SolutionsToCh6 - SOLUTION (6.1D) Known: A poem appears in...

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Unformatted text preview: SOLUTION (6.1D) Known: A poem appears in Chapter Six titled: "The One-Hoss Shay“. Find: Write a one paragraph summary of the poem "The One-Hoss Shay". Analysis: A "chaise" should be designed and built in a logical way such that it has uniform strength throughout and after a hundred years, looks okay, feels okay, and shows little trace of age, has no loss of strength and the strength is uniform throughout, the same as before, and then breaks down and wears out "all at once and nothing first" in a heap or mound (and the modern version would require that the heap or mound be immediately ready for recycling). SOLUTION (6.2) Known: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: 7075—T65 1 Aluminum su = 78 ksi sy = 70 ksi KIC = 60 ksi in0'5 2w = 6 in. t = 0.035 in. 20 = 1 in. Assumptions: 1. The crack length is a small fraction of the plate width. 2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength. Yielding has occurred within one small volume of the material at the crack root. 3. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K1 equals or exceeds the fracture toughness KIC for the material. 6-1 Analysis: 1 . 2. _ KIC _ _ - From Eq. (6.2), (Sig—1'84E —1.81/0_5_ —47.14k31 Since the area equals 2wt, P = og(2wt) = 47.14(6)(0.035) = 9,899 lb I Comment: P/A stress based on the net area, t(2w — 2c), is 56.57 ksi which is less than Sy = 70 ksi; hence, the second assumption is satisfied. SOLUTION (6.3) Known: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: D6AC Steel at —40 OF s1, = 227 ksi sy = 197 ksi KIc = 100 ksi inO'S 2w = 6 in. t= 0.06 in. 2c = 1 in. P=? Assumptions: 1. The crack length is a small fraction of the plate width. 2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength. 3. Yielding has occurred within one small volume of the material at the crack root. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K1 equals or exceeds the fracture toughness KIC for the material. Analysis: _ K c _ 100 _ - l. FromE. 6.2, _——I—____ _ q < )og 1.8m mm 78 567 ks1 2. Since the area equals 2wt, P = og(2wt) = 78,567(6)(0.06) = 28,284 lb I 6-2 Comment: P/A stress based on the net area, t(2w - 2c), is 94.28 ksi which is less than Sy = 197 ksi; hence, the second assumption is satisfied. SOLUTION (6.4) Known: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: D6AC Steel at room temperature 311 = 220 ksi y = 190 ksi Km = 115 ksi 1110-5 2w = 6 in. t= 0.06 in. 2c = 1 in. P=? Assumptions: 1. The crack length is a small fraction of the plate width. 2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength. 3. Yielding has occurred within one small volume of the material at the crack root. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K1 equals or exceeds the fracture toughness KIC for the material. Analysis: 1. From Eq. (6.2), cg = 1%: % = 90.353 ksi 2. Since the area equals 2wt, P = (Sg(2wt) = 90,353(6)(0.06) = 32,527 117 I Comment: The P/A stress based on the net area, t(2w - 2c), is 108.4 ksi which is less than Sy = 190 ksi. Hence the second assumption is satisfied. SOLUTION (6.5) Known: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: 4340 Steel at room temperature su = 260 ksi Sy = 217 ksi Igc = 115 ksi in0'5 2w = 6 in. t: 0.06 in. 2c 2 l in. P=? Assumptions: 1 . The crack length is a small fraction of the plate width. 