chapter 9
DESCRIPTION
Chapter 9. Deflection of Beams. 9.1 Introduction . -- Concerning about the “deflection” of a beam -- Special interest: the maximum deflection -- Design: to meet design criteria. 9.1 Introduction . (4.21). M = bending moment E = modulus I = moment of inertia. - PowerPoint PPT PresentationTRANSCRIPT
Chapter 9Deflection of Beams
9.1 Introduction
-- Concerning about the “deflection” of a beam
-- Special interest: the maximum deflection
-- Design: to meet design criteria
1 MEI
1 ( )M xEI
2
2
( )d y M xd x EI
9.1 Introduction
(4.21)
M = bending moment
E = modulus I = moment of inertia
If M is not a constant, i.e. M=M(x)
(9.1)
or
-- will be explained in Sec. 9.3
dydx
2
2
( )d y M xd x EI
y = y(x)
1 PxEI
(9.1)
(9.2)
9.2 Deformation of a Beam under Transverse Loading
1 ( )M xEI
Since M(x) = -Px
Example: a beam subjected to transverse loads
Moment Diagram and the Deformed Configuration:
Mmax occurs at C
In addition to M(x) and 1/, we need further information on:
1. Slope at various locations
2. Max deflection of a beam
3. Elastic curve: y = y(x)
9.3 Equation of the Elastic Curve
10( )
xdyEI M x dx Cdx
2
2
( )d y M xd x EI
(9.4)
(9.5)
2
2
32 2
1
[1 ( ) ]
d yd xdydx
2
2
1 d yd x
2
2
( )d y M xd x EI
9.3 Equation of the Elastic Curve
The curvature of a plane curve at Point Q(x,y) is
For small slope dy/dx 0, hence
2 0( )dydx
Therefore,
Fially,
(9.2)
(9.3)
(9.4)
2
2( )d y M x
d x EI -- the governing diff. equation of a
beam
-- the governing diff. equation of the “elastic curve”
EI = flexural rigidity
tan ( )dy xdx
2
2( )d y M x
d x EI
10( )
xdyEI M x dx Cdx
Integrating Eq. (9.4) once
(9.4)
10( ) ( )
xEI x M x dx C
Since,
It follows,
(9.5’)
(9.5)
1 20 0[ ( ) ]x x
EIy M x C dx C
1 20 0( )
x xEIy dx M x dx C x C (9.6)
10( )
xdyEI M x dx Cdx
(9.5)
Integrating Eq. (9.5) once again,
1 20 0
1( ) ( )x x
y x dx M x dx C x CEI
or
C1 & C2 are determined from the B.C.s
Boundary conditions:
-- the conditions imposed on the beam by its supports
Examples:
Three types of Statically Determinate Beams:
1. Simply supported beams: 2. Overhanging beams:
3. Cantilever beams:
9.4 Direct Determination of the Elastic Curve from the Load Distribution
2
2
( )d y M xd x EI
4
4
( )d y w xdx EI
3
3
1 ( )d y dV w xdx EI dx EI
(9.4)
(9.31)3
31 ( )d y dM V x
dx EI dx EI
(9.32)
3
13
2
1 22
21
2 3
31
22 3 4
12
16
12
( ) ( )
( ) ( )
( ) ( )
( ) ( )
d yEI V x w x dx Cdxd yEI M x dx w x dx C dx CdxdyEI EI x dx dx w x dx C xdx
C x C
EIy x dx dx dx w x dx C x
C x C x C
4
4 ( )d yEI w xdx
(9.33)
9.5 Statically Determinate Beams
0 0 0x y AF F M
Applying equations of equilibrium:
Conclusion:
-- This is a statically indeterminate problem.
-- Because the problem cannot be solved by means of equations of equilibrium
(9.37)
9.5 Statically Determinate Beams
By adding (1) deflection y = y(x) and = (x), the problem can be solved.
i.e. five unknowns with six equations
Statically indeterminate to the 1st degree:
-- one redundant support
Statically indeterminate to the 2nd degree:
-- two redundant supports
9.6 Using Singular functions to Determine the Slope and Deflection of a Beam
03 1( )4 4PV x P x L
3 1( )4 4PM x P x L
2
2
3 14 4
d y PEI x P x Ldx
(9.44)
(9.45)
22
13 1 18 2 4
dyEI EI Px P x L Cdx
33
1 21 1 18 6 4
EIy Px P x L C x C
3
21 10 0 0 06 4P L C
33
11 1 308 6 4PL P L C L
31
124
C wL
(9.46)
(9.47)
9.7 Method of Superposition
9.8 Application of Superposition to Statically Indeterminate Beams
9.9 Moment-Area Theorems
2
2
d d y Mdx dx IE
Md dxEI
D D
C C
x
x
Md dxEI
D
C
x
D C x
M dxEI
(9.54)
(9.55)
area under (M/EI) diagram
between C and D
DC
1dt x d
1Mdt x dxEI
1D
C
x
C xD
Mt x dxEI
1(area between C and D)CD
t x
2(area between C and D)DC
t x
(9.56)
(9.57)
(9.59)
(9.60)
9.10 Application To Cantilever Beams And Beams With Symmetric Loads
9.11 Bending-Moment Diagrams By Parts
9.12 Application Of Moment-area Theorems To Beams With Unsymmetric Loadings
BA
A
t
L
D A DA
or EF= BA
EF HB x tx L L
D D BA A
xy ED EF t tL
(9.61)
(9.62)
(9.63)
9.13 Maximum Deflection
9.14 Use Of Moment-Area Theorems With Statically Indeterminate Beams