27
Lecture Note – 6
Moment-Curvature (M-φ) Relation - I
M-φcharacteristics considering IS: 456:
The actual moment-curvature relationship of R.C. prismatic section is obtained from
stress-strain diagram of concrete and steel. Starting from the basic equations, expressions
for the axial force and moment carrying capacity of the section are calculated in nondimensional form.
The equation of the stress strain diagram of the parabolic part for concrete is
2
σ
ε ⎛ε ⎞
= 2 − ⎜ ⎟ …………………………………(1)
σ cu
ε0 ⎝ ε0 ⎠
subject to the following limiting conditions: ( Figs. 1 and 2)
1. at ε = 0 , σ = 0,
2. at ε = ε0 , σ = σcu,
dσ
= 0.
3. at ε = ε0,
dε
The permissible compressive stress in concrete are considered as: σcu = 0.446fck
Under any loading condition, the section undergoes strains and consequent stresses. A
linear strain distribution over the depth of cross-section is assumed for stress distribution
in concrete and steel(Figs.1 and 2), giving rise to two possible cases:
1. 0 ≤ ε4 ≤ ε0 and
2. ε0 ≤ ε 4 ≤ εu
2
⎡
ε
⎛ ε ⎞ ⎤
∴ σ = 0.446 f ck ⎢2
−⎜
⎟ ⎥ For ε < 0.002
⎢⎣ 0.002 ⎝ 0.002 ⎠ ⎥⎦
=0.446 f ck for 0.002 ≤ ε ≤ 0.0035
28
Fe250
Fig. 1 Stress strain diagram for steel (Fe 250)
Fig. 2 Stress strain diagram for concrete by IS code
29
εy =
0.87 f y
Fig 3 Stress strain diagram of steel (Fe415)
For Fe415
For Fe500
ε
σ (MPa)
ε
σ (MPa)
0.00000
0.00144
0.00163
0.00192
0.00241
0.00276
≥ 0.0038
0.0
288.7
306.7
324.8
342.8
351.8
360.9
0.00000
0.00174
0.00195
0.00226
0.00277
0.00312
≥ 0.00417
0.0
347.8
369.6
391.3
413.0
423.9
434.8
Es
+ 0.002
30
Case – I
Case - II
Xc
Xt D
Xc D
Xt D
Xc D
D
D
(i) Cross section
(i) Cross section
ε3 ε 4
ε1
ε2
ε0 ε3 ε 4
ε1 ε 2
0 ≤ ε4 ≤ ε0
(ii) Strain diagram
(ii) Strain diagram
f3
f 2 ≤ f sy
ε0 ≤ ε4 ≤ εu
σ cu
f4
f 2 ≤ f sy
kD
(iii) Stress diagram
k 'D
kD
(iii) Stress diagram
C1 C2
T
(iv) Force diagram
f3
C1 C3
T
(iv) Force diagram
Fig. 3 Stress strain distribution for case I and case II
C2
f4
31
Lecture Note – 7
Moment-Curvature (M-φ) Relation - II
Case – I ( 0 ≤ ε4 ≤ ε0 )
According to Fig. 3,
k=
ε4
ε2 + ε4
=
x
d
Where k is the ratio of the neutral axis depth to the effective depth.
ε1 and ε4 are the extreme fiber strains in cross section.
For the given section, tensile force for the tensile reinforcement,
T = Pt bDf2 = Pt bDEsε2
Where Es ε2 ≤ 0.87 f y
Pt = Percentage of tensile steel
b = Width of the section
D = Overall depth of the section
Es = Modulus of elasticity of steel
f2 = Stress in tensile steel
ε2 = strain in tensile steel
f y = Characteristic strength of steel
Compressive force due to parabolic part of the stress-strain diagram of concrete :
⎡ε
⎛ 250 ⎞ 2 ⎤
C1 = σ cu kbD × 10 3 ⎢ 4 − ⎜
⎟ε 4 ⎥
⎣2 ⎝ 3 ⎠ ⎦
Where σ cu = Ultimate 28 – days cube strength of concrete in compression.
