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While removing a wheel to change a tire, a driver applies forces \(P=100 \mathrm{N}\) at the ends of two of the arms of a lug wrench (see figure). The wrench is made of steel with shear modulus of elasticity \(G=78\) GPa. Each arm of the wrench is \(255 \mathrm{mm}\) long and has a solid circular cross section of diameter \(d=12 \mathrm{mm}\) (a) Determine the maximum shear stress in the arm that is turning the lug nut \((\operatorname{arm} A)\) (b) Determine the angle of twist (in degrees) of this same arm.

Short Answer

Expert verified
(a) Maximum shear stress is 60.18 MPa. (b) Angle of twist is 0.29 degrees.

Step by step solution

01

Calculate Polar Moment of Inertia

The polar moment of inertia for a solid circular cross section is given by the formula: \[ J = \frac{\pi}{32} d^4 \] Substitute the diameter: \[ J = \frac{\pi}{32} (0.012)^4 = 2.545 \times 10^{-9} \, \text{m}^4 \]
02

Calculate Torque

Since the force is applied at the end of the arm, the torque \(T\) is given by:\[ T = P \times L \] Where \(L\) is the length of the arm:\[ T = 100 \, \text{N} \times 0.255 \, \text{m} = 25.5 \, \text{Nm} \]
03

Calculate Maximum Shear Stress

The shear stress \(\tau\) in a circular shaft subjected to a torque is given by:\[ \tau = \frac{T \cdot c}{J} \] Where \(c\) is the radius:\[ \tau = \frac{25.5 \, \text{Nm} \times 0.006 \, \text{m}}{2.545 \times 10^{-9} \, \text{m}^4} = 60.18 \, \text{MPa} \]
04

Determine Angle of Twist

The angle of twist \(\theta\) of a shaft is given by:\[ \theta = \frac{T \cdot L}{J \cdot G} \] Where \(G\) is the shear modulus:\[ G = 78 \, \text{GPa} = 78 \times 10^9 \, \text{Pa} \]\[ \theta = \frac{25.5 \, \text{Nm} \times 0.255 \, \text{m}}{2.545 \times 10^{-9} \, \text{m}^4 \times 78 \times 10^9 \, \text{Pa}} = 0.005078 \, \text{radians} \] Now convert radians to degrees:\[ \theta = 0.005078 \times \frac{180}{\pi} \approx 0.29 \text{ degrees} \]

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Key Concepts

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

Shear Stress
Shear stress is a critical concept in the study of mechanics of materials. It describes the stress component parallel to the surface of a material. Imagine you're pushing a book across a table; the effort required to start its motion is akin to overcoming shear stress. In mechanical terms, shear stress (\(\tau\)) in a circular shaft is determined by how much torque is applied and the geometry of the shaft.
For a solid circular cross section, shear stress is calculated using the formula:
  • \(\tau = \frac{T \cdot c}{J}\)
where \(T\) is torque, \(c\) is the radius of the circular section, and \(J\) is the polar moment of inertia.
In our lug wrench problem, applying a force at the end of each arm creates torque, wrenching the shaft, and generating shear stress. Recognizing the components and relationships in the formula helps in understanding how these forces act on mechanical structures. This knowledge aids in designing materials that can withstand specific stress levels without failing.
Angle of Twist
The angle of twist refers to the rotational displacement observed in a shaft when subjected to torque. Consider twisting a towel to wring it; the degree to which different parts twist relative to each other is akin to the angle of twist.
This angle is crucial for determining how much a shaft can rotate without causing damage or failure.
  • The mathematical representation is: \(\theta = \frac{T \cdot L}{J \cdot G}\)
where \(\theta\) is the angle of twist, \(T\) is torque, \(L\) is the length, \(J\) is the polar moment of inertia, and \(G\) is the shear modulus.
For the wrench used in this exercise, small angular rotations like the resulting \(0.29\) degrees ensure the functionality of the tool without compromising its structural integrity. This aspect is particularly significant in designing mechanical tools and components, ensuring they perform effectively under varied conditions.
Polar Moment of Inertia
The polar moment of inertia (\(J\)) is an essential property indicating an object's ability to resist torsional deformation. Think of it as a measure of how the distribution of material in a cross-sectional area contributes to its stiffness against twisting.
For circular shafts, the formula is:
  • \(J = \frac{\pi}{32} \cdot d^4\)
where \(d\) is the diameter. Note how the diameter greatly influences the polar moment, elevated to the fourth power, emphasizing the importance of geometry in mechanical applications.
In the lug wrench scenario, knowing the polar moment helps ascertain how well the tool can withstand the applied torque without excessive twisting, thereby ensuring safe and efficient mechanical operations.
Torque Calculation
Torque is the measure of rotational force applied to an object, akin to using a doorknob: turning it involves applying torque. Accurately calculating torque is crucial to ensure mechanical components function efficiently and safely.
The formula for torque when force is applied at a point is:
  • \(T = P \times L\)
where \(P\) is the force applied, and \(L\) is the distance from the pivot point or axis.
In many mechanical contexts, such as our lug wrench problem, the proper calculation and application of torque allow for secure fastening or loosening of components. Understanding torque's role aids in ensuring the appropriate force is applied, safeguarding against overtightening or insufficient torque that might lead to failure of mechanical connections.

