Example 8.1.1

Given: A welded plate carries a force of P = 50 kN. The width of the plate W = 200 mm, and the thickness, t = 50 mm. The weld is at 30° from the vertical.

Req'd: The Normal Stress and Average Shear Stress in the weld.

Sol'n:
Step 1: The axial stress in the plate is:

Here, s_y and t_xy are both zero, so the equations simplify to:

s_x' = 0.5 s_x(1+cos2q) = Stress acting Normal to Weld
s_y' = 0.5 s_x (1-cos2q)
t_x'y' = -0.5 s_x (sin2q) = Ave. Shear Stress Parallel to Weld

Or, the Normal and Shear Stresses acting on the weld are:

Note: the Principal of Invariance requires that:

s_x' + s_y'  =  s_x + s_y =   3.75 MPa + 1.25 MPa = 5.00 MPa

Example 8.1.2

Given: An element is subjected to the following stress:
s_x = 10 ksi; s_y = 20 ksi, and t_xy = 5 ksi.

Req'd:
(a) If the element is rotated q = 15°, what are the new stresses.
(b) What are the Principal Stresses and their Angles.
(c) What is the Maximum In-Plane Shear Stress, t_max, the angles of the vectors that are normal to the faces on which they act, and the associated normal stresses.

Stress Element

Sol'n:
Step 1: The Stress Transformation equations are:

The Maximum In-Plane Shear Stress is given by:

and act on element faces that have outward pointing vectors rotated by qs from the x-axis as defined by:

Thus:

• A "positive" t_max means the shear stress causes a COUNTERCLOCKWISE rotation on the face defined by q_s. ... the shear stress on the x_s face is positive.
• A "negative" t_max means the shear stress causes a CLOCKWISE rotation on the face defined by q_s... the shear stress on the x_s face is negative.