We are away from the regions of stress concentration

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Unformatted text preview: the radial coordinate measured in inches. The shaft material has a stress–strain relationship given by τ = 12,000γ − 120,000γ 2 ksi. Determine the equivalent internal torque. (See Problem 3.147). 5.18 5.2 THEORY OF TORSION OF CIRCULAR SHAFTS In this section we develop formulas for deformation and stress in a circular shaft. We will follow the procedure in Section 5.1 but now with variables in place of numbers. The theory will be developed subject to the following limitations: 1. 2. 3. 4. The length of the member is significantly greater than the greatest dimension in the cross section. We are away from the regions of stress concentration. The variation of external torque or change in cross-sectional areas is gradual except in regions of stress concentration. External torques are not functions of time; that is, we have a static problem. (See Problems 5.55 and 5.56 for dynamic problems.) 5. The cross section is circular. This permits us to use arguments of axisymmetry in deducing deformation. Figure 5.14 shows a circular shaft that is loaded by external torques T1 and T2 at each end and an external distributed torque t(x), which has units of torque per unit length. The radius of the shaft R(x) varies as a function of x. We expect that the internal torque T will be a function of x. φ1 and φ2 are the angles of rotation of the imaginary cross sections at x1 and x2, respectively. The objectives of the theory are: 1. To obtain a formula for the relative rotation φ2 – φ1 in terms of the internal torque T. 2. To obtain a formula for the shear stress τxθ in terms of the internal torque T. y T2 r x z Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Figure 5.14 Circular shaft. x2 To account for the variations in t(x) and R(x) we will take Δ x = x2 − x1 as an infinitesimal distance in which these quantities can be treated as constants. The deformation behavior across the cross section will be approximated. The logic shown in Figure 5.15 and discussed in Section 3.2 will be used to develop the simplest theory for the torsion of circular shafts members. Assumptions will be identified as we move from one step to the next. These assumptions are the points at which complexities can be added to the theory, as discussed in the examples and Stretch Yourself problems. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts Figure 5.15 5.2.1 5 215 The logic of the mechanics of materials. Kinematics In Example 5.1 the shear strain in a bar was related to the rotation of the disc that was attached to it. In Example 5.2 we remarked that a shaft could be viewed as an assembly of bars. Three assumptions let us simulate the behavior of a cross section as a rotating rigid plate: Assumption 1 Plane sections perpendicular to the axis remain plane during deformation. Assumption 2 On a cross section, all radial lines rotate by equal angles during deformation. Assumption 3 Radial lines remain straight during deformation. Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm (a) (b) Figure 5.16 Torsional deformation: (a) original grid; (b) deformed grid. (Courtesy of Professor J. B. Ligon.) Figure 5.16 shows a circular rubber shaft with a grid on the surface that is twisted by hand. The edges of the circles remain vertical lines during deformation. This observation confirms the validity of Assumption 1. Axial deformation due to torsional loads is called warping. Thus, circular shafts do not warp. Shafts with noncircular cross section warp, and this additional deformation leads to additional complexities. (See Problem 5.53). January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 216 The axisymmetry of the problem implies that deformation must be independent of the angular rotation. Thus, all radials lines must behave in exactly the same manner irrespective of their angular position, thus, Assumptions 2 and 3 are valid for circular shafts. Figure 5.17 shows that all radial lines rotate by the same angle of twist φ. We note that if all lines rotate by equal amounts on the cross section, then φ does not change across the cross section and hence can only be a function of x φ = φ(x) (5.2) Sign Convention: φ is considered positive counterclockwise with respect to the x axis. Ao,Bo —Initial position A1,B1 —Deformed position B1 Ao A1 Bo Figure 5.17 Equal rotation of all radial lines. The shear strain of interest to us is the measure of the angle change between the axial direction and the tangent to the circle in Figure 5.16. If we use polar coordinates, then we are interested in the change in angle which is between the x and θ directions— in other words, γxθ. Assumptions 1 through 3 are analogous to viewing each cross section in the shaft as a rigid disc that rotates about its own axis. We can then calculate the shear strain as in Example 5.2, provided we have small deformation and strain. Assumption 4 Strains are small. We consider a shaft with radius ρ and length Δx in which the right sectio...
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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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