Reinforced Concrete Design
Concrete Properties
Modulus of Elasticity
For concrete weighing from 1,500 to 2500 kg/m3
Ec w 0.043 fc'
1.5
c
fc’ = 28 day compressive strength in MPa
Wc = unit weight of concrete in kg/m3
For Normal weight concrete
Ec 4700 fc'
fc’ = 28 day compressive strength in MPa
Wc = unit weight of concrete in kg/m3
Reinforcing Bars
GRADES AND STRENGTH OF REINFORCING BARS
(Non Prestressed)
GRADE
Min.Yeild
Strength
(ksi)
Min.Yeild Min
Strength Tensile
(MPa)
Strength
(ksi)
Min
Tensile
Strength
(MPa)
A615
40
60
40
60
276
414
70
90
483
620
A616
50
60
50
60
345
414
80
90
552
620
A617
40
60
40
60
276
414
70
90
483
620
A615
60
60
414
80
552
ASTM
SPECS
Modulus of Elasticity of steel shall be taken as 200,000 MPa
CONCRETE PROTECTION FOR REINFORCEMENT
Minimum
(
NON
PRESTRESSED)
CAST IN PLACE CONCRETE
Cover,mm
a) Concrete cast against and permanently exposed to earth
75
b) Concrete exposed to earth or weather:
20 mm bars through 36 mm bars
16 mm bar, W31 or D31 wire, and smaller
50
40
c) Concrete not exposed to earth or weather or in contact with ground:
Slabs, walls, joist
32 mm bar and smaller
Beams,columns
Primary reinforcement ,ties,stirrups,spirals
Shells,folded plates members:
20 mm bars and larger
16 mm bars,W31or D31 wire,or smaller
20
40
20
15
PRECAST IN PLACE CONCRETE
( manufactured under plant Conditions)
Minimum
Cover,mm
a) Concrete exposed to earth or weather:
Wall Panels
32 mm bar and smaller
Other members
20 mm bars through 32 mm bars
16 mm bars,W31or D31 wire,or smaller
20
40
30
b) Concrete not exposed to earth or weather or
in contact with ground:
Slabs, walls, joist
32 mm bar and smaller
Beams,columns
Primary reinforcement
ties,stirrups,spirals
Shells,folded plates members:
20 mm bars and larger
16 mm bars,W31or D31 wire,or smaller
15
10
15
10
ULTIMATE STRENGTH DESIGN(USD)
Required
Strength ( Load Factors)
1.Required strength U to resist dead load D and live load L
shall at least be equal to : U =1.4D +1.7L
If resistance to structural effects of specified wind load
W, are included in the design,the following combinations of D,L and W
shall be investigated to determine the greatest required strength U
a) U =0.75(1.4D + 1.7L + 1.7W)
Where load combination shall include full value and zero
value of L to determine the most severe condition, and
b) U = 0.9D +1.3L
For any combination of D, L and W required strength U
shall not be less than
c) U =1.4D +1.7L
2.
If resistance to structural effects of specified earthquake load E, are
included in the design,the following combinations of D,L and E shall
be investigated to determine the greatest required strength U
a) U =1.32D + 1.1(f1)L + 1.1E
Where load combination shall include full value and zero value of
L to determine the most severe condition, and
b) U = 0.99D +1.1E
For any combination of D, L and E required strength U shall not be less
than
c) U =1.4D +1.7L
4. If resistance to earth pressure H, are included in the design,the
following combinations of D,L and H shall be investigated to
determine the greatest required strength U
a) U =1.4D + 1.7L + 1.7H
Where D or L reduces the effect of H
b) U = 0.9D +1.7H
For any combination of D, L and E required strength U shall not be
less
than
c) U =1.4D +1.7L
3.
5. If resistance to loadings due to weight and pressure offluids
with well defined densities and controllable height F are
included in the design, such loading shall have a factor of 1.4
and be added to all loading combinations that include live load.
6. If resistance to impact effects are taken into account in
design,such effect shall be included with live load L.
