2103-231 Mechanics of Material Chapter 12. Deflection of Beams
2103-231
Mechanics of Material
Chapter 12. Deflection of Beams
รศ.ดร.ไพโรจน์ สิ งหถนัดกิจ
ภาควิชาวิศวกรรมเครื่ องกล คณะวิศวกรรมศาสตร์
จุฬาลงกรณ์มหาวิทยาลัย
This lecture note is prepared for my in-class lecture. It is intended to be incomplete!
Failure of beam
!1!
@
Deflection exceed
To find
deflection
!1!
Double
Q
Moment
!3!
limit
stress exceed
limit
.
integration
-
Mechanics of Material
method
area
superposition
2103-231
.
Chapter 12. Deflection of Beams
รศ.ดร.ไพโรจน์ สิ งหถนัดกิจ
ภาควิชาวิศวกรรมเครื่ องกล คณะวิศวกรรมศาสตร์
จุฬาลงกรณ์มหาวิทยาลัย
This lecture note is prepared for my in-class lecture. It is intended to be incomplete!
12.1 Elastic curve
Recall:
M
Idd
M
.
'
'
a
'
.
B
Original configuration
O=tg=
-
tf
TIE
-
E
-
:e
¥±#t*%n€
¥t @=
.
N.A
-
A
-
.
{
AB
=
E
B
=
ltd
Deformed configuration
I
qy
-
far
)
T
Md
Elastic
curvet
y=fcx )
2
12.2 Slope and displacement by integration
From calculus: radius of curvature ρ can be represent in form of parameters x
and y as
}
y
=¥÷
d#Y
1-
y=f(x)
: @#y]k
3
Mcx )
'
.
.
FI
x
MET
Recall
:
ddI×¥
-
wan
,
ddmy
=
÷
'
it
=
date
Van
#dEe=nxsEIg÷y=.wa£
or
EIda},.=rf
tYYht9
3
12.2 Slope and displacement by integration
KEEL
.
Thus, the differential equations for elastic curve are:
d2y
EI 2 = M ( x )
dx
or
d3y
EI 3 = V ( x )
dx
or
d4y
EI 4 = − w( x )
dx
4
12.2 Slope and displacement by integration
Approach: Integrate the differential equation
twice to obtained
y = f (x) or y = v (x)
In each integration, a constant ci is produced, so boundary, continuity or
symmetry conditions are required
d2y
EI 2 = M ( x )
dx
Boundary condition:
Simple support:
y
y
x at
x
n=o
Wako
y
Fixed support
#¥¥
x
@
w=O
,
VCO
)=0
IF't
5
12.2 Slope and displacement by integration
Continuity condition
P
1
YMH
Vzct
YAD
=
de
y
kra
2
Min
'
)
AIK )
"
=
:
Symmetry condition: If load is applied in the middle of the beam, it can be shown that
E
"
=
via
=o
6
Example
5-28. A beam is loaded and supported as shown in the figure. Derive the equation of
elastic curve, and also determine the slope and deflection at the left end.
.
B.d.
vc4=o
dva )
Mc×1+Pu=0
=
-
Ph
EIVCM
pp
tva ,7
MCX )
Bt
.
=
-1¥ -14mHz
dd÷c4=o EIco)=
:
.
.vcxi=±±fEn3tIlh+¥)
0=
)=o
3
Paid
-
,
4=112
2
.
vu
:
Pu
#d£×=t¥+4
a
MCH
n
-
0
%
.
EId↳m=mc×)=
.
"
tax
P
:
-
PL
:y-±
'
'Fa'
at
.
to
7
.
Example
:
//#
Le
a)
y
.net#fEn3tIrn+BIj=E,=th43in-2D-
b)
stpeatnro
slope
:
--d€=¥±tp¥+kI )
.int#DFoterco=oPz,=C-zdI=t13EI
8
BC
.
EIY
)=o
KCO
Kato
viC¥=kt
Example
riche )=vzCtE )
9.3-19 Derive the equations of the deflection curve
for a simple beam AB. Also, determine the
deflection δC at the midpoint of the beam.
