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Chapter 2: Problem 9

An aluminum wire having a diameter \(d=2 \mathrm{mm}\) and length \(L=3.8 \mathrm{m}\) is subjected to a tensile load \(P\) (see figure). The aluminum has modulus of elasticity \(E=\) \(75 \mathrm{GPa}\) If the maximum permissible elongation of the wire is \(3 \mathrm{mm}\) and the allowable stress in tension is \(60 \mathrm{MPa}\), what is the allowable load \(P_{\max } ?\)

Short Answer

Expert verified
The allowable load is 185.317 N.

Step by step solution

01

Calculate Cross-sectional Area

First, we need to calculate the cross-sectional area of the wire. The formula for the area \(A\) of a circle is:\[ A = \frac{\pi d^2}{4} \]where \(d = 2 \text{ mm} = 0.002 \text{ m}\). Substituting the value of diameter:\[ A = \frac{\pi (0.002)^2}{4} = 3.14159 \times 10^{-6} \text{ m}^2 \]
02

Calculate Allowable Load Based on Stress

Next, we find the allowable load based on the maximum allowable stress. The formula for stress \(\sigma\) is:\[ \sigma = \frac{P}{A} \]Rearrange it to solve for \(P\):\[ P = \sigma \times A \]Given the allowable stress \(\sigma = 60 \text{ MPa} = 60 \times 10^6 \text{ N/m}^2\), the allowable load is:\[ P = 60 \times 10^6 \times 3.14159 \times 10^{-6} = 188.495 \text{ N} \]
03

Calculate Allowable Load Based on Elongation

Now calculate the allowable load based on the maximum elongation using the formula:\[ \Delta L = \frac{P L}{A E} \]Rearrange to solve for \(P\):\[ P = \frac{\Delta L \times A \times E}{L} \]Given \(\Delta L = 3 \text{ mm} = 0.003 \text{ m}\), \(E = 75 \text{ GPa} = 75 \times 10^9 \text{ N/m}^2\), and \(L = 3.8 \text{ m}\):\[ P = \frac{0.003 \times 3.14159 \times 10^{-6} \times 75 \times 10^9}{3.8} = 185.317 \text{ N} \]
04

Determine the Maximum Allowable Load

To ensure the wire does not exceed either stress or elongation limits, we take the smaller load from the two calculations:1. Based on stress: \(188.495 \text{ N}\)2. Based on elongation: \(185.317 \text{ N}\)Therefore, the allowable load is \(P_{\max} = 185.317 \text{ N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Axial Stress
Axial stress is a measure of the internal forces acting within a material. It results from an external load and represents the stress along the length of an object like a wire or rod.
Stress is calculated by dividing the force by the cross-sectional area over which it acts. The formula is \[ \sigma = \frac{P}{A} \]where:
  • \( \sigma \) is the axial stress (in pascals, \( N/m^2 \))
  • \( P \) is the applied load (in newtons, \( N \))
  • \( A \) is the cross-sectional area (in square meters, \( m^2 \))
A material's ability to withstand stress without breaking is crucial for ensuring structural integrity. For example, in the given problem, the maximum allowable axial stress is 60 MPa. Exceeding this stress could result in the wire failing or deforming.
Modulus of Elasticity
The modulus of elasticity, also known as Young's modulus, is a measure of a material's ability to deform elastically (i.e., non-permanently) when a force is applied. It is expressed as the ratio of tensile stress to strain and is a fundamental property that indicates the stiffness of the material.
Young's modulus is typically denoted by \( E \):\[ E = \frac{\sigma}{\varepsilon} \]where:
  • \( \sigma \) is the axial stress (in pascals, \( N/m^2 \))
  • \( \varepsilon \) is the strain (a dimensionless quantity)
In the exercise, the aluminum wire's modulus of elasticity is given as 75 GPa, or \( 75 \times 10^9 \) N/m\(^2\). This property helps in determining how much the wire will stretch (elongate) for a given tensile load, making it crucial for calculating the maximum permissible elongation.
Tensile Load
Tensile load refers to the force that attempts to stretch or lengthen an object. It's an important factor when determining how much load a structure can withstand before reaching its limit, where failure or permanent deformation might occur.
In the given calculus, tensile load \( P \) was determined based on two criteria:
  • The allowable axial stress: Ensures the load does not exceed the stress capacity outlined for the material.
  • The maximum permissible elongation: Keeps the elongation within safe limits based on the wire's physical properties and the modulus of elasticity.
By using tensile load calculations, one can precisely gauge these limits to maintain structural safety and performance.
Cross-sectional Area
The cross-sectional area of an object refers to the area of the specific section when you cut straight through a structure. In the context of wires or rods, it often refers to the area of the circle at any cross-section of its length.
To find the cross-sectional area \( A \) of a wire with diameter \( d \), the formula \[ A = \frac{\pi d^2}{4} \]is used, where \( \pi \) is a constant approximately equal to 3.14159.
For the aluminum wire with a diameter of 2 mm (or 0.002 m), the calculated area is approximately \( 3.14159 \times 10^{-6} \) m\(^2\). Understanding this area is crucial, as it directly influences the wire's capacity to withstand stress and strain while remaining structurally sound during load application.

