the one from yahoo ans does not work A steel rod 0.350 M long and an aluminum ro
ID: 1899269 • Letter: T
Question
the one from yahoo ans does not workA steel rod 0.350 M long and an aluminum rod 0.250 M long, both with the same diameter, are placed end to end between rigid supports with no initial stress in the rods. The temperature of the rods is now raised by 60.0 C.
What is the stress in each rod? (Hint: The length of the combined rod remains the same, but the lengths of the individual rods do not. If the length is permitted to change by an amount dL when its temperature changes by dT the stress is equal to F/A= Y(dL/L}-alpha dT) .)
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Explanation / Answer
To solve the problem you need the following material properties - Bulk modulus [1] E1 = 200 GPa for steel E2 = 69 GPa for aluminum - Coefficient of linear thermal expansion [2] a1 = 12×10?6 °C?¹ for steel a2 = 23×10?6 °C?¹ for aluminum Each rod would undergo a tensile strain due to temperature: e1 = a1·?T = 12×10?6 °C?¹ · 60°C = 7.2×10?4 e2 = a2·?T = 23×10?6 °C?¹ · 60°C = 1.38×10?³ The total elongation of the composite rod would be ?L = L1·e1 + L2·e2 = 0.25m·72×10?5 + 0.35·138×10?5 = 6.63×10?4m Since the rods are fitted in rigid gap you need compressive strain to keep the total length constant. The strain of reach rod is given by: e1' = s1/E1 e2' = s2/E2 Although the total length of the composite rod keeps constant, these strains might differ from the thermal strain due to the different Young's modules (the steel rod is harder to compress). In other words the length of the rods might change but the total length does not. Since total length of the rod is constant, the magnitude of the total thermal expansion of two rods equals the magnitude of the mechanical compression, i.e. ?L = ?L' = L1·e1' + L2·e2' In statical equilibrium the force acting along the composite rode are constant. Since both rods have the same diameter the stress is also the same: s1 = s2 e1'·E1 = e2'·E2 e2 = e1'·E1/E2 Hence, ?L = L1·e1' + L2·e1'·E1/E2 e1' = ?L /( L1 + L2·E1/E2) = 6.63×10?4m /( 0.25m + 0.35m·200GPa/69GPa) = 5.24×10?4 So the stress in the rod is: s = s1 = 5.24×10?4 · 200GPa = 104.8MPaRelated Questions
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