## Plane State of Stress in LEHI Behavior### Problem Statement:For a LEHI (Linear Elastic Hookean Isotropic) behavior, prove that a plane state of stress requires:\[\sigma_{zz} = 0 = \frac{E}{(1+\nu)(1+2\nu)} \left[ (1-\nu)\varepsilon_{zz} + \nu (\varepsilon_{xx} + \varepsilon_{yy}) \right],\]or\[\varepsilon_{zz} = - \frac{\nu}{1-\nu} (\varepsilon_{xx} + \varepsilon_{yy}).\]### Explanation:In the context of solid mechanics and material science, a plane state of stress often implies that certain stress components in a given direction, such as the $z$-axis in this case, are zero or can be expressed in relation to other stress or strain components.The equations provided derive the condition necessary for such a stress state to exist, applying the assumptions of isotropy and linear elasticity governed by Hooke's Law. This situation frequently occurs in thin structures where out-of-plane stresses are negligible compared to the in-plane stresses.- **Parameters:** - $ \sigma_{zz} $: Stress in the $z$-direction. - $ \varepsilon_{zz}, \varepsilon_{xx}, \varepsilon_{yy} $: Strain components in the $z$, $x$, and $y$ directions respectively. - $ E $: Young's Modulus, a measure of the material's stiffness. - $ \nu $: Poisson's Ratio, indicating the negative ratio of transverse to axial strain.Understanding and applying these concepts are crucial in the fields of mechanical and structural engineering, where predicting how materials deform under various stress conditions is fundamental.

Answered: 2.20 For a LEHI behavior, show that a…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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## Plane State of Stress in LEHI Behavior

### Problem Statement:
For a LEHI (Linear Elastic Hookean Isotropic) behavior, prove that a plane state of stress requires:

\[
\sigma_{zz} = 0 = \frac{E}{(1+\nu)(1+2\nu)} \left[ (1-\nu)\varepsilon_{zz} + \nu (\varepsilon_{xx} + \varepsilon_{yy}) \right],
\]

or

\[
\varepsilon_{zz} = - \frac{\nu}{1-\nu} (\varepsilon_{xx} + \varepsilon_{yy}).
\]

### Explanation:

In the context of solid mechanics and material science, a plane state of stress often implies that certain stress components in a given direction, such as the $z$-axis in this case, are zero or can be expressed in relation to other stress or strain components.

The equations provided derive the condition necessary for such a stress state to exist, applying the assumptions of isotropy and linear elasticity governed by Hooke's Law. This situation frequently occurs in thin structures where out-of-plane stresses are negligible compared to the in-plane stresses.

- **Parameters:**
- $ \sigma_{zz} $: Stress in the $z$-direction.
- $ \varepsilon_{zz}, \varepsilon_{xx}, \varepsilon_{yy} $: Strain components in the $z$, $x$, and $y$ directions respectively.
- $ E $: Young's Modulus, a measure of the material's stiffness.
- $ \nu $: Poisson's Ratio, indicating the negative ratio of transverse to axial strain.

Understanding and applying these concepts are crucial in the fields of mechanical and structural engineering, where predicting how materials deform under various stress conditions is fundamental.

Expert Solution

Step 1: Step 1

Given data,

For LEHI behavior,

show that

$sigma subscript z z end subscript equals 0 equals fraction numerator E over denominator open parentheses 1 plus v close parentheses open parentheses 1 plus 2 v close parentheses end fraction open square brackets open parentheses 1 minus v close parentheses epsilon subscript z z end subscript plus v open parentheses epsilon subscript x x end subscript plus epsilon subscript y y end subscript close parentheses close square brackets o r epsilon subscript z z end subscript equals negative fraction numerator v over denominator 11 minus v end fraction open parentheses epsilon subscript x x end subscript plus epsilon subscript y y end subscript close parentheses$

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