7. Where structural effects T of differential settlement, creep,
shrinkage,expansion of creep compensating concrete or
temperature change maybe significant in design, required
strength U shall be a least equal to
a) U =0.75(1.4D + 1.7L + 1.4T)
but required strength U shall not be less than
b) U =1.4D +1.7L
STRENGTH REDUCTION FACTOR Ф
STRENGTH REDUCTION FACTOR Ф shall be as follows:
1. Flexure, without axial load
0.90
2. Shear and Torsion
0.85
3. Bearing on concrete except on
Post tension anchorage zone
0.70
4. Post tension anchorage zone
0.85
5. Axial tension and axial tension with flexure
0.90
6. Axial load and axial load with flexure
Both axial load and moment shall be multiplied by Ф
7. Axial compression and axial compression with flexure
a) Members with spiral reinforcement
0.75
b) Other reinforced members
0.70
except that for low values of axial compression Ф shall be
permitted to increase in accordance with the following:
For members in which fy does not exceed 415 MPa with symmetric
reinforcement,and with ( h –d’-d)/h not less than 0.7, Ф shall be permitted to
increase linearly to 0.9, as ФPn decreases from 0.10fc’Ag to zero.
For other reinforced members, Ф shall be permitted to increase linearly to 0.9, as
ФPn decreases from 0.10fc’Ag or ФPb whichever is smaller to zero.
DESIGN AND ANALYSIS FOR FLEXURE (BEAMS)
Basic Assumptions
1. Strain in concrete and the reinforcement shall be assumed
directly proportional to the distance from the neutral axis ,except, for
deep flexural members with overall depth-to-clear span ratio greater than
2/5 for continuous spans and 4/5 for simple span a non linear distribution
of strain shall be considered.
2. Maximum usable strain at extreme concrete compression fiber
shall
be 0.003
3. Stress in reinforcement below specified yield strength fy for grade of
reinforcement used shall be taken Es times steel strain. For strain greater
than corresponding to fy, stress in the reinforcement shall be considered
independent of strain and equal to fy.
4. Tensile strength of concrete shall be neglected in axial and flexural
calculations.
5. Relationship between concrete compressive stress distribution
and concrete strain shall be assumed to be
rectangular.trapezoidal,parabolic or any other assumed shape that
result in prediction of strength in substantial agreement with results of
comprehensive tests.
6. Requirements of 5 may be considered satisfied by an equivalent
rectangular stress distribution defined by the following:
Concrete stress distribution of 0.85fc’ shall be assumed uniformly
distributed over an equivalent compression zone bounded by the
edges of the cross section and a straight line located parallel to the
neutral axis at a distance “a” from the fiber of maximum compressive
strain.
Distance c from fiber of maximum strain to the neutral axis shall be
measured in a direction perpendicular to the neutral axis.
Compression Zone
(stress
in concrete)
(maximum usable strain of concrete)
0.85fc’
0.003
a
c
NA
a 1c
εs (strain of steel)
7. Factor β1 shall be taken as follows:
β1 = 0.85
if fc’≤ 30 MPa
β1 = 0.85 - 0.008( fc’- 30) if fc’ > 30 MPa but β1 shall not be less
than 0.65
SYMBOLS AND NOTATIONS
a = depth of equivalent rectangular stress block, mm
c = distance from extreme compression face, mm
As = area of non prestressed tension reinforcement, mm2
As’ = area of non prestressed compression reinforcement, mm2
b = width of compression face of the member, mm
bw = width of the web, mm
d = distance from extreme compression face to center of tension
reinforcement, mm
d’ = distance from extreme compression face to center of
compression reinforcement, mm
fc’ = specified compressive strength of concrete, MPa
fy = specified yield strength of non prestressed reinforcement,MPa
fs = calculated tensile stress in reinforcement at service loads,MPa
fs’ = calculated compressive stress in reinforcement at service loads,MPa
Mu = factored moment at section; ultimate moment capacity,design
strength
Mn = nominal moment capacity
Ф = strength reduction factor
pb = reinforcement ratio producing balance strain condition
p = ratio of non prestressed tension reinforcement = As/bd
p’ = ratio of non prestressed compression reinforcement = As’/bd
pmin = minimum required ratio of non prestressed tension
reinforcement
pmin = 1.4/fy
Ec = modulus of elasticity of concrete,MPa
Es = modulus of elasticity of reinforcement,MPa
bf = flange width of T –beams
t = flange thickness of T- beams
Balanced strain condition
This exist at a cross section when tension reinforcement reaches
the strain corresponding to its yield strength fy just as concrete
compression reaches its assumed ultimate strain of 0.003.
