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; Problem 16.7 Determine the deﬂection “U as a function
of a: and conﬁrm the results in Appendix E. Solution: The expression for the bending moment in the beam is:
M r eF{L m 3:}
Integrating to ﬁnd the expressions for o’ and 1):
E11)” : WM : F(L 7 :c)
EIU’:F[L$— +A Eru=F[L§—%3] +Am+B Using the boundary condition at a: = O, r) = 'u’ = O, we see that
A = B = 0. So the deﬂection at a: = 1.15 m is: ANS: L swim; 7m} “z E: 2 set Problem 16.8 Determine the deﬂection o as a function
of :i: and conﬁrm the results in Appendix E. Solution: The bending moment along the entire length of the
beam is —M. Starting the integration process with Equation (16—
4):
EI'U” = wM 5 M0
EI'U’ = Mon: + A [1]
2
E11): M0;— +A$+B [2] We see that the boundary conditions are:
atm:0,'u:v':0. Substituting the boundary conditions into Equations [1] and [2], we
see that A = B = D, so Equation [1] becomes: '2 ANS: 'U = “3% Problem 16.9 Determine the deﬂection o as a function
of m and conﬁrm the results in Appendix E. Solution: The expression for the bending moment is: M=MT(m—L) Starting the integration process with Equation (16—4): E11)” = —M = Mfg — L) EIv’=—j%1(§—Lm)+l4 EIv=L£Q [iﬁE—ig—g]+Ax+B The boundary conditions are: atm:0,v:04B:U atm:L,o:O%A:M'GL/3
So the equation for deﬂection of the beam is: Mg [:33 L32] M0 L3 H=EIL 6__ 2 3E1 Problem 16.16 Determine the deﬂection o as a func
tion of a: and conﬁrm the results in Appendix E. Solution:
The bending moment for the sections 0 — a and a, — L of the beam
are: Mu_.a : w—F{a. m m) Ma—L m 0
Looking ﬁrst at section 0 — a and beginning with Equation (164): EI'uﬁLa — WIN/Iowa : F(a —~ as) m2
Elvﬁvﬂ : F (on: — +14 (11122 $3 EIUDWQ=F(—'2_—E) +A$+B Now considering the intervai a w L: EIoa,L = 0 The boundary conditions are: amp—0,90 av'D_GI—U—+ADBMD {1] at m : “1 'UDWH. : wu—L [2] atm : a, MFGk : v;_L [3] From boundary condition [2], we get: 03 (I3
F[3 mg] maC+D {4] From boundary condition [3]: F(a2—?)=O [5] Solving Equations [4] and [5] together: anal—F
2 Using these values for C and D, the expressions for 2 , 2
ANS: vain = g—E—Iﬁa — cc) 'uﬂﬁL = gﬁ 3n:— n.) M}
.C‘ Problem 16.24 The titanium beam has elastic modu—
lus E = 16 x 106 lb/sqin. Determine the axial position
m at which the magnitude of the deﬂection is a maximum. Solution:
The expression for the bending moment is: 3 .
M = _@ [55—  Lac]
ti 'L
Starting the integration process with Equation (164): ED)” = —M = 1% — Lm] 4 2
Ehgg=1a[:—L—Lei+A [11
$5 EIU:%Q[W*L§3]+A:C+B The boundary conditions are: atm=0,o=D—}B=0 at$=L,n={l—>A= Substituting this value for A into Equation [1}. the expression for the
steps of the beam is: 'v'  —w0 360LEI Setting Equation [2] equal to zero and soivin g for :n, we get the location
of the maximum deﬂcction. a: = 0.519 (7L4 w 30L2m2 + 15:61) {2] L = 0.519(70 in)
ANS: z : 36.4 in Problem 16.25 The prismatic beam has elastic modu lus E and moment of inertia I. Determine the reactions
at A and B. Solution:
Summing vertical forces on the beam (Figure A): EFy = 0 = —Ay + By {1]
Summing moments on the beam:
2M:0:MA+ByL“—‘MG [2]
The bending moment at any place on the beam is (Figure B):
M : MA + Aux
Beginning with Equation (16—4):
E10" 2 —M = —MA ~Aym 2
Ag :1: EIo'=—MAm—— +0 _ 2 A 3
Enszgiu 7g" +C$+D The boundary conditions for the beam are: atm=0,va’=O—>C=D=G —M L2 AL3
atm:L,o:0ﬁ\4~§;wm 3’23 m0 Solving Equations [1], [2] and [3] together, we get: ANS: A1 :% tBszsé‘f TMA=£§1 Free Body Diagram:
MA Problem 16.31 The prismatic beam has elastic modu
lus E and moment of inertia I. Determine the reactions
at A. (Assume that the supports exert no axial forces on the beam.)
