Torsion

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138 CHAPTER 5. TORSION A stress function of the following form is assumed x2 Φ = A a b B i 3 a A Figure 5.12: A bar with an elliptical cross-section. 2 + x3 2 − 1 , b where A is an unknown constant. The boundary condition, eq. (5.67), is clearly satisfied since Φ = 0 along C. Substituting this in the governing differential equation, eq. (5.68), it follows that this is satisfied for the following value of the constant, A A The stress function then becomes 2 2 + a2 b2 Φ = − a2 b 2 a 2 + b 2 = −2Gκ1; =⇒ A = − a2 b 2 x2 The torque can now be computed from eq. (5.73) as M1 = − 2a2b2 a2 x2 Gκ1 + b2 a Note that A dA = πab; A a A 2 + x3 2 − 1 b 2 x 2 2 dA = πa3 b 4 ; + i 2 a2 Gκ1. + b2 x3 2 − 1 b A Gκ1. (5.74) dA. x 2 3 dA = πab3 4 ; are the area, and the second moments of area about ī2 and ī3, respectively, for the ellipse. Using these, the torque then can be written as M1 = G πa3 b 3 a 2 + b 2 κ1 = J κ1, where the torsional stiffness of the elliptical section is now defined as J = G πa3b3 a2 . (5.75) + b2 Using these results, the stress function can be expressed in terms of the applied torque Φ = − 1 x2 2 x3 2 + − 1 M1. πab a b The stress distribution is found from eqs. (5.64) τ12 = − 2x3 πab 3 M1; τ13 = 2x2 πa 3 b M1,
5.3. TORSION OF BARS WITH ARBITRARY CROSS-SECTIONS 139 and the maximum shear stress occurs at the extreme values of x2 and x3 which occur at the boundary of the section. The shear stress distribution is shown in fig. (5.13). Thus, at points A and B the stresses are τ B 12 = − 2M1 πab 2 ; τ A 13 = 2M1 πa 2 b . Somewhat surprisingly, the maximum shear stress occurs at the end of the minor axis of the ellipse, i.e. at point B where |τmax| = 2M1 . πab2 This is in contrast to the torsion of a bar with circular cross-section where the maximum shear stress occurs at points with the largest radius. Fig. (5.13b) shows a contour plot of the stress along with superposed arrows showing the direction of the maximum shear stress. For this section, as for the circular section, the directions are parallel to the section outline when considered along any radial line. 12 B 13 (a) Shear stress magnitude i 3 A i 2 i 3 axis i2 axis (b) Shear contours Figure 5.13: Shear stress and warping distributions on elliptic cross section. Finally, the warping function can be obtained by integrating eq. (5.65). Substituting the calculated stress function (5.74) into these equations yields ∂Ψ ∂x2 = − a2 − b2 a2 x3; + b2 ∂Ψ ∂x3 = − a2 − b2 a2 x2. + b2 The first equation can be integrated with respect to x2 and the second with respect to x3 to yield and the second with respect to x2 to yield Ψ = − a2 − b 2 a 2 + b 2 x2x3 + f(x3) Ψ = − a2 − b 2 a 2 + b 2 x2x3 + g(x2). These two solutions for Ψ must be equal,and this is only possible if f(x2) = g(x3) = 0 which yields the result Ψ = − a2 − b2 a2 x2x3. + b2 The warping displacement, u1(x2, x3), can now be written by substituting this result for Ψ into eq. (5.50) a u1(x2, x3) = −κ1 2 − b2 a2 + b2 x2x3. (5.76) This result shows that the warping is zero on the ī2 and ī3 axes, and that it is negative in the first quadrant (where both x2 and x3 are positive). It then alternates sign in the second, third and fourth quadrants.
Page 1 and 2: Chapter 5 Torsion In the previous c
Page 3 and 4: 5.1. TORSION OF CIRCULAR CYLINDERS
Page 9 and 10: 5.2. STRENGTH UNDER COMBINING LOADI
Page 11 and 12: 5.3. TORSION OF BARS WITH ARBITRARY
Page 15: 5.3. TORSION OF BARS WITH ARBITRARY
Page 25 and 26: 5.4. TORSION OF A THIN RECTANGULAR
Page 27 and 28: 5.5. TORSION OF THIN-WALLED OPEN SE
Page 29 and 30: 5.5. TORSION OF THIN-WALLED OPEN SE
Page 31 and 32: 5.6. THE MEMBRANE ANALOGY 153 the s
Page 33 and 34: 5.6. THE MEMBRANE ANALOGY 155 b t i

Torsion

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