Stress Diagram
0.85fc’
b
Compression
Zone
ab
C = 0.85fc’abb
d
Strain Diagram
0.003
cb
T = Asbfy
Asb = balance steel area
s
fy
Es
From the Stress Diagram
∑ Fx = 0
C=
T= A f
0.85fc’a
b
b
sb y
EQ.1
divide both sides of EQ.1 by bdfy
0.85 fc' ab Asb
df y
bd
Let
Asb
pb
bd
then
0.85 fc' ab
pb
df y
From the strain diagram
cb
0.003
d 0.003 s
cb
ab
1
Substituting EQ 3 in EQ 2
s
EQ. 2
fy
200000
ab
600 1
d 600 fy
0.85 fc ' 1 600
pb
(600 f y ) f y
EQ. 3
Maximum permissible tensile steel ratio
pmax = 0.75pb
This limitation is to ensure that the steel reinforcement
will yield first to ensure ductile failure.
Minimum permissible tensile steel ratio
pmin = 1.4/fy
The provision for minimum amount of reinforcement applies to
beams which for architectural and other reasons are much larger in
cross section as required by strength consideration. With very small
amount of tensile reinforcement, the computed moment strength as
a reinforced concrete member
is smaller than that of the
corresponding plain concrete section computed from its modulus of
rupture. Failure in this case is quite sudden.
Overreinforced beam
A design in which the steel reinforcement is more than that required for
balanced strain condition. If the beam is overeinforced, the steel will not
yield before failure. As the load is increased, deflections are not
noticeable although the compression concrete is highly stressed, and
failure occurs suddenly without warning to the user of the structure.
Underreinforced beam
A design in which the steel reinforcement is lesser than that required for
balanced strain condition. If the ultimate load is approached , the steel will
begin to yield although the compression concrete is understressed. As the
load is increased, the steel will continue to elongate, resulting into
appreciable deflections and large visible cracks in the tensile concrete.
Failure under this condition is ductile and will give warning to the user of
the structure to decrease the load or apply remedial measure.
SPACING LIMITS OF REINFORCEMENT
Beams
The minimum clear spacing between parallel bars in a layer should be
db( bar diameter) but not less 25 mm.
Where parallel reinforcement is placed in two or more layers,bars in
the upper layer should be directly placed above bars in the bottom
layer with clear distance between layers not less than 25 mm.
Columns
In spirally reinforced or tied reinforced compression members, clear
distance between longitudinal reinforcement shall not be less than
1.5db nor 40 mm.
Walls and Slabs
In walls and slabs other than concrete joist construction, primary
reinforcement shall be spaced not farther than three times the slab or
wall thickness nor 450 mm.
MINIMUM THICKNESS OF NON-PRESTRESSED BEAMS
AND ONE WAY SLABS UNLESS DEFLECTIONS ARE
COMPUTED
Member
Solid
One-Way
Slab
Beams or ribbed
one way slab
Simply Supported
L/20
One end continuous
Both ends continuous
Cantilever
L/24
L/28
L/10
L/18.5
L/21
L/8
L/16
Span Length L in millimeters
Values given shall be used directly for members with normal density concrete
(Wc = 2300 kg/m3) and Grade 60 (415 MPa) reinforcement. For other conditions,
the values shall be modified as follows:
For structural lightweight concrete having unit weights of 1500 -2000 kg/m3 the
values shall be multiplied by ( 1.65 – 0.0005 Wc) but not less than 1.09, where Wc
is the unit mass in kg/m3.
