Substituting these values into equation e2 we obtain

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Unformatted text preview: shafts of Example 5.5. PLAN We can find the values of RH and RS in terms of the cross-sectional area A. We can then substitute these radii in the formulas for polar area moment to obtain the polar area moments in terms of A. SOLUTION We can calculate the radii RH and RS in terms of the cross sectional area A as 2 2 AH = π [ ( 2 RH ) – RH ] = A A 2 R H = ----3π or 2 AS = π RS = A and A 2 R S = -π or (E1) The polar area moment of inertia for a hollow shaft with inside radius Ri and outside radius Ro can be obtained as J= ∫A ρ 2 dA = Ro ∫R ρ i 2 π4 ( 2 πρ ) d ρ = -- ρ 2 Ro Ri π4 4 = -- ( R o – R i ) 2 (E2) For the hollow shaft Ro = 2RH and Ri = RH, whereas for the solid shaft Ro = RS and Ri = 0. Substituting these values into Equation (E2), we obtain the two polar area moments. 2 2 15 4 15 A 2 5 A π 4 4 J H = -- [ ( 2 R H ) – R H ] = ----- π R H = ----- π ⎛ ----- ⎞ = -- ----2 2 2 ⎝ 3 π⎠ 6π π4 πA2 A -J S = -- R S = -- ⎛ -- ⎞ = ----2 2 ⎝ π⎠ 2π and (E3) Dividing JH by JS we obtain JH ----- = 5 = 1.67 -3 JS (E4) ANS. As J H > J S the polar moment for the hollow shaft is greater than that of the solid shaft for the same amount of material. COMMENT 1. The hollow shaft has a polar moment of inertia of 1.67 times that of the solid shaft for the same amount of material. Alternatively, a hollow shaft will require less material (lighter in weight) to obtain the same polar moment of inertia. This reduction in weight is the primary reason why metal shafts are made hollow. Wooden shafts, however, are usually solid as the machining cost does not justify the small saving in weight. EXAMPLE 5.6 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm A solid circular steel shaft (Gs = 12,000 ksi) of variable diameter is acted upon by torques as shown in Figure 5.25. The diameter of the shaft between wheels A and B and wheels C and D is 2 in., and the diameter of the shaft between wheels B and C is 4 in. Determine: (a) the rotation of wheel D with respect to wheel A; (b) the magnitude of maximum torsional shear stress in the shaft; (c) the shear stress at point E. Show it on a stress cube. 2 in kips 8 in kips 8.5 in kips A E 60 i 24 in n. x 2.5 in kips B 60 i n C Figure 5.25 January, 2010 Geometry of shaft and loading in Example 5.6. D 30 i n M. Vable Mechanics of Materials: Torsion of Shafts 5 221 PLAN By making imaginary cuts in sections AB, BC, and CD and drawing the free-body diagrams we can find the internal torques in each section. (a) We find the relative rotation in each section using Equation (5.12). Summing the relative rotations we can obtain φD – φA. (b) We find the maximum shear stress in each section using Equation (5.10), then by comparison find the maximum shear stress τmax in the shaft. (c) In part (b) we found the shear stress in section BC. We obtain the direction of the shear stress either using the subscript or intuitively. SOLUTION The polar moment of inertias for each segment can be obtained as ππ4 4 J AB = J CD = ----- ( 2 in. ) = -- in. 32 2 π4 4 J BC = ----- ( 4 in. ) = 8 π in. 32 (E1) We make an imaginary cuts, draw internal torques as per our sign convention and obtain the free body diagrams as shown in Figure 5.25. We obtain the internal torques in each segment by equilibrium of moment about shaft axis: T AB + 2 π in. · kips = 0 or T BC + 2 π in. · kips – 8 π in. · kips = 0 T CD + 2.5 π in. · kips = 0 (a) 2 in kips (b) 2 in kips or T AB = – 2 π in. · kips (E2) T BC = 6 π in. · kips or (E3) T CD = – 2.5 π in. · kips (E4) (c) 8 in kips TCD 2.5 in kips TAB A A D TBC B Figure 5.26 Free-body diagrams in Example 5.6 after an imaginary cut in segment (a) AB, (b) BC, and (c) CD. (a) From Equation (5.12), we obtain the relative rotations of the end of segments as T AB ( x B – x A ) ( – 2 π in. · kips ) ( 24 in. ) –3 φ B – φ A = ------------------------------ = -------------------------------------------------------- = – 8 ( 10 ) rad 4 G AB J AB ( 12 ,000 ksi ) ( π ⁄ 2 in. ) (E5) T BC ( x C – x B ) ( 6 π in. · kips ) ( 60 in. ) –3 φ C – φ B = ------------------------------- = ---------------------------------------------------- = 3.75 ( 10 ) rad 4 G BC J BC ( 12 ,000 ksi ) ( 8 π in. ) (E6) T CD ( x D – x C ) ( – 2.5 π in. · kips ) ( 30 in. ) –3 φ D – φ C = ------------------------------- = ----------------------------------------------------------- = – 12.5 ( 10 ) rad 4 G CD J CD ( 12 ,000 ksi ) ( π ⁄ 2 in. ) (E7) Adding Equations (E5), (E6), and (E7), we obtain the relative rotation of the section at D with respect to the section at A: –3 –3 φ D – φ A = ( φ B – φ A ) + ( φ C – φ B ) + ( φ D – φ C ) = ( -8 + 3.75 – 12.5 ) ( 10 ) rad = – 16.75 ( 10 ) rad ANS. (E8) φ D – φ A = 0.01675 rad cw (b) The maximum torsional shear stress in section AB and CD will exist at ρ = 1 and in BC it will exist at ρ = 2. From Equation (5.10) we can obtain the maximum shear stress in each segment: (E9) T...
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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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