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Unformatted text preview: Sec_3.5-3.7.qxd 288 9/27/08 1:10 PM Page 288 CHAPTER 3 Torsion Problem 3.5-2 A hollow steel bar (G
a maximum shear strain
respectively. 640 max 10 80 GPa) is twisted by torques T (see figure). The twisting of the bar produces
rad. The bar has outside and inside diameters of 150 mm and 120 mm, (a) Determine the maximum tensile strain in the bar.
(b) Determine the maximum tensile stress in the bar.
(c) What is the magnitude of the applied torques T ? Solution 3.5-2 Hollow steel bar G 80 GPa max d2 150 mm d1 IP p4
32 2 640 10 6 rad (b) MAXIMUM TENSILE STRESS d 4)
1 120 mm max max (120 mm)4] T
6 ; 51.2 MPa Torsion formula: tmax (a) MAXIMUM TENSILE STRAIN
320 * 10 max (c) APPLIED TORQUES 29.343 * 106 mm4 gmax
2 10 6) (80 GPa)(640 max 51.2 MPa p
32 âmax G ; 2IPtmax
2IP 2(29.343 * 106 mm4)(51.2 MPa)
150 mm 20,030 N # m
20.0 kN # m Problem 3.5-3 A tubular bar with outside diameter d2 4.0 in. is twisted by torques T
action of these torques, the maximum tensile stress in the bar is found to be 6400 psi.
(a) Determine the inside diameter d1 of the bar.
(b) If the bar has length L 48.0 in. and is made of aluminum with shear modulus G
twist (in degrees) between the ends of the bar?
(c) Determine the maximum shear strain max (in radians)? ; 70.0 k-in. (see figure). Under the 4.0 106 psi, what is the angle of ...
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