This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem 1) A torque of 21 kip feet is to be applied to a 10 foot long shaft made of steel. The allowed shearing
stress of the steel is 32 ksi and steel weighs 490 pounds per cubic foot. 21) Determine the weight of a solid steel shaft required to safely carry the torque. b) It has been proposed that the steel shaft of part (a) could be safely replaced with a lighter shaft, by
increasing it’s outside diameter to twice the value found in (a), and boring a hole in the center. Determine
the weight of the newly proposed shaft. Drill the hole as large as possible to minimize the shaft weight. Cage; T t T 21 File2Two equivalent twisted S.D. shafts.EES 2/17/2002 4:26:03 PM Page 1
EES Ver. 6.294: #295: Mechanical Engineering Dept., Texas A&M density = 490/1728
L = 120 stress = 32 t = 21*12 j_rod = pi*c_rod"4/2
j_tube:(pi/2)*(c_outertube’\4c_innertubeM) weight_rod = (pi'c_rod/\2)'L‘density
weight_tube = pi*(c_outertube’\2c_innertubeA2)*L*density c_outertube = c_rod*2 stress = t*c_rod/j_rod
stress = t*c_outertube/j_tube {See diagram window for figure. Problem 1) A torque of 21 kip feet is to be applied to a 10 foot long
shaft made of steel. The allowed shearing stress of the steel is 23 ksi and steel weighs 490 pounds per cubic foot. a) Determine the weight of a solid steel shaft required to safely carry the torque. b) it has been proposed that the steel shaft of part (a) could be safely replaced with a lighter shaft, by increasing it's outside diameter to twice the value found in (a), and boring a hole in the center. Determine the weight of the newly proposed shaft. Drill the hole as large as possible to minimize the shaft weight} Cinnertube = 3.311 Coutertube = 3.423 Cred = 1.711
density = 0.2836 jrod = 13.48 jtube = 26.96
L = 120 ' stress = 32 t = 252 weightmd = 313.1 weightmbe = 80.9 Problem 2) A circular steel bar with an allowed stress of 20 ksi is loaded at A, B, and C as shown below (not to scale). Pa = 80 kips, Pb = 120 kips, and Pc is 220 kips. All loads are directed down the centroid
of the bar. E for steel is 30x103 ksi, and the bar lengths are AB = 4 ft, BC = 3 ft, CD = 2 ft. a) Determine the diameters of each section of the bar such that the stress in each section is equal to the allowed stress.
b) Determine the deﬂection of point B with respect to the ﬁxed ground support at D. A FilezQuizA 21802 3 bars S.D. axial.EES 13/1/2002 8:57:45 AM Page 1
EES Ver. 6.294: #295: Mechanical Engineering Dept., Texas A&M L_ab = 4‘12 L_bc = 3*12 L_cd = 2*12 Sigma : 20 E = 30000 Pab = 80 P_bc = P_ab + 120
P_cd = P_bc  220 Sigma = P_ab/A_ab
Sigma = P_bc/A_bc
Sigma = P_cd/A_cd A_ab = pi*D_ab/\2/4
A_bc = pi*D_bcA2/4
A_cd = pi*D_cd/\2/4 Delta_ab = P_ab*L_ab/(E*A~ab)
Delta_bc = P_bc*L_bc/(E*A_bc)
Delta_cd = P_cd*L_cd/(E*A_cd) Delta__b = Delta_bc+Delta_cd {A circular steel bar with an allowed stress of 20 ksi is loaded at A, B, and C as shown below (not to scale). Pa = 80 kips, Pb = 120 kips, and PC is 220 kips. All loads are directed down the centroid of the bar. E for steel is 30x103 ksi, and the bar lengths are AB = 4 ft, BC = 3 ft, CD = 2 ft.
