Continuous Rc Beams

OUTLINE

CONTINUOUS R. C. BEAMS

• Idealizations in structural analysis1. Span2. Rigidity of Supports 3. Moment of Inertia (MI) of spans4. Loading 5. Elastic methods of analysis

1. SPAN


Idealization in span calculation• Width of simple supports is taken as zero

Concept of Effective span • For simply supported beam or slab• For continuous beam or slab• For cantilever• For frame members

1. SPAN


Effective span for simply supported beams or members not built integrally with supports• Clear span + Effective depth le = ln + d

OR• Centre-to-centre of supports le = l

Whichever is LESS

1. SPAN


Effective span for continuous beams or slabs with support width < clear span/ 12• Clear span + Effective depth le = ln + d

OR• Centre-to-centre of supports le = l

Whichever is LESS

1. SPAN


Effective span for continuous beams or slabs with support width > clear span/ 12 OR > 600 mm

For end span with fixed endeffective span = clear span le = ln

For end span with free endle = ln + d/2 = ln + support width/2

1. SPAN


For end span with free endle = ln + d/2 = ln + support width/2

1. SPAN


Effective span for cantilever span is For a cantilever span, the effective length is length to face of support + d/2

For cantilever span of continuous beam, length to centre of support

le = ln + d/2 le = l

ln l

1. SPAN


2. Supports


Idealization in span calculation• Supports are unyielding

Support sinking or rotation -> secondary forces • Redistribution of BMs occurs• BM can become positive over continuous

supports, and positive BM at centre increases• Beam fails at support and then at centre

Redistribution of BMs caused by support movement has to accounted in structural Analysis

3. STIFFNES

S


Idealization in stiffness• MI remains constant throughout the span

Moment of Inertial (MI) at a section depends on• Amount and location of reinforcement • Extent of cracking• Shape of concrete in compression

Relative stiffness (I/L)should be calculated consistently• Gross section MI (Or)• Transformed section MI (Or)• Cracked section MI

4. LOADIN

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Idealization in loading• The loads on all spans are known (fixed) values

But,• The load values are only characteristic values• Live load may or may not be present on a span• Change in loading on any portion can affect

BMs and SF s on a given span

4. LOADIN

G


So,• BMD is also characteristic (i.e., it has

variability)• Live load patterns that give worst effect should

be considered• Appropriate substitute frames are to be

selected for the stress-resultant .

4. LOADIN

G


Live Load Patterns• For max. + ve moment in span S, load the span

under consideration and also the alternate spans

S

4. LOADIN

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P

Live Load Patterns• For max. + ve moment at support P, unload

spans on either side of the support P and load the alternate spans on either side

4. LOADIN

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P

Live Load Patterns• For max. - ve moment at support P and max.

SF, Load adjacent spans on both sides of the support P and also the spans alternate to these

4. LOADIN

G


S

Live Load Patterns• For max. - ve moment in span S, load the spans

adjoining the span S

OUTLINE


• Moment redistribution in indeterminate beams• Stages leading to collapse• Equilibrium equations govern all stages• Load Vs moments• Design problem •Where moment redistribution can be done?• How much moment can be redistributed• Numerical problems

OUTLINE


• Advantages of moment redistribution•Assumptions• in elastic analysis• in moment redistribution• Limitations on moment redistribution• IS456 about moment redistribution• Use of IS code coefficients (simplified method for moment redistribution analysis)• Design for redistributed BMD


Distribution (shape) of BMD

6PL/32

5PL/32

L L

P P

Elastic Bending Moment Diagram

-+ +

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Redistribution (reshaping) of BMD

6PL/32 = Mp-

5PL/32

L L

P P

At first Plastic Hinge formation

-+ +

Plastic hinge

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Redistribution (reshaping) of BMD

6PL/32 = Mp-

Mp+

L L

Pu

At collapse (Second Plastic Hinge formation)

-+ +

Pu

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Equilibrium conditions

L L

P

At collapse (Second Plastic Hinge formation)

P

M+

M-

R1 = 2M+/L

2P- 2R1

P- R1

P- R1

M+

2M+ + M- = PL/2

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Relation between moments and load2M+ + M- = PL/2