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Most popular questions from this chapter

A flat bar of width \(b\) and thickness \(t\) has a hole of diameter \(d\) drilled through it (see figure). The hole may have any diameter that will fit within the bar. What is the maximum permissible tensile load \(P_{\max }\) if the allowable tensile stress in the material is \(\sigma_{t} ?\)

A steel wire and an aluminum alloy wire have equal lengths and support equal loads \(P\) (see figure). The moduli of elasticity for the steel and aluminum alloy are \(E_{s}=206 \mathrm{GPa}\) and \(E_{a}=76 \mathrm{GPa},\) respectively. (a) If the wires have the same diameters, what is the ratio of the elongation of the aluminum alloy wire to the elongation of the steel wire? (b) If the wires stretch the same amount, what is the ratio of the diameter of the aluminum alloy wire to the diameter of the steel wire? (c) If the wires have the same diameters and same load \(P,\) what is the ratio of the initial length of the aluminum alloy wire to that of the steel wire if the aluminum alloy wire stretches 1.5 times that of the steel wire? (d) If the wires have the same diameters, same initial length, and same load \(P,\) what is the material of the upper wire if it elongates 1.7 times that of the steel wire?

A circular steel rod \(A B\) of diameter \(d=15 \mathrm{mm}\) is stretched tightly between two supports so that initially the tensile stress in the rod is \(60 \mathrm{MPa}\) (see figure). An axial force \(P\) is then applied to the rod at an intermediate location \(C\) (a) Determine the plastic load \(P_{P}\) if the material is elastoplastic with yield stress \(\sigma_{Y}=290 \mathrm{MPa}\) (b) How is \(P_{P}\) changed if the initial tensile stress is doubled to 120 MPa?

A rigid steel plate is supported by three posts of high-strength concrete each having an effective crosssectional area \(A=40,000 \mathrm{mm}^{2}\) and length \(L=2 \mathrm{m}\) (see figure). Before the load \(P\) is applied, the middle post is shorter than the others by an amount \(s=1.0 \mathrm{mm}\) Determine the maximum allowable load \(P_{\text {allow if }}\) if the allowable compressive stress in the concrete is \(\sigma_{\text {allow }}=20\) MPa. (Use \(E=30\) GPa for concrete.)

A steel bar of square cross section \((50 \mathrm{mm} \times 50 \mathrm{mm})\) carries a tensile load \(P\) (see figure) The allowable stresses in tension and shear are \(125 \mathrm{MPa}\) and \(76 \mathrm{MPa}\), respectively. Determine the maximum permissible load \(P_{\max }\).
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