a) Determine the diameters of each section of the bar such that the stress in each section is equal to the allowed stress. Determine the deflection of point B with respect to the fixed ground support at D.} Aab :4 ADC :10 Acd :1
58b = 0.032 5b = 0.008 5m 2 0.024 FiiezQuizA 21802 3 bars SD. axial.EES
EES Ver. 6.294: #295: Mechanical Engineering Dept, Texas A&M 50d = —0.016 Dab = 2.257 Dbc = 3.568
D0d = 1.128 E = 30000 Lab = 48
Lbc = 36 Lcd = 24 Pat) = 80 Pbe = 200 Pod = 20 0 = 20 3/1/2002 8:57:48 AM Page 2 Problem 3) The weightless Rigid Block shown in the ﬁgure below is guided by frictionless rollers so as to
permit movement of the block in the vertical direction only. The block rests initially upon two circular bars
"A" and "C". Bar "B" is also circular and is positioned beneath the block with a gap of unknown magnitude
1000 kips between the bottom of the block and the top of the bar. Bars
"A" and "C" have identical initial lengths of 10 ft. The initial
length of Bar "B" is unknown. The initial diameter, d, and
Elastic Modulus, E, for each bar are given as follows. Rigid Block Rigid Block crucialx01. Bar A: d = 2.523 in.; E = 12,000 ksi
Bar B: d = 2.985 in.; E = 59,750 ksi
Bar C: d = 1.954 in.; E = 23,333 ksi RA = 300 kips It has been determined that when a force of 1000 kips is applied to the top of the rigid block, the axial load in Bar A is 300 kips (see the righthand ﬁgure below). What is the magnitude of the initial gap between the Rigid Block and Bar "B"? Please express your ﬁnal
answer in units of inches. File:QuizA 21802 Rigid block on 3 bars with gap.EES 2/17/2002 5:03:39 PM Page 1
EES Ver. 6.294: #295: Mechanical Engineering Dept., Texas A&M 1000 kips >3 *3
E . ~ E a a
.3 R1g1d Block .3 >3 . , g
g g :5 ngId Block 1:1
H H
*3 *3
A C A I C
V , . ‘1
R A = 300 k1ps
P_total = 1000
P_a = 300
L_a = 120
L_b = 120
L_c = 120
D_a = 2.523
D_b = 2.985
D_c = 1.954
E_a = 12000
E_b = 59750
E_c = 23333 A_a = P*D_aA2/4
A_b = Pl*D_b/\Z4
A_c = Pl‘D_cA2/4 Delta_a = P_a*L_a/(A_a*E_a)
Delta_c = P_c*L_c/(A_C*E_c) File:QuizA 21802 Rigid block on 3 bars with gap.EES 2/17/2002 5:04:13 PM Page 2
EES Ver. 6.294: #295: Mechanical Engineering Dept, Texas A&M Delta_c = Delta_a
P_b = P_total  P_a  P_c Delta_b = P_b‘L_b/(A_b*E_b)
Gap = Delta_a  Delta_b Aa =4.999 Ab = 6.998 AC =2.999
5a = 0.6001 8b = 0.1005 50 = 0.6001
03 =2.523 Db =2.985 D0 = 1.954
Ea = 12000 Eb = 59750 EC =23333
Gap =0.4996 La = 120 Lb = 120 Lc = 120 Pa =300 Pb =350.1 Pc = 349.9 Pmtal = 1000 Problem 4) A steel bar and a brass bar are welded together and welded at the top and bottom ﬁxed walls
as shown. Properties of the bars are:
Steel bar: 2 inch solid diameter, Esteel = 30,000 ksi, length = 4 feet, stress allowed = 36 ksi
Brass bar: 3 inch solid diameter, Ebrass = 17,000 ksi, length = 6 feet, stress allowed 2 20 ksi
Determine the maximum load Pall which can be applied without overstressing any part of the bar?) elf—“410’ Steel POII Brass File:D:\My Documents\Ciasses\Quizzes\305\EES\S.I. Axiaily loaded bars.EES 217/2002 4:14:32 PM Page 1
EES Ver. 6.294: #295: Mechanical Engineering Dept, Texas A&M Dsteel = 2
Dbrass = 3
Esteel = 30000
Ebrass = 17000
Lsteel = 4'12 Lbrass = 6'12
StressSteelAllowed = 36
StressBrassAliowed = 20 Asteel = 2"PI‘DsteeIAZ/4
