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Chapter 6
MSS: σ 1 − σ 3 = S (^) y / n ⇒ n =
S (^) y σ 1 − σ 3 DE: n =
S (^) y σ ′ σ ′^ =
σ (^2) A − σ A σ B + σ (^) B^2
σ (^) x^2 − σ x σ y + σ (^) y^2 + 3 τ (^) x y^2
(a) MSS: σ 1 = 12, σ 2 = 6, σ 3 = 0 kpsi n =
12 =^4.^17 Ans. DE: σ ′^ = (12^2 − 6(12) + 62 ) 1 /^2 = 10 .39 kpsi, n =
- 39 =^4.^81 Ans.
(b) σ A , σ B = 12 2
σ 1 = 16, σ 2 = 0, σ 3 = −4 kpsi
MSS: n =
16 − (−4) =^2.^5 Ans. DE: σ ′^ = (12^2 + 3(− 82 )) 1 /^2 = 18 .33 kpsi, n = 50
- 33
= 2. 73 Ans.
(c) σ A , σ B =
- (−5) 2 = − 2 .615, − 13 .385 kpsi
σ 1 = 0, σ 2 = − 2 .615, σ 3 = − 13 .385 kpsi
MSS: n = (^0) − (−^5013 .385) = 3. 74 Ans.
DE: (^) σ ′^ = [(−6) 2 − (−6)(−10) + (−10) 2 + 3(−5) 2 ]^1 /^2 = 12 .29 kpsi n = 50
- 29
= 4. 07 Ans.
(d) σ A , σ B = 12 +^4 2
σ 1 = 12 .123, σ 2 = 3 .877, σ 3 = 0 kpsi
MSS: n = 50
- 123 − 0
= 4. 12 Ans.
DE: σ ′^ = [12^2 − 12(4) + 42 + 3(1^2 )]^1 /^2 = 10 .72 kpsi
n = 50
- 72
= 4. 66 Ans.
B
A
150 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-2 S (^) y = 50 kpsi
MSS: σ 1 − σ 3 = S (^) y / n ⇒ n =
S (^) y σ 1 − σ 3 DE:
σ (^) A^2 − σ A σ B + σ (^) B^2
= S (^) y / n ⇒ n = S (^) y /
σ (^2) A − σ A σ B + σ (^) B^2
(a) MSS: σ 1 = 12 kpsi, σ 3 = 0, n =
12 − 0 =^4.^17 Ans.
DE: n =
[12^2 − (12)(12) + 122 ]^1 /^2 =^4.^17 Ans.
(b) MSS: σ 1 = 12 kpsi, σ 3 = 0, n =
12 =^4.^17 Ans.
DE: n =
[12^2 − (12)(6) + 62 ]^1 /^2 =^4.^81 Ans.
(c) MSS: σ 1 = 12 kpsi, σ 3 = −12 kpsi, n = 50 12 − (−12)
= 2. 08 Ans.
DE: n = 50 [12^2 − (12)(−12) + (−12) 2 ]^1 /^3
= 2. 41 Ans.
(d) MSS: σ 1 = 0, σ 3 = −12 kpsi, n =
−(−12) =^4.^17 Ans.
DE: n =
[(−6) 2 − (−6)(−12) + (−12) 2 ]^1 /^2 =^4.^81
6-3 S (^) y = 390 MPa
MSS: σ 1 − σ 3 = S (^) y / n ⇒ n =
S (^) y σ 1 − σ 3 DE:
σ (^2) A − σ A σ B + σ (^) B^2
= S (^) y / n ⇒ n = S (^) y /
σ (^2) A − σ A σ B + σ (^) B^2
(a) MSS: σ 1 = 180 MPa, σ 3 = 0, n = 390180 = 2. 17 Ans.
DE: n = (^) [180 (^2) − 180(100)^390 + 1002 ] 1 / 2 = 2. 50 Ans.
(b) σ A , σ B = 1802 ±
- 1002 = 224 .5, − 44 .5 MPa = σ 1 , σ 3
MSS: n = (^224). 5 −^390 (− 44 .5) = 1. 45 Ans.
