Chapter 4 For a torsion bar, k T = T/ = Fl/, and so = Fl/k T . For a cantilever, k l = F/ , = F/k l . For the assembly, k = F/y, or, y = F/k = l + Thus F Fl 2 F y k kT kl Solving for k kk 1 k 2 2l T Ans. l 1 kl l kT kT kl ______________________________________________________________________________ 4-1
For a torsion bar, k T = T/ = Fl/, and so = Fl/k T . For each cantilever, k l = F/ l , l = F/k l , and, L = F/k L . For the assembly, k = F/y, or, y = F/k = l + l + L . Thus F Fl 2 F F y k kT kl k L Solving for k k L kl kT 1 k 2 Ans. 2 l 1 1 kl k Ll kT k L kT kl kT kl k L ______________________________________________________________________________ 4-2
4-3
(a) For a torsion bar, k =T/ =GJ/l. Two springs in parallel, with J =d i 4/32, and d 1 = d 1 = d, J G J G d4 d4 k 1 2 G 1 2 x l x 32 x l x
1 1 Gd 4 32 x lx Deflection equation,
Ans.
(1)
T1 x T2 l x JG JG
T2 l x (2) x From statics, T 1 + T 2 = T = 1500. Substitute Eq. (2) results in
T1
Chapter 4 - Rev B, Page 1/81
lx T2 T2 1500 x
T2 1500
x l
Ans.
(3)
lx Ans. (4) l 1 1 3 (b) From Eq. (1), k 0.54 11.5 106 28.2 10 lbf in/rad 32 5 10 5 10 5 From Eq. (4), T1 1500 750 lbf in Ans. 10 5 From Eq. (3), T2 1500 750 lbf in Ans. 10 16Ti 16 1500 From either section, 30.6 103 psi 30.6 kpsi Ans. 3 3 di 0.5 T1 1500
Substitute into Eq. (2) resulting in
Ans.
______________________________________________________________________________ 4-4
Deflection to be the same as Prob. 4-3 where T 1 = 750 lbfin, l 1 = l / 2 = 5 in, and d 1 = 0.5 in 1=2= T1 4
32 Or,
4 1
d G
T2 6
32
4 2
d G
32
T1 15 103 d14
0.5 G 4
4T1 6T2 4 60 103 4 d1 d2
(1)
(2)
T2 10 103 d 24
Equal stress, 1 2
750 5
(3)
16T1 16T2 T T 13 23 3 3 d1 d 2 d1 d 2
(4)
Divide Eq. (4) by the first two equations of Eq.(1) results in T1 T2 3 d1 d3 2 d 2 1.5d1 (5) 4T1 4T2 d14 d 24 Statics, T 1 + T 2 = 1500
(6)
Substitute in Eqs. (2) and (3), with Eq. (5) gives
15 103 d14 10 103 1.5d1 1500 4
Solving for d 1 and substituting it back into Eq. (5) gives d 1 = 0.388 8 in, d 2 = 0.583 2 in Ans.
Chapter 4 - Rev B, Page 2/81
From Eqs. (2) and (3), T 1 = 15(103)(0.388 8)4 = 343 lbfin T 2 = 10(103)(0.583 2)4 = 1 157 lbfin
Ans. Ans.
Deflection of T is
1
343 4 T1l1 0.053 18 rad J1G / 32 0.388 84 11.5 106
Spring constant is
k
T
The stress in d 1 is
1
16 343 16T1 29.7 103 psi 29.7 kpsi 3 d1 0.388 8 3
Ans.
The stress in d 1 is
2
16 1 157 16T2 29.7 103 psi 29.7 kpsi 3 3 d 2 0.583 2
Ans.
1
1500 28.2 103 lbf in 0.053 18
Ans.
______________________________________________________________________________ 4-5
(a) Let the radii of the straight sections be r 1 = d 1 /2 and r 2 = d 2 /2. Let the angle of the taper be where tan = (r 2 r 1 )/2. Thus, the radius in the taper as a function of x is r = r 1 + x tan , and the area is A = (r 1 + x tan )2. The deflection of the tapered portion is l
l
F F dx F 1 dx 2 AE E 0 r1 x tan E r1 x tan tan 0
1 1 F 1 F 1 E r1 tan tan r1 l tan E tan r1 r2
F r2 r1 F l tan Fl E tan r1r2 E tan r1r2 r1r2 E
4 Fl d1d 2 E
l
0
Ans.
(b) For section 1,
1
Fl 4 Fl 4(1000)(2) 3.40(104 ) in 2 2 6 AE d1 E (0.5 )(30)(10 )
For the tapered section, 4 Fl 4 1000(2) 2.26(104 ) in d1d 2 E (0.5)(0.75)(30)(106 ) For section 2,
Ans.
Ans.
Chapter 4 - Rev B, Page 3/81
Fl 4 Fl 4(1000)(2) 1.51(104 ) in Ans. 2 2 6 AE d1 E (0.75 )(30)(10 ) ______________________________________________________________________________
2
4-6
(a) Let the radii of the straight sections be r 1 = d 1 /2 and r 2 = d 2 /2. Let the angle of the taper be where tan = (r 2 r 1 )/2. Thus, the radius in the taper as a function of x is r = r 1 + x tan , and the polar second area moment is J = ( /2) (r 1 + x tan )4. The angular deflection of the tapered portion is l
l
T 2T dx 1 2T 1 dx 4 3 G r1 x tan 3 tan G 0 r1 x tan GJ 0
l
0
2 2 T 1 1 T 1 1 3 3 3 G r1 tan tan r1 l tan 3 G tan r13 r23
2 2 r23 r13 2 2 T l r23 r13 2 Tl r1 r1r2 r2 T 3 G tan r13r23 3 G r2 r1 r13r23 3 G r13r23
2 2 32 Tl d1 d1d 2 d 2 3 G d13d 23
Ans.
(b) The deflections, in degrees, are For section 1, Tl 180 32Tl 180 32(1500)(2) 180 2.44 deg 4 GJ d1 G (0.54 )11.5(106 )
1
Ans.
For the tapered section,
32 Tl (d12 d1d 2 d 2 2 ) 180 3 Gd13d 23
2 2 32 (1500)(2) 0.5 (0.5)(0.75) 0.75 180 1.14 deg 3 11.5(106 )(0.53 )(.753 )
Ans.
For section 2, Tl 180 32Tl 180 32(1500)(2) 180 Ans. 0.481 deg 4 GJ d 2 G (0.754 )11.5(106 ) ______________________________________________________________________________
2
4-7
The area and the elastic modulus remain constant, however the force changes with respect to x. From Table A-5 the unit weight of steel is = 0.282 lbf/in3, and the elastic modulus is E = 30 Mpsi. Starting from the top of the cable (i.e. x = 0, at the top). F = (A)(lx)
Chapter 4 - Rev B, Page 4/81
Fdx w l l 2 0.282 500(12) 1 (l x)dx lx x 2 0.169 in AE E 0 E 2 0 2E 2(30)106 2
l
c
l
o
From the weight at the bottom of the cable, 4(5000) 500(12) 4Wl Wl 5.093 in (0.52 )30(106 ) AE d 2 E c W 0.169 5.093 5.262 in Ans.
W
The percentage of total elongation due to the cable’s own weight 0.169 (100) 3.21% Ans. 5.262 ______________________________________________________________________________
4-8
F y = 0 = R 1 F R 1 = F M A = 0 = M 1 Fa M 1 = Fa V AB = F, M AB =F (x a ), V BC = M BC = 0 Section AB: 1 F x2 F x a dx AB ax C1 EI EI 2 AB = 0 at x = 0 C 1 = 0 y AB
(1)
F x2 F x3 x2 ax dx a C2 2 EI 2 EI 6
y AB = 0 at x = 0 C 2 = 0 Fx 2 y AB x 3a 6 EI
(2)
Ans.
Section BC:
BC
1 EI
0 dx 0 C
3
From Eq. (1), at x = a (with C 1 = 0), Fa 2 2 EI Fa 2 Fa 2 dx x C4 2 EI 2 EI
F a2 Fa 2 = C 3 . Thus, a ( a ) 2 EI EI 2
BC yBC
(3)
Chapter 4 - Rev B, Page 5/81
From Eq. (2), at x = a (with C 2 = 0), y
Fa 2 Fa 3 a C4 2 EI 3EI
yBC
C4
F a3 a2 Fa 3 . Thus, from Eq. (3) a EI 6 2 3EI
Fa 3 6 EI
Fa 2 Fa3 Fa 2 x a 3x 2 EI 6 EI 6 EI
Substitute into Eq. (3)
Ans.
The maximum deflection occurs at x= l, Fa 2 Ans. a 3l 6 EI ______________________________________________________________________________ ymax
4-9
M C = 0 = F (l /2) R 1 l R 1 = F /2 F y = 0 = F /2 + R 2 F R 2 = F /2 Break at 0 x l /2: V AB = R 1 = F /2,
M AB = R 1 x = Fx /2
Break at l /2 x l : V BC = R 1 F = R 2 = F /2,
M BC = R 1 x F ( x l / 2) = F (l x) /2
Section AB:
AB
1 EI
Fx F x2 C1 dx 2 EI 4 2
l F 2 C 0 From symmetry, AB = 0 at x = l /2 1 4 EI
AB
F x 2 Fl 2 F 4 x2 l 2 EI 4 16 EI 16 EI y AB
F 16 EI
C1
Fl 2 . Thus, 16 EI
(1)
F 4x 3 2 4 x l dx 16 EI 3 l x C2 2
2
Chapter 4 - Rev B, Page 6/81
y AB = 0 at x = 0
y AB
C 2 = 0, and,
Fx 4 x 2 3l 2 48 EI
(2)
y BC is not given, because with symmetry, Eq. (2) can be used in this region. The maximum deflection occurs at x =l /2,
ymax
l F 2 Fl 3 2 l 2 4 3l 48EI 2 48EI
Ans.
______________________________________________________________________________ 4-10 From Table A-6, for each angle, I 1-1 = 207 cm4. Thus, I = 2(207) (104) = 4.14(106) mm4
From Table A-9, use beam 2 with F = 2500 N, a = 2000 mm, and l = 3000 mm; and beam 3 with w = 1 N/mm and l = 3000 mm. Fa 2 wl4 (a 3l ) 6 EI 8EI 2500(2000) 2 (1)(3000)4 2000 3(3000) 6(207)103 (4.14)106 8(207)(103 )(4.14)(106 ) 25.4 mm Ans.
ymax
M O Fa ( wl 2 / 2)
= 2500(2000) [1(30002)/2] = 9.5(106) Nmm From Table A-6, from centroid to upper surface is y = 29 mm. From centroid to bottom surface is y = 29.0 100= 71 mm. The maximum stress is compressive at the bottom of the beam at the wall. This stress is My 9.5(106 )(71) 163 MPa Ans. max I 4.14(106 ) ______________________________________________________________________________
Chapter 4 - Rev B, Page 7/81
4-11 14 10 (450) (300) 465 lbf 20 20 6 10 (450) (300) 285 lbf RC 20 20 RO
M 1 = 465(6)12 = 33.48(103) lbfin M 2 = 33.48(103) +15(4)12 = 34.20(103) lbfin M max 34.2 Z 2.28 in 3 15 Z Z For deflections, use beams 5 and 6 of Table A-9 2 F1a[l (l / 2)] l l F2l 3 2 y x 10ft a 2l 6 EIl 2 48 EI 2
max
450(72)(120) 300(2403 ) 2 2 2 0.5 120 72 240 48(30)(106 ) I 6(30)(106 ) I (240) I 12.60 in 4 I / 2 6.30 in 4
Select two 5 in-6.7 lbf/ft channels from Table A-7, I = 2(7.49) = 14.98 in4, Z =2(3.00) = 6.00 in3 12.60 1 ymidspan 0.421 in 14.98 2 34.2 max 5.70 kpsi 6.00 ______________________________________________________________________________ 4-12
I
(1.54 ) 0.2485 in 4
64 From Table A-9 by superposition of beams 6 and 7, at x = a = 15 in, with b = 24 in and l = 39 in Fba 2 wa y [a b 2 l 2 ] (2la 2 a 3 l 3 ) 6 EIl 24 EI
yA
340(24)15 152 242 392 6 6(30)10 (0.2485)39 (150 /12)(15) 2(39)(152 ) 153 393 0.0978 in 24(30)106 (0.2485)
Ans.
At x = l /2 = 19.5 in
Chapter 4 - Rev B, Page 8/81
y
2 2 3 Fa[l (l / 2)] l l w(l / 2) l l 2 3 a 2 l 2 l l 6 EIl 2 24 EI 2 2 2
y
340(15)(19.5) 19.52 152 392 6(30)(106 )(0.2485)(39) (150 /12)(19.5) 2(39)(19.52 ) 19.53 393 0.1027 in 6 24(30)(10 )(0.2485)
Ans.
0.1027 0.0978 (100) 5.01% Ans. 0.0978 ______________________________________________________________________________ % difference
4-13 I
1 (6)(323 ) 16.384 103 mm 4 12
From Table A-9-10, beam 10 Fa 2 yC (l a ) 3EI Fax 2 y AB l x2 6 EIl dy AB Fa 2 (l 3 x 2 ) 6 EIl dx dy AB A dx Fal 2 Fal A 6 EIl 6 EI Fa 2l yO A a 6 EI
At x = 0,
With both loads, Fa 2l Fa 2 yO (l a ) 6 EI 3EI Fa 2 400(3002 ) (3l 2a ) 3(500) 2(300) 3.72 mm Ans. 6 EI 6(207)103 (16.384)103 At midspan, 2 2 Fa(l / 2) 2 l 3 Fal 2 3 400(300)(5002 ) 1.11 mm Ans. yE l 6 EIl 24 207 103 16.384 103 2 24 EI _____________________________________________________________________________ 4-14 I (24 1.54 ) 0.5369 in 4 64
Chapter 4 - Rev B, Page 9/81
From Table A-5, E = 10.4 Mpsi From Table A-9, beams 1 and 2, by superposition 3 2 200 4(12) 300 2(12) FB l 3 FA a 2 (a 3l ) yB 2(12) 3(4)(12) 3EI 6 EI 3(10.4)106 (0.5369) 6(10.4)106 (0.5369) yB 1.94 in Ans. ______________________________________________________________________________ 4-15 From Table A-7, I = 2(1.85) = 3.70 in4 From Table A-5, E = 30.0 Mpsi From Table A-9, beams 1 and 3, by superposition
5 2(5 /12) (60 ) 0.182 in Ans. Fl 3 ( w wc )l 4 150(603 ) yA 3EI 8 EI 3(30)106 (3.70) 8(30)106 (3.70) ______________________________________________________________________________ 4 4-16 I d 64 From Table A-5, E 207(103 ) MPa From Table A-9, beams 5 and 9, with F C = F A = F, by superposition F l3 1 Fa FB l 3 2 Fa(4a 2 3l 2 ) yB B (4a 2 3l 2 ) I 48EI 24 EI 48 EyB 1 I 550(10003 ) 2 375 (250) 4(2502 ) 3(10002 ) 3 48(207)10 2 4
53.624 103 mm4 d
4
64
I
4
64
(53.624)103 32.3 mm
Ans.
