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 span/depth ratio of
simply supported beam is < 2,
continuous beam < 2.5,
it is classified as deep beam.
 Such structures are found in transfer girders and in
shear wall structures that resist lateral forces in
buildings. It is also found in some of the industrial
buildings.
 The traditional principles of stress analysis are
neither suitable nor adequate to determine the
strength of reinforced concrete deep beams.
 In deep beams, the bending stress distribution
across any transverse section deviates appreciably
from straight line distribution assumed in the
elementary beam theory.
 The behaviour of a deep beam depends also on
how they are loaded & special considerations should
be given to this aspect in design.
 Here cracking will occur at one-third to one-half of
the ultimate load.
 In the single span beam supporting a concentrated
load at mid span, the compressive stresses act
roughly parallel to the lines joining the
load and the supports and the tensile stresses act
parallel to the bottom of the beam.
 The flexural stresses at the bottom is constant
over much of the span.
 The figure shows the crack pattern and the
truss analogy of the same.
Each of the three tension AB, CD and EF ties
have cracked and at failure shaded region
would crush or the anchorage zones at E and F
would fail.
simplified truss model
 A single span beam supporting a uniform load
acting on the top has the stress trajectories ,
crack pattern and simplified truss as shown.
 A single span beam supporting a uniform load
acting on the lower face of the beam has the
stress trajectories , crack pattern and simplified
truss as shown.
 The compression trajectories form an arch with
the loads hanging from it. The crack pattern
shows that the load is transferred upward by
reinforcement until it acts on the compression
arch, which then transfers the load down to the
supports.
 The force in the longitudinal tension ties will be
constant along the length of the deep beam.
This is the reason that the steel must be
anchored at the joints over the reaction, failure
of which is a major cause of distress
 Parameters influencing deep beam behavior
are:
 Width of support = C
 Overall depth of beam = D
 Effective span = L
 Width / Thickness of beam = t
 Type of loading, uniform = w
The Min. thickness of deep beams should be based on two
considerations:
1. It should be thick enough to prevent buckling with
respect to its span & height i.e.
where t = thickness of beam.
2. The thickness should be such that the concrete itself
should be able to carry a good amount of the shear force
that acts in the beam without the assistance of any steel.
50
&
25 

t
L
t
D
 Z = 0.6L or 0.6D
i.e. Z = 0.6L when L/D < 1
& Z = 0.6D when L/D > 1
 From those values,
Mu = As.fs.Z where fs = 0.87 fy
 The greater value of As is taken as tension steel.
  
   L
fy
Mu
L
fy
Mu
A
fyD
Mu
D
fy
Mu
Z
f
Mu
A
s
s
s
.
9
.
1
6
.
0
87
.
0
or
9
.
1
6
.
0
87
.
0
.





Deep beams without holes
 Design procedure :
t
ft
av
D
Vc
fck
ft
D
av
Co
D
t
ft
D
av
C
Vc
Vs
Vc
Vu
Du
t
Vu
Tv
.
)
35
.
0
(
72
.
0
becomes
equation
the
Hence,
strength.
tensile
the
,
5
.
0
Depth
span
Shear
concrete
wt.
normal
for
0.72
to
equal
eff.
C
where
.
.
35
.
0
1
c)
b)
T
Tv
where
.
Shear
Nominal
Determine
a)
1
1
max


















 When designing for shear, it is assumed that
concrete itself should carry at least 65% of the
ultimate shear.
 This is ensured by choosing a suitable thk. of beam
by the following formula :
ft
av
D
Vu
t
)
35
.
0
(
72
.
0
65
.
0


 Shear capacities of tension steel & nominal web steel
should also be taken into account in resisting the
shear. Their shear capacity is calculated as :
where,
C2 = 225 N/mm2 for Fe415
As = Area of tension steel provided.


n
D
y
As
C
Vs
1
2
2 Sin
1
. 
α = Angle between the bar considered & the critical
diagonal crack.
y1= Depth from the top of the beam to the point where
bar intersects the critical diagonal crack line.
n = Number of bars including tension steel cut by the
assumed crack line
D = Total depth of beam
 Vertical steel Av and Horizontal steel Ah
 Horizontal steel bars acts as shear reinforcement
and also overcome the effects of shrinkage &
temperature.
 The amounts specified in IS:456-2000 are :
a) Vertical steel shall be 0.12% for Fe415, the bar
diameter shall not be more than 14mm and spacing
not more than 3x thk. of beam or 450mm.
b) Horizontal steel shall be 0.20% for Fe415, the bar
diameter shall not be more than 16mm & spacing
not more than 3x thk. of beam or 450mm.
c) Necessary side reinforcement should also be
provided.
Detailing of tension steel:
 In deep beams, the tension steel is placed in a zone
of depth equal to (0.25D-0.05L) adjacent to the face
of beam.
 No curtailment of the bars. It should be bent
upwards at the ends to obtain adequate anchorage
& embedment .
1. Determine whether the given beam is deep or not.
2. Check its thickness.
3. Design for flexure.
4. Design for minimum web steel & its distribution in the
beam.
5. Design for shear.
6. Check for bearing pressure at support & point loading
for local failures.
7. Detailing (BRITISH PRACTICE)