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Need for BM Envelope

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Elastic BMDFactored BMDRedistributed BMD


Distribution of BMs (BMD)

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Redistributed BMD

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MOMEN

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Steps in structural analysis for redistribution of moments

1. Perform Elastic Structural Analysis to get elastic BMD2. Mark the points of contra-flexure3. Reduce the maximum moment by x% 4. Perform equilibrium analysis for getting redistributed BMD5. Obtain the points of contra-flexure 6. Draw envelopes of (x times elastic BMD + redistributed BMD)

MOMEN

T REDISTRIBUTION


Illustrative Example

Q. Draw the design bending moment diagram of the beam of a RC fixed beam clamped at both ends and carrying ultimate uniformly distributed load of 24 kN/m with full redistribution of 30 per cent as per IS 456.

MOMEN

T REDISTRIBUTION


Illustrative Example

MOMEN

T REDISTRIBUTION


Illustrative Example - Question

Q. Draw envelope of the design moments of the two-span continuous beam shown in Fig. 15.38.7a, carrying characteristic live load of 35 kN/m in addition to its characteristic self-weight. The cross-section of the beam is 300 mm × 700 mm.

MOMEN

T REDISTRIBUTION


Illustrative Example – Load cases to be considered

Three load cases: 1. DL on both spans + LL on both spans2. DL on both spans + LL on span AB only3. DL on both spans + LL on span BC only

Load Case 1

Load Case 2

Load Case 3 is a mirror image of Load case 2 . SO need not be considered separately

MOMEN

T REDISTRIBUTION


Illustrative Example – Case 1 – Elastic and redistributed BMDs

MOMEN

T REDISTRIBUTION


Illustrative Example – Load Case 1 – Points of contra-flexure

MOMEN

T REDISTRIBUTION


Illustrative Example – Load Case 2 – loading and elastic BMDs

MOMEN

T REDISTRIBUTION


Illustrative Example – Load Case 2 – Elastic BMD using MD

Member AB BA BC CB D.F 1.0 0.5 0.5 1.0 F.E.M + 322.0 - 322.0 + 28.0 - 28.0 Balanced Moment - 322.0 + 147.0 + 147.0 + 28.0 Carry over Moment - 161.0 + 14.0 - Balanced Moment - + 73.5 + 73.5 - Total Moment 0.0 - 262.5 + 262.5 0.0

MOMEN

T REDISTRIBUTION


Illustrative Example – Load Case 2 – redistributed BMDs

MOMEN

T REDISTRIBUTION


Illustrative Example – Load Case 1 & 2 – BMD Envelop

The envelop is drawn from x times elastic BMD s of case

1,2 and 3 +

redistributed BMD of case 1,2 and 3)

MOMEN

T REDISTRIBUTION


1. Gives realistic picture of the load carrying capacity of indeterminate structures

2. Results in economical designs3. Designer gets some freedom to choose the design moments. This

helps to avoid congested reinforcement designs

Advantages of moment redistribution

MOMEN

T REDISTRIBUTION


1. Ultimate moments of resistance of the section are Mu

+ and Mu-

2. No premature shear failure occurs before the collapse load

3. Moment curvature relation is idealized bilinear curve

4. All sections have the same constant value of EI up to collapse

5. Self-weight of the beam is negligible

Assumptions used in the concept of moment redistribution

1. Sufficient plastic rotations at critical sections2. The extent of cracking should not affect the serviceability requirements

Conditions to be fulfilled to apply moment redistribution

MOMEN

T REDISTRIBUTION


1. Moment of inertia (MI) does not vary within a span2. Supports do not yield3. Supports are knife-edged (i.e support width = zero)

Assumptions used in Elastic analysis of beams

1. In calculating the ratios of MI of adjacent spans, follow the same principle (i.e use either gross section or transformed section for all spans)

2. Support settlements are to be accounted in calculating elastic BMD 3. The concept of effective length is to be used (Cl. 22.2 of IS456-2000)

Conditions to be fulfilled to use elastic analysis


IS Code stipulation

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Simplified method based on IS Code coefficients

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Continuous Rc Beams

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