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Chapter 6: Problem 5

A steel bar \(100 \mathrm{~mm}\) (4.0 in.) long and having a square cross section \(20 \mathrm{~mm}(0.8 \mathrm{in} .)\) on an edge is pulled in tension with a load of 89,000 \(\mathrm{N}\left(20,000 \mathrm{lb}_{\mathrm{f}}\right)\) and experiences an elongation of \(0.10 \mathrm{~mm}\left(4.0 \times 10^{-3}\right.\) in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.

Short Answer

Expert verified
#tag_title#Step 5: Plug in the values and calculate (Continued)#tag_content# Plugging in the given values into the equation, we get: $$ E = \frac{\frac{89,000 \mathrm{~N}}{(20 \mathrm{~mm})^2}}{\frac{0.10 \mathrm{~mm}}{100 \mathrm{~mm}}} $$ Solving for E: $$ E = \frac{89,000 \mathrm{~N}}{400 \mathrm{~mm}^2} \times \frac{100 \mathrm{~mm}}{0.10 \mathrm{~mm}} $$ $$ E = \frac{89,000 \mathrm{~N}}{400 \mathrm{~mm}^2} \times 1000 $$ $$ E = 222.5 \times 10^6 \mathrm{~N/mm}^2 $$ So the elastic modulus of the steel is \(222.5 \times 10^6 \mathrm{~N/mm}^2\).

Step by step solution

01

Identify the given parameters

Given parameters are: - Length of the steel bar, \(L = 100 \mathrm{~mm}\) - Cross-sectional area, \(A = (20 \mathrm{~mm})^2\) - Applied force, \(F = 89,000 \mathrm{~N}\) - Elongation, \(\Delta L = 0.10 \mathrm{~mm}\)
02

Calculate the stress

To begin, let's calculate the stress in the bar, $$ \text{Stress} = \frac{\text{Applied Force}}{\text{Cross-sectional Area}} $$ $$ \sigma = \frac{F}{A} $$
03

Calculate the strain

Next, we need to find the strain in the steel bar. We can use the formula, $$ \text{Strain} = \frac{\text{Elongation}}{\text{Original Length}} $$ $$ \epsilon = \frac{\Delta L}{L} $$
04

Find the elastic modulus

Now that we have stress and strain, we can use Hooke's Law to find the elastic modulus. Hooke's Law states that stress is proportional to strain, and the proportionality constant is the elastic modulus (also known as Young's modulus). Mathematically, it can be written as: $$ \sigma = E \times \epsilon $$ Rearranging the formula to isolate the elastic modulus (E), we get: $$ E = \frac{\sigma}{\epsilon} $$
05

Plug in the values and calculate

Now, let's plug in the values we found earlier for stress and strain, and calculate the elastic modulus: $$ E = \frac{F/A}{\Delta L/L} $$ With the given parameters, $$ E = \frac{\frac{89,000 \mathrm{~N}}{(20 \mathrm{~mm})^2}}{\frac{0.10 \mathrm{~mm}}{100 \mathrm{~mm}}} $$ Now, calculate the elastic modulus E.

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Most popular questions from this chapter

A cylindrical rod \(100 \mathrm{~mm}\) long and having a diameter of \(10.0 \mathrm{~mm}\) is to be deformed using a tensile load of \(27,500 \mathrm{~N}\). It must not experience either plastic deformation or a diameter reduction of more than \(7.5 \times 10^{-3} \mathrm{~mm}\). Of the materials listed as follows, which are possible candidates? Justify your choice(s). $$ \begin{array}{lccc} \hline \begin{array}{c} \text { Material } \end{array} & \begin{array}{c} \text { Modulus } \\ \text { of Elasticity } \\ \text { (GPa) } \end{array} & \begin{array}{c} \text { Yield } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { Poisson's } \\ \text { Ratio } \end{array} \\ \hline \text { Aluminum alloy } & 70 & 200 & 0.33 \\ \text { Brass alloy } & 101 & 300 & 0.34 \\ \text { Steel alloy } & 207 & 400 & 0.30 \\ \text { Titanium alloy } & 107 & 650 & 0.34 \\ \hline \end{array} $$

A specimen of aluminum having a rectangular cross section \(10 \mathrm{~mm} \times 12.7 \mathrm{~mm}(0.4\) in. \(\times 0.5\) in.) is pulled in tension with \(35,500 \mathrm{~N}\) \(\left(8000 \mathrm{lb}_{\mathrm{f}}\right)\) force, producing only elastic deformation. Calculate the resulting strain.

A cylindrical specimen of an alloy \(8 \mathrm{~mm}\) (0.31 in.) in diameter is stressed elastically in tension. A force of \(15,700 \mathrm{~N}\) (3530 lb_{f } \()\) produces a reduction in specimen diameter of \(5 \times 10^{-3} \mathrm{~mm}\left(2 \times 10^{-4}\right.\) in.). Compute Poisson's ratio for this material if its modulus of elasticity is \(140 \mathrm{GPa}\left(20.3 \times 10^{6} \mathrm{psi}\right)\).

Estimate the Brinell and Rockwell hardnesses for the following: (a) The naval brass for which the stressstrain behavior is shown in Figure \(6.12\). (b The steel alloy for which the stress-strain behavior is shown in Figure \(6.21\).

A cylindrical specimen of a titanium alloy having an elastic modulus of \(107 \mathrm{GPa}\) (15.5 \(\times\) \(10^{6} \mathrm{psi}\) ) and an original diameter of \(3.8 \mathrm{~mm}\) (0.15 in.) will experience only elastic deformation when a tensile load of \(2000 \mathrm{~N}\) \(\left(450 \mathrm{lb}_{\mathrm{f}}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.42\) \(\mathrm{mm}(0.0165\) in.).
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