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Unformatted text preview: Mechanics of Aircraft structures C.T. Sun 3.1 Show that there is no warping in a bar of circular crosssection. Solution: (a) SaintVenant assumed that as the shaft twists the plane crosssections are warped but the projections on the xy plane rotate as a rigid body, then, zy u θ − = zx v θ = (3.1.1) ) , ( y x w θψ = where ) , ( y x ψ is some function of x and y, called warping function, and θ is the angle of twist per unit length of the shaft and is assumed to be very small. (b) From the displacement field above, it is easy to obtain that = = = = xy zz yy xx γ ε ε ε So from the stressstrain relationship, we have = = = = xy zz yy xx τ σ σ σ Therefore the equilibrium equations reduce to = ∂ ∂ + ∂ ∂ y x yz xz τ τ This equation is identically satisfied if the stresses are derived from a stress function ) , ( y x φ , so that y xz ∂ ∂ = φ τ , x yz ∂ ∂ − = φ τ (3.1.2) (c) From the displacement field and stressstrain relationship, we can obtain y x w z u x w xz θ γ − ∂ ∂ = ∂ ∂ + ∂ ∂ = (3.1.3) x y w z v y w yz θ γ + ∂ ∂ = ∂ ∂ + ∂ ∂ = (3.1.4) So it forms the compatibility equation θ γ γ 2 = ∂ ∂ − ∂ ∂ y x xz yz , or in terms of Prandtl stress function θ φ φ G y x 2 2 2 2 2 − = ∂ ∂ + ∂ ∂ (3.1.5) (d) Boundary conditions, = ds d φ , or . const = φ But for a solid sections with a single contour boundary, this constant can be chosen to be zero. Then we have the boundary condition = φ on the lateral surface of the bar. (e) For a bar with circular crosssection, assume the Prandtl stress function as 3.1.1 Mechanics of Aircraft structures C.T. Sun ) 1 ( 2 2 2 2 − + = a y a x C φ which satisfies the boundary conditions stated above. Substitute φ into (3.1.5), we obtain θ G a C 2 2 1 − = Then ) ( 2 2 2 2 a y x G − + − = θ φ Using (3.1.2), we have y y G xz θ φ γ − = ∂ ∂ = 1 , and x x G yz θ φ γ = ∂ ∂ − = 1 Comparing with (3.1.3) and (3.1.4), we have y y x w xz θ θ γ − = − ∂ ∂ = => = ∂ ∂ x w . Thus, ) ( y f w = x x y w yz θ θ γ = + ∂ ∂ = => = ∂ ∂ y w , Thus, ) ( x g w = Hence we conclude . This means that the crosssection remains plane after torsion. In other words, there is no warping. const w = Therefore can be verified, and it successfully expresses the statement. ) , ( = y x w ANS 3.1.2 Mechanics of Aircraft structures C.T. Sun 3.2 Show that the Prandtl stress function for bars of circular solid sections is also valid for bars of hollow circular sections as shown in Fig. 3.34. Find the torsion constant in terms of the inner radius and outer radius , and compare with the torsion constant obtained using (3.59) for thinwalled sections. What is the condition on the wall thickness for the approximate to be within 1 percent of the exact ?...
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This note was uploaded on 01/18/2012 for the course AAE 352 taught by Professor Chen during the Fall '08 term at Purdue.
 Fall '08
 Chen

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