Unformatted text preview: Sec_3.8.qxd 9/27/08 1:13 PM Page 309 SECTION 3.8 Statically Indeterminate Torsional Members 309 Solution 3.8-9 Bar with a hole
IPA Polar moment of inertia at left-hand end IPB Polar moment of inertia at right-hand end fB TB(L/2)
GIPA L 50 in. L/2 25 in. d2 outer diameter L
2IPB diameter of hole
x Torque applied at distance x Find x so that TA ‹
IPB TB TB T0 ‹ TA 0 B 3L
4IPA soLVE FOR x: EQUILIBRIUM
TA T0 L
4IPA fB 2.4 in.
T0 (2) Substitute Eq. (1) into Eq. (2) and simplify: 3.0 in.
GIPB TB T0
2 (1) REMOVE THE SUPPORT AT END B x
d2 1 d1 4
c2 + a b d
d2 a d1 4
d2 ; SUBSTITUTE NUMERICAL VALUES:
B 2.4 in. 4
c2 + a
3.0 in. 30.12 in. ; Angle of twist at B Problem 3.8-10 A solid steel bar of diameter d1 25.0 mm is
enclosed by a steel tube of outer diameter d3 37.5 mm and inner
diameter d2 30.0 mm (see figure). Both bar and tube are held
rigidly by a support at end A and joined securely to a rigid plate
at end B. The composite bar, which has a length L 550 mm,
is twisted by a torque T 400 N m acting on the end plate.
(a) Determine the maximum shear stresses 1 and 2 in the bar and
(b) Determine the angle of rotation (in degrees) of the end plate,
assuming that the shear modulus of the steel is G 80 GPa.
(c) Determine the torsional stiffness kT of the composite bar.
(Hint: Use Eqs. 3-44a and b to find the torques in the bar and tube.) Tube
T Bar End
plate L d1
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