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For the case of plane stress, show that Hooke's law can be written as \\[\sigma_{x}=\frac{E}{\left(1-\nu^{2}\right)}\left(\boldsymbol{\epsilon}_{x}+\nu \boldsymbol{\epsilon}_{y}\right), \quad \sigma_{y}=\frac{E}{\left(1-\nu^{2}\right)}\left(\boldsymbol{\epsilon}_{y}+\nu \boldsymbol{\epsilon}_{x}\right)\\]

Short Answer

Expert verified
The given expressions for stress in x and y directions in terms of strain can be derived from Hooke's Law for plane stress by multiplying the complete expression by the term \(\frac{1}{1 - \nu^2}\). As a result, the provided equations are verified to be correct under the condition of plane stress.

Step by step solution

01

Recall Hooke's Law for Plane Stress

For plane stress, Hooke's law gives the following relationships: \[ \sigma_x = E(\epsilon_x + \nu \epsilon_y)\] \[\sigma_y = E(\epsilon_y + \nu \epsilon_x)\] where \(\sigma_x\) and \(\sigma_y\) are stresses in x and y-axis, E is the modulus of elasticity, \(\nu\) is Poisson's ratio, and \(\epsilon_x\) and \(\epsilon_y\) are strains in x and y-axis respectively.
02

Validate the Given Expressions

The task is to validate the provided expressions, which are: \[ \sigma_x = \frac{E}{(1 - \nu^2)}(\epsilon_x + \nu \epsilon_y) \] \[ \sigma_y = \frac{E}{(1 - \nu^2)}(\epsilon_y + \nu \epsilon_x) \] Now, multiply both sides of the standard Hooke's law relationships (from step 1) by \frac{1}{1 - \nu^2}. This will give the required forms of the expressions.
03

Verify the Final Expressions

Multiplying the Hooke's Law by \frac{1}{1 - \nu^2}, we get: \[ \frac{\sigma_x}{1 - \nu^2} = \frac{E}{1 - \nu^2} (\epsilon_x + \nu \epsilon_y) \] \[ \frac{\sigma_y}{1 - \nu^2} = \frac{E}{1 - \nu^2} (\epsilon_y + \nu \epsilon_x) \] Therefore, this validates the given expressions for \(\sigma_x\) and \(\sigma_y\). Expression 2 is equivalent to the original relation from Hooke's Law.

