BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
HISTORICAL BACKGROUND
CONCRETE
a mixture of sand, gravel, crushed rock or other
aggregates held together in a rock-like mass with a
paste of cement and water.
ADMIXTURES
materials added to concrete to change certain
characteristics such as workability, durability and time
of hardening.
WORKING STRESS DESIGN (WSD)
1. Design - given the load, determine the size
an English bricklayer who obtained a patent for
Portland cement
JOSEPH MONIER
a Frenchman who invented reinforced concrete a
received a patent for the const. of concrete basins and
tubs and reservoirs reinforced w/ wire mesh and iron
wire in 1867.
DESIGN METHODS:
1. WSD - Working Stress Design, Alternate
Stress Design,or Straight-Line Design
Ultimate Stress Design or Strength Design
2. Investigation - given the size, determine the load
b
- 0.45 fc' ( beams/slabs/footings)
- 0.25 fc' ( columns)
fc' - specified compressive strength of conc.
at 28 days curing (MPa)
gconc.-
unit weight of concrete
- 23.54 KN/m 3
Ec - modulus of elasticity of concrete
- 4700
fc' (MPa)
STEEL :
fs - allowable tensile stress of steel (MPa)
fs - 0.50 fy ( beams/slabs/footings)
fs - 0.40 fy ( columns)
fy - yield stress of steel (MPa)
gsteel - unit weight of steel
- 77 KN/m3
Ec - modulus of elasticity of concrete
fc =
fc = 0.45 fc'
C
kd
h
1. Crushing of Concrete - when the strain concrete
d-kd
reaches the ultimate strain of 0.003 mm/mm.
steel "fs" reaches the yield stress "fy"
3. Simultaneous crushing of concrete and
Yielding of Steel
As
z
Beam Section
nAs
Transformed
Section
fc/n
T
1. Overreinforced - when failure is due to crushing of
concrete.
2. Underreinforced - when failure is initiated by yielding
of steel.
3. Balanced Design - when failure is caused by
simultaneous crushing of concrete and
yielding of steel
where:
Mc - resisting moment
of concrete
Stress of Steel
fs = Ms (d - kd)
n
IN.A.
where:
Ms - resisting moment
of steel
Stress Diagram
where:
TYPES OF DESIGN
Mc (kd)
IN.A.
N.A.
Compressive force of Concrete
C = 1/2 fc kd b
h = overall depth of the beam (mm)
z = steel covering (measure from the centroid of bar)
d = effective depth of the beam (mm)
d = h -z
As = area of the reinforcement ( square millimeters)
fc' = compressive strength of concrete (MPa)
fs = tensile strength of steel (MPa)
b = base of the beam (mm)
n = modular ratio(always a whole number)
n = Es /Ec
Location of the neutral axis (kd)
Tensile force of Steel
T = As fs
Moment Arm ( jd )
d = jd + kd/3
j = 1 - k/3
S MN.A. = 0
FACTORED LOAD COMBINATION
fc - allowable compressive stress of conc.
b
MODES OF FAILURE IN BENDING
PROPERTIES OF MATERIALS:
CONCRETE:
Stress of Concrete
DESIGN OF BEAMS FOR FLEXURE
2. Yielding of Steel - when the actual tensile stress of
JOSEPH ASPDEN
2. USD -
TYPES OF PROBLEMS
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
b(kd)(kd/2) - nAs (d - kd) = 0
kd = --------
(NSCP C101-01)
Constant ( k )
k =
Moment of Inertia of the Transformed Section
1. U = 1.4DL + 1.7LL
2. U = 0.75(1.4DL + 1.7LL + 1.7 W)
U = 0.90DL + 1.3W
3. U = 1.1DL + 1.3LL + 1.1E)
U = 0.90DL + 1.1E
4. U = 1.4DL + 1.7LL + 1.7H
U = 0.90DL
> (# 1)
I N.A.= (1/3)(b)(kd)³ + nAs (d - kd)²
n
n + fs/fc
( For Design Only )
k = 2rn + (rn)² - rn (For Investigation)
Only )
Resisting Moment of Concrete:
> (# 1)
Steel Ratio
Mc = C(jd)
> (# 1)
Mc = fc/2 (b)(kd)(jd)
Mc = (1/2)(fc)(kj)(bd²)
DL - Dead Load
E - Earthquake Load
LL - Live Load
H - Earth Pressure
r =
As
bd
Resisting Moment of Steel:
W - Wind Load
Ms = T jd
Ms = As fs jd
- 200,000 MPa
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REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
ULTIMATE STRENGTH DESIGN (USD)
A. BEAMS (FLEXURAL STRESS)
2. Doubly Reinforced Rectangular Beam
(Reinforced in Tension/Compression)
1. Singly Reinforced Rectangular Beam
(Reinforced in Tension Only)
b
0.85 fc'
C = 0.85 fc'ab
a
c
b
0.85 fc'
As'
C'
C
a
d
d
Mn
N.A.
