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Chapter 12: Problem 43

Two parallel oppositely directed forces, each \(4000 \mathrm{~N}\), are applied tangentially to the upper and lower faces of a cubical metal block \(25 \mathrm{~cm}\) on a side. Find the angle of shear and the displacement of the upper surface relative to the lower surface. The shear modulus for the metal is 80 GPa.

Short Answer

Expert verified
The shear angle is \(8 \times 10^{-7}\) radians and the displacement is \(0.2\,\mu\text{m}\).

Step by step solution

01

Understand the Problem

We are tasked with finding two quantities: the angle of shear and the displacement of the upper surface relative to the lower surface of a cubical block. The cube is subjected to two parallel yet oppositely directed shear forces, each measuring \(4000\,\text{N}\), with the cube having side lengths of \(25\,\text{cm}\). The shear modulus, a measure of the material's rigidity, is provided as \(80\,\text{GPa}\).
02

Convert Units and Calculate Shear Stress

First, convert the side of the cube from centimeters to meters: \(25\,\text{cm} = 0.25\,\text{m}\). The area over which each force is applied is \(A = 0.25\,\text{m} \times 0.25\,\text{m} = 0.0625\,\text{m}^2\). The shear stress (\(\tau\)) is given by force over area: \(\tau = \frac{4000\,\text{N}}{0.0625\,\text{m}^2} = 64000\,\text{N/m}^2\).
03

Calculate Shear Strain Using Shear Modulus

The shear modulus (G) is related to shear stress (\(\tau\)) and shear strain (\(\gamma\)) by the equation: \(\tau = G \cdot \gamma\). Rearrange to find \(\gamma\): \(\gamma = \frac{\tau}{G} = \frac{64000\,\text{N/m}^2}{80 \times 10^9\,\text{N/m}^2} = 8 \times 10^{-7}\).
04

Find the Angle of Shear

The shear angle is approximately equal to the shear strain when the angle is small. Thus, the angle of shear equals \(\gamma = 8 \times 10^{-7}\) radians.
05

Calculate Displacement of Upper Surface

For small angles, the displacement \(\Delta x\) can be calculated as \(\Delta x = \gamma \cdot L\), where \(L\) is the side length of the cube. Therefore, \(\Delta x = 8 \times 10^{-7} \times 0.25\,\text{m} = 2 \times 10^{-7}\,\text{m}\) or \(0.2\,\mu\text{m}\).

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Key Concepts

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

Shear Strain
When a force is applied tangentially to a surface, it causes deformation in the material, which is termed as shear strain. In simple terms, shear strain is the change in angle between two sides of an object due to shear forces. It is denoted by the symbol \(\gamma\) and is dimensionless because it represents an angular measure, usually given in radians.

To calculate shear strain in the context of the given exercise, we use the formula relating shear stress (\(\tau\)) and shear modulus (G): \(\gamma = \frac{\tau}{G}\). This formula expresses the extent of deformation as a ratio of applied force's effect to the material's resistance to deformation. Understanding shear strain helps assess the lasting change in the material's structure under tangential force.
Shear Modulus
The shear modulus, often symbolized by the letter G, is a property of materials that depicts how they deform under shear stress. It is sometimes referred to as the modulus of rigidity. The shear modulus measures a material's ability to endure shape changes without changing volume.

In the given exercise, the shear modulus is provided as 80 GPa, indicating a very rigid material since the value is quite high. You can think of the shear modulus as the material's resistance to being twisted or sheared. A high shear modulus implies less deformation when a shear force is applied, while a low shear modulus suggests that the material is more susceptible to such changes.
Angle of Shear
The angle of shear is a way to describe the distortion angle due to applied shear forces. This angle is very small in most cases and, when related to materials engineering, is approximated by the shear strain since \(\gamma\) indicates the relative alteration of two lines initially at right angles.

In this problem, the calculated angle of shear is \(8 \times 10^{-7}\) radians, which is incredibly tiny and highlights the slight yet significant deformation under the forces applied. Understanding this angle provides insight into how much of a rotation or angular deformation a material undergoes under specific stress conditions without involving actual rotation.
Displacement Calculation
Displacement calculation in shear stress scenarios sneaks a peek at how much one face of a material has moved relative to another under tangential force. For calculating this, especially when dealing with small angles, we use the formula \(\Delta x = \gamma \cdot L\), where \(L\) is the original length parallel to the force direction.

In the exercise, the displacement \(\Delta x\) is found to be \(2 \times 10^{-7}\) meters or \(0.2\,\mu\)m. This reveals that even with a high shear modulus, the material exhibits a small, precise movement between its layers. This displacement helps engineers assess the limits of material deformation under operational conditions which is crucial for structural integrity.

