/*! 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 59 Two metal beams are joined toget... [FREE SOLUTION] | 91Ó°ÊÓ

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Two metal beams are joined together by four rivets, as the drawing indicates. Each rivet has a radius of \(5.0 \times 10^{-3} \mathrm{m}\) and is to be exposed to a shearing stress of no more than \(5.0 \times 10^{8}\) Pa. What is the maximum tension \(\overrightarrow{\mathbf{T}}\) that can be applied to each beam, assuming that each rivet carries one-fourth of the total load?

Short Answer

Expert verified
The maximum tension \( T \) on the beam is calculated as \( 500 \times 10^3 \pi \) N.

Step by step solution

01

Understand the Problem

We need to determine the maximum tension (\( \overrightarrow{\mathbf{T}} \)) that can be applied to each beam. Each of the four rivets experiences one-quarter of the total load, and each rivet must not exceed a shearing stress of \( 5.0 \times 10^{8} \) Pa.
02

Calculate the Area of Each Rivet

The area \( A \) of a circle (which is the cross-section of the rivet) is given by the formula \( A = \pi r^2 \). Here, \( r = 5.0 \times 10^{-3} \) meters. Thus, the area is:\[ A = \pi (5.0 \times 10^{-3})^2 = \pi \times 25.0 \times 10^{-6} \]}{
03

Determine Maximum Shearing Force Per Rivet

The maximum shearing stress is given by \( \tau = \frac{F}{A} \), where \( F \) is the force and \( A \) is the area. We need to find \( F \) such that \( \tau = 5.0 \times 10^{8} \) Pa:\[ F = \tau \times A = 5.0 \times 10^{8} \times \pi \times 25.0 \times 10^{-6} \]},
04

Calculate Total Maximum Tension on Beam

Since each rivet carries \( \frac{1}{4} \) of the total load, the total maximum tension \( T \) is \( 4F \):\[ T = 4 \times (5.0 \times 10^{8} \times \pi \times 25.0 \times 10^{-6}) \]

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Key Concepts

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

Maximum Tension
When we discuss maximum tension in the context of joined beams with rivets, we refer to the greatest force that can be applied to the beams without causing damage. As each rivet in this set-up carries one-quarter of the total load, understanding how this distribution impacts the maximum tension is vital.

The concept addresses the point where each rivet meets its stress limit, which in this exercise, is given as a shearing stress of not more than \(5.0 \times 10^{8}\) Pa. To maintain structural integrity when tension acts on the beams, none of the rivets should exceed this maximum stress level. By calculating how much force each rivet can withstand (given the stress limit), we ensure the total tension on the beam is evenly distributed among the rivets.

Knowing that the maximum tension equates to four times the force each rivet can handle helps prevent failure. Thus, understanding the maximum tension of a rivet system is crucial in real-world applications to ensure that distributed forces do not exceed the system's tolerance.
Rivet Force Distribution
When you join two beams with rivets, the way the force is distributed among the rivets is crucial. In this problem, rivet force distribution ensures that each rivet carries an equal share of the total force or load applied to the beams.

Each rivet in the setup, according to the exercise, carries one-fourth of the total load. This means that rather than expecting a single rivet to bear all the force, the load is effectively spread across all four rivets equally. This equal distribution of force helps in maintaining balance and minimizing the risk of any single rivet failing due to excessive load.

This concept is important in engineering because it influences how materials and connections are designed and evaluated. It emphasizes the importance of calculating each rivet's load to ensure no single point receives more stress than it can handle, maintaining structural safety and performance.
Cross-sectional Area of a Circle
The cross-sectional area of a circle is a fundamental computation when dealing with structures like rivets. This calculation helps you determine how much material is available to resist forces, like shear stress in this scenario.

We calculate this area using the formula \( A = \pi r^2 \), where \( r \) is the radius of the circle. In the exercise, the radius of each rivet is given as \(5.0 \times 10^{-3}\) meters. Substituting this into the formula, the cross-sectional area \( A \) is computed as \( \pi (5.0 \times 10^{-3})^2 = \pi \times 25.0 \times 10^{-6} \).

This value is crucial because it determines the amount of shearing force a rivet can resist. By knowing the area, one can apply the stress formula \( \tau = \frac{F}{A} \) to find the maximum force \( F \). The cross-sectional area thus serves as a key factor in understanding how much load can be safely managed by the rivet.

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Most popular questions from this chapter

A 0.60-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 5.70 to 4.80 m/s. What is the spring constant of the spring?

A spring is hung from the ceiling. A \(0.450-\mathrm{kg}\) block is then attached to the free end of the spring. When released from rest, the block drops \(0.150 \mathrm{m}\) before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

55m A person who weighs \(670 \mathrm{N}\) steps onto a spring scale in the bathroom, and the spring compresses by \(0.79 \mathrm{cm} .\) (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by \(0.34 \mathrm{cm} ?\)

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of \(7.0 \mathrm{rad} / \mathrm{s}\). The drawing indicates the position of the block when the spring is unstrained. This position is labeled " \(x=0 \mathrm{m}\)." The drawing also shows a small bottle located \(0.080 \mathrm{m}\) to the right of this position. The block is pulled to the right, stretching the spring by \(0.050 \mathrm{m},\) and is then thrown to the left. In order for the block to knock over the bottle, it must be thrown with a speed exceeding \(v_{0} .\) Ignoring the width of the block, find \(v_{0}\).

A spring is resting vertically on a table. A small box is dropped onto the top of the spring and compresses it. Suppose the spring has a spring constant of 450 N/m and the box has a mass of 1.5 kg. The speed of the box just before it makes contact with the spring is 0.49 m/s. (a) Determine the magnitude of the spring’s displacement at an instant when the acceleration of the box is zero. (b) What is the magnitude of the spring’s displacement when the spring is fully compressed?

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