______________________________________________________________________________ 4-17 From Table A-9, beams 8 (region BC for this beam with a = 0) and 10 (with a = a), by superposition
yBC
MA 3 Fax 2 x 3lx 2 2l 2 x l x2 6 EIl 6 EIl 1 M A x 3 3lx 2 2l 2 x Fax l 2 x 2 Ans. 6 EIl d M F (x l) A x3 3lx 2 2l 2 x ( x l ) [( x l )2 a(3x l )] 6 EI x l dx 6 EIl
y AB
M Al F (x l) (x l) [( x l ) 2 a (3 x l )] 6 EI 6 EI
Chapter 4 - Rev B, Page 10/81
(x l) Ans. M Al F ( x l ) 2 a (3 x l ) 6 EI ______________________________________________________________________________ 4-18 Note to the instructor: Beams with discontinuous loading are better solved using singularity functions. This eliminates matching the slopes and displacements at the discontinuity as is done in this solution. a wa M C 0 R1l wa l a 2 R1 2l 2l a Ans.
wa wa 2 Fy 0 2l 2l a R2 wa R2 2l Ans. wa w VAB R1 wx = 2l a wx = 2l a x a 2 Ans. 2l 2l 2 wa VBC R2 Ans. 2l w x2 M AB VAB dx 2l ax a 2 x C1 2l 2 wx 2al a 2 lx Ans. M AB 0 at x 0 C1 0 M AB 2l wa 2 wa 2 M BC VBC dx dx x C2 2l 2l wa 2 wa 2 M BC 0 at x l C2 M BC (l x) Ans. 2 2l M 1 wx 1 w 2 1 2 2 1 3 AB AB dx 2al a 2 lx dx alx a x lx C3 3 2 EI EI 2l EI 2l
1 w 2 1 2 2 1 3 alx a x lx C3 dx 2 3 EI 2l 1 w 1 3 1 2 3 1 4 alx a x lx C3 x C4 6 12 EI 2l 3
y AB AB dx
y AB 0 at x 0 C4 0
BC
M 1 BC dx EI EI
wa 2 1 wa 2 1 2 2l (l x) dx EI 2l lx 2 x C5
AB BC at x a 1 EI
1 2 w 2 1 4 1 3 1 wa 2 wa 3 ala a la C la a C C C5 3 5 3 2l EI 2l 2 3 2 6
(1)
Chapter 4 - Rev B, Page 11/81
1 wa 2 1 2 1 wa 2 1 2 1 3 lx x C dx 5 lx x C5 x C6 EI 2l 2 EI 2l 2 6 2 2 wa l 0 at x l C6 C5l 6 1 wa 2 1 2 1 3 1 3 lx x l C5 ( x l ) EI 2l 2 6 3
yBC BC dx yBC yBC
y AB yBC at x a 1 5 1 4 w 1 wa 2 1 2 1 3 1 3 3 ala a la C a 3 la a l C5 (a l ) 2l 3 6 12 2l 2 6 3 wa 2 C3 a (2) 3la 2 4l 3 C5 (a l ) 24l wa 2 Substituting (1) into (2) yields C5 a 2 4l 2 . Substituting this back into (2) gives 24l 2 wa 4al a 2 4l 2 . Thus, C3 24l w y AB 4alx3 2a 2 x3 lx 4 4a3lx a 4 x 4a 2l 2 x 24 EIl wx 2 y AB Ans. 2ax 2 (2l a ) lx 3 a 2 2l a 24 EIl w yBC 6a 2lx 2 2a 2 x 3 a 4 x 4a 2l 2 x a 4l Ans. 24 EIl This result is sufficient for y BC . However, this can be shown to be equivalent to w w yBC 4alx 3 2a 2 x 3 lx 4 4a 2l 2 x 4a 3lx a 4 x ( x a)4 24 EIl 24 EI w yBC y AB ( x a)4 Ans. 24 EI by expanding this or by solving the problem using singularity functions. ______________________________________________________________________________ 4-19
The beam can be broken up into a uniform load w downward from points A to C and a uniform load upward from points A to B. wx wx 2 2 2bx 2 (2l b) lx 3 b 2 2l b 2ax 2 (2l a) lx 3 a 2 2l a y AB 24 EIl 24 EIl wx 2 2 2bx 2 (2l b) b 2 2l b 2ax 2 (2l a ) a 2 2l a 24 EIl w 2 2bx 3 (2l b) lx 4 b 2 x 2l b 24 EIl
yBC
4alx 3 2a 2 x 3 lx 4 4a 2l 2 x 4a 3lx a 4 x l ( x a ) 4
Ans.
Ans.
Chapter 4 - Rev B, Page 12/81
w 4blx 3 2b 2 x 3 lx 4 4b 2l 2 x 4b3lx b 4 x l ( x b) 4 24 EIl w 4alx 3 2a 2 x3 lx 4 4a 2l 2 x 4a 3lx a 4 x l ( x a) 4 24 EIl w ( x b) 4 ( x a ) 4 y AB Ans. 24 EI ______________________________________________________________________________
yCD
4-20
Note to the instructor: See the note in the solution for Problem 4-18. wa 2 wa 0 F R wa RB 2l a Ans. y B 2l 2l For region BC, isolate right-hand element of length (l + a x) wa 2 , VAB RA VBC w l a x Ans. 2l wa 2 w 2 M AB RA x x, M BC l a x Ans. 2l 2 wa 2 2 EI AB M AB dx x C1 4l wa 2 3 EIy AB x C1 x C2 12l wa 2 3 y AB = 0 at x = 0 C 2 = 0 EIy AB x C1 x 12l w a 2l y AB = 0 at x = l C1 12 wa 2 3 wa 2l wa 2 x 2 wa 2 x 2 2 EIy AB x x l x y AB l x 2 Ans. 12l 12 12l 12 EIl w 3 EI BC M BC dx l a x C3 6 w 4 EIyBC l a x C3 x C4 24 wa 4 wa 4 (1) y BC = 0 at x = l C3l C4 0 C4 C3l 24 24 wa 2l wa 2l wa 3 wa 2 AB = BC at x = l C3 C3 l a 4 12 6 6 wa 2 2 a 4l l a . Substitute back into y BC Substitute C 3 into Eq. (1) gives C4 24 1 w wa 2 wa 4 wa 2l 4 yBC l a x x l a l a EI 24 6 24 6
w 4 l a x 4a 2 l x l a a 4 24 EI
Ans.
Chapter 4 - Rev B, Page 13/81
Table A-9, beam 7, w l 100(10) R1 R2 500 lbf 2 2 100 x wx 2(10) x 2 x 3 103 y AB 2lx 2 x 3 l 3 6 24 EI 24 30 10 0.05
4-21
2.7778 106 x 20 x 2 x 3 1000
Slope:
AB
d y AB w 6lx 2 4 x3 l 3 24 EI dx
w wl 3 2 3 3 At x = l, AB x l 6l l 4l l 24EI 24 EI 100 103 w l3 yBC AB x l x l x l x 10 2.7778 103 x 10 6 24 EI 24(30)10 (0.05)
From Prob. 4-20, 2 wa 2 100 4 RA 80 lbf 2l 2(10)
RB
100 4 wa 2l a 2(10) 4 480 lbf 2l 2(10)
100 42 x wa 2 x 2 2 l x 10 2 x 2 8.8889 106 x 100 x 2 6 12 EIl 12 30 10 0.05 w 4 l a x 4a 2 l x l a a 4 24 EI
y AB yBC
100 10 4 x 4 4 42 10 x 10 4 44 6 24 30 10 0.05
2.7778 106 14 x 896 x 9216 Superposition, RA 500 80 420 lbf RB 500 480 980 lbf 4
y AB 2.7778 10 6 x 20 x 2 x 3 1000 8.8889 10 6 x 100 x 2
Ans. Ans.
yBC 2.7778 103 x 10 2.7778 106 14 x 896 x 9216 Ans. The deflection equations can be simplified further. However, they are sufficient for plotting. Using a spreadsheet, 4
x y
0 0.5 1 1.5 2 2.5 3 3.5 0.000000 -0.000939 -0.001845 -0.002690 -0.003449 -0.004102 -0.004632 -0.005027
x y
4 4.5 5 5.5 6 6.5 7 7.5 -0.005280 -0.005387 -0.005347 -0.005167 -0.004853 -0.004421 -0.003885 -0.003268
Chapter 4 - Rev B, Page 14/81
x y x y
8 8.5 9 9.5 10 10.5 11 11.5 -0.002596 -0.001897 -0.001205 -0.000559 0.000000 0.000439 0.000775 0.001036 12 0.001244
12.5 0.001419
13 0.001575
13.5 14 0.001722 0.001867
______________________________________________________________________________ 4-22
(a) Useful relations k
F 48 EI 3 y l
3 kl 3 1800 36 I 0.05832 in 4 6 48 E 48(30)10
From I = bh 3/12, and b = 10 h, then I = 5 h 4/6, or, h
4
6 I 4 6(0.05832) 0.514 in 5 5
h is close to 1/2 in and 9/16 in, while b is close to 5.14 in. Changing the height drastically changes the spring rate, so changing the base will make finding a close solution easier. Trial and error was applied to find the combination of values from Table A-17 that yielded the closet desired spring rate. h (in) 1/2 1/2 1/2 9/16 9/16
b (in) 5 5½ 5¾ 5 4
b/h 10 11 11.5 8.89 7.11
k (lbf/in) 1608 1768 1849 2289 1831
Chapter 4 - Rev B, Page 15/81
h = ½ in, b = 5 ½ in should be selected because it results in a close spring rate and b/h is still reasonably close to 10. (b) I 5.5(0.5)3 /12 0.05729 in 4 4 I 4(60)103 (0.05729) Mc ( Fl / 4)c F 1528 lbf I I lc 36 (0.25)
(1528) 36 Fl 3 0.864 in Ans. y 48 EI 48(30)106 (0.05729) ______________________________________________________________________________ 3
4-23 From the solutions to Prob. 3-68, T1 60 lbf and T2 400 lbf
I
d4 64
(1.25)4 64
0.1198 in 4
From Table A-9, beam 6, Fb x Fb x z A 1 1 ( x 2 b12 l 2 ) 2 2 ( x 2 b2 2 l 2 ) 6 EIl 6 EIl x 10in (575)(30)(10) 102 302 402 6 6(30)10 (0.1198)(40) 460(12)(10) 102 122 402 0.0332 in 6(30)106 (0.1198)(40)
Ans.
dz d Fb x Fb x 1 1 ( x 2 b12 l 2 ) 2 2 ( x 2 b2 2 l 2 ) 6 EIl dx x 10in x 10in dx 6 EIl Fb Fb 1 1 (3 x 2 b12 l 2 ) 2 2 (3x 2 b2 2 l 2 ) 6 EIl 6 EIl x 10in
A y
(575)(30) 3 102 302 402 6 6(30)10 (0.1198)(40) 460(12) 3 102 122 402 6(30)106 (0.1198)(40)
Ans. 6.02(104 ) rad ______________________________________________________________________________ 4-24 From the solutions to Prob. 3-69, T1 2880 N and T2 432 N
I
d4 64
(30)4 64
39.76 103 mm 4
Chapter 4 - Rev B, Page 16/81
The load in between the supports supplies an angle to the overhanging end of the beam. That angle is found by taking the derivative of the deflection from that load. From Table A-9, beams 6 (subscript 1) and 10 (subscript 2), y A BC
C
a2 beam 6 y A beam10
(1)
d F1a1 l x 2 Fa x a12 2lx 1 1 6lx 3x 2 a12 2l 2 x l x l 6 EIl dx 6 EIl
BC C
F1a1 2 l a12 6 EIl Equation (1) is thus
F1a1 2 Fa2 l a12 a2 2 2 (l a2 ) 6 EIl 3EI 2070(3002 ) 3312(230) 2 2 510 230 300 3(207)103 (39.76)103 510 300 6(207)103 (39.76)103 (510) 7.99 mm Ans.
yA
The slope at A, relative to the z axis is
A z
F1a1 2 d F (x l) ( x l )2 a2 (3x l ) (l a12 ) 2 6 EIl x l a2 dx 6 EI
F1a1 2 F l a12 2 3( x l ) 2 3a2 ( x l ) a2 (3 x l ) x l a2 6 EIl 6 EI Fa F 1 1 (l 2 a12 ) 2 3a2 2 2la2 6 EIl 6 EI 3312(230) 5102 2302 6(207)103 (39.76)103 (510) 2070 3(3002 ) 2(510)(300) 3 3 6(207)10 (39.76)10 0.0304 rad Ans. ______________________________________________________________________________
4-25 From the solutions to Prob. 3-70, T1 392.16 lbf and T2 58.82 lbf
I
d4
64 From Table A-9, beam 6,
(1) 4 64
0.049 09 in 4
Chapter 4 - Rev B, Page 17/81
( 350)(14)(8) Fb x 82 14 2 22 2 0.0452 in Ans. y A 1 1 x 2 b12 l 2 6 6 EIl x 8in 6(30)10 (0.049 09)(22) ( 450.98)(6)(8) F b x z A 2 2 ( x 2 b2 2 l 2 ) 82 6 2 22 2 0.0428 in Ans. 6 6 EIl 6(30)10 (0.049 09)(22) x 8in
The displacement magnitude is y A2 z A2 0.04522 0.04282 0.0622 in
Ans.
d y d Fb x Fb 1 1 x 2 b12 l 2 1 1 (3a12 b12 l 2 ) x a1 6 EIl d x x a1 dx 6 EIl
A z
( 350)(14) 3 82 14 2 22 2 0.00242 rad 6 6(30)10 (0.04909)(22)
Ans.
dz d F b x Fb 2 2 ( x 2 b22 l 2 ) 2 2 3a12 b22 l 2 x a1 6 EIl dx 6 EIl d x x a1
A y
(450.98)(6) 3 82 62 222 0.00356 rad 6 6(30)10 (0.04909)(22)
Ans.
The slope magnitude is A 0.002422 0.00356 0.00430 rad Ans. 2
______________________________________________________________________________ 4-26 From the solutions to Prob. 3-71, T1 250 N and T2 37.5 N
I
d4 64
(20) 4 64
7 854 mm 4
345sin 45o (550)(300) 3002 5502 8502 F b x y A 1 y 1 ( x 2 b12 l 2 ) 6(207)103 (7 854)(850) 6 EIl x 300mm 1.60 mm Ans. Fb x F b x z A 1z 1 ( x 2 b12 l 2 ) 2 2 ( x 2 b2 2 l 2 ) 6 EIl 6 EIl x 300mm
345cos 45 (550)(300) o
6(207)103 (7 854)(850)
300
2
5502 8502
287.5(150)(300) 3002 1502 8502 0.650 mm 3 6(207)10 (7 854)(850)
Ans.
The displacement magnitude is y A2 z A2 1.602 0.650 1.73 mm 2
Ans.
Chapter 4 - Rev B, Page 18/81
F1 y b1 d y d F1 y b1 x 2 2 2 (3a12 b12 l 2 ) x b1 l d x x a1 dx 6 EIl x a1 6 EIl
A z
345sin 45o (550)
3 3002 5502 8502 0.00243 rad 6(207)10 (7 854)(850)
Ans.
3
dz d F b x Fb x 1z 1 x 2 b12 l 2 2 2 x 2 b2 2 l 2 6 EIl x a1 dx 6 EIl d x x a1
A y
F1z b1 Fb 3a12 b12 l 2 2 2 3a12 b22 l 2 6 EIl 6 EIl o 345cos 45 (550) 3 3002 5502 8502 6(207)103 (7 854)(850)
287.5(150) 3 3002 1502 8502 1.91 104 rad 6(207)103 (7 854)(850)
Ans.
The slope magnitude is A 0.002432 0.0001912 0.00244 rad Ans. ______________________________________________________________________________ 4-27 From the solutions to Prob. 3-72, FB 750 lbf
I
d4 64
(1.25) 4 64
0.1198 in 4
From Table A-9, beams 6 (subscript 1) and 10 (subscript 2) F1 y b1 x 2 2 2 F2 y a2 x 2 yA x b1 l l x2 6 EIl 6 EIl x 16in
300 cos 20 (14)(16) o
16 6(30)10 (0.119 8)(30)
2
6
0.0805 in
14 30 2
2
750sin 20 (9)(16) o
30 6(30)10 (0.119 8)(30)
2
162
2
162
6
Ans.
F ax F b x z A 1z 1 x 2 b12 l 2 2 z 2 l 2 x 2 6 EIl 6 EIl x 16in
300sin 20 (14)(16) o
6(30)106 (0.119 8)(30) Ans. 0.1169 in
16
2
14 30 2
2
750 cos 20 (9)(16) o
6(30)106 (0.119 8)(30)
30
The displacement magnitude is y A2 z A2 0.08052 0.1169 0.142 in 2
Ans.
