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# 6a - CHAPTER 6 6.1 This problem asks that we derive...

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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 cross-sectional area A that is subjected to a tensile force P . Also represented is a plane that is oriented at an angle θ referenced to the plane perpendicular to the tensile axis; the area of this plane is A' = A /cos θ . In addition, the forces normal and parallel to this plane are labeled as P' and V' , respectively. Furthermore, on the left-hand 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 σ , the normal stress to this plane σ ' , as well as the shear stress τ ' 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 σ = P A τ = V A For static equilibrium in the x' direction the following condition must be met: F x' = 0 which means that

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P' - P cos θ = 0 Or that P' = P cos θ Now it is possible to write an expression for the stress σ ' in terms of P' and A' using the above expression and the relationship between A and A' [Figure (a)]: σ ' = P' A' = Pcos θ A cos θ = P A cos 2 θ However, it is the case that P / A = σ ; and, after make this substitution into the above expression, we have Equation (6.4a)--that is σ ' = σ cos 2 θ Now, for static equilibrium in the y' direction, it is necessary that F y ' = 0 = - V' + Psin θ Or V' = Psin θ We now write an expression for τ ' as τ ' = V' A' And, substitution of the above equation for V' and also the expression for A' gives
τ ' = V' A' = Psin θ A cos θ = P A sin θ cos θ = σ sin θ cos θ which is just Equation (6.4b). 6.2 (a) Below are plotted curves of cos 2 θ (for σ ' ) and sin θ cos θ (for τ ' ) versus θ . (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 cross-sectional area is just (10 mm) x (12.7 mm) = 127 mm 2 (= 1.27 x 10 -4

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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 ε = σ 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 10 -3 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 = π d o 2 2 where d o is the original diameter. Combining Equations (6.1), (6.2), and (6.5) and solving for l o leads to l o = E π d o 2 l 4F = 107 x 10 9 N /m 2 ( 29 ( π ) 3.8 x 10 - 3 m ( 29 2 0.42 x 10 - 3 m ( 29 (4)(2000 N) = 0.25 m = 250 mm (10 in.) 6.5 This problem asks us to compute the elastic modulus of steel.
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