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Today’s Objective : 14th September

To:

a) Understand the context, concept and derivation for deflection and slope.

b) Be able to calculate max and specific values.

DEFLECTION OF BEAMS

Deflections of Beams

Serviceability of beams

Deflection limits for beams appearance (sagging) fitness for purpose (machinery, pipe grades) structural (avoid unintended load paths)

General Guidelines

Limit total deflections to span/250

Different design codes give guidance on the limits for deflections

To limit deflections, we need to be able to calculate them!

250maxLv ≤

deflection (v)

The elastic curve – deformed shape of the beam

The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of the beam is called the elastic curve

v

v

Revision- Geometry of the beam deformations

The deflection, slope and curvature of a beam are related by:Deflection = vSlope = dv/dx = θCurvature = d2v/dx2 = dθ /dx = KThe curvature is the inverse of the radius, K = 1/R

ie. for a segment of the beam dx

dx = Rdθ hence,1/R = dθ /dx

Consider the deflected shape of the beam given below

dθ

dx

R

Geometry of the beam deformations: Strain

At a distance y from the neutral axis (N.A.)

Strain:

LL∆=εdsdsds )'( −=

θθθ

RdRddyR −−

=)(

Ry

−=

Moment-Curvature Relationship

We have shown that the strain ɛ at a distance y from the N.A.

ɛ=-y/R (1)Hooke’s law

ɛ=σ/E (2)From the flexure formula, the stress σ at the point under bending is given by

σ=-My/I (3)Substituting (3) into (2), the strain can be expressed by

ɛ=-My/EI (4)Substituting (4) into (1), the moment-curvature relationship can be found as

1/R=M/EI or K=M/EI

EIM

dxvdK == 2

2Curvature:

∫∫ === Mdx

vdEIdxdvEIEI 2

2

θSlope:

∫∫ ∫∫== Mdx

vdEIEIv 2

2

Deflection:

•We need to be able to define the bending moment as a function of the distance from one end of the beam

•The constants of integration need to be determined by evaluating the slope or deflection at particular points –boundary condition

Slope and Displacement by integration

E – Young’s modulus which is a constant for a given material

I – Moment of inertia computed about the neutral axis

Boundary condition

Roller:

Pin:

Fixed:

At a point on the beam where the value of slope or deflection is known, these values are called boundary conditions. For example:

v =0

v =0

v =0

θ =0

Example 1The simply supported beam supports the uniform distributed loading. Determine the equation of the elastic curve, slope and deflection at point C and the maximum deflection of the beam. E=200GPa, I=50(106)mm4.

Reaction forces

F.B.D:

∑ = :0BM 02/101010010 =××+×− AR+

N500=AR

∑ = :0yF+ 10100×=+ BA RR

N500=BR

Moment FunctionA free-body diagram of the segment and coordinate x is shown as

∑ = :0oM+

05002

100)( =×−×+ xxxxM

xxxM 50050)( 2 +−=

Slope and deflection by integration

xxxMdx

vdEI 50050)( 22

2

+−==

123 250

350 Cxx

dxdvEIEI ++−==θ

According to Moment-Curvature relationship, we have

Integrating twice, we obtain the slope and deflection

Slope:

Deflection: 2134

3250

625 CxCxxEIv +++−=

Applying boundary conditions

2134 00

32500

6250 CCEI +++−=

To determine the constants C1 and C2, the boundary conditions will be applied:

Condition 1:

At point A, x = 0 and v = 0. Thus

02 =C

Condition 2: At point B, x =10m and v = 0.

010103

250106250 1

34 +++−= CEI 7.41661 −=C

Slope and deflection at point CWith the values of C1 and C2, the functions of slope and deflection are

7.41662503

50 23 −+−== xxdxdvEIEIθSlope:

Deflection: xxxEIv 7.41663

250625 34 −+−=

7.4166225023

50 23 −×+×−=CEIθ

3300−=

( ) )10(3.3)10)(10(5010200

33003300 41269

−− −=

×−=−=

EICθ

At Point C, x = 2m

4.773327.416623

2502625 34 −=×−+×−=CEIv m)10(73.7 4−−=Cv

?

The maximum deflection

7.41662503

500 23 −+−= xx

5.0=x

-2073.25.07.41665.03

2505.0625 34

max =×−×+×−=EIv

The maximum deflection occurs at the point where the slope θ=dv/dx =0. Assuming θ=0 in the slope function:

Finding x between 0 and 10m

Thus, the maximum deflection is

m)10(07.22073.2 3max

−−=−

=EI

v

Alternatively, by inspection of the elastic curve, the maximum deflection occurs in the middle where the slope is zero.

Summary of results

xxxEIv 7.41663

250625 34 −+−=

xEI

xEI

xEI

v 7.41663250

625 34 −+−=

)10(3.3 4−−=Cθ

1) Elastic curve:

or

m)10(73.7 4−−=Cv

m)10(07.2 3max

−−=v

5.0=x

2) Slope and deflection at point C

3) The maximum deflection

at

Positive Sign Convention

When applying double integration equations, the positive signs for slope and deflection are

Sign convention-cont.The sign convention is helpful to verify your calculation, for example

vA is negativeθA is negative

vB is negativeθB is positive

Q: By inspection, slope and deflection at point A, B and C are positive or negative under the given coordinate?

