http://syaifulsipil96.blogspot.com/
[email protected]
CHAPTER
10
THE FLEXURAL AND SHEAR DESIGN OF DEEP BEAM
10.1 INTRODUCTION Deep beam is a beam having large depth/thickness ratio and shear span depth ratio less than 2.5 for concentrated load and less than 5.0 for distributed load. Because the geometry of deep beams, they behavior is different with slender beam or intermediate beam.
The structural element that might be classified as deep beam are :
Transfer Girder, is a girder that carry all the vertical load without any vertical element below the girder.
Pile Cap, is a structural element that connect the vertical element with the deep foundation such as bored pile.
Vertical Wall, wall slab under vertical load can be designed as deep beam.
10.2 BEHAVIOR OF DEEP BEAM The followings are the major different of deep beam element compared wth ordinary beam based on the design assumption, as follows :
Two-Dimensional Action, because of the dimension of deep beam they behave as twodimensional action rather than one-dimensional action.
Plane Section Do Not Remain Plane, the assumption of plane section remain plane cannot be used in the deep beam design. The strain distribution is not longer linear.
Shear Deformation, the shear deformation cannot be neglected as in the ordinary beam. The stress distribution is not linear even in the elastic stage. At the ultimate limit state the shape of concrete compressive stress block is not parabolic shape again.
The followings are the major behavior of deep beam element, as follows :
Cracking of deep beam will occur at
The distribution of tensile stress at bottom fiber is constant over the span. In other word the
1 f' 3 c
or
1
2 f 'c
value of tensile stress at bottom fiber at support and at mid span is only little different, for this reason in deep beam the tension reinforcement must be extend to the end of support although that region is small bending moment region (in ordinary beam we can cut off the tension reinforcement and not all of the tension reinforcement in mid span is extended to the end of support, practically only two for anchor the stirrups.
The maximum tensile stress at the bottom fiber is far exceed the magnitude of compressive stress.
10 - 1
http://syaifulsipil96.blogspot.com/
[email protected]
MID SPAN SUPPORT
FIGURE 10.1
SUPPORT STRESS
STRESS DISTRIBUTION OF DEEP BEAM
The cracks is vertical follows the direction of compression trajectory, in deep beam we must provide both vertical stirrups and horizontal stirrups.
COMPRESSIVE ARC ACTION
FIGURE 10.2
CRACKS OF DEEP BEAM
10.3 FLEXURAL DESIGN OF DEEP BEAM 10.3.1
GENERAL
The flexural design for deep beam is not described in the ACI code, the method explained in this section is from Euro – International Concrete Committee (CEB).
10 - 2
http://syaifulsipil96.blogspot.com/
10.3.2
[email protected]
CEB DESIGN OF DEEP BEAM
The flexural design procedure is for simply supported beams and for continuous beams. TABLE 10.1
FLEXURAL DESIGN OF DEEP BEAM
TYPE
SIMPLY SUPPORTED BEAMS
CONTINUOUS BEAMS
Moment
Mn = A sfy (jd)
Mn = A sfy (jd)
Strength Positive
A s+ =
Reinforcement Negative
–
Reinforcement As Minimum
Mu + φfy ( jd)
A s min =
f 'c 4 fy
bw d ≥
1 .4 bw d fy
jd = 0.2(L + 2h) ⇒ 1 ≤
Lever Arm
jd = 0.6L ⇒
L <2 h
A s+ =
Mu + φfy ( jd)
A s− =
Mu − φfy ( jd)
A s min =
f 'c 1 .4 bw d ≥ bw d fy 4 fy
jd = 0.2(L + 1.5h) ⇒ 1 ≤
L <1 h
jd = 0.5L ⇒
L ≤ 2 .5 h
L <1 h
Positive Reinforcement
y = 0.25h − 0.05L < 0.20h
y = 0.25h − 0.05L < 0.20h
–
⎛L ⎞ A s1− = 0.5⎜ − 1⎟ A s ⎝h ⎠
Distribution Negative Reinforcement
A s2 − = A s − A s1
Distribution
where : jd
= lever arm
Mu
+
= positive ultimate flexure moment
Mu
-
= negative ultimate flexure moment
+
= positive reinforcements area
As
-
= negative reinforcements area
h
= beam depth
f’c
= concrete cylinder strength
(MPa)
fy
= yield strength of reinforcements
(MPa)
As
Where L is taken the minimum of effective span measured center to center of supports or 1.15 Ln. In simply supported beams, the positive tension reinforcement is distributed in the lower of beam section along the distance : y = 0.25h − 0.05L < 0.20h
[10.1]
10 - 3
[email protected]
h
http://syaifulsipil96.blogspot.com/
y
POSITIVE REINFORCEMENTS
Ln L FIGURE 9.9
DISTRIBUTION OF POSITIVE REINFORCEMENTS IN SIMPLY SUPPORTED BEAMS
In continuous beams the distribution of positive reinforcements is similar as in the simply supported
0.2h
beam, the difference is the distribution of negative reinforcements.
