1. BENDING DEFLECTIONS

2.  My My  Geometry of deformation 1. Strain definition: z
2. From picture: w z x
3. Substitution of 2 into 1
4. Linear strain
5. Plane cross-section hypothesis 

3. Normal stresses and beam curvature
1. Normal strain:
2. From Hooke law:
3. Normal stress:
4. From comparison of 2 and 3:
5. Beam curvature 

4. 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)

5. w>0, w’<0, w’’>0 w>0, w’>0, w’’>0
Sign convention
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>0, w’>0, w’’>0
w(x) x
w(x) x
w>0, w’>0, w’’>0
w<0, w’<0, w’’<0

6. Sign convention
The sign of M(x) follows adopted convention (M is positive when „undersides” are under tension):
M>0
M<0
M<0
M>0

7. w(x) x w(x) My My x Sign convention
Combination of the previous two conventions will result in following form of the bending deflection equation:
PROVIDED that the positive direction of w axis will coincide with positive direction of bending moment axis pointing towards „undersides”:
w(x) x
w(x) My My x or
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!

8. x w w(x) 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(x) x w
The next integration results in finding beam deflection:
To determine the values of integration constants C and D we need to formulate boundary conditions

9. w=0 w=0 w’=0 Integration of the deflection equation
The boundary conditions have to represent supports of beam:
w=0 A
w=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.

10. 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.

11. w1=w2 w2=w3 w3=w4 w4=w5 w5=w6 w6=0 w1=0 w’2=w’3 w’3=w’4 w’5=w’6 w’6=0
Example of the boundary conditions adjustment n 1 2 3 4 5 6
w1=w2
w2=w3
w3=w4
w4=w5
w5=w6
w6=0
w1=0
w’2=w’3
w’3=w’4
w’5=w’6
w’6=0
w’1=w’2
Independent integration
Adjustment of first derivatives
Adjustment of the deflections

12. The Conjugate Beam Method (Mohr fictitious beam method)

13. Why not to use the same method to determine the deflections?
Mohr method
STATICS
DEFLECTIONS ? ?
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 .
Why not to use the same method to determine the deflections?

14. wR(x)MF(x) , w’R  QF(x)
STATICS - Fictitious domain
DEFLECTIONS - Real domain
I F
CF= CR , DF=DR
wR(x)MF(x) , w’R  QF(x)
T H E N

15. wR(x)MF(x) , w’R  QF(x)
Fundamental requirement to be satisfied is that fictitious and real beams have the same length (0 ≤ xR ≤ l, 0 ≤ xF ≤ l).
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.

16. Example #1
Example #2
Supports (KBC)R
Real beam
Deflection wR
Rotation w’R
Bending moment MF
Shear force QF
Fictitious beam
Supports (KBC)F

17. Example #3
Example #4
Supports (KBC)R
Real beam
Deflection wR
Rotation w’R
Bending moment MF
Shear force QF
Fictitious beam
Supports (KBC)F

18. Real beam
BELKA FIKCYJNA
Built-in support
Hinge
Pin-pointed support
Free end
Fictitious beam

19. Bending moments using the supperposition principle

20. Elementary but troublesome !
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

21. Mohr method applied to the evaluation of deflections and rotations in chosen pointsEJ x w
wA=
Pa3/3EJ
w’A=
Pa2/2EJ B A a
wB=
[Pa2/2EJ][l-a/3]
w’B=
Pa2/2EJ l Pa
Bending moments for the fictitious beam
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]
2a/3
a/3
(Pa/EJ)(a/2)
Shear forces for the fictitious beam:
QAF=QBF=w’A=w’B=
(Pa/EJ)(1/2)a=
Pa2/2EJ
l-a/3 w
3rd order parabola
Straight line

22. A P a
a = l P
Special case: a=l EJ x B
wB= Pl3/3EJ = wmax
w’B= Pl2/2EJ w
Bending moment for the fictitious beam Pl
MBF=wB=
Pl/EJ)(1/2)l[2l/3]=
Pl3/3EJ
Pl/EJ
l/3
2l/3
Pl/EJ)(1/2)l
Shear forces for the fictitious beam
QAF=QBF=w’A=w’B=
Pl/EJ)(1/2)l=
Pl2/2EJ

23. Special example: end beam deflection as the function of force positionw EJ B A P x
wmax
1,0
0,75
0,63
0,5
0,31
0,25
0,01
0,61
0,25
0,5
0,75
1,0
wmax w
wB=
[Pa2/2EJ][l-a/3]

24. Area and centre of gravity for parabolic shapes
Parabola of nth degree
Area:
A= ab/(n+1) a
Horizontal tangent
Position of the centre of gravity:
c= b/(n+2) c b n A c 1 2 3 … ab
ab/2
ab/3
ab/4 …
b/2
b/3
b/4
b/5 …

25. General bending equation
Let us differentiate twice the equation:
making use of M-Q-q relationships
First differentiation yields:
Second differentiation:
Double integration of the equation w”= - M/EJ allows for finding rotations and deflections:

26. The overall picture of bending problem
General bending equation
DIFFERENTIATION
The overall picture of bending problem
Loading
Shear force
I N T E G R A T I O N
Bending moment;
Curvature
Rotation
Deflection

27. If EJ changes along beam axis EJ(x), then differential equation for displacement becomes:
Therefore the points in which bending stiffness changes are also a characteristic point.

28. Deflection of a beam of variable stifness
EJ1 K w P a1 a2
>
EJ2 x EJ l P K w x l
wK=
Pl3/3EJ Pl Pl
Pl/EJ2
{Pl/EJ2} {1-J2/J1}
Pl/EJ
Pl/EJ1
Pa2/EJ2
{Pa2/EJ2} {1-J2/J1}
{Pa2} {1-J2/J1}
Pa2/EJ1
EJ2
OR… a1
Pl/EJ2
EJ2
{P(l-a2)/a1} {1-J2/J1}
wK= ?
{Pl/EJ2} {1-J2/J1}

29. > EJ1 K w P a1 a2 EJ2 x EJ l P K w x l wK= Pl3/3EJ Pl Pl Pl/EJ2
{Pl/EJ2} {1-J2/J1}
Pl/EJ
Pl/EJ1
Pa2/EJ2
{Pa2/EJ2} {1-J2/J1}
{Pa2} {1-J2/J1}
Pa2/EJ1
EJ2
OR… a1
Pl/EJ2
EJ2
{P(l-a2)/a1} {1-J2/J1}
{Pl/EJ2} {1-J2/J1}

30. stop