# Direction & magnitude of normal stress

• Rhaegal
In summary, for the state of stress shown in the figure, the normal stress acting on the plane of maximum shear stress can be found using the equation σn = σcos2θ, where σ is the maximum stress of 100MPa and θ is the angle of the plane of maximum shear stress. Using this equation, the normal stress is found to be 25 MPa (Tensile). However, this can also be solved using an analytical approach by considering the stress tensor and unit normal vector to the plane. At the angle of maximum shear stress, the normal stress is equal to the average of the normal stresses on the two orthogonal planes.
Rhaegal

## Homework Statement

For the state of stress shown in the figure, normal stress acting on the plane of maximum shear stress is?

## Homework Equations

Normal stress σn = σcos2θ

3. The Attempt at a Solution

Plane of maximum shear stress is 45o relative to the max stress of 100MPa.

So the total normal stress is 75MPa (compression).

I have assume the direction of normal stresses in both the cases with respect to applied stress.
Have I assumed the direction in both cases correctly?

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Rhaegal said:
Have I assumed the direction in both cases correctly?
No. You have not correctly taken into account that one is positive and the other negative.

I suggest looking at Mohr’s circle.

Orodruin said:
No. You have not correctly taken into account that one is positive and the other negative.

I suggest looking at Mohr’s circle.

Thanks for your suggestion. I have solved it using Mohr's cirlcle and found the normal stress to be 25 MPa (Tensile).

But the problem is to be used solved using analytical approach in order to get an intuitive grasp on stress transformations. So can you please tell me how to correctly assume the direction of normal stresses with respect to my earlier approach?

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Well, in one case it is negative and in the other positive. You get ##100\cos^2(\theta) - 50 \sin^2(\theta)## and for ##\theta = \pi/4## this translates to ##(100-50)/2 = 25## MPa.

Rhaegal
For this system, the stress tensor is given in dyadic notation by: $$\vec{\ \sigma}=\sigma_{xx}\vec{i_x}\vec{i_x}+\sigma_{yy}\vec{i_y}\vec{i_y}$$A unit normal vector to a plane of arbitrary orientation is given by: $$\vec{\ n}=\cos{\theta}\vec{i_x}+\sin{\theta}\vec{i_y}$$where ##\theta## is the angle relative to the x axis. If we dot the stress tensor with the unit normal, we obtain the traction vector on the plane: $$\vec{\ t}=\vec{\ \sigma}\centerdot \vec{\ n}=\sigma_{xx}\cos{\theta}\vec{i_x}+\sigma_{yy}\sin{\theta}\vec{i_y}$$The shear component (shear stress) of this vector on the plane is obtained by dotting the traction vector with the unit tangent to the plane: $$\vec{\ t}\centerdot \vec{i_t}=\vec{\ t}\centerdot (-\sin{\theta}\vec{i_x}+\cos{\theta}\vec{i_y})=(\sigma_{yy}-\sigma_{xx})\sin{\theta}\cos{\theta}=\frac{(\sigma_{yy}-\sigma_{xx})}{2}\sin{2\theta}$$The normal stress component of the traction vector is obtained by dotting the traction vector with the unit normal: $$\vec{\ t}\centerdot \vec{\ n}=\vec{\ t}\centerdot (\cos{\theta}\vec{i_x}+\sin{\theta}\vec{i_y})=\sigma_{xx}\cos^2{\theta}+\sigma_{yy}\sin^2{\theta}=\frac{(\sigma_{xx}+\sigma_{yy})}{2}+\frac{(\sigma_{xx}-\sigma_{yy})}{2}\cos{2\theta}$$The shear stress has maximum magnitude at ##2\theta=\pi/2##. At this value of theta, the normal stress is equal to ##\frac{(\sigma_{xx}+\sigma_{yy})}{2}##

Rhaegal

## 1. What is normal stress and how is it defined?

Normal stress is a type of mechanical stress that is exerted on a material in a direction perpendicular to the surface. It is defined as the force per unit area acting on a material.

## 2. How is the direction of normal stress represented?

The direction of normal stress is represented by a vector that is perpendicular to the surface on which the stress is acting. The direction of the vector is always outward from the surface for tensile stress and inward for compressive stress.

## 3. What factors affect the magnitude of normal stress?

The magnitude of normal stress is affected by the amount of force applied, the area over which the force is applied, and the material's ability to resist deformation. The larger the force or the smaller the area, the higher the magnitude of normal stress will be.

## 4. How is normal stress measured in practical applications?

Normal stress is typically measured using a device called a strain gauge, which measures the amount of deformation in a material due to the applied force. The magnitude of normal stress can then be calculated using the relationship between force, area, and deformation.

## 5. What are some real-world examples of normal stress?

Normal stress can be found in many everyday objects and structures, such as buildings, bridges, and car frames. It is also a crucial factor in the design of airplanes and other vehicles that experience high forces and pressures. Additionally, normal stress is important in understanding the strength and stability of bones and other biological tissues in the human body.

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