Compressive force due to presence of compression steel is,
= Pc bD Es ε3
C2 = Pc bD f3
Where Es ε3 ≤ 0.87f y
Pc = Percentage of compressive steel
f3 = Stress in compressive steel
ε3 = Strain in compressive steel
Using non-dimensional force parameter one may get,
32
Pf
N C1 + C 2 − T
⎡ε ' ⎛ 1 ⎞ ⎤ P f
=
= k ⎢ 4 − ⎜ ⎟ε ' 24 ⎥ + c 3 − t 2 ………….(2)
N0
σ cu bD
σ cu
⎣ 2 ⎝ 12 ⎠ ⎦ σ cu
Es ε3 ≤ 0.87 f y
Es ε2 ≤ 0.87 f y
Axial thrust
Ultimate axial thrust ( = σ cu bD)
υ=
Where f3
f2
N
N0
=
=
=
=
Similarly, nondimensional moment parameter,
M
M
µ=
=
M 0 σ cu bD 2
=
k2
2
⎡⎛ 2 ⎞
⎛1⎞ 2
⎧1 1
⎫ ⎤ (1 − X t − X c ) Pc f 3
...(3)
⎢⎜ 3 ⎟ ε '4 − ⎜ 8 ⎟ ε '4 + {2(1 − X t )ε '1 − ε '4 } ⎨ 2 − 12 ε '4 ⎬ ⎥ +
σ cu
⎝ ⎠
⎩
⎭⎦
⎣⎝ ⎠
Where ε ' 4 =
ε4 x 10 3
ε '1 = ε1 x 10 3
M = Bending moment
M0 = Ultimate bending moment (σ cu bD2 )
Case II ( ε0 ≤ ε 4 ≤ εu )
According to Fig. 3, k ' =
ε0
ε2 + ε4
=
ε0
k
ε4
For the given section compressive force due to parabolic part of the stress-strain diagram
2
k’ σ cubD
of concrete is: C1 =
3
Compressive force due to compression steel, if present, is: C2 = Pc bD f3
Compressive force due to straight part of the stress-strain diagram: C3 = σ cu(k-k’)bD
Tensile force due to tensile steel: T = Pt bDf2
From the above values of C1, C2 , C3 and T, one may get
P f
Pf
1
N
= k − k '+ c 3 − t 2 …………….……………(4)
ν =
σ cu
σ cu
3
N0
and,
µ
=
M ⎛ 5 ⎞ '2 ⎛ 2 ⎞
⎛1⎞
= ⎜ ⎟ k + ⎜ ⎟ k ' ( (1 − X t ) − k ) + ⎜ ⎟ ( K − K ')( 2 × (1 − X t ) + K '− K ) +
M 0 ⎝ 12 ⎠
⎝ 3⎠
⎝ 2⎠
(1 − X t − X c ) pC f 3
σ cu
………………….……(5)
To satisfy the equations (2), (3), (4) and (5) following conditions are to be fulfilled.
1. Es ε3 ≤ 0.87 f y
33
2. Es ε2 ≤
0.87 f y
Putting different values of ν in the equations (2) and (4) and assigning some specific
values of ε4 (ε4 ≤ 0.0035), corresponding values are calculated. The strain of concrete
being known or assumed, the rotation capacity of the structure is calculated.
Corresponding to this rotation capacity, moment carrying capacity of the structure is
calculated from the equations (3) and (5). For specific values of compression and tensile
steel, the value of ∆ε (ε1+ε4) and µ ( = M / M0 ) are calculated.
Governing equations for M − Φ characteristics considering ACI Code:
Reference: P. Desayi and S. Krishnan, “Equation for the stress – strain curve of
concrete” Journal of American Concrete Institute, Vol. 61, pp. 345-350, 1964.
The equation of the stress-strain curve given by ACI Code is
=
Eε
⎛ε ⎞
1+ ⎜ ⎟
⎝ ε0 ⎠
Where, E =
2σ u
ε0
2
=
⎛ε ⎞
2σ u ⎜⎜ ⎟⎟
⎝ ε 0 ⎠ .......................................... (6)
⎛ε ⎞
1 + ⎜⎜ ⎟⎟
⎝ ε0 ⎠
2
= initial tangent modulus, assumed to be twice the secant modulus at
Stress(N/mm2)
maximum stress σ u , ε0 = strain at maximum stress.