As
d - a/2
b1 = 0.85 - 0.008 (fc' - 30)
C. SHEAR STRESS AND DIAGONAL
TENSION
d'
As' fs'
C'
d - d'
Mu2
As2
r fy
where
If r > rb steel does not yield , proceed to
step IV
III. r < rb
r fy
fc'
2
Mu = f fc' bd w ( 1 - 0.59 w )
IV. r > rb (fy = fs)
b
c
Vc =1/6 fc' bd
(Resisting Moment)
fc' w ( 1 - 0.59 w)
fy
2Ru
10.85 fc'
r bd
1st STAGE
Forces:
C 1 = 0.85 fc' ab
T = As 1 fs
Resisting Moment:
( rb )
rb = 0.85 fc'b1 600
fy ( 600 + fy)
Mu1 = f As1 fs (d-a/2)
2nd STAGE
S=
Forces:
C' = As' fs'
T1 = As 2 fs (d-d')
d-c
Resisting Moment:
f As'fs' (d-d')
f
Mu2 = As2 fs(d-d')
Mu2 =
T = As fy
Es = 200,000
fs/Es
=
d-c
If Vs < 1/3 fc' bd ,Smax = d/2 or 600mm
If Vs > 1/3 fc' bd , Smax = d/4 or 300mm
Av min = bS/3fy
TOTAL :
T = T1 + T2
Maximum and Minimum Steel Ratio
rmax = 0.75 r
b
rmin = 1.4 / fy
Vu =
factored or ultimate shear
Vc =
shear force provided by conc.
Vn =
nominal shear
Avmin = area of steel to resist shear
0.003
c
; fs = 600
d-c
c
Solve for c by summing up forces along hor.
T = C ; a = b1 c
2
600 As (d-c) = 0.85 b1 f'c b c
Use quadratic formula to solve for "c"
Then, solve for fs and "a" with known "c"
fs = 600
A S = As 1 + As2
e = fs/Es
Solve for fs from the strain diagram:
Av fy d
Vs
NSCP/ACI Code Specs:
MU= MU1 + MU2
DESIGN AND CONSTRUCTION
Mn
d - a/2
Spacing of Stirrups:
Mu1 = f 0.85fc'ab (d-a/2)
Balanced Steel Ratio
As
0.003
c
d
N.A.
2nd STAGE
2
0.85 fc'
C = 0.85 fc'ab
a
T2 = As 2 fs'
fc'
r = 0.85 fc' 1 -
CECC-3
fy ( 600 + fy)
If r < rb steel yields , proceed to step III
f = 0.85
Vn = Vc + Vs
rb
rb = 0.85 fc' b1600
Vu = f Vn
As'
As fy
0.85 fc' b
bd
II. Check if steel yields by computing
w =
T1 = As 1 fs
1st STAGE
but should not be less than 0.65
As =
f = strength reduction factor
r = As
I. Solve for
r = ratio of tension reinforcement = As/bd
rb= balance steel ratio
Mu1
As1
For fc' < 30 MPa , use b1 = 0.85
For fc' > 30 MPa ,
Ru =
steel area (As)
Mu = factored moment at section, (N-mm)
a/2
d - a/2
Mu = f Ru bd
A. Computing Mu with given tension
Mn = nominal moment, (N-mm)
a = b1 c
w =
FOR SINGLY REINFORCED BEAM
fc' = specified compressive stress of concrete (MPa)
0.85 fc'
b
C
a =
a = depth of equivalent stress block
As = area of tension reinforcement, square millimeters
b = width of the compression face of member
c = distance from extreme compression fiber to N.A. (mm)
d = distance from extreme compression fiber to
centroid of tension reinforcement (mm)
fy = specified yield strength of steel (MPa)
T = As fy
Elevation
Beam Section
T
As
where:
d' = thickness of concrete cover measured from extreme
tension fiber to center of the bar or wire, (mm)
N.A.
