Machinery's Handbook, 31st Edition
218 DEFLECTIONS x -axis. The shear stress caused by direct shear loading, Case 5, has a uniform distribution. However, the shear stress caused by torsion loading, Case 6, has a zero value at the axis and a maximum value at the surface farthest from the axis. Deflections.— For direct tension and direct compression loading on members with uni form cross sections, deflection can be calculated using Equation (17). For direct tension loading, e is an elongation; for direct compression loading, e is a contraction. Deflection is in inches when the load F is in pounds, the length L over which deflection occurs is in inches, the cross-sectional area A is in square inches, and the modulus of elasticity E is in pounds per square inch. The angular deflection of members with uniform circular cross sections subject to torsion loading can be calculated with Equation (18). (17) (18) The angular deflection θ is in radians when the applied torsion T is in inch-pounds, the length L over which the member is twisted is in inches, the modulus of rigidity G is in pounds per square inch, and the polar moment of inertia J is in in. 4 Metric SI units can be used in Equations (17) and (18), where F = force in newtons (N); L = length over which deflection or twisting occurs in meters; A = cross-sectional area in meters 2 ; E = the modulus of elasticity in N/m 2 ; θ = radians; T = the applied torsion in newton-meters (N·m); G = modulus of rigidity, in pascals (Pa); and J = the polar moment of inertia in meters 4 . If the load ( F ) is applied as a weight, it should be noted that the weight of a mass M kilograms is Mg newtons, where g = 9.81 m/s 2 . Mil limeters can be used in the calculations in place of meters, provided the treatment is consistent throughout. Combined Stresses.— A member may be loaded in such a way that a combination of sim ple stresses acts at a point. Three general cases are shown in the accompanying illustra- tion, Fig. 11. Strength data is widely available for tensile performance, and the von Mises stress method and other methods can be used to calculate effective tensile stress at a point under combined loading. von Mises Effective Tensile Stress: This method of calculating an equivalent tensile stress for combined stress scenarios is applicable to ductile material. The von Mises effective tensile stress, s ', can be directly compared to the tensile yield strength of the material to predict performance for two-dimensional loading (Equation 19) and three-dimensional loading (Equation 20): (19) σ σ σ σσ τ ´ = + − + x y x y xy 2 2 2 3 (20) Superposition of Stresses: This method may be used in cases where the deformations are small and within the elastic range. Fig. 11 at (1) illustrates a common situation that results in simple stresses combining by superposition at points a and b . The equal and opposite forces F 1 will cause a compressive stress σ 1 = − F 1 / A . Force F 2 will cause a bend- ing moment M to exist in the plane of points a and b . The resulting stress σ 2 = ± M / Z . The combined stress at point a , (21) and at b , (22) where the minus sign indicates a compressive stress and the plus sign a tensile stress. Thus, the stress at a will be compressive and at b either tensile or compressive, depending on which term in the equation for σ b ′ has the greatest value. e FL AE = ---- θ TL GJ = ---- σ σ σ σ σ σ σ τ τ σ ) ( ) ( ) ( − + − + − + + + x y y z z x xy yz zx 2 2 2 2 2 2 6 2 ´ ( ) = A F Z M a 1 σ ′ =− − A F Z M b 1 σ ′ =− +
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