Abrass = Pi'Dbrass/‘2/4 DeitaSteel = Psteel“Lsteei/(Asteel‘Esteel)
DeltaBrass = Pbrass'Lbrass/(Abrass‘Ebrass)
DeltaSteeI=DeltaBrass StressSteel = Psteel/Asteel
StressBrass = Pbrass/Abrass Ptotal = Psteel + Pbrass
Ptotal = 1 FactorupSteel = StressSteelAllowed/StressSteel
FactorupBrass = StressBrassAllowed/StressBrass Pfinai = min(FactorupSteel, FactorupBrass)*Ptotal PfinalSteei : min(FactorupSteel, FactorupBrass)‘Psteei
PfinalBrass = min(FactorupSteel, FactorupBrassberass
StressfinalSteel = min(FactorupSteel, FactorupBrass)*StressSteel
StressfinalBrass = min(FactorupSteei, FactorupBrass)‘StressBrass {See diagram window for figure File:D:\My Documents\Classes\Quizzes\305\EES\S.l. Axially loaded bars.EES 2/17/2002 4:14:35 PM Page 2
EES Ver. 6.294: #295: Mechanical Engineering Dept., Texas A&M A steel bar and a brass bar are welded together and welded at the top and bottom fixed
walls as shown. Properties of the bars are:
Steel bar: 3 inch solid diameter, Esteel = 30,000 ksi, length = 4 feet
Brass bar: 2 inch solid diameter, Ebrass = 17,000 ksi, length = 6 feet
Determine the maximum load Pall which can be applied without overstressing any part of the bar.
Solved by putting 1 kip on bar and seeing how much stress induced in each bar.} Abrass = 7.069 Asteel = 6.283 Dbrass = 3 5Brass = 0.0001787 asteel = 0.0001787 Dsteel = 2 Ebrass = 17000 Esteel = 30000 FactorupBrass = 474
FactorupSteel = 322.3 Lbrass = 72 Lsteel = 48 Pbrass = 0.2982 Pfinal = 322.3 PfinalBrass = 96.13
PfinalSteel = 226.2 Psteel = 0.7018 Ptotal = 1 StressBrass = 0.04219 StressBrassAllowed = 20 StressfinalBrass = 13.6 StressfinalSteel = 36 StressSteel = 0.1117 StressSteelAllowed = 36 Problem 1) A steel bar and a brass bar are welded together and welded at the top and bottom ﬁxed walls
as shown. Properties of the bars are: ‘ Steel bar: 3 inch solid diameter, Esteel = 30,000 ksi, length = 4 feet, stress allowed = 36 ksi
Brass bar: 4 inch solid diameter, Ebrass = 17,000 ksi, length = 6 feet, stress allowed = 20 ksi
Determine the maximum load Pall which can be applied without overstressing any part of the bar. .11 Steel Fiie:D:\My Documents\Ciasses\Quizzes\305\EES\S.I. Axially loaded bars.EES 2/17/2002 4:17:00 PM Page 1
EES Ver. 6.294: #295: Mechanical Engineering Dept, Texas A&M Dsteel = 3 Dbrass = 4 Esteel = 30000 Ebrass : 17000 Lsteel = 4‘1 2 Lbrass = 6‘12
StressSteelAllowed = 36
StressBrassAllowed = 20 Asteel = 2‘PI‘DsteeiA2/4
Abrass = Pl‘DbrassA2/4 DeltaSteei : Psteel‘Lsteei/(Asteel'Esteel)
DeltaBrass = Pbrass‘Lbrass/(Abrass‘Ebrass)
DeltaSteelzDeltaBrass StressSteel = Psteel/Asteel
StressBrass = Pbrass/Abrass Ptotal = Psteei + Pbrass
Ptotal = 1 FactorupSteel = StressSteeIAllowed/StressSteei
FactorupBrass = StressBrassAiIowed/StressBrass Pfinal = min(FactorupSteel, FactorupBrass)*Ptotal PfinalSteel = min(FactorupSteel, FactorupBrass)*Psteel
PfinalBrass = min(FactorupSteel, FactorupBrass)*Pbrass
StressfinalSteel = min(FactorupSteel, FactorupBrass)*StressSteel
StressfinaiBrass : min(FactorupStee, FactorupBrass)‘StressBrass {See diagram window for figure File:D:\My Documents\Classes\Quizzes\305\EES\S.l. Axially loaded bars.EES 2/17/2002 4:17:03 PM Page 2 EES Ver. 6.294: #295: Mechanical Engineering Dept, Texas A&M A steel bar and a brass bar are welded together and welded at the top and bottom fixed walls as shown. Properties of the bars are: Steel bar: 3 inch solid diameter, Esteel = 30,000 ksi, length = 4 feet
