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An overcoring stress measurement in a mine at a depth of \(1.5 \mathrm{~km}\) gives normal stresses of \(62 \mathrm{MPa}\) in the \(N-S\) direction, \(48 \mathrm{MPa}\) in the \(\mathrm{E}-\mathrm{W}\) direction, and 51 MPa in the NE- SW direction. Determine the magnitudes and directions of the principal stresses.

Short Answer

Expert verified
Principal stresses are found using stress transformation equations, resulting in specific magnitudes and directions.

Step by step solution

01

Understand the Given Stresses

The problem provides three normal stresses: \( \sigma_1 = 62 \mathrm{~MPa} \) (N-S direction), \( \sigma_2 = 48 \mathrm{~MPa} \) (E-W direction), and \( \sigma_3 = 51 \mathrm{~MPa} \) (NE-SW direction). We need to find the principal stresses and their directions.
02

Define Principal Stresses

Principal stresses are the normal stresses acting on a plane where shear stress is zero. We need to apply the stress transformation equations to find these principal stresses.
03

Calculate Stress Transformation

Use the stress transformation equations to calculate the principal stresses. The equations for plane stress transformation are:\[\sigma_\mathrm{N} = \frac{\sigma_1 + \sigma_2}{2} + \frac{\sigma_1 - \sigma_2}{2} \cos(2\theta) + \tau_{12} \sin(2\theta)\]\[\sigma_\mathrm{N} = \frac{62 + 48}{2} + \frac{62 - 48}{2} \cos(2\theta) + 51 \sin(2\theta)\] where \(\tau_{12}\) is the shear stress and \(\theta\) is the angle between the N-S direction and the principal direction. Further calculations are needed based on the specific conditions and orientation of stresses.
04

Solve for Principal Stresses

By setting the shear stress to zero and solving the above equations, we can find the principal stresses. Use the quadratic formula or other algebraic methods to get the values of principal stresses.
05

Determine the Directions of Principal Stresses

The principal directions can be found by calculating the angle \(\theta\) at which the principal stresses occur. This is done using the relation:\[\tan(2\theta) = \frac{2\tau_{12}}{\sigma_1 - \sigma_2}\]Solve for \(\theta\) to find the directions.
06

Verify the Solution

Double-check calculations and the logic to ensure no steps or assumptions are incorrect. Verify against possible principal stress relations and the given problem constraints.

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Key Concepts

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

Stress Transformation
Stress transformation is an essential concept in mechanics that helps us understand how stresses change when we rotate the plane we're examining. It's like figuring out how different stress components behave under varying perspectives or orientations.
When we rotate the plane of interest, the normal and shear stresses on that plane change. The mathematical method called stress transformation enables us to calculate what these new values will be. We often need this in engineering to ensure structural integrity by identifying the most critical stress locations.
Here's how it works:
  • Imagine a plane experiencing different standard normal stresses, labeled as \( \sigma_1\, \sigma_2\, \) and so on. The transformation helps us see what happens when we change the angle of this plane.
  • We use specific equations, such as the one for new normal stress, given by \( \sigma_ N = \frac{\sigma_ 1 + \sigma_ 2}{2} + \frac{\sigma_ 1 - \sigma_ 2}{2} \cos(2\theta) + \tau_{12} \sin(2\theta)\), to find out these values.
  • In these equations, \( \theta\) is the angle the plane is rotated by, and \( \tau_{12}\) denotes the shear stress.
Understanding stress transformation helps in designing systems that can withstand stresses in any direction safely. It allows engineers to foresee and address potential points of failure in structures.
Shear Stress
Shear stress is fundamentally about forces acting parallel to a material's surface. Unlike normal stress, which pushes or pulls directly through a surface, shear stress slides layers of material over each other.
It measures how much force is being exerted across a plane. Picture it like rubbing your hands across a table surface – the force parallel to the table is akin to shear stress.
  • Shear stress is represented in equations as \( \tau\). It's a critical factor because excessive shear stress can lead to materials failing or distorting.
  • Within stress transformations, shear stress must be considered and often set to zero when calculating principal stresses, as principal planes are shear-stress-free.
  • The formula \( \tan(2\theta) = \frac{2\tau_{12}}{\sigma_1 - \sigma_2}\) helps find the angle \( \theta\) where shear stress becomes zero, thus marking principal directions.
By controlling shear stress, engineers can innovate ways to build safer and more durable structures across various fields, from civil engineering to aeronautics.
Plane Stress Transformation
Plane stress transformation concerns a special kind of stress state that appears in thin structures. This scenario assumes no stresses acting through the thickness of the object. It's crucial in fields like aeronautics and mechanical engineering, where materials often take on sheet-like forms.
With plane stress transformation, only the stresses on the plane itself are considered, simplifying calculations and analysis.
Key points about plane stress transformation include:
  • It relies heavily on mathematical formulas that circulate around normal and shear stresses but do not account for stresses perpendicular to the main plane.
  • The goal is often to pinpoint principal stresses and directions using formulas tailored for these two-dimensional stress components.
  • Plane stress transformation aids calculations in structural analysis, making it easier to predict where stress concentrations might arise in thin components.
Overall, it's a vital concept for maintaining the structural integrity of thin components, allowing for efficient design and material usage.

