By plotting the internal torque as a function of x we

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Unformatted text preview: O ⎛ ---- ⎞ ⎝ L 2⎠ φ → O( ) ⎛ FL ⎞ T -------- → O ⎜ -----------⎟ → O ( ) → checks F ⎜ ---- 3⎟ Gr 3 -L ⎠ ⎝ L2 T FL F --- → O ⎛ ------ ⎞ → O ⎛ ---- ⎞ → checks ⎝ L3 ⎠ ⎝ L 2⎠ r3 F τ → O ⎛ ---- ⎞ ⎝ L 2⎠ COMMENT 1. The direction of the shear stress can be determined using subscripts or intuitively, as shown in Figure 5.35. (b) (a) Negative x x Figure 5.35 Shear stress opposing counterclockwise motion of left segment B Direction of shear stress in Example 5.9: (a) by subscripts; (b) by inspection. EXAMPLE 5.10 A uniformly distributed torque of q in.·lb/in. is applied to an entire shaft, as shown in Figure 5.36. In addition to the distributed torque a concentrated torque of T = 3qL in.·lb is applied at section B. Let the shear modulus be G and the radius of the shaft r. In terms of q, L, G, and r, determine: (a) The rotation of the section at C. (b) The maximum shear stress in the shaft. T 3qL in lb L q in lb in C Figure 5.36 Shaft and loading in Example 5.10. L 2L Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm PLAN (a) The internal torque in segments AB and BC as a function of x must be determined first. Then the relative rotation in each section is found by integrating Equation (5.9). (b) Since J and ρmax are constant over the entire shaft, the maximum shear stress will exist on a section where the internal torque is maximum. By plotting the internal torque as a function of x we can determine its maximum value. Figure 5.37 shows the free body diagrams after imaginary cuts are made at x distance from A and internal torques drawn as per our sign convention. We replace the distributed torque by an equivalent torque that is equal to the distributed torque intensity multiplied by the length of the cut shaft (the rectangular area). From equilibrium of moment about the shaft axis in Figure 5.37 we obtain the internal torques: T AB + 3 qL – q ( 3 L – x ) = 0 T BC – q ( 3 L – x ) = 0 3 or or T AB = – q x (E1) T BC = q ( 3 L – x ) (E2) O( ) represents the dimension of the quantity on the left. F represents dimension for the force. L represents the dimension for length. Thus shear modulus, which has dimension of force (F) per unit area (L2), is represented as O(F/L2 ). January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts T (a) 5 227 3qL in lb L q in in T 3qL in lb L q (3 L x) C C TAB TAB 3L 3L x q(3L x) x (b) q in lb in C C TBC TBC 3L 3L x x Figure 5.37 Free-body diagrams in Example 5.10 after imaginary cut in segment (a) AB, and (b) BC. Integrating Equation (5.9) for each segment we obtain the relative rotations of segment ends as T AB –q x ⎛ dφ⎞ = ----------------- = ------------------⎝ d x⎠ AB G AB J AB G π r4 ⁄ 2 or φB ∫φ A d φ = –∫ T BC q(3L – x) ⎛ dφ⎞ = ---------------- = ---------------------⎝ d x⎠ BC G BC J BC Gπr4 ⁄ 2 2 2q x φ C – φ B = ------------ ⎛ 3 Lx – ----⎞ 2⎠ G π r4 ⎝ 3L L x B= L 2 qx------------ d x G π r4 x A =0 or φC ∫φ dφ = B x C =3 L ∫x =L B 2 L qx 2 φ B – φ A = – -----------Gπr4 or 2 0 qL 2 = – -----------4 Gπr (E3) 2q(3L – x) ------------------------- d x or Gπr4 2 L 2q (3L) 4 qL = ------------ 9 L 2 – ------------- – 3 L 2 + ---- = -----------4 4 2 2 Gπr Gπr (E4) (a) Adding Equations (E3) and (E4), we obtain the rotation of the section at C with respect to the section at A: qL 2 2 4 qL φ C – φ A = – ------------ + -----------G π r 4 G π r4 (E5) ANS. ⎛ 3 qL 2 ⎞ φ C – φ A = = ⎜ ------------ ⎟ ccw 4 ⎝Gπr ⎠ (b) Figure 5.38 shows the plot of the internal torque as a function of x using Equations (E1) and (E2). The maximum torque will occur on a section just to the right of B. From Equation (5.10) the maximum torsional shear stress is T max ρ max ( 2 qL ) ( r ) τ max = ---------------------- = --------------------4 J πr ⁄ 2 (E6) ANS. 4 qL --------3 πr τ max = T 2qL A Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Figure 5.38 Torque diagram in Example 5.10. B L C 3L x qL Dimension check: The dimensional consistency (see footnote 12) of our answers is checked as follows: FL q → O ⎛ ------ ⎞ → O ( F ) ⎝ L⎠ φ → O( ) r → O(L) L → O(L) F G → O ⎛ ---- ⎞ ⎝ L 2⎠ FL 2 qL 2 -------- → O ⎛ ----------------------- ⎞ → O ( ) → checks 4 ⎝ ( F ⁄ L 2 ) L 4⎠ Gr Fτ → O ⎛ ---- ⎞ ⎝ L 2⎠ qL FL F ------ → O ⎛ ------ ⎞ → O ⎛ ---- ⎞ → checks ⎝ L3 ⎠ ⎝ L 2⎠ r3 COMMENT 1. A common mistake is to write the incorrect length of the shaft as a function of x in the free-body diagrams. It should be remembered that the location of the cut is defined by the variable x, which is measured from the common origin for all segments. Each cut produces two parts, and we are free to choose either part. January, 2010 M. Vable 5.2.7* Mechanics of Materials: Torsion of Shafts 5 228 General Approach to Distributed Torque Distributed torques are usually due to inertial forces or frictional forces acti...
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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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