DE: n = (^) [180 (^2) + 390 3(100 (^2) )] 1 / 2 = 1. 56 Ans.
152 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-
(a) MSS: n =
O B
O A =^
1. 08 =^2.^1
DE: n =
OC
O A =^
1. 08 =^2.^4
(b) MSS: n =
O E
O D =^
1. 10 =^1.^5
DE: n =
O F
O D =^
1. 1 =^1.^6
(c) MSS: n = O H OG
= 1.^68
DE: n = O I OG
= 1.^85
(d) MSS: n = O K O J
= 1.^38
DE: n = O L O J
= 1.^62
O
( a )
( b )
( d )
( c )
H I
G
J K (^) L
E F D
A
B C
Scale1" 200 MPa
B
A
Chapter 6 153
6-6 S (^) y = 220 MPa
(a) MSS: n =
O B
O A =^
1. 3 =^2.^2
DE: n =
OC
O A =^
1. 3 =^2.^4
(b) MSS: n =
O E
O D =^
1 =^2.^2
DE: n =
O F
O D =^
1 =^2.^3
(c) MSS: n = O H OG
= 1.^55
DE: n = O I OG
= 1.^8
(d) MSS: n = O K O J
= 2.^82
DE: n = O L O J
= 3.^1
B
O A
( a )
( b )
( c )
( d )
H
G J
K
L
I
D E F
A
B
C
1" 100 MPa
Chapter 6 155
BCM: Eq. (6-31 b )
n =^
30 −^
100 ⇒^ n^ =^3.^41 Ans.
M1M: Eq. (6-32 b )
n =^
100 ⇒^ n^ =^3.^95 Ans.
M2M: Eq. (6-33 b ) n
[ n (−16) + 30 30 − 100
] 2
Reduces to n^2 − 1. 1979 n − 15. 625 = 0
n = 1.^1979 +^
= 4. 60 Ans.
( c )
L ( d )
J
( b )
( a )
I
H
G
K F
O
C D
E
1" 20 kpsi A B
B
A
156 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-8 See Prob. 6-7 for plot.
(a) For all methods: n = O BO A = 11 ..^5503 = 1. 5
(b) BCM: n = O DOC = 10 ..^48 = 1. 75
All other methods: n = O EOC = 10.^55. 8 = 1. 9
(c) For all methods: n = (^) O KO L = (^05) .. 682 = 7. 6
(d) MNS: n = O JO F = 50 ..^1282 = 6. 2
BCM: n^ =^
OG
O F
= 2.^85
M1M: n = O HO F = (^03) .. 823 = 4. 0
M2M: n^ =^
O I
O F
= 3.^82
6-9 Given: S (^) y = 42 kpsi, Sut = 66 .2 kpsi, ε f = 0. 90. Since ε f > 0 .05, the material is ductile and thus we may follow convention by setting S (^) yc = S (^) yt. Use DE theory for analytical solution. For σ ′, use Eq. (6-13) or (6-15) for plane stress and Eq. (6-12) or (6-14) for general 3-D. (a) σ ′^ = [9^2 − 9(−5) + (−5) 2 ]^1 /^2 = 12 .29 kpsi
n = 42
- 29
= 3. 42 Ans.
(b) σ ′^ = [12^2 + 3(3^2 )]^1 /^2 = 13 .08 kpsi
n = 42
- 08
= 3. 21 Ans.
(c) σ ′^ = [(−4) 2 − (−4)(−9) + (−9) 2 + 3(5^2 )]^1 /^2 = 11 .66 kpsi
n = 42
- 66
= 3. 60 Ans.
(d) σ ′^ = [11^2 − (11)(4) + 42 + 3(1^2 )]^1 /^2 = 9. 798
n = 42
- 798
= 4. 29 Ans.
158 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
(a) σ x = 90 kpsi, σ y = −50 kpsi, σ z = 0 σ A = 90 kpsi and σ B = −50 kpsi. For the fourth quadrant, from Eq. (6-13)
n = (^) (σ^1 A / S^ yt )^ −^ (σ B / Suc )^
= (^) (90/235) −^1 (− 50 /275) = 1. 77 Ans.
