Question

A system of bars consists of a steel bar (AB) and an aluminum bar (BC). Bar AB has a cross-section of 2 inches by 0.75 inches and has a modulus of elasticity of 29000 ksi and a coefficient of thermal expansion of 6.5X10^-6 /F. Bar BC has a cross-section of 3 inches by 0.75 inches and has a modulus of elasticity of 10600 ksi and a coefficient of thermal expansion of 12.8X10^-6 /F. A gap of 0.03 inches exists between the left wall and point A when the bars are at a temperature of 70 degrees F. Both bars are then subjected to a temperature increase of the same magnitude.

Determine:

a) the temperature at which the stress in bar AB will be equal to 30 ksi.           Gap = 0.03 inches.

b) the change in length of both the aluminum and steel bar.

The temperature in the system of bars is _____(F)

The change in length of the aluminum bar is _____ (in.)

The change in length of the steel bar is ______ (in.)

B С A 2 in. 3 in. 32 in. 44 in. Gap

can someone help me with this strengths problem?

Answer 1

given System of Bars. B A Steel Aluminium 214 for Steel AB. area. A,= Qin Xo7sinches A, 1-sin' ( = 32in Modily of Eleuticity E, = 29000ksi 3in length 32in ——— чч in gap Coefficient of thermal expansio a= 6.510*% for Aluminium BC area. A = 2.25 in? A₂ = 340.ts in a Mochelley of Elasticity Ex = 10600 ksi, Lenyta hauvin. Coethicient of thermal expansion L₂ = 12.88100/of fp = 0.03 in. at T= 70°F stress in AB DAR = 30 ksi part a) Temperature at which. for this storess force developed due to support will be Rus DABA, Pa = Boki) (l's in) Ra=uskips. us tips So stress in BC, uskips TBCE = ☆ Chc= 20 ksì Az 2.25in? So let the temp. be T at which is occus- So, Techz 0.03 1,0, (T-T.) + Las (T-T;) - Toronto - (32)(65409) (7- 20) + (4) (12-89104) (T-70) – (3) (6) Solving this we get 20) (44) -0.03 24000 10600 259.473°F

part (bi Change in longte of Bar. for Aluminium AL= '₂QAT Bc La E₂ Ale (44)12-8810(259:473 -40) - (20) (44) 10600 Al 0.02369 inches 0.0237 1 for Stel Alst = 42,4T - TABL = (32) (6-580°) ( 259473-7) - 30x32 29060 Alst = 0.0063 inchen ALAI + Alg = 0.02369 t 0.0063 0.02ggg -0.03 inch So okay
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