# hw22solutions - AE 221 – Aerospace Structures II –...

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Unformatted text preview: AE 221 – Aerospace Structures II – Spring 2004 Chapter 2.2 – Beam Torsion – Solution 1. Relevant information M t = 400 N*m Length = 4 m E = 70,000 N/mm 2 ν = .3 a = .02 m b = .0125 m i. ( ) ( ) Pa 10 36923 mm N 26923 3 . 1 2 70000 1 2 6 2 × = = + = + = ν µ E ii. The torsional constant, J is defined as ( ) 2 2 2 2 b a b a J + = π In this instance, the torsional constant is ( ) ( ) ( ) ( ) 4 8 2 3 2 3 3 3 m 10 8247 . 8 10 5 . 12 10 20 10 5 . 12 10 20 − − − − − × = × + × × × = π J iii. Twist per unit length is J M t µ ϑ = plugging in the values for mu and the torsional constant, J ( ) ( ) m deg 646 . 9 m rad 16836 . 10 8247 . 8 10 26923 400 8 6 = = × × = − ϑ Thus, the total twist is equal to rad 67344 . 58 . 38 4 = = ⋅ o ϑ iv. The location of maximum stress is at (0, b ± ) Thus, the shear stress at that point is MPa 487 . 81 0125 . 02 . 800 2 2 2 max = ⋅ ⋅ = = π π σ ab M t xs yz b a a b J M u t 2 2 2 2 − − ⋅ = µ The maximum will be on the boundary 2 2 1 a y b z − = 2 2 2 2 2 2 1 a y a y ab b a a b J M u t − − − ⋅ = ⇒ µ u is maximum at 2 1 2 1 = ⇒ = b z a y Therefore 2 2 2 2 2 max ab b a a b J M u t − − ⋅ = µ Plugging in the appropriate values found previously, mm 0092 . max = u 2a. It happens that the solution for the ring is the same as that for the CIS. The exact solution is: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = 2 2 2 2 2 2 1 2 ) ( 1 2 ) , ( e e e e e a r a r a z a y a z y θ µ ψ θ µ ψ This is because it satisfies the DE on the ring and it satisfies the b.c’s ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = = = = 2 2 2 1 2 ) ( ) ( e i e i e a a a a r a r θ µ ψ ψ The torsional constant J is ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = = ∫ 2 2 2 2 2 2 2 1 1 1 2 e i e i a a e e t a a a a dr a r r a M J i e π π µθ Integrating, ( ) 4 4 2 i e a a J − = π b. Assume a linear approximation ( ) r a h k i − = ψ along c i , n=-r h k n = ∂ ∂ ⇒ ψ The length of the contour c i =2 π a To find k, we use the following relationship ∫ = ∂ ∂ i A n µθ ψ 2 Therefore, h a k i µθ = Plugging into the equation for the torsional constant, [ ] 2 2 2 e i i a a h a J + = π c. To obtain J 2 from J 1 ( )( ) 2 2 2 2 1 2 i e e i a a a a J − + = π Note that a m = “averaged radius” = ( ) 2 i e a a + Thus, ( ) h a a a J m e i ⋅ ⋅ + = 2 2 2 2 1 π ( ) 2 2 1 e i m a a a J + = π h The two solutions are the same if a i = a m . 3. Using the thin walled open cross section approximation, we get, f or the torsional constant, ( ) 3 2 3 t b a J + = The maximum stress is at: ( ) ( ) b a t M J t M t t xs 2 3 3 max + = = σ 4. Using the membrane analogy, a. Since ψ must be 0 along the whole boundary, it is expected that the moment M t , which is proportional to the “volume” defined by ψ will be much smaller in the open case than in the closed case. open case than in the closed case....
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hw22solutions - AE 221 – Aerospace Structures II –...

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