2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength. 3. Yielding has occurred within one small volume of the material at the crack root. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor KI equals or exceeds the fracture toughness KIC for the material. Analysis: 1. From Eq. (6.2), cg = 112:6 = 1.;11/%§ = 90.353 ksi 2. Since the area equals 2wt, P = Og(2wt) = 90,353(6)(0.06) = 32,527 lb I Comment: The P/A stress based on the net area, t(2w - 2c), is 108.4 ksi which is less than Sy = 217 ksi. Hence the second assumption is satisfied. 6-4 SOLUTION (6.6) Known: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: Ti—6Al—4V (annealed) titanium alloy u = 130 ksi Sy = 120 ksi KIc = 110 ksi 1110-5 2w 2 6 in. t = 0.06 in. 20 = 1 in. P=? Assumptions: 1. The crack length is a small fraction of the plate width. 2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength. 3. Yielding has occurred within one small volume of the material at the crack root. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K1 equals or exceeds the fracture toughness KIC for the material. Analysis: 1. From Eq. (6.2), 6g: = 1.3%.? = 86.42 ksi 2. Since the area equals 2wt, P = cg(2wt) = 86,420(6)(0.06) = 31,112 lb I Cement: The P/A stress based on the net area, t(2w — 2c), is 103.6 ksi which is less than Sy = 120 ksi. Hence the second assumption is satisfied. 6—5 SOLUTION (6.7) Known: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: Leaded beryllium copper 311 = 98 ksi Sy= 117 ksi KIC = 70 ksi 1110-5 2w 2 8 in. t: 0.05 in. 2c = 1.5 in. P:? Assumptions: 1. The crack length is a small fraction of the plate width. 2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength. 3. Yielding has occurred within one small volume of the material at the crack root. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor KI equals or exceeds the fracture toughness KIC for the material. Analysis: 1. From Eq. (6.2), cg = 11%;: Tia—74%? = 44.9 ksi 2. Since the area equals 2wt, P = og(2wt) = 44,900(8)(0.05) = 17,960 lb I Cement: The P/A stress based on the net area, t(2w - 2c), is 55.26 ksi which is less than Sy = 117 ksi. Hence the second assumption is satisfied. SOLUTION (6.8) Known: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: Leaded brass su = 55 ksi Sy = 42 ksi KIC = 35 ksi in0-5 2W = 8 in. t: 0.05 in. 2c = 1.5 in. P=? Assumptions: 1 The crack length is a small fraction of the plate width. The tensile stress based on the net area (minus the area of the crack) is less than 2. the yield strength. 3. Yielding has occurred within one small volume of the material at the crack root. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor KI equals or exceeds the fracture toughness KIC for the material. Analysis: 1. From Eq. (6.2), cg = 515—: L = 22.45 ksi 1.8m 1,840.75 2. Since the area equals 2wt, P = og(2wt) = 22,450(8)(0.05) = 8980 lb I Comment: The P/A stress based on the net area, t(2w - 2c), is 27.6 ksi which is less than Sy = 42 ksi. Hence the second assumption is satisfied. SOLUTION (6.9) Known: A thick plate having a central crack is loaded in tension to a known gross- area stress. Find: Determine the critical crack depth, acr, at which rapid fracture will occur for 7075-T651 aluminum. Schematic and Given Data: 7075-T65 1 Aluminum 2w = 6 in. t: 1 in. a/2c = 0.25 cg = 0.73 sy Assumptions: 1. The 7075-T651 aluminum is at room temperature. 