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Unformatted text preview: rdinates at the cross section
was found to be γ x θ = 0.08 ρ , where ρ is in meters. (a)Write expressions for τxθ as a function of ρ and plot the shear strain and shear
stress distributions across both cross sections. (b) For each of the cross sections determine the statically equivalent internal torques. x 120 mm Figure 5.12 Homogeneous and composite cross sections in Example 5.4. x 80 mm
120 mm PLAN
(a) Using Hooke’s law we can find the shear stress distribution as a function of ρ in each material. (b) Each of the shear stress distributions can be substituted into Equation (5.1) and the equivalent internal torque obtained by integration. SOLUTION
(a) From Hooke’s law we can write the stresses as
9 2 (E1) 9 2 (E2) ( τ x θ ) brass = [ 40 ( 10 ) N/m ] ( 0.08 ρ ) = 3200 ρ MPa
( τ x θ ) steel = [ 80 ( 10 ) N/m ] ( 0.08 ρ ) = 6400 ρ MPa Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm For the homogeneous cross section the stress distribution is as given in Equation (E1), but for the composite section it switches between
Equation (E2) and Equation (E1), depending on the value of ρ. We can write the shear stress distribution for both cross sections as a
function of ρ, as shown below.
Homogeneous cross section:
τ x θ = 3200 ρ M Pa 0.00 ≤ ρ < 0.06
x (E3) () x (MPa) x 256
4 (a) (b) (c) Figure 5.13 Shear strain and shear stress distributions in Example 5.4: (a) shear strain distribution; (b) shear stress distribution in homogeneous cross section; (c) shear stress distribution in composite cross section. January, 2010 (MPa) M. Vable Mechanics of Materials: Torsion of Shafts 5 211 Composite cross section:
⎧ 6400 ρ M Pa
τxθ = ⎨
⎩ 3200 ρ M Pa 0.00 ≤ ρ < 0.04 m (E4) 0.04 m < ρ ≤ 0.06 m The shear strain and the shear stress can now be plotted as a function of ρ, as shown in Figure 5.13(b). The differential area dA is the area
of a ring of radius ρ and thickness dρ, that is, dA = 2 πρ d ρ . Equation (5.1) can be written as
T= 0.06 ∫0 ρτ x θ ( 2 πρ d ρ ) (E5) Homogeneous cross section: Substituting Equation (E3) into Equation (E5) and integrating, we obtain the equivalent internal torque.
T= 4 ρ
ρ [ 3200 ρ ( 10 ) ] ( 2 πρ d ρ ) = [ 6400 π ( 10 ) ] ⎛ ---- ⎞
⎝ 4⎠ 0.06 ∫0 0.06
3 = 65.1 ( 10 ) N ⋅ m (E6) 0 T = 65.1 kN·m
Composite cross section: Writing the integral in Equation (E5) as a sum of two integrals and substituting Equation (E3) we obtain the
equivalent internal torque.
ρτ x θ ( 2 πρ d ρ ) = 0.04 ∫0 ρτ x θ ( 2 πρ d ρ ) + 0.06 ∫0.04 ρτx θ ( 2 πρ d ρ ) Tsteel
T steel = T brass = T brass
4 0.04 ∫0 6
ρ [ 6400 ρ ( 10 ) ] ( 2 πρ d ρ ) = ( 12800 π ) ( 10 ) ⎛ ---- ⎞
4 0.06 (E7) ⎧
⎩ 0.06 ∫0 ⎧
⎩ T= 6
∫0.04 ρ [ 3200 ρ ( 10 ) ] ( 2 πρ d ρ ) = ( 6400 π ) ( 10 ) ⎛ -----⎞
⎝ 4⎠ 0.04
3 (E8) 3 (E9) = 25.7 ( 10 ) N ⋅ m = 25.7 kN·m
0.06 = 52.3 ( 10 ) N ⋅ m = 52.3 kN·m
0.04 (E10) T = T steel + T brass = 25.7 kN·m + 52.3 kN·m ANS. T = 78 kN·m COMMENTS
1. The example demonstrates that although the shear strain varies linearly across the cross section, the shear stress may not. In this
example we considered material non homogeneity. In a similar manner we can consider other models, such as elastic–perfectly plastic, or material models that have nonlinear stress–strain curves.
2. The material models dictate the shear stress distribution across the cross section, but once the stress distribution is known, Equation
(5.1) can be used to find the equivalent internal torque, emphasizing that Equation (5.1) does not depend on the material model. PROBLEM SET 5.1
5.1 A pair of 48-in. long bars and a pair of 60-in. long bars are symmetrically attached to a rigid disc at a radius of 2 in. at one end and
built into the wall at the other end, as shown in Figure P5.1. The shear strain at point A due to a twist of the rigid disc was found to be 3000
μrad. Determine the magnitude of shear strain at point D. Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm T B Figure P5.1 48 in C
60 in 5.2 If the four bars in Problem 5.1 are made from a material that has a shear modulus of 12,000 ksi, determine the applied torque T on the
rigid disc. The cross sectional areas of all bars are 0.25 in.2.
If bars AB in Problem 5.1 are made of aluminum with a shear modulus Gal = 4000 ksi and bars CD are made of bronze with a shear
modulus Gbr = 6500 ksi, determine the applied torque T on the rigid disc. The cross-sectional areas of all bars are 0.25 in.2. 5.3 January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 212 5.4 Three pairs of bars are symmetrically attached to rigid discs at the radii shown in Figure P5.4. The discs were observed to rotate by
angles φ 1 = 1.5 ° , φ 2 = 3.0 ° , and φ 3 = 2.5 ° in the direction of the applied torques T1, T2, and T3, respectively. The shear modulus of the
bars is 40 ksi and cross-sectional area is 0.04 in.2. Determine the applied torques.
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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