Question

1 Stress tensor example 3.3-1 Given the following stress tensor in MPa: 21 -63 42 [0]= -63 O 84 | 42 84 -21] B (0,1,0) A (1.00) • Find the stress vector on the plane with normal: î = 1(2ê, – 3ê, +6ê,) HINT ") = 0;n; • Find the stress vector on the plane parallel to the plane ABC: HINT FIND PLANE EQUATION AND NORMAL - HIND TUE Stross VECTOR ON THE PLANE PARAUEL TO THE PLANE ABC. SOLUTION THE EQUATION OF A PLANE IS ax + by + cz =d WHERE ITS NORMAL IS ñ = [a b c]. SINCE AB = (1,0,0)-(0,1,0) = (1-10) CB = (0,0,2)-(0,1,0) = (0-1,2 AB X CB = T-2 -2 -17 15 PERPENIBICULAR TO THE PLANE Thus : -2% -29-12=d is The EQUATION OF THE PLANE SUBSTITUTING A (1,0,0) INTO IT YIELDS d=-2 so THAT: 2x +2y +2 =2 EQUATION OF THE PLANE ABC û zêntzertêz - 160 Vz+ 2 + 7 = 3/22, +382 +ês) OUTWARD NORMAL WE CAN NOW COMPUTE THE STRASS VECTOR : 1 [21 63 427 | z zi LŽ Ž / |-63 o 84 = 1-14 -14 77 | 42 th -21 J

Here are a question and the solution for its part b. I didn't understand how the vector multiplication of AB and CB gave us the normal to the plane? And also the questions ask about the stress vector parallel to the plane but we're calculating the normal vector perpendicular to the plane. Don't we need normal vector parallel to the plane?

Answer 1

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