610.05.2005 Dr. N. SureshDr. N. Suresh Department of Civil EngineeringNational Institute of Engineering,Mysore

Definition of Beam

Flexural Stiffness

Types of Beams

3 - Moment Equation

Learning Outcomes

What is a beam?

A (usually) horizontal structural member that is subjected

to a load that tends to bend it

Examples of beams

Engineering examples

Floor joists and rafters I beams

Biological examples

Tree branches

Vertebral column and neck

Insect thorax/abdomen exoskeleton

Flexural stiffness (EI)

Indicates how resistant a structure is to bending

Depends on the material making up the structure and on its shape

(((( ))))(((( ))))area ofmoment Secondmodulus sYoung'====EI

Simply Supported Beams

Types of Beams

Cantilever Beam

Continuous Beam

Single Overhang Beam

Double Overhang Beam

Single Overhang Beam with Internal Hinge

Fixed Beam

Continuous beam

Simply supported beam

Cantilever beam

A cantilever beam can be thought of as half of a fixed beam turned upside down

See?

ANALYSIS OF CONTINUOUS BEAMS (using 3-moment equation)

Stability of structure If the equilibrium and geometry of structure is maintained under the action of forces than the structure is said to be stable.

External stability of the structure is provided by the reaction at the supports. Internal stability is provided by proper design and geometry of the member of the structure.

Statically determinate and indeterminate structures A structure whose reactions at the support can be determined

using available condition of equilibrium is called statically

determinate otherwise it is called statically indeterminate.

Ex:

W A B

HA HB

VA VB MA MB

End moments

FIXED BEAM

W W

A

RA

RB

A C

RC

No. of unknowns = 6 No. of eq . Condition = 3 Therefore statically indeterminate Degree of indeterminacy =6-3 = 3 No. of unknowns = 3 No. of equilibrium Conditions = 2 Therefore Statically indeterminate Degree of indeterminacy = 1

Advantages of fixed ends or fixed supports 1. Slope at the ends is zero.

2. Fixed beams are stiffer, stronger and more stable than SSB. 3. In case of fixed beams, fixed end moments will reduce the

BM in each section.

4. The maximum deflection is reduced.

Bending moment diagram for fixed beam Ex:

W

4WL

2L

2L

+

+

M M

Continuous beams Beams placed on more than 2 supports are called continuous beams. Continuous beams are used when the span of the beam is very large, deflection under each rigid support will be equal zero.

BMD for Continuous beams BMD for continuous beams can be obtained by superimposing the fixed end moments diagram over the free bending moment diagram.

Three - moment Equation for continuous beams OR CLAPERONS THREE MOMENT EQUATION Ex:

FREE B.M.

1x

a1 a2

8

2WL

L2 L2

A

B

C

N

N

4WL

2x

++++

++++++++

22

2C

22

2

11

1B

11

1A IE

LMIE

LIE

LM2IE

LM

++++

====

2

BC

1

BA

222

22

111

11

LL6

LIExa6

LIExa6

The above equation is called generalized 3-moments Equation.

MA, MB and MC are support moments E1, E2 Youngs modulus of Elasticity of 2 spans. I1, I2 M O I of 2 spans, a1, a2 Areas of free B.M.D.

21 xandx Distance of free B.M.D. from the end supports, or outer supports. (A and C) A, B and C are sinking or settlements of support from their initial position.

Normally Youngs modulus of Elasticity will be same through out than the equation reduces to

+

++

2

2C

2

2

1

1B

1

1A I

LMI

LI

LM2I

LM

+

=

2

BC

1

BA

22

22

11

11

LL6

LIxa6

LIxa6

If the supports are rigid then A = B = C = 0

+

++

2

2C

2

2

1

1B

1

1A I

LMI

LI

LM2I

LM

22

22

11

11

LIxa6

LIxa6

=

If the section is uniform through out

( ) 2C21B1A LMLLM2LM +++

2

22

1

11

Lxa6

Lxa6

=

1.

If the end supports or simple supports then MA = MC = 0

2.

MC = - WL3 Overhang portion the support moment near the overhang can

be computed directly.

A

B

C

N

N

D

L1 L2 L3

A

B

C

N

N

3.

If the end supports are fixed assume an extended span of zero length and apply 3-moment equation.

Zero Span

A

B

C

A1 D

Zero Span

i) Bending Moment Diagram for an Eccentric Load

In this case centroid lies as shown in figure

a b W

LWab

a b

3a+

3b+

ii) Bending Moment Diagram for Two Load at equidistant

a b

Wa Wa

2L

x =

L