Strain( ×10−3 )
Fig 4 Stress strain diagram for concrete (ACI Code)
The values of stresses and strains are taken as per ACI code.
34
Strain distribution over the depth of section is considered as linear. According to Fig. 4,
ε4
k=
ε2 + ε4
Where
k = Ratio of the natural axis depth to the effective depth.
(ε2 and ε4 are the extreme fiber strains in cross section.)
For the given section, tensile force for the tensile reinforcement,
T = Pt bdf2 = Pt bdEsε2
Where Es ε2 ≤ 0.87 f y
Pt = Percentage of tensile steel
b = Width of the section
D = Overall depth of the section
Es = Modulus of elasticity of steel
f2 = Stress in tensile steel
ε2 = Strain in tensile steel
f y = Characteristic strength of steel
Compressive force due to parabolic part of the stress-strain diagram of concrete :
2
σ u kbd ⎡ ⎛ ε ⎞ ⎤
C1 =
ln ⎢1 + ⎜⎜ ⎟⎟ ⎥
ε0
⎢⎣ ⎝ ε 0 ⎠ ⎥⎦
Where σu = Ultimate 28 – days cube strength of concrete in compression.
Compressive force due to the presence of compression steel, is
C2 = Pc bd f3 = Pcbd Esε3
Where Es ε3 ≤ 0.87f y
Pc = Percentage of compressive steel
f3 = Stress in compressive steel
ε3 = Strain in compressive steel
Using non-dimensional force parameter,
2
ε 0 ⎡ ⎛ ε 4 ⎞ ⎤ Pc f 3 Pt f 2
N
C1 + C 2 − T
−
=
= k ln ⎢1 + ⎜⎜ ⎟⎟ ⎥ +
………….(7)
ν=
ε4 ⎢ ⎝ ε0 ⎠ ⎥ σ u
σu
σ u bd
No
⎣
⎦
Where f3 = Es ε3 ≤ 0.87 f y
f2 = Es ε2 ≤ 0.87 f y
N = Axial thrust
N0 = Ultimate axial thrust ( = σ cu bD)
Similarly, nondimensional moment parameter,
35
µ
=
=
M
M
=
M 0 σ cu bD 2
(1 − X t ) k ln ⎡⎢1 + ⎛ ε 4 ⎞
ε4
⎣⎢
⎤ 2 k 2ε 2 ⎡
⎤ (1 − X c − X t ) Pc Es
2
−1 ⎛ ε 4 ⎞
0
⎢ε 4 − ε 0 tan ⎜
⎜ ⎟ ⎥−
⎟⎥ +
2
ε4 ⎣
σu
⎝ ε 0 ⎠ ⎦⎥
⎝ ε 0 ⎠⎦
……………………………(8)
2
Where,
M = Bending moment
M0 = Ultimate bending moment (σ cu bD2 )
To satisfy the equations (7) and (8) following conditions are to be fulfilled.
1. Es ε3 ≤ 0.87 f y
2. Es ε2 ≤ 0.87 f y
36
Lecture Note – 8
Behavior of RC Member: Flexure
Equivalent Compression Block
Total compression, C= Area of (ABFE-EFD) = 0.446 f ck xu b − 0.446 f ck
Thus, C = 0.364 f ck xu b ≈ 0.36 f ck xu b
Taking moment about EF
0.36 f ck xu b( x − k2 xu ) = −0.446 f ck xu b.
xu
(0.57 xu ) 2
− 0.45 f ck
b
2
12
⇒ k 2 = 0.416 ≈ 0.42
To find depth of NA
T=C ⇒ f st Ast = 0.36 f ck xb
⇒ xu =
f st Ast
0.36 f ck b
For balanced section,
xu 0.87 f st Ast
----------------------(i)
=
d 0.36 f ck bd
0.57
xu b
3
37
Limiting value of Xu /d
fy
ε su =
+ 0.002 ;
1.15 Es
Again,
Thus,
Where Ε s = 2 × 10 5 Mpa
xu
ε cu
0.0035
=
=
d ε cu + ε su 0.0035 + ε su
xu
can be obtained as given below:
d
Type of steel
ε su
Fe250
Fe415
Fe500
0.0031
0.0038
0.0042
xu
d
0.53
0.48
0.46
The compressive force due to concrete will be: C = 0.36 f ck xu d = 0.36 f ck
Where, F = 0.36
xu
d
Type of steel
Fe250
F
0.192
Fe415
0.175
Fe500
0.167
Moment capacity
M u = 0.87 f y Ast ( L. A.)