For fy other 415 MPa, the values shall be multiplied by
( 0.4 +fy/700)
FLEXURAL ANALYSIS : BEAMS REINFORCED FOR TENSION
Case I : Steel yields at failure (pmax ≥ p , fs ≥ fy)
Stress Diagram
b
0.85fc’
Compression
zone
a
d
C = 0.85fc’ab
(d – a/2)
Mu
As
T = Asfy
Depth of concrete stress block
a
As f y
0.85 fc' b
Ultimate moment capacity
Mu = Ф 0.85fc’ab(d – a/2)
Mu = Ф Asfy (d – a/2)
General Procedure for Analysis : Case I
Given: b,d,As,fc’,fy
Required
: Mrequirements
1. Check
for ductility
u
0.85 fc ' 1 600
p = As/bd
pb
(600 f y ) f y
β1 = 0.85
if fc’≤ 30 MPa
β1 = 0.85 - 0.008( fc’- 30) if fc’ > 30 MPa but β1 shall not
be less than 0.65
pmax = 0.75pb
pmin =1.4/fy
As f y
pmin ≤ p≤ pmax
a
2. Solve for the depth of the concrete stress block
0.85 fc' b
3. Check for minimum depth if necessary
4. Determine MU
Mu = Ф 0.85fc’ab(d – a/2) or Mu = Ф Asfy (d – a/2)
Units:
If As is in mm2, fc’ and fy in MPa, a,b and d in mm then Mu is in
N.mm. Dividing this by 106 changes N.mm to kN.m
5. Solve for any other requirement if there are any.
Problems
CE Board May 1981
A rectangular beam with b =250 mm and d =460 mm is reinforced for
tension only with 3 – 25 mm bars. The beam is simply supported on a
span of 6 m and carries a uniform dead load of 12 kN/m. Calculate the
uniform live load the beam can carry. Concrete weighs 23 kN/m3 and
steel covering is 60 mm. fc’ = 20.7 MPa, fy =276 MPa. Also check for
minimum depth requirement.
Solution
3 ( 25) 2
As
1472.62mm 2
4
A
1472.62
p s
0.013
bd
250( 460)
1.4 1.4
pmin
0.005
fy
276
pb
0.85 fc' 1 600 0.85(20.7)0.85(600)
0.03711
600 f y f y
(600 276)276
pmax 0.75 pb 0.75(0.03711) 0.0278 p
Steel yields at failure
As f y
1472.62(276)
a
92.4mm
0.85 fc' b 0.85(20.7)250
92.4
0.9(1472.62)276(460
)
a
2 151.37kN .m
M u As f y (d )
2
(10) 6
Wu L2
Mu
8
2
Wu (6)
151.37
8
Wu 33.64kN / m
Wu 1.4WD 1.7WL
460
60
Weight of the beam
WB=bDWc
WB 0.25(0.52)23 3kN / m
Total dead load
WD 12 3 15kN / m
250
33.64 1.4(15) 1.7WL
WL 7.42kN / m
Minimum required depth
d min
fy
L
6000
276
(0.4
)
(0.4
) 297.85mm 520mm
16
700
16
700
CE Board May 1985
A 350 mm x 500 mm rectangular beam is reinforced for tension only with 5 of
28 mm diameter bars. The beam has an effective depth of 446 mm. The beam
carries a uniform dead load of 4.5 kN/m ( including its own weight), a uniform
live load of 3 kN/m, and a concentrated dead load of P and 2P as shown in the
figure.
