Unformatted text preview: Problem 9.23 The deﬂection curve for a cantilever beam AB
(see ﬁgure) is given by the following equation: 2
v = — JUL—(top — 10sz + 5m — x3)
1201.5! Describe the load acting on the beam.
Probe. 9.23 and 9.24 Solution 9.23 Cantilever beam (1013—10sz + 51.3 —x3) Take four consecutive derivatives and obtain: an ._ _ﬂ ._
v LEI“ x} From Eq. (912c): q= —Efv’”' =qo(l —£) 4— The load is a downward triangular load of maximum
intensity qcr 4— Problem 9.39 Derive the equation of the deﬂection curve
for a simple beam AB loaded by a couple M0 at the lefthand
support (see ﬁgure). Also. determine the maximum def] ection M0
6“. (Note: Use the second—order differential equation of the G
deﬂection curve.) Solution 9.39 Simple beam (couple MO) BENDINGvMOMENT aoumon (Bo. 912a) Mammm DEFLECTION
n .. 3 _ i I = _£ 2 _
51v —M M00 L) it “ﬂat. 6Lx+3x1)
: I} s l ' = 0 d 1 r 
Eh =Ma(x__2z)+cl 1: VL 1 anvgove orx.
= — — 4—.
x2 x3 ‘1 ( 3 )
51v =Mo(— )+ C.x + C2
2 61 Substitute xl into the equation for v:
3.0 v(0} = D C2 = i} MDL 3m 2 _ (v):_xl
as. v(L] = 0 C, = —— = .M'JL .—
3 9V§El Mo":
6 LE! [These results agree with Case 7. Thble G2.) v= (2L1—3Lx+x2) — Fraulein 9.312 The beam shown in the ﬁgure has a roller support at A and a guided support at B. The guided support permits vertical movement
but no rotation. Derive the equation of the deﬂection curve and determine the
deﬂection 55 at end 3 due to the uniform load of intensity q. (Note.
Use the secondorder differential equation of the deflecti on curve.) Solution 9.312 Béam with a guided support  REACTIONS AND DEFLECI'ION CURVE BENDINGMOMENT EQUATION (EQ. 9—123)
qxz
Elv" = M = qu — 7
quz qxj
a: ' — — — — +
2 6 C1
L3
3 c m) = 0 C1 = 5’?
qu’ go: (11.31
E! = — — — — — +
v 6 24 3 C2 4x 3
= m — + 4—
v 24E:(3L 4sz r")
SqL‘
5 =— L = 4—
“ ( ) 24E] Problem 9.45 The distributed load acting on a cantilever beam AB has an intensity 9 given by the expression go cos 17x {213. where go is
the maximum intensity of the load (see ﬁgure). Derive the equation of the deﬂection curve, and then determine
the deﬂection 53 at die free end. Use the fourthorder differential
equation of the deﬂection curve (the load equation]. Solution 9.45 Cantilever beam (cosine load) LOAD EQUA‘I'ION (EQ. 912 B) 3.0. 3 v'{0) = 0 C1 = 0
17x 4 2
En." = —q = —qn cos— = _ E qoLxs — qoL :2
21. u Elv 4., 1? cos 2L + 3” n + C;
M _ ' m
Elv  qo(:) SIRE + C, 16%;}
3.12.4 v(0) =0 c4: 4
11'
L
BvCl EN” = V Elvmm =0 Cl=2q1: 90" (48 3 m‘ 48 3 379m: 1913)
= — L —— L + — ‘—
2 2 v MAE! cosm
2L mt anx
Elv” = qo(—) 005— + + C2 ‘
7r 2,; L
w a, = —v{L) = 24" (a? — 24) 4— 3716‘s: 29013 {These results agnse with Case 10. Table G 1.) 34:. 2 En!" =M EIv"(L) =0 C2 =  ﬂ 2L 17x grub?" _ 2%sz 3
Elv' = 1.1.4?) Sing + n 1r 4 C3 ...
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