Solution to Problem 269 Thermal Stress

Problem 269
As shown in Fig. P-269, there is a gap between the aluminum bar and the rigid slab that is supported by two copper bars. At 10°C, Δ = 0.18 mm. Neglecting the mass of the slab, calculate the stress in each rod when the temperature in the assembly is increased to 95°C. For each copper bar, A = 500 mm2, E = 120 GPa, and α = 16.8 µm/(m·°C). For the aluminum bar, A = 400 mm2, E = 70 GPa, and α = 23.1 µm/(m·°C).
 

Figure P-269

 

Solution to Problem 267 Thermal Stress

Problem 267
At a temperature of 80°C, a steel tire 12 mm thick and 90 mm wide that is to be shrunk onto a locomotive driving wheel 2 m in diameter just fits over the wheel, which is at a temperature of 25°C. Determine the contact pressure between the tire and wheel after the assembly cools to 25°C. Neglect the deformation of the wheel caused by the pressure of the tire. Assume α = 11.7 μm/(m·°C) and E = 200 GPa.
 

Relationship Between Arithmetic Mean, Harmonic Mean, and Geometric Mean of Two Numbers

For two numbers x and y, let x, a, y be a sequence of three numbers. If x, a, y is an arithmetic progression then 'a' is called arithmetic mean. If x, a, y is a geometric progression then 'a' is called geometric mean. If x, a, y form a harmonic progression then 'a' is called harmonic mean.
 

Let AM = arithmetic mean, GM = geometric mean, and HM = harmonic mean. The relationship between the three is given by the formula
 

$ AM \times HM = GM^2 $

 

Below is the derivation of this relationship.
 

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