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Ch6_solution

# Ch6_solution - Solutions for Chapter 6 MSEG 302 V(a A force...

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Unformatted text preview: Solutions for Chapter 6 - MSEG 302 V/(a) A force of 100,000 N is applied to a 10 mm X 20 mm iron bar having a yield strength of 400 MPa and a tensile strength of 480 MPa. Determine whether the bar will plastically deform and whether the bar will experience necking. Solution: First detennine the stress acting on the wire: a- : PM = 100.000 N I (10 mm)(20 mm) = 500 Wmm2 = 500 MPa Because a' is greater than the yield strength of 400 MPa. the wire will plastically deform. Because a- is greater than the tensile strength of 480 MPa. the wire will also neck. (b) Calculate the maximum force that a 0.2-in. diameter rod of A1203. having a yield strength of 35,000 psi, can withstand with no plastic deformation. Express your answer in pounds and Newtons. smitten: F = m = (35.000 psanI4)(0.2 in.)2 = 1100 lb / F: (1100 le4.448 Nab): 4891 N u/ A force of 20.000 N will cause a 1 cm x 1 cm bar of magnesium to stretch from 10 cm to 10.045 cm. Calculate the modulus of elasticity. both in GPa and psi. Solution: The strain 8 is c = (10045 cm — 10 cm)” 0 cm = 0.0045 cm/cm -\___, The stress a- is o- =20,000Nl(10mm)(10mm) =200 Nimm2 =200MPa L.) E: o-le = 200 MPa 2' 0.0045 cmfcm = 44.444 MPa = 44.4 GPa 1...———- E = (44.444 MPa)(l45 psb’M'Pa) = 6.44 x 106 mi L-- W The following data were collected from a standard 0.505—in.-diarneter test specimen of a copper alloy (initial length (E0) = 2.0 in): W (lb) (16.) (psi) (inJinJ 0 2.00000 0 0.0 3.000 2.0016? 15.000 0.000335 6,000 2.00333 30,000 0.001665 7.500 2.00417 37,500 0.002085 9,000 2.0090 45.000 0.0045 10.500 2.040 52,500 0.02 12,000 2.26 60.000 0.13” 12,400 2.50 (max load) 62,000 0.25 11.400 3.02 (fracture) 57,000 0.51 After fracture. the gage length is 3.014 in. and the diameter is 0.374 in. Plot the data and calculate the 0.2% offset yield strength along with (a) the tensile strength, {b} the modulus of elasticity, (c) the % elongation. (d) the % reduction in area, (1:) the engineering stress at fracture. (f) the true stress at fracture. and (g) the modulus of resilience. Solution: 0' = F ! (1r.ht){0.505)2 = no.2 e = (Z — 2) I 2 0.2% offset (.0 Stress (1:51} ID 0.002 0.01 0.02 Streln (in.lin.) 0.2% offset yield strength = 45,000 psi v“ (a) tensile strength = 62.000 psi v/ (b) a = (30,000 — 0) I (0.001665 — 0)=18 x10‘5 psi 5/ (0} 9'0 elongation = x 100 = 503% J 0000.505)? — (1041(0374)? W (e) engineering stress at fracture = 57,000 psi L/ (0 true stress at fracture = 11.0001b;’{'_ir1'.9‘4)(0.37'4)2 = 103.770 psi L/ 9 (S) From the graph. yielding begins at about 37,500 psi. Thus: 52(yield strength)(strain at yield) = l/2(3'i'.500)(0.002085) = 39.1 psi (0} as reduction in area = x 100 = 45.2% 1/" W A bar of A1203 that is 0.25 in. thick, 0.5 in. wide. and 9 in. long is tested in a three-point bending apparatus. with the supports located 6 in. apart. The deﬂection of the center of the bar is measured as a function of the applied load. The data are shown below. Determine the ﬂexural strength and the ﬂexural modulus. Force Deﬂection Stress (1b) (in) (1350 14.5 0.0025 4,176 28.9 0.0050 8.323 43 .4 0.0075 12.499 5?.9 0.0100 16.675 315.0 0.0149 (fracture) 24.768 Solution: stress = 3LF/2wh2 (6-15} = (3)05 in.)F l(2)(0.§ in.)(0.25 in)? = 2831-” ' 25 20 15 Stress (ksi) 10 0.005 0.010 0.015 DeﬂectIOn tin.) The ﬂexural strength is the stress at fracture. or 24,768 psi. / The ﬂexural modulus can be calculated from the linear curve; picking the ﬁrst point as an example: FL3 (14.5 lb)(6 in.)3 FM=—— =—————=40x106 i I." 4wh36 (4)(o.5 in.)(o.25in.)3(0.0025 in.) 135 ...
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