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Unformatted text preview: SOLUTION (6.1D)
Known: A poem appears in Chapter Six titled: "The OneHoss Shay“. Find: Write a one paragraph summary of the poem "The OneHoss 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. 61 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 62 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 11105 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. 64 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 11105 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 11105 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 in05 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
7075T651 aluminum. Schematic and Given Data: 7075T65 1
Aluminum 2w = 6 in.
t: 1 in.
a/2c = 0.25 cg = 0.73 sy Assumptions:
1. The 7075T651 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 68 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 ksrV 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. 611 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 maximumnormal—stress theory (b) the maximumshearstress theory (c) the maximumdistortion—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 maximumnormalstress theory:
SF = 60/20 = 3.0
(b) For the maximumshear—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:
 maximumdistortion—energy theory is best
 maximumshearstress theory may be acceptable
0 maximumnormal—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—normalstress theory (b) the maximumshear—stress theory (c) the maximumdistortionenergy theory Schematic and Given Data: Assumption: The material is homogeneous. Analysis: _
1. From the above Mohrcircle,
“cum = (25 + 15)/2 = 20 ksi
2. (a) For the maximumnormalstress theory:
SF = 60/25 = 2.4 (b) For the maximumshearstress theory:
SF = 30km = 30/20 = 1.5 (c) For the maximumdistortionenergy 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—distortionenergy theory is b_es_t 6—19  maximumshear—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 maximumshear—stress
theory and the maximum—distortionenergy 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. (6.56.8) 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 maximumshearstress
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 620 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 maximumnormalstress theory, 626 Sy Sy 60,000 SF = 2 = 6m = 76;; = 16(5000)
m3
Solving for (1, gives d = 0.95 in. I (b) For the maximumshear—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 distortionenergy 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 distortionenergy theory is the most accurate,
followed by the maximumshear—stress theory . If the shaft were a brittle
material, then the normal—stresstheory 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. 627 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 maximumnormalstress theory, S S
SF=2=—Y—=_y_=_§9£99_
Cm Txy w 1rd3 Solving for (1, gives (1 = 1.01 in.
(b) For the maximumshear—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 628 (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—shearstress theory (most steel shafts are ductile). If
the shaft were a brittle material, then the normalstresstheory 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 = 370 629 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. 630 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 maximumshearstress theory, (b) the maximumdistortionenergy theory. Schematic and Given Data: Analysis:
(a) For the maximum—shearstress 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—distortionstress theory, using Eq.(6.5)
66 = lgﬁozol)2+(o3—ol)2+(o3oz)2]“2 = iZ—Z—{(80)2+(180)2+(100)2]“2 = 156 MPa The yield strength is Sy = 2.5(Ge) = 390 MPa I 63 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 maximumnormal—stress theory (b) the maximum—shearstress theory (0) the maximumdistortionenergy theory Schematic and Given Data: (52 = 25,000 psi iEEE=E=E=E=E= —»61 = 45,000 Psi Analysis:
1. Maximumnormalstress theory
For 0'1 2 45,000 psi, Sy = 63,300 psi SF 2 Sy/(51 = 63,300/45,000 = 1.4 I 2. Maximumshearstress 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 maximumnormalstress theory (b) the maximumshear—stress theory (c) the maximumdistortionenergy theory Schematic and Given Data: (5y = 25,000 psi T 'txy = 15,000 psi ——>
——> ox = 45,000 psi Analysis:
1. Maximum—normalstress 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. Maximumdistortionenergy 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. 634 ...
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 Spring '06
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