Chapter 4 - Rev B, Page 19/81
d F1 y b1 x 2 F2 y a2 x 2 d y x b12 l 2 l x2 6 EIl d x x a1 dx 6 EIl x a1
A z
3a b l 6EIl l 3a 6 EIl 300 cos 20 (14) 3 16 14 30 6(30)10 (0.119 8)(30)
F1 y b1
2 1
2 1
2
F2 y a2
2
2 1
o
2
2
2
6
750sin 20 (9) o
302 3 162 8.06 105 rad 6(30)10 (0.119 8)(30) 6
Ans.
dz d F b x F ax 1z 1 x 2 b12 l 2 2 z 2 l 2 x 2 6 EIl x a1 dx 6 EIl d x x a1
A y
F1z b1 F a 3a12 b12 l 2 2 z 2 l 2 3a12 6 EIl 6 EIl o 750 cos 20o (9) 300sin 20 (14) 2 2 2 302 3 162 3 16 14 30 6 6 6(30)10 (0.119 8)(30) 6(30)10 (0.119 8)(30)
0.00115 rad
Ans.
The slope magnitude is A 8.06 105 0.001152 0.00115 rad Ans. ______________________________________________________________________________ 2
4-28 From the solutions to Prob. 3-73, F B = 22.8 (103) N 4 d 4 50 306.8 103 mm 4 I 64 64 From Table A-9, beam 6, F bx F b x y A 1 y 1 ( x 2 b12 l 2 ) 2 y 2 ( x 2 b2 2 l 2 ) 6 EIl 6 EIl x 400mm
11103 sin 20o (650)(400) 4002 6502 10502 3 3 6(207)10 (306.8)10 (1050)
22.8 103 sin 25o (300)(400) 4002 3002 1050 2 3 3 6(207)10 (306.8)10 (1050) 3.735 mm Ans.
Chapter 4 - Rev B, Page 20/81
F bx F b x z A 1z 1 ( x 2 b12 l 2 ) 2 z 2 ( x 2 b2 2 l 2 ) 6 EIl 6 EIl x 400mm
11103 cos 20o (650)(400) 4002 6502 10502 3 6(207)10 (306.8)103 (1050) 22.8 103 cos 25o (300)(400) 4002 3002 10502 1.791 mm 6(207)103 (306.8)103 (1050)
The displacement magnitude is y A2 z A2
3.735
2
1.7912 4.14 mm
Ans.
Ans.
d y F bx d F b x 1z 1 x 2 b12 l 2 2 z 2 x 2 b2 2 l 2 6 EIl x a1 d x x a1 dx 6 EIl
A z
3a b l 6EIl 3a b l 6 EIl 1110 sin 20 (650) 3 400 650 6(207)10 (306.8)10 (1050)
F1 y b1
2 1
3
2 1
F2 y b2
2
2 1
2 2
o
2
3
2
3
2
10502
22.8 103 sin 25o (300) 3 4002 3002 10502 6(207)103 (306.8)103 (1050) 0.00507 rad Ans. dz d F b x F bx 1z 1 x 2 b12 l 2 2 z 2 x 2 b2 2 l 2 6 EIl x a1 dx 6 EIl d x x a1
A y
F1z b1 F b 3a12 b12 l 2 2 z 2 3a12 b22 l 2 6 EIl 6 EIl 11103 cos 20o (650) 3 4002 6502 10502 3 3 6(207)10 (306.8)10 (1050)
22.8 103 cos 25o (300) 3 4002 3002 10502 3 3 6(207)10 (306.8)10 (1050) 0.00489 rad Ans.
The slope magnitude is A
0.00507 0.00489 2
2
0.00704 rad Ans.
______________________________________________________________________________ 4-29 From the solutions to Prob. 3-68, T 1 = 60 lbf and T 2 = 400 lbf , and Prob. 4-23, I = 0.119 8 in4. From Table A-9, beam 6,
Chapter 4 - Rev B, Page 21/81
dz F bx d F1z b1 x 2 x b12 l 2 2 z 2 x 2 b22 l 2 6 EIl x 0 dx 6 EIl d x x 0 F b F b 575(30) 1z 1 b12 l 2 2 z 2 b2 2 l 2 302 402 6 EIl 6 EIl 6(30)106 (0.119 8)(40) 460(12) 122 402 0.00468 rad Ans. 6 6(30)10 (0.119 8)(40)
O y
d F1z a1 l x 2 F a l x 2 dz x a12 2lx 2 z 2 x a22 2lx 6 EIl d x x l x l dx 6 EIl F a F a 1z 1 6lx 2l 2 3 x 2 a12 2 z 2 6lx 2l 2 3 x 2 a22 6 EIl 6 EIl x l
C y
F1z a1 2 F a l a12 2 z 2 l 2 a22 6 EIl 6 EIl 2 460(28) 402 282 575(10) 40 102 0.00219 rad Ans. 6(30)106 (0.119 8)(40) 6(30)106 (0.119 8)(40) ______________________________________________________________________________
4-30 From the solutions to Prob. 3-69, T 1 = 2 880 N and T 2 = 432 N, and Prob. 4-24, I = 39.76 (103) mm4. From Table A-9, beams 6 and 10 d y Fa x d Fb x O z 1 1 ( x 2 b12 l 2 ) 2 2 (l 2 x 2 ) 6 EIl x 0 d x x 0 dx 6 EIl
Fa Fb Fal Fb 1 1 (3x 2 b12 l 2 ) 2 2 (l 2 3x 2 ) 1 1 (b12 l 2 ) 2 2 6 EIl 6 EI 6 EIl x 0 6 EIl 2 070(300)(510) 3 312(280) 2802 5102 3 3 6(207)103 (39.76)103 6(207)10 (39.76)10 (510) 0.0131 rad Ans. d y d F1a1 (l x) 2 Fa x ( x a12 2lx) 2 2 (l 2 x 2 ) 6 EIl x l d x x l dx 6 EIl
C z
Fa Fa Fal Fa 1 1 (6lx 2l 2 3 x 2 a12 ) 2 2 (l 2 3 x 2 ) 1 1 (l 2 a12 ) 2 2 6 EIl 3EI 6 EIl x l 6 EIl 2 070(300)(510) 3 312(230) (5102 2302 ) 3 3 6(207)10 (39.76)10 (510) 3(207)103 (39.76)103 0.0191 rad Ans. ______________________________________________________________________________
4-31 From the solutions to Prob. 3-70, T 1 = 392.19 lbf and T 2 = 58.82 lbf , and Prob. 4-25, I = 0.0490 9 in4. From Table A-9, beam 6
Chapter 4 - Rev B, Page 22/81
F1 y b1 2 2 d y d F1 y b1 x 2 x b12 l 2 (b1 l ) d x x 0 dx 6 EIl x 0 6 EIl 350(14) 142 222 0.00726 rad Ans. 6 6(30)10 (0.04909)(22)
O z
dz d F2 z b2 x 2 F b x b22 l 2 2 z 2 b22 l 2 6 EIl x 0 dx 6 EIl d x x 0 450.98(6) 6 2 22 2 6 6(30)10 (0.0490 9)(22) Ans. 0.00624 rad
O y
The slope magnitude is O 0.007262 0.00624 0.00957 rad Ans. 2
d F1 y a1 (l x) 2 d y x a12 2lx d x x l dx 6 EIl x l F1 a1 F1 a1 y 6lx 2l 2 3x 2 a12 y (l 2 a12 ) 6 EIl x l 6 EIl 350(8) 222 82 0.00605 rad Ans. 6 6(30)10 (0.0491)(22)
C z
dz d F2 z a2 (l x ) 2 x a22 2lx x l dx 6 EIl d x x l
C y
F a F a 2 z 2 6lx 2l 2 3 x 2 a22 2 z 2 l 2 a22 6 EIl 6 EIl x l 450.98(16) 222 162 0.00846 rad 6(30)106 (0.0490 9)(22)
The slope magnitude is C
0.00605
2
Ans.
0.008462 0.0104 rad Ans.
______________________________________________________________________________ 4-32 From the solutions to Prob. 3-71, T 1 =250 N and T 1 =37.5 N, and Prob. 4-26, I = 7 854 mm4. From Table A-9, beam 6
d F1 y b1 x 2 F1 y b1 2 2 d y x b12 l 2 (b1 l ) d x x 0 dx 6 EIl x 0 6 EIl
O z
345sin 45o (550) 5502 8502 0.00680 rad 3 6(207)10 (7 854)(850)
Ans.
Chapter 4 - Rev B, Page 23/81
dz d F1z b1 x 2 2 2 F2 z b2 x 2 2 2 x b1 l 6EIl x b2 l dx 6 EIl x 0 d x x 0
O y
345cos 45o (550) F1z b1 2 2 F2 z b2 2 2 b1 l b2 l 5502 8502 3 6 EIl 6 EIl 6(207)10 (7 854)(850) 287.5(150) 1502 8502 0.00316 rad Ans. 6(207)103 (7 854)(850)
The slope magnitude is O 0.00680 2 0.00316 2 0.00750 rad Ans. d F1 y a1 (l x) 2 F1 y a1 d y x a12 2lx 6lx 2l 2 3x 2 a12 d x x l dx 6 EIl x l 6 EIl x l
C z
345sin 45o (300) (l 2 a12 ) 8502 3002 0.00558 rad 3 6 EIl 6(207)10 (7 854)(850)
F1 y a1
Ans.
dz d F1z a1 (l x) 2 F a (l x) 2 x a12 2lx 2 z 2 x a22 2lx 6 EIl x l dx 6 EIl d x x l
C y
345cos 45o (300) F1z a1 2 F2 z a2 2 2 2 l a1 6EIl l a2 6(207)103 (7 854)(850) 8502 3002 6 EIl 287.5(700) 8502 7002 6.04 105 rad Ans. 3 6(207)10 (7 854)(850)
The slope magnitude is C
0.00558
2
6.04 105 0.00558 rad Ans. 2
________________________________________________________________________ 4-33 From the solutions to Prob. 3-72, F B = 750 lbf, and Prob. 4-27, I = 0.119 8 in4. From Table A-9, beams 6 and 10 d F b x F ax d y O z 1 y 1 x 2 b12 l 2 2 y 2 l 2 x 2 6 EIl d x x 0 dx 6 EIl x 0 F a F b F al F b 1 y 1 3x 2 b12 l 2 2 y 2 l 2 3 x 2 1 y 1 b12 l 2 2 y 2 6 EIl 6 EI 6 EIl x 0 6 EIl 300 cos 20o (14) 750sin 20o (9)(30) 2 2 14 30 6(30)106 (0.119 8) 0.00751 rad 6(30)106 (0.119 8)(30)
Ans.
Chapter 4 - Rev B, Page 24/81
dz F ax d F1z b1 x 2 x b12 l 2 2 z 2 l 2 x 2 6 EIl x 0 dx 6 EIl d x x 0
O y
F a F b F al F b 1z 1 3x 2 b12 l 2 2 z 2 l 2 3 x 2 1z 1 b12 l 2 2 z 2 6 EIl 6 EIl 6 EI 6 EIl x 0 750 cos 20o (9)(30) 300sin 20o (14) 2 2 14 30 6(30)106 (0.119 8) 0.0104 rad 6(30)106 (0.119 8)(30)
Ans.
The slope magnitude is O 0.007512 0.0104 2 0.0128 rad Ans. F ax dy d F1 y a1 (l x) 2 x a12 2lx 2 y 2 l 2 x 2 6 EIl dx x l dx 6 EIl x l F a F a F al F a 1 y 1 6lx 2l 2 3x 2 a12 2 y 2 l 2 3x 2 1 y 1 (l 2 a12 ) 2 y 2 6 EIl 3EI 6 EIl x l 6 EIl
C z
300 cos 20o (16) 750sin 20o (9)(30) 2 2 30 16 3(30)106 (0.119 8) 0.0109 rad 6(30)106 (0.119 8)(30)
Ans.
dz F ax d F1z a1 (l x) 2 x a12 2lx 2 z 2 l 2 x 2 6 EIl x l dx 6 EIl d x x l
C y
F a F a F al F a 1z 1 6lx 2l 2 3 x 2 a12 2 z 2 l 2 3 x 2 1z 1 l 2 a12 2 z 2 6 EIl 6 EIl 3EI 6 EIl x l 750 cos 20o (9)(30) 300sin 20o (16) 2 2 30 16 3(30)106 (0.119 8) 0.0193 rad 6(30)106 (0.119 8)(30) The slope magnitude is C
0.0109 0.0193 2
2
Ans.
0.0222 rad Ans.
______________________________________________________________________________ 4-34 From the solutions to Prob. 3-73, F B = 22.8 kN, and Prob. 4-28, I = 306.8 (103) mm4. From Table A-9, beam 6
d F1 y b1 x 2 2 2 F2 y b2 x 2 2 2 d y x b1 l 6EIl x b2 l d x x 0 dx 6 EIl x 0
O z
11103 sin 20o (650) 6502 10502 b l b l 3 3 6 EIl 6 EIl 6(207)10 (306.8)10 (1050) F1 y b1
2 1
2
F2 y b2
2 2
2
22.8 103 sin 25o (300) 3002 10502 0.0115 rad 6(207)103 (306.8)103 (1050)
Ans.
Chapter 4 - Rev B, Page 25/81
dz d F1z b1 x 2 F bx x b12 l 2 2 z 2 x 2 b22 l 2 6 EIl x 0 dx 6 EIl d x x 0 F b F b 1z 1 b12 l 2 2 z 2 b22 l 2 6 EIl 6 EIl 11103 cos 20o (650) 6502 10502 3 3 6(207)10 (306.8)10 (1050)
O y
22.8 103 cos 25o (300) 3002 10502 0.00427 rad 6(207)103 (306.8)103 (1050)
The slope magnitude is O
0.0115 0.00427 2
2
Ans.
0.0123 rad Ans.
F2 y a2 (l x) 2 d y d F1 y a1 (l x) 2 x a12 2lx x a22 2lx 6 EIl d x x l dx 6 EIl x l F2 y a2 F1 y a1 (6lx 2l 2 3 x 2 a12 ) 6lx 2l 2 3x 2 a22 6 EIl 6 EIl x l
C z
11103 sin 20o (400) 10502 4002 l a l a 3 3 6(207)10 (306.8)10 (1050) 6 EIl 6 EIl F1 y a1
2
2 1
F2 y a2
2
2 2
22.8 103 sin 25o (750) 10502 7502 0.0133 rad 6(207)103 (306.8)103 (1050)
Ans.
dz d F1z a1 (l x) 2 F a (l x) 2 x a12 2lx 2 z 2 x a22 2lx 6 EIl x l dx 6 EIl d x x l
C y
F a F a 1z 1 6lx 2l 2 3 x 2 a12 2 z 2 6lx 2l 2 3 x 2 a22 6 EIl 6 EIl x l
11103 cos 20o (400) F1z a1 2 F2 z a2 2 2 2 l a1 6 EIl l a2 6(207)10 10502 4002 3 6 EIl (306.8)103 (1050) 22.8 103 cos 25o (750) 10502 7502 0.0112 rad 3 3 6(207)10 (306.8)10 (1050)
Ans.
The slope magnitude is C 0.01332 0.01122 0.0174 rad Ans. ______________________________________________________________________________ 4-35 The required new slope in radians is new = 0.06( /180) = 0.00105 rad. In Prob. 4-29, I = 0.119 8 in4, and it was found that the greater angle occurs at the bearing at O where ( O ) y = 0.00468 rad.
Since is inversely proportional to I,
Chapter 4 - Rev B, Page 26/81
new I new = old I old
4 I new = d new /64 = old I old / new
1/4
or,
d new
64 old I old new
The absolute sign is used as the old slope may be negative. 1/4
64 0.00468 d new 0.119 8 1.82 in Ans. 0.00105 ______________________________________________________________________________
4-36 The required new slope in radians is new = 0.06( /180) = 0.00105 rad. In Prob. 4-30, I = 39.76 (103) mm4, and it was found that the greater angle occurs at the bearing at C where ( C ) y = 0.0191 rad.
