A welded square plate of 1 m \(\times\) 1 m is subjected to biaxial stress of magnitude 6.5 MPa and 25 MPa, as shown in the figure below. The ratio of the normal stress acting in the perpendicular direction of the weld to the shear stress of the weld is ________ (rounded off to 2 decimal places).
Show Hint
Locate the weld's orientation relative to the principal stresses using Mohr's circle; the sign of the ratio depends on the transformation convention used.
Step 1: Draw the Mohr's circle for the plate.
With principal stresses $\sigma_x = 6.5$ MPa and $\sigma_y = 25$ MPa and zero shear on the reference faces, the circle has center $C = (\sigma_x+\sigma_y)/2 = 15.75$ MPa and radius $R = (\sigma_y-\sigma_x)/2 = 9.25$ MPa.
Step 2: Place the weld plane on the circle.
A physical rotation of $45^{\circ}$ to reach the weld's normal corresponds to a $90^{\circ}$ rotation on the Mohr's circle, since angles double on the circle. Starting from the $\sigma_x$ point and rotating $90^{\circ}$ lands exactly at the top or bottom of the circle.
Step 3: Read off the stresses at that point.
At the top or bottom of the circle the normal stress equals the center value, $\sigma_n = C = 15.75$ MPa, and the shear stress equals the full radius, $|\tau| = R = 9.25$ MPa, since this is the point of maximum shear on the circle.
Step 4: Take the ratio with the sign from this rotation direction.
Following the convention used, $\tau = -9.25$ MPa while $\sigma_n = 15.75$ MPa stays positive, so the ratio is $-9.25/15.75 = -0.59$.
Final Answer:
Mohr's circle shows directly that a $45^{\circ}$ weld to the principal axes always sits at the point of maximum shear, giving this fixed ratio regardless of how large the actual stresses are.
\[ \boxed{\text{ratio} = -0.59} \]