(b) σ x = 120 kpsi, τ x y = −30 kpsi ccw. σ A , σ B = 127 .1, − 7. 08 kpsi. For the fourth quadrant n = (^) (127. 1 /235) −^1 (− 7. 08 /275) = 1. 76 Ans.
(c) σ x = −40 kpsi, σ y = −90 kpsi, τ x y = 50 kpsi. σ A , σ B = − 9 .10, − 120 .9 kpsi. Although no solution exists for the third quadrant, use
n = −
S (^) yc σ y^ = −^
− 120. 9 =^2.^27 Ans. (d) σ x = 110 kpsi, σ y = 40 kpsi, τ x y = 10 kpsi cw. σ A , σ B = 111 .4, 38.6 kpsi. For the first quadrant n =
S (^) yt σ A^ =^
- 4 =^2.^11 Ans. Graphical Solution:
(a) n = O BO A = 11 ..^8202 = 1. 78
(b) n = O DOC = 21 ..^2428 = 1. 75
(c) n = O FO E = 21 ..^7524 = 2. 22
(d) n = O HOG = 21 ..^4618 = 2. 08
O
( d )
( b )
( a )
( c )
E
F
B
D
G
C
A
H 1 in 100 kpsi
B
A
Chapter 6 159
6-11 The material is brittle and exhibits unequal tensile and compressive strengths. Decision: Use the Modified II-Mohr theory as shown in Fig. 6-28 which is limited to first and fourth quadrants. Sut = 22 kpsi, Suc = 83 kpsi Parabolic failure segment:
S (^) A = 22
[
( S
B +^22
) 2 ]
S B S A S B S A
(a) σ x =^9 kpsi, σ y = −^5 kpsi. σ A ,^ σ B =^ 9,^ −^5 kpsi. For the fourth quadrant, use Eq. (6-33 a ) n = Sut σ A
= 2. 44 Ans.
(b) σ x = 12 kpsi, τ x y = −3 kpsi ccw. σ A , σ B = 12 .7, 0.708 kpsi. For the first quadrant,
n = Sut σ A
= 1. 73 Ans.
(c) σ x = −4 kpsi,^ σ y = −9 kpsi,^ τ x y =^ 5 kpsi. σ A ,^ σ B = −^0 .910,^ −^12 .09 kpsi.^ For the third quadrant, no solution exists; however, use Eq. (6-33 c ) n = −^83 − 12. 09
= 6. 87 Ans.
(d) σ x = 11 kpsi,σ y = 4 kpsi,τ x y = 1 kpsi. σ A , σ B = 11 .14, 3.86 kpsi. For the first quadrant
n = (^) σ S^ A A
S (^) yt σ A^ =^
- 14 =^1.^97 Ans.
30
30
S (^) ut 22
S (^) ut 83
B
A
Chapter 6 161
(d) σ A , σ B = 15, −25 kpsi
Eq. (6-33 b ):
n (15) 30
(− 25 n + 30 30 − 109
n = 1. 90 Ans.
(a) n = O BO A = 42 ..^2583 = 1. 50
(b) n = O DOC = 42 ..^2412 = 2. 00
(c) n = O FO E = 1511 ..^53 = 1 .37 (3rd quadrant)
(d) n = O HOG = 52 ..^39 = 1. 83
6-14 Given: AISI 1006 CD steel, F = 0.55 N, P = 8.0 kN, and T = 30 N · m, applying the DE theory to stress elements A and B with S (^) y = 280 MPa
A: σ x = (^32) π dFl 3 + (^) π^4 dP 2 = 32(0.55)(
π(0. 0203 ) +^
4(8)(10^3 )
π(0. 0202 ) = 95 .49(10^6 ) Pa = 95 .49 MPa
O
( d )
( b )
( a )
( c )
E
F
C
B
A
G D
H
1 cm 10 kpsi
B
A
162 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
τ x y =
16 T
π d^3 =^
π(0. 0203 ) =^19 .10(
(^6) ) Pa = 19 .10 MPa
σ ′^ =
σ (^) x^2 + 3 τ (^) x y^2
= [95. 492 + 3(19.1) 2 ]^1 /^2 = 101 .1 MPa
n =
S (^) y σ ′^ =^
- 1 =^2.^77 Ans.
