HW4 Solutions

HW4 Solutions - Mechanics of Aircraft structures C.T Sun...

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Unformatted text preview: Mechanics of Aircraft structures C.T. Sun 2.7 Find the principal stresses and corresponding principal directions for the stresses given in Problem 2.6. Check the result with other methods such as Mohr’s circle. Solution: (a) The stress given in problem 2.6 is ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 3 2 2 4 ] [ ij σ , To find the principal stresses, we require that [ ] [ ] = − I ij σ σ , or 3 2 2 4 = − − − σ σ σ Expanding the determinant yields ( ) 8 7 2 = + − − σ σ σ , the solutions of σ are = σ , or 2 17 7 ± = σ , (which are 1.43845 and 5.56155) --- ANS (i) When 1 = σ We have the equations ⎪ ⎩ ⎪ ⎨ ⎧ = ⋅ = + = + 3 2 2 4 z y x y x n n n n n , and also we have ( ) ( ) ( ) 1 2 2 2 = + + z y x n n n So the solutions can be obtained uniquely as ⎪ ⎩ ⎪ ⎨ ⎧ = = = 1 z y x n n n , and is the corresponding principal direction ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 1 ) 1 ( z y x n n n--- ANS (ii) When 43845 . 1 2 = σ We have the equations ⎪ ⎩ ⎪ ⎨ ⎧ = − = + = + 43845 . 1 56155 . 1 2 2 56155 . 2 z y x y x n n n n n , and also we have ( ) ( ) ( ) 1 2 2 2 = + + z y x n n n 2.7.1 Mechanics of Aircraft structures C.T. Sun Therefore we have the corresponding principal direction ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ − = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 78821 . 61541 . ) 2 ( z y x n n n--- ANS (iii) When 56155 . 5 3 = σ We have the equations ⎪ ⎩ ⎪ ⎨ ⎧ = − = − = + − 56155 . 5 56155 . 2 2 2 56155 . 1 z y x y x n n n n n , and also we have ( ) ( ) ( ) 1 2 2 2 = + + z y x n n n Therefore we have the corresponding principal direction ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 61541 . 78821 . ) 3 ( z y x n n n--- ANS (b) Comparing with Mohr’s circle Since the stresses associated with z are all zero, we know one principal stress is 0, and its corresponding principal direction is . So here we can use the 2D analysis on the x-y plane just for other principal values. ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ 1 ) 1 ( z y x n n n (4,2) (3,-2) max σ min σ σ τ According to the Mohr’s circle, we have the radius of the circle 2.7.2 Mechanics of Aircraft structures C.T. Sun ( ) ( ) ( ) 06155 . 2 2 2 3 4 2 1 2 2 = + + − = r , The central coordinate of the circle is ( ) ( ) , 5 . 3 2 2 2 , 2 3 4 , = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + = c c y x Therefore we have the maximum and minimum stresses, respectively, 56155 . 5 06155 . 2 5 . 3 max = + = + = r x c σ 43845 . 1 06155 . 2 5 . 3 min = − = − = r x c σ These are the same as we obtained above....
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HW4 Solutions - Mechanics of Aircraft structures C.T Sun...

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