Transcript Title

SM2-06: Bending deflections
BENDING DEFLECTIONS
M.Chrzanowski: Strength of Materials
1/29
SM2-06: Bending deflections
Geometry of deformation

1. Strain
definition:
z

 x  lim
 0
 z
My z

My

2. From picture:
w


M.Chrzanowski: Strength of Materials
x
3. Substitution of 2
into 1
4. Linear strain
5. Plane crosssection hypothesis


  
 x  lim
 0
x 
z
c
1
z


x  c z

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SM2-06: Bending deflections
Normal stresses and beam curvature
z
1. Normal strain:
x 
2. From Hooke law:
 x  E   x  E z
3. Normal stress:
4. From comparison of
2 and 3:
5. Beam curvature 
M.Chrzanowski: Strength of Materials

x 
My
Jy

My
Jy
E
1


z
1

My
EJ y
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SM2-06: Bending deflections
Differential equation for beam deflection

1. Curavture-bending moment
relationship:
2. Formula for curvature of a
curve:
3. Differential equation for beam
deflection w(x)
M.Chrzanowski: Strength of Materials

My
EJ y
d 2 w / dx2
1  dw / dx 
2 3/ 2
My
EJ y
 d 2 w / dx2  w"
 w"
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SM2-06: Bending deflections
Sign convention
The signs of
w(x), w’(x) i w’’(x) depend upon co-ordinate system:
w(x)
w(x)
x
w>0, w’<0, w’’>0
x
x
w>0, w’>0, w’’>0
x
w(x)
w(x)
w>0, w’>0, w’’>0
M.Chrzanowski: Strength of Materials
w<0, w’<0, w’’<0
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SM2-06: Bending deflections
Sign convention
The sign of M(x) follows adopted convention (M is positive when „undersides”
are under tension):
M>0
M<0
M.Chrzanowski: Strength of Materials
M<0
M>0
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SM2-06: Bending deflections
Sign convention
Combination of the previous two conventions will result in following form of the
bending deflection equation:
M y x 
EJ y
  w" x 
PROVIDED that the positive direction of w axis will coincide with positive
direction of bending moment axis pointing towards „undersides”:
w(x)
My
x
x
My
or
w(x)
In the opposite case of the two directions discordance the minus sign in the
bending equation has to be replaced by the positive sign
The direction of x-axis only does not change the sign in the equation of
deflection.
However, one has to remember that in this case the first derivative of
deflection w’ will change the sign!
M.Chrzanowski: Strength of Materials
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SM2-06: Bending deflections
Integration of the deflection equation
To find beam deflection one has to
integrate the deflection equation twice.
The first integration yields a tangent to
the beam axis and, therefore, rotation
of the beam cross-section
w"  
w'   
My
EJ y
My
EJ y

The next integration results in
finding beam deflection:
w   ( 
My
EJ y
dx  C
x
w
  arctg(w' )
w(x)
dx)dx  Cx  D
To determine the values of integration constants C and D we need to
formulate boundary conditions
M.Chrzanowski: Strength of Materials
8/29
SM2-06: Bending deflections
Integration of the deflection equation
The boundary conditions have to represent supports of beam:
w=0
w=0 w=0
w=0 w=0
w=0
A
B
w=0 w’=0
w=0 w’=0
w=0 w’=0
NOTICE: As we do not take into account normal forces all three cases
shown in the row A or B are equivalent with respect to deflections
calculation.
M.Chrzanowski: Strength of Materials
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SM2-06: Bending deflections
Integration of the deflection equation
In a general case when moment bending equation cannot be given by in the
analytical form for the whole beam it has to be formulated and integrated for all
characteristic sections of a beam
As a consequence we need to find 2n constants of integration (where n denotes
number of characteristic sections) and to write down 2n-2 (2 boundary
conditions correspond to beam supports) additional adjustment conditions at the
neighbouring characteristic sections
This procedure yields the set of 2n linear algebraic equations. The solution of this
set can be cumbersome and it is advisable only if need an analytical form of a
bending deflection for the whole beam.
M.Chrzanowski: Strength of Materials
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SM2-06: Bending deflections
Example of the boundary conditions adjustment
n
1
2
w1=w2
w2=w3
w1=0
w’2=w’3
w’1=w’2
4
3
w3=w4
w’3=w’4
6
5
w4=w5
w5=w6
w6=0
w’5=w’6
w’6=0
Independent
integration
Adjustment of
first derivatives
Adjustment of
the deflections
M.Chrzanowski: Strength of Materials
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SM2-06: Bending deflections
The Conjugate Beam Method
(Mohr fictitious beam method)
M.Chrzanowski: Strength of Materials
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SM2-06: Bending deflections
Mohr method
STATICS
dM  x 
 Qz  x 
dx
DEFLECTIONS
?
dQ z  x 
  q x 
dx
?
d 2 M ( x)
  q x 
2
dx
d 2 w x 
M x 

