39bwe then write the equilibrium equation as 0 lim t

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ng on the surface of the shaft. The internal torque T becomes a function of x when a shaft is subjected to a distributed external torque, as seen in Example 5.10. If t (x) is a simple function, then we can find T as a function of x by drawing a free-body diagram, as we did in Example 5.10. However, if the distributed torque t (x) is a complex function (see Problems 5.39, 5.61, and 5.62), it may be easier to use the alternative solution method described in this section. Consider an infinitesimal shaft element that is created by making two imaginary cuts at a distance dx from each other, as shown in Figure 5.39a. t(x) dx (a) T (b) T dT t (xA) Text TA dx Figure 5.39 (a) Equilibrium of an infinitesimal shaft element. (b) Boundary condition on internal torque. By equilibrium moments about the axis of the shaft, we obtain ( T + d T ) + t ( x ) d x – T = 0 or dT + t(x) = 0 dx (5.14) Equation (5.14) represents the equilibrium equation at any section x. It assumes that t (x) is positive counterclockwise with respect to the x axis. The sign of T obtained from Equation (5.14) corresponds to the direction defined by the sign convention. If t (x) is zero in a segment of a shaft, then the internal torque is constant in that segment. Equation (5.14) can be integrated to obtain the internal torque T. The integration constant can be found by knowing the value of the internal torque T at either end of the shaft. To obtain the value of T at the end of the shaft (say, point A), a free-body diagram is constructed after making an imaginary cut at an infinitesimal distance ε from the end, as shown in Figure 5.39b.We then write the equilibrium equation as lim [ T ext – T A – t ( x A ) ε ] = 0 ε→0 or T A = T ext (5.15) Equation (5.15) shows that the distributed torque does not affect the boundary condition on the internal torque. The value of the internal torque T at the end of the shaft is equal to the concentrated external torque applied at the end. Equation (5.14) is a differential equation. Equation (5.15) is a boundary condition. A differential equation and all the conditions necessary to solve it is called the boundary value problem. EXAMPLE 5.11 The external torque on a drill bit varies linearly to a maximum intensity of q in.·lb/in., as shown in Figure 5.40. If the drill bit diameter is d, its length L, and the modulus of rigidity G, determine the relative rotation of the end of the drill bit with respect to the chuck. Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm PLAN The relative rotation of section B with respect to section A has to be found. We can substitute the given distributed torque in Equation (5.14) and integrate to find the internal torque as a function of x. We can find the integration constant by using the condition that at section B the internal torque will be zero. We can substitute the internal torque expression into Equation (5.9) and integrate from point A to point B to find the relative rotation of section B with respect to section A. L x A x q ⎛ -- ⎞ in. ⋅ lb/in. ⎝ L⎠ Figure 5.40 Distributed torque on a drill bit in Example 5.11. January, 2010 B M. Vable Mechanics of Materials: Torsion of Shafts 5 229 SOLUTION The distributed torque on the drill bit is counterclockwise with respect to the x axis. Thus we can substitute t (x) = q(x/L) into Equation (5.14) to obtain the differential equation shown as Equation (E1). At point B, that is, at x = L, the internal torque should be zero as there is no concentrated applied torque at B.The boundary condition is shown as Equation (E2). The boundary value problem statement is • Differential Equation dT x + q -- = 0 dx L T(x = L) = 0 • (E1) (E2) Boundary Condition Integrating Equation (E1) we obtain 2 x T = – q ----- + c 2L (E3) Substituting Equation (E2) into Equation (E3) we obtain the integration constant c as 2 L – q ------ + c = 0 2L qL c = -----2 or (E4) Substituting Equation (E4) into Equation (E3) we obtain internal torque as q22 T = ------ ( L – x ) 2L (E5) Substituting Equation (E5) into Equation (5.9) and integrating we obtain the relative rotation of the section at B with respect to the section at A as 2 φB 2 dφ (q ⁄ 2L)(L – x ) = --------------------------------------4 dx G π d ⁄ 32 ∫φ or 16 q d φ = ---------------4 π GLd A xB = L ∫x = 0 ( L 2 2 – x ) dx 3 or A 16 q x -2 φ B – φ A = ---------------- ⎛ L x – ----⎞ 4 3⎠ π GLd ⎝ L (E6) 0 2 ANS. 32 qL φ B – φ A = ---------------- ccw 4 3 π Gd Dimension check: The dimensional consistency (see footnote 12) of our answer is checked as follows: FL q → O ⎛ ------ ⎞ → O ( F ) ⎝ L⎠ d → O(L) F G → O ⎛ ---- ⎞ ⎝ L 2⎠ L → O(L) φ → O( ) 2 qL FL 2 ---------4 → O ⎛ ----------------------- ⎞ → O ( ) → checks ⎝ ( F ⁄ L 2 ) L 4⎠ Gd COMMENTS 1. No free-body diagram was needed to find the internal torque because Equation (5.14) is an equilibrium equation. It is therefore valid at each and every section of the shaft. 2. We could have obtained the internal torque by integrating Equation (5.14) from L to x as follows: T ∫T = 0 dT B x x x q22 t...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online