2. The crack length is critical when the value of the stress intensity factor K exceeds KIc. Analysis: 1. From Eq. (6.4), and setting K = KIC ch 0.39 — 0.053(csg/sy)2 2 aCf = 6 3 Since 6g = 0.73 Sy _ KIc 2 _ 2 act—(0‘73 Sy) [0.39 0.053(0.73)] 2 aCI = 0.6 6-8 2. Using Table 6.1 to find Sy and K10, for 7075—T651 Aluminum: KIC = 27 ker‘ in., Sy = 70 ksi, ac, = 0.101 in. I Comments: 1. Eq. (6.4) is appropriate if Zoo/t > 6, a/2c = about 0.25, (o/c > 3, a/t < 0.5, and 6g < 0.8 Sy. For this problem, these conditions are satisfied. 2. An important design requirement of internally pressurized members is that a crack be able to propagate through the full wall thickness (thereby causing a leak that can be readily detected) without becoming unstable and leading to total fracture. SOLUTION (6.10) Known: A thick plate having a central crack is loaded in tension to a known gross- area stress. Find: Determine the critical crack depth, a“, at which rapid fracture will occur for D6AC steel at room temperature. Schematic and Given Data: D6AC Steel at room temperature 2w = 6 in. t: 1 in. a/2c = 0.25 cg = 0.73 sy Assumption: The crack length is critical when the value of the stress intensity factor K exceeds KIC. Analysis: 1. From Eq. (6.4), and setting K = KIC 2 KIc 0.39 — 0.053(cg/sy)2 Cyg acr = Since 6g = 0.73 Sy aa=( KR fb39—om3mjafi 0.73 Sy 2 %r=06%ék) Sy 2. Using Table 6.1 to find Sy and K10, for D6AC steel at room temperature: KIC = 70 ksr-V in., Sy = 190 ksi, aCir = 0.092 in. I Comments: 1. Eq. (6.4) is appropriate if 2(0/t > 6, a/2c = about 0.25, (D/C > 3, a/t < 0.5, and (5g < 0.8 Sy. For this problem, these conditions are satisfied. 2. An important design requirement of internally pressurized members is that a crack be able to propagate through the full wall thickness (thereby causing a leak that can be readily detected) without becoming unstable and leading to total fracture. SOLUTION (6.11) Known: A thick plate having a central crack is loaded in tension to a known gross— area stress. Find: Determine the critical crack depth, a“, at which rapid fracture will occur for D6AC steel at —40 OF. Schematic and Given Data: D6AC Steel at ~40 °F 2w = 6 in. t: l in. 23/20 = 0.25 g=073sy 6—10 Assumption: The crack length is critical when the value of the stress intensity factor K exceeds KIC. Analysis: 1. From Eq. (6.4), and setting K = KIC 2 KIC 0.39 - 0.053(og/Sy)2 a = 0 CI g Since 6g = 0.73 Sy KIC 2 _ 5 2 0.73 8y) [0.39 0.0 3(0.73)] K 2 ac, = 0.68{s—1yc) 2. Using Table 6.1 to find Sy and KIC for D6AC steel at -40 0F: KIC = 45 ksnl' Sy = 197 ksi, aCI = 0.0354 in. I 21or = Comments: 1. Eq. (6.4) is appropriate if Zm/t > 6, a/2c = about 0.25, (o/c > 3, a/t < 0.5, and (5g < 0.8 Sy. For this problem, these conditions are satisfied. 2. An important design requirement of internally pressurized members is that a crack be able to propagate through the full wall thickness (thereby causing a leak that can be readily detected) without becoming unstable and leading to total fracture. 