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Chapter 4
(a)
k =
F
y ;^ y^ =^
F
k 1 +^
F
k 2 +^
F
k 3
so k =
(1/ k 1 ) + (1/ k 2 ) + (1/ k 3 ) Ans.
(b) F = k 1 y + k 2 y + k 3 y k = F / y = k 1 + k 2 + k 3 Ans.
(c) (^) 1 k =^
k 1 +^
k 2 + k 3^ k^ =
k 1 +^
k 2 + k 3
4-2 For a torsion bar, k (^) T = T /θ = Fl /θ, and so θ = Fl / k (^) T. For a cantilever, k (^) C = F /δ, δ = F / k (^) C. For the assembly, k = F / y , y = F / k = l θ + δ
So y = F k
= Fl
2 k (^) T
+ F
k (^) C
Or k = 1 ( l^2 / k (^) T ) + (1/ k (^) C )
Ans.
4-3 For a torsion bar, k = T /θ = GJ / l where J = π d^4 /32. So k = π d^4 G /(32 l ) = K d^4 / l. The springs, 1 and 2, are in parallel so
k = k 1 + k 2 = K d
4 l 1
4 l 2
= K d^4
x
l − x
And θ = T k
= T
K d^4
x +^
l − x
Then T = k θ = K d
4 x
θ + K d
(^4) θ l − x
k 2 k 1 k 3
F k 2
k 1
k 3
y
F k 1 k 2 k 3 y
Chapter 4 71
Thus T 1 = K d
4 x
θ; T 2 = K d
(^4) θ l − x If x = l /2, then T 1 = T 2. If x < l /2, then T 1 > T 2 Using τ = 16 T /π d^3 and θ = 32 T l /( G π d^4 ) gives
T = π d
(^3) τ 16 and so θall = 32 l G π d^4
· π d
(^3) τ 16
= 2 l τall Gd Thus, if x < l /2, the allowable twist is θall = 2 x τall Gd
Ans.
Since k = K d^4
x
l − x
= π^ Gd
4 32
x
l − x
Ans.
Then the maximum torque is found to be
T max = π d
(^3) x τall 16
x
l − x
Ans.
4-4 Both legs have the same twist angle. From Prob. 4-3, for equal shear, d is linear in x. Thus, d 1 = 0. 2 d 2 Ans.
k = π^ G 32
[
( 0. 2 d 2 )^4
- 2 l
- 8 l
]
= π^ G 32 l
- 258 d^42
Ans.
θall = 2(0.^8 l )τall Gd 2
Ans.
T max = k θall = 0. 198 d 23 τall Ans.
4- A = π r^2 = π( r 1 + x tan α)^2
d δ = Fdx AE
= Fdx E π( r 1 + x tan α) 2
δ = F π E
∫ (^) l
0
dx ( r 1 + x tan α) 2
= F π E
tan α( r 1 + x tan α)
) l
0
= F π E
r 1 ( r 1 + l tan α)
l
x
dx
F
F
r 1
Chapter 4 73
4- M = M 1 = M (^) B
E I dydx = M (^) B x + C 1 , dydx = 0 at x = 0, C 1 = 0
E I y = M^ B^ x
2 2 +^ C^2 ,^ y^ =^ 0 at^ x^ =^ 0,^ ^ C^2 =^0
y = M^ B^ x
2 2 E I Ans.
4-
ds =
dx^2 + dy^2 = dx
( (^) dy dx
Expand right-hand term by Binomial theorem [ 1 +
( (^) dy dx
) 2 ]^1 /^2
( (^) dy dx
Since dy / dx is small compared to 1, use only the first two terms, d λ = ds − dx
= dx
[
( (^) dy dx
) 2 ]
− dx
dy dx
dx
λ =
∫ (^) l
0
( (^) dy dx
dx Ans.
This contraction becomes important in a nonlinear, non-breaking extension spring.
