### Problem Statement1. The strut is supported by a pin at C and a steel guy wire AB. The wire AB has a diameter of 0.2 inches. The maximum normal stress that the steel in the wire can withstand is \(\sigma_{\text{max}} = 18 \text{ ksi.}\) Will the wire fail? Show your work and your reasoning.### DiagramThe diagram illustrates a strut supported at point C and by a guy wire AB. The following details are provided:- **Wire AB**: Oriented at a \(60^\circ\) angle to the horizontal axis.- **Force Distribution**: A triangular distributed load of \(200 \text{ lb/ft}\) acts along the horizontal span from point C to point B.- **Distance CB**: The span length CB is 9 feet.### Solution Steps1. **Calculate the Load on the Wire**: - Integrate the distributed load over the 9-foot span to find the total load applied horizontally.2. **Determine the Force in Wire AB**: - Use trigonometry to resolve forces, considering the angle of \(60^\circ\).3. **Calculate the Stress in the Wire**: - Use the formula for stress: \(\sigma = \frac{F}{A}\), where \(A\) is the cross-sectional area of the wire. - Cross-sectional area, \(A = \pi \left(\frac{0.2}{2}\right)^2\).4. **Compare with Maximum Allowable Stress**: - Determine if the calculated stress exceeds \(\sigma_{\text{max}} = 18 \text{ ksi}\).This approach will ascertain whether the wire can safely support the load without failure.

Given Data: The diameter of wire AB is: d=0.2 in=0.0166 ft The maximum normal stress that the steel…

The strut is supported by a pin at C and a steel guy wire AB. The wire AB has a diameter of 0.2 in. The maximum normal stress that the steel in the wire can withstand is omax = 18 ksi. Will the wire fail? Show your work and your reasoning.

The strut is supported by a pin at C and a steel guy wire AB. The wire AB has a diameter of 0.2 in. The maximum normal stress that the steel in the wire can withstand is omax = 18 ksi. Will the wire fail? Show your work and your reasoning.

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

### Problem Statement

1. The strut is supported by a pin at C and a steel guy wire AB. The wire AB has a diameter of 0.2 inches. The maximum normal stress that the steel in the wire can withstand is \(\sigma_{\text{max}} = 18 \text{ ksi.}\) Will the wire fail? Show your work and your reasoning.

### Diagram

The diagram illustrates a strut supported at point C and by a guy wire AB. The following details are provided:

- **Wire AB**: Oriented at a \(60^\circ\) angle to the horizontal axis.
- **Force Distribution**: A triangular distributed load of \(200 \text{ lb/ft}\) acts along the horizontal span from point C to point B.
- **Distance CB**: The span length CB is 9 feet.

### Solution Steps

1. **Calculate the Load on the Wire**:
- Integrate the distributed load over the 9-foot span to find the total load applied horizontally.

2. **Determine the Force in Wire AB**:
- Use trigonometry to resolve forces, considering the angle of \(60^\circ\).

3. **Calculate the Stress in the Wire**:
- Use the formula for stress: \(\sigma = \frac{F}{A}\), where \(A\) is the cross-sectional area of the wire.
- Cross-sectional area, \(A = \pi \left(\frac{0.2}{2}\right)^2\).

4. **Compare with Maximum Allowable Stress**:
- Determine if the calculated stress exceeds \(\sigma_{\text{max}} = 18 \text{ ksi}\).

This approach will ascertain whether the wire can safely support the load without failure.

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