Stress (mechanics) - Maximum and Minimum Shear Stresses

Maximum and Minimum Shear Stresses

The maximum shear stress or maximum principal shear stress is equal to one-half the difference between the largest and smallest principal stresses, and acts on the plane that bisects the angle between the directions of the largest and smallest principal stresses, i.e. the plane of the maximum shear stress is oriented from the principal stress planes. The maximum shear stress is expressed as

Assuming then

The normal stress component acting on the plane for the maximum shear stress is non-zero and it is equal to

Derivation of the maximum and minimum shear stresses
The normal stress can be written in terms of principal stresses as $\begin{align} \sigma_\mathrm{n} &= \sigma_{ij}n_in_j \\ &=\sigma_1n_1^2 + \sigma_2n_2^2 + \sigma_3n_3^2\\ \end{align} \,\!$ Knowing that, the shear stress in terms of principal stresses components is expressed as $\begin{align} \tau_\mathrm{n}^2 &= \left( T^{(n)} \right)^2-\sigma_\mathrm{n}^2 \\ &=\sigma_1^2n_1^2+\sigma_2^2n_2^2+\sigma_3^2n_3^2-\left(\sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2\right)^2 \\ &=(\sigma_1^2-\sigma_2^2)n_1^2+(\sigma_2^2-\sigma_3^2)n_2^2+\sigma_3^2-\left^2 \\ &= (\sigma_1-\sigma_2)^2n_1^2n_2^2+(\sigma_2-\sigma_3)^2n_2^2n_3^2+(\sigma_1-\sigma_3)^2n_1^2n_3^2 \\ \end{align} \,\!$ The maximum shear stress at a point in a continuum body is determined by maximizing subject to the condition that This is a constrained maximization problem, which can be solved using the Lagrangian multiplier technique to convert the problem into an unconstrained optimization problem. Thus, the stationary values (maximum and minimum values)of occur where the gradient of is parallel to the gradient of . The Lagrangian function for this problem can be written as $\begin{align} F\left(n_1,n_2,n_3,\lambda\right) &= \tau^2 + \lambda \left(g\left(n_1,n_2,n_3\right) - 1 \right) \\ &= \sigma_1^2n_1^2+\sigma_2^2n_2^2+\sigma_3^2n_3^2-\left(\sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2\right)^2+\lambda\left(n_1^2+n_2^2+n_3^2-1\right)\\ \end{align} \,\!$ where is the Lagrangian multiplier (which is different from the use to denote eigenvalues). The extreme values of these functions are thence These three equations together with the condition may be solved for and By multiplying the first three equations by and, respectively, and knowing that we obtain Adding these three equations we get $\begin{align} \left-2\left(\sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2\right)\sigma_\mathrm{n}+\lambda\left(n_1^2+n_2^2+n_3^2\right)&=0 \\ \left-2\sigma_\mathrm{n}^2+\lambda&=0 \\ \left-2\sigma_\mathrm{n}^2+\lambda &=0 \\ \lambda &= \sigma_\mathrm{n}^2-\tau_\mathrm{n}^2 \\ \end{align}\,\!$ this result can be substituted into each of the first three equations to obtain $\begin{align} \frac{\partial F}{\partial n_1} = n_1\sigma_1^2-2n_1\sigma_1\left(\sigma_1 n_1^2+\sigma_2 n_2^2+\sigma_3 n_3^2\right)+\left(\sigma_\mathrm{n}^2-\tau_\mathrm{n}^2\right) n_1 &= 0 \\ n_1\sigma_1^2-2n_1\sigma_1\sigma_\mathrm{n}+\left(\sigma_\mathrm{n}^2-\tau_\mathrm{n}^2\right) n_1 &= 0 \\ \left(\sigma_1^2-2\sigma_1\sigma_\mathrm{n}+\sigma_\mathrm{n}^2-\tau_\mathrm{n}^2\right) n_1 &= 0 \\ \end{align}\,\!$ Doing the same for the other two equations we have A first approach to solve these last three equations is to consider the trivial solution . However this options does not fulfill the constrain . Considering the solution where and, it is determine from the condition that, then from the original equation for it is seen that . The other two possible values for can be obtained similarly by assuming and and Thus, one set of solutions for these four equations is: $\begin{align} n_1&=0,\,\,n_2&=0,\,\,n_3&=\pm1,\,\,\tau_\mathrm{n}&=0 \\ n_1&=0,\,\,n_2&=\pm1,\,\,n_3&=0,\,\,\tau_\mathrm{n}&=0 \\ n_1&=\pm1,\,\,n_2&=0,\,\,n_3&=0,\,\,\tau_\mathrm{n}&=0 \\ \end{align}\,\!$ These correspond to minimum values for and verifies that there are no shear stresses on planes normal to the principal directions of stress, as shown previously. A second set of solutions is obtained by assuming and . Thus we have To find the values for and we first add these two equations $\begin{align} \sigma_2^2-\sigma_3^2-2\sigma_2\sigma_n+2\sigma_2\sigma_n&=0 \\ \sigma_2^2-\sigma_3^2-2\sigma_n\left(\sigma_2-\sigma_3\right)&=0 \\ \sigma_2+\sigma_3&=2\sigma_n \\ \end{align}\,\!$ Knowing that for and we have $\begin{align} \sigma_2+\sigma_3&=2\sigma_n \\ \sigma_2+\sigma_3&=2\left(\sigma_2n_2^2 + \sigma_3n_3^2\right) \\ \sigma_2+\sigma_3&=2\left(\sigma_2n_2^2 + \sigma_3\left(1-n_2^2\right)\right)&=0 \\ \end{align}\,\!$ and solving for we have Then solving for we have and $\begin{align} \tau_\mathrm{n}^2&=(\sigma_2-\sigma_3)^2n_2^2n_3^2 \\ \tau_\mathrm{n}&=\frac{\sigma_2-\sigma_3}{2}\end{align}\,\!$ The other two possible values for can be obtained similarly by assuming and and Therefore the second set of solutions for, representing a maximum for is Therefore, assuming, the maximum shear stress is expressed by and it can be stated as being equal to one-half the difference between the largest and smallest principal stresses, acting on the plane that bisects the angle between the directions of the largest and smallest principal stresses.

