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Chapter 25: Problem 2803

A solid cylindrical steel column is \(4 \mathrm{~m}\) long and \(9 \mathrm{~cm}\) in diameter $\cdot\left(\mathrm{Y}_{\text {steel }}=1.9 \times 10^{11} \mathrm{Nm}^{-2}\right)$. The decrease in length of the column, while carrying a load of \(80000 \mathrm{~kg}\) is........ (A) \(3.2 \mathrm{~mm}\) (B) \(1.8 \mathrm{~mm}\) (C) \(4.4 \mathrm{~mm}\) (D) \(2.6 \mathrm{~mm}\)

Short Answer

Expert verified
The decrease in length of the column while carrying a load of $80000 \mathrm{~kg}$ is approximately \(1.8 \mathrm{~mm}\) (Option B).

Step by step solution

01

Calculate Area of Cross-Section

We are given the diameter of the cylindrical column as \(9 cm\). To calculate the area of the cross-section A, let's first convert diameter to radius. radius \(r = \frac{diameter}{2} = \frac{9}{2} cm\) Now, the area of the cross-section can be calculated as: A = \(\pi r^2\) A = \(\pi\left(\frac{9}{2}\right)^2 cm^2\) A \(= \pi\left(\frac{81}{4}\right) cm^2\) Now, convert area A to \(m^2\): A \(= \pi\left(\frac{81}{4} \cdot 10^{-4}\right) m^2\)
02

Calculate Stress

A force of \(80000 kg\) is applied on the column, we need to convert this to Newtons (N) using the gravitational acceleration (\(g = 9.81 m/s^2\)). Force F = mass × acceleration_due to gravity F = \(80000 kg × 9.81 m/s^2\) F = \(784800 N\) Now, calculate the stress (σ) applied on the column using the formula: Stress σ = \(\frac{Force (F)}{Area (A)}\) σ = \(\frac{784800 N}{\pi\left(\frac{81}{4} \cdot 10^{-4}\right) m^2}\)
03

Calculate Strain

To calculate the strain (ε) in the column, use Young's modulus (Y): Strain ε = \(\frac{Stress (σ)}{Young's Modulus (Y)}\) ε = \(\frac{\frac{784800 N}{\pi\left(\frac{81}{4} \cdot 10^{-4}\right) m^2}}{1.9 \times 10^{11} Nm^{-2}}\)
04

Find Decrease in Length

Lastly, find the decrease in length of the column by multiplying the strain (ε) and the initial length (L) Decrease in length = Strain (ε) × Initial Length (L) Decrease in length = ε × \(4 m\) Decrease in length = \(\frac{\frac{784800 N}{\pi\left(\frac{81}{4} \cdot 10^{-4}\right) m^2}}{1.9 \times 10^{11} Nm^{-2}} \times 4 m\) Plug in the values and evaluate the expression to find the decrease in length (in meters) and then convert it to millimeters. The result will be the correct answer among the options provided.

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Most popular questions from this chapter

A uniform rod of length \(L\) and density \(\rho\) is being pulled along A smooth floor with a horizontal acceleration \(\alpha\) what is the magnitude of the stress at the transverse cross-section through the midpoint of the rod? (A) \((1 / 3) \operatorname{L} \rho \alpha\) (B) \(\operatorname{L} \rho \alpha\) (C) \((2 / 3) \mathrm{L} \rho \rho\) (D) \(\\{(\operatorname{L} \rho \alpha) / 2\\}\)

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