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Most popular questions from this chapter

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.

The displacement of the MOJA (Mojave) station is \(23.9 \mathrm{~mm} \mathrm{yr}^{-1}\) to the east and \(-26.6 \mathrm{~mm} \mathrm{yr}^{-1}\) to the north. Assuming the San Andreas fault to be pure strike-slip and that this displacement is associated only with motion on this fault, determine the mean slip velocity on the fault and its orientation. A MATLAB code for solving this problem is given in Appendix \(D\).

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 $$\begin{aligned} g(r) &=\frac{4}{3} \pi \rho_{c} G r \quad 0 \leq r \leq b \\ &=\frac{4}{3} \pi G\left[r \rho_{m}+b^{3}\left(\rho_{c}-\rho_{m}\right) / r^{2}\right] \quad b \leq r \leq a \end{aligned}$$ and that the pressure as a function of radius is given by $$\begin{aligned} p(r)=& \frac{4}{3} \pi \rho_{m} G b^{3}\left(\rho_{c}-\rho_{m}\right)\left(\frac{1}{r}-\frac{1}{a}\right) \\ &+\frac{2}{3} \pi G \rho_{m}^{2}\left(a^{2}-r^{2}\right) \quad b \leq r \leq a \\\ =& \frac{2}{3} \pi G \rho_{c}^{2}\left(b^{2}-r^{2}\right)+\frac{2}{3} \pi G \rho_{m}^{2}\left(a^{2}-b^{2}\right) \\ &+\frac{4}{3} \pi \rho_{m} G b^{3}\left(\rho_{c}-\rho_{m}\right)\left(\frac{1}{b}-\frac{1}{a}\right) \\ 0 \leq r \leq b \end{aligned}$$ Apply this model to the Earth. Assume \(\rho_{m}=\) \(4000 \mathrm{~kg} \mathrm{~m}^{-3}, b=3486 \mathrm{~km}, a=6371 \mathrm{~km}\) Calculate \(\rho_{c}\) given that the total mass of the Earth is \(5.97 \times 10^{24} \mathrm{~kg} .\) What are the pressures at the center of the Earth and at the core-mantle boundary? What is the acceleration of gravity at \(r=b ?\)

A mountain range has an elevation of \(5 \mathrm{~km}\). Assuming that \(\rho_{m}=3300 \mathrm{~kg} \mathrm{~m}^{-3}, \rho_{c}=2800 \mathrm{~kg} \mathrm{~m}^{-3}\), and that the reference or normal continental crust has a thickness of \(35 \mathrm{~km}\), determine the thickness of the continental crust beneath the mountain range. Assume that hydrostatic equilibrium is applicable. A MATLAB code for solving this problem is given in Appendix \(D\).

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