Derivation of the maximum and minimum shear stresses

The normal stress can be written in terms of principal stresses as

$\begin{align} \sigma_\mathrm{n} &= \sigma_{ij}n_in_j \\ &=\sigma_1n_1^2 + \sigma_2n_2^2 + \sigma_3n_3^2\\ \end{align} \,\!$

Knowing that, the shear stress in terms of principal stresses components is expressed as

$\begin{align} \tau_\mathrm{n}^2 &= \left( T^{(n)} \right)^2-\sigma_\mathrm{n}^2 \\ &=\sigma_1^2n_1^2+\sigma_2^2n_2^2+\sigma_3^2n_3^2-\left(\sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2\right)^2 \\ &=(\sigma_1^2-\sigma_2^2)n_1^2+(\sigma_2^2-\sigma_3^2)n_2^2+\sigma_3^2-\left^2 \\ &= (\sigma_1-\sigma_2)^2n_1^2n_2^2+(\sigma_2-\sigma_3)^2n_2^2n_3^2+(\sigma_1-\sigma_3)^2n_1^2n_3^2 \\ \end{align} \,\!$

The maximum shear stress at a point in a continuum body is determined by maximizing subject to the condition that

This is a constrained maximization problem, which can be solved using the Lagrangian multiplier technique to convert the problem into an unconstrained optimization problem. Thus, the stationary values (maximum and minimum values)of occur where the gradient of is parallel to the gradient of .

The Lagrangian function for this problem can be written as

$\begin{align} F\left(n_1,n_2,n_3,\lambda\right) &= \tau^2 + \lambda \left(g\left(n_1,n_2,n_3\right) - 1 \right) \\ &= \sigma_1^2n_1^2+\sigma_2^2n_2^2+\sigma_3^2n_3^2-\left(\sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2\right)^2+\lambda\left(n_1^2+n_2^2+n_3^2-1\right)\\ \end{align} \,\!$

where is the Lagrangian multiplier (which is different from the use to denote eigenvalues).