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
d-c
;
c
a = b1c
Finally, solve for Mu:
M u = f 0.85fc'ab (d-a/2) or
M u = f As fs (d-a/2)
= 2 Asteel
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REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
B. Computing the required tension steel
area (As) of beam with given Mu
IV. Verify of compression will yield.
ec
b
rmax and M umax
I. Solve for
c
rmax = 0.75 0.85 fc' b1600
=
fy ( 600 + fy)
As'
fs' = 600 c - d'
c
where:
es' = fs'/Es
c-d
If fs' < fy, proceed to step V.
es
MU= MU1 + MU2
MU = f As1 fy (d-a/2) +
If fs' > fy, proceed to step V.
2
M umax = f fc' bd w ( 1 - 0.59 w )
then, proceed to step II.
d' C2 = As'fs'
(compression steel yields)
II. Solve for
r
C1
NA
As' = As2
c
2
r = 0.85 fc' 1 -
1-
fy
2Ru
0.85 fc'
C2 = T2
fs' = 600
As' fs' = As 2 fy
= ____
Reinforced Beam with given Mu.
C
d
As'
C
d
Mu1
d'
C'
d-d'
d-a/2
Mu = f 0.85fc'ab (d-a/2) +
T1 = As 1 fy
f As' fs' (d-d')
T2 = As fy
I. Assume Compression steel yield
T2 = As 2 fs'
I. Solve for As 1 = r maxbd
II. Solve for "a" and "c": C 1 = T1
0.85 f'c ab = As1 fy ; a = ____
a = b1 c ; c = ______
III. Solve for MU1, MU2 and As2
of the beam
IV. Spacing of stirrups:
Spacing, S = Av fy d
Vs
If S < 25mm, increase the value of Av.
either by bigger bar or shear area.
Maximum spacing, s:
If Vs < 1/3 fc' bd ,Smax = d/2 or 600mm
d-d'
Mu2
T1 = As 1 fs
a/2
d'
As
d-a/2
As
C'
a/2
Solve for c by quadratic formula
Solve for fs' and "a"
Solve for Mu :
As' fs'
0.85 fc'
a
As' fs'
0.85 fc'
b
c - d'
0.85 fc' b1 c b + As' 600 c = As fy
Beam with given As and As'
As'
If VS > 2/3 fc' bW d , adjust the size
0.85 fc' ab + As' fs' = As fy
a = b1 c
B. Computing Mu of a Doubly Reinforced
b
If VS < 2/3 fc' bW d , proceed to IV.
[ C1 + C2 = T ]
r bd = _________
A. Computing As and As' of a Doubly
2. Vs = Vn - Vc = Vu /f - V c
c - d'
c
From stress diagram.
As' = ________
FOR DOUBLY REINFORCED BEAM
If Vu < f Vc , but Vu > 1/2 f Vc
proceed to to Step V
1. Vn = Vu /f
T = As fy
(compression steel will not yield)
(Solve for Ru)
Vu > f Vc , stirrups is necessary,
proceed to to Step III.
III. Calculate the shear strength Vs to be
provided by the stirrup.
d - a/2
VI. fs' < fy, then use fs'
Mu = f Ru bd
As =
a
Vc =1/6 fc' bd
If
If Vu < 1/2 fV c , stirrups are not needed
0.85 fc'
V. fs' > fy, then use fs' = fy
If Mu > Mumax design as Doubly Reinforced
f As' fy(d-d')
V. Since fs' < fy, assumption is wrong
If fs' < fy, proceed to step VI.
If Mu < Mumax design as Single Reinforced
II. Calculate the shear strength provided
by concrete, Vc
IV. Since fs' > fy, compression steel yields
Es = 200,000
c - d'
fs' = 600
c
fc'
VERTICAL STIRRUP DESIGN
I. Compute the factored shear force, Vu
If fs' > fy, proceed to step IV.
