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STATICALLY INDETERMINATE BEAMS

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.

2. TYPES OF STATICALLY INDETERMINATE BEAMS

- Propped Cantilever Beam

- Fixed – End Beam

- Continuous Beam

(more than one span)

There are 4 ways of solving ...view middle of the document...

apply moment – area theorems to find redundant reactions

8. solve for remaining reactions using equations of equilibrium

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

EXAMPLE No. 2

GIVEN:

The beam shown.

FIND:

Reactions at supports using the Moment – Area Method.

SOLn:

10.4 METHOD OF SUPERPOSITION

1. pick redundant reaction(s)

2. remove redundant reaction(s) to leave a statically determinate released structure

3. determine deflections due to loads on released structure

4. apply redundant reaction(s) as loads

5. determine deflections due to redundant reaction(s)

6. sum the deflections for the total deflection – this is the superposition principle

7. 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.

FBD

Because the beam is continuous across B: [pic]____________ ( 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 )

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

Arclength : ∆ = ( L

From 2nd M-A Th: EI∆ = Ax

Substituting in arclength:

GENERIC Moment Diagrams for EXTERNAL LOADS:

Substituting the above terms for...

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