2
dx
EJ
This equation (7) is „integrated”
on the basis of M(x) i Q(x)
definitions when external loading
q(x) and boundary conditions are
known .
M.Chrzanowski: Strength of Materials
Why not to use the same
method to determine the
deflections?
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SM2-06: Bending deflections
STATICS - Fictitious domain


d 2M F
 qF x 
2
dx
DEFLECTIONS - Real domain
dx


dM F
 QF    qF  x dx  CF dx

dx


d 2 wR
 2   M R x  dx
dx
EJ
dwR
M R x 

w
'


dx  C R

R

dx
EJ
M F x      qF x dx dx  CF  x  DF


dx
 M x  
wR x       R dxdx  CR  x  DR
 EJ

M R x 


qF x 
EJ
IF
CF= CR , DF=DR
THEN
wR(x)MF(x) , w’R  QF(x)
M.Chrzanowski: Strength of Materials
14/29
SM2-06: Bending deflections
Fundamental requirement to be satisfied is that fictitious and real beams have the
same length (0 ≤ xR ≤ l, 0 ≤ xF ≤ l).
M R x 
qF  x  
EJ
From the condition:
[1/m]
follows, that the only loading of the fictitious beams will be continuous loading of
the dimension [Nm/(Nm-2m4)]=[ m-1] distributed exactly like bending moment
distribution for the real beam. Therefore, the bending moment and shear force
distributions in the fictitious beams cannot contain any discontinuities (no loading
in form of concentrated moments or forces exists).
To satisfy the conditions:
CF= CR , DF=DR
the kinematical conditions have to bet set upon the fictitious beam in such a way
that in characteristic points will be:
wR(x)MF(x) , w’R  QF(x)
So, if for the real beam wR=0 in a given point, then for the fictitious beam has to be
MF=0 in this point. Similarly if w’R =0 then QF =0 etc.
M.Chrzanowski: Strength of Materials
15/29
SM2-06: Bending deflections
Fictitious beam
Real beam
Example #1
Example #2
Supports (KBC)R
Deflection wR
0
0
0
0
Rotation w’R
0
0
0
0
Bending moment MF
0
0
0
0
Shear force QF
0
0
0
0
Supports (KBC)F
M.Chrzanowski: Strength of Materials
16/29
SM2-06: Bending deflections
Real beam
Example #3
Supports (KBC)R
Deflection wR
Rotation w’R
Fictitious beam
Example #4
0
0
Bending moment MF
0
Shear force QF
0
0 0
w' L  w' P
0 0
0
QL  QP
0
0
0
0
0
0
0
wL  wP
0
wL  wP
0
ML  MP
0
QL  QP
0
0
0
0
0
Supports (KBC)F
M.Chrzanowski: Strength of Materials
17/29
Real beam
SM2-06: Bending deflections
w0
w'  0
wL  wP  0
w' L  w' P  0
Fictitious beam
Built-in support Hinge
MF  0
QF  0
w0
wL  wP  0
Pin-pointed support
MF  0
QL  QP  0
w0
w'  0
Free end
MF  0
QF  0
ML  MP 0
QL  QP  0
M.Chrzanowski: Strength of Materials
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Bending moments using the supperposition principle
SM2-06: Bending deflections
M.Chrzanowski: Strength of Materials
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SM2-06: Bending deflections
Real beam
w
Bending moment
for the fictitious
beam
Fictitious beam
loading
Fictitious
beam
Elementary but troublesome !
Bending moment
diagram for
fictitious beam
≡ deflections of
the real beam
M.Chrzanowski: Strength of Materials
20/29
SM2-06: Bending deflections
Mohr method applied to the evaluation of
deflections and rotations in chosen points
P
EJ
x
A
a
w
l
B
wA= Pa3/3EJ
w’A= Pa2/2EJ
wB= [Pa2/2EJ][l-a/3]
w’B= Pa2/2EJ
Bending moments for the fictitious beam
Pa
MAF=wA= (Pa/EJ)(a/2)(2a/3)= Pa3/3EJ
Pa/EJ
MBF=wB= (Pa/EJ)(a/2)[l-a/3]= [Pa2/2EJ][l-a/3]
a/3
(Pa/EJ)(a/2)
Shear forces for the fictitious beam:
2a/3
l-a/3
QAF=QBF=w’A=w’B= (Pa/EJ)(1/2)a=Pa2/2EJ
Straight line
w
3rd order parabola
M.Chrzanowski: Strength of Materials
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SM2-06: Bending deflections
P
EJ
P
Special case: a=l
x
A
a
w
a=l
Pl
B
wB= Pl3/3EJ = wmax
w’B= Pl2/2EJ
Bending moment for the fictitious
beam
MBF=wB= Pl/EJ)(1/2)l[2l/3]= Pl3/3EJ
Pl/EJ
Shear forces for the fictitious beam
l/3
2l/3
QAF=QBF=w’A=w’B= Pl/EJ)(1/2)l= Pl2/2EJ
Pl/EJ)(1/2)l
M.Chrzanowski: Strength of Materials
22/29
SM2-06: Bending deflections
Special example: end beam deflection as
the function of force position
P
EJ
x
w
B
A
a
l
f  
Pl3  a  3a 2
wB 
1   2
3EJ  3l  2l

