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Chapter 7: Problem 5

An aluminum bar \(125 \mathrm{~mm}(5.0\) in.) long and having a square cross section \(16.5 \mathrm{~mm}(0.65 \mathrm{in} .)\) on an edge is pulled in tension with a load of 66,700 \(\mathrm{N}\left(15,000 \mathrm{lb}_{f}\right)\) and experiences an elongation of \(0.43 \mathrm{~mm}\left(1.7 \times 10^{-2}\right.\) in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.

Short Answer

Expert verified
Question: Calculate the modulus of elasticity of an aluminum bar with dimensions 16.5 mm x 16.5 mm x 125 mm when a force of 66,700 N is applied, resulting in an elongation of 0.43 mm. Answer: The modulus of elasticity of the aluminum bar is approximately 68.27 x 10^9 Pa.

Step by step solution

01

Calculate the cross-sectional area of the square bar

The cross-sectional area (A) of a square is given by the formula: \(A = a^2\), where \(a\) is the edge of the square. In our case, \(a = 16.5 \mathrm{mm}\), so we have: \(A = (16.5 \mathrm{mm})^2 = 272.25 \mathrm{mm}^2\)
02

Calculate the stress on the aluminum bar

Stress (σ) is the force applied (F) divided by the cross-sectional area (A) of the bar: \(σ = \dfrac{F}{A}\) In our case, we have \(F = 66,700 \mathrm{N}\) and \(A = 272.25 \mathrm{mm}^2\). Plugging these values into the equation, we get: \(σ = \dfrac{66,700 \mathrm{N}}{272.25 \mathrm{mm}^2} = 234,678.9 \dfrac{\mathrm{N}}{\mathrm{mm}^2}\)
03

Calculate the strain experienced by the aluminum bar

Strain (ε) is the change in length (ΔL) divided by the original length (L) of the bar: \(ε = \dfrac{ΔL}{L}\) In our case, we have \(ΔL = 0.43 \mathrm{mm}\) and \(L = 125 \mathrm{mm}\). Plugging these values into the equation, we get: \(ε = \dfrac{0.43 \mathrm{mm}}{125 \mathrm{mm}} = 0.00344\)
04

Calculate the modulus of elasticity of the aluminum

Now we can use the formula for the modulus of elasticity (E): \(E = \dfrac{σ}{ε}\) Plugging in the values we calculated for stress and strain, we get: \(E = \dfrac{234,678.9 \dfrac{\mathrm{N}}{\mathrm{mm}^2}}{0.00344} \approx 68.27 \times 10^9 \mathrm{Pa}\) So, the modulus of elasticity of the aluminum is approximately \(68.27 \times 10^9 \mathrm{Pa}\).

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Most popular questions from this chapter

In Section \(2.6\), it was noted that the net bonding energy \(E_{N}\) between two isolated positive and negative ions is a function of interionic distance \(r\) as follows: $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ where \(A, B\), and \(n\) are constants for the particular ion pair. Equation \(6.31\) is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity \(E\) is proportional to the slope of the interionic force-separation curve at the equilibrium interionic separation; that is,Derive an expression for the dependence of the modulus of elasticity on these \(A, B\), and \(n\) parameters (for the two-ion system), using the following procedure: 1\. Establish a relationship for the force \(F\) as a function of \(r\), realizing that $$ F=\frac{d E_{N}}{d r} $$ 2\. Now take the derivative \(d F / d r\). 3\. Develop an expression for \(r_{0}\), the equilibrium separation. Because \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N}\)-versus- \(r\) curve (Figure 2.10b), take the derivative \(d E_{N} / d r\), set it equal to zero, and solve for \(r\), which corresponds to \(r_{0}\) - 4\. Finally, substitute this expression for \(r_{0}\) into the relationship obtained by taking \(d F / d r\).

A cylindrical specimen of steel having a diameter of \(15.2 \mathrm{~mm}(0.60\) in.) and length of 250 \(\mathrm{mm}(10.0 \mathrm{in} .)\) is deformed elastically in tension with a force of \(48,900 \mathrm{~N}\left(11,000 \mathrm{lb}_{e}\right)\). Using the data contained in Table \(6.1\), determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Will the diameter increase or decrease?

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are \(30.00\) and \(30.04 \mathrm{~mm}\), respectively, and its final length is \(105.20 \mathrm{~mm}\), compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are \(65.5\) and \(25.4\) GPa, respectively.

(a) Compare planar densities (Section \(3.11\) and Problem 3.60) for the \((100),(110)\), and (111) planes for FCC. (b) Compare planar densities (Problem 3.61) for the (100), (110), and (111) planes for BCC.

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is \(2.4 \mathrm{MPa}\).
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