= 0.87 f y Ast ( d − 0.416 xu )
= 0.87 f y Ast d (1 − 0.416
= 0.87 f y Ast d [1 − 0.416
= 0.87 f y Ast d [1 − 1.005
xu
)
d
0.87 f y Ast
0.36 f ck bd
f y Ast
f ck bd
]
] [From eqn (i)]
xu
bd = Ff ck bd
d
38
= 0.87 f y pbd 2 [1 −
∴
pf y
f ck
]
⎡ pf y ⎤
Mu
f
p
=
0.87
⎢1 −
⎥
y
bd 2
f ck ⎦
⎣
Mu in terms of concrete
M u = 0.36 f ck xu b(d − 0.416 xu )
= 0.36 f ck
xu
d
x
⎛
⎜1 − 0.416 u
d
⎝
⎞ 2
⎟bd
⎠
Thus, M u = Kf ck bd 2 where, K = 0.36
Type of steel
Fe250
Fe415
Fe500
xu
d
xu ⎞
⎛
⎜ 1 − 0.416 d ⎟
⎝
⎠
K
0.15
0.14
0.13
Minimum depth for given M u
M u = Kf ck bd 2
⇒d =
Mu
Kf ck b
For Fe250, K = 0.15, d =
6.66M u
f ck b
For Fe415, K = 0.14, d =
7.1M u
f ck b
Expression for Steel Area for Balanced Singly Reinforced Section
From the equilibrium condition: T = C
∴
Ast 0.36 f ck xu
f
=
⋅ = k / ck
bd 0.87 f y d
fy
⇒ 0.87 f y Ast = 0.36 f ck bx u
39
Now, assuming p =
Ast
f
p
× 100 ; ∴
= k / ck
bd
100
fy
fy
x
0.36 ×100 xu
⋅ = 41.3 u
0.87
d
d
Thus, p
f ck
= 100k / =
Thus the percentage of balanced reinforcement pB will be as shown in the table below:
Steel
x/d
Fe250
Fe415
Fe500
0.53
0.48
0.46
⎛f
⎞
p⎜ y
⎟
f
ck ⎠
⎝
21.97
19.82
18.87
pB (for M20)
1.758
0.955
0.755
Where p B = Balanced percentage of steel
Expression for x /d for a given b, d & Mu
M u = 0.36 f ck
x⎛
x⎞ 2
⎜ 1 − 0.42 ⎟ bd
d⎝
d⎠
2
⎛ x⎞
⎛ x ⎞ 6.68M u
⇒ ⎜ ⎟ − 2.4⎜ ⎟ +
=0
2
⎝d⎠
⎝ d ⎠ f ck bd
⎡
6.68M u ⎤
x
2
⇒ = 1.2 − ⎢(1.2) −
⎥
d
f ck bd 2 ⎦
⎣
Thus,
1
2
x
can be found out from M u , f ck , b & d
d
Doubly Reinforced Beam
This type of beam is necessary if
1. Depth is restricted for architectural point of view
2. Moment is high
3. Moving load
4. Ductlity is required
5. Seismic resistant
6. At support of continuous beam.
The strain in steel can be calculated from the relation of: ε s = 0.0035(1 − d
/
x
)
40
Two cases may arise regarding the stress in reinforcement
1. Strain of compression steel is reached at yield strain.
2. Strain of concrete is reached to yield strain of 0.0035 & strain of compression
steel is below of yield strain.
Analysis & Design of Doubly Reinforced Beam
1. Strain compatibility method using basic equation.
2. Steel Beam Theory.
3. Use of design aids (SP-16)
Strain Compatibility Method
Step1: Choose the value of x. Assume the concrete fails at a compressive strain of 0.0035.