As1
h3
0.6h h
As2
FIGURE 9.10
DISTRIBUTION OF NEGATIVE REINFORCEMENTS IN CONTINUOUS BEAMS
As1 is distributed along height h1=0.2h and As2 is distributed along h2=0.6h.
Reinforcements in zone h3 are come from the tension reinforcements that continued from the mid span to the support section.
10.3.3
STEP – BY – STEP PROCEDURE
The followings are the step – by – step procedure used in the flexural design for deep beam, as follows :
Classified the structure as simply supported beam or continuous beam.
Calculate the approximate lever arm jd.
TYPE
Lever Arm
10 - 4
SIMPLY SUPPORTED BEAMS
jd = 0.2(L + 2h) ⇒ 1 ≤ jd = 0.6L ⇒
L <2 h
L <1 h
CONTINUOUS BEAMS
jd = 0.2(L + 1.5h) ⇒ 1 ≤ jd = 0.5L ⇒
L ≤ 2 .5 h
L <1 h
http://syaifulsipil96.blogspot.com/
[email protected]
Calculate the required positive or negative reinforcement As+, As-.
TYPE
SIMPLY SUPPORTED BEAMS
Positive
A s+ =
Reinforcement Negative
A s+ =
Mu+ φfy ( jd)
A s− =
Mu− φfy (jd)
Check the required steel bars area with minimum steel bars area Asmin.
TYPE
SIMPLY SUPPORTED BEAMS
As Minimum
Mu + φfy ( jd) –
Reinforcement
CONTINUOUS BEAMS
A s min =
f 'c 4 fy
bw d ≥
1. 4 bw d fy
CONTINUOUS BEAMS
A s min =
f 'c 1 .4 bw d ≥ bw d fy 4 fy
Choose the number of bars and the reinforcement is distributed as follows : TYPE
SIMPLY SUPPORTED BEAMS
CONTINUOUS BEAMS
Positive
y = 0.25h − 0.05L < 0.20h
y = 0.25h − 0.05L < 0.20h
–
⎛L ⎞ A s1− = 0.5⎜ − 1⎟ A s ⎝h ⎠
Reinforcement
Negative Reinforcement
A s2 − = A s − A s1
10.4 SHEAR DESIGN OF DEEP BEAM 10.4.1
GENERAL
The shear design of deep beam is similar as shear design of ordinary beam, the difference is only at the concrete shear strength, limitation of ultimate shear force and horizontal and vertical stirrups distribution.
