Download - BENDING DEFLECTIONS
1/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
BENDING DEFLECTIONS
2/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
z
wz
x
1
c
0
limx1. Strain definition:
z 2. From picture:
zcx 5. Plane cross-section hypothesis
z
x 4. Linear strain
z
x0
lim
3. Substitution of 2 into 1
MyMy
Geometry of deformation
3/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
z
x 1. Normal strain:
zxx EE 2. From Hooke law:
zJ
M
y
yx 3. Normal stress:
1
EJ
M
y
y 4. From comparison of 2 and 3:
y
y
EJ
M
1
5. Beam curvature
Normal stresses and beam curvature
4/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
"wEJ
M
y
y
y
y
EJ
M1. Curavture-bending moment
relationship:
"//1
/ 222/32
22
wdxwddxdw
dxwd
2. Formula for curvature of a
curve:
3. Differential equation for beam
deflection w(x)
Differential equation for beam deflection
5/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
The signs of w(x), w’(x) i w’’(x) depend upon co-ordinate system:
w(x)
x
w(x)
x
w>0, w’<0, w’’>0
w(x)
x
w(x)x
Sign convention
w>0, w’>0, w’’>0
w>0, w’>0, w’’>0 w<0, w’<0, w’’<0
6/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
The sign of M(x) follows adopted convention (M is positive when „undersides” are under tension):
M>0
M>0
M<0
M<0
Sign convention
7/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
Combination of the previous two conventions will result in following form of the bending deflection equation: xw
EJ
xM
y
y "
PROVIDED that the positive direction of w axis will coincide with positive direction of bending moment axis pointing towards „undersides”:
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!
In the opposite case of the two directions discordance the minus sign in the bending equation has to be replaced by the positive sign
Sign convention
w(x)x
My
w(x)
xMy
or
8/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
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
CdxEJ
Mw
y
y '
DCxdxdxEJ
Mw
y
y )(
The next integration results in finding beam deflection:
y
y
EJ
Mw "
w(x)
x
)'(warctg
w
To determine the values of integration constants C and D we need to formulate boundary conditions
Integration of the deflection equation
9/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
The boundary conditions have to represent supports of beam:
w=0w=0 w=0 w=0w=0 w=0
A
w=0 w’=0 w=0 w’=0w=0 w’=0
B
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.
Integration of the deflection equation
10/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
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.
Integration of the deflection equation
11/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
Example of the boundary conditions adjustment
n 1 2 3 4 5 6
w1=w2
w’1=w’2
w2=w3
w’6=0
w4=w5 w5=w6 w6=0
w’2=w’3 w’5=w’6w’3=w’4
w3=w4
w1=0
Independent integration
Adjustment of first derivatives
Adjustment of the deflections
12/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
The Conjugate Beam Method(Mohr fictitious beam method)
13/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
STATICS DEFLECTIONS
xQdx
xdMz
xqdx
xdQz
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 .
EJ
xM
dx
xwd
2
2
Why not to use the same method to determine the
deflections?
xqdx
xMd
2
2 )(
?
?
Mohr method
14/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
STATICS - Fictitious domain DEFLECTIONS - Real domain
FFFF DxCdxdxxqxM
dx xqdx
MdF
F 2
2
dx FFFF CdxxqQ
dx
dM
RR
RR DxCdxdx
EJ
xMxw
dx
RR
RR Cdx
EJ
xMw
dx
dw '
dx EJ
xM
dx
wd RR 2
2
EJ
xMxq R
F I F
T H E N
CF= CR , DF=DR
wR(x)MF(x) , w’R QF(x)
15/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
EJ
xMxq R
F
wR(x)MF(x) , w’R QF(x)
From the condition:
CF= CR , DF=DR
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).
[1/m]
To satisfy the conditions:
the kinematical conditions have to bet set upon the fictitious beam in such a way that in characteristic points will be:
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.
Fundamental requirement to be satisfied is that fictitious and real beams have the same length (0 ≤ xR ≤ l, 0 ≤ xF ≤ l).
16/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
Fic
titi
ou
s b
eam
Deflection wR
Rotation w’R
Rea
l b
eam
Bending moment MF
Shear force QF
Supports (KBC)F
ww0
0 0
0 0
0
0 0
0
Example #1 Example #2
Supports (KBC)R
0
0
0
0
0
0
0
0
17/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
Fic
titi
ou
s b
eam
Deflection wR
Rotation w’R
Rea
l b
eam
Bending moment MF
Shear force QF
Supports (KBC)F
ww0
Example #3 Example #4
Supports (KBC)R
0
0
0
0 0
0
0
0
0
0
0
0
0
0
0
PL ww ''
PL QQ
0
0PL ww
0PL ww
PL MM
PL QQ 0
0
0
0
0
0
18/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
0 PL ww
0 PL QQ
0FM0FQ
0FM
0 PL MM
0 PL QQ
0'' PL ww
0FM0FQ
Built-in support Free endPin-pointed supportHingeBELKA FIKCYJNA
0'w 0 PL ww0w 0w 0w
0'w
Rea
l b
eam
Fic
titi
ou
s b
eam
19/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
Ben
ding
mom
ents
usi
ng t
he s
uppe
rpos
ition
prin
cipl
e
20/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
Bending moment diagram for fictitious beam
Real beam
Bending moment for the fictitious beam
Fictitious beam loading
Elementary but troublesome !
