Explanation Given information: A soft material is placed within the confines of a rigid cylinder which rests on a rigid support. Poisson’s ration is ν = 0.3 . Calculation: Assuming that ε x = 0 and ε y = 0 . Apply the normal strain in x direction as shown below. ε x = 1 E ( σ x − ν ( σ y + σ z ) ) Here, E is the modulus of elasticity, σ x is the normal stress in x direction, ν is the Poisson’s ratio, σ y is the normal stress in y direction, and σ z is the normal stress in z direction. Substitute 0 for ε x . 0 = 1 E ( σ x − ν ( σ y + σ z ) ) σ x − ν ( σ y + σ z ) = 0 (1) Apply the normal strain in y direction as shown below. ε y = 1 E ( σ y − ν ( σ x + σ z ) ) Substitute 0 for ε y . 0 = 1 E ( σ y − ν ( σ x + σ z ) ) σ y − ν ( σ x + σ z ) = 0 (2) Equating Equations (1) and (2). σ x − σ y = ν ( σ y + σ z ) − ν ( ) σ x − σ y = ν ( σ y + σ z − σ x − σ z ) σ x − σ y = − ν ( σ x − σ y ) ( σ x − σ y ) + ν ( σ x − σ y ) = 0 ( 1 + ν ) ( σ x − σ y ) = 0 σ x − σ y = 0 σ x = σ y Substitute σ x for σ y in Equation (1). σ x − ν ( σ x + σ z ) = 0 ( 1 − ν ) σ x − ν σ z = 0 ( 1 − ν ) σ x = ν σ z σ x = ν ( 1 − ν ) σ z Hence, σ y = ν ( 1 − ν ) σ z . Apply the normal strain in z direction as shown below. ε z = 1 E ( σ z − ν ( σ x + σ y ) ) Substitute ν ( 1 − ν ) σ z for σ x and ν ( 1 − ν ) σ z for σ y

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

See similar textbooks

Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

Videos

Question

100%

Chapter 10.6, Problem 10.58P

To determine

The increase factor by which the stiffness of the material, or the apparent modulus of elasticity.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.6, Problem 10.57P

Chapter 10.7, Problem 10.59P

Students have asked these similar questions

How do i solve this problem?

Q4/ A compressor is driven motor by mean of a flat belt of thickness 10 mm and a width of 250 mm. The motor pulley is 300 mm diameter and run at 900 rpm and the compressor pulley is 1500 mm diameter. The shaft center distance is 1.5 m. The angle of contact of the smaller pulley is 220° and on the larger pulley is 270°. The coefficient of friction between the belt and the small pulley is 0.3, and between the belt and the large pulley is 0.25. The maximum allowable belt stress is 2 MPa and the belt density is 970 kg/m³. (a) What is the power capacity of the drive and (b) If the small pulley replaced by V-grooved pulley of diameter 300 mm, grooved angle of 34° and the coefficient of friction between belt and grooved pulley is 0.35. What will be the power capacity in this case, assuming that the diameter of the large pulley remain the same of 1500 mm.

You are tasked with designing a power drive system to transmit power between a motor and a conveyor belt in a manufacturing facility as illustrated in figure. The design must ensure efficient power transmission, reliability, and safety. Given the following specifications and constraints, design drive system for this application: Specifications: Motor Power: The electric motor provides 10 kW of power at 1,500 RPM. Output Speed: The output shaft should rotate at 150 rpm. Design Decisions: Transmission ratio: Determine the necessary drive ratio for the system. Shaft Diameter: Design the shafts for both the motor and the conveyor end. Material Selection: Choose appropriate materials for the gears, shafts. Bearings: Select suitable rolling element bearings. Constraints: Space Limitation: The available space for the gear drive system is limited to a 1-meter-long section. Attribute 4 of CEP Depth of knowledge required Fundamentals-based, first principles analytical approach…

Chapter 10 Solutions

Mechanics of Materials

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.

The increase factor by which the stiffness of the material, or the apparent modulus of elasticity.

Solution Summary: The author explains the increase factor by which the stiffness of the material, or the apparent modulus of elasticity, is 1.35.

Concept explainers

Videos

The increase factor by which the stiffness of the material, or the apparent modulus of elasticity.

Want to see the full answer?

Chapter 10 Solutions