Given: A welded plate supports 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: Determine the Normal Stress and Average Shear Stress in the weld.

Step 2. The Stress Transformation Equations are:

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

       s_x' = 0.5 s_x(1+cos2q) = stress 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 for the normal stresses is satisfied:

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

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

Req'd:
(a) If the element is rotated q = 15°, determine the new stresses.
(b) Determine the Principal Stresses and their Angles.
(c) Determine 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. For the element rotated by 15° :

s_x' = 13.2 ksi,   s_y' = 16.8 ksi,  t_x'y' = 6.83 ksi

Step 2. Principal Stresses and Principal Angles.

The Principal Stresses are:

and occur at angles rotated byq_p:

For the element:

s_I  = 22.1 ksi at q_I = 67.5°       s_II = 7.93 ksi at q_II = 113°

Step 3. Maximum Shear Stress.

The Maximum In-Plane Shear Stress is:

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

Thus:

t_max,1 = 7.07 ksi on face:   q_s,1 = 22.5°

t_max,2 = –7.07 ksi on face:   q_s,2 = 113°

s_s,x = s_s,y = s_ave = 15 ksi

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.