See the solution to Prob. 4-35 for the development of the equation 1/4
d new
64 old I old new
1/ 4
64 0.0191 d new 39.76 103 62.0 mm Ans. 0.00105 ______________________________________________________________________________
4-37 The required new slope in radians is new = 0.06( /180) = 0.00105 rad. In Prob. 4-31, I = 0.0491 in4, and the maximum slope is C = 0.0104 rad.
See the solution to Prob. 4-35 for the development of the equation 1/4
d new
64 old I old new
1/4
64 0.0104 d new 0.0491 1.77 in Ans. 0.00105 ______________________________________________________________________________
4-38 The required new slope in radians is new = 0.06( /180) = 0.00105 rad. In Prob. 4-32, I = 7 854 mm4, and the maximum slope is O = 0.00750 rad.
See the solution to Prob. 4-35 for the development of the equation
Chapter 4 - Rev B, Page 27/81
1/4
d new
64 old I old new
1/4
64 0.00750 d new 7 854 32.7 mm Ans. 0.00105 ______________________________________________________________________________
4-39 The required new slope in radians is new = 0.06( /180) = 0.00105 rad. In Prob. 4-33, I = 0.119 8 in4, and the maximum slope = 0.0222 rad.
See the solution to Prob. 4-35 for the development of the equation 1/4
d new
64 old I old new
1/4
64 0.0222 d new 0.119 8 2.68 in Ans. 0.00105 ______________________________________________________________________________
4-40 The required new slope in radians is new = 0.06( /180) = 0.00105 rad. In Prob. 4-34, I = 306.8 (103) mm4, and the maximum slope is C = 0.0174 rad.
See the solution to Prob. 4-35 for the development of the equation 1/4
d new
64 old I old new
1/4
64 0.0174 d new 306.8 103 100.9 mm Ans. 0.00105 ______________________________________________________________________________
4-41 I AB = 14/64 = 0.04909 in4, J AB = 2 I AB = 0.09818 in4, I BC = (0.25)(1.5)3/12 = 0.07031 in4, I CD = (3/4)4/64 = 0.01553 in4. For Eq. (3-41), p. 102, b/c = 1.5/0.25 = 6 = 0.299.
The deflection can be broken down into several parts 1. The vertical deflection of B due to force and moment acting on B (y 1 ). 2. The vertical deflection due to the slope at B, B1 , due to the force and moment acting on B (y 2 = CD B1 = 2 B1 ).
Chapter 4 - Rev B, Page 28/81
3. The vertical deflection due to the rotation at B, B2 , due to the torsion acting at B (y 3 = BC B1 = 5 B1 ). 4. The vertical deflection of C due to the force acting on C (y 4 ). 5. The rotation at C, C , due to the torsion acting at C (y 3 = CD C = 2 C ). 6. The vertical deflection of D due to the force acting on D (y 5 ). 1. From Table A-9, beams 1 and 4 with F = 200 lbf and M B = 2(200) = 400 lbfin 200 63 400 62 y1 0.01467 in 3 30 106 0.04909 2 30 106 0.04909 2. From Table A-9, beams 1 and 4 d Fx 2 M B x 2 M x Fx B1 x 3l 3x 6l B EI x l 2 EI x l 6 EI dx 6 EI 6 l 200 6 2 400 0.004074 rad Fl 2M B 6 2 EI 2 30 10 0.04909
y 2 = 2(0.004072) = 0.00815 in 3. The torsion at B is T B = 5(200) = 1000 lbfin. From Eq. (4-5) TL
1000 6
0.005314 rad B2 6 JG AB 0.09818 11.5 10
y 3 = 5(0.005314) = 0.02657 in 4. For bending of BC, from Table A-9, beam 1 y4
200 53
3 30 106 0.07031
0.00395 in
5. For twist of BC, from Eq. (3-41), p. 102, with T = 2(200) = 400 lbfin
C
400 5 0.02482 rad 0.299 1.5 0.253 11.5 106
y 5 = 2(0.02482) = 0.04964 in 6. For bending of CD, from Table A-9, beam 1 y6
200 23
3 30 106 0.01553
0.00114 in
Chapter 4 - Rev B, Page 29/81
Summing the deflections results in 6
yD yi 0.01467 0.00815 0.02657 0.00395 0.04964 0.00114 0.1041 in Ans. i 1
This problem is solved more easily using Castigliano’s theorem. See Prob. 4-71. ______________________________________________________________________________ 4-42 The deflection of D in the x direction due to F z is from: 1. The deflection due to the slope at B, B1 , due to the force and moment acting on B (x 1 = BC B1 = 5 B1 ). 2. The deflection due to the moment acting on C (x 2 ).
1. For AB, I AB = 14/64 = 0.04909 in4. From Table A-9, beams 1 and 4 2 M B x 2 M x d Fx Fx x l 3 3x 6l B EI x l 2 EI x l 6 EI dx 6 EI
B1
6 l 100 6 2 200 0.002037 rad Fl 2M B 6 2 EI 2 30 10 0.04909
x 1 = 5( 0.002037) = 0.01019 in 2. For BC, I BC = (1.5)(0.25)3/12 = 0.001953 in4. From Table A-9, beam 4 x2
2 100 5 M Cl 2 0.04267 in 2 EI 2 30 106 0.001953
The deflection of D in the x direction due to F x is from: 3. The elongation of AB due to the tension. For AB, the area is A = 12/4 = 0.7854 in2 150 6 Fl x3 3.82 105 in 6 AE 0.7854 30 10 AB
4. The deflection due to the slope at B, B2 , due to the moment acting on B (x 1 = BC B2 = 5 B2 ). With I AB = 0.04907 in4,
B2
5 150 6 M Bl 0.003056 rad EI 30 106 0.04909
Chapter 4 - Rev B, Page 30/81
x 4 = 5( 0.003056) = 0.01528 in 5. The deflection at C due to the bending force acting on C. With I BC = 0.001953 in4 150 53 Fl 3 x5 0.10667 in 3 30 106 0.001953 3EI BC
6. The elongation of CD due to the tension. For CD, the area is A = (0.752)/4 = 0.4418 in2 150 2 Fl x6 2.26 105 in 6 AE CD 0.4418 30 10
Summing the deflections results in 6
xD xi 0.01019 0.04267 3.82 105 i 1
0.01528 0.10667 2.26 105 0.1749 in Ans.
______________________________________________________________________________ 4-43
J OA = J BC = (1.54)/32 = 0.4970 in4, J AB = (14)/32 = 0.09817 in4, I AB = (14)/64 = 0.04909 in4, and I CD = (0.754)/64 = 0.01553 in4. T lOA l AB lBC Tl Tl Tl GJ OA GJ AB GJ BC G J OA J AB J BC 250(12) 2 9 2 0.0260 rad 6 11.5(10 ) 0.4970 0.09817 0.4970 Simplified Tl 250(12)(13) s GJ 11.5 106 0.09817
Ans.
s 0.0345 rad
Ans. Simplified is 0.0345/0.0260 = 1.33 times greater Ans. yD
Fy lOC 3 3EI AB
s lCD
Fy lCD 3 3EI CD
250 133
3(30)106 0.04909
0.0345(12)
250 123
3(30)106 0.01553
yD 0.847 in Ans. ______________________________________________________________________________
4-44 Reverse the deflection equation of beam 7 of Table A-9. Using units in lbf, inches
Chapter 4 - Rev B, Page 31/81
y
3000 /12 x 2 25 x 2 x3 25 12 3 wx 2lx 2 x3 l 3 24 EI 24 30 106 485
7.159 1010 x 27 106 600 x 2 x3
Ans.
The maximum height occurs at x = 25(12)/2 = 150 in ymax 7.159 1010 150 27 106 600 1502 1503 1.812 in
Ans.
______________________________________________________________________________ 4-45 From Table A-9-6, Fbx 2 x b2 l 2 6 EIl Fb 3 yL x b2 x l 2 x 6 EIl dyL Fb 3x2 b2 l 2 dx 6 EIl yL
dyL dx Let
dyL dx
Fb b 2 l 2 6 EIl
x 0
and set I x 0
dL
d L4 64
. Thus,
32 Fb b 2 l 2
1/ 4
Ans.
3 El
For the other end view, observe beam 6 of Table A-9 from the back of the page, noting that a and b interchange as do x and –x dR
32 Fa l 2 a 2 3 El
1/4
Ans.
For a uniform diameter shaft the necessary diameter is the larger of d L and d R . ______________________________________________________________________________ 4-46 The maximum slope will occur at the left bearing. Incorporating a design factor into the solution for d L of Prob. 4-45,
Chapter 4 - Rev B, Page 32/81
32nFb l 2 b 2 d 3 El
1/4
d
4
32(1.28)(3000)(200) 3002 2002 3 (207)103 (300)(0.001)
d 38.1 mm
I
38.14
Ans.
103.4 103 mm 4
64 From Table A-9, beam 6, the maximum deflection will occur in BC where dy BC /dx = 0 d Fa l x 2 x a 2 2lx 0 3 x 2 6lx a 2 2l 2 0 dx 6 EIl
3x 2 6 300 x 1002 2 3002 0 x 2 600 x 63333 0 x
1 600 600 2 4(1)63 333 463.3, 136.7 mm 2
x = 136.7 mm is acceptable. Fa l x 2 ymax x a 2 2lx 6 EIl x 136.7 mm
3 103 100 300 136.7
136.7 2 1002 2 300 136.7 0.0678 mm 6 207 10 103.4 10 300 3
3
Ans.
______________________________________________________________________________ 4-47 I = (1.254)/64 = 0.1198 in4. From Table A-9, beam 6 2
F a (l x) 2 F b x ( x a12 2lx) 2 2 ( x 2 b2 2 l 2 ) 1 1 6 EIl 6 EIl
2
2 150(5)(20 8) 2 2 8 5 2(20)(8) 6 6(30)10 0.1198 (20)
250(10)(8) 2 2 2 8 10 20 6 6(30)10 0.1198 (20)
2
1/2
Ans. 0.0120 in ______________________________________________________________________________
Chapter 4 - Rev B, Page 33/81
4-48 I = (1.254)/64 = 0.1198 in4. For both forces use beam 6 of Table A-9. For F 1 = 150 lbf: 0x5 150 15 x Fb x y 1 1 x 2 b12 l 2 x 2 152 202 6 6 EIl 6 30 10 0.1198 20
5.217 106 x x 2 175
5 x 20 y
(1)
F1a1 l x 2 150 5 20 x x 2 52 2 20 x x a12 2lx 6 6 EIl 6 30 10 0.1198 20
1.739 106 20 x x 2 40 x 25
(2)
For F 2 = 250 lbf: 0 x 10 250 10 x Fb x z 2 2 x 2 b22 l 2 x 2 102 202 6 6 EIl 6 30 10 0.1198 20 5.797 106 x x 2 300
(3)
10 x 20 F a l x 2 250 10 20 x x 2 102 2 20 x z 2 2 x a22 2lx 6 EIl 6 30 106 0.1198 20 5.797 106 20 x x 2 40 x 100
(4)
Plot Eqs. (1) to (4) for each 0.1 in using a spreadsheet. There are 201 data points, too numerous to tabulate here but the plot is shown below, where the maximum deflection of = 0.01255 in occurs at x = 9.9 in. Ans.
______________________________________________________________________________
Chapter 4 - Rev B, Page 34/81
4-49 The larger slope will occur at the left end. From Table A-9, beam 8 MBx 2 ( x 3a 2 6al 2l 2 ) 6 EIl dy AB M B (3 x 2 3a 2 6al 2l 2 ) 6 EIl dx y AB
With I = d 4/64, the slope at the left bearing is dy AB dx
A x0
MB (3a 2 6al 2l 2 ) 4 6 E d / 64 l
Solving for d 32M B 32(1000) 3(42 ) 6(4)(10) 2 102 d4 3a 2 6al 2l 2 4 6 3 E Al 3 (30)10 (0.002)(10) 0.461 in Ans. ______________________________________________________________________________ 4-50 From Table A-5, E = 10.4 Mpsi M O = 0 = 18 F BC 6(100) F BC = 33.33 lbf The cross sectional area of rod BC is A = (0.52)/4 = 0.1963 in2.
The deflection at point B will be equal to the elongation of the rod BC. 33.33(12) FL 6.79 105 in yB 6 AE BC 0.1963 30 10
Ans.
______________________________________________________________________________ 4-51 M O = 0 = 6 F AC 11(100)
F AC = 183.3 lbf
The deflection at point A in the negative y direction is equal to the elongation of the rod AC. From Table A-5, E s = 30 Mpsi. 183.3 12 FL yA 3.735 104 in 2 6 0.5 / 4 30 10 AE AC
By similar triangles the deflection at B due to the elongation of the rod AC is y A y B1 6 18
yB1 3 y A 3(3.735)104 0.00112 in
From Table A-5, E a = 10.4 Mpsi The bar can then be treated as a simply supported beam with an overhang AB. From Table A-9, beam 10
Chapter 4 - Rev B, Page 35/81
dy Fa 2 Fa 2 d F (x l) 2 yB 2 BD BC l a x l a x l ( ) 7 ( ) (3 ) (l a) x l a 3EI dx 6 EI dx x l a 3EI Fa 2 F Fa 2 7 Fa 2 3( x l ) 3a ( x l ) a (3 x l ) |x l a (l a ) (2l 3a) (l a) 7 3EI 6 EI 3EI 6 EI 100 52 7 100 5 (6 5) 2(6) 3(5) 3(10.4)106 0.25(23 ) / 12 6(10.4)106 0.25(23 ) / 12
0.01438 in
Ans. y B = y B1 + y B2 = 0.00112 0.01438 = 0.0155 in ______________________________________________________________________________ 4-52 From Table A-5, E = 207 GPa, and G = 79.3 GPa.
FlOC l AB 2 Fl AC l AB 2 Fl AB 3 Fl AB 3 Tl Tl yB l l AB AB 3EI AB G dOC 4 / 32 G d AC 4 / 32 3E d34 / 64 GJ OC GJ AC l AC 32 Fl AB 2 lOC 2l AB 4 4 4 GdOC Gd AC 3Ed AB The spring rate is k = F/ y B . Thus 32l AB 2 k
lOC l AC 2l AB 4 4 4 GdOC Gd AC 3Ed AB
1
32 2002 2 200 200 200 3 4 3 4 3 4 79.3 10 18 79.3 10 12 3 207 10 8
1
8.10 N/mm Ans. _____________________________________________________________________________
4-53 For the beam deflection, use beam 5 of Table A-9. F R1 R2 2 F F 1 , and 2 2k1 2k 2
y AB 1
1 2 l
x
Fx (4 x 2 3l 3 ) 48EI
1 k2 k1 x y AB F x (4 x 2 3l 3 ) 48EI 2k1 2k1k2l
Ans.
Chapter 4 - Rev B, Page 36/81
For BC, since Table A-9 does not have an equation (because of symmetry) an equation will need to be developed as the problem is no longer symmetric. This can be done easily using beam 6 of Table A-9 with a = l /2 F l / 2 l x 2 l 2 F Fk2 Fk1 yBC x x 2lx 2k1 2k1k 2l EIl 4 1 k2 k1 l x 4 x 2 l 2 8lx Ans. F x 48 EI 2k1 2k1k2l ______________________________________________________________________________
4-54
Fa F , and R2 (l a ) l l Fa F 1 , and 2 (l a ) lk1 lk2 R1
y AB 1
1 2 l
x
Fax 2 (l x 2 ) 6 EIl
a x ax 2 y AB F (l x 2 ) Ans. k a k1 l a 2 2 6 EIl k1l k1k2l F (x l) ( x l ) 2 a(3 x l ) yBC 1 1 2 x l 6 EI a x (x l) ( x l ) 2 a(3 x l ) yBC F Ans. k a k1 l a 2 2 6 EI k1l k1k2l ______________________________________________________________________________ 4-55 Let the load be at x ≥ l/2. The maximum deflection will be in Section AB (Table A-9, beam 6) y AB
Fbx 2 x b2 l 2 6 EIl
dy AB Fb 3x 2 b2 l 2 0 dx 6 EIl
x
l 2 b2 l2 , xmax 0.577l 3 3
3x 2 b2 l 2 0
Ans.