6-11 SOLUTION (6.15) Known: A machine component with given critical stresses is ductile, with yield strengths in tension and compression of 60 ksi. Find: Determine the safety factor according to: (a) the maximum-normal—stress theory (b) the maximum-shear-stress theory (c) the maximum-distortion—energy theory Schematic and Given Data: Assumption: The material is homogeneous. Analysis: 1. From the above Mohr—circle, Tmax = (20 + 15)/2 = 17.5 ksi 2. (a) For the maximum-normal-stress theory: SF = 60/20 = 3.0 (b) For the maximum-shear—stress theory: SF = 150/me 2 30/175 = 1.72 (c) For the maximum—distortion—energy theory: SF = 60/S', where from Eq. (6.6) S' = [(512 - 6162 + 622]“2 = [(20)2 - (20)(—15) + (— 15)2]1/2 = 30.5 thus, SF 2 60/305 = 1.97 3. The existence of a yield strength implies a ductile material for which: - maximum-distortion—energy theory is best - maximum-shear-stress theory may be acceptable 0 maximum-normal—stress theory is not appropriate 6—18 SOLUTION (6.16) Known: A machine component with given critical stresses is ductile, with yield strengths in tension and compression of 60 ksi. Find: Determine the safety factor according to: (a) the maximum—normal-stress theory (b) the maximum-shear—stress theory (c) the maximum-distortion-energy theory Schematic and Given Data: Assumption: The material is homogeneous. Analysis: _- 1. From the above Mohr-circle, “cum = (25 + 15)/2 = 20 ksi 2. (a) For the maximum-normal-stress theory: SF = 60/25 = 2.4 (b) For the maximum-shear-stress theory: SF = 30km = 30/20 = 1.5 (c) For the maximum-distortion-energy theory: SF = 60/S', where from Eq. (6.6) S' = [612 - 6162 + 622]“2 = [(25)2 — (25)(-15) + (- 15)2]1/2 = 35.0 thus, SF = 60/350 = 1.71 3. The existence of a yield strength implies a ductile material for which: 0 maximum—distortion-energy theory is b_es_t 6—19 - maximum-shear—stress theory may be acceptable - maximum—normal—stress theory is not appropriate SOLUTION (6.17) Known: The solutions to problems (a) 4.31, (b) 4.34, (c) 4.36, (d) 4.38, (e) 4.40, (f) 4.41, (g) 4.43, (h) 4.47, and (i) 4.51 are given. Find: Determine the tensile strength a ductile material must have in order to provide a safety factor of 2 with respect to initial yielding at the locations investigated in the above listed problems. Determine the answer using both the maximum-shear—stress theory and the maximum—distortion-energy theory. Assumption: The materials are homogeneous. Analysis: MA r. m Per 1: Theory (Seq From Per D.E. Theory rmax S! for SF = 2 Eqns. (6.5-6.8) 83! for SF 2 2 (a) 97.5 MPa (@ "8") 390 MPa 185.4 MPa 371 MPa (b) 128 MPa (@ "a") 512 MPa 241.4 MPa 483 MPa (c) 37.2 ksi 148.8 ksi 68.1 ksi 136 ksi (d) 17 ksi 68 ksi 31.9 ksi 64 ksi (e) 34.3 MPa 137.2 MPa 66.0 MPa 132 MPa (f) 278 MPa 1112 MPa 500 MPa 1000 MPa (g) 110 MPa 440 MPa 193.6 MPa 387 MPa (h) 200 MPa 800 MPa 346 MPa 692 MPa (i) 350 MPa 1400 MPa 608 MPa 1216 MPa SOLUTION (6.18) Known: The solutions to problems (a) 4.31, (b) 4.34, (c) 4.36, (d) 4.38, (e) 4.40, (t) 4.41, (g) 4.43, (h) 4.47, and (i) 4.51 are given. Find: Determine the tensile strength a ductile material must have in order to provide a safety factor of 1.5 with respect to initial yielding at the location(s) investigated in the above listed problems. Determine the answer using both the maximum-shear-stress theory and the maximum—distortion—energy theory. Assumption: The materials are homogeneous. Analysis: Answer Answer Per 1 Theory oeq From Per D.E. Theory rm” SE for SF = 1.5 Eqns. (6.5—6.8) S)! for SF = 1.5 (a) 97.5 MPa (@ "3") 293 MPa 185.4 MPa 278 MPa (b) 128 MPa (@ "a") 384 MPa 241.4 MPa 362 MPa (0) 37.2 ksi 112 ksi 68.1 ksi 102 ksi (d) 17 ksi 51 ksi 31.9 ksi 47.9 ksi (e) 34.3 MPa 102.9 MPa 66.0 MPa 99 MPa (f) 