Question

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).

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.

Answer 1

Correct Answer: -0.59

Step 1: Note the plate's stress state.
The plate carries \(\sigma_x = 6.5\) MPa (horizontal faces) and \(\sigma_y = 25\) MPa (vertical faces), with no shear stress on these faces, so these are the principal stresses.

Step 2: Locate the weld's orientation.
The weld runs along the diagonal of the square, which sits at \(45^{\circ}\) to both \(\sigma_x\) and \(\sigma_y\). A plane at \(45^{\circ}\) to the principal directions is exactly the plane of maximum in-plane shear stress.

Step 3: Use the stress transformation equations at \(\theta = 135^{\circ}\), the normal to the weld.
\(\sigma_n = \dfrac{\sigma_x+\sigma_y}{2} + \dfrac{\sigma_x-\sigma_y}{2}\cos2\theta = \dfrac{6.5+25}{2} + \dfrac{6.5-25}{2}\cos270^{\circ} = 15.75 + 0 = 15.75\) MPa, since \(\cos270^{\circ}=0\).
\(\tau = -\dfrac{\sigma_x-\sigma_y}{2}\sin2\theta = -\dfrac{6.5-25}{2}\sin270^{\circ} = -(-9.25)(-1) = -9.25\) MPa.

Step 4: Form the required ratio.
The normal stress perpendicular to the weld is \(15.75\) MPa, and the shear stress along the weld carries a sign of \(-9.25\) MPa under this convention.
Ratio \(= \tau/\sigma_n = -9.25/15.75 = -0.59\).

Final Answer:
Because the weld sits exactly at \(45^{\circ}\) to the principal stresses, it always carries the maximum possible shear stress relative to the local normal stress. \[ \boxed{\text{ratio} = -0.59} \]