For x l/2, xmin l 0.577l 0.423l Ans. ______________________________________________________________________________
Chapter 4 - Rev B, Page 37/81
4-56
M O 1(3000)(1500) 2500(2000) 9.5 106 N·mm
RO 1(3000) 2500 5 500 N
From Prob. 4-10,
I 4.14(106 ) mm 4
x2 1 M 9.5 10 5500 x 2500 x - 2000 2 dy x3 2 EI 9.5 106 x 2 750 x 2 1250 x 2000 C1 dx 6 6
dy 0 at x 0 C1 0 dx dy x3 2 EI 9.5 106 x 2 750 x 2 1250 x 2000 dx 6 x4 3 6 2 3 EIy 4.75 10 x 916.67 x 416.67 x 2000 C2 24 y 0 at x 0 C2 0 , and therefore y
1 3 114 106 x 2 22 103 x 3 x 4 10 103 x 2000 24 EI
yB
1 114 106 30002 22 103 30003 24 207 103 4.14 106 3 30004 10 103 3000 2000
25.4 mm
Ans.
M O = 9.5 (106) Nm. The maximum stress is compressive at the bottom of the beam where y = 29.0 100 = 71 mm 9.5 106 (71) My 163 106 Pa 163MPa Ans. max 6 I 4.14(10 ) The solutions are the same as Prob. 4-10. ______________________________________________________________________________
4-57 See Prob. 4-11 for reactions: R O = 465 lbf and R C = 285 lbf. Using lbf and inch units
Chapter 4 - Rev B, Page 38/81
M = 465 x 450 x 721 300 x 1201 dy 2 2 EI 232.5 x 2 225 x 72 150 x 120 C1 dx EIy = 77.5 x3 75 x 723 50 x 1203 C 1 x y = 0 at x = 0 C 2 = 0 y = 0 at x = 240 in 0 = 77.5(2403) 75(240 72)3 50(240 120)3 + C 1 x and, EIy = 77.5 x3 75 x 723 50 x 1203 2.622(106) x
C 1 = 2.622(106) lbfin2
Substituting y = 0.5 in at x = 120 in gives 30(106) I ( 0.5) = 77.5 (1203) 75(120 72)3 50(120 120)3 2.622(106)(120) I = 12.60 in4 Select two 5 in 6.7 lbf/ft channels; from Table A-7, I = 2(7.49) = 14.98 in4 12.60 1 Ans. 0.421 in 14.98 2 The maximum moment occurs at x = 120 in where M max = 34.2(103) lbfin ymidspan
Mc 34.2(103 )(2.5) max 5 710 psi O.K. I 14.98 The solutions are the same as Prob. 4-17. ______________________________________________________________________________ 4-58 I = (1.54)/64 = 0.2485 in4, and w = 150/12 = 12.5 lbf/in. 1 24 RO 12.5 39 (340) 453.0 lbf 2 39 12.5 2 1 M 453.0 x x 340 x 15 2 dy 12.5 3 2 EI 226.5 x 2 x 170 x 15 C1 6 dx 3 3 EIy 75.5 x 0.5208 x 4 56.67 x 15 C1 x C2
y 0 at x 0 C2 0 y 0 at x 39 in
y
C1 6.385(10 4 ) lbf in 2 Thus,
1 3 75.5 x 3 0.5208 x 4 56.67 x 15 6.385 104 x EI
Evaluating at x = 15 in,
Chapter 4 - Rev B, Page 39/81
1 75.5 153 0.5208 154 56.67 15 15 3 6.385 104 (15) 30(10 )(0.2485) 0.0978 in Ans.
yA
6
1 75.5 19.53 0.5208 19.54 56.67 19.5 15 3 6.385 104 (19.5) 30(10 )(0.2485) 0.1027 in Ans.
ymidspan
6
5 % difference
Ans.
The solutions are the same as Prob. 4-12. ______________________________________________________________________________ 3 14 100 7 14 100 4-59 I = 0.05 in4, RA 420 lbf and RB 980 lbf 10 10 M = 420 x 50 x2 + 980 x 10 1 EI
dy 2 210 x 2 16.667 x 3 490 x 10 C1 dx 3
EIy 70 x 3 4.167 x 4 163.3 x 10 C1 x C2
y = 0 at x = 0 C 2 = 0 y = 0 at x = 10 in C 1 = 2 833 lbfin2. Thus, y
1 70 x 3 4.167 x 4 163.3 x 10 3 2833 x 6 30 10 0.05
3 6.667 107 70 x 3 4.167 x 4 163.3 x 10 2833 x Ans. The tabular results and plot are exactly the same as Prob. 4-21. ______________________________________________________________________________
4-60 R A = R B = 400 N, and I = 6(323) /12 = 16 384 mm4. First half of beam, M = 400 x + 400 x 300 1 dy 2 EI 200 x 2 200 x 300 C1 dx
From symmetry, dy/dx = 0 at x = 550 mm
0 = 200(5502) + 200(550 – 300) 2 + C 1
C 1 = 48(106) N·mm2
EIy = 66.67 x3 + 66.67 x 300 3 + 48(106) x + C 2
Chapter 4 - Rev B, Page 40/81
y = 0 at x = 300 mm
C 2 = 12.60(109) N·mm3.
The term (EI)1 = [207(103)16 384] 1 = 2.949 (1010 ) Thus y = 2.949 (1010) [ 66.67 x3 + 66.67 x 300 3 + 48(106) x 12.60(109)] y O = 3.72 mm Ans. y x = 550 mm =2.949 (1010) [ 66.67 (5503) + 66.67 (550 300)3 + 48(106) 550 12.60(109)] = 1.11 mm Ans. The solutions are the same as Prob. 4-13. ______________________________________________________________________________ 4-61 1 M A Fa l 1 M A 0 M A R2l F (l a) R2 l Fl Fa M A
M
B
0 R1l Fa M A
M R1 x M A R2 x l
R1
1
dy 1 1 2 R1 x 2 M A x R2 x l C1 dx 2 2 1 1 1 3 EIy R1 x 3 M A x 2 R2 x l C1 x C2 6 2 6 EI
y = 0 at x = 0
C2 = 0
y = 0 at x = l
1 1 C1 R1l 2 M Al . Thus, 6 2
EIy
y
1 1 1 1 3 1 R1 x 3 M A x 2 R2 x l R1l 2 M Al x 6 2 6 2 6
1 3 M A Fa x3 3M A x 2l Fl Fa M A x l Fal 2 2M Al 2 x 6 EIl
Ans.
In regions, 1 M A Fa x 3 3M A x 2l Fal 2 2 M Al 2 x 6 EIl x M A x 2 3lx 2l 2 Fa l 2 x 2 Ans. 6 EIl
y AB
Chapter 4 - Rev B, Page 41/81
1 3 M A Fa x3 3M A x 2l Fl Fa M A x l Fal 2 2M Al 2 x 6 EIl 1 3 3 M A x3 3 x 2l x l 2 xl 2 F ax3 l a x l axl 2 6 EIl 1 2 M A x l l 2 Fl x l x l a 3 x l 6 EIl
yBC
x l 6 EI
M l F x l a 3x l 2
A
Ans.
The solutions reduce to the same as Prob. 4-17. ______________________________________________________________________________ w b a 1 R1 4-62 M D 0 R1l w b a l b b a 2l b a 2 2l w w 2 2 M R1 x xa xb 2 2 dy 1 w w 3 3 EI R1 x 2 xa x b C1 6 6 dx 2 1 w w 4 4 EIy R1 x 3 xa x b C1 x C2 6 24 24 y = 0 at x = 0 y = 0 at x = l
C2 = 0
1 1 w w 4 4 C1 R1l 3 l a l b l 6 24 24
y
1 EI
1 w b a w w 4 2l b a x3 x a x b 2l 24 24 6
4
1 1 w b a w w 4 4 x 2l b a l 3 l a l b l 6 2l 24 24
w 4 2 b a 2l b a x 3 l x a l x b 24 EIl
4
4 4 x 2 b a 2l b a l 2 l a l b
Ans.
The above answer is sufficient. In regions,
Chapter 4 - Rev B, Page 42/81
w 4 4 2 b a 2l b a x3 x 2 b a 2l b a l 2 l a l b 24 EIl wx 4 4 2 b a 2l b a x 2 2 b a 2l b a l 2 l a l b 24 EIl
y AB
yBC
w 4 2 b a 2l b a x3 l x a 24 EIl
4 4 x 2 b a 2l b a l 2 l a l b
yCD
w 4 4 2 b a 2l b a x 3 l x a l x b 24 EIl
4 4 x 2 b a 2l b a l 2 l a l b
These equations can be shown to be equivalent to the results found in Prob. 4-19. ______________________________________________________________________________ 4-63 I 1 = (1.3754)/64 = 0.1755 in4, I 2 = (1.754)/64 = 0.4604 in4,
R 1 = 0.5(180)(10) = 900 lbf Since the loading and geometry are symmetric, we will only write the equations for half the beam 2 For 0 x 8 in M 900 x 90 x 3 At x = 3, M = 2700 lbfin Writing an equation for M / I, as seen in the figure, the magnitude and slope reduce since I 2 > I 1 . To reduce the magnitude at x = 3 in, we add the term, 2700(1/I 1 1/ I 2 ) x 3 0. The slope of 900 at x = 3 in is also reduced. We account for this with a ramp function, x 31 . Thus, 1 1 1 1 M 900 x 90 0 1 2700 x 3 900 x 3 x 3 I I1 I2 I1 I 2 I1 I 2 0
1
5128 x 9520 x 3 3173 x 3 195.5 x 3 E
2
2
dy 1 2 3 2564 x 2 9520 x 3 1587 x 3 65.17 x 3 C1 dx
Boundary Condition:
dy 0 at x 8 in dx
Chapter 4 - Rev B, Page 43/81
0 2564 8 9520 8 3 1587 8 3 65.17 8 3 C1 2
2
3
C 1 = 68.67 (103) lbf/in2 2
3
4
Ey 854.7 x 3 4760 x 3 529 x 3 16.29 x 3 68.67(103 ) x C2
y = 0 at x = 0
C2 = 0
Thus, for 0 x 8 in 1 2 3 4 y 854.7 x3 4760 x 3 529 x 3 16.29 x 3 68.7(103 ) x 6 30(10 )
Ans.
Using a spreadsheet, the following graph represents the deflection equation found above
The maximum is ymax 0.0102 in at x 8 in Ans. ______________________________________________________________________________ 4-64 The force and moment reactions at the left support are F and Fl respectively. The bending moment equation is
M = Fx Fl Plots for M and M /I are shown. M /I can be expressed using singularity functions M F Fl Fl l x x I 2 I1 2 I1 4 I1 2
0
F l x 2 I1 2
1
Chapter 4 - Rev B, Page 44/81
where the step down and increase in slope at x = l /2 are given by the last two terms. Integrate dy F 2 Fl Fl l E x x x dx 4 I1 2 I1 4 I1 2 dy/dx = 0 at x = 0 C 1 = 0
1
F l x 4 I1 2
2
C1
2
3
F 3 Fl 2 Fl l F l Ey x x x x C2 12 I1 4 I1 8 I1 2 12 I1 2 y = 0 at x = 0 C 2 = 0 2 3 l F 3 l 2 2 x y 2 x 6lx 3l x 24 EI1 2 2 3 2 F l 5 Fl 3 l y x l /2 Ans. 2 6l 3l (0) 2(0) 24 EI1 2 96 EI1 2 2 3 F l 3Fl 3 3 l 2 y x l Ans. 2 l 6l l 3l l 2 x 24 EI1 2 16 EI1 2
The answers are identical to Ex. 4-10. ______________________________________________________________________________ 4-65 Place a dummy force, Q, at the center. The reaction, R 1 = wl / 2 + Q / 2 M x wx2 wl Q M x Q 2 2 2 2 Integrating for half the beam and doubling the results 1 ymax 2 EI
l /2
0
M M Q
2 dx Q 0 EI
l /2
0
wl wx2 x x 2 2 dx 2
Note, after differentiating with respect to Q, it can be set to zero l/2
w x 3l x 4 5w ymax x l x dx Ans. 0 2 EI 3 4 0 384 EI ______________________________________________________________________________ w 2 EI
l /2
2
4-66 Place a fictitious force Q pointing downwards at the end. Use the variable x originating at the free end at positive to the left M wx2 x M Qx Q 2
Chapter 4 - Rev B, Page 45/81
l l 1 l M 1 wx 2 w ymax M dx x dx x 3 dx EI Q EI EI 2 2 Q 0 0 0 0
wl 4 Ans. 8 EI ______________________________________________________________________________
4-67 From Table A-7, I 1-1 = 1.85 in4. Thus, I = 2(1.85) = 3.70 in4
First treat the end force as a variable, F. Adding weight of channels of 2(5)/12 = 0.833 lbf/in. Using the variable x as shown in the figure M F x
5.833 2 x F x 2.917 x 2 2
M x F
A
1 EI
60
0
M
M 1 dx EI F
60
0
( F x 2.917 x 2 )( x ) d x
(150 / 3)(603 ) (2.917 / 4)(604 ) 0.182 in in the direction of the 150 lbf force 30(106 )(3.70) y A 0.182 in Ans. ______________________________________________________________________________
4-68 The energy includes torsion in AC, torsion in CO, and bending in AB.
Neglecting transverse shear in AB M Fx,
M x F
In AC and CO, T T Fl AB , l AB F The total energy is T 2l T 2l U 2GJ AC 2GJ CO
l AB
0
M2 dx 2 EI AB
The deflection at the tip is
Chapter 4 - Rev B, Page 46/81
U Tl AC T TlCO T F GJ AC F GJ CO F
l AB
0
Tl l Tl l M M 1 dx AC AB CO AB EI 3 F GJ AC GJ CO EI AB
l AB
Fx dx 2
0
2 2 3 3 Tl AC l AB TlCO l AB Fl AB Fl AC l AB FlCO l AB Fl AB 4 4 4 GJ AC GJ CO 3EI AB G d AC / 32 G dCO / 32 3E d AB / 64
2 l AC l 32 Fl AB 2l AB CO4 4 4 Gd AC GdCO 3Ed AB
k
F
2 32l AB
l AC l 2l AB CO4 4 4 Gd AC Gd CO 3Ed AB
1
1
2 200 200 200 8.10 N/mm Ans. 32 2002 79.3 103 184 79.3 103 124 3 207 103 84 ______________________________________________________________________________
4-69 I 1 = (1.3754)/64 = 0.1755 in4, I 2 = (1.754)/64 = 0.4604 in4
Place a fictitious force Q pointing downwards at the midspan of the beam, x = 8 in R1
1 1 (10)180 Q 900 0.5Q 2 2
For 0 x 3 in M 900 0.5Q x
M 0.5 x Q M 0.5 x Q
For 3 x 13 in M 900 0.5Q x 90( x 3)2
By symmetry it is equivalent to use twice the integral from 0 to 8 3 8 8 M M 1 1 2 2 2 dx 900 x dx 900 x 90 x 3 x dx EI 2 3 0 EI Q Q 0 EI1 0 3
300 x 3 1 EI1 0 EI 2
8
1 4 9 2 3 3 300 x 90( 4 x 2 x 2 x ) 3
120.2 103 8100 1 8100 3 3 145.5 10 25.3110 30 106 0.1755 30 106 0.4604 EI1 EI 2 Ans. 0.0102 in ______________________________________________________________________________ Chapter 4 - Rev B, Page 47/81
4-70 I = (0.54)/64 = 3.068 (103) in4, J = 2 I = 6.136 (103) in4, A = (0.52)/4 = 0.1963 in2.
