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8/10/2019 Deflection of Beam- IsM
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DEFLECTION OF BEAM
x
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BE M DEFLECTIONS
As a beam is loaded,Different regions are subjected to V and M
The beam will deflect
Recall the curvature equation
x x
The slope (q) & deflection (y) at any spanwise location
can be derived
R
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e
Bending
e
e
e
CT
Shear
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DESIGN OF BEAMS
The cross section of a beam has to be designed in such away that it is strong enough to limit the bending moment and
shear force that are developed in the beam. This criterion is
known as the STRENGTH CRITERION of design .
Another criterion for beam design is that the maximum
deflection of the beam must not exceed a given permissible
limit and the beam must be stiff enough to resist the
deflection caused due to loading. This criterion is known as„STIFFNESS CRITERION of design”
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Definitions:-
(i) DEFLECTION :-The vertical distance in transverse direction
between positions of axis before and after loading at the
section of the beam, is defined as the deflection of beam at
that section.
(ii) ELASTIC CURVE OR DEFLECTION CURVE:-The neutral
axis in its deflected position after loading of the beam is knownas its elastic curve or deflection curve.
(iii) SLOPE:-The slope of the beam at any section is defined as
the angle (in radians) of inclination of the tangent drawn at
that section to the axis in its deflected position afterloading, measured w. r. t. the un-deformed axis.
(iv) FLEXURAL RIGIDITY(EI):-The product of modulus of elasticity
and Moment of Inertia is known as Flexural rigidity.
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ASSUMPTIONS MADE IN THE DEFLECTION:-
(i) Axis of the beam is horizontal before loading.
(ii) Deflection due to S.F. is negligible.
(iii) (a) Simple Bending equation M/I=σ/y=E/R is
applicable and all the assumptions made in simple bending
theory are valid.
(b) Material of the beam is homogenous, isotropic and obey
Hook‟s law.
(c) The modulus of elasticity is same in compression as well as
in tension.
(d) Plane section remain plane before and after bending
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DIFFERENTIAL EQUATION FOR DEFLECTION AND
SLOPE OF BEAMS
Below is shown the arc of the neutral axis of a beam
subject to bending.
For small angle dy/dx = tan θ = θ
The curvature of a beam is identified as dθ /ds =
1/R
In the figure δθ is small ,and δx=~ δs;
i.e ds /dx =1 6
x
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Then dy /ds =sinθ
dx /ds= cosθ
dy/dx =tanθ
Differentiating equation (a) w.r.t. x,
we get
Sec2 θ.(dθ/dx)=d2y/dx2 Therefore,
dθ/dx =(d2y/dx
2)/sec
2θ –-(2)
Other Method
R = ds/dθ = ( ds /dθ) ×(dx/dx) =(ds /dx)/(dθ/dx)
= secφ/(d θ /dx) -----(1)
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From equation(1),
R= secθ /(dθ /dx) = sec3θ /(d
2y/dx2)
1/R=(d2y/dx
2)/sec
3θ =(d2y/dx2)/ (sec2θ)3/2
= (d2y/dx2)/ (1+tan2θ)3/2
=(d2y/dx
2)/[1+(dy/dx)
2]3/2
In any practical case of bending of beams, the slope (dy/dx) is
very small (because curve is almost flat); hence (dy/dx)2 can be
ignored
so, 1/R=d2y/dx
2
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From the pure bending equation, we know that.
Integrating between selected limits. Thedeflection between limits is obtained by further
integration.
9
R
E
I
M
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SIGN CONVENTIONS:
Linear horizontal distance x: positive when measured from left
to right
Vertical distance or deflection y is positive when measured
above the axis and is negative when measured below theaxis of beam.
NOTE :
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Deflection at support A and B are zero and maximum at the
middle ofSpan. slope θ is maximum at A and B and zero at middle of Span
(at point of symmetry); At point of maximum deflection slope is
zero
θB
C A B D
θ A A
NOTE : SUPPORT or BOUNDARY CONDITIONS:
(i) Simply supported beams:
Deflection at support A and B are zero but more at free end and
also at the centre of span . Slope is maximum at supports B and
A
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(ii) Cantilever Beam:
Deflection and slope both are zero at fixed support.
