This preview shows page 1. Sign up to view the full content.
Unformatted text preview: llow (ksi) γ (lb/in.3) Steel 12,000 18 0.285 Aluminum 4000 10 0.100 5.59 Table P5.59 shows the measured radii of the solid tapered shaft shown in Figure P5.59, at several points along the axis of the shaft.
The shaft is made of aluminum (G = 28 GPa) and has a length of 1.5 m. Determine: (a) the rotation of the free end with respect to the wall
using numerical integration; (b) the maximum shear stress in the shaft. 4 2 2 2 2 Equations of elasticity show that the warping function satisfies the Laplace equation, ∂ ψ ⁄ ∂ y + ∂ ψ ⁄ ∂ z = 0. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 TABLE P5.59 T = 35 kN ⋅ m
m x
(m) R(x)
(mm) TABLE P5.59
x
(m) R(x)
(mm) 100.6 0.8 60.1 92.7 0.9 60.3 0.2 82.6 1.0 59.1 0.3 79.6 1.1 54.0 0.4 75.9 1.2 54.8 0.5 68.8 1.3 54.1 0.6 68.0 1.4 49.4 0.7 Figure P5.59 0.0
0.1 x 238 65.9 1.5 50.6 Let the radius of the tapered shaft in Problem 5.59 be represented by the equation R(x) = a + bx. Using the data in Table P5.59 determine the constants a and b by the leastsquares method and then find the rotation of the section at B by analytical integration. 5.60 5.61 Table P5.61 shows the values of distributed torque at several points along the axis of the solid steel shaft (G = 12,000 ksi) shown in
Figure P5.61. The shaft has a length of 36 in. and a diameter of 1 in. Determine (a) the rotation of end A with respect to the wall using numerical integration; (b) the maximum shear stress in the shaft.
TABLE P5.61
x
(in.) TABLE P5.61 t (x)
(in.·lb/in.) x
(in.) t (x)
(in.·lb/in.) 0 93.0 21 588.8 3 146.0 24 700.1 6 214.1 27 789.6 9 260.4 30 907.4 12 335.0 33 1040.3 15 424.7 36 1151.4 18 492.6 Let the distributed torque t ( x ) in Problem 5.61 be represented by the equation t ( x ) = a + bx + cx2. Using the data in Table P5.61
determine the constants a, b, and c by the leastsquares method and then find the rotation of the section at B by analytical integration. 5.62 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm QUICK TEST 5.1 January, 2010 Time: 20 minutes/Total: 20 points M. Vable Mechanics of Materials: Torsion of Shafts 5 239 Answer true or false and justify each answer in one sentence. Grade yourself with the answers given in Appendix E. 1. Torsional shear strain varies linearly across a homogeneous cross section.
2. Torsional shear strain is a maximum at the outermost radius for a homogeneous and a nonhomogeneous cross section.
3. Torsional shear stress is a maximum at the outermost radius for a homogeneous and a nonhomogeneous cross section.
4. The formula τ x θ = T ρ ⁄ J can be used to find the shear stress on a cross section of a tapered shaft.
5. The formula φ 2 – φ 1 = T ( x 2 – x 1 ) ⁄ G J can be used to find the relative rotation of a segment of a tapered shaft.
6. The formula τ x θ = T ρ ⁄ J can be used to find the shear stress on a cross section of a shaft subjected to distributed
torques.
7. The formula φ 2 – φ 1 = T ( x 2 – x 1 ) ⁄ G J can be used to find the relative rotation of a segment of a shaft subjected to distributed torques.
8. The equation T = ∫A ρτx θ dA cannot be used for nonlinear materials. 9. The equation T = ∫A ρτx θ dA can be used for a nonhomogeneous cross section. 10.Internal torques jump by the value of the concentrated external torque at a section. 5.3 STATICALLY INDETERMINATE SHAFTS In Chapter 4 we saw the solution of statically indeterminate axial problems require equilibrium equations and compatibility
equations. This is equally true for statically indeterminate shafts. The primary focus in this section will be on the solution of statically indeterminate shafts that are on a single axis. However, equilibrium equations and compatibility equations can also be
used for solution of shafts with composite cross section, as will be demonstrated in Example 5.14. The use of equilibrium equations and compatibility equations to shafts on multiple axis is left as exercises in Problem Set 5.3.
T Figure 5.43 Statically indeterminate shaft. B Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Figure 5.43 shows a statically indeterminate shaft. In statically indeterminate shafts we have two reaction torques, one at the
left and the other at the right end of the shaft. But we have only one static equilibrium equation, the sum of all torques in the x
direction should be zero. Thus the degree of static redundancy is 1 and we need to generate 1 compatibility equation. We shall
use the continuity of φ and the fact that the sections at the left and right walls have zero rotation. The compatibility equation state
that the relative rotation of the right wall with respect to the left wall is zero. Once more we can use either the displacement
method or the force method: 1. In the displacement method, we can use the rotation of the sections as the unknowns. If torque is applied at several
sections along the shaft, then the rotation of each of the sections is treated as an unknown.
2. In the force method, we can use either the reaction torque as the unknown or the internal torques in the sections as the
unknowns....
View
Full
Document
 Spring '10
 SMITH
 Torsion

Click to edit the document details