Statically Indeterminate Beams

Mechanics of Solids 1 / 12Chapter 10

STATICALLY INDETERMINATE BEAMS

10.1 INTRODUCTION

Most of the structures encountered in real-life are Statically Indeterminate.

Statically Indeterminate Beams: No. of Reactions > No. of Eqns. of Equilibrium

Degree of Static Indeterminacy = No. of Reactions in excess of the No. of Eqns of Equilibrium.

Static Redundants = excess reactions; must be selected for each particular case.

Assumption throughout this chapter is that the beams are made of Linearly Elastic Materials.

10.2 TYPES OF STATICALLY INDETERMINATE BEAMS

- Propped Cantilever Beam

- Fixed – End Beam

- Continuous Beam(more than one span)

There are 4 ways of solving these types of problems.1. Use of the deflection curve2. Moment – Area Method3. Superposition (Flexibility Method)4. Indeterminate Beams Tables (handout)

We will examine No. 1 & 3, above.

P

Ay

Ax

MA

By

A B

P

Ay

Ax

MA

By

A B

MB

Bx

P

Ay

Ax

By

A B

Cy

C


10.3 ANALYSIS BY THE DIFFERENTIAL EQUATIONS OF THE DEFLECTION CURVE

1. pick redundant reaction2. express other reactions in terms of the redundant reaction3. write diff. eqn. of the deflection curve4. integrate to obtain general solution5. apply B.C. to obtain constants of integration & the redundant reaction6. solve for the remaining reactions using equations or equilibrium

This method is useful for:- simple loading conditions- beams of only one span (not good method for continuous beam)


EXAMPLE No. 1

GIVEN:The beam shown.

FIND:Reactions at supports using the deflection curve.

SOLn:

q

A B

L


MOMENT – AREA METHOD (just mention)

1. pick redundant reaction(s)2. remove redundant reaction(s) to leave a statically determinate beam3. apply loads on released structure4. draw M / EI diagram for these loads5. apply redundant reactions as loads6. draw M / EI diagram for redundant reactions7. apply moment – area theorems to find redundant reactions8. solve for remaining reactions using equations of equilibrium

Once reactions are determined, stresses and deflections can be calculated.

OPTIONAL


EXAMPLE No. 2


FIND:Reactions at supports using the Moment – Area Method.

SOLn:

OPTIONAL

q lb / ft

A B

2L ftL ft


10.4 METHOD OF SUPERPOSITION

1. pick redundant reaction(s)2. remove redundant reaction(s) to leave a statically determinate released structure3. determine deflections due to loads on released structure4. apply redundant reaction(s) as loads5. determine deflections due to redundant reaction(s) 6. sum the deflections for the total deflection – this is the superposition principle7. solve for the redundant reaction(s)8. solve for the remaining reaction(s) using the eqns. Of equilibrium

continuous beams:

This beam has 4 supports and 2 eqns of equilibrium. Thus there are 2 redundant forces.

If all loads are vertical and there are no axial deformations,then all reactions will be vertical.

The number of redundant forces is:

No. of Redundant Forces = No. of Supports – 2

We can analyze continuous beams by any of the previous methods but only superposition is practical.

HINT: when there are more than 2 supports, select the bending moments in the beam at the intermediate supports as the redundants.

Let’s see how this is done.

2 Eqns of Equilibrium

A B C

LA

IA

LB

IB


FBD

Because the beam is continuous across B: ____________ ( 1 )

SUPERPOSITION deflection due to LOADS + deflection due to REDUNDANTS

BL = due to MA + due to MB + due to loads in AB ____________ ( 2a )

BR = due to MB + due to MC + due to loads in BC ____________ ( 2b )

BL due to MA: BL1 = due to MB: BL2 =

BR due to MB: BR1 = BR due to MC: BR2 =

SUBSTITUTING (the above terms for BL1 , BL2 , BR1 , BR2 into EQNs ( 2a ) & ( 2b ):

BL = + + BL3 due to LOADS in AB _______ ( 3a )

BR = + + BR3 due to LOADS in BC _______ ( 3b )

A B

RA

MBMA

RBLBL

CB

RBR

MCMB

RCBR

MALA

6 E IA

MBLA

3 E IA

Case 7Appendix GTable G – 2 Page 909

MBLB

3 E IB

MCLB

6 E IB

MALA

6 E IA

MBLA

3 E IA

MBLB

3 E IB

MCLB

6 E IB


We can use 2nd Moment – Area Theorem ( pg 628 )

Arclength : ∆ = L

From 2 nd M-A Th: EI∆ = Ax

Substituting in arclength:

GENERIC Moment Diagrams for EXTERNAL LOADS:

Substituting the above terms for BL3 & BR3 into EQNs ( 3a ) & ( 3b ); using EQN ( 1 ) and rearranging:

If IA = IB I:

If LA = LB L:

xA

AA

xB

AB

L

L

∆

AxE I

∆ =

AxE I

L =

= AxE I L

3 – Moment equation

___________ ( 4 )


PROCEDURE: write one 3-moment equation for each intermediate support. This provides as many equations as redundant moments. solve simultaneously for moments.

ASSUMPTION MADE:The 2 extreme ends of the continuous beam were simply supported.

For fixed ends, see Text Example 10 – 4, Pg 724.

What happens if one or both are a fixed support?ANS: the number of redundant moments will be increased.

SOLUTION:Replace the fixed support by an additional span having an infinite moment of inertia. This prevents rotation, which is what a fixed support does.

FIG 10 – 15 / pg 724

Moments found at points 1, 2, and 3 will be the same as in the original beam.

To find remaining reactions, the equations of equilibrium must be applied to each beam section. For the example we just did;

RB = RBL + RBR

RBL is made up of 3 parts , , external load

RBR is made up of 3 parts , , external load

Must do this at each support.

MA

LA

- MB

LA

MC

LB

- MB

LB


EXAMPLE No. 1

GIVEN:The beam shown. I’s are equal.

FIND:Reactions at supports.

SOLn:

1 2 3

R1 R3R2

2m 2m 2m 3m

8 kN10 kN/m10 kN/m

8 kN m


EXAMPLE No. 2


FIND:Reactions at the supports using superposition method.

SOLn:

q lb / ftBA

2LL


INDETERMINATE BEAM TABLES (handout)

Statically Indeterminate Beams

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