Explanation Given information: The torques T 1 = 20 kN ⋅ m and T 2 = 50 kN ⋅ m . The angle of twist at end A ( ϕ A / C ) is 0. Calculation: Let the torque in the segment AB and segment BC of the shaft be T A B and T B C . Sketch the internal torque developed in the segment AB as shown in Figure 1. Sketch the internal torque developed in the segment BC as shown in Figure 2. Refer Figure 1 and Figure 2. The value of internal torque T A B is − 20 kN ⋅ m and T B C is 30 kN ⋅ m . The polar moment of inertia for outer radius 100 mm and inner radius 80 mm in the segment AB is, J A B = π 2 [ ( 100 mm ) 4 − ( 80 mm ) 4 ] = 92.7398 × 10 6 mm 4 The polar moment of inertia for radius 100 mm in the segment BC is, J B C = π 2 [ ( 100 mm ) 4 ] = 157.0796 × 10 6 mm 4 Determine the angle of twist of the shaft using the relation: ϕ = ∑ T L J G (1) Here, T is the torque applied, L is the length of the shaft, J is the polar moment of inertia, and G is the shear modulus of elasticity of the material. Rearrange Equation (1) for the angle of twist of end A ( ϕ A / C ) ϕ A / C = ∑ T L J G = 1 G [ T A B L A B J A B + T B C L B C J B C ] (2) Refer the properties of red brass C83400 copper. The value of shear modulus of elasticity of red brass C83400 copper is 37 × 10 9 Pa . Substitute 0 for ϕ A / C , − 20 kN ⋅ m for T A B , L for L A B , 92

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

See similar textbooks

Question

Chapter 5.4, Problem 5.64P

To determine

The distance L for the angle of twist of end A is 0.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5.4, Problem 5.63P

Chapter 5.4, Problem 5.65P

Students have asked these similar questions

Problem 6 (Optional, extra 6 points) 150 mm 150 mm 120 mm 80 mm 60 mm PROBLEM 18.103 A 2.5 kg homogeneous disk of radius 80 mm rotates with an angular velocity ₁ with respect to arm ABC, which is welded to a shaft DCE rotating as shown at the constant rate w212 rad/s. Friction in the bearing at A causes ₁ to decrease at the rate of 15 rad/s². Determine the dynamic reactions at D and E at a time when ₁ has decreased to 50 rad/s. Answer: 5=-22.01 +26.8} N E=-21.2-5.20Ĵ N

Problem 1. Two uniform rods AB and CE, each of weight 3 lb and length 2 ft, are welded to each other at their midpoints. Knowing that this assembly has an angular velocity of constant magnitude c = 12 rad/s, determine: (1). the magnitude and direction of the angular momentum HD of the assembly about D. (2). the dynamic reactions (ignore mg) at the bearings at A and B. 9 in. 3 in. 03 9 in. 3 in. Answers: HD = 0.162 i +0.184 j slug-ft²/s HG = 2.21 k Ay =-1.1 lb; Az = 0; By = 1.1 lb; B₂ = 0.

Problem 5 (Optional, extra 6 points) A 6-lb homogeneous disk of radius 3 in. spins as shown at the constant rate w₁ = 60 rad/s. The disk is supported by the fork-ended rod AB, which is welded to the vertical shaft CBD. The system is at rest when a couple Mo= (0.25ft-lb)j is applied to the shaft for 2 s and then removed. Determine the dynamic reactions at C and D before and after the couple has been removed at 2 s. 4 in. C B Mo 5 in 4 in. Note: 2 rotating around CD induced by Mo is NOT constant before Mo is removed. and ₂ (two unknowns) are related by the equation: ₂ =0+ w₂t 3 in. Partial Answer (after Mo has been removed): C-7.81+7.43k lb D -7.81 7.43 lb

Chapter 5 Solutions

Mechanics of Materials

Ch. 5.3 - Determine the internal torque at each section and...Ch. 5.3 - Determine the. internal torque at each section and...Ch. 5.3 - The solid and hollow shafts are each subjected to...Ch. 5.3 - The motor delivers 10 hp to the shaft. If it...Ch. 5.3 - The solid circular shaft is subjected to an...Ch. 5.3 - The hollow circular shaft is subjected to an...Ch. 5.3 - The shaft is hollow from A to B and solid from B...Ch. 5.3 - Determine the maximum shear stress in the...Ch. 5.3 - Determine the maximum shear stress in the shaft at...Ch. 5.3 - Determine the shear stress a: point A on the...

Knowledge Booster