/*! 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 137 A torque is applied to the shaft... [FREE SOLUTION] | 91Ó°ÊÓ

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A torque is applied to the shaft having a radius of \(80 \mathrm{mm} .\) If the material obeys a shear stress-strain relation of \(\tau=500 \gamma^{1 / 4}\) MPa, determine the torque that must be applied to the shaft so that the maximum shear strain becomes 0.008 rad.

Short Answer

Expert verified
The torque that must be applied to the shaft for the maximum shear strain to become 0.008 rad is calculated by first finding the maximum shear stress from the given stress-strain relation, converting units appropriately, and then using this along with the given radius in the formula for torque. The calculated value is the required torque.

Step by step solution

01

Apply the expression for shear stress

First, we apply the given shear stress-strain relation \( \tau = 500 \gamma^{1 / 4} \) MPa. Here, \( \tau \) is the shear stress and \( \gamma \) is the shear strain. Plugging the given maximum shear strain of 0.008 rad into this equation, we find the corresponding maximum shear stress to be \( \tau_{max} = 500 (0.008)^{1 / 4} \) MPa.
02

Convert units

It's very important to note that 1 MPa is equal to \( 10^6 \) N/m^2, and we should transform our \( \tau_{max} \) into the basic unit to be compatible with the formula for torque. Therefore, \( \tau_{max} = 500 (0.008)^{1 / 4} \times 10^6 \) N/m^2.
03

Apply the formula for torque

The formula characterizing the relationship between torque, shear stress and radius is \( T = \tau_{max} r^2 \pi \), where r is the radius. Note that the radius is given in mm, so it should be converted into meters, \( r = 80/1000 \) m. Substituting the values, we get the torque \( T = \tau_{max} (80/1000)^2 \pi \) N.m.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque Calculation
Torque is an essential concept in material mechanics, especially when dealing with shafts and rotating systems. It measures how much a force acting on an object causes that object to rotate. The formula to calculate torque \( T \) when dealing with circular shafts involves the maximum shear stress \( \tau_{max} \) and the radius \( r \) of the shaft:

\[T = \tau_{max} \cdot r^2 \cdot \pi\]This indicates that torque is directly proportional to the square of the radius. Therefore, even small increases in the radius can significantly affect the amount of torque required. When calculating torque, it's also critical to use consistent units to ensure accuracy.
  • Shear stress must be in N/m² (Pascal).
  • The radius should always be in meters.
Converting units is an essential step; for example, if the radius is given in millimeters, such as 80 mm, it needs to be converted to meters (80/1000 m) to fit into the equation properly.

Understanding this basic calculation helps in designing and assessing shafts and other rotating structures, ensuring they can withstand applied forces without failing.
Material Mechanics
Material mechanics is a branch of mechanical engineering that deals with how materials react under stresses such as tension, compression, and shear. A key aspect of material mechanics is understanding the relationship between stress and strain in materials.

Materials have certain properties that dictate how they respond to various forces:
  • Elasticity: Determines how a material returns to its original shape after deformation.
  • Plasticity: Indicates how a material undergoes permanent deformation.
  • Strength: The ability of a material to withstand an applied stress without failure.
  • Hardness: Resistance of a material to deformation or penetration.
In the context of the exercise, material mechanics focuses on the shear stress-strain relationship within the shaft. The specific relation \( \tau = 500 \cdot \gamma^{1/4} \) MPa describes how shear stress \( \tau \) changes with shear strain \( \gamma \). Knowing these relationships helps in predicting how a material will behave under different conditions and is crucial for ensuring safety and performance in engineering applications.
Shear Strain
Shear strain is a measurement of deformation reflecting the change in shape that a material undergoes when subjected to shear stress. This concept is critical in understanding how materials respond to forces that cause them to slide parallel past each other.

In our example, the shear strain \( \gamma \) is linked to shear stress through the relationship given by \( \tau = 500 \cdot \gamma^{1/4} \) MPa. Here, this relationship indicates a nonlinear behavior where shear stress doesn't increase proportionally with shear strain.
  • The exponent \( 1/4 \) in the equation suggests materials with this behavior can be quite ductile, showing significant deformation before failure.
  • Unlike linear elastic materials that obey Hooke's Law, this nonlinear relationship highlights the complexity of real-world materials.
The maximum shear strain given as \( 0.008 \) rad indicates the extent to which the material will deform under maximum allowable conditions. By managing shear strain, engineers can predict how the material will behave during operation, allowing them to design components that remain safe and functional under service loads.

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Most popular questions from this chapter

The tubular drive shaft for the propeller of a hovercraft is \(6 \mathrm{m}\) long. If the motor delivers \(4 \mathrm{MW}\) of power to the shaft when the propellers rotate at 25 rad \(/\) s, determine the required inner diameter of the shaft if the outer diameter is \(250 \mathrm{mm}\). What is the angle of twist of the shaft when it is operating? Take \(\tau_{\text {allow }}=90 \mathrm{MPa}\) and \(G=75 \mathrm{GPa}\)

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? \(G=75 \mathrm{GPa}\)

A circular shaft having a diameter of 4 in. is subjected to a torque of 250 kip \(\cdot\) in. If the material is elastic perfectly plastic, with \(\tau_{Y}=16\) ksi, determine the radius of the elastic core.

The drive shaft \(A B\) of an automobile is made of a steel having an allowable shear stress of \(\tau_{\text {allow }}=8\) ksi. If the outer diameter of the shaft is 2.5 in. and the engine delivers 200 hp to the shaft when it is turning at 1140 rev / min, determine the minimum required thickness of the shaft's wall.

The shaft has a radius \(c\) and is subjected to a torque per unit length of \(t_{0},\) which is distributed uniformly over the shaft's entire length \(L\). If it is fixed at its far end \(A\), determine the angle of twist \(\phi\) of end \(B\). The shear modulus is \(G\)

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