T t ext t1 a t2 2 b 2 25 c d 25 x 6 6 figure 531

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Unformatted text preview: 29a to guide us, we can draw the torque diagram. SOLUTION Let LA be an imaginary extension on the left side of the shaft, as shown in Figure 5.30. Clearly the internal torque in the imaginary section LA is zero, that is, T1 = 0. The torque at A is in the same direction as the torque Text shown on the template in Figure 5.29a. Using the template equation, we subtract the value of the applied torque to obtain a value of –2π in.·kips for the internal torque T2 just after wheel A. This is the starting value in the internal torque diagram. 2 in kips 8 in kips L 8.5 in kips A 2.5 in kips B Figure 5.30 Imaginary extensions of the shaft in Example 5.7. C R D We approach wheel B with an internal torque value of –2π in.·kips, that is, T1 = –2π in.·kips. The torque at B is in the opposite direction to the torque shown on the template in Figure 5.29a we add 8π in·kips to obtain a value of +6π in.·kips for the internal torque just after wheel B. T 6 T ext 6 T1 A T2 Figure 5.31 Torque diagram in Example 5.7. T2 1 2 B C 2 2.5 D x 2.5 T We approach wheel C with a value of 6π in.·kips and note that the torque at C is in the same direction as that shown on the template in Figure 5.29a. Hence we subtract 8.5π in.·kips as per the template equation to obtain –2.5π in.·kips for the internal torque just after wheel C. The torque at D is in the same direction as that on the template, and on adding we obtain a zero value in the imaginary extended bar DR as expected, for the shaft is in equilibrium. From Figure 5.31 the internal torque values in the segments are Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm T AB = – 2 π in.· kips T BC = 6 π in.· kips T CD = – 2.5 π in.· kips (E1) To obtain the relative rotation of wheel D with respect to wheel A, we substitute the torque values in Equation (E1) into Equation (5.13): T AB ( x B – x A ) T BC ( x C – x B ) T CD ( x D – x C ) φ D – φ A = ------------------------------ + ------------------------------- + ------------------------------- or G AB J AB G BC J BC G CD J CD ( – 2 π in.· kips ) ( 24 in. ) ( 6 π in.· kips ) ( 60 in. ) ( – 2.5 π in.· kips ) ( 30 in. ) –3 φ D – φ A = -------------------------------------------------------- + ---------------------------------------------------- + ----------------------------------------------------------- = 16.75 ( 10 ) rad 4 4 4 ( 12 ,000 ksi ) ( π ⁄ 2 in. ) ( 12 ,000 ksi ) ( 8 π in. ) ( 12 ,000 ksi ) ( π ⁄ 2 in. ) ANS. (E2) φ D – φ A = 0.01675 rad cw COMMENT 1. We could have created the torque diagram using the template shown in Figure 5.29b and the template equation. It may be verified that we obtain the same torque diagram. This shows that the direction of the applied torque Text on the template is immaterial. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 224 EXAMPLE 5.8 A 1-m-long hollow shaft in Figure 5.32 is to transmit a torque of 400 N·m. The shaft can be made of either titanium alloy or aluminum. The shear modulus of rigidity G, the allowable shear stress τallow, and the density γ are given in Table 5.1. The outer diameter of the shaft must be 25 mm to fit existing attachments. The relative rotation of the two ends of the shaft is limited to 0.375 rad. Determine the inner radius to the nearest millimeter of the lightest shaft that can be used for transmitting the torque. TABLE 5.1 Material properties in Example 5.8 G (GPa) τallow (MPa) γ (Mg/m3) Titanium alloy 36 450 4.4 Aluminum 28 150 2.8 Material 25 m m 1m Figure 5.32 Shaft in Example 5.8. PLAN The change in inner radius affects only the polar moment J and no other quantity in Equations 5.10 and 5.12. For each material we can find the minimum polar moment J needed to satisfy the stiffness and strength requirements. Knowing the minimum J for each material we can find the maximum inner radius. We can then find the volume and hence the mass of each material and make our decision on the lighter shaft. SOLUTION We note that for both materials ρmax = 0.0125 m and x2 – x1 = 1 m. From Equations 5.10 and 5.12 for titanium alloy we obtain limits on JTi shown below. ( 400 N ⋅ m ) ( 1 m ) ( Δφ ) Ti = --------------------------------------------- ≤ 0.375 rad 9 2 [ 36 ( 10 ) N/m ] J Ti –9 J Ti ≥ 29.63 ( 10 ) m or ( 400 N ⋅ m ) ( 0.0125 m ) 6 2 ( τ max ) Ti = --------------------------------------------------------- ≤ 450 ( 10 ) N/m J Ti 4 (E1) –9 J Ti ≥ 11.11 ( 10 ) m or 4 (E2) Using similar calculations for the aluminum shaft we obtain the limits on JAl: ( 400 N ⋅ m ) ( 1 m ) ( Δφ ) Al = ---------------------------------------------------- ≤ 0.375 rad 9 2 [ 28 ( 10 ) N/m ] × J Al ( 400 N ⋅ m ) ( 0.0125 m ) 6 2 ( τ max ) Al = -------------------------------------------------------- ≤ 150 ( 10 ) N/m J Al –9 –9 J Al ≥ 38.10 ( 10 ) m or or 4 –9 J Al ≥ 33.33 ( 10 ) m 4 (E3) 4 (E4) –9 4 Thus if J Ti ≥ 29.63 ( 10 ) m , it will meet both conditions in Equations (E1) and (E2). Similarly if...
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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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