Memtuedumavablemom2ndhtm ab d 200 mm 160 mm 120 mm

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Unformatted text preview: nted from: http://www.me.mtu.edu/~mavable/MoM2nd.htm (c) A1 B B1 C CD D D1 D E1 EF F1 EF E F CD C1 200 mm 160 mm 120 mm Figure 5.9 Approximate deformed shapes for Method 1 in Example 5.3 of segments (a) AB, (b) CD, and (c) EF. Using Figure 5.9a we can find the shear strain in AB as AA 1 = r AB φ 1 = ( 20 mm ) ( 0.0436 ) = 0.872 mm BB 1 = r AB φ 2 = ( 20 mm ) ( 0.0262 ) = 0.524 mm AA 1 + BB 1 0.872 mm + 0.524 mm tan γ AB ≈ γ AB = ------------------------- = ------------------------------------------------------AB 200 mm (E2) (E3) The shear strain is positive as the angle γAB represents a decrease of angle from right angle. ANS. Using Figure 5.9b we can find the shear strain in CD as January, 2010 γ AB = 6980 μ rad M. Vable Mechanics of Materials: Torsion of Shafts CC 1 = r CD φ 2 = ( 15 mm ) ( 0.0262 ) = 0.393 mm 5 209 DD 1 = r CD φ 3 = ( 15 mm ) ( 0.0262 ) = 0.393 mm (E4) CC 1 + DD 1 0.393 mm + 0.393 mm tan γ CD ≈ γ CD = --------------------------- = ------------------------------------------------------CD 160 mm (E5) The shear strain is negative as the angle γCD represents an increase of angle from right angle. γ CD = – 4913 μ rad ANS. Using Figure 5.9c we can find the shear strain in EF as EE 1 = r EF φ 3 = ( 10 mm ) ( 0.0262 ) = 0.262 mm FF 1 = r EF φ 4 = ( 10 mm ) ( 0.0567 ) = 0.567 mm (E6) FF 1 – E E 1 0.567 mm – 0.262 mm tan γ EF ≈ γ EF = -------------------------- = -----------------------------------------------------EF 120 mm (E7) The shear strain is negative as the angle γEF represents an increase of angle from right angle. γ EF = – 2542 μ rad ANS. METHOD 2: PLAN We assign a sign to the direction of rotation, calculate the relative deformation of the right disc with respect to the left disc, and analyze the entire shaft. We draw an approximate deformed shape of the entire shaft, as shown in Figure 5.10. Let the counterclockwise rotation with respect to the x axis be positive and write each angle with the correct sign, φ 1 = – 0.0436 rad φ 2 = 0.0262 rad φ 3 = – 0.0262 rad φ 4 = – 0.0567 rad (E8) 1 2 Positive x Figure 5.10 Shear strain calculation by Method 2 in Example 5.3. 3 A1 A AB B C B1 C1 EF D1 D CD E 4 F We compute the relative rotation in each section and multiply the result by the corresponding section radius to obtain the relative movement of two points in a section. We then divide by the length of the section as we did in Example 5.2. Δφ AB = φ 2 – φ 1 = 0.0698 Δφ CD = φ 3 – φ 2 = – 0.0524 Δφ EF = φ 4 – φ 3 = – 0.0305 r AB Δ φ AB ( 20 mm ) ( 0.0698 ) γ AB = ---------------------- = --------------------------------------------- = 0.00698 rad AB ( 200 mm ) (E9) r CD Δ φ CD ( 15 mm ) ( – 0.0524 ) γ CD = ----------------------- = ------------------------------------------------ = – 0.004913 rad CD 160 mm (E10) r EF Δ φ EF ( 10 mm ) ( – 0.0305 ) γ EF = ---------------------- = ------------------------------------------------ = – 0.002542 rad EF 120 mm (E11) ANS. γ AB = 6980 μ rad γ CD = – 4913 μ rad γ EF = – 2542 μ rad COMMENTS Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 1. Method 1 is easier to visualize, but the repetitive calculations can be tedious. Method 2 is more mathematical and procedural, but the repetitive calculations are easier. By solving the problems by method 2 but spending time visualizing the deformation as in method 1, we can reap the benefits of both. 2. We note that the shear strain in each section is directly proportional to the radius and the relative rotation of the shaft and inversely proportional to its length. 5 .1.1 Internal Torque Example 5.1 showed that the shear stress τxθ can be replaced by an equivalent torque using an integral over the cross-sectional area. In this section we formalize that observation. Figure 5.11 shows the shear stress distribution τxθ that is to be replaced by an equivalent internal torque T. Let ρ represent the radial coordinate, that is, the radius of the circle at which the shear stress acts. The moment at the center due to the shear stress on the differential area is ρτ x θ d A. By integrating over the entire area we obtain the total internal torque at the cross section. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts T= 5 ∫A ρ dV = ∫A ρτx θ dA dV x 210 (5.1) dA T x Figure 5.11 Statically equivalent internal torque. Equation (5.1) is independent of the material model as it represents static equivalency between the shear stress on the entire cross section and the internal torque. If we were to consider a composite shaft cross section or nonlinear material behavior, then it would affect the value and distribution of τxθ across the cross section. But Equation (5.1), relating τxθ and T, would remain unchanged. Examples 5.4 will clarify the discussion in this paragraph. EXAMPLE 5.4 A homogeneous cross section made of brass and a composite cross section of brass and steel are shown in Figure 5.12. The shear moduli of elasticity for brass and steel are GB = 40 GPa and GS = 80 GPa, respectively. The shear strain in polar coo...
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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.

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