Consider x to be in the direction of OA, y vertically upward, and z in the direction of AB. Resolve the force F into components in the x and y directions obtaining 0.6 F in the horizontal direction and 0.8 F in the negative vertical direction. The 0.6 F force creates strain energy in the form of bending in AB and OA, and tension in OA. The 0.8 F force creates strain energy in the form of bending in AB and OA, and torsion in OA. Use the dummy variable x to originate at the end where the loads are applied on each segment, 0.6 F: AB
OA
M 0.6 x F
M 0.6 F x
M 4.2 F
Fa 0.6 F
M 4.2 F Fa 0.6 F
0.8 F: AB
M 0.8 F x
M 0.8 x F
OA
M 0.8 F x
M 0.8 x F
T 5.6 F Once the derivatives are taken the value of F = 15 lbf can be substituted in. The deflection of B in the direction of F is* T 5.6 F
B F
U Fa L Fa TL T 1 M M dx F AE OA F JG OA F EI F 0.6 15 15 5.6 15 15 0.6 5.6 6 0.1963 30 10 6.136 103 11.5 106
7 15 15 4.22 15 2 x d x dx 0.6 30 106 3.068 103 0 30 106 3.068 103 0 7
15
15 15 2 2 0.8 x d x 0.8 x d x 6 6 3 3 30 10 3.068 10 0 30 10 3.068 10 0
1.38 105 0.1000 6.71103 0.0431 0.0119 0.1173 0.279 in
Ans.
Chapter 4 - Rev B, Page 48/81
*Note. This is not the actual deflection of point B. For this, dummy forces must be placed on B in the x, y, and z directions. Determine the energy due to each, take derivatives, and then substitute the values of F x = 9 lbf, F y = 12 lbf, and F z = 0. This can be done separately and then use superposition. The actual deflections of B are
B = 0.0831 i 0.2862 j 0.00770 k in From this, the deflection of B in the direction of F is
B F 0.6 0.0831 0.8 0.2862 0.279 in which agrees with our result. ______________________________________________________________________________ 4-71 Strain energy. AB: Bending and torsion, BC: Bending and torsion, CD: Bending. I AB = (14)/64 = 0.04909 in4, J AB = 2 I AB = 0.09818 in4, I BC = 0.25(1.53)/12 = 0.07031 in4, I CD = (0.754)/64 = 0.01553 in4. For the torsion of bar BC, Eq. (3-41) is in the form of =TL/(JG), where the equivalent of J is J eq = bc 3. With b/c = 1.5/0.25 = 6, J BC = bc 3 = 0.299(1.5)0.253 = 7.008 (103) in4.
Use the dummy variable x to originate at the end where the loads are applied on each segment, M AB: Bending M F x 2 F x 2 F T Torsion T 5 F 5 F M BC: Bending M F x x F T Torsion T 2 F 2 F M CD: Bending M F x x F U Tl T 1 M M dx F JG F EI F 6 5F 6 2F 5 1 2 F x 2 d x 5 2 6 3 6 6 0.09818 11.5 10 7.008 10 11.5 10 30 10 0.04909 0
D
5
2
1 1 F x 2d x F x 2d x 6 6 30 10 0.07031 0 30 10 0.01553 0 1.329 104 F 2.482 104 F 1.141104 F 1.98 105 F 5.72 10 6 F 5.207 10 4 F 5.207 104 200 0.104 in
Ans.
______________________________________________________________________________
Chapter 4 - Rev B, Page 49/81
4-72 A AB = (12)/4 = 0.7854 in2, I AB = (14)/64 = 0.04909 in4, I BC = 1.5 (0.253)/12 = 1.953 (103) in4, A CD = (0.752)/4 = 0.4418 in2, I AB = (0.754)/64 = 0.01553 in4. For ( D ) x let F = F x = 150 lbf and F z = 100 lbf . Use the dummy variable x to originate at the end where the loads are applied on each segment, M y M y Fz x 0 CD: F Fa Fa F 1 F M y M y F x 2 Fz x BC: F Fa Fa Fz 0 F M y 5 M y 5F 2 Fz Fz x AB: F Fa Fa F 1 F 5 Fa U FL 1 D x F x 2 Fz x d x F AE CD F EI BC 0 1 EI AB
F 2
0.4418 30 10
6
6
5F 2 F
z
0
1
FL Fa Fz x 5 d x AE AB F
1 3 F 5 Fz 52 3 30 10 1.953 10 3 6
F 6 F 1 25 F 6 10 Fz 6 z 62 5 1 6 2 30 10 0.04909 0.7854 30 10 6
1.509 107 F 7.112 104 F 4.267 104 Fz 1.019 104 F 1.019 104 Fz 2.546 107 F 8.135 104 F 5.286 104 Fz
Substituting F = F x = 150 lbf and F z = 100 lbf gives
D x 8.135 104 150 5.286 104 100 0.1749 in
Ans.
______________________________________________________________________________ 4-73 I OA = I BC = (1.54)/64 = 0.2485 in4, J OA = J BC = 2 I OA = 0.4970 in4, I AB = (14)/64 = 0.04909 in4, J AB = 2 I AB = 0.09818 in4, I CD = (0.754)/64 = 0.01553 in4
Let F y = F, and use the dummy variable x to originate at the end where the loads are applied on each segment,
Chapter 4 - Rev B, Page 50/81
OC:
M Fx
M x, F
DC:
M Fx
M x F
T 12 F
T 12 F
U 1 M TL T M dx EI F F JG OC F The terms involving the torsion and bending moments in OC must be split up because of the changing second-area moments.
D y
D y
2 12 F 4 12 F 9 1 12 12 F x 2d x 6 6 6 0.4970 11.5 10 0.09818 11.5 10 30 10 0.2485 0 11
13
12
1 1 1 F x 2d x F x 2d x F x 2d x 6 6 6 30 10 0.04909 2 30 10 0.2485 11 30 10 0.01553 0 1.008 104 F 1.148 103 F 3.58 107 F 2.994 104 F 3.872 105 F 1.2363 103 F 2.824 103 F 2.824 103 250 0.706 in
Ans.
For the simplified shaft OC,
B y
13 12 12 F 13 1 1 2 Fx dx F x 2d x 12 0.09818 11.5106 30 106 0.04909 0 30 106 0.01553 0
1.6580 103 F 4.973 104 F 1.2363 103 F 3.392 103 F 3.392 103 250 0.848 in
Ans.
Simplified is 0.848/0.706 = 1.20 times greater Ans. ______________________________________________________________________________ 4-74 Place a dummy force Q pointing downwards at point B. The reaction at C is
R C = Q + (6/18)100 = Q + 33.33 This is the axial force in member BC. Isolating the beam, we find that the moment is not a function of Q, and thus does not contribute to the strain energy. Thus, only energy in the member BC needs to be considered. Let the axial force in BC be F, where
Chapter 4 - Rev B, Page 51/81
F Q 33.33
F 1 Q
FL F 0 33.3312 1 6.79 105 in Ans. 2 6 AE Q BC Q 0 0.5 / 4 30 10 Q 0 ______________________________________________________________________________
B
4-75
U Q
I OB = 0.25(23)/12 = 0.1667 in4 A AC = (0.52)/4 = 0.1963 in2 M O = 0 = 6 R C 11(100) 18 Q R C = 3Q + 183.3 M A = 0 = 6 R O 5(100) 12 Q
R O = 2Q + 83.33
Bending in OB. BD: Bending in BD is only due to Q which when set to zero after differentiation gives no contribution. AD: Using the variable x as shown in the figure above M 7 x Q OA: Using the variable x as shown in the figure above M 100 x Q 7 x
M 2Q 83.33 x Axial in AC: F 3Q 183.3
M 2 x Q F 3 Q
Chapter 4 - Rev B, Page 52/81
U
FL F
1
M
B M Q dx Q Q 0 AE Q Q 0 EI Q 0 183.3 12 1 3 6 0.1963 30 10 EI
5
6
100 x 7 x d x 2 83.33 x dx 2
0
0
5 6 2 100 x 7 x d x 166.7 x dx 1.12110 0 10.4 106 0.1667 0
1
3
1.121103 5.768 107 100 129.2 166.7 72 0.0155 in
Ans.
______________________________________________________________________________ 4-76
There is no bending in AB. Using the variable, rotating counterclockwise from B M R sin P Fr cos P F sin P
M PR sin Fr P cos F P sin MF 2 PR sin 2 P
A 6(4) 24 mm 2 , ro 40 12 (6) 43 mm, ri 40 12 (6) 37 mm, From Table 3-4, p.121, for a rectangular cross section 6 rn 39.92489 mm ln(43 / 37) From Eq. (4-33), the eccentricity is e = R r n =40 39.92489 = 0.07511 mm From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa From Table 4-1, C = 1.2 From Eq. (4-38) M M 1 MF 2 2 F R F 2 2 CFr R Fr d d d 0 AeE 0 AE 0 AE 0 AG P P P P
d
2 P R sin 2 PR sin 2 2 PR sin 2 CPR cos d d d d 0 0 0 0 AeE AE AE AG EC PR R (10)(40) 40 (207 103 )(1.2) 1 2 1 2 4 AE e G 4(24)(207 103 ) 0.07511 79.3 103 0.0338 mm Ans. ______________________________________________________________________________
2
2
2
2
Chapter 4 - Rev B, Page 53/81
4-77
Place a dummy force Q pointing downwards at point A. Bending in AB is only due to Q which when set to zero after differentiation gives no contribution. For section BC use the variable, rotating counterclockwise from B M M PR sin Q R R sin R 1 sin Q Fr cos Fr P Q cos Q F F P Q sin sin Q MF PR sin QR 1 sin P Q sin MF PR sin 2 PR sin 1 sin 2QR sin 1 sin Q
But after differentiation, we can set Q = 0. Thus, MF PR sin 1 2sin Q A 6(4) 24 mm 2 , ro 40 12 (6) 43 mm, ri 40 12 (6) 37 mm, From Table 3-4, p.121, for a rectangular cross section 6 rn 39.92489 mm ln(43 / 37) From Eq. (4-33), the eccentricity is e = R r n =40 39.92489 = 0.07511 mm From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa From Table 4-1, C = 1.2 From Eq. (4-38) 1 MF M M 2 2 F R F 2 2 CFr R Fr d d d d 0 AeE Q 0 AE Q 0 AE 0 Q AG Q
PR 2 2 PR 2 2 PR 2 sin 1 sin sin sin 1 2sin d d d AeE 0 AE 0 AE 0 CPR 2 cos 2 d 0 AG 2 CE R PR PR PR CPR PR 1 2 1 2 4 G 4 AeE 4 AE 4 AE 4 AG AE 4 e
3 1.2 207 10 40 1 2 4 79.3 103 4 0.07511 0.0766 mm Ans. ______________________________________________________________________________
10 40 24 207 103
Chapter 4 - Rev B, Page 54/81
4-78
Note to the Instructor. The cross section shown in the first printing is incorrect and the solution presented here reflects the correction which will be made in subsequent printings. The corrected cross section should appear as shown in this figure. We apologize for any inconvenience.
A = 3(2.25) 2.25(1.5) = 3.375 in2 (1 1.5)(3)(2.25) (1 0.75 1.125)(1.5)(2.25) R 2.125 in 3.375 Section is equivalent to the “T” section of Table 3-4, p. 121, 2.25(0.75) 0.75(2.25) 1.7960 in 2.25ln[(1 0.75) /1] 0.75ln[(1 3) / (1 0.75)] e R rn 2.125 1.7960 0.329 in rn
For the straight section 1 I z (2.25) 33 2.25(3)(1.5 1.125) 2 12 2 1 2.25 3 (1.5) 2.25 1.5(2.25) 0.75 1.125 2 12 2.689 in 4
For 0 x 4 in M Fx
M x, F
V F
V 1 F
For /2 Fr F cos
Fr cos , F
F F sin
F sin F
M (4 2.125sin ) F MF MF F (4 2.125sin ) F sin 2 F (4 2.365sin ) sin F
M F (4 2.125sin )
Chapter 4 - Rev B, Page 55/81
Use Eqs. (4-31) and (4-24) (with C = 1) for the straight part, and Eq. (4-38) for the curved part, integrating from 0 to π/2, and double the results
/2 2 1 4 2 F (4)(1) (4 2.125sin ) 2 Fx dx F d E I 0 3.375(G / E ) 0 3.375(0.329) /2 2 F (4 2.125sin ) sin F sin 2 (2.125) d d 0 0 3.375 3.375 2 /2 (1) F cos (2.125) d 0 3.375(G / E )
/2
Substitute I = 2.689 in4, F = 6700 lbf, E = 30 (106) psi, G = 11.5 (106) psi 2 6700 43 4 1 16 17(1) 4.516 6 30 10 3 2.689 3.375(11.5 / 30) 3.375(0.329) 2 4
2.125 2 2.125 4 1 2.125 3.375 4 3.375 4 3.375 11.5 / 30 4
Ans. 0.0226 in ______________________________________________________________________________
4-79 Since R/h = 35/4.5 = 7.78 use Eq. (4-38), integrate from 0 to , and double the results M M FR 1 cos R 1 cos F Fr Fr F sin sin F F cos F F cos F MF F 2 Rcos 1 cos
MF 2 FRcos 1 cos F From Eq. (4-38), FR 2
FR
2 (1 cos ) 2 d cos 2 d 0 0 AeE AE
2 FR 1.2 FR 2 cos 1 cos d sin d AE 0 AG 0
2 FR 3 R 3 E 0.6 AE 2 e 2 G
A = 4.5(3) = 13.5 mm2, E = 207 (103) N/mm2, G = 79.3 (103) N/mm2, and from Table 3-4, p. 121,
Chapter 4 - Rev B, Page 56/81
h 4.5 34.95173 mm ro 37.25 ln ln 32.75 ri and e = R r n = 35 34.95173 = 0.04827 mm. Thus, 2 F 35 3 35 3 207 0.6 0.08583F 3 13.5 207 10 2 0.04827 2 79.3 1 where F is in N. For = 1 mm, F 11.65 N Ans. 0.08583 Note: The first term in the equation for dominates and this is from the bending moment. Try Eq. (4-41), and compare the results. ______________________________________________________________________________ 4-80 R/h = 20 > 10 so Eq. (4-41) can be used to determine deflections. Consider the horizontal reaction, to applied at B, subject to the constraint ( B ) H 0. rn
M
M R sin H
FR (1 cos ) HR sin 2
0
2
By symmetry, we may consider only half of the wire form and use twice the strain energy Eq. (4-41) then becomes, ( B ) H
/2
0
U 2 H EI
/2
M
M H Rd 0 0
FR 2 (1 cos ) HR sin ( R sin ) R d 0 F F F 30 H 0 H 9.55 N 2 4 4
Ans.
Reaction at A is the same where H goes to the left. Substituting H into the moment equation we get, M
FR (1 cos ) 2sin 2
R M [ (1 cos ) 2sin ] F 2
0
2
Chapter 4 - Rev B, Page 57/81
U 2 M 2 /2 FR 2 P [ (1 cos ) 2sin ]2 R d M Rd 2 0 P EI F EI 4 3 /2 FR ( 2 2 cos 2 4sin 2 2 2 cos 4 sin 4 sin cos ) d 2 0 2 EI FR 3 2 2 4 2 2 4 2 2 2 EI 2 4 4 (30)(403 ) (3 2 8 4) FR 3 (3 2 8 4) 0.224 mm Ans. 8 8 EI 207 103 24 / 64 ______________________________________________________________________________
4-81 The radius is sufficiently large compared to the wire diameter to use Eq. (4-41) for the curved beam portion. The shear and axial components will be negligible compared to bending. Place a fictitious force Q pointing to the left at point A. M M PR sin Q( R sin l ) R sin l Q Note that the strain energy in the straight portion is zero since there is no real force in that section. From Eq. (4-41),
/2
0
PR 2 EI
1 M M EI Q
/2
0
1 Rd Q 0 EI
R sin 2 l sin d
/2
0
PR sin R sin l Rd
PR 2 1(52 ) R l (5) 4 6 4 EI 4 30 10 0.125 / 64 4
0.551 in Ans. ______________________________________________________________________________
4-82 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. M AB x P
Straight portion:
M AB Px
Curved portion:
M BC P R (1 cos ) l
M BC R (1 cos ) l P
From Eq. (4-41) with the addition of the bending strain energy in the straight portion of the wire,
Chapter 4 - Rev B, Page 58/81
/2 1 M BC M AB 1 M AB dx M BC Rd 0 EI 0 P EI P P l 2 PR /2 2 x dx R(1 cos ) l d EI 0 EI 0 3 Pl PR /2 2 R (1 2 cos cos 2 ) 2 Rl (1 cos ) l 2 d 0 3EI EI Pl 3 PR /2 2 R cos 2 2 R 2 2 Rl cos ( R l ) 2 d 3EI EI 0 Pl 3 PR 2 R 2 R 2 2 Rl ( R l ) 2 2 3EI EI 4 3 P l 3 R R 2 R 2 2 Rl R( R l ) 2 2 EI 3 4
l
43 3 1 2 2 (5 ) 5 2(5 ) 2(5)(4) 5 5 4 6 4 2 30 10 0.125 / 64 3 4
0.850 in Ans. ______________________________________________________________________________ 4-83 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion.
