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

9th Edition

ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

See similar textbooks

Videos

Question

Chapter 10.6, Problem 10.52P

To determine

The temperature rise that will cause the joint to fail.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.6, Problem 10.51P

Chapter 10.6, Problem 10.53P

Students have asked these similar questions

a 300n girl and an 400n boy stand on a 16m platform supported by posts A and B. The platform itself weighs 200N. What are the forces exerted by the supports on the platform?

C A cylindrical piece of steel 38 mm (1½ in.) in diameter is to be quenched in moderately agi- tated oil. Surface and center hardnesses must be at least 50 and 40 HRC, respectively. Which of the following alloys satisfy these requirements: 1040, 5140, 4340, 4140, and 8640? Justify your choice(s).

Using the isothermal transformation diagram for a 1.13 wt% C steel alloy (Figure 10.39), determine the final microstructure (in terms of just the microconstituents present) of a small specimen that has been subjected to the following time-temperature treatments. In each case assume that the specimen begins at 920°C (1690°F) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Rapidly cool to 250°C (480°F), hold for 103 s, then quench to room temperature. (b) Rapidly cool to 775°C (1430°F), hold for 500 s, then quench to room temperature. (c) Rapidly cool to 400°C (750°F), hold for 500 s, then quench to room temperature. (d) Rapidly cool to 700°C (1290°F), hold at this temperature for 105 s, then quench to room temperature. (e) Rapidly cool to 650°C (1200°F), hold at this temperature for 3 s, rapidly cool to 400°C (750°F), hold for 25 s, then quench to room temperature. (f) Rapidly cool to 350°C (660°F), hold for…

Chapter 10 Solutions

Mechanics Of Materials, Si Edition

Ch. 10.3 - Prove that the sum of the normal strains in...Ch. 10.3 - The state of strain at the point on the arm has...Ch. 10.3 - Prob. 10.3P Ch. 10.3 - Prob. 10.4P Ch. 10.3 - 10-5. The state of strain at the point on the gear...Ch. 10.3 - Use the strain transformation equations and...Ch. 10.3 - Use the strain transformation equations and...Ch. 10.3 - Prob. 10.8P Ch. 10.3 - Use the strain transformation equations to...Ch. 10.3 - Use the strain- transformation equations to...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.