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Most popular questions from this chapter

In an unhealthy, dusty cement mill, there were \(2.6 \times 10^{9}\) dust particles (sp \(\mathrm{gr}=3.0\) ) per cubic meter of air. Assuming the particles to be spheres of \(2.0 \mu \mathrm{m}\) diameter, calculate the mass of dust \((a)\) in a \(20 \mathrm{~m} \times 15 \mathrm{~m} \times 8.0 \mathrm{~m}\) room and (b) inhaled in each average breath of \(400-\mathrm{cm}^{3}\) volume.

A thin, semitransparent film of gold \(\left(\rho=19300 \mathrm{~kg} / \mathrm{m}^{3}\right)\) has an area of \(14.5 \mathrm{~cm}^{2}\) and a mass of \(1.93 \mathrm{mg} .(a)\) What is the volume of \(1.93 \mathrm{mg}\) of gold? \((b)\) What is the thickness of the film in angstroms, where \(1 \AA=10^{-10} \mathrm{~m} ?(c)\) Gold atoms have a diameter of about \(5 \AA\). How many atoms thick is the film?

An electrolytic tin-plating process gives a tin coating that is \(7.50 \times\) \(10^{-5}\) thick. How large an area can be coated with \(0.500 \mathrm{~kg}\) of tin? The density of tin is \(7300 \mathrm{~kg} / \mathrm{m}^{3}\). The volume of \(0.500\) kg of tin is given by \(\rho=m / V\) to be $$ V=\frac{m}{\rho}=\frac{0.500 \mathrm{~kg}}{7300 \mathrm{~kg} / \mathrm{m}^{3}}=6.85 \times 10^{-5} \mathrm{~m}^{3} $$ The volume of a film with area \(A\) and thickness \(d\) is \(V=A d\). Solving for \(A\), we find $$ A=\frac{V}{d}=\frac{6.85 \times 10^{-5} \mathrm{~m}^{3}}{7.50 \times 10^{-7} \mathrm{~m}}=91.3 \mathrm{~m}^{2} $$ as the area that can be covered.

Determine the fractional change in volume as the pressure of the atmosphere \(\left(1 \times 10^{5} \mathrm{~Pa}\right)\) around a metal block is reduced to zero by placing the block in vacuum. The bulk modulus for the metal is \(125 \mathrm{GPa}\)

A 15-kg ball of radius \(4.0 \mathrm{~cm}\) is suspended from a point \(2.94 \mathrm{~m}\) above the floor by an iron wire of unstretched length \(2.85 \mathrm{~m}\). The diameter of the wire is \(0.090 \mathrm{~cm}\), and its Young's modulus is 180 GPa. If the ball is set swinging so that its center passes through the lowest point at \(5.0 \mathrm{~m} / \mathrm{s}\), by how much does the bottom of the ball clear the floor? Discuss any approximations that you make. Call the tension in the wire \(F_{T}\) when the ball is swinging through the lowest point. Since \(F_{T}\) must supply the centripetal force as well as balance the weight, $$ F_{T}=m g+\frac{m v^{2}}{r}=m\left(9.81+\frac{25}{r}\right) $$ all in proper SI units. This is complicated, because \(r\) is the distance from the pivot to the center of the ball when the wire is stretched, and so it is \(\mathrm{r}_{0}+\Delta r\), where \(r_{0}\), the unstretched length of the pendulum, is $$ r_{0}=2.85 \mathrm{~m}+0.040 \mathrm{~m}=2.89 \mathrm{~m} $$ and where \(\Delta r\) is as yet unknown. However, the unstretched distance from the pivot to the bottom of the ball is \(2.85 \mathrm{~m}+\) \(0.080 \mathrm{~m}=2.93 \mathrm{~m}\), and so the maximum possible value for \(\Delta r\) is $$ 2.94 \mathrm{~m}-2.93 \mathrm{~m}=0.01 \mathrm{~m} $$ We will therefore incur no more than a \(1 / 3\) percent error in \(r\) by using \(r=r_{0}=2.89 \mathrm{~m}\). This gives \(F_{T}=277 \mathrm{~N}\). Under this tension, the wire stretches by $$ \Delta L=\frac{F L_{0}}{A Y}=\frac{(277 \mathrm{~N})(2.85 \mathrm{~m})}{\pi\left(4.5 \times 10^{-4} \mathrm{~m}\right)^{2}\left(1.80 \times 10^{11} \mathrm{~Pa}\right)}=6.9 \times 10^{-3} \mathrm{~m} $$ Hence, the ball misses by \(2.94 \mathrm{~m}-(2.85+0.0069+0.080) \mathrm{m}=0.0031 \mathrm{~m}=3.1 \mathrm{~mm}\) To check the approximation we have made, we could use \(r=\) \(2.90 \mathrm{~m}\), its maximum possible value. Then \(\Delta L=6.9 \mathrm{~mm}\), showing that the approximation has caused a negligible error.
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