9 beam deflection 2
TRANSCRIPT
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MECHANICS OF
MATERIALS
Third Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. e!ol"
Le#ture Notes$
J. !alt Oler
Te%as Te#h &ni'ersit(
CHAPTER
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9Deflection of Beams
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MECHANICS OF MATERIALST h i r d Beer ) Johnston ) e!ol"
Deformation of a Beam Under Transverse Loading
• Relationship between bending moment and
curvature for pure bending remains valid for
general transverse loadings.
EI
x M )(1 = ρ
• Cantilever beam subected to concentrated
load at the free end!
EI
Px−=
ρ
1
• Curvature varies linearl" with x
• #t the free end A! ∞== A A
ρ ρ
!$1
• #t the support B! PL
EI B
B
=≠ ρ ρ
!$1
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Deformation of a Beam Under Transverse Loading• &verhanging beam
• Reactions at A and C
• 'ending moment diagram
• Curvature is ero at points where the bending
moment is ero! i.e.! at each end and at E .
EI x M )(1 = ρ
• 'eam is concave upwards where the bending
moment is positive and concave downwards
where it is negative.
• a*imum curvature occurs where the moment
magnitude is a ma*imum.
• #n e+uation for the beam shape or elastic curve
is re+uired to determine ma*imum deflection
and slope.T
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MECHANICS OF MATERIALSTh i r d Beer ) Johnston ) e!ol"
Equation of the Elastic Curve
• rom elementar" calculus! simplified for beam
parameters!
2
2
2%2
2
2
1
1
dx
yd
dx
dy
dx
yd
≈
+
= ρ
• ubstituting and integrating!
( )
( )
( ) 21$$
1
$
2
21
C xC dx x M dx y EI
C dx x M dx
dy EI EI
x M dx
yd EI EI
x x
x
++=
+=≈
==
∫ ∫
∫ θ
ρ
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MECHANICS OF MATERIALSTh i r d Beer ) Johnston ) e!ol"
Equation of the Elastic Curve
( ) 21$$
C xC dx x M dx y EI
x x
++= ∫ ∫
• Constants are determined from boundar"
conditions
• 0hree cases for staticall" determinant beams!
impl" supported beam$!$ == B A y y
&verhanging beam$!$ == B A y y
Cantilever beam$!$ == A A y θ
• ore complicated loadings re+uire multiple
integrals and application of re+uirement for
continuit" of displacement and slope.
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MECHANICS OF MATERIALSTh i r d Beer ) Johnston ) e!ol"
Direct Determination of the Elastic Curve From the
Load Distriution
• or a beam subected to a distributed load!
( ) ( ) xwdx
dV
dx
M d xV
dx
dM −===2
2
• 3+uation for beam displacement becomes
( ) xwdx
yd EI dx
M d −==,
,2
2
( ) ( )
,%2
221%
11 C xC xC xC
dx xwdxdxdx x y EI
++++
−= ∫ ∫ ∫ ∫ • 4ntegrating four times "ields
• Constants are determined from boundar"
conditions.
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!taticall" #ndeterminate Beams• Consider beam with fi*ed support at A and roller
support at B.
• rom free-bod" diagram! note that there are four
un6nown reaction components.
• Conditions for static e+uilibrium "ield
$$$ =∑=∑=∑ A y x M F F
0he beam is staticall" indeterminate.
( ) 21$$
C xC dx x M dx y EI
x x
++= ∫ ∫
• #lso have the beam deflection e+uation!
which introduces two un6nowns but providesthree additional e+uations from the boundar"
conditions7
$!#t$$!$#t ===== y L x y x θ
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!am$le Prolem %&'
ft,ft1/6ips/$
psi1$29in52%81, ,
===×==×
a L P
E I W
or portion AB of the overhanging beam!
(a) derive the e+uation for the elastic
curve! (b) determine the ma*imum
deflection!
(c) evaluate ymax.
&:04&;7
•
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!am$le Prolem %&'
&:04&;7
•
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MECHANICS OF MATERIALSTh i r d Beer ) Johnston ) e!ol"
!am$le Prolem %&'
PaLC LC L L
a P y L x
C y x
1
1$7$!at
$7$!$at
11%
2
=+−===
===
• 4ntegrate differential e+uation twice and appl"
boundar" conditions to obtain elastic curve.
21%
12
1
2
1
C xC x L
a P y EI
C x L
a P
dx
dy EI
++−=
+−=
x L
a P
dx
yd EI −=
2
2
−=%2
L
x
L
x
EI
PaL y
PaLx x L
a P y EI
L x
EI PaL
dxdy PaL x
La P
dxdy EI
1
1
%1
121
%
22
+−=
−=+−=
ubstituting!
