15 printed from httpwwwmemtuedumavablemom2ndhtm plan

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Unformatted text preview: shows that the only nonzero stress component is τxs. It can be assumed uniform in the n direction because the tube is thin. Figure 5.50c shows a free-body diagram of a differential element with an imaginary cut through points A and B. By equilibrium of forces in the x direction we obtain τA ( tA d x ) = τB ( tB d x ) τA tA = τB tB (5.24b) qA = qB January, 2010 (5.24a) (5.24c) M. Vable Mechanics of Materials: Torsion of Shafts 5 249 The quantity q = τ xs t is called shear flow5 and has units of force per unit length. Equation (5.24c) shows that shear flow is uniform at a given cross section. We can replace the shear stresses (shear flow) by an equivalent internal torque, as shown in Figure 5.51. The line OC is perpendicular to the line of action of the force dV, which is in the tangent direction to the arc at that point. Noting that the shear flow is a constant, we take it outside the integral sign, T= ∫ ∫ ° d T = °q ( h d s ) ∫ ° = q 2 dA E = 2 qA E T q = -------2 AE or (5.25) We thus obtain Tτ xs = ---------- (5.26) 2 tA E where T is the internal torque at the section containing the point at which the shear stress is to be calculated, AE is the area enclosed by the centerline of the tube, and t is the thickness at the point where the shear stress is to be calculated. Area enclosed AE OAB dAE dV q ds B ds A C O dT h x 1 (h ds) 2 dAE h dV h(q ds) perpendicular distance from origin to force dF O T x Figure 5.51 Equivalency of internal torque and shear stress (flow). The thickness t can vary with different points on the cross section provided the assumption of thin-walled is not violated. If the thickness varies, then the shear stress will not be constant on the cross section, even though the shear flow is constant. EXAMPLE 5.15 A semicircular thin tube is subjected to torques as shown in Figure 5.52. Determine: (a) The maximum torsional shear stress in the tube. (b) The torsional shear stress at point O. Show the results on a stress cube. 70 in kips A B Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Figure 5.52 Thin-walled tube in Example 5.15. 1 8 50 in kips x in 20 in kips O C 5 in D O 3 16 in x Cross section PLAN From Equation 5.26 we know that the maximum torsional shear stress will exist in a section where the internal torque is maximum and the thickness minimum. To determine the maximum internal torque, we make cuts in AB, BC, and CD and draw free-body diagrams by taking the right part of each cut to avoid calculating the wall reaction. SOLUTION 5 This terminology is from fluid mechanics, where an incompressible ideal fluid has a constant flow rate in a channel. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 250 Figure 5.53 shows the free-body diagrams after making an imaginary cut and taking the right part. TAB 70 in kips TBC 50 in kips 50 in kips 20 in kips TAB TAB 50 40 70 20 in kips 0 TCD 20 in kips T TBC 50 30 0 20 in kips T TCD in kips 20 20 0 in kips Figure 5.53 Internal torque calculations in Example 5.15. (a) The maximum torque is in AB and the minimum thickness is 1 -8 2 2 in. The enclosed area is A E = π ( 5 in. ) ⁄ 2 = 12.5 π in. . From Equation 5.26 we obtain ( 40 π in.· kips ) τ max = -------------------------------------------2 -( 12.5 π in. ) ( 1 in. ) 8 (E1) τ max = 25.6 ksi ANS. (b) At point O the internal torque is TBC and t = 3----16 in. We obtain the shear stress at O as ( 30 π in.· kips ) τ O = ----------------------------------------------2 3( 12.5 π in. ) ( ----- in. ) 16 (E2) τ O = 12.8 k si ANS. Figure 5.54 shows part of the tube between sections B and C. Segment BO would rotate counterclockwise with respect to segment OC. The shear stress must be opposite to this possible motion and hence in the clockwise direction, as shown. The direction on the other surfaces can be drawn using the observation that the symmetric pair of shear stress components either point toward the corner or away from it. 70 in kips 50 in kips B Figure 5.54 Direction of shear stress in Example 5.15. O O O x C x COMMENT 1. The shear flow in the cross-section containing point O is a constant over the entire cross-section. The magnitude of torsional shear stress at point O however will be two-thirds that of the value of the shear stress in the circular part of the cross-section because of the variation in wall thickness. PROBLEM SET 5.4 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Torsion of thin-walled tubes 5.94 Calculate the magnitude of the maximum torsional shear stress if the cross section shown in Figure P5.94 is subjected to a torque T = 100 in.·kips. t R Figure P5.94 January, 2010 1 4 R 2 in 4 in in 2 in M. Vable Mechanics of Materials: Torsion of Shafts 5 251 5.95 Calculate the magnitude of the maximum torsional shear stress if the cross section shown in Figure P5.95 is subjected to a torque T = 900 N·m. t 3 mm R t Figure P5.95 50 mm t 5 mm 6 mm 100 mm 5.96 Calculate the magnitude of the maximum torsional shear stress if the cross section shown in Figure P5.96 is subje...
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This note was uploaded on 04/08/2010 for the course ENGR 232 taught by Professor Smith during the Spring '10 term at Aarhus Universitet.

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