### Understanding Stress and Strain in a Polymer Bar#### Problem Statement:The polymer bar shown in the figure below has a width of \( b = 58 \, \text{mm} \), a depth of \( d = 92 \, \text{mm} \), and a height of \( h = 282 \, \text{mm} \). At a compressive load of \( P = 120 \, \text{kN} \), the bar height contracts by \( \Delta h = -2.30 \, \text{mm} \), and the bar depth elongates by \( \Delta d = 0.39 \, \text{mm} \). At this load, the stress in the polymer bar is less than its proportional limit. Determine:(a) The modulus of elasticity, \( E \).(b) Poisson’s ratio, \( \nu \).(c) The change in the bar width \( b \).#### Diagram Explanation:The diagram provided shows a polymer bar positioned vertically, with the following labels:- **Rigid Plate**: A plate applies a compressive load \( P \) on the top face of the bar.- **b**: The width of the bar at the face parallel to the direction of the load.- **d**: The depth of the bar at the face perpendicular to the direction of the load.- **h**: The height of the bar, which is the distance between the rigid plate (top) and the rigid base (bottom).The load \( P \) is applied downwards, leading to changes in dimensions due to compression: height decreases by \( \Delta h \) and depth increases by \( \Delta d \).#### Calculation Inputs:- Apply the concepts of stress (\( \sigma \)) and strain (\( \epsilon \)) to determine the modulus of elasticity (\( E \)) and Poisson’s ratio (\( \nu \)).- Assess the change in bar width \( b \) due to the applied load.#### Input Fields:(a) \( E = \) [Input Box] GPa(b) \( \nu = \) [Input Box](c) \( \Delta b = \) [Input Box] mm#### Steps to Solve:1. **Calculate the stress in the bar**: \[ \sigma = \frac{P}{A} \] where \( A = b \

Width of b = 58 mmDepth of d = 92 mmHeight of h = 282 mmCompressive load of P = 120kNThe bar height…

The polymer bar shown in the figure below has a width of b = 58 mm, a depth of d = 92 mm, and a height of h = 282 mm. At a = 0.39 mm. At compressive load of P = 120 kN, the bar height contracts by Ah = -2.30 mm, and the bar depth elongates by Ad this load, the stress in the polymer bar is less than its proportional limit. Determine: (a) the modulus of elasticity. (b) Poisson's ratio. (c) the An bor width h

The polymer bar shown in the figure below has a width of b = 58 mm, a depth of d = 92 mm, and a height of h = 282 mm. At a = 0.39 mm. At compressive load of P = 120 kN, the bar height contracts by Ah = -2.30 mm, and the bar depth elongates by Ad this load, the stress in the polymer bar is less than its proportional limit. Determine: (a) the modulus of elasticity. (b) Poisson's ratio. (c) the An bor width h

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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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 load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.

Question

### Understanding Stress and Strain in a Polymer Bar

#### Problem Statement:
The polymer bar shown in the figure below has a width of \( b = 58 \, \text{mm} \), a depth of \( d = 92 \, \text{mm} \), and a height of \( h = 282 \, \text{mm} \). At a compressive load of \( P = 120 \, \text{kN} \), the bar height contracts by \( \Delta h = -2.30 \, \text{mm} \), and the bar depth elongates by \( \Delta d = 0.39 \, \text{mm} \). At this load, the stress in the polymer bar is less than its proportional limit. Determine:

(a) The modulus of elasticity, \( E \).

(b) Poisson’s ratio, \( \nu \).

(c) The change in the bar width \( b \).

#### Diagram Explanation:
The diagram provided shows a polymer bar positioned vertically, with the following labels:
- **Rigid Plate**: A plate applies a compressive load \( P \) on the top face of the bar.
- **b**: The width of the bar at the face parallel to the direction of the load.
- **d**: The depth of the bar at the face perpendicular to the direction of the load.
- **h**: The height of the bar, which is the distance between the rigid plate (top) and the rigid base (bottom).

The load \( P \) is applied downwards, leading to changes in dimensions due to compression: height decreases by \( \Delta h \) and depth increases by \( \Delta d \).

#### Calculation Inputs:
- Apply the concepts of stress (\( \sigma \)) and strain (\( \epsilon \)) to determine the modulus of elasticity (\( E \)) and Poisson’s ratio (\( \nu \)).
- Assess the change in bar width \( b \) due to the applied load.

#### Input Fields:
(a) \( E = \) [Input Box] GPa

(b) \( \nu = \) [Input Box]

(c) \( \Delta b = \) [Input Box] mm

#### Steps to Solve:

1. **Calculate the stress in the bar**:
\[
\sigma = \frac{P}{A}
\]
where \( A = b \

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