ENGR 220 Homework Problems: Week 5 - Chapter 6 1. A specimen of aluminum having a rectangular cross section 10 mm x 12.7 mm (0.4 in. x 0.5 in.) is pulled in tension with 35,500 N (8000 lbf) force, producing only elastic deformation. Calculate the resulting strain. e = (35500)/(0.01*0.0127)/(69*10^9) = 0.004051 2. A cylindrical rod of copper (E = 110 GPa, 16 x 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6660 N (1500 lbf). If the length of the rod is 380 mm (15.0 in.), what must be the diameter to allow an elongation of 0.50 mm (0.020 in.)? do = sqrt(4 *. 38*6660/(pi*110*10^9 *. 5*10^-3)) do = 7.654mm 3. A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its origina length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 GPa and 39.7 GPa, respectively. 105GPa = 2G(1+u) u = 0.32242 ed = d-d0/d0 = 0.00124844 Lo = L(/1-ed/u) = 75.25mm 4. A cylindrical specimen of aluminum having a diameter of 12.8 mm and a gauge length of 50.800 mm pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (e). Load N 0 7330 15.100 23,100 50.952 30,400 51.003 34,400 51.054 38.400 51.308 41,300 51.816 52.832 44,800 46,200 53.848 47.300 54.864 47.500 55.880 46,100 56.896 44,800 57.658 42.600 36,400 Length mm 50.800 50.851 50.902 58.420 59.182 Fracture (a) Plot the data as engineering stress versus engineering strain using a computer program (such
Stress-strain chart 400.000 350.000 300.000 250.000 Stress (MPa) 200.000 150.000 100.000 50.000 0.000 0.000 0.020 0.040 0.060 0.080 0.100 0.120 0.140 0.160 0.180 Strain(mm/mm) -Series1 as Excel). (b) Compute the modulus of elasticity. E = 200-0/0.003-0 = 66.67 GPa (c) Determine the yield strength at a strain offset of 0.002. for yield strength, the 0.002 strain offset line is shown as dashed line in the given graph below. it intersect the stress strain curve at approximately 283 MPa Stress-strain chart 300.000 280.000 260.000 240.000 220.000 200.000 180.000 160.000 Stress (MPa) 140.000 120.000 100.000 80.000 60.000 40.000 20.000 0.000 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.018 Strain(mm/mm) 0.020 (d) Determine the tensile strength of this alloy. the maximum value of stress in the stress column of table will give the tensile strength of the alloy. Sut = 369.136 MPa (e) What is the approximate ductility, in percent elongation? The ductility, in percent elongation, is the strain at fracture pt. multiplied by a hundred. the total fracture strain at fracture is 0.165; elastic strain is about 0.005 thus plastic strain at fracture = 0.165 - 0.005 = 0.160 and the approximate ductility in percent approximation will be = 0.160*100 = 16%