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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 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 = (^) AF oE
= 35,500 N
(1.27 x 10 −^4 m^2 ) (69 x 10 9 N/ m^2 )
= 4.1 x 10-
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 do is the original diameter. Combining Equations (6.1), (6.2), and (6.5) and solving for lo leads to
lo = E π d (^) o^2 ∆ l 4F
(107 x 10 9 N / m^2 )(π) 3.8 x 10( −^3 m)
2
(0.42 x 10 −^3 m)
(4)(2000 N)
= 0.25 m = 250 mm (10 in.)
6.7 (a) This portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation ( Fy ). Taking the yield strength to be 345 MPa, and employment of Equation (6.1) leads to
Fy = σy A o = (345 x 10 6 N/m^2 ) (130 x 10 -6^ m^2 )
= 44,850 N (10,000 lbf)
(b) The maximum length to which the sample may be deformed without plastic deformation is determined from Equations (6.2) and (6.5) as
li = l (^) o ⎛⎝⎜ 1 + σ E⎞⎠⎟
= (76 mm) 1 + 345 MPa 103 x 10 3 MPa
= 76.25 mm (3.01 in.)
6.13 This problem asks that we rank the magnitudes of the moduli of elasticity of the three hypothetical metals X, Y, and Z. From Problem 6.12, it was shown for materials in which the bonding energy is dependent on the interatomic distance r according to Equation (6.36), that the modulus of elasticity E is proportional to
E ∝ − 2A A nB
⎝⎜^
3 /(1 − n) +^
(n)(n + 1)B A nB
⎝⎜^
(n + 2) /(1 − n)
Or
∆ l = 4 Flo π d (^) o^2 E
(4)(48, 800 N) 200 x 10( −^3 m)
(π) 19 x 10( −^3 m)
2
(69 x 10 9 N/ m^2 )
= 0.50 mm (0.02 in.)
(b) We are now called upon to determine the change in diameter, ∆ d. Using Equation (6.8)
ν = − εx εz =^ −
∆d / d (^) o ∆l/ lo
From Table 6.1, for Al, ν = 0.33. Now, solving for ∆d yields
∆d = − ν ∆ld (^) o lo =^ −^
(0.33)(0.50 mm)(19 mm) 200 mm
= -1.57 x 10-2 mm (-6.2 x 10-4 in.)
The diameter will decrease.
6.22 Elastic deformation is time-independent and nonpermanent, anelastic deformation is time- dependent and nonpermanent, while plastic deformation is permanent.
6.29 This problem calls for us to make a stress-strain plot for aluminum, given its tensile load- length data, and then to determine some of its mechanical characteristics. (a) The data are plotted below on two plots: the first corresponds to the entire stress-strain curve, while for the second, the curve extends just beyond the elastic region of deformation.
(b) The elastic modulus is the slope in the linear elastic region as
E = ∆ σ∆ε = 200 MPa0.0032^ − −0 MPa 0 = 62.5 x 10^3 MPa = 62.5 GPa (9.1 x 10 6 psi)
(c) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain curve at approximately 285 MPa (41,000 psi ). (d) The tensile strength is approximately 370 MPa (53,500 psi), corresponding to the maximum stress on the complete stress-strain plot.
(b) The elastic modulus is the slope in the linear elastic region as
E = ∆ σ∆ε = 50 MPa0.001^ − −0 MPa 0 = 50 x 10^3 MPa = 50 GPa (7.3 x 10 6 psi)
(c) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain curve at approximately 150 MPa (21,750 psi). (d) The tensile strength is approximately 240 MPa (34,800 psi), corresponding to the maximum stress on the complete stress-strain plot. (e) From Equation (6.14), the modulus of resilience is just
Ur =
σy^2 2E
which, using data computed in the problem, yields a value of
Ur =
(150 x 10 6 N/ m^2 )
2
(2) 50 x 10 ( 9 N/ m 2 )
= 2.25 x 10^5 J/m^3 (32.6 in. - lb f /in.^3 )
(f) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by one- hundred. The total fracture strain at fracture is 0.110; subtracting out the elastic strain (which is about 0.003) leaves a plastic strain of 0.107. Thus, the ductility is about 10.7%EL.