4-10 y = C x^2 (4 lx − x^2 − 6 l^2 ) where C =
w 24 E I dy dx =^ C x (12 lx^ −^4 x
(^2) − 12 l (^2) ) = 4 C x (3 lx − x (^2) − 3 l (^2) )
( (^) dy dx
= 16 C^2 (15 l^2 x^4 − 6 lx^5 − 18 x^3 l^3 + x^6 + 9 l^4 x^2 )
λ =
∫^ l
0
( (^) dy dx
dx = 8 C^2
∫^ l
0
(15 l^2 x^4 − 6 lx^5 − 18 x^3 l^3 + x^6 + 9 l^4 x^2 ) dx
= 8 C^2
14 l
7
( (^) w 24 E I
) 2 (^9
14 l
7
( (^) w E I
l^7 Ans.
y ds dy dx
74 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
4-11 y = C x (2 lx^2 − x^3 − l^3 ) where C =
w 24 E I dy dx =^ C (6 lx
(^2) − 4 x (^3) − l (^3) )
( (^) dy dx
= C^2 (36 l^2 x^4 − 48 lx^5 − 12 l^4 x^2 + 16 x^6 + 8 x^3 l^3 + l^6 )
λ =
∫^ l
0
( (^) dy dx
dx = 1 2
C^2
∫^ l
0
(36 l^2 x^4 − 48 lx^5 − 12 l^4 x^2 + 16 x^6 + 8 x^3 l^3 + l^6 ) dx
= C^2
70 l
7
( (^) w 24 E I
70 l
7
( (^) w E I
l^7 Ans.
I = 2(5.56) = 11 .12 in^4
y max = y 1 + y 2 = −
w l^4 8 E I +^
Fa^2 6 E I ( a^ −^3 l ) Here w = 50 / 12 = 4 .167 lbf/in, and a = 7(12) = 84 in, and l = 10(12) = 120 in.
y 1 = − 4 .167(120)^
4 8(30)(10^6 )(11.12)
= − 0 .324 in
y 2 = −600(84)^
2 [3(120) − 84]
6(30)(10^6 )(11.12)
= − 0 .584 in
So y max = − 0. 324 − 0. 584 = − 0 .908 in Ans. M 0 = − Fa − (w l^2 /2) = −600(84) − [4.167(120) 2 /2] = −80 400 lbf · in c = 4 − 1. 18 = 2 .82 in
σmax = − M y I
= −(−80 400)(−^2 .82)
(10−^3 )
= − 20 .4 kpsi Ans. σmax is at the bottom of the section.
76 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
4-
I = 121 (9)(35^3 ) = 32. 156(10^3 ) mm^4
From Table A-9- yC = −
Fa^2 3 E I ( l^ +^ a ) dy (^) AB dx =^
Fa 6 E I l ( l
(^2) − 3 x (^2) )
Thus, θ A =
Fal^2 6 E I l =^
Fal 6 E I
y (^) D = −θ A a = −
Fa^2 l 6 E I With both loads,
y (^) D = − Fa
(^2) l 6 E I
− Fa
2 3 E I
( l + a )
= − Fa
2 6 E I (3 l^ +^2 a )^ = −^
500(250^2 )
6(207)(10^9 )(32.156)(10^3 ) [3(500)^ +^ 2(250)](
= − 1 .565 mm Ans.
y (^) E =
2 Fa ( l /2) 6 E I l
[
l^2 −
( (^) l 2
) 2 ]
Fal^2 8 E I
= 500(250)(
8(207)(10^9 )(32.156)(10^3 ) =^0 .587 mm^ Ans.
4-16 a = 36 in, l = 72 in, I = 13 in 4 , E = 30 Mpsi
y = F^1 a
2 6 E I
( a − 3 l ) − F^2 l
3 3 E I
= 400(36)^
6(30)(10^6 )(13)
− 400(72)^
3 3(30)(10^6 )(13) = − 0 .1675 in Ans.
4-17 I = 2(1.85) = 3 .7 in^4
Adding the weight of the channels, 2(5)/ 12 = 0 .833 lbf/in,
y (^) A = −
w l^4 8 E I −^
Fl^3 3 E I = −^
8 ( 30 )( 106 )( 3. 7 ) −^
= − 0 .1378 in Ans.
A
a
D (^) C
F A E B (^) a
Chapter 4 77
4-
I = π d^4 / 64 = π( 2 )^4 / 64 = 0 .7854 in^4 Tables A-9-5 and A-9-
y = −
F 2 l^3 48 E I +^
F 1 a 24 E I (4 a
(^2) − 3 l (^2) )
48(30)(10^6 )(0.7854) +^
24(30)(10^6 )(0.7854) = −^0 .0134 in^ Ans.
4- (a) Useful relations k =
F
y =^
48 E I
l^3
I = kl
3 48 E =^
48(30)10^6 =^0 .1843 in
4
From I = bh^3 / 12
h = 3
b Form a table. First, Table A-17 gives likely available fractional sizes for b : 8 12 , 9 , 9 12 , 10 in For h : 1 2 ,^
16 ,^
8 ,^
16 ,^
For available b what is necessary h for required I?