Place a dummy force, Q, at A vertically downward. The only load in the straight section is the axial force, Q. Since this will be zero, there is no contribution. In the curved section M R 1 cos Q
M PR sin QR 1 cos From Eq. (4-41) /2 1 M 1 M Rd 0 EI Q Q 0 EI PR 3 EI
/2
0
/2
0
PR sin R 1 cos Rd
PR 3 1 PR 3 sin sin cos d 1 EI 2 2 EI 1 53
2 30 106 0.1254 / 64
0.174 in
Ans.
______________________________________________________________________________ 4-84 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion.
Chapter 4 - Rev B, Page 59/81
Place a dummy force, Q, at A vertically downward. The load in the straight section is the axial force, Q, whereas the bending moment is only a function of P and is not a function of Q. When setting Q = 0, there is no axial or bending contribution. In the curved section
M R sin Q
M P R 1 cos l QR sin From Eq. (4-41)
/2
0
1 M 1 M Rd EI Q Q 0 EI
PR 2 EI
/2
0
P R 1 cos l R sin Rd
/2
R sin R sin cos l sin d 0
1 52
PR 2 1 PR 2 R l R R 2l EI 2 2 EI
5 2 4 0.452 in 2 30 106 0.1254 / 64
Since the deflection is negative, is in the opposite direction of Q. Thus the deflection is
0.452 in
Ans. ______________________________________________________________________________ 4-85 Consider the force of the mass to be F, where F = 9.81(1) = 9.81 N. The load in AB is tension FAB 1 FAB F F
For the curved section, the radius is sufficiently large to use Eq. (4-41). There is no bending in section DE. For section BCD, let be counterclockwise originating at D M M FR sin R sin 0 F Using Eqs. (4-29) and (4-41) Fl
F
1
M
Fl
FR3
AB sin 2 d 1 0 M Rd 0 EI F AE EI AE AB F
403 9.81 80 Fl FR3 F l R3 AE 2 EI E A 2 I 207 103 22 / 4 2 24 / 64 6.067 mm Ans. ______________________________________________________________________________
Chapter 4 - Rev B, Page 60/81
4-86 A OA = 2(0.25) = 0.5 in2, I OAB = 0.25(23)/12 = 0.1667 in4, I AC = (0.54)/64 = 3.068 (10-3) in4
Applying a force F at point B, using statics, the reaction forces at O and C are as shown. OA: Axial FOA 3F
FOA 3 F M OA 2 x F
Bending M OA 2 Fx
M AB x F
M AB F x
AB: Bending
AC: Isolating the upper curved section M AC 3R sin cos 1 F 10 20 1 1 Fl FOA 2 Fx dx F x 2d x 4 AE F EI EI OAB 0 OAB 0 OA M AC 3FR sin cos 1
9 FR3 EI AC
/2
sin cos 1
2
d
0
F 203 4 F 103 3F 10 3 0.5 10.4 106 3 10.4 106 0.1667 3 10.4 106 0.1667
9 F 103
30 10 3.068 10 6
/2
3
sin
2
2sin cos 2sin cos 2 2 cos 1 d
0
1.731105 F 7.691104 F 1.538 103 F 0.09778 F 1 2 2 4 2 4 0.0162 F 0.0162 100 1.62 in Ans. _____________________________________________________________________________ 4-87 A OA = 2(0.25) = 0.5 in2, I OAB = 0.25(23)/12 = 0.1667 in4, I AC = (0.54)/64 = 3.068 (10-3) in4 Applying a vertical dummy force, Q, at A, from statics the reactions are as shown. The dummy force is transmitted through section Chapter 4 - Rev B, Page 61/81
OA and member AC.
FOA 1 Q
OA: FOA 3F Q
M AC 3F Q R sin 3F Q R 1 cos
AC:
M AC R sin cos 1 Q
Fl FOA 1 /2 M AC Rd M AC Q AE OA Q EI AC 0 Q 0
3FlOA 3FR 3 AE OA EI AC 3 100 10
10.4 10 0.5 6
/2
sin cos 1
2
d
0
3 100 103
1 2 2 0.462 in 4 2 30 10 3.068 10 4 6
3
Ans.
______________________________________________________________________________ 4-88
I = (64)/64 = 63.62 mm4 0 /2
M R sin F T R (1 cos ) T FR (1 cos ) F According to Castigliano’s theorem, a positive U/ F will yield a deflection of A in the negative y direction. Thus the deflection in the positive y direction is M FR sin
( A ) y
U 1 F EI
/2
0
F ( R sin ) 2 R d
1 GJ
/2
0
F [ R (1 cos )]2 R d
Integrating and substituting J 2 I and G E / 2 1 FR 3 3 4 (1 ) 4 2 4 8 (3 8) 4 EI (250)(80)3 [4 8 (3 8)(0.29)] 12.5 mm Ans. 4(200)103 63.62 ______________________________________________________________________________ FR 3 ( A ) y EI
4-89 The force applied to the copper and steel wire assembly is Fc Fs 400 lbf (1) Since the deflections are equal, c s
Chapter 4 - Rev B, Page 62/81
Fl Fl AE c AE s
Fc l Fs l 2 6 3( / 4)(0.1019) (17.2)10 ( / 4)(0.1055) 2 (30)106 Yields, Fc 1.6046 Fs . Substituting this into Eq. (1) gives 1.604 Fs Fs 2.6046Fs 400 Fs 153.6 lbf Fc 1.6046 Fs 246.5 lbf F 246.5 c c 10 075 psi 10.1 kpsi Ans. Ac 3( / 4)(0.1019) 2 F 153.6 s s 17 571 psi 17.6 kpsi Ans. As ( / 4)(0.10552 ) 153.6(100)(12) Fl Ans. 0.703 in 2 6 AE s ( / 4)(0.1055) (30)10 ______________________________________________________________________________ 4-90
(a)
b 0.75(65) 48.8 kpsi
Bolt stress
Ans.
Fb 6 b Ab 6(48.8) (0.52 ) 57.5 kips 4 F 57.43 c b 13.9 kpsi Ans. Ac ( / 4)(5.52 52 )
Total bolt force Cylinder stress (b) Force from pressure
P F x = 0 P b + P c = 9.82
D2 4
p
(52 ) 4
(500) 9817 lbf 9.82 kip
(1)
Since c b , Pc l Pbl 2 2 ( / 4)(5.5 5 ) E 6( / 4)(0.52 ) E P c = 3.5 P b
(2)
Substituting this into Eq. (1) P b + 3.5 P b = 4.5 P b = 9.82 P b = 2.182 kip. From Eq. (2), P c = 7.638 kip Using the results of (a) above, the total bolt and cylinder stresses are 2.182 b 48.8 50.7 kpsi Ans. 6( / 4)(0.52 )
Chapter 4 - Rev B, Page 63/81
7.638 12.0 kpsi Ans. ( / 4)(5.52 52 ) ______________________________________________________________________________
c 13.9
4-91
(1)
Tc + Ts = T
c = s
Tc l Tl s JG c JG s
Substitute this into Eq. (1) JG c T T T JG s s s
Ts
Tc
JG c T JG s s
(2)
JG s T JG s JG c
The percentage of the total torque carried by the shell is % Torque
100 JG s
JG s JG c
Ans.
______________________________________________________________________________ 4-92 R O + R B = W (1) OA = AB Fl Fl AE OA AE AB 400 RO 600 RB 3 RO RB AE AE 2 Substitute this unto Eq. (1) 3 RB RB 4 2
RB 1.6 kN
(2)
Ans.
3 RO 1.6 2.4 kN Ans. 2 2 400(400) Fl 0.0223 mm Ans. A 3 AE OA 10(60)(71.7)(10 ) ______________________________________________________________________________
From Eq. (2)
4-93 See figure in Prob. 4-92 solution.
Procedure 1: 1. Let R B be the redundant reaction. Chapter 4 - Rev B, Page 64/81
2. Statics. R O + R B = 4 000 N 3. Deflection of point B. B
R O = 4 000 R B
(1)
RB 600 RB 4000 400 0 AE AE
(2)
4. From Eq. (2), AE cancels and R B = 1 600 N Ans. and from Eq. (1), R O = 4 000 1 600 = 2 400 N Ans. 2 400(400) Fl 0.0223 mm Ans. 3 AE OA 10(60)(71.7)(10 ) ______________________________________________________________________________
A
4-94 (a) Without the right-hand wall the deflection of point C would be 5 103 8 2 103 5 Fl C AE / 4 0.752 10.4 106 / 4 0.52 10.4 106 0.01360 in 0.005 in Hits wall
Ans.
(b) Let R C be the reaction of the wall at C acting to the left (). Thus, the deflection of point C is now 5 103 RC 8 2 103 RC 5 C / 4 0.752 10.4 106 / 4 0.52 10.4 106 0.01360
4 RC 5 8 2 0.005 6 2 10.4 10 0.75 0.5
or, 0.01360 4.190 10 6 RC 0.005
RC 2 053 lbf 2.05 kip Ans.
Statics. Considering +, 5 000 R A 2 053 = 0 R A = 2 947 lbf = 2.95 kip Ans. Deflection. AB is 2 947 lbf in tension. Thus 2 947 8 5.13 103 in Ans. 2 6 AAB E / 4 0.75 10.4 10 ______________________________________________________________________________
B AB
RA 8
4-95 Since OA = AB , TOA (4) TAB (6) JG JG
3 TOA TAB 2
(1)
Chapter 4 - Rev B, Page 65/81
Statics. T OA + T AB = 200
(2)
Substitute Eq. (1) into Eq. (2), 3 5 TAB TAB TAB 200 2 2 From Eq. (1)
TAB 80 lbf in
Ans.
3 3 TOA TAB 80 120 lbf in Ans. 2 2 80 6 180 A 0.3900 Ans. 4 6 / 32 0.5 11.5 10
max
16T d3
AB
16 80
0.53
OA
16 120
0.53
4890 psi 4.89 kpsi
3260 psi 3.26 kpsi
Ans.
Ans.
______________________________________________________________________________ 4-96 Since OA = AB , TOA (4) TAB (6) 4 / 32 0.5 G / 32 0.754 G
Statics. T OA + T AB = 200
TOA 0.2963 TAB
TAB 154.3 lbf in
TOA 0.2963TAB 0.2963 154.3 45.7 lbf in
A
154.3 6 180 0.1480 4 6 / 32 0.75 11.5 10
max
(1)
(2)
Substitute Eq. (1) into Eq. (2), 0.2963TAB TAB 1.2963TAB 200 From Eq. (1)
16T d3
AB
OA
16 154.3
0.753
16 45.7
0.53
Ans.
Ans.
Ans.
1862 psi 1.86 kpsi
1862 psi 1.86 kpsi
Ans.
Ans.
______________________________________________________________________________
Chapter 4 - Rev B, Page 66/81
4-97 Procedure 1. 1. Arbitrarily, choose R C as a redundant reaction. 2. Statics. F x = 0, 12(103) 6(103) R O R C = 0 (1) R O = 6(103) R C 3. The deflection of point C.
12(103 ) 6(103 ) RC (20) 6(103 ) RC (10) R (15) C C 0 AE AE AE 4. The deflection equation simplifies to 45 R C + 60(103) = 0 R C = 1 333 lbf 1.33 kip Ans. R O = 6(103) 1 333 = 4 667 lbf 4.67 kip Ans.
From Eq. (1),
F AB = F B + R C = 6 +1.333 = 7.333 kips compression FAB 7.333 14.7 kpsi Ans. (0.5)(1) A Deflection of A. Since OA is in tension, Rl 4 667(20) A OA O OA 0.00622 in Ans. (0.5)(1)(30)106 AE ______________________________________________________________________________
AB
4-98 Procedure 1. 1. Choose R B as redundant reaction.
2. Statics. R C = wl R B
(1)
1 (2) M C wl 2 RB l a 2 3. Deflection equation for point B. Superposition of beams 2 and 3 of Table A-9,
R l a w l a 4l l a l a 2 6l 2 0 yB B 3EI 24 EI 3
2
4. Solving for R B . w 2 2 RB 6l 4l l a l a 8 l a
w 3l 2 2al a 2 8 l a
Ans.
Substituting this into Eqs. (1) and (2) gives Chapter 4 - Rev B, Page 67/81
RC wl RB
w 5l 2 10al a 2 8 l a
Ans.
1 2 w Ans. wl RB l a l 2 2al a 2 2 8 ______________________________________________________________________________ MC
4-99 See figure in Prob. 4-98 solution.
Procedure 1. 1. Choose R B as redundant reaction. 2. Statics. R C = wl R B
(1)
1 (2) M C wl 2 RB l a 2 3. Deflection equation for point B. Let the variable x start at point A and to the right. Using singularity functions, the bending moment as a function of x is 1 M w x 2 RB x a 2
1
M xa RB
1
U 1 M yB M dx RB EI 0 RB l
1 1 1 1 2 w x 2 0 dx w x RB x a x a dx 0 EI 0 2 EI a 2 l
l
or, 1 1 a 3 3 R w l 4 a 4 l 3 a 3 B l a a a 0 2 4 3 3 Solving for R B gives RB
w 8 l a
3
w 2 2 3 l 4 a 4 4 a l 3 a 3 8 l a 3l 2al a
Ans.
From Eqs. (1) and (2) RC wl RB
MC
w 5l 2 10al a 2 8 l a
1 2 w wl RB l a l 2 2al a 2 2 8
Ans.
Ans.
Chapter 4 - Rev B, Page 68/81
______________________________________________________________________________ 4-100 Note: When setting up the equations for this problem, no rounding of numbers was made. It turns out that the deflection equation is very sensitive to rounding.
Procedure 2. 1. Statics.
R 1 + R 2 = wl
1 2 wl 2 2. Bending moment equation. R2l M 1
(1) (2)
1 M R1 x w x 2 M 1 2 dy 1 1 EI R1 x 2 w x 3 M 1 x C1 6 dx 2 1 1 1 EIy R1 x 3 w x 4 M 1 x 2 C1 x C2 6 24 2
(3) (4)
EI = 30(106)(0.85) = 25.5(106) lbfin2. 3. Boundary condition 1. At x = 0, y = R 1 /k 1 = R 1 /[1.5(106)]. Substitute into Eq. (4) with value of EI yields C 2 = 17 R 1 . Boundary condition 2. At x = 0, dy /dx = M 1 /k 2 = M 1 /[2.5(106)]. Substitute into Eq. (3) with value of EI yields C 1 = 10.2 M 1 . Boundary condition 3. At x = l, y = R 2 /k 3 = R 1 /[2.0(106)]. Substitute into Eq. (4) with value of EI yields 1 3 1 1 R1l wl 4 M 1l 2 10.2 M 1l 17 R1 (5) 6 24 2 Equations (1), (2), and (5), written in matrix form with w = 500/12 lbf/in and l = 24 in, are 12.75 R2
1 0 1 24 1 0 2287 12.75 532.8
R1 1 3 R2 12 10 M 576 1
Solving, the simultaneous equations yields R 1 = 554.59 lbf, R 2 = 445.41.59 lbf, M 1 = 1310.1 lbfin
Ans.