Slope and deflections are maximum at free end
θ increases from
point A towards B.
B
θmax
ymax
A
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Methods for finding slope and deflection of
beams:
(i) Double or Direct integration method
(ii) Macaulay’s method
(iii) Area moment method
(iv) Conjugate beam method
(v) Unit load method
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We have from differential equation of flexure,
EI d2y/dx2=M
Integrating w. r. t.. x both sides, we get
EI (dy /dx) =∫M dx + C1
Integrating again w .r .t. x both sides ,we get
EI (y) = ∫ ∫ ( M dx) dx + C1(x) + C2
where C1 and C2 are constant of integration
Thus, integrating the differential equation w .r .t. x, we get the
equation for slope (dy /dx) ,and integrating it twice w. r .t. x, we
get the equation for deflection ( y).
DOUBLE INTEGRATION METHODS
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The two constants of integration can be determined
using the known boundary conditions of deflectionand slope at the supports
The method of double or direct integration is
convenient only in few cases of loadings. Ingeneral case of loading, where several loads are
acting the B.M. expression for each of the different
regions of loading is different .Then the method of
double integration will become extremely lengthyand laborious. Therefore ,it is not generally used.
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Case--1(i): Determine the slope and deflection equation for
the beam loaded as shown in fig.
(ii) Find the maximum deflection and maximum slope.
Solution:
B.M. at section 1-1
M= - P( x)
EI d
2
y/dx
2
= M = - P (x)EI (dy /dx ) = -P (x
2/2) + C1
EI y = -P (x3/6) + C1x + C2
P1
1
x
A B
θ A l
y A
At fixed end, when x = L ,(dy /dx) =0
(iii) What will be the deflection and slope at free end when
P=6kN, L=3m, E=210GPa, I=16x104cm4.
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Therefore,
- (PL2/2) + C1= 0
C1= PL
2
/2 At x = L, y = 0,
-PL3/6+(PL
2/2) L+C2=0
or,C2= PL3/6 - PL
3/2
= PL3/6[1-3]= - PL3/3Therefore ,C2= - PL
3/3
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Equation of slope; EI (dy/ dx) =-Px2 /2 + PL
2 /2-----(1)
Equation of deflection ,EI (y)=-Px3 /6 + PL
2x/2 - PL
3 /3-----(2)
Maximum deflection :
When x=0 (at free end) ,then from equation (2),
EI (y)=-0+0-PL3 /3
ymax= -PL3 /3EI
Maximum Slope:
Slope is maximum at free end (at x=0).hence fromequation(1),
EI (dy/ dx) = -0 + PL2 /2
(dy /dx) max = PL2 /2EI
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Deflection at free end (i.e; at A):= y A = PL3/3EI
(iii)
θ A=PL2/2EI
slope θ A=(dy/dx) at A
radian5
83
233
100357.8
)1016()10210(2
)103(106
)1016()10210(3
)103(10683
333
= 0.161mm
P
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Case (2) : When cantilever is subjected to an u .d.
L. of intensity w unit/m run over entire span
Here A is the origin. Take a
section X-X at a distance x
from A.
B.M at distance x from A= Mx = EI d2y/dx
2
=-w.x.x/2=-wx2/2
Integrating once,
EI (dy/dx) =-wx3/6 + C1 ------------------------(1)
where C1 is constant of integration
B A
Lx
w unit /m
X
X
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Applying boundary conditions:-
at x =L, dy/dx=0 from equation(1)
0=-wL3/6 + C1 C1 = wL3/6therefore,
EI dy/dx=-wx3/6+wL
3/6---------(2)
Integrating once again,
EI y=-wx4/24 + wL3.x/6 +C2 ---------------- (3)
where C2 is 2nd constant of integration
Applying boundary condition;
at x=L, y=0
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0=-wL4/24+wL
4/6+C2
Therefore,
C2=wL4/24-wL
4/6
C2=-wL4/8.
