Explanation Given information: The diameter of the solid shaft is 25 mm. The power delivered by the motor is 12 kW with turning of 50 rev sec . Gears A , B , and C removes power of 3 kW, 4 kW, and 5 kW. Calculation: The expression for angular velocity of the shaft ( ω ) is shown below: ω = 2 π f Here, the frequency of shaft’s rotation is f . Substitute 50 rev sec for f . ω = 2 π ( 50 rev sec ) = 314.1593 rad sec The expression for the power transmitted ( P ) by the shaft is shown below: P = T ω Here, the applied torque is T and the angular velocity of the shaft is ω . Rearrange Equation to find the torque at F . T F = P F ω Here, the power at motor F is P F . Substitute 12 kW for P F and 314.1593 rad sec for ω . T F = 12 kW 314.1593 rad sec = 12 kW ( 10 3 W 1 kW ) 314.1593 rad sec = 38.20 N ⋅ m Find the torque at A . T A = P A ω Here, the power at gear A is P A . Substitute 3 kW for P A and 314.1593 rad sec for ω . T A = 3 kW 314.1593 rad sec = 3 kW ( 10 3 W 1 kW ) 314.1593 rad sec = 9.549 N ⋅ m Find the torque at B . T B = P B ω Here, the power at gear B is P B . Substitute 4 kW for P B and 314.1593 rad sec for ω

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

See similar textbooks

Concept explainers

Design of Shafts and Shaft Components

The shaft is basically defined as the machine element that rotates, is in circular cross-section, it is majorly used to transfer the power from one part of the system to another, or from a power generating machine to a power absorbing machine. The shaft …

Question

Chapter 10.3, Problem 34P

To determine

Find the absolute maximum shear stress in the shaft.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.3, Problem 33P

Chapter 10.4, Problem 9FP

Students have asked these similar questions

A spring package with two springs and an external force, 200N. The short spring has a loin of 35 mm. Constantly looking for spring for short spring so that total compression is 35 mm (d). Known values: Long spring: Short spring:C=3.98 N/mm Lo=65mmLo=87.4mmF=c·fTotal compression is same for both spring. 200 = (3.98(c1) × 35) + (c₂ × 35) 200 = 139.3 + 35c₂ 200 - 139.3 = 35c₂ 60.7 = 35c₂ c₂ = 60.7/35 Short spring (c₂) = 1.73 N/mm According to my study book, the correct answer is 4.82N/mm What is wrong with the calculating?

What is the reason for this composition?

Homework: ANOVA Table for followed design B AB Dr -1 -1 1 (15.18,12) 1 -1 -1 (45.48.51) -1 1 -1 (25,28,19) 1 1 (75.75,81)

Chapter 10 Solutions

Statics and Mechanics of Materials (5th Edition)

Ch. 10.3 - Determine the internal torque at each section and...Ch. 10.3 - Determine the internal torque at each section and...Ch. 10.3 - Prob. 3PP Ch. 10.3 - Prob. 4PP Ch. 10.3 - Prob. 1FP Ch. 10.3 - The hollow circular shaft is subjected to an...Ch. 10.3 - Prob. 3FP Ch. 10.3 - Prob. 4FP Ch. 10.3 - Determine the maximum shear stress in the shaft at...Ch. 10.3 - Prob. 6FP

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.

Find the absolute maximum shear stress in the shaft.

Solution Summary: The author calculates the absolute maximum shear stress in the shaft. The power delivered by the motor is 12 kW with turning of 50

Concept explainers

Find the absolute maximum shear stress in the shaft.

Want to see the full answer?

Chapter 10 Solutions