Fictitious beam
w
≡ deflections of the real beam
21/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
Mohr method applied to the evaluation of deflections and rotations in chosen points
l
aw
xEJ
BA
Pa Bending moments for the fictitious beam
MBF=wB=
2a/3a/3
(Pa/EJ)(a/2) l-a/3
wA=
w’B=
w’A=
wB=
MAF=wA=
QAF=QB
F=w’A=w’B=
Shear forces for the fictitious beam:
(Pa/EJ)(a/2)(2a/3)= Pa3/3EJ
(Pa/EJ)(a/2)[l-a/3]= [Pa2/2EJ][l-a/3]
(Pa/EJ)(1/2)a=Pa2/2EJ
Pa3/3EJ Pa2/2EJ
[Pa2/2EJ][l-a/3] Pa2/2EJ
Straight line
3rd order parabola
P
Pa/EJ
w
22/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
B
w
xEJ
A
P
a
Pl
Bending moment for the fictitious beam
MBF=wB=
l/3 2l/3
Pl/EJ)(1/2)l
w’B= Pl2/2EJ
QAF=QB
F=w’A=w’B=
Shear forces for the fictitious beam
Pl/EJ)(1/2)l[2l/3]= Pl3/3EJ
Pl/EJ)(1/2)l= Pl2/2EJ
a = l
P
wB= Pl3/3EJ = wmax
Pl/EJ
Special case: a=l
23/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
[Pa2/2EJ][l-a/3]
x
l
a
w
EJ
BA
P
wwB=
2
22
3
3
31
2 l
ll
l
a
EJ
PawB
2
23
2
3
31
3 l
a
l
a
EJ
PlwB
23
2
3
31
3
EJ
PlwB
0,50,25
0,75
1,0
f
0,25
0,75
0,5
1,0
l
a
0,01
0,31
0,63
0,61
Special example: end beam deflection as the function of force position
wmax
fwB wmax
24/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
Area and centre of gravity for parabolic shapes
Parabola of nth degree
Horizontal tangenta
b
c
A= ab/(n+1)
Area:
Position of the centre of gravity:
c= b/(n+2)
An c
0
1
2
3
…
ab
ab/2
ab/3
ab/4
…
b/2
b/3
b/4
b/5
…
25/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
y
y
EJ
M
dx
wd
2
2
Let us differentiate twice the equation:
zy Q
dx
dM q
dx
dQz making use of M-Q-q relationships
y
y
EJ
M
dx
dw
dx
d"
y
y
EJdx
dM
dx
wd 13
3
y
z EJQ
dx
wd 13
3
First differentiation yields:
yz EJ
Qdx
d
dx
wd
dx
d 13
3
yEJ
qdx
wd 14
4
Second differentiation:
CdxEJ
Mw
y
y ' DCxdxdxEJ
Mw
y
y )(
Double integration of the equation w”= - M/EJ allows for finding rotations and deflections:
General bending equation
26/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
yz EJ
Qdx
wd 13
3
yEJq
dx
wd 14
4
CdxEJ
Mw
y
y ' DCxdxdxEJ
Mw
y
y )(y
y
EJ
M
dx
wd
2
2
DIF
FE
RE
NT
IAT
IONI N T E G R A T I O N
Loading
RotationBending moment; Curvature
Shear force
Deflection
The overall picture of bending problem
General bending equation
27/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
xEJ
xM
dx
xwd
2
2Therefore the points in which bending stiffness changes are also a characteristic point.
If EJ changes along beam axis EJ(x), then differential equation for displacement becomes:
28/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
Deflection of a beam of variable stifness
EJ2
Pa2/EJ2
Pa2/EJ1
wK= Pl3/3EJ
x
l
{Pl/EJ2} {1-J2/J1}
wK= ?
Pl
Pl/EJ2
xEJ
l
P
K
w
Pl
Pl/EJ1
Pl/EJ2
OR…
EJ2
{Pa2/EJ2} {1-J2/J1}
{Pl/EJ2} {1-J2/J1}
EJ1
K
w
P
a1 a2
> EJ2
a1
{P(l-a2)/a1} {1-J2/J1}
{Pa2} {1-J2/J1}
Pl/EJ
29/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
EJ2
Pa2/EJ2
Pa2/EJ1
wK= Pl3/3EJ
x
l
{Pl/EJ2} {1-J2/J1}
Pl
Pl/EJ2
xEJ
l
P
K
w
Pl
Pl/EJ1
Pl/EJ2
OR…
EJ2
{Pa2/EJ2} {1-J2/J1}
{Pl/EJ2} {1-J2/J1}
EJ1
K
w
P
a1 a2
> EJ2
a1
{P(l-a2)/a1} {1-J2/J1}
{Pa2} {1-J2/J1}
Pl/EJ
30/29M.Chrzanowski: Strength of Materials
SM2-06: Bending deflections
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