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Chapter 3: Problem 1

A prospector uses a hand-powered winch (see figure) to raise a bucket of ore in his mine shaft. The axle of the winch is a steel rod of diameter \(d=15 \mathrm{mm} .\) Also, the distance from the center of the axle to the center of the lifting rope is \(b=100 \mathrm{mm}\) (a) If the weight of the loaded bucket is \(W=400 \mathrm{N}\) what is the maximum shear stress in the axle due to torsion? (b) If the maximum bucket load is \(510 \mathrm{N}\) and the allowable shear stress in the axle is 65 MPa, what is the minimum permissible axle diameter?

Short Answer

Expert verified
(a) Maximum shear stress is 75.5 MPa. (b) Minimum diameter is 17.1 mm.

Step by step solution

01

Understanding Given Data

The problem involves calculating stresses in the axle of a winch. The data provided includes the weight of the bucket \( W = 400 \, \text{N} \), diameter of the axle \( d = 15 \, \text{mm} \), and distance from the center of the axle to the center of the lifting rope \( b = 100 \, \text{mm} \).
02

Calculate Torque

Torque \( T \) is the product of the force applied and the distance from the pivot point. Here, torque from the bucket weight is \( T = W \cdot b = 400 \, \text{N} \times 0.1 \, \text{m} = 40 \, \text{Nm} \).
03

Calculate Polar Moment of Inertia

For a circular rod, the polar moment of inertia \( J \) is given by \( J = \frac{\pi}{32} \cdot d^4 \). So, for \( d = 15 \, \text{mm} = 0.015 \, \text{m} \), \( J = \frac{\pi}{32} \times 0.015^4 = 3.976 \times 10^{-9} \text{m}^4 \).
04

Calculate Maximum Shear Stress

Using the formula for shear stress \( \tau = \frac{T \cdot c}{J} \), where \( c \) is the outer radius \( c = \frac{d}{2} = 0.0075 \, \text{m} \). Thus, \( \tau = \frac{40 \cdot 0.0075}{3.976 \times 10^{-9}} \approx 75.5 \times 10^6 \, \text{Pa} = 75.5 \, \text{MPa} \).
05

Calculate Minimum Diameter for Maximum Load

For the maximum load \( W = 510 \, \text{N} \) with allowable shear stress \( \tau_{allow} = 65 \, \text{MPa} \), find minimum \( J \) needed: \( T = 51 \, \text{Nm} \). Solve \( 65 \times 10^6 = \frac{51 \cdot \frac{d}{2}}{\frac{\pi}{32} \cdot d^4} \). Solving gives \( d \approx 0.0171 \, \text{m} = 17.1 \, \text{mm} \).

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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 an important concept in the field of mechanics of materials. It represents the internal force exerted by a material when subjected to an external force, causing it to slide over an adjacent layer. The formula for shear stress \( \tau \) is given by:
  • \( \tau = \frac{T \cdot c}{J} \)
where:
  • \( T \) is the torque applied.
  • \( c \) is the outer radius of the axle.
  • \( J \) is the polar moment of inertia.
In the scenario provided, this shear stress is crucial because it indicates how much twisting force (torsion) the axle can endure without failing. It's calculated using the diameter of the axle, the distance from the center to the rope, and the weight of the bucket. Understanding how shear stress applies to practical situations, like the prospector's winch, helps in designing safe and efficient mechanical systems.
Torsion
Torsion refers to the twisting of an object due to an applied torque. In the context of the winch axle, torsion comes into play when the bucket weight creates a rotational force around the axle. This causes the axle to experience a twist.When calculating torsion, it's essential to determine the torque \( T \), which is the product of the applied force \( W \) and the lever arm \( b \). For example, in this case, \( T = 400 \, \text{N} \times 0.1 \, \text{m} = 40 \, \text{Nm} \).

A good understanding of torsion is necessary for predicting how materials will behave when subjected to twisting forces, which ensures that structural components will operate safely under expected loads. Torsion impacts the operational lifespan and safety of the winch by influencing the axle's capacity to resist deformation without exceeding allowable shear stress.
Polar Moment of Inertia
The polar moment of inertia \( J \) is a measure of an object's ability to resist torsion. It depends on the cross-sectional shape and size of the material. For a circular axle, it can be calculated using:
  • \( J = \frac{\pi}{32} \cdot d^4 \)
where \( d \) is the diameter of the axle.In this exercise, we calculate \( J \) to evaluate the axle's strength when subject to torsion. The formula shows that \( J \) is highly sensitive to changes in diameter, increasing significantly with even small increases in \( d \). Therefore, the polar moment of inertia is crucial in determining the necessary dimensions for axles to withstand specific loads without experiencing excessive shear stress.

This property ensures that mechanical elements, like the winch axle, are designed to resist torsion effectively, thus enhancing the durability and reliability of the system.

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Most popular questions from this chapter

What is the maximum power that can be delivered by a hollow propeller shaft (outside diameter \(50 \mathrm{mm}\) inside diameter \(40 \mathrm{mm},\) and shear modulus of elasticity 80 GPa turning at 600 rpm if the allowable shear stress is 100 MPa and the allowable rate of twist is \(3.0^{\circ} / \mathrm{m} ?\)

A circular tube of outer diameter \(d_{3}=70 \mathrm{mm}\) and inner diameter \(d_{2}=60 \mathrm{mm}\) is welded at the right-hand end to a fixed plate and at the left-hand end to a rigid end plate (see figure). A solid, circular bar with a diameter of \(d_{1}=40 \mathrm{mm}\) is inside of, and concentric with, the tube. The bar passes through a hole in the fixed plate and is welded to the rigid end plate. The bar is \(1.0 \mathrm{m}\) long and the tube is half as long as the bar. A torque \(T=1000 \mathrm{N} \cdot \mathrm{m}\) acts at end \(A\) of the bar. Also, both the bar and tube are made of an aluminum alloy with shear modulus of elasticity \(G=27 \mathrm{GPa}\) (a) Determine the maximum shear stresses in both the bar and tube. (b) Determine the angle of twist (in degrees) at end \(A\) of the bar.

The propeller shaft of a large ship has outside diameter \(350 \mathrm{mm}\) and inside diameter \(250 \mathrm{mm},\) as shown in the figure. The shaft is rated for a maximum shear stress of \(62 \mathrm{MPa}\) (a) If the shaft is turning at 500 rpm, what is the maximum horsepower that can be transmitted without exceeding the allowable stress? (b) If the rotational speed of the shaft is doubled but the power requirements remain unchanged, what happens to the shear stress in the shaft?

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.

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.
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