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Chapter 1: Problem 15

Determine the diameter of the largest circular hole that can be punched into a sheet of polystyrene \(6 \mathrm{mm}\) thick, knowing that the force exerted by the punch is \(45 \mathrm{kN}\) and that a \(55-\mathrm{MPa}\) average shearing stress is required to cause the material to fail.

Short Answer

Expert verified
The largest hole diameter is approximately 43.58 mm.

Step by step solution

01

Understand the Problem

We need to determine the largest diameter of a circular hole that can be punched into a polystyrene sheet, given the force, thickness, and shearing stress.
02

Identify the Formula

The formula for the shearing stress \( \tau \) is given by \( \tau = \frac{F}{A} \), where \( F \) is the force applied and \( A \) is the area. In this problem, we want to use this formula to find the area associated with the shearing.
03

Calculate the Shearing Area

The shearing area for a circular hole in the sheet is the lateral surface area of the cylinder created by the punch, given by \( A = \pi d t \), where \( d \) is the diameter of the punch and \( t \) is the thickness of the sheet (\( t = 6 \mathrm{mm} \) or \( 0.006 \mathrm{m} \)).
04

Set Up the Equation

Substitute \( A = \pi d t \) into the shearing stress formula: \[ \tau = \frac{F}{\pi d t} \] With \( F = 45000 \mathrm{N} \), \( \tau = 55000000 \mathrm{Pa} \), and \( t = 0.006 \mathrm{m} \), substitute these values into the equation.
05

Solve for Diameter \( d \)

Rearrange the equation to solve for \( d \): \[ d = \frac{F}{\pi \tau t} \] Substitute the known values to find: \[ d = \frac{45000}{\pi \times 55000000 \times 0.006} \approx 0.04358 \mathrm{m} \]
06

Convert Diameter to Millimeters

Multiply the diameter in meters by 1000 to convert it to millimeters: \[ 0.04358 \mathrm{m} \times 1000 = 43.58 \mathrm{mm} \] Thus, the largest diameter of the hole is approximately 43.58 mm.

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Key Concepts

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

Material Failure
Material failure occurs when a material can no longer withstand the stresses and forces applied to it, leading to it breaking or deforming. This understanding is crucial in engineering and construction as it helps predict when a material might fail and thus ensure safety and integrity.
There are different types of material failure, including:
  • Tensile Failure: When the material is pulled apart.
  • Compressive Failure: When the material is squished or compressed.
  • Shearing Failure: When a force causes one part of a material to slide past another, as is the focus in this exercise.
Knowing the shearing stress a material can handle helps in determining the limits beyond which it will fail. For instance, in this problem, polystyrene fails at a shearing stress of 55 MPa, which is critical in calculating how large a hole we can punch into it without causing failure.
Circular Hole Punch
A circular hole punch is a tool designed to create perfectly round holes in materials by cutting through them with a force applied perpendicularly. In mechanics, the understanding of how these tools work requires knowledge of the forces and stresses involved.
For this exercise, the goal is to determine the largest diameter of a hole that can be punched without surpassing the material's shearing stress limit. The hole punch creates a cylindrical hole, and its effectiveness is influenced by:
  • The diameter of the hole to be punched.
  • The thickness of the material being punched.
  • The mechanical properties of the material, such as its maximum shear stress capacity.
By calculating the stress exerted by the punch and comparing it to the material's limit, we can ensure the operation stays within safe parameters.
Polystyrene Properties
Polystyrene is a synthetic polymer known for its lightweight and relatively weak mechanical properties in comparison to metals and other materials. It is often used in various applications such as packaging, insulation, and disposable cups due to its versatility and ease of manufacturing.
Here are some important properties of polystyrene concerning mechanical stress and material failure:
  • Shear Strength: Polystyrene can withstand an average shearing stress of 55 MPa, beyond which it is prone to tearing or failure.
  • Density and Weight: Being lightweight, polystyrene is easy to handle but this also means it may not perform well under high loads.
  • Thermal Sensitivity: Its properties can change with exposure to heat, becoming softer and thereby reducing its ability to handle stress.
Understanding these properties is essential when applying force to punch holes in polystyrene, as they determine the maximum allowable stress without causing material failure.
Force Calculations
Force calculations are necessary to ensure that an operation like hole punching stays within safe stress limits and does not lead to material failure. The exercise involves utilizing the relationship between force, area, and shearing stress to find the maximum hole size possible.
Key formulae used in these calculations include:
  • Shearing Stress Formula: \( \tau = \frac{F}{A} \) where \( \tau \) is the shearing stress, \( F \) is the force, and \( A \) is the shearing area.
  • Area of Shearing Surfaces: For a circular punch, \( A = \pi d t \) where \( d \) is the diameter and \( t \) is the thickness.
By rearranging these equations and substituting the given values, we calculated the diameter \( d \) as \( d = \frac{45000}{\pi \times 55000000 \times 0.006} \) to find an approximate diameter of 43.58 mm, which adheres to the stress limitations and ensures no failure occurs.

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Most popular questions from this chapter

Link \(A C\) is made of a steel with a 65 -ksi ultimate normal stress and has a \(\frac{1}{4} \times \frac{1}{2}\) -in. uniform rectangular cross section. It is connected to a support at \(A\) and to member \(B C D\) at \(C\) by \(\frac{3}{4}\) -in. -diameter pins, while member \(B C D\) is connected to its support at \(B\) by a \(\frac{5}{16}\) -in.- diameter pin. All of the pins are made of a steel with a 25 -ksi ultimate shearing stress and are in single shear. Knowing that a factor of safety of 3.25 is desired, determine the largest load \(\mathbf{P}\) that can be applied at \(D\). Note that link \(A C\) is not reinforced around the pin holes.

Link \(A B,\) of width \(b=2\) in. and thickness \(t=\frac{1}{4}\) in., is used to support the end of a horizontal beam. Knowing that the average normal stress in the link is -20 ksi and that the average shearing stress in each of the two pins is 12 ksi determine ( \(a\) ) the diameter \(d\) of the pins, (b) the average bearing stress in the link.

A steel loop \(A B C D\) of length 5 ft and of \(\frac{3}{8}\) -in. diameter is placed as shown around a 1 -in.-diameter aluminum rod \(A C\). Cables \(B E\) and \(D F\), each of \(\frac{1}{2}\) -in. diameter, are used to apply the load \(Q\). Knowing that the ultimate strength of the steel used for the loop and the cables is \(70 \mathrm{ksi}\), and that the ultimate strength of the aluminum used for the rod is 38 ksi, determine the largest load \(\mathbf{Q}\) that can be applied if an overall factor of safety of 3 is desired.

Two steel plates are to be held together by means of \(16-\mathrm{mm}\) diameter high-strength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the average normal stress must not exceed \(200 \mathrm{MPa}\) in the bolts and \(130 \mathrm{MPa}\) in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design.

An aircraft tow bar is positioned by means of a single hydraulic cylinder connected by a 25 -mm-diameter steel rod to two identical arm-and-wheel units \(D E F\). The mass of the entire tow bar is \(200 \mathrm{kg},\) and its center of gravity is located at \(G\). For the position shown, determine the normal stress in the rod.
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