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 freebody 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 freebody 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 freebody 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...
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 Spring '10
 SMITH
 Shear Stress, Torsion, M. Vable

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