(b) I = 9 ( 0. 625 )^3 / 12 = 0 .1831 in^4
k = (^48) lE I 3 = 48(30)(
483 =^ 2384 lbf/in F = 4 σ^ I cl
= 4394 lbf
y =
F
k =^
2384 =^1 .84 in^ Ans.
choose 9"×^5 8
Ans.
b
3
b 8.5 0. 9.0 0.626 ← 9.5 0. 10.0 0.
Chapter 4 79
z (^) O B = 6 FbxE I l ( x^2 + b^2 − l^2 ), where b = 15 ",
x = 12", l = 48", I = 0 .2485 in^4
Then, z (^) A = 502(15)(12)(1446(30)(10 (^6) )(0^ +.2485)(48)^225 −^ 2304) = − 0 .081 44 in
For z (^) B use x = 33"
z (^) B = 502 (^15 )(^33 )(^1089 +^225 −^2304 ) 6 ( 30 )( 106 )( 0. 2485 )( 48 ) = − 0 .1146 in Therefore, by superposition z (^) A = + 0. 1182 − 0. 0814 = + 0 .0368 in Ans. z (^) B = + 0. 1103 − 0. 1146 = − 0 .0043 in Ans.
4- (a) Calculate torques and moment of inertia T = (400 − 50)(16/2) = 2800 lbf · in (8 T 2 − T 2 )(10/2) = 2800 ⇒ T 2 = 80 lbf, T 1 = 8(80) = 640 lbf I =
π 64 (1.^25
(^4) ) = 0 .1198 in 4
Due to 720 lbf, flip beam A-9-6 such that y^ AB →^ b^ =^9 ,^ x^ =^0 ,^ l^ =^20 ,^ F^ = −720 lbf
θ B = dydx
x = 0
= − 6 FbE I l (3 x^2 + b^2 − l^2 )
6(30)(10^6 )(0.1198)(20)
(0 + 81 − 400) = − 4 .793(10−^3 ) rad
yC = − 12 θ B = − 0 .057 52 in Due to 450 lbf, use beam A-9-10,
yC = − Fa
2 3 E I
( l + a ) = − 450(144)(32) 3(30)(10^6 )(0.1198)
= − 0 .1923 in
720 lbf 450 lbf O 9"^ 11"^ 12"
y
A
B C
R (^) O
R (^) B
O^ A^ B C R 1 R 2
12"
502 lbf 21" 15"
z
x
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Adding the two deflections, yC = − 0 .057 52 − 0. 1923 = − 0 .2498 in Ans. (b) At O : Due to 450 lbf: dy dx
x = 0
Fa 6 E I l ( l
(^2) − 3 x (^2) )
x = 0
Fal 6 E I
θ O = −
6(30)(10^6 )(0.1198)(20) +^
6(30)(10^6 )(0.1198) =^0 .010 13 rad^ =^0.^5805
◦
At B : θ B = − 4 .793(10−^3 ) + (^) 6(30)(10450(12) (^6) )(0.1198)(20) [20^2 − 3(20^2 )]
= − 0 .014 81 rad = 0. 8485 ◦
I = 0. 1198
= 1 .694 in^4
d =
( 64 I
π
[64(1.694)
π
] 1 / 4
= 2 .424 in
Use d = 2 .5 in Ans.
I =
π 64 (^2.^5
(^4) ) = 1 .917 in 4
yC = − 0. 2498
= − 0 .015 61 in Ans.
(a) l = 36(12) = 432 in
y max = −
5 w l^4 384 E I = −^
384(30)(10^6 )(5450)
= − 1 .16 in The frame is bowed up 1.16 in with respect to the bolsters. It is fabricated upside down and then inverted. Ans. (b) The equation in xy -coordinates is for the center sill neutral surface
y =
w x 24 E I (2 lx
(^2) − x (^3) − l (^3) ) Ans.
y
x l
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4-24 Incorporating a design factor into the solution for d (^) L of Prob. 4-23,
d =
[ (^32) n 3 π El ξ Fb ( l
(^2) − b (^2) )
] 1 / 4
∣∣(mm 10−^3 ) kN mm
3 GPa mm
103 (10−^9 )
109 (10−^3 )
1 / 4
d = 4
32(1.28)(3.5)(150)|(250^2 − 1502 )|
3 π(207)(250)(0.001)
10 −^12
= 36 .4 mm Ans.