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Deep beams

  • 1.
  • 2.  span/depth ratio of simply supported beam is < 2, continuous beam < 2.5, it is classified as deep beam.  Such structures are found in transfer girders and in shear wall structures that resist lateral forces in buildings. It is also found in some of the industrial buildings.
  • 3.
  • 4.  The traditional principles of stress analysis are neither suitable nor adequate to determine the strength of reinforced concrete deep beams.  In deep beams, the bending stress distribution across any transverse section deviates appreciably from straight line distribution assumed in the elementary beam theory.
  • 5.  The behaviour of a deep beam depends also on how they are loaded & special considerations should be given to this aspect in design.  Here cracking will occur at one-third to one-half of the ultimate load.  In the single span beam supporting a concentrated load at mid span, the compressive stresses act roughly parallel to the lines joining the
  • 6. load and the supports and the tensile stresses act parallel to the bottom of the beam.  The flexural stresses at the bottom is constant over much of the span.  The figure shows the crack pattern and the truss analogy of the same.
  • 7.
  • 8. Each of the three tension AB, CD and EF ties have cracked and at failure shaded region would crush or the anchorage zones at E and F would fail. simplified truss model
  • 9.  A single span beam supporting a uniform load acting on the top has the stress trajectories , crack pattern and simplified truss as shown.
  • 10.
  • 11.  A single span beam supporting a uniform load acting on the lower face of the beam has the stress trajectories , crack pattern and simplified truss as shown.
  • 12.
  • 13.  The compression trajectories form an arch with the loads hanging from it. The crack pattern shows that the load is transferred upward by reinforcement until it acts on the compression arch, which then transfers the load down to the supports.  The force in the longitudinal tension ties will be constant along the length of the deep beam. This is the reason that the steel must be anchored at the joints over the reaction, failure of which is a major cause of distress
  • 14.
  • 15.  Parameters influencing deep beam behavior are:  Width of support = C  Overall depth of beam = D  Effective span = L  Width / Thickness of beam = t  Type of loading, uniform = w
  • 16. The Min. thickness of deep beams should be based on two considerations: 1. It should be thick enough to prevent buckling with respect to its span & height i.e. where t = thickness of beam. 2. The thickness should be such that the concrete itself should be able to carry a good amount of the shear force that acts in the beam without the assistance of any steel. 50 & 25   t L t D
  • 17.  Z = 0.6L or 0.6D i.e. Z = 0.6L when L/D < 1 & Z = 0.6D when L/D > 1
  • 18.  From those values, Mu = As.fs.Z where fs = 0.87 fy  The greater value of As is taken as tension steel.       L fy Mu L fy Mu A fyD Mu D fy Mu Z f Mu A s s s . 9 . 1 6 . 0 87 . 0 or 9 . 1 6 . 0 87 . 0 .     
  • 20.  Design procedure : t ft av D Vc fck ft D av Co D t ft D av C Vc Vs Vc Vu Du t Vu Tv . ) 35 . 0 ( 72 . 0 becomes equation the Hence, strength. tensile the , 5 . 0 Depth span Shear concrete wt. normal for 0.72 to equal eff. C where . . 35 . 0 1 c) b) T Tv where . Shear Nominal Determine a) 1 1 max                  
  • 21.  When designing for shear, it is assumed that concrete itself should carry at least 65% of the ultimate shear.  This is ensured by choosing a suitable thk. of beam by the following formula : ft av D Vu t ) 35 . 0 ( 72 . 0 65 . 0  
  • 22.  Shear capacities of tension steel & nominal web steel should also be taken into account in resisting the shear. Their shear capacity is calculated as : where, C2 = 225 N/mm2 for Fe415 As = Area of tension steel provided.   n D y As C Vs 1 2 2 Sin 1 . 
  • 23. α = Angle between the bar considered & the critical diagonal crack. y1= Depth from the top of the beam to the point where bar intersects the critical diagonal crack line. n = Number of bars including tension steel cut by the assumed crack line D = Total depth of beam
  • 24.  Vertical steel Av and Horizontal steel Ah  Horizontal steel bars acts as shear reinforcement and also overcome the effects of shrinkage & temperature.  The amounts specified in IS:456-2000 are : a) Vertical steel shall be 0.12% for Fe415, the bar diameter shall not be more than 14mm and spacing not more than 3x thk. of beam or 450mm.
  • 25. b) Horizontal steel shall be 0.20% for Fe415, the bar diameter shall not be more than 16mm & spacing not more than 3x thk. of beam or 450mm. c) Necessary side reinforcement should also be provided.
  • 26. Detailing of tension steel:  In deep beams, the tension steel is placed in a zone of depth equal to (0.25D-0.05L) adjacent to the face of beam.  No curtailment of the bars. It should be bent upwards at the ends to obtain adequate anchorage & embedment .
  • 27. 1. Determine whether the given beam is deep or not. 2. Check its thickness. 3. Design for flexure. 4. Design for minimum web steel & its distribution in the beam. 5. Design for shear. 6. Check for bearing pressure at support & point loading for local failures. 7. Detailing (BRITISH PRACTICE)
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