# HWsolution9 - AE 321 – Homework#9 Solution 1(a For a bar...

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Unformatted text preview: AE 321 – Homework #9 Solution 1. (a) For a bar having a non-circular cross-section subjected to torsion, the stress functions, y x , must satisfy the following relation (obtained from the compatibility equations) 2 x 2 2 y 2 2 (1) where is the material shear modulus and is the angle of twist per unit length of the bar. Then 2 x 2 C 4 x 2 3 h 2 x 2 3 h C 6 x 2 h and 2 y 2 C 6 x 2 h substituting the previous results in (1) we have 2 x 2 2 y 2 C 6 x 2 h C 6 x 2 h 4 hC 2 C 1 2 h (2) substituting (2) in (1) we have 1 2 h x 3 y 2 3 h x 3 y 2 3 h x 1 3 h (3) Now, let’s applied the boundary condition (torque) at the end surface T 2 x , y dA A (4) substituting (3) in (4) we obtain 3 15 3 1 3 2 3 3 2 3 2 2 2 4 3 2 3 9 3 2 3 3 h T dydx h x h y x h y x h T h h h x Therefore 15 3 T h 4 (5) substituting (5) in (3) we obtain the stress function for the torsion of a long cylinder of solid triangular cross section subjected to a torque T at its end face x , y 15 3 2 T h 5 x 3 y 2 3 h x 3 y 2 3 h x 1 3 h (6) (b.1) Stress components xx yy zz xy xz y 15 3 T h 5 y 3 x h and yz x 15 3 T 2 h 5 2 hx 3 x 2 3 y 2 (b.2) Strain components (constitutive equations) ij 1 E 1 ij ij kk xx yy zz xy xz 1 E xz 15 3 T 1 Eh 5 y 3 x h and yz 1 E yz 15 3 T 1 2 Eh 5 2 hx 3 x 2 3 y 2 (b.3) Displacement components xx u x u u y , z (7) yy v y v v x , z (8) zz w z w w x , y (9) xy 1 2 u y v x u y v x (10) substituting (7) and (8) in (10) x z x v y z y u , , z g z yf u and z p z xf v (11) xz 1 2 u z w x A 3 xy hy where A 15 3 T 1 Eh 5 (12) substituting (11) in (12) we have ) 3 ( 2 ) , ( hy xy A x y x w...
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HWsolution9 - AE 321 – Homework#9 Solution 1(a For a bar...

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