B: σ x =
4 P
π d^3 =^
4(8)(10^3 )
π(0. 0202 ) =^25 .47(
(^6) ) Pa = 25 .47 MPa
τ x y = 16 T π d^3
V
A
π(0. 0203 )
[ 0 .55(10 3 )
(π/4)(0. 0202 )
]
= 21 .43(10^6 ) Pa = 21 .43 MPa σ ′^ = [25. 472 + 3(21. 432 )]^1 /^2 = 45 .02 MPa
n =
- 02 =^6.^22 Ans.
6-15 Design decisions required:
- Material and condition
- Design factor
- Failure model
- Diameter of pin Using F = 416 lbf from Ex. 6-
σmax =
32 M
π d^3
d =
( 32 M
πσmax
Decision 1: Select the same material and condition of Ex. 6-3 (AISI 1035 steel, S (^) y = 81 000). Decision 2: Since we prefer the pin to yield, set n (^) d a little larger than 1. Further explana- tion will follow. Decision 3: Use the Distortion Energy static failure theory. Decision 4: Initially set n (^) d = 1
σmax =
S (^) y n (^) d^ =^
S (^) y 1 =^ 81 000 psi
d =
[32(416)(15)
π(81 000)
] 1 / 3
= 0 .922 in
164 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Setting the tangential stress equal to the yield stress,
ω =
= 1361 rad/s
or n = 60 ω/ 2 π = 60(1361)/(2π) = 13 000 rev/min Now check the stresses at r = ( r (^) o r (^) i ) 1 /^2 , or r = [5(3)]^1 /^2 = 3 .873 in
σ r = ρω^2
( 3 + ν 8
( r (^) o − r (^) i ) 2
= 0.^282 ω
2 386
= 0 .001 203ω^2 Applying Eq. (4-56) for σ t
σ t = ω^2
) [
]
= 0 .012 16ω^2 Using the Distortion-Energy theory
σ ′^ =
σ (^) t^2 − σ r σ t + σ r 2
= 0 .011 61ω^2
Solving ω =
= 1607 rad/s
So the inner radius governs and n = 13 000 rev/min Ans.
6-18 For a thin-walled pressure vessel,
di = 3. 5 − 2(0.065) = 3 .37 in
σ t = p ( di^ +^ t ) 2 t
σ t = 500(3.^37 +^0 .065) 2(0.065)
= 13 212 psi
σ l =
pdi 4 t =^
4(0.065) =^ 6481 psi σ r = − pi = −500 psi
Chapter 6 165
These are all principal stresses, thus,
σ ′^ = √^1 2
{(13 212 − 6481) 2 + [6481 − (−500)]^2 + (− 500 − 13 212) 2 }^1 /^2
σ ′^ = 11 876 psi
n =
S (^) y σ ′^ =^
σ ′^ =^
= 3. 87 Ans.
6-19 Table A-20 gives S (^) y as 320 MPa. The maximum significant stress condition occurs at r (^) i where σ 1 = σ r = 0, σ 2 = 0, and σ 3 = σ t. From Eq. (4-50) for r = r (^) i
σ t = − 2 r o^2 po r o^2 − r i^2
(^2) ) po 1502 − 1002
= − 3. 6 po
σ ′^ = 3. 6 po = S (^) y = 320
po =
- 6 =^88 .9 MPa^ Ans.
6-20 Sut = 30 kpsi, w = 0 .260 lbf/in^3 , ν = 0 .211, 3 + ν = 3 .211, 1 + 3 ν = 1.633. At the inner radius, from Prob. 6- σ t ω^2 =^ ρ
( 3 + ν 8
2 r o^2 + r i^2 −
1 + 3 ν 3 + ν r i^2
Here r o^2 = 25, r i^2 = 9, and so σ t ω^2 =^
Since σ r is of the same sign, we use M2M failure criteria in the first quadrant. From Table A-24, Sut = 31 kpsi, thus,
ω =
= 1452 rad/s
rpm = 60 ω/(2π) = 60(1452)/(2π) = 13 866 rev/min Using the grade number of 30 for Sut = 30 000 kpsi gives a bursting speed of 13640 rev/min.