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Key Concepts

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

Modulus of Elasticity
Imagine a sponge that you press down with your finger. The way it compresses and then bounces back is related to a property that materials science calls the 'modulus of elasticity', often represented by the symbol 'E'. This property is an indicator of a material's stiffness or rigidity. When we apply stress to a material, causing it to strain or deform, the ratio of stress to strain is the modulus of elasticity. In the context of the provided exercise, the modulus of elasticity is crucial in determining how much a material will deform under plane stress conditions.
In simple terms, materials with a high modulus of elasticity—like steel or diamond—don't deform much under load, they're very stiff. Meanwhile, materials with a low modulus, like rubber, are more flexible. We can formulate this in an equation as \( E = \frac{\sigma}{\epsilon} \), where \( \sigma \) represents stress and \( \epsilon \) represents strain. So, for any given material, understanding 'E' allows us to predict how it will behave when forces are applied to it in the plane of the material, as is the case with the plane stress scenario in Hooke's law.
Poisson's Ratio
When materials are stretched or compressed, something interesting happens: they tend to thin or thicken in the directions perpendicular to the applied load. This behavior is characterized by 'Poisson's ratio', symbolized by \( u \). Poisson's ratio is a measure of the transverse strain to axial strain. For instance, if you pull a rubber band, it not only gets longer but also narrower. This narrowing effect is precisely what Poisson's ratio describes.
To understand the concept further, consider the formula \( u = -\frac{\epsilon_{transverse}}{\epsilon_{axial}} \). The negative sign indicates that the material tends to contract in the transverse direction when stretched in the axial direction, and vice versa. Poisson's ratio is dimensionless and typically ranges between 0 and 0.5 for most materials. In the mentioned exercise, the Poisson's ratio, along with the modulus of elasticity, contributes to defining the relationship between stress and strain in two perpendicular directions under plane stress conditions.
Plane Stress
The term 'plane stress' refers to a state of stress where the stress in one particular direction is zero, typically the thickness direction in thin plates. In essence, we can imagine a scenario where a large sheet of material is pulled or compressed in only two dimensions—its plane. This simplification is especially useful in many engineering applications because it models the behavior of elements like beams, sheets, or walls that are slim compared to their length and width.
Under plane stress conditions, we mainly focus on the stresses in the two-dimensional plane, symbolized as \( \sigma_x \) and \( \sigma_y \) in our exercise. Plane stress hence simplifies complex three-dimensional stress states into a more manageable two-dimensional problem, without losing the essence of what happens within the material's plane. When solving problems involving thin-walled structures, considering plane stress is often an accurate approximation that streamlines calculations and understanding.
Stress-Strain Relationship
The stress-strain relationship is a fundamental concept in materials science and mechanical engineering that describes how an object deforms under the application of force. Stress \( (\sigma) \) essentially defines the internal forces per unit area within a material, whereas strain \( (\epsilon) \) measures the deformation of the material as a result of this stress. The stress-strain relationship is best visualized in a graph, where the amount of strain is plotted against the applied stress, revealing properties like elasticity, plasticity, and ultimate strength of the material.
Particularly, the linear part of the stress-strain curve is where Hooke's law applies—indicating a direct proportionality between stress and strain for small deformations. This linearity is where the modulus of elasticity, E, is the slope of the curve, and the limit of this proportional behavior marks what we call the 'elastic limit' of the material. In the provided exercise, understanding the stress-strain relationship allows us to use Hooke's law for plane stress to describe how the material will behave when subjected to loads in its plane, cementing a central principle in solid mechanics.

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Most popular questions from this chapter

If a solid shaft having a diameter \(d\) is subjected to a torque \(\mathbf{T}\) and moment \(\mathbf{M}\), show that by the maximumshear-stress theory the maximum allowable shear stress is \(\tau_{\text {allow }}=\left(16 / \pi d^{3}\right) \sqrt{M^{2}+T^{2}} .\) Assume the principal stresses to be of opposite algebraic signs.

The spherical pressure vessel has an inner diameter of \(2 \mathrm{m}\) and a thickness of \(10 \mathrm{mm} .\) A strain gauge having a length of \(20 \mathrm{mm}\) is attached to it, and it is observed to increase in length by \(0.012 \mathrm{mm}\) when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which \(E_{\mathrm{st}}=200 \mathrm{GPa}\) and \(\nu_{\mathrm{st}}=0.3\).

The gas tank is made from \(A-36\) steel and has an inner diameter of \(1.50 \mathrm{m}\). If the tank is designed to withstand a pressure of \(5 \mathrm{MPa}\), determine the required minimum wall thickness to the nearest millimeter using (a) the maximum-shear-stress theory, and (b) maximum- distortion-energy theory. Apply a factor of safety of 1.5 against yielding.

The state of strain on an element has components \(\boldsymbol{\epsilon}_{x}=-400\left(10^{-6}\right), \boldsymbol{\epsilon}_{y}=0, \gamma_{x y}=150\left(10^{-6}\right) .\) Determine the equivalent state of strain on an element at the same point oriented \(30^{\circ}\) clockwise with respect to the original element. Sketch the results on this element.

The strain gauge is placed on the surface of a thinwalled steel boiler as shown. If it is 0.5 in. long, determine the pressure in the boiler when the gauge elongates \(0.2\left(10^{-3}\right)\) in. The boiler has a thickness of 0.5 in. and inner diameter of 60 in. Also, determine the maximum \(x, y\) in- plane shear strain in the material. \(E_{\mathrm{st}}=29\left(10^{3}\right) \mathrm{ksi}, \nu_{\mathrm{st}}=0.3\).

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