278 MPa 834 MPa 500 MPa 750 MPa (g) 110 MPa 330 MPa 193.6 MPa 290 MPa (h) 200 MPa 600 MPa 346 MPa 519 MPa (1) 350 MPa 1050 MPa 608 MPa 912 MPa M 6-20 SOLUTION (6.19) Known: The solutions to problems (a) 4.31, (b) 4.34, (c) 4.36, (d) 4.38, (e) 4.40, (f) 4.41, (g) 4.43, (h) 4.47, and (i) 4.51 are given. Find: Use the modified Mohr theory to determine the ultimate tensile strength that would be required of a brittle material in order to provide a safety factor of 4 to a member subjected to the same state(s) of stress as the above listed problems. If overloaded to failure, what would be the orientation of the brittle crack in each case? Assumptions: 1. The materials are homogeneous. 2. The ultimate compressive strength is 3.5 times the ultimate tensile strength. Analysis: ANSWER ANSWER g; §§ for §F=4 Crack Orientation (a) 174 MPa (@ "8") 696 MPa 193° C.W. from a transverse plane (b) 225 MPa (@ "a") 900 MPa 20° C.C.W. from a transverse plane (c) 58 ksi 232 ksi 283° C.C.W. from a transverse plane ((1) 30 ksi 120 ksi 19° C .W. from a transverse plane (6) 63 MPa 252 MPa 16.5° C.W. from a transverse plane (f) 556 MPa 2224 MPa 51.5° C.W. from a transverse plane (g) 220 MPa 880 MPa 675° C.C.W. from a transverse plane (h) 400 MPa 1600 MPa Longitudinal (i) 600 MPa 2400 MPa Longitudinal SOLUTION (6.20) Known: The solutions to problems (a) 4.31, (b) 4.34, (c) 4.36, (d) 4.38, (e) 4.40, (t) 4.41, (g) 4.43, (h) 4.47, and (i) 4.51 are given. Find: Use the modified Mohr theory to determine the ultimate tensile strength that would be required of a brittle material in order to provide a safety factor of 3.5 to a member subjected to the same state(s) of stress as the above listed problems. If overloaded to failure, what would be the orientation of the brittle crack in each case? Assumptions: 1. The materials are homogeneous. 2. The ultimate compressive strength is 3.5 times the ultimate tensile strength. Analysis: ANSWER ANSWER Q1 S__” for SF==3.5 (gragk Orientation (a) 174 MPa (@ "S") 609 MPa 19.30 C.W. from a transverse plane (b) 225 MPa (@ "a") 788 MPa 20° C.C.W. from a transverse plane (c) 58 ksi 203 ksi 283° C.C.W. from a transverse plane (d) 30 ksi 105 ksi 19° C.W. from a transverse plane (8) 63 MPa 221 MPa 16.50 C.W. from a transverse plane (f) 556 MPa 1946 MPa 51.5° C.W. from a transverse plane g) 220 MPa 770 MPa 67.5o C.C.W. from a transverse plane (h) 400 MPa 1400 MPa Longitudinal (i) 600 MPa 2100 MPa Longitudinal 6—21 Analysis: 1. From Eq. (6.8), the distortion energy stress is ca = (0x2 + 3 Txy2)1/2 = (502 + 3(100)2)1/2 Ge = 1803 MPa 2. The safety factor is SF = Sy/Ge = 400/1803 = 2.22 I Comment: The maximum shear stress theory predicts a safety factor SF=_SS_Y=___S_Y._/_2____=___2m__= 194 "m T2 + (0x ' 6y)2 \(1002 + 252 xy 2 SOLUTION (6.25) Known: A round shaft of known strength and specified safety factor is loaded with a known torque. Find: Determine the shaft diameter. Schematic and Given Data: T = 5000 1b in. T = 5000 1b in. m =Tr/J =16T/1rd3 Assumption: The shaft material is homogeneous. Analysis: (a) For the maximum-normal-stress theory, 6-26 Sy Sy 60,000 SF = 2 = 6m = 76;; = 16(5000) m3 Solving for (1, gives d = 0.95 in. I (b) For the maximum-shear—stress theory, Ssy _ Sy/Z _ 30,000 SF = 2 = Tm ' Txy ' 16(5000) m3 Solving for d, gives d = 1.19 in. I (c) From Eq. (6.8), Ce = 1/0 + 3132 = 151. For the distortion-energy theory, SF _ 2 __§ _ 60,000 _ _ 0e _ 1B 16(5000) 7rd3 Solving for (1, gives (1 = 1.14 in. I Cements: 1. If the shaft is a ductile material, the distortion-energy theory is the most accurate, followed by the maximum-shear—stress theory . If the shaft were a brittle material, then the normal—stress-theory would be the most appropriate of the three theories used in the analysis. 