Boundary condition for a symmetry structure

Due to symmetry, only half of the structure needs to be calculated for slope and deflection. In such cases, the slope at the symmetry axis is zero. This boundary condition will greatly simplify our calculation.

Symmetry condition:

2at0 Lxdxdv ==

Previous example using the symmetry condition

2134 00

32500

6250 CCEI +++−=

xxxMdx

vdEI 50050)( 22

2

+−==

123 52505

3500 CEI +×+×−=

123 250

350 Cxx

dxdvEIEI ++−==θ

2134

3250

625 CxCxxEIv +++−=

From previous slides, we obtained

Boundary conditions:1) At point A, x = 0 and v = 0.

02 =∴C

2) When x = 5m and θ = 0.

xxxEIv 7.41663

250625 34 −+−=

7.41661 −=∴C

Function of elastic curve:

-Same to the previous result

Continuity conditionsIf a single x coordinate cannot be used to express the equation for the slope or deflection, then continuity conditions must be used to evaluate some of integration constants.

Continuity conditions – cont.Each coordinate is only valid within the regions

and ax ≤≤ 10 )(2 baxa +≤≤

Continuity conditions require

)()( 21 avav =)()( 21 aa θθ =

Each coordinate is only valid within the regions

and ax ≤≤ 10 bx ≤≤ 20Continuity conditions require

)()( 21 avav =)()( 21 aa θθ −=

Example 2

Determine the equations of the elastic curve for the beam using x1 and x2coordinates. Specify the slope at A and the maximum deflection. EI is constant.

Reaction Forces

∑ = :0AM+ 03

23

=×−×−×LPLPLRB

PRB =

∑ = :0yF+ 0=−−+ PPRR BA

PRA =

Moment Functions30 1 Lx ≤≤When

∑ = :0M+

0)( 11 =− PxxM

11)( PxxM =

23 2 LxL ≤≤When

∑ = :0M+

0)3

()( 222 =−+−LxPPxxM

3)( 2

PLxM =

Slope and deflection by integrationThus, for 30 1 Lx ≤≤

23 2 LxL ≤≤

1121

12

)( PxxMdx

vdEI ==

3)( 22

2

22 PLxM

dxvdEI ==

211311 6

CxCxPEIv ++=

For

423222 6

CxCxPLEIv ++=

Note: There are four constants of integration: C1, C2, C3 and C4. To determine the above four constants, four additional equations are needed by applying boundary conditions

------(1)

----------(2)

------(3)

--------(4)

12111

1

1

2)( CxPxEI

dxdvEI +== θ

32222

2

3)( CxPLxEI

dxdvEI +== θ

Boundary conditions

Because only half of the structure is considered, so only the boundary condition at point A will be applied, that is,

When 01 =x , 01 =v

213 00

60 CCPEI +×+×=

So 02 =C ------------------------------------------(5)

. From eqn.(2)

Boundary condition due to symmetry

Due to the symmetry of the structure, we know

when 2/2 Lx = , 02

2 =dxdv

From eqn.(3),

3230 CLPLEI +×=

6

2

3PLC −= ------------------------------------------(6)

Continuity conditions

From eqn.(1) and eqn.(3), slope continuity condition is

when 3/21 Lxx == , continuity conditions will be applied

)3

()3

( 21LEILEI θθ =

63332

2

1

2 PLLPLCLP−×=+

×

9

2

1PLC −=

From eqn.(2) and eqn.(4), deflection continuity condition is

4

2223

36363936CLPLLPLLPLLP

+×−

×=×−

×

)3

()3

( 21LEIvLEIv =

162

3

4PLC =

------------------------------------------(7)

-----(8)

Slope and deflection

)30( 1 Lx ≤≤

)23( 2 LxL ≤≤

)30( 1 Lx ≤≤

)23( 2 LxL ≤≤

1

2311 96

xPLxPEIv −=

16266

3

2

2222

PLxPLxPLEIv +−=

92)(

22111

PLxPxEI −=θ

63)(

2

222PLxPLxEI −=θ

Slope:

Deflection:

Slope at point A:99

02

222 PLPLPEI A −=−×=θ

EIPL

A 9

2

−=θ

1622626

322

maxPLLPLLPLEIv +

×−

×=

The max v occurs at x2=L/2

EIPLv

64823 3

max −=

Steps for DOUBLE INTEGRATION

Select appropriate coordinator(s) and establish the equation(s) for M(x)

Integrate to get slope, EIθ. Integrate again to get deflection, EIv. Apply the boundary conditions and continuity

conditions to θ and v to evaluate the integration constants.

)(xM

Why is deflection calculation important?

a) Deflection is another important set off calculation in structural analysis.

b) Loads create bending and bending causes deflection

c) Deflection is an important service condition

d) Deflection is not important at all

What does dv/dx represent?

a) Deflectionb) The elastic curvec) The curvatured) The slope

What is the purpose of boundary conditions?

a) Allows to calculate or eliminate constants

b) Defines the equations at boundariesc) Is a part of all equationsd) Not useful at all