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Two bars are unstressed and have lengths of 25 cm and 30 cm at 20^(@)C as shown in Figure. Bar(1) is of aluminium and bar(2) is of steel. The cross-sectional areas of the bars are 20 cm^(2) for aluminium and 10 cm^(2) for steel. Assuming that the top and bottom supports are rigid, compute the stress in each member when the temperature is 70^(@) C. Take Y_(a) = 0.70 xx 10^(5)N//mm^(2). Y_(s) = 2.1 xx 10^(5)N//mm^(2). alpha_(a) = 24 xx 10^(-6) //^(@) C and alpha_(2) = 12 xx 10^(-6) //^(@) C |
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Answer» Solution :Given Data : Length of ALUMINIUM, `" " L_(a) = 25 ` cm = 250 mm Length of steel , `"" L_(s) ` = 30 cm = 300 mm Cross - sectional area of aluminium `A_(s)= 20 cm^(2) = 2000 mm^(2)` . Cross - sectional area of steel, `A_(s) = 10 cm^(2) = 1000 mm^(2) . Y_(a) = 0.7 xx 10^(5) N//mm^(2)` `Y_(s) = 2.1 xx 10^(5) N//mm^(2)` `alpha_(a) = 24 xx 10^(-6) //^(@)C and alpha_(s) = 12 xx 10^(-6) //^(@) C ` Due to rise in temperature, both bars will tend to increase in length. As the SUPPORTS are rigid, both of them will be under compression, but the total length will remain unchanged. HENCE, contraction of the two bars due to compressive stress = ELONGATION of the two bars due to rise in temperature Let `simga_(s) = ` compressive stress in steel `simga_(a)`= compressive stress in aluminium T = Rise in temperature = 70 - 20 `= 50^(@)` C Contraction of the two bars due to compressive stress = `(sigma_(s))/(2.1 xx 10^(5)) xx 300 + (sigma_(a))/(0.7 xx 10^(5)) xx 250 `..... (1) Elongation of the two bars to rise of temperature `alpha_(s) . T. L_(s) + alpha_(a). T. L_(a).` `= 12 xx 10^(-6) xx 50 xx 300 + 24 xx 10^(-6) xx 50 xx 250` = `10^(-6) [ 48 xx 10^(4)] ` = 0.48 mm .......... (2) Equating equations (1) and (2) , we get `(sigma_(s))/(2.1 xx 10^(5)) xx 300 + (sigma_(a))/(0.7 xx 10^(5)) xx 250` `142.86 S_(s) + 357 S_(a) = `48000 Also, force in steel = force in aluminium `singa_(s) xx A_(s)""sigma_(a) xx A_(a) " " sigma_(s) xx 1000 = sigma_(s) xx 2000` `sigma_(s) = 2 sigma_(a) `.......... (3) From equation (3) substituting the value of `sigma_(s) = 2 sigma_(a),` we get `142.86 xx 2 sigma_(a) + 357 sigma_(a) = 48,000` `alpha_(a) ` = 74.68 N/`mm^(2)` Ans. `sigma_(s)` = 149.36 N/`mm^(2)` Ans. |
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