2. A steel shaft with Sy = 60,000 psi would have an elongation in 2 in. of approximately 20%, and hence would be a ductile material. Indeed, most steel shafts are ductile. 3. Good test data pertaining to actual material and torsion loading would be recommended to improve the failure theory prediction. 6-27 SOLUTION (6.26) Known: A round shaft of known strength and specified safety factor is loaded with a known torque. Find: Determine the shaft diameter. Schematic and Given Data: T = 6000 lb in. zxy =Tr/J =16T/1td3 Assumption: The shaft material is homogeneous. Analysis: (a) For the maximum-normal-stress theory, S S SF=2=—-Y—=_y_=_§9£99_ Cm Txy w 1rd3 Solving for (1, gives (1 = 1.01 in. (b) For the maximum-shear—stress theory, Ssy _ Sy/z _ 30,000 SF 2 2 = rm xxy ‘ 16(6000) 3 1rd Solving for d, gives (1 = 1.27 in. I 6-28 (c) From Eq. (6.8), O'e = 1/0 + 31:2 = 15 1:. For the distortion—energy theory, S . 60 000 SF = 2 = —’ = ———’—— Ge «5 16(6000) m3 Solving for (1, gives d = 1.21 in. I Comments: 1. If the shaft is a ductile material, the distortion—energy theory is the most accurate, followed by the maximum—shear-stress theory (most steel shafts are ductile). If the shaft were a brittle material, then the normal-stress-theory would be the most appropriate of the three theories used in the analysis. 2. Good test data pertaining to actual material and torsion loading would be recommended to improve the failure theory prediction. SOLUTION (6.27) Known: A round steel bar of given strength is subjected to known tensile, torsional, bending and transverse shear stresses. Find: (a) Draw a sketch showing the maximum normal and shear stress, and (b) determine the safety factor for yield failure. Schematic and Given Data: P/A + Mc/I = 70 + 300 = 370 —Tc/J + 4V/3A = 200 + 170 = 370 6-29 s y = 800 MPa, Tensile Test Assumption: The location T (top) is subjected to torsion and bending tension, but no transverse shear. Location S (side) is on a neutral bending axis and is on the side where 4V/3A and Tc/J are additive. Analysis: 1. The Mohr circle plot shows "S" has the higher shear stress, and "T" the higher tensile stress. These locations are 900 apart. The safety factor with respect to initial yielding according to the maximum—shear- stress theory is _ 3/2 _ 400 _ SF _ Tmax — 370 — 1.08 I From Eq. (6.8), 68 = 43(370) = 641 MPa The safety factor according to the distortion energy theory is _ 3y _ 800 _ SF _ Ge _ 641 1.25 I Comments: 1 . The effect of tensile force P/A is not considered while calculating the stresses on location S. If it is considered the safety factors according to the maximum shear stress theory and distortion energy theory are 1.076 and 1.24 respectively. The maximum shear stress theory is more conservative in predicting failure than the distortion energy theory. 