Because of the higher modulus of rigidity of titanium

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Unformatted text preview: J Al ≥ 38.10 ( 10 ) m , it will meet both conditions in Equations (E3) and (E4). The internal diameters DTi and DAl can be found as follows: π 4 4 –9 J Ti = ----- ( 0.025 – D Ti ) ≥ 29.63 ( 10 ) 32 D Ti ≤ 17.3 ( 10 ) m π 4 4 –9 J Al = ----- ( 0.025 – D Al ) ≥ 38.10 ( 10 ) 32 D Al ≤ 7.1 ( 10 ) m –3 (E5) –3 (E6) Rounding downward to the closest millimeter, we obtain –3 D Ti = 17 ( 10 ) m –3 D Al = 7 ( 10 ) m (E7) Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm We can find the mass of each material from the material density as 6 3π 2 2 2 M Ti = [ 4.4 ( 10 ) g/m ] -- ( 0.025 – 0.017 ) m ( 1 m ) = 1161 g 4 (E8) 6 3π 2 2 2 M Al = [ 2.8 ( 10 ) g/m ] -- ( 0.025 – 0.007 ) m ( 1 m ) = 1267 g 4 (E9) From Equations (E8) and (E9) we see that the titanium alloy shaft is lighter. ANS. A titanium alloy shaft should be used with an inside diameter of 17 mm. COMMENTS 1. For both materials the stiffness limitation dictated the calculation of the internal diameter, as can be seen from Equations (E1) and (E3). 2. Even though the density of aluminum is lower than that titanium alloy, the mass of titanium is less. Because of the higher modulus of rigidity of titanium alloy we can meet the stiffness requirement using less material than for aluminum. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 225 3. If in Equation (E5) we had 17.95(10–3) m on the right side, our answer for DTi would still be 17 mm because we have to round downward to ensure meeting the less-than sign requirement in Equation (E5). EXAMPLE 5.9 The radius of a tapered circular shaft varies from 4r units to r units over a length of 40r units, as shown in Figure 5.33. The radius of the uniform shaft shown is r units. Determine (a) the angle of twist of wheel C with respect to the fixed end in terms of T, r, and G; (b) the maximum shear stress in the shaft. 2.5 T A Figure 5.33 Shaft geometry in Example 5.9 T C B x 40 r 10 r PLAN (a) We can find the relative rotation of wheel C with respect to wheel B using Equation (5.12). For section AB we obtain the polar moment J as a function of x and integrate Equation (5.9) to obtain the relative rotation of B with respect to A. We add the two relative rotations and obtain the relative rotation of C with respect to A. (b) As per Equation (5.10), the maximum shear stress will exist where the shaft radius is minimum (J is minimum) and T is maximum. Thus by inspection, the maximum shear stress will exist on a section just left of B. SOLUTION We note that R is a linear function of x and can be written as R ( x ) = a + bx . Noting that at x = 0 the radius R = 4 r we obtain a = 4 r . Noting that x = 40 r the radius R = r we obtain b = – 3 r ⁄ ( 40 r ) = – 0.075 . The radius R can be written as R ( x ) = 4 r – 0.075 x (E1) Figure 5.34 shows the free body diagrams after imaginary cuts have been made and internal torques drawn as per our sign convention. By equilibrium of moment about the shaft axis we obtain the internal torques: (E2) T BC = T T AB + 2.5 T – T = 0 or (E3) T AB = – 1.5 T (b) T (a) TBC C 2.5T T B TAB C Figure 5.34 Free-body diagrams in Example 5.9 after imaginary cut in segment (a) BC (b) AB The polar moment of inertias can be written as Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm π π π J BC = -- r 4 J AB = -- R 4 = -- ( 4 r – 0.075 x ) 4 2 2 2 (a) We can find the relative rotation of the section at C with respect to the section at B using Equation (5.12): T BC ( x C – x B ) T ( 10 r ) 6.366 T φ C – φ B = ------------------------------ = ------------------------ = ---------------G BC J BC G ( π ⁄ 2 ) r4 Gr 3 (E4) (E5) Substituting Equations (E3) and (E4) into Equation (5.9) and integrating from point A to point B, we can find the relative rotation at the section at B with respect to the section at A: T AB – 1.5 T ⎛ dφ⎞ = ----------------- = ------------------------------------------------------⎝ d x⎠ AB G AB J AB G ( π ⁄ 2 ) ( 4 r – 0.075 x ) 4 3 T 111 φ B – φ A = – ------- ----- --------------- ---------------------------------G π – 3 – 0.075 ( 4 r – 0.075 x ) 3 or 40 r 0 φB ∫φ dφ = A xB ∫x A 3T – ------------------------------------------- dx or G π ( 4 r – 0.075 x ) 4 T -1 1T= – --------------------- --- – ------------ = – 4.178 -------0.075 G π r 3 ( 4 r ) 3 Gr 3 (E6) Adding Equations (E5) and (E6), we obtain T φ C – φ A = -------- ( 6.366 – 4.178 ) Gr 3 January, 2010 (E7) M. Vable Mechanics of Materials: Torsion of Shafts 5 ANS. 226 T φ C – φ A = 2.2 -------- ccw Gr 3 (b) Just left of the section at B we have JAB = πr4/2 and ρmax = r. Substituting these values into Equation (5.10), we obtain the maximum torsional shear stress in the shaft as 0.955 T – 1.5 T r (E8) τ max = ---------------- = – ---------------4 r3 πr ⁄ 2 ANS. 3 τ max = 0.955 ( T ⁄ r ) Dimension check: The dimensional consistency3 of the answer is checked as follows: T → O ( FL ) r → O(L) F G...
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