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Unformatted text preview: e rotation of the section at B with respect to the rotation of the section at A is given by
1
φ B – φ A = GJ xB ∫x (5.17) ( x – x B ) t ( x ) dx A 5.49 A composite shaft made from n materials is shown in Figure P5.49. Gi and Ji are the shear modulus of elasticity and polar moment
of inertia of the ith material. (a) If Assumptions from 1 through 6 are valid, show that the stress ( τ x θ ) i in the ith material is given Equation
(5.18a), where T is the total internal torque at a cross section. (b) If Assumptions 8 through 10 are valid, show that relative rotation φ 2 – φ 1
is given by Equation (5.18b). (c) Show that for G1=G2=G3....=Gn=G Equations (5.18a) and (5.18b) give the same results as Equations (5.10)
and (5.12). Gi ρ T
( τ x θ ) i = n
Gj Jj ∑ j =1 T(x – x ) 2
1
φ 2 – φ 1 = n ∑ j=1 Gj Jj (5.18a) (5.18b) Figure P5.49
A circular solid shaft of radius R is made from a nonlinear material that has a shear stressshear strain relationship given by τ = Gγ0.5.
Assume that the kinematic assumptions are valid and shear strain varies linearly with the radial distance across the crosssection. Determine
the maximum shear stress and the rotation of section at B in terms of external torque Text, radius R, material constant G, and length L. 5.50 Text
A B
L Figure P5.50 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 5.51 A hollow circular shaft is made from a nonlinear materials that has the following shear stressshear strain relation τ = Gγ2. Assume
that the kinematic assumptions are valid and shear strain varies linearly with the radial distance across the crosssection. In terms of internal
torque T, material constant G, and R, obtain formulas for (a) the maximum shear stress τ max and (b) the relative rotation φ 2 – φ 1 of two
crosssections at x1 and x2. R Figure P5.51 2R 5.52 A solid circular shaft of radius R and length L is twisted by an applied torque T. The stress–strain relationship for a nonlinear maten
rial is given by the power law τ = G γ . If Assumptions 1 through 4 are applicable, show that the maximum shear stress in the shaft and the
relative rotation of the two ends are as follows:
T(n + 3)
(n + 3)T 1 ⁄ n
τ max = L
Δ φ = 3
3+ n
2πR
2 π GR
Substitute n = 1 in the formulas and show that we obtain the same results as from Equations 5.10 and 5.12. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 237 5.53 The internal torque T and the displacements of a point on a cross section of a noncircular shaft shown in Figure P5.53 are given by
the equations below
d Vy dV xy dA y (5.19a) v = – xz
T dφ
dx (5.19b) w = xy z x Figure P5.53 dφ
dx u = ψ ( y, z ) y dφ
dx (5.19c) x Torsion of noncircular shafts.
T= ∫A ( y τxz – z τxy ) dA (5.20) where u, v, and w are the displacements in the x, y, and z directions, respectively; d φ ⁄ dx is the rate of twist and is considered constant.
ψ ( x, y ) is called the warping function4 and describes the movement of points out of the plane of cross section. Using Equations (2.12d) and
(2.12f) and Hooke’s law, show that the shear stresses for a noncircular shaft are given by ∂ψ ⎞ dφ
τ xy = G ⎛
–z
⎝ ∂y
⎠ dx 5.54 ∂ψ ⎞ dφ
+y
τ xz = G ⎛
⎝∂z
⎠ dx (5.21) Show that for circular shafts, ψ ( x, y ) = 0, the equations in Problem 5.53 reduce to Equation (5.9). 5.55 Consider the dynamic equilibrium of the differential element shown in Figure P5.55, where T is the internal torque, γ is the material
2
2
density, J is the polar area moment of inertia, and ∂ φ ⁄ ∂ t is the angular acceleration. Show that dynamic equilibrium results in Equation
(5.22)
2 J
T T t2 dx
2 ∂φ dT ∂t Figure P5.55 Dynamic equilibrium. 5.56 dx 2 =c 2 2 ∂φ
∂x where c = 2 G
γ (5.22) dx Show by substitution that the solution of Equation (5.23) satisfies Equation (5.22):
ωx
ωx
φ = ⎛ A cos  + B sin  ⎞ × ( C cos ω t + D sin ω t )
⎝
c
c⎠ (5.23) where A, B, C, and D are constants that are determined from the boundary conditions and the initial conditions and ω is the frequency of
vibration. Computer problems
5.57 A hollow aluminum shaft of 5 ft in length is to carry a torque of 200 in.·kips. The inner radius of the shaft is 1 in. If the maximum
torsional shear stress in the shaft is to be limited to 10 ksi, determine the minimum outer radius to the nearest 1 in.
8 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 5.58 A 4ftlong hollow shaft is to transmit a torque of 100 in.·kips. The relative rotation of the two ends of the shaft is limited to
0.06 rad. The shaft can be made of steel or aluminum. The shear modulus of rigidity G, the allowable shear stress τallow , and the specific
weight γ are given in Table P5.58. The inner radius of the shaft is 1 in. Determine the outer radius of the lightest shaft that can be used for transmitting the torque and the corresponding weight.
TABLE P5.58
Material G (ksi) τa...
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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.
 Spring '10
 SMITH
 Torsion

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