6-30 SOLUTION (6.28) Known: A steel member has a specified safety factor and given stresses. Find: Determine the tensile yield strength with respect to initial yielding according to: (a) the maximum-shear-stress theory, (b) the maximum-distortion-energy theory. Schematic and Given Data: Analysis: (a) For the maximum—shear-stress theory, with 61 = 100 MPa, 62 = 20 MPa, and 63 = -80 MPa, we have tmx = (100 + 80)/2 = 90 MPa Sy = (2.5)(2)('tmax) = 450 MPa I (b) For the maximum—distortion-stress theory, using Eq.(6.5) 66 = lgfioz-ol)2+(o3—ol)2+(o3-oz)2]“2 = iZ—Z—{(-80)2+(-180)2+(-100)2]“2 = 156 MPa The yield strength is Sy = 2.5(Ge) = 390 MPa I 6-3 1 SOLUTION (6.29) Known: A downhold oil tool has known biaxial static stresses, an ultimate tensile strength of 97,000 psi and a yield strength of 63,300 psi. Find: Determine the safety factor according to: (a) the maximum-normal—stress theory (b) the maximum—shear-stress theory (0) the maximum-distortion-energy theory Schematic and Given Data: (52 = 25,000 psi iEEE=E=E=E=E= —»61 = 45,000 Psi Analysis: 1. Maximum-normal-stress theory For 0'1 2 45,000 psi, Sy = 63,300 psi SF 2 Sy/(51 = 63,300/45,000 = 1.4 I 2. Maximum-shear-stress theory For 01 = 45,000 psi, 62 = 25,000 psi, 63 = 0 ‘Cmax = (0 + 45,000)/2 = 22,500 psi SF = SSy/Tmax = 31,650/22,500 = 1.41 I 3. Maximum—distortion—energy theory Ge = (6% + 63 - 6162)“2 = [(45,000)2+(25,000)2—(45,000)(25,OOO)]“2 = 39,051 psi The safety factor is SF = Sy/(Se = 63,300/39,051 = 1.62 I 6—32 SOLUTION (6.30) Known: A lawn mower component has known stresses, an ultimate tensile strength of 97,000 psi, and a yield strength of 63,300 psi. Find: Determine the safety factor according to: (a) the maximum-normal-stress theory (b) the maximum-shear—stress theory (c) the maximum-distortion-energy theory Schematic and Given Data: (5y = 25,000 psi T 'txy = 15,000 psi ——> —-—> ox = 45,000 psi Analysis: 1. Maximum—normal-stress theory From, Eq. (4.16) m = (03%) + [(nyfi (ME = (45,000 + 25,000) + 20,000 2 2 = 53,028 psi )2+ (15 ,000)2]% SF = Sy/ ($1 = 63,300/53,028 = 1.19 I 2. Maximum—shear—stress theory From, Eq. (4.18) o'x .. (5y )2]1/2 2 tmax = [fly + ( 20,000 2 2 1/2 = [(15,000)2 +( = 18,028 psi 6—33 _ Sy _ 63,300 _ ' SF‘m“ra§To2_s>‘1'8 ' 3. Maximum-distortion-energy theory From, Eq. (6.7) m=b%o&mq+kgm Ge = [(45,000)2 + (25,000)2 — (45,000)(25,000) + 3(15,000)2]1’2 = 46,904 psi SF = Sy/(Se = 63,300/46,904 = 1.35 I SOLUTION (6.3 1 D) Known: The web site http://www.mecheng.asme.0rg/database/STAT/ MASTER.HTML lists statistics shareware programs. Find: Select a user—friendly interactive statistics shareware program under the topic statistics, data analysis, and data manipulation tools. Analysis: 1. On the date searched, the web site http://www.mecheng.asme. org/database/STAT/MASTER.HTML under the topic "Statistics, data analysis, and data manipulation tools“ listed eleven pages of programs in reverse chronological order. 2. A selected list of five of these programs is given below: ‘ Merlin ' ESTAT21.ZIP TS l STZOZIP MODSTAT3ZIP STAT_1.ZIP 3. The exercise of downloading shareware programs for evaluation and testing is left for the student. 4. The exercise of writing a memorandum discussing the shareware evaluated and tested is left for the student. Comment: We recommend discussing the shareware using the following categories: (a) usefulness, (b) ease of use, (0) relative cost, and ((1) accuracy and correctness. 6-34 ...
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SolutionsToCh6 - SOLUTION (6.1D) Known: A poem appears in...

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