Then draw the strain distribution of the section.
Step2: Calculate strain & corresponding stress in the tensile steel.
T = f st Ast
Step3: Calculate strain & corresponding stress in the compressive steel.
C s = f sc Asc
Step4: Calculate the compressive force in the concrete block C c = 0.36 f ck xb
Step5: Calculate total compressive force as C = C s + C c
Step6: Check if T = C , then assumed neutral axis depth x is OK. Otherwise choose
another suitable x so that T = C
Step7: M u = C c (d − k 2 x ) + C s (d − d / )
41
Lecture Note – 9
Behavior of RC Member: Flexure
Steel Beam Theory
Mu1 =Maximum moment of the concrete beam can carry.
Mu2 =Moment capacity of the steel Beam
The total moment will be: M u = M u1 + M u 2 where M u 2 = f sc Asc ((d − d / )
Singly reinforced
beam reaches
ultimate strain in
t
Steel Beam Theory
Steps for analysis
Step1: Calculate M u for concrete failure as a singly reinforced beam. Thus,
M u1 = 0.149 f ck bd 2 (Fe250)
= 0.138 f ck bd 2 (Fe415)
= 0.133 f ck bd 2 (Fe500)
Step2:
Determine the balanced “p”
⎛ 21.97 f ck
pt = ⎜
⎜
fy
⎝
⎞
⎟ for Fe250
⎟
⎠
42
⎛ 19.82 f ck
=⎜
⎜
fy
⎝
⎞
⎟ for Fe415
⎟
⎠
⎛ 18.87 f ck ⎞
⎟ for Fe500
=⎜
⎟
⎜
f
y
⎠
⎝
p bd
Ast1 = t
100
Step3: Determine f sc , Then C s = f sc Asc
Step4: Find additional moment, M u 2 = f sc Asc ( d − d / )
Step5: Find total moment from compression failure M u = M u1 + M u2 = Muc
Step6: Find Ast 2 by tension failure; Ast 2 = Ast − Ast1
(
)
Step7: Find the total moment by tension failure, M u = M u1 + Ast 2 (0.87 f y ) d − d / = M ut
Step8: M u = lesser value of M uc & M ut
For given value of Mu, b, d and grade of concrete and steel, find out Asc & Ast
Step-1. Find M ul = kf ck bd 2 for concrete failure.
Step-2. Calculate Ast1 considering balanced reinforced section.
(
Step-3. Find M u 2 = M u − M u1 = 0.87 f y Ast 2 d − d /
Step-4. Calculate Ast 2 =
M u2
0.87 f y d − d /
(
)
)
∴ Ast = Ast1 + Ast 2
Step-5. Find f sc which depends on d ' . Find the corresponding stresses from table F of
d
SP-16.
⎛ d/ ⎞
Otherwise ε sc = 0.0035 ⎜1 − ⎟ . The value of fsc then can be
⎝ xu ⎠
found from the stress-strain diagram of steel.
43
Step-6. Find Asc =
Ast 2 (0.87 f y )
f sc
Thus Ast & Asc can be found out
Design Aids (SP-16)
• Tables are easier than chart
• Charts 19 &20 give Ast2 value for (d-d/) and Fe250 only.
• For other grade of steel, it is to be modified by table G on page 13 of SP16
• Table (45-56) give values for direct design of doubly reinforced beams.
A
p c = sc × 100
bd
A
pt = st × 100 where , Ast = Ast1 + Ast 2
bd
p t = p t1 + p t 2
p c & p t are obtained from the following expression.
We know,
M u = M u (lim) +
Or,
pt 2
bd (0.87 f y )(d − d / )
100
/
M u M u (lim) pt 2
=
+
(0.87 f y )(1 − d )
2
2
d
100
bd
bd
Now,
p t = p t1(lim) + p t 2
and
⎛ 0.87 f y
p c = pt 2 ⎜⎜
⎝ f sc
⎞
⎟⎟
⎠
Maximum & minimum Tension steel in Beam:
Maximum steel
M u = 0.87 f y
fy p ⎞ 2
p ⎛
⎜1 −
⎟ bd
f ck 100 ⎠
100 ⎝
Here M u increases with p in a parabolic relation.
