Solution Manual For Materials For Civil And Construction Engineers, 3rd Edition

Solution Manual For Materials For Civil And Construction Engineers, 3rd Edition makes textbook problem-solving easy with a comprehensive guide that explains every step.

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1CHAPTER 1. MATERIALS ENGINEERING CONCEPTS1.2.Strength at rupture =45 ksiToughness = (45 x 0.003) / 2 =0.0675 ksi1.3.A = 0.36 in2= 138.8889 ksiA= 0.0035 in/inL= -0.016667 in/inE = 39682 ksi= 0.211.4.A = 201.06 mm2= 0.945 GPaA= 0.002698 m/mL= -0.000625 m/mE =350.3GPa= 0.231.5.A =d2/4 = 28.27 in2-150,000 / 28.27 in2= -5.31 ksiE == 8000 ksiA/ E = -5.31 ksi / 8000 ksi = -0.0006631 in/inL =Ao= -0006631 in/in (12 in) = -0.00796 inLf=L + Lo= 12 in0.00796 in =11.992 in= -L/A= 0.35Ld / do= -A= -0.35 (-0.0006631 in/in) = 0.000232 in/ind =Lo= 0.000232 (6 in) = 0.00139 indf=d + do= 6 in + 0.00139 in =6.00139 in1.6.A =d2/4 = 0.196 in22= 10.18 ksi (Less than the yield strength. Within the elasticregion)E == 10,000 ksiA/ E = 10.18 ksi / 10,000 ksi = 0.0010186 in/inL =Ao= 0.0010186 in/in (12 in) = 0.0122 inLf=L + Lo= 12 in + 0.0122 in =12.0122 in= -L/A= 0.33Ld / do= -A= -0.33 (0.0010186 in/in) = -0.000336 in/ind =Lo= -0.000336 (0.5 in) = -0.000168 indf=d + do= 0.5 in - 0.000168 in = 0.49998 in12 in6 in12 in6 in

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