Module 8 Indeterminate Beams_2013

INDETERMINATE BEAMS INTRODUCTION

STRENGTH OF MATERIALS 3B (SLR 3B21)

• Chapter 8 in “Mechanics of Engineering Materials” – Benham, Crawford and Armstrong • Chapter 12 in “Mechanics of Materials” – Hibbeler (12.6-12.9)



• Statically determinacy depends on the number of support reactions. So far only statically determinate beams have been considered having two unknown reactions which can be solved using conditions of equilibrium ∑ F =0 and ∑M=0.

• Only forces perpendicular to beams are considered for SF, BM and beam deflections making the third condition, ∑ F = 0, of no use for this purpose, only for end thrusts.

• Therefore, with statically determinate beams: Two equations resolve two supports.

• On the other hand, statically indeterminate beams have more than two supports, so other techniques have to be employed to solve all the support moment and forces.

INDETERMINATE BEAMS

INTRODUCTION


STATICALLY DETERMINATE BEAMS

F

F



• A structural member of any type is said to be statically indeterminate if the number of unknown reactions exceeds the available number of equilibrium equations.

• The additional support reactions on the beam or structure that are not needed to keep it in stable equilibrium are called redundants.

• The number of these redundants is referred to as the degree of indeterminacy.



STATICALLY INDETERMINATE BEAMS

1 DOI

1 DOI

2 DOI



As can be seen, these beams have more than two reactions. Therefore they require more equations to be fully resolved. These additional equations can be provided by considering: 1. The basic differential equation of deflection for the beam, i.e. direct integration; 2. The method of superposition; 3. The area-moment method. These approaches will be considered in turn.

INDETERMINATE BEAMS DOUBLE INTEGRATION METHOD


This method, as the name implies, requires two integrations of the differential equation:

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

= 𝑀𝑀𝐸𝐸𝐸𝐸

• Once the integrations have been done, the resulting equation has two constants of integration and unknown redundancies which can all be determined from the boundary conditions.

• Any outstanding redundancies can be determined from equilibrium and compatibility conditions.

• The starting point therefore is to identify the redundancy that would be incorporated into the internal momentum equation and be determined from the boundary conditions.*

* - Section 12.7 Hibbeler [Example 12.17 p.629]



Example [Example 12.18 p.630]

Consider the example illustrated below.

A B w [kg/m]

L RA RB

MA MB

+ - -

BMD



SOLUTION

From geometrical and loading symmetry, we note that MA = MB and RA = RB. From equilibrium of forces in the vertical direction as well we get:

Substituting this into the differential equation we have:

2wxM

2wLxM

dxydEI

2

A2

2++−=−=

2LwRR BA ==

Integrating for the first time yields:

A6

wxxM4

wLxdxdyEI

3

A

2+++−=



Integrating for the second leads to:

BAx24

wx2

xM12

wLxEIy42

A3

++++−=

The unknown terms in this equation include the integration constants A and B and the redundancy MA. These can be obtained from the boundary conditions:

0yiv.)

0dxdyiii.)

0y ii.)

0dxdyi.)

Lx

Lx

0x

0x

=

=

=

=

=

=

=

=



Applying these boundary conditions gives:

24wLMM 0,B 0,A

2

BA ====

Maximum displacement occurs at x = L/2 and is:

IE384Lwyy

4

2Lxmax == =



Example

Next consider the example illustrated below where an point load is applied in the place of a distributed load.

a

L

RA RB

F

MA MB



SOLUTION You will notice that a single expression for bending moment does not apply over the whole length of the beam (unless the origin is moved to the load point) so, keeping origin at A, Macaulay's method is used..

axFxRMMdx

ydEI AA2

2−+−=−=

Integrating once gives:

Aax2F

2xRxM

dxdyEI 2

2

AA +−+−=

Integrating again yields:

BAxax6F

6xR

2xMyIE 3

3

A

2

A ++−+−=



Boundary conditions for the problem are:

i.) At x = 0, y = 0

ii.) At x = L, y = 0

iii.) At x = 0, dy/dx = 0

iv.) At x = L, dy/dx = 0

Applying condition i.) we have B = 0.

Applying condition iii.) we have A = 0.

Now applying condition ii.) we have: 3

3

A

2

A aL6F

6LR

2LM0 −+−=

INDETERMINATE BEAMS

DOUBLE INTEGRATION METHOD


Applying condition iv.) we get:

22

AA aL2F

2LRLM0 −+−=

Solving these two equations simultaneously for MA and RA we get:

( ) ( )( )3

2

A2

2

A LaLa2LFR

LaLFaM −+

=−

= and

The remaining reactions can be determined from equilibrium conditions. For equilibrium of forces in the vertical direction we have:

0FRR BA =−+

And for equilibrium of moments about A we have:

0LRMM- BBA =−+



Solving these two equations for RB and MB we have:

( ) ( )3

2

B2

2

B L2bLFaRand

LaLFaM +

=−

=

Once the reactions have been determined, the complete beam deflection function can be determined enabling the evaluation of the displacement at any given position along the beam.