10.4.2
BASIC DESIGN EQUATION
According to ACI code the design of deep beam due to shear force must follows the following condition : φVn ≥ Vu
[10.2]
where : Vn
= nominal shear strength
φVn
= design shear strength
φ
= strength reduction factor (0.85)
Vu
= ultimate shear force, factored shear force
10 - 5
http://syaifulsipil96.blogspot.com/
[email protected]
As shear design of ordinary beam, the shear force is resisted by the concrete component and by the shear reinforcement component, as follows :
Vn = Vc + Vs
[10.3]
where : Vn
= nominal shear strength
Vc
= concrete shear strength without shear reinforcement
Vs
= shear reinforcement (stirrup) shear strength
10.4.3
CONCRETE SHEAR STRENGTH
The concrete shear strength of deep beam is taken as : ⎛ M ⎞ 1⎛ V d⎞ Vc = ⎜⎜ 3.5 − 2.5 u ⎟⎟ ⎜⎜ f 'c + 120ρ w u ⎟⎟b w d V d 7 Mu ⎠ u ⎠ ⎝ ⎝
[10.4]
⎛ M ⎞ 1.0 ≤ ⎜⎜ 3.5 − 2.5 u ⎟⎟ ≤ 2.50 V ud ⎠ ⎝
where : Vc
= concrete shear strength
(N)
Mu
= ultimate flexure moment
(Nmm)
Vu
= ultimate shear force
(N)
f’c
= concrete cylinder strength
(MPa)
d
= effective depth
bw
= width of beam web
ρw
= longitudinal reinforcement ratio
Or the concrete shear strength can be determined as :
Vc =
1 f 'c b w d 6
[10.5]
The maximum limit of concrete shear strength is :
Vc −max =
1 f 'c b w d 2
[10.6]
The section must be enlarged if the ultimate shear force is not follows the condition below :
⎛2 ⎞ Vu ≤ φ⎜ f 'c b w d ⎟ ⎝3 ⎠ Ln for < 2. 0 d Or
10 - 6
[10.7]
http://syaifulsipil96.blogspot.com/
[email protected]
⎧1 ⎛ ⎫ L ⎞ Vu ≤ φ⎨ ⎜10 + n ⎟ f 'c b w d⎬ 18 d ⎠ ⎩ ⎝ ⎭ for 2.0 ≤
10.4.4
[10.8]
Ln ≤ 5. 0 d
STIRRUP SHEAR STRENGTH
The shear reinforcements must be provided in the deep beams follows the condition below : Vu ≤ φVc
[10.9]
The strength of horizontal and vertical shear reinforcements is :
⎧⎛ ⎛L ⎞⎞ ⎛ ⎛ L ⎞ ⎞⎫ 1+ ⎜ n ⎟ ⎟ ⎜ 11 − ⎜ n ⎟ ⎟⎪ ⎪⎜ A d ⎪⎜ A v ⎟ ⎜ ⎝ ⎠ + ⎝ d ⎠ ⎟ ⎪f d vh Vs = ⎨⎜ ⎟ ⎜ s ⎟⎬ y s 12 12 h ⎪⎜ v ⎟ ⎜ ⎟⎟⎪ ⎟ ⎜ ⎪⎜⎝ ⎠ ⎝ ⎠⎪⎭ ⎩
[10.10]
where : Vs
= horizontal and vertical stirrups shear strength
Av
= area of vertical stirrups
sv
= spacing of vertical stirrups
Ln
= clear distance of beam
d
= effective depth
Avh
= area of horizontal stirrups
sv
= spacing of horizontal stirrups
fy
= yield strength of stirrups
10.4.5
(N)
LIMITS OF SHEAR REINFORCEMENT
The minimum shear reinforcement area is : A v −min = 0.0015(bs v )
[10.11]
A vh −min = 0.0025(bsh )
where : Av-min
= minimum vertical stirrups
Avh-min
= minimum horizontal stirrups
b
= width of beam
sv
= spacing of vertical stirrups
sh
= spacing of horizontal stirrups
The maximum spacing of shear reinforcement is :
10 - 7
http://syaifulsipil96.blogspot.com/
TABLE 10.2
[email protected]
MAXIMUM SPACING OF SHEAR REINFORCEMENT
VERTICAL STIRRUPS
sv ≤
HORIZONTAL STIRRUPS
d 5
sh ≤
sv ≤ 500mm
10.4.6
d 3
sh ≤ 500mm
CRITICAL SECTION IN DEEP BEAM
The critical section to determines the ultimate shear force in the deep beam is : TABLE 10.3
CRITICAL SECTION OF DEEP BEAM DUE TO SHEAR
UNIFORM LOAD
CONCENTRATED LOAD
x = 0.15(Ln )
x = 0.50(a )
10.4.7
STEP – BY – STEP PROCEDURE
The followings are the step – by – step procedure used in the shear design for deep beam, as follows :
Determine the critical section to calculate the ultimate shear force Vu. UNIFORM LOAD
CONCENTRATED LOAD
x = 0.15(Ln )
x = 0.50(a )
Check the ultimate shear force, enlarge the section if the condition is not achieved. ⎛2 ⎞ Vu ≤ φ⎜ f 'c b w d ⎟ ⎝3 ⎠ ⎧1 ⎛ ⎫ L ⎞ Vu ≤ φ⎨ ⎜10 + n ⎟ f 'c b w d⎬ 18 d ⎠ ⎩ ⎝ ⎭
for
Ln < 2 .0 d
for 2.0 ≤
Ln ≤ 5. 0 d
Calculate the concrete shear strength Vc ⎛ M ⎞ 1⎛ V d⎞ Vc = ⎜⎜ 3.5 − 2.5 u ⎟⎟ ⎜⎜ f 'c + 120ρ w u ⎟⎟b w d Vud ⎠ 7 ⎝ Mu ⎠ ⎝ ⎛ M ⎞ 1.0 ≤ ⎜⎜ 3.5 − 2.5 u ⎟⎟ ≤ 2.50 Vud ⎠ ⎝
If Vu < 0.5φVc then no shear reinforcements needed, but for practical reason provide minimum shear reinforcement. A vh −min = 0.0025(bsh ) A v −min = 0.0015(bsv )
10 - 8
If Vu > φVc then provide the shear reinforcements.