fc’ =34.5MPa, fy = 414 MPa. Determine the following :
a) Ultimate moment capacity in kN.m
b) the maximum value of P in kN
Figure
2P
2m
P
2m
2m
350
As
5 (28) 2
As
3078.76mm 2
4
As
3078.76
p
0.0197
446
bd 350(446)
1.4
p min
0.00338
414
1 0.85 0.008( fc '30)
1 0.85 0.008(34.5 30) 0.814
0.85 fc ' 1 600 0.85(34.5)0.814(600)
pb
0.03412
600 f y f y
(600 414)414
pmax 0.75 pb 0.75(0.03412) 0.0256 p
Steel yields at failure
a
As f y
0.85 fc ' b
3078.76(414)
124.18mm
0.85(34.5)350
124.18
0.9(3078.76)414(446
)
a
2
M u As f y (d )
440.4kN .m
6
2
(10)
2P
2m
P
2m
2m
w 1.4wD 1.7 wL 1.4(4.5) 1.7(3) 11.4kN / m
1.4(2 P) 2.8P
1.4 P
2.8P
1.4P
M
11.4 kN/m
A
2m
0
R(4) 2.8P (2) 1.4 P(2) 11.4(6)1 0
R 0.7 P 17.1
B
2m
B
2m
R
Maximum Positive Moment
M u R (2) 11.4(2)1
M u (0.7 P 17.1)2 11.4(2)
440.4 1.4 P 17.1(2) 11.4(2)
440.4 11.4(2) 17.1(2)
P
306.43kN
1 .4
Maximum Negative Moment
M u 1.4 P(2) 11.4(2)1
M u 2.8 P 22.8
440.4 2.8 P 22.8
440.4 22.8
P
149.14kN
2 .8
Use P =149.14 kN
Case II : Steel does not yield at failure ( pmax < p , fs < fy )
General Procedure for Analysis : Case II
Given: b,d,As,fc’,fy
Required : Mu
1. Check for ductility requirements
p = As/bd
β1 = 0.85
if fc’≤ 30 MPa
β1 = 0.85 - 0.008( fc’- 30) if fc’ > 30 MPa but β1 shall not
pb
be
less
0.65
0.85
fcthan
' 1 600
(600 f y ) f y
pmax = 0.75pb
p > pmax
2. Using the stress and strain diagram solve for fs and a.
Stress Diagram
Strain Diagram
b
0.003
0.85fc’
C=0.85fc’ab
d
d-a/2
Mu
c
NA
d-c
As
T =Asfs
From the stress Diagram :
∑F =0
C=T
εs
0.85fc’ab = Asfs
EQ 1
From the strain
diagram :
s
d c
0.003
c
substitute
EQ 2
fs
s
200000
in EQ 2, combine with EQ1 to solve for fs and a
a
c
1
3. Determine Mu
Mu = Ф 0.85fc’ab(d – a/2)
Mu = Ф Asfs (d – a/2)
Problem:
A rectangular beam has b =300 mm, d = 500 mm, As = 6 of 32 mm,
fc’ =27.6 MPa,fy =414 MPa. Calculate the ultimate moment capacity.
6 (32) 2
As
4825.49mm 2
4
300
500
As
As
4825.49
p
0.032
bd 300(500)
1.4
p min
0.00338
414
0.85 fc ' 1 600 0.85(27.6)0.85(600)
pb
0.02850
600 f y f y
(600 414)414
pmax 0.75 pb 0.75(0.02850) 0.0214 p
∑F =0
C=T
0.85fc’ab = Asfs
0.85(27.6)300a 4825.49 f s
f s 1.46a EQ.1
Tension Steel
does not yield at failure
s
d c
0.003
c
fs
200000(0.003)
fs
600
500
a
0.85
fs
s
200000
a
c
1
500
a
a
1
1
a
0.85
500(0.85) a
fs
0.85
a
600
0.85
425 a
f s 600(
) EQ.2
a
EQ.1 EQ.2
600
1.46a
(425 a )
a
1.46a 2 255000 600a
a 2 410.96a 174657.53 0
410.96 (410.76) 4(174657.53)
a
260.22mm
2
2
600( 425 260.22)
fs
379.94MPa 414 MPa
260.22
260.22
0.9(4825.49)379.94(500
)
a
2
M u As f s (d )
2
(10) 6
M u 610.34kN .m
Plate #1 : Beams Reinforced for tension
1.
