The formulas were developed empirically to meet a

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Unformatted text preview: cted to a torque T = 15 kN·m. t 6 mm 100 mm Figure P5.96 100 mm 5.97 A tube of uniform thickness t and cross section shown in Figure P5.97 has a torque T applied to it. Determine the maximum torsional shear stress in terms of t, a, and T. 60 Figure P5.97 60 a 5.98 A tube of uniform thickness t and cross section shown in Figure P5.98 has a torque T applied to it. Determine the maximum torsional shear stress in terms of t, a, and T. a Figure P5.98 a 5.99 A tube of uniform thickness t and cross section shown in Figure P5.99 has a torque T applied to it. Determine the maximum torsional shear stress in terms of t, a, and T. Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm a Figure P5.99 5.100 The tube of uniform thickness t shown in Figure P5.100 has a torque T applied to it. Determine the maximum torsional shear stress in terms of t, a, b, and T. b a Figure P5.100 January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 252 5.101 A hexagonal tube of uniform thickness is loaded as shown in Figure P5.101. Determine the magnitude of the maximum torsional shear stress in the tube T4 m T3 m T2 T1 m 1000 N m 100 mm Figure P5.101 t 5.102 4 mm A rectangular tube is loaded as shown in Figure P5.102. Determine the magnitude of the maximum torsional shear stress. T4 2 in kips T3 3 in kips T1 2 in kips 6 in Figure P5.102 4 in 5.103 The three tubes shown in Problems 5.97 through 5.99 are to be compared for the maximum torque-carrying capability, assuming that all tubes have the same thickness t, the maximum torsional shear stress in each tube can be τ, and the amount of material used in the cross section of each tube is A. (a) Which shape would you use? (b) What is the percentage torque carried by the remaining two shapes in terms of the most efficient structural shape? 5.5* CONCEPT CONNECTOR Like so much of science, the theory of torsion in shafts has a history filled with twists and turns. Sometimes experiments led the way; sometimes logic pointed to a solution. As so often, too, serendipity guided developments. The formulas were developed empirically, to meet a need—but not in the mechanics of materials. Instead, a scientist had a problem to solve in electricity and magnetism, and torsion helped him measure the forces. It was followed with an analytical development of the theory for circular and non-circular shaft cross sections that stretched over a hundred years. The description of the history concludes with an experimental technique used in the calculation of torsional rigidity, even for shafts of arbitrary shapes. Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 5 .5.1 History: Torsion of Shafts It seems fitting that developments begin with Charles-Augustin Coulomb (Figure 5.55). Coulomb, who first differentiated shear stress from normal stress (see Section 1.6.1), also studied torsion, in which shear stress is the dominant component. In 1781 Coulomb started his research in electricity and magnetism. To measure the small forces involved, he devised a very sensitive torsion balance. A weight was suspended by a wire, and a pointer attached to the weight indicated the wire’s angular rotation. Figure 5.55 Charles-Augustin Coulomb. The design of this torsion balance led Coulomb to investigate the resistance of a wire in torsion. He assumed that the resistance torque (or internal torque T) in a twisted wire is proportional to the angle of twist φ. To measure the changes, he twisted the January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 253 wire by a small angle and set it free to oscillate, like a pendulum. After validating his formula experimentally, thus confirming his assumption, he proceeded to conduct a parametric study with regard to the length L and the diameter D of the wire and devel4 oped the following formula T = ( μ D ⁄ L ) φ , where μ is a material constant. If we substitute d φ ⁄ dx = φ ⁄ L and 4 J = π D ⁄ 32 into Equation 5.9 and compare our result with Coulomb’s formula, we see that Coulomb’s material constant is μ = π G ⁄ 32 . Coulomb’s formula, although correct, was so far only an empirical relationship. The analytical development of the theory for circular shafts is credited to Alphonse Duleau. Duleau, born in Paris the year of the French revolution, was commissioned in 1811 to design a forged iron bridge over the Dordogne river, in the French city of Cubzac. Duleau had graduated from the École Polytechnique, one of the early engineering schools. Founded in 1794, it had many pioneers in the mechanics of materials among its faculty and students. At the time, there was little or no data on the behavior of bars under the loading conditions needed for bridge design. Duleau therefore conducted extensive experiments on tension, compression, flexure, torsion, and elastic stability. He also compared bars of circular, triangular, elliptical, and rectangular cross section. In 1820 he published his results. In this paper Duleau developed Coulomb...
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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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