http://syaifulsipil96.blogspot.com/
[email protected]
Calculate the ultimate shear force carried by the stirrups Vs. Vs =
Vu − Vc φ
Choose the vertical and horizontal stirrups until the condition achieved. ⎧⎛ ⎛L ⎞⎞ ⎛ ⎛ L ⎞ ⎞⎫ 1+ ⎜ n ⎟ ⎟ ⎜ 11 − ⎜ n ⎟ ⎟⎪ ⎪⎜ ⎪⎜ A ⎝ d ⎠ ⎟ + ⎜ A vh ⎝ d ⎠ ⎟ ⎪f d Vs = ⎨⎜ v ⎟ ⎜ s ⎟⎬ y s 12 12 v h ⎪⎜ ⎟ ⎜ ⎟⎟⎪ ⎟ ⎜ ⎪⎜⎝ ⎠ ⎝ ⎠⎪⎭ ⎩
Check the spacing of shear reinforcement sv and sh.
VERTICAL STIRRUPS
sv ≤
d 5
sv ≤ 500mm
HORIZONTAL STIRRUPS
sh ≤
d 3
sh ≤ 500mm
If necessary check the chosen shear reinforcements for the basic design equation for shear design.
Vn = Vc + Vs ⎧⎛ ⎛L ⎞⎞ ⎛ ⎛ L ⎞ ⎞⎫ 1+ ⎜ n ⎟ ⎟ ⎜ 11 − ⎜ n ⎟ ⎟⎪ ⎪⎜ A d ⎪⎜ A v ⎟ ⎜ ⎝ ⎠ + ⎝ d ⎠ ⎟ ⎪f d vh Vs = ⎨⎜ ⎟ ⎜ s ⎟⎬ y s 12 12 h ⎪⎜ v ⎟ ⎜ ⎟⎟⎪ ⎟ ⎜ ⎪⎜⎝ ⎠ ⎝ ⎠⎪⎭ ⎩
The design procedure above is repeats until the basic design equation for shear design is achieved.
10.5 APPLICATIONS APPLICATION 01 – FLEXURAL DESIGN OF SIMPLY SUPPORTED DEEP BEAM
275
10.5.1
470
50
500
PROBLEM
Design the flexural reinforcement of simply supported deep beam above.
10 - 9
http://syaifulsipil96.blogspot.com/
[email protected]
MATERIAL
Concrete strength
= K – 300
Steel grade
= Grade 400
Concrete cylinder strength
= f 'c = 0.83 × 30 = 24.9 MPa β1 = 0.85
DIMENSION
b
= 500
mm
h
= 2750 mm
Concrete cover
= 50
d
= 2700 mm
mm
DESIGN FORCE
⎛1 ⎞ ⎛1 ⎞ Mu = 1.4⎜ qL2 ⎟ = 1.4⎜ × 6000 × 52 ⎟ = 26250 kgm ⎝8 ⎠ ⎝8 ⎠
DEEP BEAM CHECKING Ln 4700 = = 1.74 d 2700
1.0 ≤ 1.74 ≤ 5.0
Î Deep Beam Action
LEVER ARM jd = 0.2(L + 2h) = 0.2(5000 + 2(2750 )) = 2100 mm
POSITIVE REINFORCEMENT
Mu = 262500000 Nmm A s+ =
Mu+ 262500000 2 = = 348 mm φfy ( jd) 0.9 × 400(2100 )
A s min =
A s min =
f 'c 4 fy
bw d =
24.9 2 500 × 2700 = 4211 mm 4 × 400
1 .4 1 .4 2 bw d = 500 × 2700 = 4725 mm fy 400
A s = 4725 mm2 ⎛1 ⎞ ⎛1 ⎞ Use 10D25, A s = 10⎜ πD2 ⎟ = 10⎜ π × 252 ⎟ = 4906 mm2 ⎝4 ⎠ ⎝4 ⎠
10 - 10
http://syaifulsipil96.blogspot.com/
[email protected]
43.75
275
DISTRIBUTION OF POSITIVE REINFORCEMENT
5D25
50 y = 0.25h − 0.05L = 0.25(2750 ) − 0.05(5000 ) = 437.5 < 0.20h = 0.20(2750 ) = 550
y = 437.5 mm
The longitudinal positive reinforcement must be distributed at the lower base of the beam with a distance 437.5 mm from the bottom fiber. We place 5D25 at each face of the section.
