Explanation Given information: Hooke’s law can be written as σ x = E ( 1 − ν 2 ) ( ε x + ν ε y ) and σ y = E ( 1 − ν 2 ) ( ε y + ν ε x ) . Explanation: For the case of plane stress σ z = 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 σ z . ε x = 1 E ( σ x − ν ( σ y + 0 ) ) = 1 E ( σ x − ν σ y ) E ε x = σ x − ν σ y Multiply both sides of the Equation by ν . ν E ε x = ( σ x − ν σ y ) ν = ν σ x − ν 2 σ y (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 σ z . ε y = 1 E ( σ y − ν ( σ x + 0 ) ) = 1 E ( σ y − ν σ x ) E ε y = σ y − ν σ x (2) Add Equation (1) and (2)

Statics and Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (5th Edition)

5th Edition

ISBN: 9780134301006

Author: Russell C. Hibbeler

Publisher: PEARSON

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

Chapter 14.11, Problem 109P

To determine

Show that the plane stress Hooke’s law can be written as

σx=E(1−ν2)(εx+νεy) and σy=E(1−ν2)(εy+νεx).

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.11, Problem 108P

Chapter 14.11, Problem 110P

Students have asked these similar questions

= MMB 241- Tutorial 1.pdf 1/3 80% + + MMB 241 - Dynamics of Particles Tutorial 1 Topic: Kinematics of Particles:- Displacement, velocity and acceleration of particles. QUESTIONS 1. If v = (4t+5) m/s, where t is in seconds, determine a when t = 2 s. 2. If s = (2t³) m, where t is in seconds, determine v when t = 2 s. 3. If a = 2 m/s², determine v at s = 4 m if v = 3 m/s at s = 0. 4. If a = 4 m/s², determines when t = 3 s if v = 2 m/s and s = 2 m when t=0. 5. A particle moves along a straight line such that its position is defined by s = (t² - 6t + 5) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s. 6. A particle is moving with a velocity of vo when s=0 and t=0. If it is subjected to a deceleration of a = -kv³ where k is a constant, determine its velocity and position as functions of time. 7. A particle travels along a straight-line path such that in 4s it moves from an initial position SA = -8 m to a position SB = +3 m. Then in…

3-141

(3-113)

Chapter 14 Solutions

Statics and Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (5th Edition)

Ch. 14.3 - In each ease, the state of stress x, y, xy...Ch. 14.3 - Given the state of stress shown on the element,...Ch. 14.3 - Determine the normal stress and shear stress...Ch. 14.3 - Prob. 2FP Ch. 14.3 - Determine the equivalent state of stress on an...Ch. 14.3 - Prob. 4FP Ch. 14.3 - The beam is subjected to the load at its end....Ch. 14.3 - Prob. 6FP Ch. 14.3 - Prove that the sum of the normal stresses x+y=x+y...Ch. 14.3 - Determine the stress components acting on the...

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.

Show that the plane stress Hooke’s law can be written as σ x = E ( 1 − ν 2 ) ( ε x + ν ε y ) and σ y = E ( 1 − ν 2 ) ( ε y + ν ε x ) .

Solution Summary: The author explains that the plane stress Hooke’s law can be written as underset_sigmax=E(1-nu

Concept explainers

Videos

Show that the plane stress Hooke’s law can be written as

σx=E(1−ν2)(εx+νεy) and σy=E(1−ν2)(εy+νεx).

Want to see the full answer?

Chapter 14 Solutions

Show that the plane stress Hooke’s law can be written as σ x = E ( 1 − ν 2 ) ( ε x + ν ε y ) and σ y = E ( 1 − ν 2 ) ( ε y + ν ε x ) .

Solution Summary: The author explains that the plane stress Hooke’s law can be written as underset_sigmax=E(1-nu

Concept explainers

Videos

Show that the plane stress Hooke’s law can be written as σx=E(1−ν2)(εx+νεy) and σy=E(1−ν2)(εy+νεx).

Want to see the full answer?

Chapter 14 Solutions

Show that the plane stress Hooke’s law can be written as

σx=E(1−ν2)(εx+νεy) and σy=E(1−ν2)(εy+νεx).