A welded square plate of 1 m × 1 m is subjected to biaxial stress of magnitude 6.5 MPa and 25 MPa as shown in figure. The ratio of normal stress acting in the perpendicular direction of weld to shear stress of weld is __________. (Round off to two decimal places)

The correct answer is 0.59.

1. Analysis of the stress state from the given figure:
Based on the provided illustration of the square plate (1 m × 1 m):
• The horizontal stress (${\sigma }_x$) is compressive, as indicated by the arrows pointing inward on the left and right faces of the plate. Thus, ${\sigma }_x=-6.5\text{MPa}$.
• The vertical stress (${\sigma }_y$) is tensile, as indicated by the arrows pointing outward on the top and bottom faces of the plate. Thus, ${\sigma }_y=25\text{MPa}$.
• There is no shear stress applied directly to the outer boundaries, so the shear stress component in the xy-coordinate system is ${\tau }_{xy}=0$.

2. Determination of the weld plane orientation:
The plate is a square of 1 m × 1 m, and the weld runs diagonally from the bottom-left corner to the top-right corner. Therefore, the weld line forms an angle of 45° with the horizontal axis. The normal to this weld plane is oriented at an angle of:

$\theta =-45°$

with respect to the horizontal x-axis (or $135°$).

3. Calculation of normal stress perpendicular to the weld (${\sigma }_n$):
The formula for normal stress on an inclined plane is:

${\sigma }_n=\frac{{\sigma }_x+{\sigma }_y}{2}+\frac{{\sigma }_x-{\sigma }_y}{2}\cos \left(2\theta \right)+{\tau }_{xy}\sin \left(2\theta \right)$

Substituting the known values (${\sigma }_x=-6.5\text{MPa}$, ${\sigma }_y=25\text{MPa}$, ${\tau }_{xy}=0$, and $2\theta =-90°$):

${\sigma }_n=\frac{-6.5+25}{2}+\frac{-6.5-25}{2}\cos (-90°)+0$

Since $\cos (-90°)=0$:

${\sigma }_n=\frac{18.5}{2}+0=9.25\text{MPa}$

4. Calculation of shear stress along the weld ($\tau$):
The formula for shear stress on an inclined plane is:

$\tau =-\frac{{\sigma }_x-{\sigma }_y}{2}\sin \left(2\theta \right)+{\tau }_{xy}\cos \left(2\theta \right)$

Substituting the known values:

$\tau =-\frac{-6.5-25}{2}\sin (-90°)+0$

Since $\sin (-90°)=-1$:

$\tau =-(-15.75)\times (-1)=-15.75\text{MPa}$

The magnitude of the shear stress is therefore $\left|\tau \right|=15.75\text{MPa}$.

5. Finding the ratio of normal stress to shear stress:
Now, we compute the ratio of the normal stress acting in the perpendicular direction of the weld to the magnitude of the shear stress along the weld:

$\text{Ratio}=\frac{{\sigma }_n}{\left|\tau \right|}=\frac{9.25}{15.75}\approx 0.5873$

Rounding to two decimal places gives 0.59.