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Most popular questions from this chapter

A high-strength steel drill rod used for boring a hole in the earth has a diameter of \(12 \mathrm{mm}\) (see figure). The allowable shear stress in the steel is \(300 \mathrm{MPa}\) and the shear modulus of elasticity is \(80 \mathrm{GPa}\) (a) What is the minimum required length of the rod so that one end of the rod can be twisted \(30^{\circ}\) with respect to the other end without exceeding the allowable stress? (b) If the shear strain in part (a) is limited to \(3.2 \times 10^{-3},\) what is the minimum required length of the drill rod?

A hollow circular tube \(A\) fits over the end of a solid circular bar \(B\), as shown in the figure. The far ends of both bars are fixed. Initially, a hole through bar \(B\) makes an angle \(\beta\) with a line through two holes in tube \(A\). Then bar \(B\) is twisted until the holes are aligned, and a pin is placed through the holes. When bar \(B\) is released and the system returns to equilibrium, what is the total strain energy \(U\) of the two bars? (Let \(I_{P A}\) and \(I_{P B}\) represent the polar moments of inertia of bars \(A\) and \(B,\) respectively. The length \(L\) and shear modulus of elasticity \(G\) are the same for both bars.)

A uniformly tapered tube \(A B\) of hollow circular cross section is shown in the figure. The tube has constant wall thickness \(t\) and length \(L\). The average diameters at the cnds are \(d_{A}\) and \(d_{B}=2 d_{A}\). The polar moment of inertia may be represented by the approximate formula \\[ I_{P}=\pi d^{3} t / 4[\text { see Eq. }(3-21)] \\] Derive a formula for the angle of twist \(\phi\) of the tube when it is subjected to torques \(T\) acting at the ends.

A generator shaft in a small hydroelectric plant turns at 120 rpm and delivers \(38 \mathrm{kW}\) (see figure). (a) If the diameter of the shaft is \(d=75 \mathrm{mm},\) what is the maximum shear stress \(\tau_{\max }\) in the shaft? (b) If the shear stress is limited to \(28 \mathrm{MPa}\), what is the minimum permissible diameter \(d_{\min }\) of the shaft?

A thin-walled steel tube having an elliptical cross section with constant thickness \(t\) (see figure) is subjected to a torque \(T=5.5 \mathrm{kN} \cdot \mathrm{m}\) Determine the shear stress \(\tau\) and the rate of twist \(\theta\) (in degrees per meter) if \(G=83 \mathrm{GPa}, t=5 \mathrm{mm}, a=75 \mathrm{mm}\) and \(b=50 \mathrm{mm}\). (Note: See Appendix D, Case \(16,\) for the properties of an ellipse.)

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