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Unformatted text preview: king an imaginary cut in AB. Figure 5.49b shows the decomposition of a composite shaft as two homogenous shafts. (a) TAB (b) TAB Ts TBr 75 kN-m B Δφ Δφ Figure 5.49 Δφ (a) Free body diagram (b) Composite shaft as two homogenous shafts in Example 5.14. From Figure 5.49 we obtain the equilibrium equation, 3 T AB = T s + T Br = 75 kN ⋅ m = 75 ( 10 ) N ⋅ m (E3) Using Equation (5.12) we can write the relative rotation of section at B with respect to A for the two material as Ts ( xB – xA ) Ts ( 2 ) –6 Δφ = φ B – φ A = -------------------------- = -------------------------- = 6.219 ( 10 ) T s rad 3 Gs Js 321.6 ( 10 ) (E4) T Br ( 2 ) –6 Δφ = φ B – φ A = -------------------------- = 3.0619 ( 10 ) T Br rad 3 653.2 ( 10 ) (E5) Equating Equations (E4) and (E5) we obtain (E6) T s = 2.03 T Br Solving Equations (E3) and (E4) for the internal torques give 3 T s = 24.75 ( 10 ) N ⋅ m 3 T Br = 50.25 ( 10 ) N ⋅ m (E7) Substituting Equation (7) into Equation (4), we find –6 3 φ B – φ A = 6.219 ( 10 ) ( 24.75 ) ( 10 ) = 0.1538 rad (E8) ANS. φ B – φ A = 0.1538 rad ccw The maximum torsional shear stress in each material can be found using Equation (5.10): 3 T s ( ρ s ) max [ 24.75 ( 10 ) N ⋅ m ] ( 0.04 m ) 6 2 ( τ s ) max = ---------------------- = --------------------------------------------------------------------- = 246.3 ( 10 ) N/m –6 4 Js 4.02 ( 10 ) m 3 T Br ( ρ Br ) max [ 50.25 ( 10 ) N ⋅ m ] ( 0.06 m ) 6 2 ( τ Br ) max = ---------------------------- = --------------------------------------------------------------------- = 184.6 ( 10 ) N/m –6 4 J Br 16.33 ( 10 ) m (E9) (E10) The maximum torsional shear stress is the larger of the two. ANS. τ max = 246.3 MPa COMMENT Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 1. The kinematic condition that all radial lines must rotate by equal amount for a circular shaft had to be explicitly enforced to obtain Equations (E6). We could also have implicitly assumed this kinematic condition and developed formulas for composite shafts (see Problem 5.49) as we did for homogenous shaft. We can then use these formulas to solve statically determinate and indeterminate problems (see Problem 5.82) as we have done for homogenous shafts. PROBLEM SET 5.3 Statically indeterminate shafts 5.63 A steel shaft (Gst = 12,000 ksi) and a bronze shaft (Gbr = 5600 ksi) are securely connected at B, as shown in Figure P5.63. Determine the maximum torsional shear stress in the shaft and the rotation of the section at B if the applied torque T = 50 in.·kips. T Figure P5.63 January, 2010 4 ft M. Vable Mechanics of Materials: Torsion of Shafts 5 244 5.64 A steel shaft (Gst = 12,000 ksi) and a bronze shaft (Gbr = 5600 ksi) are securely connected at B, as shown in Figure P5.63. Determine the maximum torsional shear strain and the applied torque T if the section at B rotates by an amount of 0.02 rad. 5.65 Two hollow aluminum shafts (G = 10,000 ksi) are securely fastened to a solid aluminum shaft and loaded as shown Figure P5.65. If T = 300 in.·kips, determine (a) the rotation of the section at C with respect to the wall at A; (b) the shear strain at point E. Point E is on the inner surface of the shaft. T Figure P5.65 24 in 36 in 24 in 5.66 Two hollow aluminum shafts (G = 10,000 ksi) are securely fastened to a solid aluminum shaft and loaded as shown Figure P5.65. The torsional shear strain at point E which is on the inner surface of the shaft is –250 μ. Determine the rotation of the section at C and the applied torque T that produced this shear strain. 5.67 Two solid circular steel shafts (Gst = 80 GPa) and a solid circular bronze shaft (Gbr = 40 GPa) are securely connected by a coupling at C as shown in Figure P5.67. A torque of T = 10 kN·m is applied to the rigid wheel B. If the coupling plates cannot rotate relative to one another, determine the angle of rotation of wheel B due to the applied torque. T 10 kN m Figure P5.67 5m 3m 5.68 Two solid circular steel shafts (Gst = 80 GPa) and a solid circular bronze shaft (Gbr = 40 GPa) are connected by a coupling at C as shown in Figure P5.67. A torque of T = 10 kN·m is applied to the rigid wheel B. If the coupling plates can rotate relative to one another by 0.5° before engaging, then what will be the angle of rotation of wheel B? 5.69 A solid steel shaft (G = 80 GPa) is securely fastened to a solid bronze shaft (G = 40 GPa) that is 2 m long, as shown in Figure P5.69. If Text = 10 kN · m, determine (a) the magnitude of maximum torsional shear stress in the shaft; (b) the rotation of the section at 1 m from the left wall. Text 2m Figure P5.69 1m 5.70 A solid steel shaft (G = 80 GPa) is securely fastened to a solid bronze shaft (G = 40 GPa) that is 2 m long, as shown in Figure P5.69. If the section at B rotates by 0.05 rad, determine (a) the maximum torsional shear strain in the shaft; (b) the applied torque Text. Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 5.71 Two shafts with shear moduli G1 = G and G2 = 2G are securely fastened at section B, as shown in Figure P5.71. In terms of Text, L, G,...
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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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