A cylindrical rod of steel \((E=207 \mathrm{GPa}, 30 \times\) \(\left.10^{6} \text { psi }\right)\) having a yield strength of \(310 \mathrm{MPa}\) \((45,000 \mathrm{psi})\) is to be subjected to a load of \(11,100 \mathrm{N}\left(2500 \mathrm{lb}_{\mathrm{f}}\right) .\) If the length of the rod is \(500 \mathrm{mm}(20.0 \text { in. }),\) what must be the diameter to allow an elongation of \(0.38 \mathrm{mm}(0.015 \text { in. }) ?\)

To simplify calculations, we will convert the given values to consistent units. In this case, we will use the SI units (meter, Newton, etc.). The given values are: - Load: \(F = 11,100 N\) - Length: \(L = 500 mm = 0.5 m\) - Elongation: \(\delta_L = 0.38 mm = 0.00038 m\) - Young's modulus: \(E = 207 GPa = 207\times10^9 N/m^2\) - Yield strength: \(\sigma_Y = 310 MPa = 310\times10^6 N/m^2\)

We are given the elongation and we need to find the diameter of the rod. To do this, we will first find the stress acting on the rod, using the following formula which describes the relationship between elongation, stress, Young's modulus, and the original length of the rod: \[\frac{\delta_L}{L} = \frac{\sigma}{E}\] where \(\delta_L\) is the elongation, \(L\) is the length of the rod, \(\sigma\) is the stress in the rod, and \(E\) is the Young's modulus. We can solve for the stress, \(\sigma\): \[\sigma = \frac{\delta_L}{L} \times E\] Now, substitute the given values into the formula: \[\sigma = \frac{0.00038 \text{ m}}{0.5 \text{ m}} \times 207\times10^9 N/m^2\] \[\sigma \approx 159 \times 10^6 N/m^2\] The calculated stress in the rod is approximately \(159 MPa\).

Since the steel rod will be subjected to a tensile load, we can use the stress formula to relate the stress, tensile load, and cross-sectional area. The formula is: \[\sigma = \frac{F}{A}\] where \(\sigma\) is the stress in the rod, \(F\) is the tensile load, and \(A\) is the cross-sectional area of the rod. We can solve for the cross-sectional area, \(A\): \[A = \frac{F}{\sigma}\] Substitute the given values into the formula: \[A = \frac{11,100 N}{159 \times 10^6 N/m^2}\] \[A \approx 6.98 \times 10^{-5} m^2\] The required cross-sectional area of the steel rod is approximately \(6.98 \times 10^{-5} m^2\).

Now we can use the cross-sectional area of the cylindrical rod to find its diameter. The formula for the cross-sectional area of a cylinder is: \[A = \frac{1}{4}\pi d^2\] where \(A\) is the cross-sectional area and \(d\) is the diameter of the rod. We can solve for the diameter, \(d\): \[d = 2\sqrt{\frac{A}{\pi}}\] Substitute the value of the cross-sectional area that we have calculated: \[d = 2\sqrt{\frac{6.98 \times 10^{-5} m^2}{\pi}}\] \[d \approx 0.00949 m\] The required diameter of the steel rod is approximately \(9.49 mm (0.374 \text{ in.})\) to allow an elongation of \(0.38 mm\) under the given tensile load.