8a-ex. Stress Transformation Examples
Ex. 8a.1Ex. 8a.2
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Example 8a.1

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 1. The axial stress in the plate is:
sx = P/A = P/(Wt) = 5.0 MPa

Step 2. The Stress Transformation Equations are:

Here, sy and txy are both zero, so the equations simplify to:

       sx' = 0.5 sx(1+cos2q) = stress normal to weld
       sy' = 0.5 sx (1–cos2q)
    tx'y' = –0.5 sx (sin2q) = Ave. shear stress parallel to weld

Therefore : sx' = 3.75 MPa,   sy' = 1.25 MPa,   tx'y' = –2.17 MPa

Or, the normal and shear stresses acting on the weld are:

sw = 3.75 MPa,  |tw| = 2.17 MPa

Note: the Principal of Invariance for the normal stresses is satisfied:

sx' + sy'  =  sx + sy =   3.75 MPa + 1.25 MPa = 5.00 MPa

Example 8a.2

Given: An element is subjected to the following stress:
          sx = 10 ksi;  sy = 20 ksi;  txy = 5 ksi.

(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, tmax, the angles of the vectors that are normal to the faces on which they act, and the associated normal stresses.

Stress Element

Step 1. For the element rotated by 15° :

sx' = 13.2 ksi,   sy' = 16.8 ksi,  tx'y' = 6.83 ksi

Step 2. Principal Stresses and Principal Angles.

The Principal Stresses are:

and occur at angles rotated byqp:

For the element:

sI  = 22.1 ksi at qI = 67.5°       sII = 7.93 ksi at qII = 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 qs from the x-axis:


tmax,1 = 7.07 ksi on face:   qs,1 = 22.5°

tmax,2 = –7.07 ksi on face:   qs,2 = 113°

ss,x = ss,y = save = 15 ksi

A positive tmax means the shear stress causes a counterclockwise rotation on the face defined by qs. ... the shear stress on the xs face is positive.

A negative tmax means the shear stress causes a clockwise rotation on the face defined by qs... the shear stress on the xs face is negative.

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Updated: 05/23/09 DJD