10 and noting that max 00375 mm using the limits on

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Unformatted text preview: a, respectively. Determine the maximum torque that can be applied to wheel B. T Bronze Aluminum A x 0.75 m Figure 5.46 Shaft in Example 5.13. C B 2m PLAN We will follow the procedure of Section 5.3.1 and solve for the maximum shear stress in aluminum and bronze in terms of T. We will obtain the two limiting values on T to meet the limitations on maximum shear stress and determine the maximum permissible value of T. SOLUTION We can find the polar moment of inertia and the torsional rigidities as 4 J = π ( 0.075 m ) ⁄ 32 = 3.106 × 10 –6 m 4 3 G AB J AB = 83.87 ( 10 ) N ·m 2 3 G BC J BC = 139.8 ( 10 ) N ·m 2 (E1) Step 1: Let TA, the reaction torque at A, be clockwise with respect to the x axis. We can make imaginary cuts in AB and BC and draw the free-body diagrams as shown in Figure 5.47. Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm TA TAB TA T TBC A A B Figure 5.47 Free-body diagrams in Example 5.13. Step 2: From equilibrium of moment about shaft axis in Figure 5.47 we obtain the internal torques in terms of TA and T. T AB = T A (E2) T BC = T A – T Step 3: Using Equation (5.12), we obtain the relative rotation in each segment ends as T AB ( x B – x A ) T A ( 0.75 ) –6 φ B – φ A = ------------------------------ = -------------------------- = 8.942 ( 10 ) T A 3 G AB J AB 83.87 ( 10 ) (E3) ( TA – T ) ( 2 ) T BC ( x C – x B ) –6 φ C – φ B = ------------------------------ = --------------------------- = ( 14.31 T A – 14.31 T ) ( 10 ) 3 G BC J BC 139.8 ( 10 ) (E4) Step 4: We obtain φ C – φ A by adding Equations (E3) and (E4) and equate it to zero to find TA in terms of T: –6 φ C – φ A = ( 8.942 T A + 14.31 T A – 14.31 T ) ( 10 ) = 0 January, 2010 or 14.31 T T A = -------------------------------- = 0.6154 T 8.942 + 14.31 (E5) M. Vable Mechanics of Materials: Torsion of Shafts 5 242 Step 5: We obtain the internal torques in terms of T by substituting Equation (E5) into Equation (E2): T AB = 0.6154 T (E6) T BC = – 0.3846 T The maximum shear stress in segment AB and BC can be found in terms of T using Equation (5.10) and noting that ρmax = 0.0375 mm. Using the limits on shear stress we obtain the limits on T as T AB ( ρ AB ) max ( 0.6154 T ) ( 0.0375 m ) 6 2 ( τ AB ) max = ----------------------------- = ---------------------------------------------------- = 100 ( 10 ) N/m –6 4 J AB 3.106 ( 10 ) m or T ≤ 13.46 ( 10 ) N ·m T BC ( ρ BC ) max ( 0.3846 T ) ( 0.0375 m ) 6 2 ( τ BC ) max = ------------------------------ = ---------------------------------------------------- = 120 ( 10 ) N/m –6 4 J BC 3.106 ( 10 ) m or T ≤ 25.84 ( 10 ) N ·m 3 (E7) 3 (E8) The value of T that satisfies Equations (E7) and (E8) is the maximum value we seek. ANS. T max = 13.4 kN·m. COMMENTS 1. The maximum torque is limited by the maximum shear stress in bronze. If we had a limitation on the rotation of the wheel, then we could easily incorporate it by calculating φB from Equation (E3) in terms of T. 2. We could have solved this problem by the displacement method. In that case we would carry the rotation of the wheel φB as the unknown. 3. We could have solved the problem by initially assuming that one of the materials reaches its limiting stress value, say aluminum. We can then do our calculations and find the maximum stress in bronze, which would exceed the limiting value of 120 MPa. We would then resolve the problem. The process, though correct, can become tedious as the number of limitations increases. Instead put off deciding which limitation dictates the maximum value of the torque toward the end. In this way we need to solve the problem only once, irrespective of the number of limitations. EXAMPLE 5.14 A solid steel (G = 80 GPa) shaft is securely fastened to a hollow bronze (G = 40 GPa) shaft as shown in Figure 5.48. Determine the maximum value of shear stress in the shaft and the rotation of the right end with respect to the wall. 75 kN-m A 120 80 m mm m B Figure 5.48 Composite shaft in Example 5.14. 2m PLAN The steel shaft and the bronze shaft can be viewed as two independent shafts. At equilibrium the sum of the internal torques on each material is equal to the applied torque. The compatibility equation follows from the condition that a radial line on steel and bronze will rotate by the same amount. Hence, the relative rotation is the same for each length segment. Solving the equilibrium equation and the compatibility equation we obtain the internal torques in each material, from which the desired quantities can be found. SOLUTION We can find the polar moments and torsional rigidities as Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm π4 –6 4 J S = ----- ( 0.08 m ) = 4.02 ( 10 ) m 32 3 π4 4 –6 4 J Br = ----- [ ( 0.12 m ) – ( 0.08 m ) ] = 16.33 ( 10 ) m 32 G S J S = 321.6 ( 10 ) N ⋅ m January, 2010 2 3 G Br J Br = 653.2 ( 10 ) N ⋅ m 2 (E1) (E2) M. Vable Mechanics of Materials: Torsion of Shafts 5 243 Figure 5.49a shows the free body diagram after ma...
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