Therefore, equation (3) becomes,
EI(y)=-wx4/24 + wL
3.x/6 –wL
4/8--------(4)
Maximum deflectionIt occurs at free end where x= 0
From (4),EIy=-0+0-wL4/8
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ymax=-wL4/8EI
similarly maximum slope occurs at free end where x=0
from (2),
EI (dy/dx) =-0+wL
3
/6
(dy/dx )max=wL3/6EI
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Case 3:- When simply supported beam is subjected
to a single concentrated load at mid-span.
L/2 L/2
P
P
C
A B
R A=P/2
RB=P/2
x X
X
Mx = (P/2) x
EI d2y/dx2=(P/2)x
EI dy/dx=(P/2)x2/2 + C1
Due to symmetry slope at x = L/2 is zero
C1 = -PL2/16
EI dy/dx=(P/2)x2/2 -PL2/16
Integrating again we get
EIY = (P/2)x3
/6 – (PL2
/16) x + C2
At x=0 , Y = 0
C2 = 0
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Hence EIY = (Px3/12) – (PL2/16) x
Deflection at mid span i.e. at x = L/2 is
Y = -PL3/48EI
= PL3/48EI (downward)
Slope at support, is obtained by puttingx = 0, in slope equation
EI WLWL
EI dxdy
x
A
16161
22
0
q
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Case4:- Simply supported beam of span L carrying a
uniformly distributed load of intensity w per unit run over the
whole span.
RB=WL/2
A
x
W unit / run
B
R A=WL/2
X
X
1
32
2
2
2
2
64
22
22
C
xw xwL
dx
dy
EI
xw x
wL
xd
yd EI
xw
x
wL
Mx
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Due to symmetry dy/dx = 0 at x = L/2
24
3
1
wL
C Integrating both side w.r.t. x, we get
2
343
242412
C xwl wxwLx
Y EI
At x = 0, y = 0
C2=0
xwl wxwLx
Y EI 242412
343
Hence
Maximum deflection yc which occurs at centre C is
obtained by substituting x = L/2 in the above equation
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Due to symmetry dy/dx = 0 at x = L/2
24
3
1
wL
C Integrating both side w.r.t. x, we get
2
343
242412 C x
wl wxwLx
Y EI
At x = 0, y = 0
C2=0
xwl wxwLx
Y EI 242412
343
Hence
Maximum deflection yc
which occurs at centre C is
obtained by substituting x =
L/2 in the above equation.
EI
wL
dx
dyend at Slope
downward EI wl Yc
EI
wl Yc
x
A
24
)(3845
384
5
3
0
4
4
q
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MACAULAY‟S METHOD
For a general case of loading on the beam, though the
expression for B.M. varies from region to region, the
constants of integration remain the same for all regions.
Macaulay recognized this fact and proposed a method
which is known as the Macaulay‟s method to obtain the
slope and deflection equations. It is essentially modified
method of double integration of the B.M. expression but
following a set of rules given below:-
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(1). Assuming origin of the beam at extreme left end, take a
section in the last region of the beam at a distance x from the
origin and write the expression for B.M. at that section
considering all the force on the left of section.(2). Integrate the term of the form (x-a)
n using the formula
∫(x-a)n dx=(x-a)
n+1 /n+1
where a=distance of load from origin.
(3). While finding slope and deflection in the form
(x-a)n ,if (x-a) becomes negative on substituting the value of x,
neglect the term containing the factor (x – a)n
(4). If a couple (moment) of magnitude „c‟ is acting at a distance„a‟ from the origin of the beam, then write the BM due to couple
in the form c (x-a)0.
(5). If the beam carries a U.D.L, extend it up to the extreme right
end and superimpose an UDL equal and opposite to that which
has been added while extending the given UDL on the beam.