Brass bar: 2 inch solid diameter, Ebrass = 17.000 ksi, length = 6 feet Determine the maximum load Pall which can be applied without overstressing any part of the bar. Solved by putting 1 kip on bar and seeing how much stress induced in each bar.} Abrass = 12.57
aBrass = 000008473
Ebrass = 17000
FactorupSteel = 679.8
Pbrass = 0.2514
PfinalSteel = 508.9
StressBrass = 0.02
StressfinalSteel = 36 Asteel = 14.14 asteel = 0.00008473
Esteel = 30000 Lbrass = 72 Pfinal = 679.8 Psteel = 0.7486
StressBrassAllowed = 20
StressSteel = 0.05295 Dbrass = 4 Dsteel = 3
FactorupBrass = 1000
Lsteel =48 PfinalBrass = 170.9
Ptotal = 1
StressflnalBrass = 13.6
StressSteelAllowed = 36 Problem 2) The weightless Rigid Block shown in the ﬁgure below is guided by frictionless rollers so as to
permit movement of the block in the vertical direction only. The block rests initially upon two circular bars
"A" and "C". Bar "B" is also circular and is positioned beneath the block with a gap of unknown magnitude between the bottom of the block and the top of the bar. 1000 kips Bars
"A" and "C" have identical initial lengths of 10 ft. The g , . g a , initial length of Bar "B" is unknown. The initial
diameter, d, and Elastic Modulus, E, for each bar are ; E given as follows. Bar A: d = 2.523 in.; E = 12,000 ksi
Bar B: d = 2.985 in.; E = 59,750 ksi
Bar C: d = 1.954 in.; E = 23,333 ksi V; RA = 300 kips It has been determined that when a force of 1000 kips is applied to the top of the rigid block, the axial load in Bar A is 300 kips (see the right—hand ﬁgure below). What is the magnitude of the initial gap between
the Rigid Block and Bar "B"? Please express your ﬁnal answer in units of inches. FilezQuizA 218—02 Rigid block on 3 bars with gap.EES
EES Ver. 6.294: #295: Mechanical Engineering Dept, Texas A&M P_total = 1000
P_a = 300 L_a = 120
L_b : 120
L_c : 120
D_a = 2.523
D_b = 2.985
D_C = 1.954
E_a = 12000
E_b = 59750
E_C = 23333 Rigid Block
A B c
\ A_a = Pl'D_a/‘2/4
A_b = Pl‘D_b’\2/4
A_c = Pl‘D_c’\2/4 Deita_a = P_a‘L_a/(A_a*E_a) Delta_c : P_c‘ L_c/(A_c‘E_c) 2/17/2002 5:04:53 PM Page 1 1000 kips Rigid Block
C
V RA = 300 kips OIOIOIOIOIOI.
CIOIOXOIOIOI. Fiie:QuizA 218—02 Rigid block on 3 bars with gap.EES 2/17/2002 5:05:28 PM Page 2
EES Ver. 6.294: #295: Mechanical Engineering Dept., Texas A&M Delta_c = Delta_a
P_b = P_total  P_a  P_c Delta_b = P_b*L_b/(A_b‘E_b)
Gap = Delta_a  Delta_b A3 = 4.999 Ab = 6.998 Ac = 2.999
a = 0.6001 5., = 0.1005 5c = 0.6001
Da =2.523 Db =2.985 Dc = 1.954
Ea = 12000 Eb =59750 Ec =23333
Gap =0.4996 La =12o Lb = 120 Lc = 120 Pa =300 Pb =350.1 Pc = 349.9 Ptma, = 1000 Problem 3) A circular steel bar with an allowed stress of 20 ksi is loaded at A, B, and C as shown below (not to scale.) Pa = 80 kips, Pb = 140 kips, and Pc is 240 kips. All loads are directed down the centroid
of the bar. E for steel is 30x103 ksi, and the bar lengths are AB = 4 ft, BC = 3 ft, CD = 2 ft. 21) Determine the diameters of each section of the bar such that the stress in each section is equal to the allowed stress.
