This preview shows page 1. Sign up to view the full content.
Unformatted text preview: n with respect to the left section is rotated by an angle
Δφ, as shown in Figure 5.18a. Using geometry we obtain the shear strain expression.
y (a) (b) x max
x R Figure 5.18 Shear strain in torsion. (a) Deformed shape. (b) Linear variation of shear strain. BB 1
tan γ x θ ≈ γ x θ = lim ⎛ -------- ⎞ = lim ⎛ --------- ⎞ or
AB → 0 ⎝ AB ⎠
Δx → 0 ⎝ Δx ⎠ Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm γxθ = ρ dφ
dx (5.3) where ρ is the radial coordinate of a point on the cross section. The subscripts x and θ emphasize that the change in angle is
between the axial and tangent directions, as shown in Figure 5.18a. The quantity d φ ⁄ d x is called the rate of twist. It is a function of x only, because φ is a function of x only.
Equation (5.3) was derived from purely geometric considerations. If Assumptions 1 through 4 are valid, then Equation (5.3)
is independent of the material. Equation (5.3) shows that the shear strain is a linear function of the radial coordinate ρ and
reaches the maximum value γmax at the outer surface (ρ = ρmax = R), as shown in Figure 5.18a. Equation (5.4), an alternative
form for shear strain, can be derived using similar triangles. γ max ρ
γ x θ = -----------R 5.2.2 (5.4) Material Model Our motivation is to develop a simple theory for torsion of circular shafts. Thus we make assumptions regarding material behavior that will permit us to use the simplest material model given by Hooke’s law. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 217 Assumption 5 The material is linearly elastic.1
Assumption 6 The material is isotropic. Substituting Equation (5.3) into Hooke’s law, that is, τ = G γ , we obtain τxθ = G ρ dφ
dx (5.5) Noting that θ is positive in the counterclockwise direction with respect to the x axis, we can represent the shear stress due to torsion on a stress element as shown in Figure 5.19. Also shown in Figure 5.19 are aluminum and wooden shafts that broke in torsion. The shear stress component that exceeds the shear strength in aluminum is τxθ. The shear strength of wood is weaker along
the surface parallel to the grain, which for shafts is in the longitudinal direction. Thus τθx causes the failure in wooden shafts. The
two failure surfaces highlight the importance of visualizing the torsional shear stress element.
Failure ilure surface ialuminum sshaft ue to to τxθ
surface in n aluminum haft d due
x x Figure 5.19 5.2.3 Stress element showing torsional shear stress. Failure surface in wooden shaft due
Failure surfacein wooden shaft due to x to τθx Torsion Formulas Substituting Equation (5.5) into Equation (5.1) and noting that d φ ⁄ dx is a function of x only, we obtain T= ∫A G ρ 2 dφ
dx ∫A G ρ 2 dA (5.6) To simplify further, we would like to take G outside the integral, which implies that G cannot change across the cross section.
Assumption 7 The material is homogeneous across the cross section.2 From Equation (5.6) we obtain T=G dφ
dx ∫A ρ 2 dA = GJ dφ
dx (5.7) where J is the polar moment of inertia for the cross section. As shown in Example 5.5, J for a circular cross section of radius R
or diameter D is given by ∫A ρ 2 π4
dA = -- R = ----- D
32 (5.8) dφ
GJ J= (5.9) Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Equation (5.7) can be written as The higher the value of GJ, the smaller will be the deformation φ for a given value of the internal torque. Thus the rigidity
of the shaft increases with the increase in GJ. A shaft may be made more rigid either by choosing a stiffer material (higher value
of G) or by increasing the polar moment of inertia. The quantity GJ is called torsional rigidity.
Substituting Equation (5.9) into Equation (5.5), we obtain 1
2 See Problems 5.50 through 5.52 for nonlinear material behavior.
In Problem. 5.49 this assumption is not valid. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 218 Tρ
τ x θ = ------ (5.10)
The quantities T and J do not vary across the cross section. Thus the shear stress varies linearly across the cross section with ρ as
shown in Figure 5.20. For a solid shaft, it is zero at the center where ρ = 0 and reaches a maximum value on the outer surface of
the shaft where ρ = R,.
x max R Figure 5.20 Linear variation of torsional shear stress. Let the angle of rotation of the cross section at x1 and x2 be φ1 and φ2, respectively. By integrating Equation (5.9) we can obtain the
relative rotation as:
φ2 – φ1 = φ2 ∫φ dφ = 1 x2 T ∫x ------- dx
GJ (5.11) 1 To obtain a simple formula we would like to take the three quantities T, G, and J outside the integral, which means that these
quantities should not change with x. To achieve this simplicity we make the following assumptions:
Assumption 8 The material is homogeneous between x1 and x2. (G is constant)
Assumption 9 The shaft is not tapered between x1 and x2. (J is constant)
Assumption 10 The external (and hence also the internal) torque does not change with x between x1 and x2. (T is constant) If Assumptions 8 t...
View Full Document
- Spring '10