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Show that the sum of the normal stresses on any two orthogonal planes is a constant. Evaluate the constant.

Short Answer

Expert verified
The sum is \( \sigma_x + \sigma_y \), a constant for any orthogonal planes.

Step by step solution

01

Understanding the Problem

We need to show that the sum of the normal stresses on two orthogonal planes remains constant. Here, we relate the stresses using equilibrium conditions and stress transformation equations. The problem requires finding this invariant sum under stress transformations.
02

Define Plane Stress Condition

Consider a stress element under plane stress conditions with normal stresses \( \sigma_x \) on the x-plane and \( \sigma_y \) on the y-plane, and shear stress \( \tau_{xy} \). We aim to examine stresses on planes at an angle \( \theta \) to the x-axis.
03

Apply Transformation Equations

Use the stress transformation equations: \[\sigma_{n} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos(2\theta) + \tau_{xy}\sin(2\theta)\]\[\sigma_{o} = \frac{\sigma_x + \sigma_y}{2} - \frac{\sigma_x - \sigma_y}{2}\cos(2\theta) - \tau_{xy}\sin(2\theta)\]where \( \sigma_{n} \) and \( \sigma_{o} \) are the normal stresses on planes at angle \( \theta \) and \( \theta + 90^\circ \), respectively.
04

Sum of Normal Stresses

Calculate the sum of the transformed normal stresses:\[\sigma_{n} + \sigma_{o} = \left( \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x + \sigma_y}{2} \right) = \sigma_x + \sigma_y\]Notice the terms with trigonometric functions cancel each other, resulting in a constant sum.
05

Conclusion

Thus, the sum of the normal stresses on any two orthogonal planes is the constant \( \sigma_x + \sigma_y \). This result is independent of the angle \( \theta \) chosen for the planes.

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Key Concepts

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

Normal Stresses
Normal stresses are the forces per unit area acting perpendicular to a plane within a material. These stresses can either compress or stretch the material. When looking at a stress element subjected to various forces, normal stresses can be categorized based on the plane they act. In a typical coordinate system, these are represented as \( \sigma_x \) on the x-plane and \( \sigma_y \) on the y-plane. Knowing how these stresses interact helps in predicting the behavior of materials under load.
Mathematically, normal stresses influence the calculations in stress transformation equations. Their behavior under different conditions, such as changes in angle, is what we aim to explore with tools like Mohr's Circle.
Orthogonal Planes
Orthogonal planes are two planes that intersect at right angles, or 90 degrees. In the context of stress analysis, when we talk about normal stresses on orthogonal planes, we often refer to the x and y planes in a Cartesian coordinate system. Calculating stresses on these orthogonal planes helps engineers evaluate how forces distribute within materials.
When employing stress transformation equations, we examine how normal stresses shift when looking at a new set of orthogonal planes at different angles, such as \( \theta \) and \( \theta + 90^\circ \). These transformations are key to understanding how internal forces behave.
Stress Invariants
Stress invariants are values that remain unchanged regardless of the orientation of the coordinate system in which they are measured. In stress transformation, these invariants help verify that transformations are correctly applied. For normal stresses, the sum on two orthogonal planes is an important invariant.
In the example solution, the sum of normal stresses \( \sigma_n + \sigma_o \), represented by the expression \( \sigma_x + \sigma_y \), demonstrates this invariant property. This sum does not change no matter the angle \( \theta \), showing that it is independent of plane rotation due to the balanced nature of internal forces. Recognizing these invariant sums supports consistent material behavior in different applications.
Plane Stress Conditions
Under plane stress conditions, a stress element experiences stresses only on two dimensions. Usually, this involves the x-plane and y-plane in a 2D stress scenario, with the third dimension (out of plane) assumed to have negligible stress. This assumption simplifies the complexities of stress analysis and is common in thin-walled structures like plates and shells.
In plane stress analysis, we consider the normal stresses \( \sigma_x \) and \( \sigma_y \), along with shear stress \( \tau_{xy} \). These components can then be used in stress transformation equations, allowing engineers to assess how materials respond under different loading scenarios and orientations. By focusing on plane stress conditions, we simplify the analysis while capturing the essential stress behavior needed for engineering design and evaluation.

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Most popular questions from this chapter

The state of stress at a point on a fault plane is \(\sigma_{y y}=150 \mathrm{MPa}, \sigma_{x x}=200 \mathrm{MPa}\), and \(\sigma_{x y}=0\) ( \(y\) is depth and the \(x\) axis points westward). What are the normal stress and the tangential stress on the fault plane if the fault strikes \(\mathrm{N}-\mathrm{S}\) and dips \(35^{\circ}\) to the west?

An average thickness of the oceanic crust is \(6 \mathrm{~km}\). Its density is \(2900 \mathrm{~kg} \mathrm{~m}^{-3}\). This is overlain by \(5 \mathrm{~km}\) of water \(\left(\rho_{w}=1000 \mathrm{~kg} \mathrm{~m}^{-3}\right)\) in a typical ocean basin. Determine the normal force per unit area on a horizontal plane at the base of the oceanic crust due to the weight of the crust and the overlying water.

There is observational evidence from the continents that the sea level in the Cretaceous was \(200 \mathrm{~m}\) higher than today. After a few thousand years, however, the seawater is in isostatic equilibrium with the ocean basins. What was the corresponding increase in the depth of the ocean basins? Take \(\rho_{w}=\) \(1000 \mathrm{~kg} \mathrm{~m}^{-3}\) and the density of the displaced mantle to be \(\rho_{m}=3300 \mathrm{~kg} \mathrm{~m}^{-3}\).

Consider a simple two-layer model of a planet consisting of a core of density \(\rho_{c}\) and radius \(b\) surrounded by a mantle of density \(\rho_{m}\) and thickness \(a-b\). Show that the gravitational acceleration as a function of radius is given by $$ \begin{aligned} g(r) &=\frac{4}{3} \pi \rho_{c} G r \quad 0 \leq r \leq b \\ &=\frac{4}{3} \pi G\left[r \rho_{m}+b^{3}\left(\rho_{c}-\rho_{m}\right) / r^{2}\right] \quad b \leq r \leq a \end{aligned} $$

Consider a block of rock with a height of \(1 \mathrm{~m}\) and horizontal dimensions of \(2 \mathrm{~m}\). The density of the rock is \(2750 \mathrm{~kg} \mathrm{~m}^{-3}\). If the coefficient of friction is \(0.8,\) what force is required to push the rock on a horizontal surface?

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