# Solution2 - AE 321 Solution to Practice Problems Chapter 2...

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AE 321 – Solution to Practice Problems Chapter 2: Traction and Stress 1. Determine the components of the traction on a given plane and for a given state of stress. The components of the traction at a given plane is given by j ji j ij n i n n T ! = = ) ( or T n ( ) [ ] = [ ] n [ ] , (2.1) where ij represent the symmetric stress tensor and j n represents the outward normal unit vector of the plane under consideration. (a) State of stress: ij [ ] = 0 0 0 0 0 0 " # \$ \$ \$ % & unit vector normal to the plane: n j [ ] = 1 2 ! 1 2 0 " # \$ \$ \$ \$ \$ \$ \$ % & After substituting the known values into Equation (2.1), we obtained the traction on the plane under consideration

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T n ( ) [ ] = ! 0 0 0 0 0 0 " # \$ \$ \$ % & 1 2 ( 1 2 0 " # \$ \$ \$ \$ \$ \$ \$ % & = 2 2 ( 2 2 0 " # \$ \$ \$ \$ \$ \$ \$ % & or ! T n ( ) = 2 2 " e 1 " " e 2 ( ) (b) State of stress: ij [ ] = 0 " 0 0 0 0 # \$ % % % & ( ( ( the plane under consideration is the same as in (a). Applying Equation (2.1) T n ( ) [ ] = 0 " 0 0 0 0 # \$ % % % & ( ( ( 1 " 1 0 # \$ % % % % % % % & ( ( ( ( ( ( ( = 0 0 # \$ % % % & ( ( ( or ! T n ( ) = 2 " e 2 2. Given: [ ] KPa ij ! ! ! " # \$ \$ \$ % & = 3 2 1 2 15 2 1 2 20 ( a. Compute the components of traction on the plane passing through P whose outward normal vector n makes equal angles with the coordinate axes. Determination of n : Equal angle condition: cos 3 2 1 = " = " = " e n e n e n ! ! ! ! ! ! then, 2 1 cos cos e e n ! ! ! ! + ! = "
because n ˆ is an unit vector, cos 3 cos cos cos 1 2 2 2 2 ! = + + = " = " n n n n ! ! ! ! hence, 3 1 cos ± = , or ( ) 3 2 1 3 1 ˆ e e e n ! ! ! + + = KPa 3 6 3 11 3 23 3 1 3 1 3 1 3 2 1 2 15 2 1 2 20 3 2 1 3 2 1 ! ! ! ! ! ! " # \$ \$ \$ \$ \$ \$ % & = ! ! ! " # \$ \$ \$ % & ! ! ! ! ! ! " # \$ \$ \$ \$ \$ \$ % & ! ! ! " # \$ \$ \$ % & = ! ! ! " # \$ \$ \$ % & = = T T T T T T n T j ij n i ( b. Compute the normal and tangential components of traction on this plane. Normal component: n T n nn ! = " = magnitude of normal component ( ) KPa e e e e e e nn nn 6 3 18 3 1 3 6 3 11 3 23 3 2 1 3 2 1 = = + + ! " " # \$ % % & + ( = ) ! ! ! " ! !

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Thus: ( ) ( ) ( ) 3 2 1 3 2 1 3 2 3 1 6 e e e N e e e n N nn ! ! ! " ! ! ! ! " + + = ! " # \$ % & + + = = Tangential component: [ ] 3 2 17 1 1 3 17 3 17 3 6 3 6 3 6 3 11 3 6 3 23 2 1 3 2 1 = + = ! = " " # \$ % % & ! + " " # \$ % % & ! ! + " " # \$ % % & ! = ! = S e e e e e N T S ! ! ! ! ! Thus the magnitude of S is 13.88044 KPa. 3. State of stress at point P. ! ij [ ] = 10 2 1 2 " 15 5 1 5 3 # \$ % % % & ( ( ( KPa (a) Compute components of traction on the plane passing trough P whose outward normal vector n makes equal angles with the coordinate axes.