The extreme values of these functions are

thence

These three equations together with the condition may be solved for and

By multiplying the first three equations by and, respectively, and knowing that we obtain

Adding these three equations we get

$\begin{align} \left-2\left(\sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2\right)\sigma_\mathrm{n}+\lambda\left(n_1^2+n_2^2+n_3^2\right)&=0 \\ \left-2\sigma_\mathrm{n}^2+\lambda&=0 \\ \left-2\sigma_\mathrm{n}^2+\lambda &=0 \\ \lambda &= \sigma_\mathrm{n}^2-\tau_\mathrm{n}^2 \\ \end{align}\,\!$

this result can be substituted into each of the first three equations to obtain

$\begin{align} \frac{\partial F}{\partial n_1} = n_1\sigma_1^2-2n_1\sigma_1\left(\sigma_1 n_1^2+\sigma_2 n_2^2+\sigma_3 n_3^2\right)+\left(\sigma_\mathrm{n}^2-\tau_\mathrm{n}^2\right) n_1 &= 0 \\ n_1\sigma_1^2-2n_1\sigma_1\sigma_\mathrm{n}+\left(\sigma_\mathrm{n}^2-\tau_\mathrm{n}^2\right) n_1 &= 0 \\ \left(\sigma_1^2-2\sigma_1\sigma_\mathrm{n}+\sigma_\mathrm{n}^2-\tau_\mathrm{n}^2\right) n_1 &= 0 \\ \end{align}\,\!$

Doing the same for the other two equations we have

A first approach to solve these last three equations is to consider the trivial solution . However this options does not fulfill the constrain .

Considering the solution where and, it is determine from the condition that, then from the original equation for it is seen that . The other two possible values for can be obtained similarly by assuming

and

Thus, one set of solutions for these four equations is:

$\begin{align} n_1&=0,\,\,n_2&=0,\,\,n_3&=\pm1,\,\,\tau_\mathrm{n}&=0 \\ n_1&=0,\,\,n_2&=\pm1,\,\,n_3&=0,\,\,\tau_\mathrm{n}&=0 \\ n_1&=\pm1,\,\,n_2&=0,\,\,n_3&=0,\,\,\tau_\mathrm{n}&=0 \\ \end{align}\,\!$

These correspond to minimum values for and verifies that there are no shear stresses on planes normal to the principal directions of stress, as shown previously.

A second set of solutions is obtained by assuming and . Thus we have

To find the values for and we first add these two equations

$\begin{align} \sigma_2^2-\sigma_3^2-2\sigma_2\sigma_n+2\sigma_2\sigma_n&=0 \\ \sigma_2^2-\sigma_3^2-2\sigma_n\left(\sigma_2-\sigma_3\right)&=0 \\ \sigma_2+\sigma_3&=2\sigma_n \\ \end{align}\,\!$

Knowing that for

and

we have

$\begin{align} \sigma_2+\sigma_3&=2\sigma_n \\ \sigma_2+\sigma_3&=2\left(\sigma_2n_2^2 + \sigma_3n_3^2\right) \\ \sigma_2+\sigma_3&=2\left(\sigma_2n_2^2 + \sigma_3\left(1-n_2^2\right)\right)&=0 \\ \end{align}\,\!$

and solving for we have

Then solving for we have

and

$\begin{align} \tau_\mathrm{n}^2&=(\sigma_2-\sigma_3)^2n_2^2n_3^2 \\ \tau_\mathrm{n}&=\frac{\sigma_2-\sigma_3}{2}\end{align}\,\!$

The other two possible values for can be obtained similarly by assuming

and

Therefore the second set of solutions for, representing a maximum for is

Therefore, assuming, the maximum shear stress is expressed by

and it can be stated as being equal to one-half the difference between the largest and smallest principal stresses, acting on the plane that bisects the angle between the directions of the largest and smallest principal stresses.

Read more about this topic: Stress (mechanics)

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