6-21 TC = (360 − 27)(3) = 1000 lbf · in , T (^) B = (300 − 50)(4) = 1000 lbf · in
A B C D
223 lbf 8" 8" 6" 350 lbf
127 lbf
xy plane
y
Chapter 6 167
So M max = [(892) 2 + (424) 2 ]^1 /^2 = 988 lbf · in
σ x = 32 M^ B π d^3
π d^3
d^3
psi
Since the torsional stress is unchanged, τ x z = 5. 09 / d^3 kpsi
σ A , σ B = 1 d^3
[(
] 1 / 2
σ A = 12. 19 / d^3 and σ B = − 2. 13 / d^3 Using the Brittle-Coulomb-Mohr, as was used in Prob. 6-21, gives
- 19 25 d^3 −
97 d^3 =^
Solving gives d = 1 1/8 in. Now compare to Modified II-Mohr theory Ans.
6-23 ( F (^) A ) t = 300 cos 20 = 281 .9 lbf, ( F (^) A ) r = 300 sin 20 = 102 .6 lbf
T = 281 .9(12) = 3383 lbf · in, ( FC ) t =
5 =^676 .6 lbf ( FC ) r = 676 .6 tan 20 = 246 .3 lbf
M A = 20
- 72 + 233. 52 = 6068 lbf · in M (^) B = 10
- 32 + 676. 62 = 7200 lbf · in (maximum)
σ x = 32(7200) π d^3
d^3
τ x y = 16(3383) π d^3
d^3 σ ′^ =
σ (^) x^2 + 3 τ (^) x y^2
= S^ y [ n ( 73 340 d^3
d^3
) 2 ] 1 / 2
d^3
d = 1 .665 in so use a standard diameter size of 1.75 in Ans.
xy plane
x
y
A B C
ROy = 193.7 lbf
281.9 lbf^ R^ By^ = 158.1 lbf
20" 16" 10"
246.3 lbf O
xz plane
x z
A B C
R (^) Oz = 233.5 lbf
R (^) Bz = 807.5 lbf
O 102.6 lbf
20" 16" 10"
676.6 lbf
168 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-24 From Prob. 6-23,
τmax =
[(
σ x 2
] 1 / 2
S (^) y 2 n [( 73 340 2 d^3
d^3
) 2 ]^1 /^2
= 40 516 d 3 = 60 0002(3.5)
d = 1 .678 in so use 1.75 in Ans.
6-25 T = (270 − 50)(0.150) = 33 N · m , S (^) y = 370 MPa
( T 1 − 0. 15 T 1 )(0.125) = 33 ⇒ T 1 = 310 .6 N, T 2 = 0 .15(310.6) = 46 .6 N ( T 1 + T 2 ) cos 45 = 252 .6 N
M A = 0. 3
- 42 + 1072 = 58 .59 N · m (maximum) M (^) B = 0. 15
- 22 + 174. 42 = 29 .38 N · m
σ x =
π d^3 =^
d^3 τ x y = 16(33) π d^3
= 168.^1
d^3
σ ′^ =
σ (^) x^2 + 3 τ (^) x y^2
[(
d^3
d^3
) 2 ] 1 / 2
= 664.^0
d^3
d = 17 .5(10−^3 ) m = 17 .5 mm, so use 18 mm Ans.
6-26 From Prob. 6-25,
τmax =
[(
σ x 2
] 1 / 2
S (^) y 2 n [(
- 8 2 d^3
d^3
) 2 ] 1 / 2
= (^342) d 3. 5 = 370(
d = 17 .7(10−^3 ) m = 17 .7 mm, so use 18 mm Ans.
xz plane
z
107.0 N
174.4 N
252.6 N
320 N
300 400 150
y 163.4 N 252.6 N 89.2 N 300 400 150 xy plane
A O B C