4-25 The maximum occurs in the right section. Flip beam A-9-6 and use
y =
Fbx 6 E I l ( x
(^2) + b (^2) − l (^2) ) where b = 100 mm
dy dx =^
Fb 6 E I l (3 x
(^2) + b (^2) − l (^2) ) = 0
Solving for x ,
x =
l^2 − b^2 3 =
3 =^132 .29 mm^ from right
y =
3 .5(10^3 )(0.1)(0.132 29)
6(207)(10^9 )(π/64)(0. 03644 )(0.25) [0.132 29
2 + 0. 12 − 0. 252 ](10 3 )
= − 0 .0606 mm Ans.
4-
x
y
z
F 1
a 2 b 2
b 1
a 1
F 2
3.5 kN 100
250
150
d
The slope at x = 0 due to F 1 in the xy plane is
θ x y =
F 1 b 1
b^21 − l^2
6 E I l and in the xz plane due to F 2 is
θ x z =
F 2 b 2
b^22 − l^2
6 E I l
For small angles, the slopes add as vectors. Thus
θ L =
θ x y^2 + θ x z^2
F 1 b 1
b^21 − l^2
6 E I l
F 2 b 2
b^22 − l^2
6 E I l
1 / 2
Chapter 4 83
Designating the slope constraint as ξ, we then have
ξ = |θ L | =
6 E I l
{∑ [
Fi bi
b i^2 − l^2
)] 2 }^1 /^2
Setting I = π d^4 /64 and solving for d
d =
∣∣^32
3 π El ξ
{∑ [
Fi bi
b^2 i − l^2
)] 2 }^1 /^2 ∣∣∣
1 / 4
For the LH bearing, E = 30 Mpsi, ξ = 0 .001, b 1 = 12, b 2 = 6, and l = 16. The result is d (^) L = 1 .31 in. Using a similar flip beam procedure, we get d (^) R = 1 .36 in for the RH bearing. So use d = 1 3/8 in Ans.
4-27 I =
π 64 (1.^375
(^4) ) = 0 .17546 in (^4). For the xy plane, use yBC of Table A-9-
y =
6(30)(10^6 )(0.17546)(16) [
(^2) + 42 − 2(16)8] = − 1 .115(10− (^3) ) in
For the xz plane use yAB
z = 300(6)(8) 6(30)(10^6 )(0.17546)(16)
[8^2 + 62 − 162 ] = − 4 .445(10−^3 ) in
δ = (− 1. 115 j − 4. 445 k )(10−^3 ) in |δ| = 4 .583(10−^3 ) in Ans.
4-28 d (^) L =
∣∣^32 n 3 π El ξ
{∑ [
Fi bi
b i^2 − l^2
)] 2 } 1 / 2 ∣∣∣
1 / 4
∣∣^ 32(1.5)
3 π(207)(10^9 )(250)0. 001
[3.5(150)(150^2 − 2502 )]^2
+ [2.7(75)(75^2 − 2502 )]^2
(10^3 ) 3
1 / 4
= 39 .2 mm
d (^) R =
∣∣^ 32(1.5)
3 π(207)10^9 (250)0. 001
[3.5(100)(100^2 − 2502 )]^2
+ [2.7(175)(175^2 − 2502 )]^2
(10^3 ) 3
1 / 4
= 39 .1 mm Choose d ≥ 39 .2 mm Ans.
4-29 From Table A-9-8 we have
y (^) L = M^ B^ x 6 E I l
( x^2 + 3 a^2 − 6 al + 2 l^2 )
dy (^) L dx
= M^ B
6 E I l
(3 x^2 + 3 a^2 − 6 al + 2 l^2 )
Chapter 4 85
4-
θ = T LJ G = (^) (π/32)(0(0.. 0121 F )(1 (^4) )(79.5).3)(10 (^9) ) = 9 .292(10−^4 ) F
Due to twist δ B 1 = 0 .1(θ) = 9 .292(10−^5 ) F Due to bending
δ B 2 = F L
3 3 E I
= F (0.^1
3(207)(10^9 )(π/64)(0. 0124 )
= 1 .582(10−^6 ) F
δ B = 1 .582(10−^6 ) F + 9 .292(10−^5 ) F = 9 .450(10−^5 ) F
k = 1 9 .450(10−^5 )
= 10 .58(10^3 ) N/m = 10 .58 kN/m Ans.