INDETERMINATE BEAMS

SUPERPOSITION METHOD*


This method involves breaking a complex system down into simpler 'standard‘ systems, analysing them independently then summing the resulting deflections to get the response of the complex system. As an example consider the system shown below where the system is split into two simple configurations. *-Section 12.5 Hibbeler

= + A B

ω

y’B

y’’B

RB

ω

RB RA

MA

L

INDETERMINATE BEAMS SUPERPOSITION METHOD


By super position, the total deflection at point B is given by: ''B

'BB yyy +=

And we already know that:

EI8Lωy

4'B −=

and:

EI3LRy

3B''

B +=

Therefore the total deflection at point B is given by:

EI3LR

EI8ωLy

3B

4

B +−=

INDETERMINATE BEAMS SUPERPOSITION METHOD


However we know that from compatibility of displacements, the deflection at B must be zero, hence:

0EI3LR

EI8ωL 3

B4

=+−

Solving for RB we have:

8Lω3RB =

Application of equilibrium of forces in the vertical direction gives:

8Lω5RA =

Taking moments about B gives the moment at A i.e.

8LωM

2

A =

INDETERMINATE BEAMS MOMENT AREA METHOD


Consider a general beam subjected to general loading as shown below.



You will recall that the elastic curve for the beam is given by: d2ydx2

= MEI

This can be expressed as: d

dxdydx

= MEI

where we know that dy/dx = θ, the slope at any point along the elastic curve. Hence:

dθdx

= MEI

and

dθ = MEI

dx



This means that the change in angle between any two pints in the elastic curve is equal to the area under the M/EI distribution function. This is termed the first moment area theorem and is stated as:

� dθ B

A = θB − θA = ∆θB/A = �

MEI

dx B

A

where ∆θB/A is the change in angle between point B and A. If the slope at point A is known i.e. θA , then the slope at point B is given by:

θ𝐵𝐵 = θA + ∆θB/A

If the small change in slope, ∆θ, is multiplied by an arbitrary distance from a reference state, we get a vertical displacement that we can represent by dv i.e.

dv = ∆θ dx

INDETERMINATE BEAMS

MOMENT AREA METHOD




This leads us to second moment-area theorem which can be expressed as:

� dv B

A = vB − vA = ∆vB/A = �

MEI

x dx B

A

where ∆vB/A is the change in vertical displacement between point B and A with respect to a particular reference point. This can also be expressed in the form:

vA/B = φ x�

where φ is area of the M/EI diagram between point B and A and x� is the horizontal distance to the centroid of the area from A. This can also be computed from point B.



EXAMPLE Find the maximum downward deflection of the small aluminium beam shown in the figure below due to the applied force P = 100 N. The beam’s constant flexural rigidity is EI = 60 Nm2.

P = 100 N

0.15 m 0.10 m

RA RB

MA EI = 60 Nm2

INDETERMINATE BEAMS

MOMENT AREA METHOD


SOLUTION The first part of this solution is to determine the redundant reaction so as to be able to establish the bending moment distribution. The expected elastic curve is as shown in the figure below.

P = 100 N

0.15 m 0.10 m

RA RB

MA



The tangent at the built-in end remains horizontal after the application of the load. To apply this condition using moment area theorem, we need the bending moment distribution diagram as shown below.

P = 100 N

0.15 m 0.10 m

RA RB

MA

6 Nm



Now applying the moment area approach we have: 1

6012

0.15 613

0.15 + 0.1 +12

0.1 623

0.1 +12

0.25 𝑀𝑀𝐴𝐴23

0.25 = 0

Solving this equation for MA yields: MA = −4.4 Nm

Having obtained the redundant reaction, we can then proceed to determine the other reaction using equilibrium conditions. This gives:

RA = 57.6 N and

RB = 42.4 N



The maximum deflection occurs where the tangent is zero. Once all the reactions are found we can determine the exact bending moment diagram as shown below.

P = 100 N

0.15 m 0.10 m

RA RB

MA

INDETERMINATE BEAMS

MOMENT AREA METHOD


The location of zero slope occurs when the negative moment is cancelled by the positive moment and is determined to occur at 0.148 m from A. The maximum deflection is then:

vmax = 160

12

0.074 4.2 0.074 + 23

0.074 + 12

0.074 −4.2 13

0.074

= 0.256 mm (down)

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Module 8 Indeterminate Beams_2013