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Two gage marks are placed exactly \(250 \mathrm{mm}\) apart on a 12-mm-diameter aluminum rod with \(E=73 \mathrm{GPa}\) and an ultimate strength of \(140 \mathrm{MPa}\). Knowing that the distance between the gage marks is \(250.28 \mathrm{mm}\) after a load is applied, determine \((a)\) the stress in the rod, \((b)\) the factor of safety.

Short Answer

Expert verified
The stress in the rod is 81.76 MPa, and the factor of safety is approximately 1.71.

Step by step solution

01

Calculate the Strain

To find the strain, use the formula for normal strain, which is the change in length divided by the original length. The change in length is given by the new length (250.28 mm) minus the original length (250 mm). Therefore, the strain \( \epsilon \) can be calculated as:\[ \epsilon = \frac{{250.28 - 250}}{{250}} = \frac{{0.28}}{{250}} \approx 0.00112 \]
02

Calculate the Stress

Stress is related to strain through Hooke's Law, \( \sigma = E \cdot \epsilon \). We can use this expression to find the stress \( \sigma \):\[ \sigma = 73 \text{ GPa} \times 0.00112 = 0.08176 \text{ GPa} = 81.76 \text{ MPa} \]
03

Determine the Factor of Safety

The factor of safety is the ratio of the ultimate strength to the actual stress. It can be calculated as:\[ \text{Factor of Safety} = \frac{{\text{Ultimate Strength}}}{{\text{Actual Stress}}} = \frac{{140 \text{ MPa}}}{{81.76 \text{ MPa}}} \approx 1.71 \]

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Key Concepts

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

Strain Calculation
Strain is a measure of deformation representing the elongation or compression of a material. It is crucial in the mechanics of materials to determine if the material can withstand the applied load without failing. Strain occurs when a material is subjected to a load, causing it to change its length.
In this case, the original length of the aluminum rod is 250 mm, and after applying a load, the length becomes 250.28 mm. The formula to calculate strain, denoted by \( \epsilon \), is:
  • Change in length \( \Delta L = 250.28 \, \text{mm} - 250 \, \text{mm} = 0.28 \, \text{mm}\)
  • Original length \( L_0 = 250 \, \text{mm} \)
  • Strain \( \epsilon = \frac{ \Delta L }{ L_0 } = \frac{ 0.28 }{ 250 } \approx 0.00112 \)
This value indicates how much the rod has stretched per unit length. It's important to note that strain is a dimensionless quantity and provides insight into the ductile or brittle nature of a material.
Stress Analysis
Stress is a measure of internal forces within a material when it is subjected to an external load. For engineers, performing stress analysis is essential to ensure structures can handle everyday stresses without failure.
Stress \( \sigma \) is calculated using Hooke’s Law, which states that the stress is equal to the product of the modulus of elasticity \( E \) and the strain \( \epsilon \). Given that the modulus of elasticity for aluminum, \( E \), is \( 73 \, \text{GPa} \), we calculate stress as follows:
  • \( \sigma = E \times \epsilon = 73 \, \text{GPa} \times 0.00112 \)
  • This gives \( \sigma = 81.76 \, \text{MPa} \)
Understanding stress values helps determine how close a material is to its ultimate strength. In this instance, the aluminum rod experiences a stress of 81.76 MPa, which needs to be compared against its ultimate strength to assess its safety.
Factor of Safety
The factor of safety (FoS) is a critical concept in engineering design, providing a margin between the applied stress and the material's strength to account for uncertainties in material properties, loading conditions, and potential flaws.
The factor of safety is defined as the ratio of the material's ultimate strength (the maximum stress it can withstand before failure) to the actual applied stress:
  • Ultimate strength for aluminum is \( 140 \, \text{MPa} \)
  • Calculated stress is \( 81.76 \, \text{MPa} \)
  • \( \text{FoS} = \frac{140}{81.76} \approx 1.71 \)
A factor of safety of 1.71 implies that the aluminum rod can sustain loads 1.71 times greater than the current stress before reaching its ultimate strength. Engineering applications often aim for specific FoS values depending on the criticality of the use; a higher FoS indicates a more conservative design.

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Most popular questions from this chapter

Denoting by \(\epsilon\) the "engineering strain" in a tensile specimen, show that the true strain is \(\epsilon_{t}=\ln (1+\epsilon)\)

In many situations physical constraints prevent strain from occurring in a given direction. For example, \(\epsilon_{z}=0\) in the case shown, where longitudinal movement of the long prism is prevented at every point. Plane sections perpendicular to the longitudinal axis remain plane and the same distance apart. Show that for this situation, which is known as plane strain, we can express \(\sigma_{z}, \epsilon_{x},\) and \(\epsilon_{y}\) as follows: \\[ \begin{array}{l} \sigma_{z}=\nu\left(\sigma_{x}+\sigma_{y}\right) \\ \epsilon_{x}=\frac{1}{E}\left[\left(1-\nu^{2}\right) \sigma_{x}-\nu(1+\nu) \sigma_{y}\right] \\ \epsilon_{y}=\frac{1}{E}\left[\left(1-\nu^{2}\right) \sigma_{y}-\nu(1+\nu) \sigma_{x}\right] \end{array} \\]

A block of 10 -in. length and \(1.8 \times 1.6\) -in. cross section is to support a centric compressive load \(\mathbf{P}\). The material to be used is a bronze for which \(E=14 \times 10^{6}\) psi. Determine the largest load that can be applied, knowing that the normal stress must not exceed 18 lssi and that the decrease in length of the block should be at most \(0.12 \%\) of its original length.

A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2 -kip tensile load is applied to it. Knowing that \(E=29 \times 10^{6} \mathrm{psi},\) determine \((a)\) the smallest diameter rod that should be used, ( \(b\) ) the corresponding normal stress caused by the load.

Bar \(A B\) has a cross-sectional area of \(1200 \mathrm{mm}^{2}\) and is made of a steel that is assumed to be elastoplastic with \(E=200 \mathrm{GPa}\) and \(\sigma_{Y}\) \(=250\) MPa. Knowing that the force \(\mathbf{F}\) increases from 0 to \(520 \mathrm{kN}\) and then decreases to zero, determine ( \(a\) ) the permanent deflection of point \(C,(b)\) the residual stress in the bar.

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