EXERCISE PROBLEMS
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EXERCISE PROBLEMS :
Q.(1) Figure shows a simply supported beam of span 5m carrying
two point loads. Find (1)the deflection at the section of the point
loads. (ii) Slope at A,B,C and D, (iii) maximum deflectionof the beam.
Take E=200GPa, I=7332.9 cm4
Solution:-
R A =34 KN, RB =36 KN A B
R A RB
30KN 40KN1m
5m
1.25m
x
X
X
c D
Mx = R Ax - 30(x-1) - 40(x-3.75)
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EI d2y/dx
2= 34 x -30(x-1) -40(x-3.75)
Integrating once,
EI ( dy/dx) =c1 + 34 x2/2 -30(x-1)
2/2 - 40(x-3.75)
2/2 ---(1)
Integrating once again,
---------- (2)
Support conditions:
at x=0,y=0 therefore,C2=0
at x=5m,y=0 therefore, C1=-75.06 ≈ -75.1
6
)75.3(40
6
)1(30
6
34 333
12
x x x
xC C EIy
Thus the equation for slope and deflection will be
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Thus the equation for slope and deflection will be
E I (dy / dx) =-75.1+17x2 - 15(x-1)
2-20(x-3.75)
2 -----------(3)
E I (y)= -75.1 x +17x3/3 -5(x-1)3 -20(x-3.75)3/3 ---------(4)
Total deflection at section of point loads
At C x=1m , deflection y= yC (say)
EI yC = -75.1×1 +17(1)3/3 =-69.393
Or, yC =-69.393/EI
=-69.393 /(200×106×7332.9×10-8)=- 4.73× 10-3 m = - 4.73mm
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(dy/dx)D =[17 ×(3.75)2-75.06-15 ×(2.75)2] × 1/EI
(when x=3.75m)
=50.525/EI=3.445 radians.
(dy/ dx) at B=1/EI[17 ×(5)2-75.06-15 ×(4)2-20
×(1.25)2]=78.65/EI=5.363 × 10-3rad =0.197 degree
Assuming the deflection to be maximum in the region
CD:
(at point of maximum deflection dy/dx = 0)
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say at x=x1 where dy / dx=0
From equation (3),
EI(0)=34/2(x1)2-75.06-30(x1-1)
2/2
=17 x12 -75.06 -15x1
2 +30x1 -15
=2x12+30x1-90.06=0
x1= 2.56m
The assumption that the maximum deflection is withinthe region CD is correct.
EI ymax=34(2.56)3/6-75.06 × 2.56-30(2.56-1)
3/6
Ymax=-116.13/EI=-7.918 × 10-3
m
=-7.918mm
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(Q-2)
(A) Obtain the equation for slope and elastic curve for the
beam loaded as shown in figure and find the deflection and
slope at mid-point of beam.
Take EI=15× 103 kNm2
(B) Find the slope at A,C and D
Solution:-
Reactions
R A=1/4[80×3+120]
=90KN( )
RB =80-90=-10kN=10( )
80KN 1m
120kN
mx
X
X
2m1m A
c
DB
Alternatively R =1/4[80×1-120]
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Alternatively, RB =1/4[80×1-120]
=-10KN =10 KN( )
EI (dy /dx )= C1 + 90x2
/2 - 80(x-1)2
/2 –120 (x-3) -------(1)
EI(y)= C2+C1(x) + 90 x3/6 -80(x-1)
3/6 -120(x-3)
2/2 ------(2)
Support reactions
at x = 0 ,y = 0 ,
C2= 0
MX= 90x - 80(x-1) -120(x-3)0
Mx = 90 x – 80 (x-1) – 120 (x-3)0
At 4 0 C 135
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At x=4,y=0 ,C1=-135
Equation for slope (dy/dx):-
EI (dy /dx)=-135 +90 x2/2 -40(x-1)2 -120(x-3)
Equation for deflection (y):-
EIy = -135x + 90 x3/6 -80(x-1)
3/6 -120(x-3)
2/2
To find deflection at centre (i.e. x=2m, at mid span ):-EIy=90(2)
3/6 -135(2) -80/6(2-1)
3/6 =-163.33
y=-163.33/(15×103)=-10.89×10
-3 m =-10.89 mm
To find slope at centre (i.e. x=2m, at mid span ):-
EI (dy/dx) =90× (2)2/2-135-80(2-1)
2/2 =+5
3 4
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dy/dx =5/(15×103)=3.33×10
-4 radians~ 0.019°
=19.1×10-3
degree ~ 0.02°
(b)θ A(at x=0) =-135/EI=-135/(15×103)
θC ( at x= 1m ) =[45× (1)2-135 ×1)/15 × 10
3=-90/15 ×
103radian
θD (at x=3m )= [45 ×(3)2-135-40(2)
2] =110/15 × 10
3radian
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(Q.3) Find maximum slope and maximum deflection of the
beam loaded as shown in fig.