For M20 & Fe415, M u = 3.6 p − 0.75 p 2
44
⎛x⎞
From T = C we can have: 0.36 f ck b ⎜ ⎟ d = f st Ast
⎝d ⎠
⇒
f st
x
p
=
d 0.36 f ck 100
x
= 0.58 p
d
Ductility can be measured by curvature.
For Fe415 & M20: ⇒
εc
x
ϕ
dθ
εs
Thus,
1 Strain in compression fibre 0.0035
=
=
R
Depth of NA
x
1
dθ ε c 0.0035 0.006
=φ =
=
=
=
R
dx
x 0.58 pd
pd
Thus, Ductility ∝
1
p
Hence, though M u increases with the increase of p but ductility (i.e., curvature)
decreases.
Hence, IS code (Clause 26.5.1) has put the upper limit of tension & compression steel as
4%.
Minimum Steel
Minimum steel is required to take care ductility & shrinkage of concrete.
As per clause 26.5.1
45
As 0.85
=
bd
fy
= 0.00205 = 0.2% for Fe415
= 0.0035 = 0.35% for Fe250
Necessity of minimum steel for shear
Clause:26.5.1.6
Minimum steel is necessary to
1. Prevent brittle shear failure, which can occur without shear steel.
2. Guard against any sudden failure of a beam if concrete cover bursts and the bond to the
tension is lost.
3. Prevent failure that can be caused by tension due to shrinkage and thermal stresses and
internal cracking in the beam.
4. Hold the reinforcements in place while pouring the concrete.
5. Act as the necessary ties for the compression steel and make them effective.
Minimum spacing
Asv
0.4
=
bSv 0.87 f y
⎛ AS ⎞
Sv = 2.175 f y ⎜ v ⎟
⎝ b ⎠
⎛ AS
= 902 ⎜ v
⎝ b
⎞
⎟ for Fe415
⎠
⎛ AS
= 544 ⎜ v
⎝ b
⎞
⎟ for Fe250
⎠
Maximum spacing
Sv ⊄ 0.75d for vertical stirrups
⊄ d for inclined ( 450 )
⊄ 300 in any case
Asv= total area of stirrups
Sv=Spacing
b=breadth of web at level of tension
46
Lecture Note – 10
Behavior of RC Beam: Shear
Types of shear Failure
Shear strength of RC Beam (Without web Reinforcement)
Total resistance = vcz + vay + vd
47
vcz → Shear in compression zone
vay → Aggregate interlock forces
vd → Dowel action from longitudinal bars
1.
2.
3.
4.
5.
Tensile strength of concrete Æ Affect inclined cracking load
Longitudinal reinforcement (p) ÆRestrain cracks
Increase in the depth of beamÆ Reduced shear stress at inclined cracking
Axial tension ÆDecrease inclined cracking load
Axial compressionÆ Increase inclined cracking load
Function & strength of web reinforcement
Function of web Reinforcement
• Web reinforcement is provided to ensure that the full flexural capacity is
developed.
• Acts as ‘clamps’ to keep shear cracks from widening.
• Shear resisted by v s apart from vcz , v ay & v d .
•
v s increases as cracks widen until yielding of stirrups & then stirrups provide
constant resistant.
Flexural Cracking:
Shear is resisted by Vcz , Vay , Vd & V s
Designing to Resist Shear
Shear Strength (ACI 318 Sec 11.1)
φVn ≥ Vu
[i.e., Capacity ≥ demand]
48
Vu Æfactored shear force at section
Vn ÆNormal shear strength
φ Æ0.85(Shear)-strength reduction factor
V n = Vc + V s
Vc ÆNormal shear resistance provided by concrete
Vs Æ Normal shear provided by the shear reinforcement
Shear Strength Provided by Concrete
Bending only
Vc = 2 f c bw d Eqn [11.3]
Simple formula:
≤ 3.5 f c bw d
More detailed:
Vc
⎛
⎛ V d ⎞⎞
= ⎜ 1.9 f c + 2500 pw ⎜ u ⎟ ⎟ bw d
⎜
⎟
⎝ Mu ⎠⎠
⎝
≤ 3.5 f c bw d Eqn [11.6]
Bending and Axial Compression
Simple formula
⎛
N u ⎞⎟
Vc = 2 ⎜1 +
f b d Eqn [11.4]
⎜ 2000 A ⎟ c w
g
⎠
⎝
≤ 3.5 f c bw d 1 +
Nu
500 Ag
N u is positive for compression and
Nu
Eqn [11.8]
Ag
are in psi.