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MECHANICS OF MATERIALSTh i r d Beer ) Johnston ) e!ol"
!am$le Prolem %&'
• ocate point of ero slope or point
of ma*imum deflection.
−=
%2
L
x
L
x
EI
PaL y
L L
x L
x
EI
PaL
dx
dym
m /55.$%
%1
$2
==
−==
• 3valuate corresponding ma*imum
deflection.
( )[ ]%2
ma* /55.$/55.$
−= EI
PaL y
EI
PaL y
$,2.$2
ma* =
( ) ( ) ( )
( )( ),2
ma*in52% psi1$29
in18$in,86ips/$$,2.$
×= y
in2%8.$ma* = y
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!am$le Prolem %&(
or the uniform beam! determine the
reaction at A! derive the e+uation for
the elastic curve! and determine the
slope at A. (;ote that the beam is
staticall" indeterminate to the first
degree)
&:04&;7
•
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MECHANICS OF MATERIALSTh i r d Beer ) Johnston ) e!ol"
!am$le Prolem %&(
• Consider moment acting at section D!
L
xw
x R M
M x
L
xw x R
M
A
A
D
$%2
1
$
%$
2$
−=
=−
−
=∑
L
xw x R M
dx
yd EI
A
%$
2
2
−==
• 0he differential e+uation for the elastic
curve!
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!am$le Prolem %&(
L
xw x R M
dx
yd EI A
%$
2
2−==
• 4ntegrate twice
21
/$%
1
,
$2
12$
1
2,2
1
C xC L
xw x R y EI
C L
xw x R EI
dx
dy EI
A
A
++−=
+−== θ
• #ppl" boundar" conditions7
$12$
17$!at
$2,2
17$!at
$7$!$at
21
,
$%
1
%$2
2
=++−==
=+−==
===
C LC Lw
L R y L x
C Lw
L R L x
C y x
A
Aθ
• olve for reaction at A
$%$
1
%
1 ,$
% =− Lw L R A ↑= Lw R A $1$
1
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!am$le Prolem %&(
x Lw L
xw x Lw y EI
−−
= %$
/$%
$12$
1
12$1$
1
1
( ) x L x L x EIL
w y ,%2/$ 2
12$
−+−=
• ubstitute for C1! C2! and R # in the
elastic curve e+uation!
( ),22,$ /12$
L x L x
EIL
w
dx
dy−+−==θ
EI
Lw A
12$
%$=θ
•
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MECHANICS OF MATERIALSThi r d Beer ) Johnston ) e!ol"
)ethod of !u$er$osition
>rinciple of uperposition7
•
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MECHANICS OF MATERIALShi r d Beer ) Johnston ) e!ol"
!am$le Prolem %&*
or the beam and loading shown!
determine the slope and deflection at
point B.
&:04&;7
uperpose the deformations due to L!adi"# I and L!adi"# II as shown.
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!am$le Prolem %&*
L!adi"# I
( ) EI
wL I B
%−=θ ( )
EI
wL y I B 8
,−=
L!adi"# II
( ) EI
wL II C ,8
%=θ ( ) EI
wL y II C 128,=
4n beam segment C'! the bending moment is
ero and the elastic curve is a straight line.
( ) ( ) EI
wL II C II B ,8
%== θ θ
( ) EI
wL L
EI
wL
EI
wL y II B %8,
5
2,8128
,%,
=
+=
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!am$le Prolem %&*
( ) ( ) EI
wL EI
wL II B I B B ,8
%% +−=+= θ θ θ
( ) ( ) EI
wL
EI
wL y y y II B I B B %8,
5
8
,,
+−=+=
EI wL
B,85 %=θ
EI
wL y B
%8,
,1 ,=
Combine the two solutions!
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MECHANICS OF MATERIALShi r d Beer ) Johnston ) e!ol"
A$$lication of !u$er$osition to !taticall"
#ndeterminate Beams
• ethod of superposition ma" be
applied to determine the reactions at
the supports of staticall" indeterminate
beams.
•
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MECHANICS OF MATERIALShi r d Beer ) Johnston ) e!ol"
!am$le Prolem %&+
or the uniform beam and loading shown!determine the reaction at each support and
the slope at end A.
&:04&;7• Release the ?redundant@ support at '! and find deformation.
• #ppl" reaction at B as an un6nown load to force ero displacement at B.
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!am$le Prolem %&+
•
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!am$le Prolem %&+
( ) EI
wL
EI
wLw A
%%
$,15.$2,
−=−=θ
( ) EI wL L L L
EILwL R A
%2
2 $%%98.$%%
$88.$ =
− =
θ
( ) ( ) EI
wL
EI
wL R Aw A A
%%
$%%98.$$,15.$ +−=+= θ θ θ EI
wL A
%
$$59.$−=θ
lope at end A!
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