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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 = .3 a = .02 m b = .0125 m i. ( ) ( ) Pa 10 36923 mm N 26923 3 . 1 2 70000 1 2 6 2 = = + = + = E ii. The torsional constant, J is defined as ( ) 2 2 2 2 b a b a J + = 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 = + = J iii. Twist per unit length is J M t = 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 = = = Thus, the total twist is equal to rad 67344 . 58 . 38 4 = = o 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 = = = ab M t xs yz b a a b J M u t 2 2 2 2 = 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 = 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 = 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 This is because it satisfies the DE on the ring and it satisfies the b.cs = = = = 2 2 2 1 2 ) ( ) ( e i e i e a a a a r a r 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 Integrating, ( ) 4 4 2 i e a a J = b. Assume a linear approximation ( ) r a h k i = along c i , n=r h k n = The length of the contour c i =2 a To find k, we use the following relationship = i A n 2 Therefore, h a k i = Plugging into the equation for the torsional constant, [ ] 2 2 2 e i i a a h a J + = c. To obtain J 2 from J 1 ( )( ) 2 2 2 2 1 2 i e e i a a a a J + = 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 ( ) 2 2 1 e i m a a a J + = 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 + = = 4. Using the membrane analogy, a. Since must be 0 along the whole boundary, it is expected that the moment M t , which is proportional to the volume defined by will be much smaller in the open case than in the closed case. open case than in the closed case....
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 Spring '04
 Lambros

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