More detailed
⎛ 4h − d ⎞
M m = M u − Nu ⎜
⎟ Use M m in Eqn [11.6] with no limits
⎝ 8 ⎠
Vc ≤ 3.5 f c bw d 1 +
Nu
Eqn [11.8]
500 Ag
N u is positive for compression and N u
Ag
are in psi.
⎛V d ⎞
Note ⎜⎜ u ⎟⎟ ≤ 1
⎝ Mu ⎠
49
Bending and Axial Tension
Simple formula
Vc = 0 Design shear reinforcement for all shear
⎛
Nu ⎞
f b d
= 2 ⎜1 +
⎜ 500Ag ⎟⎟ c w
⎝
⎠
N u is negative for tension and N u
are in psi.
Ag
Vc
Lightweight Concrete:
Shear Strength Provided by Shear Reinforcement
Minimum Shear Reinforcement: (11.5.5)
1
Required when Vu ≥ φ Vc
2
Except:
(a)Slabs & Footings
(b)Concrete Joist Construction (defined 8.11)
⎧ 10"
⎪
(c )Beams with h ≤ larger of ⎨ 2.5 t f
⎪1/2 b
w
⎩
50
Typical Shear Reinforcement
Stirrup - perpendicular to axis of
Members (minimum labor - more material)
Av f y d (sin α + cos α )
Vs =
ACI eqn 11-15
s
A f d
α = 90o ⇒ Vs = v y
s
Bent Bars (more labor - minimum material) see
required in 11.5.6
Vs =
Av f y d (sin α + cos α )
s
α = 45o ⇒ Vs =
1.41Av f y d
s
ACI 11-5.6
51
Design Procedure for Shear
1. Calculate Vu
2. Calculate φVc Eqn 11-3 or 11-5 (no axial force)
3. Check
Vu ≥
4. If
⎧ If yes, add web reinforcement (go to 4)
1
φVc → ⎨
2
⎩ If no, done
1
φVc ≤ Vu ≤ φVc → Provide minimum shear reinforcement
2
Av (min ) = 50
bw s
fy
A f
⎞
⎛
⎜⎜ or smax = v ys for min Av ⎟⎟
50bw
⎠
⎝
Also done
d
smax ≤ ≤ 24" (11.5.4)
2
5. If Vu ≥ φVc , calculate Vs (required)
Vu ≤ φVn = φVc + φVs
⇒ φVs = Vu − φVc ⇒ Vs =
Vu
φ
− Vc
Check, Vs ≤ 8 f c′ bw d (otherwise illegal) (11.5.4)
6. Solve for required stirrup spacing (strength) Assume # 3, #4, or #5 stirrups
Av f ys d
from 11-15
s≤
Vs
7. Check minimum steel requirement (eqn 11-13) smax =
Av f ys
50bw
8. Check maximum spacing requirement (ACI 11.5.4)
d
If Vs ≤ 4 f c′ bw d → smax ≤ ≤ 24"
2
d
If Vs ≥ 4 f c′ bw d → smax ≤ ≤ 12 " Note: Vs ≥ 8 f c′ bw d
4
9. Use smallest spacing from steps 6,7,8
Note: A practical limit to minimum stirrup spacing is 4 inches.
(Illegal)
52
Location of Maximum Shear for Beam Design
Non-pre-stressed members:
Sections located less than a distance d from face of support may be designed for same
shear, Vu, as the computed at a distance d.
Compression fan carries load directly into support.
Compression fan carries load into support
Location of Maximum Shear for Beam Design
When:
1. The support reaction introduces compression into the end regions of the member.
2. No concentrated load occurs within d from face of support.
Location of Maximum Shear for Beam Design
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