R 1 = Fb l
R 2 = Fa l
δ 1 = R^1 k 1
δ 2 = R^2 k 2
Spring deflection
y (^) S = −δ 1 +
(δ 1 −^ δ 2 l
x = − Fb k 1 l
( (^) Fb k 1 l^2
− Fa k 2 l^2
x
y (^) AB = Fbx 6 E I l
( x^2 + b^2 − l^2 ) + F x l^2
( (^) b k 1
− a k 2
− Fb k 1 l
Ans.
y (^) BC = Fa ( l^ −^ x ) 6 E I l
( x^2 + a^2 − 2 lx ) + F x l^2
( (^) b k 1
− a k 2
− Fb k 1 l
Ans.
4-33 See Prob. 4-32 for deflection due to springs. Replace Fb / l and Fa / l with w l / 2
y (^) S = − w l 2 k 1
( (^) w l 2 k 1 l
− w l 2 k 2 l
x = w x 2
k 1
k 2
− w l 2 k 1
y = w x 24 E I
( 2 lx^2 − x^3 − l^3 ) + w x 2
k 1
k 2
− w l 2 k 1
Ans.
F A a B b C l (^) R 2 1 2
R 1
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4-34 Let the load be at x > l /2. The maximum deflection will be in Section AB (Table A-9-10)
y (^) AB = 6 FbxE I l ( x^2 + b^2 − l^2 )
dy (^) AB dx =^
Fb 6 E I l (3 x
(^2) + b (^2) − l (^2) ) = 0 ⇒ 3 x (^2) + b (^2) − l (^2) = 0
x =
l^2 − b^2 3 ,^ x max^ =
l^2 3 =^0.^577 l^ Ans. For x < l / 2 x min = l − 0. 577 l = 0. 423 l Ans.
4- M (^) O = 50(10)(60) + 600(84) = 80 400 lbf · in R (^) O = 50(10) + 600 = 1100 lbf I = 11 .12 in^4 from Prob. 4- M = −80 400 + 1100 x − 4.^167 x
2 2 −^600 〈 x^ −^84 〉
1
E I dydx = −80 400 x + 550 x^2 − 0. 6944 x^3 − 300 〈 x − 84 〉^2 + C 1 dy dx =^ 0 at^ x^ =^0 ^ C^1 =^0 E I y = −402 00 x^2 + 183. 33 x^3 − 0. 1736 x^4 − 100 〈 x − 84 〉^3 + C 2 y = 0 at x = 0 C 2 = 0 y (^) B =
30(10^6 )(11.12) [−40 200(
− 0 .1736(120^4 ) − 100(120 − 84) 3 ]
= − 0 .9075 in Ans.
4-36 See Prob. 4-13 for reactions: R (^) O =^ 860 lbf,^ R^ C =^ 540 lbf
M = 860 x − 800 〈 x − 36 〉^1 − 600 〈 x − 60 〉^1 E I dy dx
= 430 x^2 − 400 〈 x − 36 〉^2 − 300 〈 x − 60 〉^2 + C 1
E I y = 143. 33 x^3 − 133. 33 〈 x − 36 〉^3 − 100 〈 x − 60 〉^3 + C 1 x + C 2 y = 0 at x = 0 ⇒ C 2 = 0 y = 0 at x = 120 in ⇒ C 1 = − 1 .2254(10^6 ) lbf · in^2 Substituting C 1 and C 2 and evaluating at x^ =^ 60,
E I y = 30(10^6 ) I
= 143 .33(60^3 ) − 133 .33(60 − 36) 3 − 1 .2254(10^6 )(60)
I = 23 .68 in^4 Agrees with Prob. 4-13. The rest of the solution is the same.
R 7'^ 10' O
50 lbf/ft^ 600 lbf
M (^) O
O (^) A B
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0 = w 12 l
( l^2 − a^2 ) l^3 − w 24
l^4 + C 1 l ⇒ C 1 = w l 24
(2 a^2 − l^2 )
y =
w 24 E I l [2( l
(^2) − a (^2) ) x (^3) − lx (^4) + 2( l + a ) 2 〈 x − l 〉 (^3) + l (^2) (2 a (^2) − l (^2) ) x ] Ans.
