This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: E
; 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 = [email protected] [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) ...
View
Full Document
 Fall '08
 CHATTERJEE

Click to edit the document details