and bar (2) is of steel. The cross-sectional area of bars are 20cm² for aluminium and 10cm² for steel. Assuming that the top and bottom supports are rigid, stress in Al and steel bars in $\frac{N}{{mm}^{2}}$ when the temperature is 70⁰C. (Nearly )
$\begin{array}{l} (Y_{a} = 0 .70 \times 10^{5} N / {mm}^{2} \cdot Y_{s} = 2 .1 \times 10^{5} N / {mm}^{2} \\ α_{a} = 24 \times 10^{- 6} /^{\circ} C and α_{s} = 12 . \times 10^{- 6} /^{\circ} C) \end{array}$

Question

Two bars are unstressed and have lengths of 25cm and 30cm at 20° as shown in Figure. Bar (1) is of aluminium and bar (2) is of steel. The cross-sectional area of bars are 20cm² for aluminium and 10cm² for steel. Assuming that the top and bottom supports are rigid, stress in Al and steel bars in $\frac{N}{{mm}^{2}}$ when the temperature is 70⁰C. (Nearly )

$\begin{array}{l} (Y_{a} = 0 .70 \times 10^{5} N / {mm}^{2} \cdot Y_{s} = 2 .1 \times 10^{5} N / {mm}^{2} \\ α_{a} = 24 \times 10^{- 6} /^{\circ} C and α_{s} = 12 . \times 10^{- 6} /^{\circ} C) \end{array}$

Infinity Learn · Accepted Answer

Contraction of the two bars due to compressive stress $$=$$ Elongation of the two bars due to rise of temperature $$SlYAl+SlYSt=($$αAlLAlΔ$$t)Al+($$αStLsΔ$$t)StForce$$ in steel $$=$$ force in aluminiumSAl×$$AAl=SSt$$×ASt

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