4-39 R (^) A = R (^) B = 500 N, and I =
(^3) = 32 .156(10 (^3) ) mm 4 For first half of beam, M = − 500 x + 500 〈 x − 0. 25 〉^1 where x is in meters
E I
dy dx = −^250 x
(^2) + 250 〈 x − 0. 25 〉 (^2) + C 1
At x = 0 .5 m, dy / dx = 0 ⇒ 0 = −250(0. 52 ) + 250(0. 5 − 0 .250) 2 + C 1 ⇒ C 1 = 46 .875 N · m^2
E I y = −
3 x
3 〈 x^ −^0.^25 〉
(^3) + 46. 875 x + C 2
y = 0 at x = 0 .25 m ⇒ 0 = −
3 0.^25
(^3) + 46 .875(0.25) + C 2 ⇒ C 2 = − 10 .417 N · m 3
∴ E I y = −
3 x
3 〈 x^ −^0.^25 〉
(^3) + 46. 875 x − 10. 42
Evaluating y at A and the center,
y (^) A =
207(10^9 )32. 156(10−^9 )
[
3 (0)^
]
= − 1 .565 mm Ans.
y | x = 0 .5m =
207(10^9 )32.156(10−^9 )
[
3 (0.^5
3 (0.^5 −^0 .25)^
3
]
= − 2 .135 mm Ans.
4-40 From Prob. 4-30, R (^) O = 160 lbf ↓, F (^) AC = 240 lbf I = 0 .1667 in^4
M = − 160 x + 240 〈 x − 6 〉^1 E I
dy dx = −^80 x
(^2) + 120 〈 x − 6 〉 (^2) + C 1
E I y = − 26. 67 x^3 + 40 〈 x − 6 〉^3 + C 1 x + C 2 y = 0 at x = 0 ⇒ C 2 = 0
y (^) A = −
( F L
AE
AC
(π/4)(1/2) 2 (10)(10^6 ) = −^1 .4668(
− (^3) ) in
at x = 6 10(10^6 )(0.1667)(− 1 .4668)(10−^3 ) = − 26 .67(6^3 ) + C 1 (6) C 1 = 552 .58 lbf · in^2
Chapter 4 89
y (^) B = 1 10(10^6 )(0.1667)
[− 26 .67(18^3 ) + 40(18 − 6) 3 + 552 .58(18)]
= − 0 .045 87 in Ans.
4-
I 1 = π 64
(1. 54 ) = 0 .2485 in^4 I 2 = π 64
(2^4 ) = 0 .7854 in^4
R 1 = 200
( 12 ) = 1200 lbf
For 0^ ≤^ x^ ≤^ 16 in,^ M^ =^1200 x^ −^
〈 x − 4 〉^2 M I
= 1200 x I 1
I 1
I 2
〈 x − 4 〉^0 − 1200
I 1
I 2
〈 x − 4 〉^1 − 100 I 2
〈 x − 4 〉^2
= 4829 x − 13 204〈 x − 4 〉^0 − 3301. 1 〈 x − 4 〉^1 − 127. 32 〈 x − 4 〉^2
E dy dx
= 2414. 5 x^2 − 13 204〈 x − 4 〉^1 − 1651 〈 x − 4 〉^2 − 42. 44 〈 x − 4 〉^3 + C 1
Boundary Condition:
dy dx
= (^0) at x = 10 in
0 = 2414 .5(10^2 ) − 13 204(10 − 4) 1 − 1651(10 − 4) 2 − 42 .44(10 − 4) 3 + C 1 C 1 = − 9. 362 ( 104 ) E y = 804. 83 x^3 − 6602 〈 x − 4 〉^2 − 550. 3 〈 x − 4 〉^3 − 10. 61 〈 x − 4 〉^4 − 9 .362(10^4 ) x + C 2 y = 0 at x = 0 ⇒ C 2 = 0 For 0 ≤ x ≤ 16 in
y = 1 30(10^6 )
[804. 83 x^3 − 6602 〈 x − 4 〉^2 − 550. 3 〈 x − 4 〉^3 − 10. 61 〈 x − 4 〉^4 − 9 .362(10^4 ) x ] Ans. at x = 10 in y | x = 10 =
30(10^6 ) [804.83(
− 10 .61(10 − 4) 4 − 9 .362(10^4 )(10)]
= − 0 .016 72 in Ans.
4-42 q = F 〈 x 〉−^1 − Fl 〈 x 〉−^2 − F 〈 x − l 〉−^1
Integrations produce V = F 〈 x 〉^0 − Fl 〈 x 〉−^1 − F 〈 x − l 〉^0 M = F 〈 x 〉^1 − Fl 〈 x 〉^0 − F 〈 x − l 〉^1 = F x − Fl
x
M I