Take E=200KN/mm2, I=60 × 10
6mm4
2m 1m
25kNm15kN
x
X
X
A B
c
EI(d2y/dx
2)=-15x -25(x-2)
0
EI(dy/dx)= C1 -15x2/2 -25(x-2) …………(1)
EI(y)= C2 + C1x -(15/2) x3/3 -25(x-2)
2/2 …(2)
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Support conditions:
slope and deflection are zero at fixed support
at x=3m,dy/dx=0 from equation (1),
EI(0)=-15(3)2/2 + C1- 25(3-2)=0
C1
=25+67.5=92.5
At C, y=0, x=3 from (2)
EI(0)=-15 (3)3/6+92.5(3) +C2 -25(3-2)
2/2
C2=12.5-277.5+67.5=-197.5
Now equation (1)
EI(dy/dx)=-15x2/2+92.5 -25(x-2)
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Now equation (1)
EI(dy/dx)=-15x2/2 +92.5 -25(x-2) ---------(1)
and
Maximum slope at A when x=0 from (1)
EI(dy/dx) A=92.5
(dy/dx) A
=92.5/EI=7.708×10-3
radian
=0.441degree
EI(y)=-15/2*x3/3+92.5x-197.5 -25(x-2)
2/2 -----(2)
Maximum Deflection at free end when x=0
EI(y) A=-197.5, yA=-197.5/EI=-16.458mm.
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Q. 3 Determine the equation for the elastic curve for
the beam loaded as shown in figure. Find the
maximum deflection.
1m1m3m
A C D B2kN/m
Solution: R A=RB = 2x3/2 = 3KN
D A1m1m 3m
C
2kN/m
x
M =3(x) -2(x-1)2/2 +2 (x-4)
2/2
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M =3(x) -2(x-1)2/2 +2 (x-4)
2/2
EI (dy/dx)=3/2(x)2 +C1 –(x-1)
3/3 +(x-4)
3/3 -------(1)
EIy=3 x3/6 +C1x +C 2 – (x-1)
4/12 +(x-4)
4/12 ------(2)
S t ti
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Support reactions:
at x=0,y=0, C2=0
at x=5m,y=0 , C1=-8.25
Equation for slope:-
EI (dy/dx)=3x2/2-8.25 –(x-1)
3/3 +(x-4)
3/3 ------(3)
Equation for the elastic curve :
EIy=x3/2 -8.25x –(x-1)
4/12 +(x-4)
4/12 ---------(4)
Due to symmetry, deflection is maximum at centre at
x=2.5m,EI ymax=(2.5)
3/2 -8.25x(2.5)-(2.5-1)
4/12=-13.23
ymax = -13.23/EI
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Practice problems:-
(Q-1) A cantilever beam of span L carries a udl of
intensity w/unit length for half of its span as shown in
figure.If E is the modulus of elasticity and I is moment of
inertia,determine the following in terms of w,L,E and I.
(i)A expression for slope(dy/dx)at free end
(ii)An expression for deflection( y ) at free end
(iii)The magnitude of upward vertical force to be applied at
free end in order to resume this end to the same horizontal
level as built in end.
