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Unformatted text preview: CHAPTER 6 6.1 This problem asks that we derive Equations (6.4a) and (6.4b), using mechanics of materials principles. In Figure (a) below is shown a block element of material of crosssectional area A that is subjected to a tensile force P . Also represented is a plane that is oriented at an angle referenced to the plane perpendicular to the tensile axis; the area of this plane is A' = A /cos . In addition, the forces normal and parallel to this plane are labeled as P' and V' , respectively. Furthermore, on the lefthand side of this block element are shown force components that are tangential and perpendicular to the inclined plane. In Figure (b) are shown the orientations of the applied stress , the normal stress to this plane ' , as well as the shear stress ' taken parallel to this inclined plane. In addition, two coordinate axis systems in represented in Figure (c): the primed x and y axes are referenced to the inclined plane, whereas the unprimed x axis is taken parallel to the applied stress. Normal and shear stresses are defined by Equations (6.1) and (6.3), respectively. However, we now chose to express these stresses in terms (i.e., general terms) of normal and shear forces ( P and V ) as = P A = V A For static equilibrium in the x' direction the following condition must be met: F x' = 0 which means that 113 P' P cos = 0 Or that P' = P cos Now it is possible to write an expression for the stress ' in terms of P' and A' using the above expression and the relationship between A and A' [Figure (a)]: ' = P' A' = P cos A cos = P A cos 2 However, it is the case that P / A = ; and, after make this substitution into the above expression, we have Equation (6.4a)that is ' = cos 2 Now, for static equilibrium in the y' direction, it is necessary that F y ' = 0 = V' + P sin Or V' = P sin We now write an expression for ' as ' = V' A' And, substitution of the above equation for V' and also the expression for A' gives 114 ' = V' A' = P sin A cos = P A sin cos = sin cos which is just Equation (6.4b). 6.2 (a) Below are plotted curves of cos 2 (for ' ) and sin cos (for ' ) versus . (b) The maximum normal stress occurs at an inclination angle of 0 . (c) The maximum shear stress occurs at an inclination angle of 45 . 6.3 This problem calls for us to calculate the elastic strain that results for an aluminum specimen stressed in tension. The crosssectional area is just (10 mm) x (12.7 mm) = 127 mm 2 (= 1.27 x 104 115 m 2 = 0.20 in. 2 ); also, the elastic modulus for Al is given in Table 6.1 as 69 GPa (or 69 x 10 9 N/m 2 ). Combining Equations (6.1) and (6.5) and solving for the strain yields = E = F A o E = 35,500 N 1.27 x 10 4 m 2 ( 29 69 x 10 9 N/ m 2 ( 29 = 4.1 x 103 6.4 We are asked to compute the maximum length of a cylindrical titanium alloy specimen that is deformed elastically in tension. For a cylindrical specimen A o =...
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 Spring '08
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