/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 The solid shaft has a linear tap... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 43

The solid shaft has a linear taper from \(r_{A}\) at one end to \(r_{B}\) at the other. Derive an equation that gives the maximum shear stress in the shaft at a location \(x\) along the shaft's axis.

Short Answer

Expert verified
The maximum shear stress in a solid, linearly tapered shaft along the shaft's axis, \(x\), is given by: \(\tau_{max} (x) = \frac{2T}{\pi r(x)^3}\), where \(r(x) = r_A - x \cdot k\), \(k = \frac{r_A - r_B }{L}\), \(T\) is the torque, \(r_A\) and \(r_B\) are the radii at the ends of the shaft, and \(L\) is the total length of the shaft.

Step by step solution

01

Write down the formula and variables

Start by writing down the formula for shear stress in a circular shaft: \(\tau = \frac{T \cdot r}{J}\). The variables are as follows: \(T\) is the torque, \(r\) is the radius (which depends on \(x\), so it's given as \(r(x)\)), and \(J\) is the polar moment of inertia, which for a solid circular cross-section is given by \(\frac{\pi}{2} r^4\).
02

Express \(\tau\) as a function of \(x\)

The radius \(r\) decreases linearly along the length of the shaft, so it can be expressed as a function of \(x\). Given the problem statement, \(r(x) = r_A - x \cdot k\), where \(k = \frac{r_A - r_B }{L}\) is the shaft's constant rate of change, and \(L\) is the length of the shaft.
03

Substitute 'r' in 'J' and '蟿'

Substitute \(r(x)\) into the formulas for \(J\) and \(\tau\) and simplify. The expressions become: \(J(x) = \frac{\pi}{2} r(x)^4\) and \(\tau(x) = \frac{T \cdot r(x)}{J(x)}\), respectively.
04

Final expression for maximum shear stress

Substitute \(J(x)\) into the expression for \(\tau(x)\), simplify and obtain a function for the maximum shear stress: \(\tau_{max} (x) = \frac{2T}{\pi r(x)^3}\). This function expresses the maximum shear stress at any point \(x\) along the shaft鈥檚 axis.

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Most popular questions from this chapter

The solid shaft of radius \(c\) is subjected to a torque \(\mathbf{T}\) at its ends. Show that the maximum shear strain in the shaft is \(\gamma_{\max }=T c / J G .\) What is the shear strain on an element located at point \(A, c / 2\) from the center of the shaft? Sketch the shear strain distortion of this element.

The steel shaft is made from two segments: \(A C\) has a diameter of 0.5 in, and \(C B\) has a diameter of 1 in. If the shaft is fixed at its ends \(A\) and \(B\) and subjected to a torque of \(500 \mathrm{lb} \cdot \mathrm{ft},\) determine the maximum shear stress in the shaft. \(G_{\mathrm{st}}=10.8\left(10^{3}\right) \mathrm{ksi}\)

The gear motor can develop \(\frac{1}{10}\) hp when it turns at 300 rev \(/\) min. If the shaft has a diameter of \(\frac{1}{2}\) in., determine the maximum shear stress in the shaft

A bar having a circular cross section of 3 in. diameter is subjected to a torque of 100 in. \(\cdot\) kip. If the material is elastic perfectly plastic, with \(\tau_{Y}=16 \mathrm{ksi}\) determine the radius of the elastic core.

A tubular shaft has an inner diameter of \(60 \mathrm{mm},\) an outer diameter of \(80 \mathrm{mm},\) and a length of \(1 \mathrm{m}\). It is made of an elastic perfectly plastic material having a yield stress of \(\tau_{Y}=150\) MPa. Determine the maximum torque it can transmit. What is the angle of twist of one end with respect to the other end if the inner surface of the tube is about to yield? \(G=75 \mathrm{GPa}\)
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