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[Ans (i)θA=WL3 /48EI (ii) ∂ A =7wL4 /384EI( ) (iii)P=7wL/128]
L/2 L/2
w/m
A B
Q (2 ) f f C
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Q- (2 ) Determine the values of deflections at points C,D
and E in the beam as shown in figure.Take E=2*105MPa ; I=
60 *108 mm
4
1m 2m
10kN/m
1m 1m
20kN30kN
A C D E B
Answers:
[ᵟC=0.0603mm(downward), ᵟ D=0.0953mm(downward)
ᵟE=0.0606mm(downward)]
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.
Q-(3) Find the position and magnitude of maximum deflection
for the beam loaded as shown in fig.
Take E=200GPa ,I=7500cm4 .
3KN/m20 KN
AD B
X
4m 4m 4mC
X
X
[Answers: ymax at 3.7 m from A=-118/EI=7.99mm
yc=-32/EI=-2.13mm]
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DOUBLE INTEGRATION METHOD
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1.
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EXAMPLE SIMPLY SUPPORTED BEAM
Consider a simply supported uniform sectionbeam with a single load F at the
centre. The beam will be deflect
symmetrically about the centre line with aslope (dy/dx) at the centre line. It is
convenient to select the origin at the centre
line.
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THE DOUBLE INTEGRATION METHOD
The constants of integration are determined byevaluating the functions for slope or
displacement at a particular point on the beam
where the value of the function is known. These values are called boundary conditions.
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2. THE DOUBLE INTEGRATION METHOD
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Consider the beam shown in Fig.Here x1 and x2 coordinates are valid within the regions
AB & BC.
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3-3 THE DOUBLE INTEGRATION METHOD
Once the functions for the slope and
deflections are obtained, they must give
the same values for slope & deflection at
point B. This is so as for the beam to be physically
continuous.
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EXAMPLE 3
The cantilevered beam shown in Fig. 8.11(a) issubjected to a couple moment Mo at its end.
Determine the equation of the elastic curve
EI is constant.
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EXAMPLE 3.3 - SOLUTION
The load tends to deflect the beam asshown in Fig. 8.9(a).
By inspection, the internal moment can be
represented throughout the beam using a
single x coordinate.
From the free-body diagram, with M acting
in +ve direction as shown in Fig. 8.11(b), we
have:
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o M M
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EXAMPLE 3.3 - SOLUTION
Applying equation 8.4 & integrating twice yields:
Using boundary conditions, dy/dx = 0 at x = 0 & y =0 at x = 0 then C1 = C2 =0
60
21
2
1
2
2
2 C xC x M y EI
C x M dx
dv EI
M dx
vd EI
o
o
o
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EXAMPLE 3.3 - SOLUTION
Substituting these values into earlierequations, we get:
Max slope & displacement occur at A (x = L)
for which
61
EI
x M y
EI
x M
dxdy
oo
2
;
/with
2
q
q
EI L M y
EI L M o
Ao
A2
;2
q
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EXAMPLE
The +ve result for q A
indicates counterclockwiserotation & the +ve result for y A indicates that it isupwards.
This agrees with results sketched in Fig. 8.11(a).
In order to obtain some idea to the actual magnitudeof the slope
Consider the beam in Fig. 8.11(a) to:
Have a length of 3.6m
Support a couple moment of 20kNmBe made of steel having Est = 200GPa
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MOMENT-AREA THEOREMS To develop the theorems, reference is made to Fig. 8.13(a).
If we draw the moment diagram for the beam & then divide itby the flexural rigidity, EI, the “M/EI diagram” shown in Fig.8.13(b) results.
By equation
63
dx
EI
M d
q
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3-4 MOMENT-AREA THEOREMS
dq on either side of the element dx = the lighter shadearea under the M/EI diagram.
Integrating from point A on the elastic curve to point B,
Fig 8.13(c), we have
This equation forms the basis for the first moment-
area theorem.
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/ dx EI
M B
A A B q
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MOMENT-AREA THEOREMS
Theorem 1: The change in slope betweenany two points on the elastic curve equals thearea of the M/EI diagram between the twopoints.