1,0
0,75
0,5
0,25
Pa2  a  3l 2
wB 
1  l 2
2 EJ  3l  3l
0,63
0,31

0,01
0,25
0,5
0,61
0,75
w
M.Chrzanowski: Strength of Materials
a
l
wmax
Pl3    3 2
wB 
1   
3EJ  3  2
wB  f   wmax
1,0
wB= [Pa2/2EJ][l-a/3]
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SM2-06: Bending deflections
Area and centre of gravity for parabolic shapes
Parabola of nth degree
Area:
A= ab/(n+1)
Horizontal tangent
a
Position of the centre of gravity:
c= b/(n+2)
c
b
M.Chrzanowski: Strength of Materials
n
A
c
0
ab
b/2
1
ab/2
b/3
2
ab/3
b/4
3
ab/4
b/5
…
…
…
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SM2-06: Bending deflections
General bending equation
Let us differentiate twice the equation:
making use of M-Q-q
relationships
First differentiation yields:
dM y
dx
dQ z
 q
dx
 Qz
My
d 2w

2
dx
EJ y
d
d My
w"  
dx
dx EJ y
dM y 1
d 3w


dx3
dx EJ y
d 3w
1
 Qz
3
dx
EJ y
Second differentiation:
d d 3w
d
1


Q
z
dx dx3
dx EJ y
d 4w
1

q
dx4
EJ y
Double integration of the equation w”=
for finding rotations and deflections:
w'   
My
EJ y
dx  C
M.Chrzanowski: Strength of Materials
- M/EJ allows
w   (
My
EJ y
dx)dx  Cx  D
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SM2-06: Bending deflections
General bending equation
Shear force
d 3w
1


Q
z
dx3
EJ y
My
d 2w

2
dx
EJ y
Bending moment;
Curvature
M.Chrzanowski: Strength of Materials
DIFFERENTIATION
Loading
d 4w
1
q
4
dx
EJ y
The overall picture
of bending problem
I NT E G RAT I O N
w'   
My
EJ y
  EJ
dx  C w   (
Rotation
My
dx)dx  Cx  D
y
Deflection
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SM2-06: Bending deflections
If EJ changes along beam axis EJ(x), then differential equation for
displacement becomes:
d 2 wx 
M x 

2
dx
EJ x 
M.Chrzanowski: Strength of Materials
Therefore the points in which
bending stiffness changes are also
a characteristic point.
27/29
Deflection of a beam of variable stifness
SM2-06: Bending deflections
EJ
P
EJ1
x
l
Pl
w
wK= Pl3/3EJ
Kx
{Pl/EJ2} {1-J2/J1}
Pa2/EJ2
{Pa2/EJ2} {1-J2/J1}
{Pa2} {1-J2/J1}
a1
a2
Pl
Pl/EJ1
EJ2
P
l
Pl/EJ2
Pl/EJ
EJ2
a1
K
w
>
Pl/EJ2
Pa2/EJ1
OR…
EJ2
{P(l-a2)/a1} {1-J2/J1}
wK= ?
M.Chrzanowski: Strength of Materials
{Pl/EJ2} {1-J2/J1}
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SM2-06: Bending deflections
EJ
P
EJ1
x
l
Pl
w
wK= Pl3/3EJ
Kx
{Pl/EJ2} {1-J2/J1}
Pa2/EJ2
{Pa2/EJ2} {1-J2/J1}
{Pa2} {1-J2/J1}
a1
a2
Pl
Pl/EJ1
EJ2
P
l
Pl/EJ2
Pl/EJ
EJ2
a1
K
w
>
Pl/EJ2
Pa2/EJ1
OR…
EJ2
{P(l-a2)/a1} {1-J2/J1}
M.Chrzanowski: Strength of Materials
{Pl/EJ2} {1-J2/J1}
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SM2-06: Bending deflections
stop
M.Chrzanowski: Strength of Materials
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