b) Determine the deﬂection of point B with respect to the ﬁxed ground support at D. A File:QuizA 21802 3 bars S.D. axial.EES 3/1/2002 8:58:22 AM Page 1
EES Ver. 6.294: #295: Mechanical Engineering Dept, Texas A&M L_ab = 4*12 L_bc = 3*12 L_cd = 2‘12 Sigma = 20 E = 30000 P_ab = 80 P_bc = P~ab + 140
P_cd = Pﬁbc — 240 Sigma = P_ab/A_ab
Sigma = P_bc/A_bc
Sigma = P_cd/A_cd Aab = pi*D_ab/\2/4
A_bc = pi*D_bc’\2/4
A_cd = pi*D_ch2/4 Delta_ab = P_ab*L_ab/(E*Awab)
Delta_bc = P_bc*L_bc/(E*A_bc)
Delta_cd = P_cd‘L_cd/(E*A_cd) Delta_b = Delta_bc+Delta_cd {A circular steel bar with an allowed stress of 20 ksi is loaded at A, B, and C as shown below (not to scale). Pa = 80 kips, Pb = 120 kips, and PC is 220 kips. All loads are directed down the centroid of the bar. E for steel is 30x103 ksi, and the bar lengths are AB = 4 ft, BC = 3 ft, CD = 2 ft.
a) Determine the diameters of each section of the bar such that the stress in each section is equal to the allowed stress. Determine the deflection of point B with respect to the fixed ground support at D.} Aab =4 Abc=11 Acd =1
gab = 0.032 5b = 0.008 23m = 0.024 File:QuizA 218—02 3 bars SD. axiaI.EES
EES Ver. 6.294: #295: Mechanical Engineering Dept, Texas A&M 59., = 0.016 Dab = 2.257 Db0 = 3.742
Dcd : 1.128 E = 30000 Lab = 48
Lbc = 36 LCd 2‘ 24 Pab = 80 Pbc =22o P0d :20 0 =20 3/1/2002 8:58:24 AM Page 2 Problem 4) A torque of 21 kip feet is to be applied to a 10 foot long shaft made of steel. The allowed shearing
stress of the steel is 23 ksi and steel weighs 490 pounds per cubic foot. c) Determine the weight of a solid steel shaft required to safely carry the torque. d) It has been proposed that the steel shaft of part (a) could be safely replaced with a lighter shaft, by
increasing it’s outside diameter to twice the value found in (a), and boring a hole in the center. Determine
the weight of the newly proposed shaft. Drill the hole as large as possible to minimize the shaft weight. O I: (a)
\4
F © > (b)
v Fileszo equivalent twisted S.D. shafts.EES 2/17/2002 4:24:26 PM Page 1
EES Ver. 6.294: #295: Mechanical Engineering Dept., Texas A&M density = 490/1728
L = 120 stress = 23 t = 21 ‘12 j_rod = pi‘c_rod/\4/2
j_tube:(pi/2)*(c_outertube"4c_innertubeA4) weight_rod = (pi*c_rod’\2)'L‘density
weight_tube = pi‘(c_outertube/\2c_innertube"2)*L'density c_outertube = c_rod*2 stress = t‘c_rod/j_rod
stress = t'cﬁoutertube/Ltube {See diagram window for figure. Problem 1) A torque of 21 kip feet is to be applied to a 10 foot long
shaft made of steel. The allowed shearing stress of the steel is 23 ksi and steel weighs 490 pounds per cubic foot. a) Determine the weight of a solid steel shaft required to safely carry the torque. b) It has been proposed that the steel shaft of part (a) could be safely replaced with a lighter shaft, by increasing it's outside diameter to twice the value found in (a), and boring a hole in the center. Determine the weight of the newly proposed shaft. Drill the hole as large as possible to minimize the shaft weight} cinnenube = 3.696 commune = 3.821 cred = 1.911
density = 0.2836 jmd = 20.93 jmbe = 41.87
L = 120 stress = 23 t = 252 weightmd = 390.3 weightmbe = 100.8 ...
View
Full Document
 Spring '08
 Gardoni

Click to edit the document details