The second moment-area theorem is basedon the relative derivation of tangents to theelastic curve.
Shown in Fig 8.12(c) is a greatly exaggeratedview of the vertical deviation dt of thetangents on each side of the differentialelement, dx.
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MOMENT-AREA THEOREMS
Fig 8.12(c)
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Since slope of elastic curve & its deflectionare assumed to be very small, it issatisfactory to approximate the length ofeach tangent line by x & the arc ds‟ by dt.
Using qs = r dt = xdq Using equation 8.2, dq = (M/EI) dx
The vertical deviation of the tangent at A
with the tangent at B can be found byintegration.
67
8.6eqn/ dx EI
M xt
B
A B A
MOMENT-AREA THEOREMS
MOMENT AREA THEOREMS
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Centroid of an area
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B.&A betweenareatheof centroidthetoA
throughaxisverticalthefromdistance
8.7eqn
dAdA
/
x
dx EI M xt
x x
B
A B A
MOMENT-AREA THEOREMS
MOMENT AREA THEOREMS
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Theorem 1
The vertical deviation of the tangent at a point
(A) on the elastic curve with the tangent
extended from another point (B) equals the“moment” of the area under the M/EI diagram
between the 2 points (A & B).
This moment is computed about point A where thederivation is to be determined.
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MOMENT-AREA THEOREMS
MOMENT AREA THEOREMS
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Provided the moment of a +ve M/EI area from A
to B is computed as in Fig. 8.14(b), it indicates
that the tangent at point A is above the tangent
to the curve extended from point B as shown inFig. 8.14(c).
-ve areas indicate that the tangent at A is below
the tangent extended from B.
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MOMENT-AREA THEOREMS
MOMENT AREA THEOREMS
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MOMENT-AREA THEOREMS
MOMENT AREA THEOREMS
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It is important to realize that the moment-area
theorems can only be used to determine the
angles and deviations between 2 tangents on
the beam‟s elastic curve. In general, they do not give a direct solution for
the slope or displacement at a point.
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MOMENT-AREA THEOREMS
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3-5 CONJUGATE-BEAM METHOD
Here the shear V compares with the slope ,the moment M compares with the displacement
Y & the external load w compares with the M/EI
diagram. To make use of this comparison we will now
consider a beam having the same length as the
real beam but referred to as the “conjugatebeam” as shown in Fig. 8.22.
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q
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3-5 CONJUGATE-BEAM METHOD
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3-5 CONJUGATE-BEAM METHOD
To show this similarity, we can write theseequation as shown
75
wdx
dV w
dx
M d
2
2
EI
M
dx
d
q
EI
M
dx
vd
2
2
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3-5 CONJUGATE-BEAM METHOD
Or intergrating
76
wdxV dxwdx M
dx EI
M
q dxdx
EI
M v
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3-5 CONJUGATE-BEAM METHOD
The conjugate beam is loaded with theM/EI diagram derived from the load w on
the real beam.
From the above comparisons, we can state2 theorems related to the conjugate beam.
Theorem 1
The slope at a point in the real beam isnumerically equal to the shear at the
corresponding point in the conjugate beam.
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3-5 CONJUGATE-BEAM METHOD
Theorem 2The displacement of a point in the real beam is
numerically equal to the moment at the
corresponding point in the conjugate beam.
When drawing the conjugate beam, it is
important that the shear & moment developed
at the supports of the conjugate beam account
for the corresponding slope and displacementof the real beam at its supports.
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3-5 CONJUGATE-BEAM METHOD For example, as shown in Table 8.2, a pin or roller support at
the end of the real beam provides zero displacement but the
beam has a non-zero slope.
Consequently from Theorem 1 & 2, the conjugate beam must
be supported by a pin or roller since this support has zero
moment but has a shear or end reaction.
When the real beam is fixed supported, both beam has a free
end since at this end there is zero shear & moment.
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3-5 CONJUGATE-BEAM METHOD
Corresponding real & conjugate beamsupports for other cases are listed in the
table
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