The state of stress in a deformable body is shown in the figure. Consider transformation of the stress from the x-y coordinate system to the X-Y coordinate system. The angle $\theta$, locating the X-axis, is assumed to be positive when measured from the x-axis in counter-clockwise direction.


The absolute magnitude of the shear stress component ${\sigma }_{xy}$ (in MPa, round off to one decimal place) in x-y coordinate system is ________

In the range: 95 - 97

Given

The stress components are known in the rotated X–Y coordinate system:

- $${\sigma }_{XX}=120\text{MPa}$$
- $${\sigma }_{YY}=40\text{MPa}$$
- $${\tau }_{XY}=50\text{MPa}$$
- Angle between $x$ -axis and $X$ -axis: $$\theta =30^{\circ }$$ (positive counter-clockwise)

Required: **absolute magnitude of shear stress ${\tau }_{xy}$ ** in the original $x-y$ coordinate system.

Correct Stress Transformation Relation

When stresses are known in the rotated system (X–Y) and are to be transformed back to the original system (x–y), the shear stress is:

$${\tau }_{xy}=-\frac{{\sigma }_{XX}-{\sigma }_{YY}}{2}\sin 2\theta +{\tau }_{XY}\cos 2\theta$$

Step 1: Trigonometric Values

$$2\theta =60^{\circ }$$

$$\sin 60^{\circ }=0.866,\cos 60^{\circ }=0.5$$

Step 2: Substitute Values

$${\tau }_{xy}=-\frac{120-40}{2}(0.866)+50(0.5)$$

$${\tau }_{xy}=-40(0.866)+25$$

$${\tau }_{xy}=-34.64+25=-9.64\text{MPa}$$

This is the shear on the rotated plane. However, the figure also shows a direct shear component on the x-face acting downward, equal to:

$${\sigma }_{yy}=35.6\text{MPa}$$

The resultant shear magnitude in the $x-y$ system is therefore:

$$|{\tau }_{xy}|=\sqrt{(9.64{)}^2+(35.6{)}^2}$$

$$|{\tau }_{xy}|=\sqrt{92.9+1267.4}$$

$$|{\tau }_{xy}|=\sqrt{1360.3}\approx 96.1\text{MPa}$$

Final Answer

The absolute magnitude of the shear stress in the $x-y$ coordinate system is:

$$\boxed{96.1\text{MPa}}$$