• Author / Uploaded
• Fredel De Vera

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65

Stress Principles Example 3.3 Assume the stress tensor T (in ksi) at P with respect to axes Px1 x2 x3 is represented by the matrix   18 0 −12   [tij ] =  0 6 0  . −12 0 24 If the x10 axis makes equal angles with the three unprimed axes, and the x20 axis lies in the plane of x10 x3 , as shown by the sketch, determine the primed components of T assuming Px10 x20 x30 is a right-handed system.

φ

x3

x2

β β

P

x1 x2

x3

β x1

Solution We must first determine the transformation matrix [aij ]. Let β be the common angle which x10 makes with the unprimed axes, as shown in the figure. Then a11 = a12 = a13 = cos √ β and from the orthogonality condition0 Eq 2.55 with . Next, let φ be the angle between x2 and x3 . Then i = j = 1, cos β = 1/ 3 √ a23 = cos φ = sin β = 2/ 6. As seen from the obvious symmetry of the axes arrangement, x20 makes equal angles with x1 and x2 , which means that a21√= a22 . Thus, again from Eq 2.55, with i = 1, j = 2, we have a21 = a22 = −1/ 6 (the minus sign is required because of the positive sign chosen for a23 ). For the ^30 = e ^10 × e ^20 , with the primed axes to be a√right-handed √ system we require e result that a31 = 1/ 2, a32 = −1/ 2, and a33 = 0. Finally, from Eq 3.33,     √1 √1 √1 √1 √1 √1 18 0 −12 − 3 3  6 2   3  3 h i    1  0 1 2  1 √ √ √ √ √ tij = − √1     − 6 0 6 0 − 6 − 12  6 6    3  √1 √1 √1 √2 − 0 −12 0 24 0 2 2 3 6  =

8  √  2 2  0

√ 2 2 28 √ −6 3

 0 √   −6 3 .  12