Free Body Diagram: Solution:
Summing vertica! forces on the beam: 2F” =0=—ng+Ay+By Summing moments about point A on the beam:
EMA = G = M A — W — — U
i 3"“
x wI v3/
5’
The bending moment equation for any point on the beam is 1110.132
2
Starting with Equation [3] in Equation (164): MzwMAJrAymi [3] womz EITJ”:—M=MA_Ay$+ Aye? + 100533 EI':M 
1: An: 2 6 +C M'AJI;2 Agata + 1:10:34
6 24 Eh): +Cw+D Using {he boundary conditions: atm:0,v:v'=0%C=D=0 MAL2 H AyL3 + 2091.34 _ atm:L,1J:04 —D [4}
2 6 24 A L2 L3
y + “’0
2 6
Solving Equations [i], [2], {4], and 15] together, we get: . L 7 7 L2
ANS. Ay=By=Eg—MA_MB_‘9%§m atsch,v'=0—>MAL— =0 {5] Problem 16.40 The prismatic beam has elastic moduw
1115 E and moment of inertia I. Determine the reactions
at A and B. (Assume that the supports exert no axial
forces on the beam.) Solution: Free Body Diagram:
Summing vertical forces on the beam 2F. =0=Ay+By——“~’%Ii {1] Summing moments about point A on the beam MA—M3+ByL—WTDL :0 [2] The bending moment in the beam in the interval 0 S m g L / 2 is: M1 : MA 7 Ayn: [3] MA M1
ﬁat} Starting with Equation [3] above in Equation 064): E
i
i
.
. EI’Ui’ = —M1 = ~MA +Ay$ A 2
Erv;=—MAm+ W + C [4}
—M m2 A x3
___.A_ + y 6
The bending moment in the interval I/Z g a: g L is: Elm = + Cm + D [5] 1
M2 : MB 7 39(L uw) + Ema? —2L:c+1:2) {5} Now using Equation [6} above and starting with Equation (i6v4): ' 1
EMS = "M2 2 —MB + By(L — m) — Ema? — 2m: + m2) 2 1 3
ED); 3 —MB:J:+By (La: — %)—§m9 (L22: * L322 + %)+0 [7] EIng = 2 6 —M332 +B (L32 :33) 1 (L2$2 133:3 $4
— y D
2 2 " ""5"" + {8] Using the boundary conditions: atm:0,111:vi:0ﬁ¥C:D:O [9] 2 6 atm:L,v2:0% 2 —M L2 L3 L3 1 L4 L4 L4
B +By( ) “(2 3 ——— —— GL 11:0 10
2 +12)Jr + [l n r_ 2 L2 1 a 3 L3 _
atmeL,v270~+u—MBL+BI, L ——2— ——2—wo L ——L +F +G_o [11] 414541;2 AyLa ﬁMBLQ (L3 L3) 1 (L4 L4 L4 ) GL
t —L2, — — B ——— —— mmw 7 M H 12
a”: [1’1 “2’ 8 +48 8 +9 8 48 2”“ 8 24+192+2+ I] t=L2l=I—> _ 777
am /’”l “2 2 +8 2 2 8 Solving Equations {1], [21, [10], [11], {12], and [13] together, we get: . _ 3 L m 132.!) L 7 L3 i L4
ANS. All—11% By_—:§— G__I£g§_. H__%®80T
MA = SwQLZ M _ 111ml.2 192 Mm 2 iMAL 11ng MMBL L2 L2 1 L3 L3 L3
7 +33; 0 MH— — 1
2 4+24)+G {3] Problem 16.53 In Problem 16.52 determine the de~
ﬁection in the beam as a function of m for D g as g L / 2. Solution:
Superimposing the displacements for W
Illll
ﬂ_& From Appendix E for 0 5 m g L/2, we get: _ won: Fm
"d _ 2431 48E] Letting n = U at a: = L /2, the reaction at B is (L3 — 2Lm2 + m3) — (3L2 — 43:2) Summing verticai forces and moments on the free body diagram: 3woL
16 The expression for deﬂection on the left—hand portion of the beam is: _ won: _ (5ng/8)m
’ 2431 48131 ANS: v = % (L3 — mm2 + 16333) A20: '0 (L3 — 2m? + 1:3) (3152 — 433) Problem 16.54 Determine the deﬂection of the beam
as a function of m. Solution:
Superimposing the deﬂection for “CM and Me r—u Io ~——1
from Appendix E and setting a. : L, we get: Mon; 2_ 2
(2L 3Lzr+m )+6LEI (L " m) (6L2 A at2 w 2L2 — $2) a Mom
GLEI (3r.2 — 31m) ...
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 Fall '08
 CHATTERJEE

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