Question

Take og = 440 kPa (Figure 1) Part A Determine the normal stress acting on the inclined plane AB Express your answer to three significant figures and include the appropriate units. 01 = 190.52 kPa Submit Previous Answers Request Answer X Incorrect; Try Again; 13 attempts remaining Figure < 1 of 1 > Part B Determine the shear stress acting on the inclined plane AB. Express your answer to three significant figures and include the appropriate units. CA* * oo ? T2y = 330 kPa < 30° Submit Previous Answers Request Answer X Incorrect; Try Again; 14 attempts remaining

Answer 1

Aim:

To determine Normal Stress and Shear Stress acting on the inclined plane AB

Solution:

As per given information,

A body is subject to stress in x - direction as

$\sigma _{x} = 440 kPa$

Plane AB is inclined at an angle of

A = 30°

There is no stress acting in y - direction

..σ, = 0

The body is not subjected to any kind of shear stress.

.. Try = 0

Normal Stress acting on the inclined plane is given by the following relationship.

Or+yy cos 20 + Try sin 20

$\mathbf{\sigma _{y^{'}}} =\frac{\sigma _{x}+\sigma _{y}}{2}-\frac{\sigma_{x}-\sigma_{y}}{2}\cos 2\theta-\tau_{xy}\sin 2\theta$but we have to find only the value of $\sigma _{x^{'}}$ .

Also we have values

$\sigma _{x} = 440 kPa$

$\sigma _{y} = 0$

$\tau _{xy} = 0$

A = 30°

Now substitute all the values in the above relationship.

$\therefore \mathbf{\sigma _{x^{'}}} =\frac{440+0}{2}+\frac{440-0}{2}\cos (2*30^{o})+0*\sin( 2*30^{o})$$\therefore \mathbf{\sigma _{x^{'}}} =\frac{440}{2}+\frac{440}{2}\cos (60^{o})+0$

$\therefore \mathbf{\sigma _{x^{'}}} =220+110$

$\therefore \mathbf{\sigma _{x^{'}}} =330 kPa$

Shear Stress acting on the inclined plane is given by the following relationship

$\mathbf{\tau _{x^{'}y^{'}}} =-\frac{\sigma_{x}-\sigma_{y}}{2}\sin 2\theta+\tau_{xy}\cos 2\theta$

Substitute all the values in the above relationship.

$\therefore \mathbf{\tau _{x^{'}y^{'}}} =-\frac{440-0}{2}\sin( 2*30^{o})+0*\cos (2*30^{o})$$\therefore \mathbf{\tau _{x^{'}y^{'}}} =-\frac{440}{2}\sin( 60^{o})+0$

$\therefore \mathbf{\tau _{x^{'}y^{'}}} =-190.5255 kPa$

$\therefore \mathbf{\tau _{x^{'}y^{'}}} =190.5255 kPa.....(magnitude)$​​​​​

Answer:

$\mathbf{\sigma _{x^{'}}} =330.000 kPa$

$\mathbf{\tau _{x^{'}y^{'}}} =190.5255 kPa$