¥otim0,setim0
h
EIvEH=Mi
EIvkm=
%qL
EIrexi.tk#+gyIgtsu+4
EIY*=tfT+8yT+4u+h
eIrEm=t÷np+9¥utb
Marie q¥+q±
SIM
oH<
12
-
s
-9dL+3zgLn
section
"cxi=Me× '
-
Mz
Kye ,÷qui+L
me 's
,
¥u<
!2!
=
µ
QL
-
8
L
-
8
9
Example
klo
vzc
-0=>4--0
)
d4=
4=0
vh÷)=v!c¥
-
QLY
Zq
LO
-14L
4=4+1
48
v
,kd=kL¥ ↳=17q ¢ 4=-g±d
384
d4= 914
384
:
.mx
%<
){jb÷*±Cl6h3t24n4t9B
jY÷±±l8htr4nt+hxi d)
.
z
¥n<
<
10
Example
µ
9.4-8 Derive the equation of the deflection curve for a
simple beam AB carrying a triangularly distributed load
of maximum intensity qo. Also, determine the
maximum deflection δmax of the beam. Use the fourthorder differential equation of the deflection curve (the
load equation).
EIda±x=-qm=
EIV
EIR
13€
qoy
#
,
1=0
Mco )=o
-
'
.
C
'=tz÷tt+q÷mu
,
,
"=t÷Y+dµ+h=mc×
.
'=foy
E±i=Tfht+8¥u
.
"=t±iu
Muto
't
t r=tfh÷+8;thtgu+h
1 BE
VCO
)=o d¢=o
04=0
q¥+d,L=o
:
,=qoL
,
.↳=
-90¥
't8÷t+↳c=o
-
7qoI
360
-
•
:
.v*=÷±ctcnw5eG¥f3%¥ )
11
Example
or
,
vcx )
=
( she
god
-
't7L4 )
loin
Arg
360 LEI
Determine
Jmax
:
vin=¥±lt÷Ittto¥ -7¥
at
fmx
vin÷
,
tygttttolhd
Solve
.
.
.
H÷=o
0.5193L
n=
Emay
-
=
=
-
VC 0.519 }L )
0.0065229014
EI
Ans
-
12
we
)=qU
VLX
"
ftp.mkibp.fm
MA
Example
9.4-4 A beam with a uniform load has a guided support at one end and spring support at the other. The
spring has stiffness k =48EI/L3. Derive the equation of the deflection curve by starting with the thirdorder differential equation (the shear-force equation). Also, determine the angle of rotation θB at support
B. (Gere&Goodno)
MCL )=0
13¥
,3=KsB=kmL
!1!
|
!2!
VKO )=o
!3!
p
R , }=
-
Begin with
EIdd1×t=VcN=
EIV
'
VCL )=
-481¥
'tC4
qL=
Egnh
KVCL )
"=
-
-8212+2
Tx ,
48EI
gu
mc↳=o
,
:
-KI+4=o
.
d
EIi=
-
-81
Gydtktrtlz
,=Kz
13
Example
'
EIV
EIV
¥
Was
ra '
-8¥ +9ha
=
=
-82nF
totted
ttf
;gf÷→
-
+4
the
dz=o
=o
:
:
+22
ra
=
.
↳
-
=
the
tofts
Its got
'E±to÷e4+g i ¥sgiY
93=44
.
=
q¥±
}
And
14
12.6 Statically indeterminate beam
P
:
(
P
µ
Free body diagram
Force and moment at the support can be determined from static equilibrium
So, it’s statically determinate problem, no problem, everyone is happy.
and
M
P
L1
:
V
,
:
M
,
E Fy
P
:(
L2
unknown
eqt
determined
be
can
v
a
R
Free body diagram
R
=0
,
EM
-
°
,
Need
VCH
one
=
more
eq
"
o
15
12.6 Statically indeterminate beam
Other statically indeterminate members
16
.
