Explanation Given information: The glued joint can resist a maximum shear stress of τ allow = 2 ksi . The thermal strain is α = 6 × 10 − 6 / ° F . The modulus of elasticity is E = 10 × 10 3 ksi . The Poisson’s ratio is ν = 0.2 . Calculation: The sides of the aluminum plate are confined in the rigid constraint along y direction. Therefore, ε y = 0 . The plate is having free expansion along x and z direction. So, the stress in x and z direction is σ x = σ z = 0 . Apply the normal strain in y direction as shown below. ε y = 1 E ( σ y − ν ( σ x + σ z ) ) + α Δ T Here, E is the modulus of elasticity, σ x is the normal stress in x direction, ν is the Poisson’s ratio, α is the thermal strain, Δ T is the increase in temperature, σ y is the normal stress in y direction, and σ z is the normal stress in z direction. Substitute 0 for ε y , 10 × 10 3 ksi for E , 6 × 10 − 6 / ° F for α , 0 for σ x , 0 for σ z , and 0.2 for ν . 0 = 1 10 × 10 3 ( σ y − 0.2 ( 0 + 0 ) ) + ( 6 × 10 − 6 ) Δ T 1 10 × 10 3 σ y = − 6 × 10 − 6 Δ T σ y = − 0.06 Δ T Sketch the stress on the element as shown in Figure 1. Apply the procedure for the construction of Mohr’s circle as shown below. Find the centre of the circle C located σ avg = σ x + σ y 2 from the origin. Plot the reference points A having coordinates A ( σ x , τ x y 2 ) . This point represents the case when the x ′ axis coincides with the x axis. Hence, θ = 0 . Connect the point A with C and from the shaded triangle find the radius R of the circle. Sketch the circle using calculated R value

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 10.6, Problem 10.52P

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The temperature rise that will cause the joint to fail.

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Chapter 10.6, Problem 10.51P

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