APPLICATION 02 – SHEAR DESIGN OF SIMPLY SUPPORTED DEEP BEAM
275
10.5.2
50
470
15000
DESIGN SHEAR FORCE DIAGRAM
10770
d=70.5
SHEAR FORCE DIAGRAM
10770
15000
500
PROBLEM
Design the web reinforcement of simply supported deep beam above.
MATERIAL
Concrete strength
= K – 300
Steel grade
= Grade 240
Concrete cylinder strength
= f 'c = 0.83 × 30 = 24.9 MPa β1 = 0.85
10 - 11
http://syaifulsipil96.blogspot.com/
[email protected]
DIMENSION
b
= 500
mm
h
= 2750 mm
Concrete cover
= 50
d
= 2700 mm
mm
DESIGN FORCE
x = 0.15Ln = 0.15 × 4700 = 705 mm Vu = 1.4(10770 ) = 15078 kg Vu = 150780 N LIMITATION CHECKING
⎛2 ⎞ ⎛2 ⎞ Vu = 150780 ≤ φ⎜ f 'c b w d ⎟ = 0.85⎜ 24.9 × 500 × 2700 ⎟ = 3817342 N ⎝3 ⎠ ⎝3 ⎠ The section is not enlarged.
CONCRETE SHEAR STRENGTH
⎛ 262500000 ⎞ M ⎞ ⎛ ⎜ 3 .5 − 2 .5 u ⎟ = ⎜ 3 .5 − 2 .5 ⎟ = 1.88 ⎜ ⎟ × 2700 ⎠ 150780 V d u ⎠ ⎝ ⎝ ρw =
As 4906 = = 0.0036 b w d 500 × 2700
⎛ V d⎞ M ⎞ 1⎛ Vc = ⎜⎜ 3.5 − 2.5 u ⎟⎟ ⎜⎜ f 'c + 120ρ w u ⎟⎟b w d Mu ⎠ V d 7 u ⎠ ⎝ ⎝ ⎧1 ⎛ ⎫ 150780 × 2700 ⎞ Vc = 1.88⎨ ⎜ 24.9 + (120 × 0.0036 ) ⎟500 × 2700 ⎬ = 2052143N 262500000 ⎠ ⎩7 ⎝ ⎭ φVc = 0.85 × 2052143 = 1744322 N 0.5φVc = 0.5 × 1744322 = 872161 N
DESIGN OF STIRRUPS
Vu = 150780 < 0.5φVc = 872161
Î Provide minimum web reinforcement
For horizontal and vertical stirrups we choose 2 legs φ10. ⎛1 ⎞ ⎛1 ⎞ A v = 2⎜ πφ2 ⎟ = 2⎜ π × 102 ⎟ = 157 mm2 ⎝4 ⎠ ⎝4 ⎠
HORIZONTAL STIRRUPS
10 - 12
VERTICAL STIRRUPS
sh
Avh-min
sv
Av-min
125
A vh = 0.0025(bsh )
200
A v = 0.0015(bsv )
A vh = 0.0025(500 × 125 )
A v = 0.0015(500 × 200 )
A vh = 156
A v = 150
http://syaifulsipil96.blogspot.com/
[email protected]
Ø10-200 Ø10-125
10 - 13