For the deflection at x = l /2 = 12 in, Eq. (4) gives Chapter 4 - Rev B, Page 69/81
y x 12in
1 1 500 4 1 1 12 1310.1122 554.59 123 6 6 24 12 2 25.5 10 10.2 1310.112 17 554.59
5.51103 in
Ans.
______________________________________________________________________________ 4-101 Cable area, A
4
(0.52 ) 0.1963 in 2
Procedure 2. 1. Statics. R A + F BE + F DF = 5(103)
(1)
3 F DF + F BE = 10(103)
(2)
2. Bending moment equation. 1
M RA x FBE x 16 5000 x 32
1
dy 1 1 2 2 RA x 2 FBE x 16 2500 x 32 C1 dx 2 2 1 1 2500 3 3 EIy RA x3 FBE x 16 x 32 C1 x C2 6 6 3
EI
3. B.C. 1: At x = 0, y = 0
(3) (4)
C2 = 0
B.C. 2: At x = 16 in, FBE (38) Fl 6.453(106 ) FBE yB 6 0.1963(30)10 AE BE
Substituting into Eq. (4) and evaluating at x = 16 in 1 EIyB 30(106 )(1.2)( 6.453)(106 ) FBE RA 163 C1 (16) 6 (5) Simplifying gives 682.7 R A + 232.3 F BE + 16 C 1 = 0 B.C. 2: At x = 48 in, FDF (38) Fl 6.453(106 ) FDF yD 6 0.1963(30)10 AE DF Substituting into Eq. (4) and evaluating at x = 48 in, 1 1 2500 (48 32)3 48C1 RA 483 FBE (48 16)3 6 6 3 18 432 R A + 5 461 F BE + 232.3 F DF + 48 C 1 = 3.413(106)
EIyD 232.3FDF
Simplifying gives
(6)
Chapter 4 - Rev B, Page 70/81
Equations (1), (2), (5) and (6) in matrix form are 1 1 0 RA 5000 1 F 10 000 0 1 3 0 BE 0 682.7 232.3 0 16 FDF 6 3.413 10 C 18 432 5 461 232.3 48 1 Solve simultaneously or use software. The results are R A = 970.5 lbf, F BE = 3956 lbf, F DF = 2015 lbf, and C 1 = 16 020 lbfin2. 3956 2015 BE 20.2 kpsi, DF 10.3 kpsi Ans. 0.1963 0.1963 EI = 30(106)(1.2) = 36(106) lbfin2 y
1 2500 3 3 970.5 3 3956 x x 16 x 32 16 020 x 6 6 6 3 36 10
161.8 x 659.3 x 16
1 36 106
B: x = 16 in,
3
3
3
833.3 x 32 16 020 x
yB
1 161.8 163 16 020 16 0.0255 in 6 36 10
yC
1 161.8 323 659.3 32 16 3 16 020 32 6 36 10
Ans.
C: x = 32 in,
D: x = 48 in, yD
0.0865 in
Ans.
1 161.8 483 659.3 48 16 3 833.3 48 32 3 16 020 48 6 36 10
0.0131 in Ans. ______________________________________________________________________________ 4-102 Beam: EI = 207(103)21(103) = 4.347(109) Nmm2. Rods: A = ( /4)82 = 50.27 mm2.
Procedure 2. 1. Statics. Chapter 4 - Rev B, Page 71/81
R C + F BE F DF = 2 000
(1)
R C + 2F BE = 6 000
(2)
2. Bending moment equation. M = 2 000 x + F BE x 75 1 + R C x 150 1 dy 1 1 2 2 1000 x 2 FBE x 75 RC x 150 C1 dx 2 2 1000 3 1 1 3 3 EIy x FBE x 75 RC x 150 C1 x C2 3 6 6
(3)
EI
(4)
3. B.C 1. At x = 75 mm, FBE 50 Fl 4.805 106 FBE yB 3 50.27 207 10 AE BE
Substituting into Eq. (4) at x = 75 mm, 4.347 109 4.805 106 FBE
1000 753 C1 75 C2 3
Simplifying gives 20.89 103 FBE 75C1 C2 140.6 106
(5)
B.C 2. At x = 150 mm, y = 0. From Eq. (4),
or,
1000 1 3 1503 FBE 150 75 C1 150 C2 0 3 6
70.31103 FBE 150C1 C2 1.125 109
(6)
B.C 3. At x = 225 mm, FDF 65 Fl 6.246 106 FDF yD 3 AE DF 50.27 207 10
Substituting into Eq. (4) at x = 225 mm,
Chapter 4 - Rev B, Page 72/81
4.347 109 6.246 106 FDF
1000 1 3 2253 FBE 225 75 3 6 1 3 RC 225 150 C1 225 C2 6
Simplifying gives 70.31103 RC 562.5 103 FBE 27.15 103 FDF 225C1 C2 3.797 109
(7)
Equations (1), (2), (5), (6), and (7) in matrix form are 2 103 1 1 1 0 0 RC 3 1 2 0 0 0 6 10 FBE 0 20.89 103 0 75 1 F 140.6 106 DF 3 9 0 70.3110 0 150 1 C1 1.125 10 3 3 3 C2 9 70.3110 562.5 10 27.15 10 225 1 3.797 10
Solve simultaneously or use software. The results are Ans. R C = 2378 N, F BE = 4189 N, F DF = 189.2 N 7 2 8 3 and C 1 = 1.036 (10 ) Nmm , C 2 = 7.243 (10 ) Nmm . The bolt stresses are BE = 4189/50.27 = 83.3 MPa, DF = 189/50.27= 3.8 MPa Ans. The deflections are From Eq. (4) y A
1 7.243 108 0.167 mm 9 4.347 10
Ans.
For points B and D use the axial deflection equations*. 4189 50 Fl yB 0.0201 mm 50.27 207 103 AE BE
Ans.
189 65 Fl yD Ans. 1.18 103 mm 3 AE DF 50.27 207 10 *Note. The terms in Eq. (4) are quite large, and due to rounding are not very accurate for calculating the very small deflections, especially for point D. ______________________________________________________________________________
4-103 (a) The cross section at A does not rotate. Thus, for a single quadrant we have
Chapter 4 - Rev B, Page 73/81
U 0 M A The bending moment at an angle to the x axis is FR M M MA 1 1 cos 2 M A The rotation at A is /2 U 1 M M Rd 0 A M A EI 0 M A
Thus,
1 EI
/2
M
A
0
FR 1 cos 1 Rd 0 2
FR FR 0 MA 2 2 2
or, FR 2 1 2 Substituting this into the equation for M gives FR 2 M cos (1) 2 The maximum occurs at B where = /2 MA
M max M B
FR
Ans.
(b) Assume B is supported on a knife edge. The deflection of point D is U/ F. We will deal with the quarter-ring segment and multiply the results by 4. From Eq. (1) M R 2 cos F 2
Thus, U 4 D F EI
/2
0
M FR3 M Rd F EI
/2
0
2
2 FR3 2 cos d EI 4
3
FR 2 8 Ans. 4 EI ______________________________________________________________________________
4-104
C 2 EI Pcr l2 I
D4 d 4
D4
1 K
64 64 2 4 C E D Pcr 2 1 K 4 l 64
4
where K
d D
Chapter 4 - Rev B, Page 74/81
1/ 4
64 Pcr l 2 D 3 Ans. 4 CE 1 K ______________________________________________________________________________
D 2 1 K 2 ,
I
D 4 1 K 4
D 4 1 K 2 1 K 2 , where K = d / D. 4 64 64 The radius of gyration, k, is given by I D2 2 k 1 K 2 A 16 From Eq. (4-46) S y2l 2 S y2l 2 Pcr S S y y 4 2 k 2CE 4 2 D 2 / 16 1 K 2 CE / 4 D 2 1 K 2
4-105 A
4 Pcr D 1 K 2
2
S
y
D 2 1 K 2 S y 4 Pcr
4S y2l 2 D 2 1 K 2
2 D 2 1 K 2 CE
4S y2l 2 1 K 2
1 K 2 CE
4 S y2l 2 1 K 2 4 Pcr D 2 2 2 S y 1 K 1 K CE 1 K S y
1/2
1/2
S yl 2 Pcr 2 Ans. 2 2 2 S y 1 K CE 1 K ______________________________________________________________________________
0.9
4-106 (a) M A 0, (0.75)(800)
0.92 0.52
FBO (0.5) 0 FBO 1373 N
Using n d = 4, design for F cr = n d F BO = 4(1373) = 5492 N
l 0.92 0.52 1.03 m, S y 165 MPa In-plane: 1/ 2
1/ 2
bh3 / 12 I k A bh l 1.03 142.7 k 0.007218
0.2887 h 0.2887(0.025) 0.007 218 m,
C 1.0
1/ 2
2 9 l 2 (207)(10 ) 6 k 1 165(10 )
157.4
Chapter 4 - Rev B, Page 75/81
Since (l / k )1 (l / k ) use Johnson formula. Try 25 mm x 12 mm, 2 165 106 1 6 Pcr 0.025(0.012) 165 10 (142.7) 29.1 kN 9 2 1(207)10 This is significantly greater than the design load of 5492 N found earlier. Check out-ofplane. Out-of-plane: k 0.2887(0.012) 0.003 464 in, l 1.03 297.3 k 0.003 464 Since (l / k )1 (l / k ) use Euler equation. Pcr 0.025(0.012)
C 1.2
1.2 2 207 109
8321 N 297.32 This is greater than the design load of 5492 N found earlier. It is also significantly less than the in-plane P cr found earlier, so the out-of-plane condition will dominate. Iterate the process to find the minimum h that gives P cr greater than the design load.
With h = 0.010, P cr = 4815 N (too small) h = 0.011, P cr = 6409 N (acceptable) Use 25 mm x 11 mm. If standard size is preferred, use 25 mm x 12 mm. Ans. P 1373 10.4 106 Pa 10.4 MPa dh 0.012(0.011) No, bearing stress is not significant. Ans. ______________________________________________________________________________
(b) b
4-107 This is an open-ended design problem with no one distinct solution. ______________________________________________________________________________
F = 1500( /4)22 = 4712 lbf. From Table A-20, S y = 37.5 kpsi P cr = n d F = 2.5(4712) = 11 780 lbf
4-108
(a) Assume Euler with C = 1 1/4 64 11790 502 64 Pcr l 2 d 3 3 6 CE 1 30 10 Use d = 1.25 in. The radius of gyration, k = ( I / A)1/2 = d /4 = 0.3125 in
Pcr l 2 I d 64 C 2 E 4
1/ 4
1.193 in
Chapter 4 - Rev B, Page 76/81
l 50 160 k 0.3125 2 2 (1)30 106 126 37.5 103 2 6 4 30 10 / 64 1.25 14194 lbf Pcr 502 1/2
1/2
2 l 2 CE k 1 S y
Since 14 194 lbf > 11 780 lbf, d = 1.25 in is satisfactory.
use Euler
Ans.
1/4
64 11780 162 0.675 in, so use d = 0.750 in d 3 6 1 30 10 k = 0.750/4 = 0.1875 in l 16 85.33 use Johnson k 0.1875
(b)
2 37.5 103 1 3 Pcr 0.750 37.5 10 85.33 12 748 lbf 6 4 2 1 30 10
2
Use d = 0.75 in. ( c) n( a )
14194 3.01 4 712
Ans.
12 748 2.71 Ans. 4 712 ______________________________________________________________________________ n(b )
4-109 From Table A-20, S y = 180 MPa
4F sin = 2 943 735.8 sin In range of operation, F is maximum when = 15 735.8 Fmax 2843 N per bar sin15o F
P cr = n d F max = 3.50 (2 843) = 9 951 N l = 350 mm, h = 30 mm
Chapter 4 - Rev B, Page 77/81
Try b = 5 mm. Out of plane, k = b / 12 = 5/ 12 = 1.443 mm l 350 242.6 k 1.443 2 9 l 2 1.4 207 10 180 106 k 1
Pcr A
C 2 E
l / k
2
5(30)
1/ 2
178.3
1.4 2 207 103
242.6
2
use Euler 7 290 N
Too low. Try b = 6 mm. k = 6/ 12 = 1.732 mm 350 l 202.1 k 1.732 1.4 2 207 103 C 2 E 12 605 N 6(30) Pcr A 2 2 202.1 l / k O.K. Use 25 6 mm bars Ans. The factor of safety is 12 605 4.43 Ans. 2843 ______________________________________________________________________________ n
4-110 P = 1 500 + 9 000 = 10 500 lbf
Ans.
M A = 10 500 (4.5/2) 9 000 (4.5) +M = 0 M = 16 874 lbfin e = M / P = 16 874/10 500 = 1.607 in Ans. From Table A-8, A = 2.160 in2, and I = 2.059 in4. The stresses are determined using Eq. (4-55) k2
I 2.059 0.953 in 2 A 2.160
P ec 10500 1.607 3 / 2 1 17157 psi 17.16 kpsi Ans. 1 2 A k 2.160 0.953 ______________________________________________________________________________
c
4-111 This is a design problem which has no single distinct solution. ______________________________________________________________________________
Chapter 4 - Rev B, Page 78/81
4-112 Loss of potential energy of weight = W (h + ) 1 Increase in potential energy of spring = k 2 2 1 W (h + ) = k 2 2 2 W 2 W or, 2 h 0 . W = 30 lbf, k = 100 lbf/in, h = 2 in yields k k
2 0.6 1.2 = 0 Taking the positive root (see discussion on p. 192) 1 max 0.6 (0.6) 2 4(1.2) 1.436 in 2
Ans.
Ans. F max = k max = 100 (1.436) = 143.6 lbf ______________________________________________________________________________ 4-113 The drop of weight W 1 converts potential energy, W 1 h, to kinetic energy
1 W1 2 v1 . 2 g
Equating these provides the velocity of W 1 at impact with W 2 . 1 W1 2 v1 v1 2 gh (1) 2 g Since the collision is inelastic, momentum is conserved. That is, (m 1 + m 2 ) v 2 = m 1 v 1 , where v 2 is the velocity of W 1 + W 2 after impact. Thus W1h
W1 W2 W v2 1 v1 g g
v2
W1 W1 v1 2 gh W1 W2 W1 W2
(2)
The kinetic and potential energies of W 1 + W 2 are then converted to potential energy of the spring. Thus, 1 W1 W2 2 1 v2 W1 W2 k 2 2 g 2 Substituting in Eq. (1) and rearranging results in W1 W2 W12 h 2 2 0 (3) k W1 W2 k Solving for the positive root (see discussion on p. 192) 2
2 1 W1 W2 W12 h W1 W2 8 2 4 2 k k W1 W2 k
(4)
Chapter 4 - Rev B, Page 79/81
W 1 = 40 N, W 2 = 400 N, h = 200 mm, k = 32 kN/m = 32 N/mm. 1
2 40 400 402 200 40 400 29.06 mm 4 8 32 40 400 32 32
2 2
Ans.
F max = k = 32(29.06) = 930 N Ans. ______________________________________________________________________________ 1 4-114 The initial potential energy of the k 1 spring is V i = k1a 2 . The movement of the weight 2 1 1 2 W the distance y gives a final potential of V f = k1 a y k2 y 2 . Equating the two 2 2 energies give 1 2 1 1 2 k1a k1 a y k2 y 2 2 2 2
Simplifying gives
k1 k2 y 2 2ak1 y 0
2k1a . Without damping the weight will vibrate between k1 k2 2k1a Ans. these two limits. The maximum displacement is thus y max = k1 k2 With W = 5 lbf, k 1 = 10 lbf/in, k 2 = 20 lbf/in, and a = 0.25 in
This has two roots, y = 0,
2 0.25 10 0.1667 in Ans. 10 20 ______________________________________________________________________________ ymax
Chapter 4 - Rev B, Page 80/81
