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A thin-walled hollow tube \(A B\) of conical shape has constant thickness \(t\) and average diameters \(d_{A}\) and \(d_{B}\) at the ends (see figure). (a) Determine the strain energy \(U\) of the tube when it is subjected to pure torsion by torques \(T\). (b) Determine the angle of twist \(\phi\) of the tube. (Note: Use the approximate formula \(I_{P} \approx \pi d^{3} t / 4\) for a thin circular ring; see Case 22 of Appendix D.)

Short Answer

Expert verified
Strain energy \(U\) is calculated by integrating the torque over the conical section, and angle of twist \(\phi\) is found by integrating the twist per unit length.

Step by step solution

01

Understand the Problem

We are tasked with finding the strain energy and angle of twist for a conical tube subjected to torsion. The tube has constant thickness \(t\) and varying diameters \(d_A\) and \(d_B\). We will use the formula for the polar moment of inertia for a thin circular ring \(I_P \approx \frac{\pi d^3 t}{4}\).
02

Express Torque in Terms of Shear Stress

The torque \(T\) applied to a shaft causes shear stresses. For a differential segment of the cone with diameter \(d(x)\), the relation between torque \(T\), shear stress \(\tau\), and polar moment of inertia \(I_P\) is given by \(T = \tau \cdot I_P/d\).
03

Define Polar Moment of Inertia

For a differential section of the cone, the average diameter \(d(x) = d_A + \frac{(d_B - d_A)x}{L}\), where \(x\) is the distance along the tube and \(L\) is the length of the tube. Using the given approximation, \(I_P(x) = \frac{\pi d(x)^3 t}{4}\).
04

Calculate Strain Energy

The strain energy \(U\) for a torsion tube is \(U = \int_{0}^{L} \frac{T^2}{2G I_P(x)} dx\). Substitute \(I_P(x)\) with the expression from step 3 and integrate over the length of the tube.
05

Calculate Angle of Twist

The angle of twist \(\phi\) is given by \(\phi = \int_{0}^{L} \frac{T dx}{G I_P(x)}\). Substitute the expression for \(I_P(x)\) and integrate.

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Key Concepts

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

Strain Energy
Strain energy is the energy stored in a material as it undergoes deformation. When a conical tube is subjected to torsion, it twists, storing strain energy. This energy comes from the internal stresses developing in the material as it resists the applied torque. Particularly, strain energy in torsion is calculated by the integral formula:
  • Strain Energy, \( U = \int_{0}^{L} \frac{T^2}{2G I_P(x)} \, dx \)
Here, \( T \) is the torque applied, \( G \) is the shear modulus of the material, \( I_P(x) \) is the polar moment of inertia at point \( x \), and \( L \) is the length of the tube. This formula sums the tiny bits of energy stored along the entire length of the tube. Calculating this energy helps in understanding how much the material can handle before failing.
Torsion
Torsion refers to the twisting of an object due to an applied torque. In the context of a conical tube, as torque is applied, each segment of the tube experiences a rotational force.
This results in shear stress that varies with the distance from the tube’s center. The torsional forces want to rotate one section of the material at one end relative to the other.
When dealing with torsion, it is crucial to relate it to material properties and geometry. This is where the polar moment of inertia, \( I_P \), and shear modulus, \( G \), come into play. Together, they help determine how a material will react when twisted.
  • Torsion formula: \( T = \tau \cdot I_P/d \), where \( \tau \) is shear stress and \( d \) is diameter.
Angle of Twist
The angle of twist, represented by \( \phi \), measures how much a material twists under the effect of torsion over its length. Think of it as the angular displacement due to the applied torque.
  • Angle of Twist formula: \( \phi = \int_{0}^{L} \frac{T \, dx}{G I_P(x)} \)
Here, \( T \) is the torque, \( G \) is the shear modulus, \( I_P(x) \) is the polar moment of inertia along the tube's length, and \( L \) is the length of the tube.
When you apply this integral, you sum up all the small rotations along the length of the tube to calculate the total twist.
By knowing \( \phi \), engineers can ensure that the material or component will not exceed acceptable deformation limits under the expected loads.
Polar Moment of Inertia
The polar moment of inertia, \( I_P \), is a geometrical property that describes how a cross-section can resist torsion. In essence, it measures the distribution of the area about an axis, giving insight into its torsional rigidity.For a conical tube with constant thickness, it follows the formula:
  • \( I_P \approx \frac{\pi d^3 t}{4} \), where \( d \) is the average diameter and \( t \) is the thickness of the tube.
This approximation is especially helpful for thin-walled sections, as they often appear in mechanical and structural applications.
The larger the polar moment of inertia, the greater the resistance to twisting. It is a critical factor in designing shafts and other components that are expected to undergo torsion. By calculating the \( I_P \) at various sections, engineers ensure that materials and components meet the necessary performance and safety requirements.

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Most popular questions from this chapter

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?

A hollow circular shaft for use in a pumping station is being designed with an inside diameter equal to 0.8 times the outside diameter. The shaft must transmit \(300 \mathrm{kW}\) at 800 rpm without exceeding the allowable shear stress of \(42 \mathrm{MPa}\) Determine the minimum required outside diameter \(d\)

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?

An aluminum bar of solid circular cross section is twisted by torques \(T\) acting at the ends (see figure). The dimensions and shear modulus of elasticity are as follows: \(L=1.4 \mathrm{m}, d=32 \mathrm{mm},\) and \(G=28 \mathrm{GPa}\) (a) Determine the torsional stiffness of the bar. (b) If the angle of twist of the bar is \(5^{\circ},\) what is the maximum shear stress? What is the maximum shear strain (in radians)? (c) If a hole of diameter \(d / 2\) is drilled longitudinally through the bar, what is the ratio of the torsional stiff- nesses of the hollow and solid bars? What is the ratio of their maximum shear stresses if both are acted on by the same torque? (d) If the hole diameter remains at \(d / 2,\) what new outside diameter \(d_{2}\) will result in equal stiffnesses of the hollow and solid bars?

A solid circular bar of steel \((G=80 \text { GPa })\) with length \(L=1.5 \mathrm{m}\) and diameter \(d=75 \mathrm{mm}\) is subjected to pure torsion by torques \(T\) acting at the ends (see figure). (a) Calculate the amount of strain energy \(U\) stored in the bar when the maximum shear stress is \(45 \mathrm{MPa}\). (b) From the strain energy, calculate the angle of twist \(\phi\) (in degrees).

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