/*! 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 10 A hollow circular tube \(A\) fit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 10

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.)

Short Answer

Expert verified
The total strain energy is the sum of strain energies in bars A and B, considering angle compatibility and equilibrium.

Step by step solution

01

Understanding Strain Energy

The strain energy in a twisted bar is the work done in twisting it through an angle of twist and is given by the formula \( U = \frac{1}{2} I_P G \theta^2 L \), where \( \theta \) is the angle of twist, \( I_P \) is the polar moment of inertia, \( G \) is the shear modulus of elasticity, and \( L \) is the length of the bar.
02

Angle of Twist Alignment

When bar \( B \) is twisted to align the holes, its initial twist angle is \( \beta \). Upon release, the system returns to equilibrium meaning the bars share the twist angle in proportion to their stiffness (\( I_P \)). At equilibrium, the internal angles, \( \theta_A \) and \( \theta_B \), are related through twisting compatibility \( \theta_A + \theta_B = \beta \).
03

Calculate Individual Strain Energies

Using the relation \( G \theta_A I_{P A} = G \theta_B I_{P B} \) (from torque equilibrium), find \( \theta_A \) and \( \theta_B \). The energy in tube \( A \) becomes \( U_A = \frac{1}{2} I_{P A} G \theta_A^2 L \), and in bar \( B \) it is \( U_B = \frac{1}{2} I_{P B} G \theta_B^2 L \). Substitute \( \theta_B = \beta - \theta_A \) in these equations.
04

Solve for Total Strain Energy

Sum the energies of both the tube and the bar: \[ U = U_A + U_B = \frac{1}{2} I_{P A} G \theta_A^2 L + \frac{1}{2} I_{P B} G (\beta - \theta_A)^2 L \]. Given the equilibrium equations and simplifying, solve for the total strain energy \( U \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Polar Moment of Inertia
The polar moment of inertia, often denoted as \( I_P \), is a crucial concept in understanding torsion in engineering components, especially in circular cross-sections. It's a measure of an object's ability to resist torsional deformation. In simpler terms, it's how much force the object can take without twisting too much.
  • For circular shapes, it's calculated as \( I_P = \frac{\pi}{32}(d_o^4 - d_i^4) \) for hollow circles, where \( d_o \) and \( d_i \) are the outer and inner diameters.
  • For solid circles, it's \( I_P = \frac{\pi}{32}d^4 \).
In exercises involving objects like our hollow tube \( A \) and solid bar \( B \), \( I_{P A} \) and \( I_{P B} \) help define each object's torsional stiffness, or how much it can twist under a certain load without bending too much.
Angle of Twist
In the context of our exercise, the angle of twist \( \theta \) refers to the rotational displacement experienced by bar \( B \) when it is twisted before aligning holes with tube \( A \). This twist occurs due to an applied torque.
  • When the system is at equilibrium, bar \( B \)'s twist angle changes from its initial state, then shares the load with tube \( A \).
  • Twist angles are crucial in determining how much energy is stored due to twisting, expressed using \( \theta_A \) and \( \theta_B \) for tube \( A \) and bar \( B \) respectively.
Shear Modulus of Elasticity
The shear modulus of elasticity, \( G \), is a material's response to shearing forces, essentially a measure of its rigidity. In simple terms, it's how much the material will deform under shear stress.
  • It is a crucial factor when calculating strain energy because it defines how a material will twist under torsional forces.
  • For both tube \( A \) and bar \( B \), knowing \( G \) allows us to understand how they behave and react under twisting forces, maintaining equilibrium at certain angles.
Twisting Compatibility
Twisting compatibility refers to the condition in systems with interconnected elements, ensuring both components share compatible twist angles at equilibrium. In our scenario, it means tube \( A \) and bar \( B \) must reach a state where their total twist is equal to the initial angle \( \beta \).
  • Your application here is defined by the equation \( \theta_A + \theta_B = \beta \).
  • This compatibility ensures equal shear stress distribution, helping solve for individual twist angles \( \theta_A \) and \( \theta_B \).
This helps effectively equate combined twist forces, allowing correct calculation of total strain energy in complex systems like the one in our exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A full quarter-circular fillet is used at the shoulder of a stepped shaft having diameter \(D_{2}=25 \mathrm{mm}\) (see figure). A torque \(T=115 \mathrm{N} \cdot \mathrm{m}\) acts on the shaft. Determine the shear stress \(\tau_{\max }\) at the stress concentration for values as follows: \(D_{1}=18,20,\) and \(22 \mathrm{mm}\) Plot a graph showing \(\tau_{\max }\) versus \(D_{1}\)

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).

When drilling a hole in a table leg, a furniture maker uses a hand-operated drill (see figure) with a bit of diameter \(d=4.0 \mathrm{mm}\) (a) If the resisting torque supplied by the table leg is equal to \(0.3 \mathrm{N} \cdot \mathrm{m}\), what is the maximum shear stress in the drill bit? (b) If the allowable shear stress in the drill bit is \(32 \mathrm{MPa}\), what is the maximum resisting torque before the drill binds up? (c) If the shear modulus of elasticity of the steel is \(G=75 \mathrm{GPa},\) what is the rate of twist of the drill bit (degrees per meter)?

A hollow aluminum tube used in a roof structure has an outside diameter \(d_{2}=104 \mathrm{mm}\) and an inside diameter \(d_{1}=82 \mathrm{mm}\) (see figure). The tube is \(2.75 \mathrm{m}\) long, and the aluminum has shear modulus \(G=28\) GPa. (a) If the tube is twisted in pure torsion by torques acting at the ends, what is the angle of twist (in degrees) when the maximum shear stress is 48 MPa? (b) What diameter \(d\) is required for a solid shaft (see figure) to resist the same torque with the same maximum stress? (c) What is the ratio of the weight of the hollow tube to the weight of the solid shaft?

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?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Rotational Dynamics

Read Explanation

Magnetism

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Force

Read Explanation

Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.