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Unformatted text preview: y z 212} wan) [21 13]
6E1] 2 2 24E] 2 2 “ 340(15)(19.5) " 6(30)(106)(0.2485)(39)
(150/12)(19.5) +———~—————g——.—————[2(39)(19.52)19.53 493] = —0.1027 in Ans.
24(30)(10 )(0.2485) y [19.52 +152 —392] % difference :4 W000) = 5.01% Ans. —0.0978 413 I = 115(6)(323) =16.384(103) m4 From Table A—910, beam 10 Fa2
=——— Z+
yC 3EI( 0)
Fax 2 2
yAB:6EIZ(Z “‘35)
ﬂﬂ=ﬂ(12_3x2)
dx 6E1] Atx=0,d:1f =6A “ Fa]2 ﬂ ﬂ
6EIZ 6E1
FaZI
6E1 With both loads, _ F612] _ Fa2
6E1 3E]
2 2
: _f_‘l_(3[ +200 : __ﬂ03g’29_2w_3.
6E1 6(207)10 (16.384)10 At midspan, _2Fa(Z/2) KEY _ _3_ M2 2 _3L 400(300)(5002)
6E1] 24 E] 24 207(103)16.384(103) 9A Yo =_6Aa: yo = (1+a) [3(500) + 2(300)] = —3.72 mm Ans. =1.11mm Ans. yE 2 414 1 = 33424 —1.54) = 0.5369 in“ Chapter 4  Rev B, Page 9/81 4100 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. T 1. Statics. R1 + R2 = wl l
l (1) w
' I I I I I I I I I I i i i ‘
ml.
R21+M1=—:—wlz (2) “<1 I
‘ i
1 R2 2. Bending moment equation. R
1 2
M = Rlx—wa M1
E19791 = lax2 — l W3  Mlx + C1 (3)
dx 2 6
1 3 1 4 1 2
EIyz—éRlx —wa “EMIIXI +C1xlC2 EI= 30(106)(0.85) = 25.5(106)1bfin2.
3. Boundary condition 1. At x = 0, y = — Rl/kt = — rel/[15006)]. Substitute into Eq. (4)
with value of E1 yields C2 = — 17 R1. Bounan condition 2. At x = o, dy /dx = — M1/k2 = — Mt/[2.5(106)]. Substitute into
Eq. (3) with value of E] yields C1 = —— 102 M1. Bounde condition 3. At x = I, y = — R2/k3 =  Jet/[2.00 06)]. Substitute into Eq. (4) with value of E] yields
42.751?2 = —6Rll *awl ~§Mll —10.2Mll—17R1 (5)
Equations (1), (2), and (5), written in matrix form with w = 500/12 lbf/in and Z = 24 in,
are
1 1 0 R1 1
0 24 1 R2 = 12 (103) 2287 12.75 ——532.8 M1 576 Solving, the simultaneous equations yields R1= 554.59 lbf, R2 = 445.41.591bf, M1 = 1310.1 lbfin Ans. For the deﬂection at x = [/2 = 12 in, Eq. (4) gives Chapter 4 — Rev B, Page 69/81 6U = 0
aMA
The bending moment at an angle 6to the x axis is
M=MA~E(l—cose) 6M =1
2 EMA
The rotation at A is
2/2
6A: 6U =—L [M W Rd6=0
aMA E1 0 (MIA
Ir/Z
Thus, ~1~ J[MA —E(l—cosﬁ)](l)Rd6 = O ::> (MA —E)£+E = 0
E1 0 2 2 2 2
or,
MA = up 2)
2 72‘
Substituting this into the equation for M gives
M = ﬂ[cost—Z—J (1)
2 it
The maximum occurs at B where 6 = 7r/2
Mmax =MB =~f§ Ans.
7r (b) Assume B is supported on a knife edge. The deﬂection of point D is a U/é’F. We will
deal with the quarterring segment and multiply the results by 4. From Eq. (1) 6M =—I:(COSH—~2;] E 2 72'
Thus,
ir/2 321/2 2 3
5D=§E=i MaﬁkdezFR [(0056—2) d6=FR {1—3
61? E10 aF E10 27 E147:
3
= FR (7238) Ans.
47rEI
4104
PcrzCﬁ:EI
l
4
I l( 4d4)=ﬂD (l—K4) whereK=i
64 64 D
2 4
Per_C7Z'2E 7TD (1__K4)
l 64 Chapter 4  Rev B, Page 74/81 2 1/4
D: “364—1111‘?‘ Ans.
n“ CE(1—K ) 4105 A:%D2(1—K2), I=—6’—;—D4 (14639:?)4 (1—K2)(1+K2), WhereK= d/D. The radius of gyration, k, is given by
2
k2=—I—=P—(1+K2)
A 16
From Eq. (446) P sz2 5312
_““L“T=Sy——T§_=Sy"—TT_—“7_
(7r/4)D2(1—K) 47: k CE 47: (D /16)(1+K )CE 45212 132 l—K2
7: D (1+K )CE 22 _ 2
7rD2(1—K2)Sy=4pcr+w
77(1+K )CE 4p 45212 (1—K2) “2
D: ———~—————~°‘ +~—————————————y
[ﬂSyO—Kz) 7r(1+K2)CE7r(1—K2)Sy] 1/2
522
zSy(1—K) 7r CE(1+K)
4106 EM —0 075 800 ————r__.__*0‘9 F 05 —0 :> F ~1373N
‘ (a) A— a ) 0.92+0.52 Bo(  )— 30‘ Using nd = 4, design for For = nd F Bo = 4(1373) = 5492 N
Z: 0.92+0.52 =1.03 m, Sy =165 MPa Inplane: 1/2 3 W
k = = [b12212] = 0.288711 = 0.2887(0025) = 0.007218 m, C =1.0 =1_'°3_.=142.7 l
75 0.007218 [1] g 2z2(207)(109) “2 21574
k 1" 165(106) ' Chapter 4  Rev B, Page 75/81 4112 Loss of potential energy of weight = W (h +6) Increase in potential energy of spring = ék62
W (h +6 ) = éké‘z or, 52 —3£Z6—2£:h = o. W: 301m; k=1001bf/in,h = 2 in yields 62—0.66—1.2=O Taking the positive root (see discussion on p. 192) am =—;[0.6+./(—0.6)2 +4(1.2)] = 1.436 in Ans. FmX = k 5mm = 100 (1.436) = 143.6 lbf Ans. 4113 The drop of weight W1 converts potential energY, W1 h, to kinetic energy—{Er}? . 2g Equating these provides the velocity of W1 at impact with W2.
Wlh = of :> 171 =t/2gh (1)
2 g Since the collision is inelastic, momentum is conserved. That is, (m1 + m2) 122 = m1 711,
where 222 is the velocity of W1 + W2 after impact. Thus W1+Wz W1
022—221 2 '02 g g _ W1
Wi+Wz til: W1 42gb (2) Wi'l'Wz The kinetic and potential energies of W1 + W2 are then converted to potential energy of
the spring. Thus, fit—WINE v§+(I/V1+W2)5=lk52
g 2 Substituting in Eq. (1) and rearranging results in W1 + W2 5 _ 2 WE 12 z
k m + W2 k
Solving for the positive root (see discussion on p. 192) 2 2
=1 2WI+W2+ 4(WHW2] +8 WI ﬂ (4)
2 k k m+W2k 52—2 0 (3) Chapter 4 ~ Rev B, Page 79/81 ...
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This note was uploaded on 03/07/2012 for the course MAE 3090 taught by Professor Dr.yang during the Spring '10 term at FIT.
 Spring '10
 Dr.Yang

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