As
Mumax = with considered factored load
r fy
w =
= ________
III. Verify if Compression steel will yield
c-d'
N.A.
r
d'
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
If Vs > 1/3 fc' bd , Smax = d/4 or 300mm
(fs' = fy)
V. If Vu < f Vc , but Vu > 1/2 f Vc
As 2 = As' = _______
As 1 = As - As' = _______
Av min = bw S /3fy
where S = d/2 or 600mm (whichever
is smaller
II. Solve for a and c:
[ C1 = T1 ]
0.85 fc' ab = As1 fy ; a = ____
a = b1 c ; c = ______
Mu1 = f As1 fy ( d-a/2 )
Mu2 = MU1 - MU
Mu2 =
CECC-3
f As2 fy(d-d') ; As2 = ____
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REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
TYPICAL RESISTANCE FACTORS ARE AS FOLLOWS:
SHEARING STRESS OF RC BEAMS
For members subjected to significant axial tension:
SITUATION
Flexure, without axial load
0.90
Axial tension and axial tension w/ flexure
0.90
Shear and torsion
0.85
Compression members, spirally reinforced
0.75
Other Compression members
0.70
Bearing on concrete
0.70
Plain Concrete: flexure, compression, shear
and bearing
0.65
Vn = Vc + Vs
rmax
=
0.75 rb
To avoid sudden tensile failure : rmin = 0.25 fc'
To control deflection:
r
>
fy
<
0.18 fc'
fy
BALANCED STEEL RATIOS
1.4
fy
1. BEAM REINFORCED FOR TENSION
fy ( 600 + fy)
where:
Checking Ductility
if
r < r ,
where:
r '=
rw = As
bw d
As'
bd
1. In T-beam construction, the flange and web
shall be built integrally or otherwise effectively
bonded together.
2. The width of slab effective as a T-beam shall
not exceed 1/4 of the span of the beam, and
effective overhanging flange on each side of the
web shall not exceed:
bw d
fy d (600-fy)
r < rlim , compression steel yields fs = fy
4. Isolated beams in which T-shape are used to
provide a flange for additional compression
area shall a flange thickness not less than 1/2
the width of the web and an effective flange
width not more than four times the width of
the web.
b
b1
a) 8 times the slab thickness and
b) 1/2 the clear distance to the next web
Mu = factored moment ocurring
simultaneously w/ Vu
3. For beams with slab on one side only, the
effective overhanging flange shall not exceed:
b2
bw
a) 1/12 the span length of the beam,
b) 6 times the slab thickness
c) 1/2 the clear distance to the next web
For members subjected to axial compression:
t > bw /2
Mm = Mu - N u
rlim = 0.85 b fc' d' 600 + r'
choose the
smallest
t
tension steel yields fs = fy
For compression steel
if
Code Requirements for T-beams
Vc = 0.30 fc' bw d
Vu d < 1.0
Mu
2. BEAM REINFORCED FOR COMPRESSION
r = rb + r '
fc' b w d 1 + Nu
14Ag
fc' + 120 rw Vu d
Mu
1) b = L/4
2) b = 16t + b w
3) b = center-center
spacing of beams
T - BEAMS
but shall not be greater than
rb = 0.85 fc'b1 600
choose the
smallest
For Symmetrical Interior Beam
For members subjected to shear and flexure:
1
7
1) b' = L/12 + b'w
2) b' = 6t + b'w
3) b' = S3 /2 + b'w
Distance of Stirrups from support:
fc' bw d
where:
Ag = gross area of section in sq.mm
Nu = factored axial load occurring with Vu
(- ) for compression, (+) for tension
Nu/Ag = expressed in MPa
Vc =
choose the
smallest
For End Beam
a. 0.50 S from face of column support
b. 0.25 S from face of beam support
For members subjected to axial compression:
Vc = 1
6
1) b = L/4
2) b = 16t + bw
3) b = S1 /2 + S2 /2 + b w
0.30N u
Ag
Nu/Ag = expressed in MPa
Nu is negative for tension
For shear reinforcement, fy < 414 MPa.
For members subjected to shear and flexure only:
Vc = 1
6
fc' bw d 1 +
For Interior Beam
where:
where:
CODE PROVISIONS: FOR DESIGN OF
SINGLY-REINFORCED BEAMS
To ensure yield failure:
Vc = 1
6
Nominal Shear Strength Provided by Concete:
Vn = nominal shear strength of RC section
Vc = nominal shear strength provided by concrete
Vs = nominal shear strength of the shear reinforcement
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
4h - d
8
b
but shall not be greater than
Vc = 0.30 fc' bw d
b1
1+
0.30N u
Ag
b'
b2
b < 4b w
b3
t
S1
Substitute Mm for Mu and Vud/Mu not limited to 1.0
bw
S2
Interior Beam
S3
bw'
End Beam
where, h = overall thickness of member
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