Concept of plastic hinge, Application of Plastic Theory to RCC beam, Is requirements and specifications,

1. Presented by
Dr. Subhash V. Patankar
Department of Civil Engineering,
Sanjivani College of Engineering, Kopargaon,
Dist:Ahmednagar, Maharashtra, India.
E-mail: patankarsubhashcivil@sanjivani.org.in,
Mobile No. : 8087482971

2. CONTENTS
1) Introduction
2) Concept of Plastic Theory
3) Redistribution of Moments
4) IS Specifications
5) Application to Design of continuous beam

3. Mp
C
T
C/S of Beam Fb=Fy Fy Fy Mp
BMD
L
Development of Plastic Hinge
M= Mp
W=0 to Wu
Concept of Plastic Hinge

4. REDISTRIBUTION OF MOMENTS
 The R.C.C structure does not behave elastically at
collapse.
 If the loads on the beam are continuously increased
beyond the service loads, initially the beam behaves
elastically till stress at any section (C/S) reaches its
yields value.
 At this stage, only outer fiber yields but sections does
not fail.
 Beyond this stage, if load further increases
continuously then stress cannot be increase beyond
yield stress but inner and inner fibers get yielded
and their plasticization take place. This stage is
called elasto-plastic stage.

5.  If load continuously increase beyond this stage then
gradually complete the plasticization of both zones.
 At this stage, both tension and compression zones
are connected to each other with very small point
(hinge) on which they rotated.
 This imaginary hinge is called plastic hinge.
 When the bending movement at a section reaches
its ultimate value then the concrete above N.A. is in
plastic stage while the concrete below N.A. being in
tension gets cracked & steel reinforcement yields.

6.  Therefore plasticization of concrete in compression &
yielding of steel in tension causes the section to rotate at
constant ultimate moment at that section & it behaves as
a hinge called as plastic hinge.
 Plastic hinge is that section of the beam where
plasticization of concrete in compression & yielding
of steel in tension zone has occurred causing rotation
of section under constant ultimate movement.
 Rotation capacity of R.C. plastic hinge
 The rotation can be increased by reducing the depth of
N.A. i.e. design the under reinforced section.

7. Application of Redistribution of moments
used in Design of RCC Beam
 The transfer of additional moment from one section to
other section having reserve moment capacity is known
as redistribution of moments.
 (i.e. the mid-span has the reserve load carrying capacity
because the moment at that section is less than the
ultimate moment)
 For a pure ductile homogeneous material like steel,
100% redistribution of moment is permissible.
 But the RCC members have limited ductility, therefore
redistribution of moment should be such that it does not
develop serious cracks in concrete & there should be
adequate ductility at the hinge points.

8. Code requirements (cl. No. 37.1.1 P. No. 68 & 69)
 Equilibrium between the internal forces and the external
loads in maintained.
 The ultimate moment of resistance (MR) provided at
any section of a member is not less than 70% of the
moments at that section obtained from elastic maximum
moment diagram covering all appropriate combinations
of load (i.e. (Mu)provided ≥0.7Meu).
 The elastic moment at any section in a member due to
particular combination of loads shall not be reduced by
more than 30% of the numerically largest moment
given anywhere by the elastic maximum moment
diagram for the particular member covering all
appropriate combination of loads.

9.  At sections where the moment capacity after
redistribution is less than that from the elastic
maximum moment diagram, the following
relationship shall be satisfied;
 The code permits redistribution of moments up to
15% for elastic theory of design without any cracks
30% for limit state method of design.

10. Cl. No. 22.4.1(P. No. 35, IS 456-2000):
Arrangement of Imposed load
1) Design dead load on all span with full design
imposed load on two adjacent spans;
2) Design dead load on all spans with full design
imposed load on alternate spans.
Cl. No. 22.7 (P. No.36, IS 456-2000):
Redistribution of moments may be done in accordance
with Cl No. 37.1.1 for limit state method and in
accordance with B-1.2 for working stress method.
To find max. B.M. for different load combination

11. To find max. B.M. at Mid Span of Beam for Span
A B C D E F
L1 L2 L3 L4 L5
To find MAB & MCD
To find MBC & MDE
A B C D
B C D E

12. To find max. B.M. at support B & E
To find max. B.M. at support C
To find max. B.M. at support D
B
D

13. Case I: To find Max. B.M. at mid-span of AB
Wu
Wd
Le1 Le2

14. Case II: To find Max. B.M. at mid-span of BC
Le2Le1
Wu
Wd

15. Case I: To find Max. B.M. at Continuous Support (B)
Le2Le1
Wu Wu

16. Bending Moment Envelop
le2le1
A B C
MB max
Mu
Mu

17. Shear Force Envelop
le1 le2
VA
VBA
VBC
VC

18. Maximum B. M. at Mid span and at Support
and Shear Force at Supports
BM at Mid Span of AB & S. F. at Support A
BM at Mid Span of BC & S. F. at Support C
BM at support B & S. F. at Support B

19. Design Steps:
1. Select suitable C/S of beam
2. Effective span of beam
3. Load on beam
4. Find the maximum B. M. for different combination of loads
5. Calculate effective flange width ( in case of flanged beam )
6. Find section moment, Mus (assume Xu≤Df)
7. Compare Mu with Mus
8. Calculate Ast at mid span of AB & BC
9. Calculate Ast at continuous support
10. Check for deflection
11. Check for shear
12. Check for development length
13. Reinforcement details

20. After redistribution the moment capacity of section
is reduced at continuous support.
So, calculate M.R. using following procedure

21. Example : Wd =20 kN/m, WL = 9 kN/m, BM calculation

26. Thank You