→
:
sp
the )
qH=q°(
Example
,
Tris
M€~Ebm
10.3-8 A fixed-end beam AB of length L supports a triangularly distributed load of maximum
intensity q0. Beginning with the fourth-order differential equation of the deflection curve (the
load equation), obtain the reactions of the beam and the equation of the deflection curve.
Lwhkmhm
[ EFy=o ]
µc×
n
:
RA
RB
,
Rat
RB
[ EMA
Ma My
,
,
8¥
=
"
]
!1!
-
Ma
Metric
-
-
,
Elastic
"
EIV
"
EIV
EIV
EIV
curve
"
foot
-
=
'
=
qoihe ,
-
focktldtt
-
-
!2!
.
"=-qcxi=
=
qqI=o
-
goat
,
nestle
-
.
Lah
)
,nth
tdilthutb
17
H
auto
+dpE+hu+d/°
,¥÷d+d¥+hn÷t↳fiHf°
EIV
'
=
EIv=
Example
Yak )
(
go
-
-
Jcoto
1¥
=
↳
via
Reaction
-
2Lz=q#
diet
)=o
-
*
4=0
v(o)=o
.
.
.
:
.
d
.
war )
Izoqol
=
-
Koh
120
=8¥
4
,
C
-
-
-
-
Fo÷
°
h3+5nL>uE+3d )
-
=
RA=Vco )
44322
=>
=
Ko )
=
'
V=EIv"
4
v(↳=o
RA
0
golhtdjlt Ioqol
zoqol
=
*
1
RBIVCL
m=EIr
Ma
=
Foqol
"
MCO )
=
)
to qoll
a
LEI
MB
=
Mcb
'
=
-
g÷y
}D¥
18
$
hem
It
. '
hat
'
.
P
n
Example
RA
6-5 A beam is loaded and supported as shown in the figure. Determined (a) the reaction at
supports A and B in terms of M and L, (b) The deflection at the middle of the span in terms of
M, L, E and I.
V=o
Ra
-
MBTM
Mah
-
=
RAL
Ran
EId£
r=
13€
Lio
vcoto
vkh=o
d,
=
ML
-
r¥i
=0
=
-
0
-
!2!
M
MCH
=
Rae
EIV '= Ranft Mntd
EIv=
VCL )
!1!
-
RAY
+21
-
M
,
tdptdc
RA= IM
2L
From
!1!
VIET
,
From @
MB=Mz
ad
19
Example
20
gmo
GMA
GMB
IVRB
Pra
RA=RB
MBTMO
.
!1!
-
Ma
-
Ratio
Example
-
!2!
10.3-10 A counterclockwise moment M0 acts at the midpoint of a fixed-end beam ACB of length L (see
figure). Beginning with the second-order differential equation of the deflection curve (the bendingmoment equation), determine all reactions of the beam and obtain the equation of the deflection curve for
the left-hand half of the beam. Then construct the shear-force and bending-moment diagrams for the
entire beam, labeling all critical ordinates. Also, draw the deflections curve for the entire beam.
M
,##µgm*
:|
y
n
PRA
Moe )=
.
.
.
Mat Ran
-
Mo
MDk+RA¥t↳
.
EIKECMA
KCO
.mn#.+r.n/EtEIYIIIYFYLYIlEIY'=mantRa=hld
⇐#ra
,=M¥i+R±k3e4u+h
-
B€YEo)=o 4=o
kw=0
,
EIV
)=O
dz=0
R#i
da=CMa-Mo)¥+R¥
'
21
Example
via
Mats
tRa¥
Malt
+
:
CMA Mo )
-
=
Raf
HE
MAGI
via ,
=
)
Raf
st C
.ra.s€
Mqz
MB
=
-
t
-
RAI
( Ma MDL
-
-
-
Ra¥
30
ka
=
=
Mol
=
}
Yfo
Ifo
Mandi +5€ RAP
RB
.
32mF
}
Ads
22
Example
q N/m
A
6-2 Determine reaction at A and B on term of q and L
B
Lm
2L m
Va
VB=
=¥sql
IJQLP
MA=±gqe
MB=÷qt
cow
cw
23
Example
24
Additional Note
25
Additional Note
26
Additional Note
27
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