A simply supported beam 6 m long is 350 mm wide has an effective
depth of 500 mm. It supports a uniform dead load of 12 kN/m and
a concentrated live load applied at the midspan. If it is reinforced
with 6 of 22 mm diameter bars, fc’ = 20.7 MPa, fy = 414 MPa, concrete
weighs 23 kN/m3, determine the maximum value of this concentrated load .
Use concrete cover of 70 mm.
2.
A rectangular beam reinforced for tension has b = 300 mm, d = 480 mm
The beam is reinforced with 7 of 25 mm bars with fc’ = 21 MPa, fy =415 MPa.
If the beam is a cantilever beam 3 m long and supports a uniform dead load
of 15 kN/m( including its own weight) applied along its entire length, calculate
the maximum value of the concentrated live load that can be applied at the
free end.
3.
A reinforced concrete beam rectangular beam 300 mm wide has an effective
depth of 450 mm and is reinforced for tension only. Assuming fc’ = 27 MPa.
fy = 350 MPa, determine the required steel area that would produce balance
strain condition.
Solution to #1
(22) 2
As 6
2280.8mm 2
4
As
2280.8
p
0.013
bd 350(500)
pmin
1.4 1.4
0.00338 p
f y 414
a
M u As f y (d )
2
0.85(20.7)0.85(600)
pb
0.02137
(600 414)414
pmax 0.75 pb 0.016 p
Tension steel
Yields at failure
As f y
2280.8(414)
a
153.28
0.85 fc' b 0.85( 20.7)350
0.9(2280.8) 414(500
(10) 6
153.28
)
2
359.78kN .m
Weight of the beam
Total Dead
WB 23(0.35)(0.57) 4.59kN / m
WD 12 4.59 16.59kN / m
WD L2
PL L
M u 1.4M D 1.7 M L 1.4
1.7
8
4
(16.59)(6) 2
PL (6)
359.78 1.4
1.7
8
4
PL 100.1kN
Solution to #2
(25) 2
As 7
3436.1mm 2
4
A
3436.1
p s
0.024
bd 300(480)
0.85fc’ab = Asfs
pb
0.85(21)0.85(600)
0.02161
(600 415)415
pmax 0.75 pb 0.0162 p
0.85(21)300a=3436.1fs
Tension steel does
Not Yield at failure
fs =1.56a
EQ.1
s
d c
0.003
c
a
d
fs
1
a
(200000)0.003
1
[(0.85)(480) a]
fs
600
a
fs
s
200000
a
c
1
( 1d a )
fs
600
a
244800 600a
fs
a
EQ. 1 = EQ.2
244800 600a
1.56a
a
a 156,923 384.6a
2
1.56a 2 244800 600a
a 2 384.6a 156,923 0
EQ 2
a 2 384.6a 156,923 0
384.6 (384.6) 2 4(156923)
a
248.04mm
2
244800 600(248.04)
fs
386.94 MPa f y
248.04
a
M u As f s (d )
2
a
M u As f s (d )
2
0.9(3436.1)386.94( 480
(10) 6
248.04
)
2
425.97 kN .m
PL
3m
15 kN/m
WD L2
M u 1.4 M D 1.7 M L 1.4
1.7 PL L
2
15(3) 2
425.97 1.4
1.7 PL 3
2
PL 65kN
FLEXURAL ANALYSIS: BEAMS REINFORCED FOR TENSION &
COMPRESSION ( DOUBLY REINFORCED BEAMS )
Criteria for adding compression reinforcement : p > 0.75pb
b
STRESS DIAGRAMS
STRAIN DIAGRAM
0.003
0.85fc’
As’
d
d’
Mu
As
a
C1=0.85fc’ab
d-a/2
As1
As’
d-d’
Mu1
T1=As1fy
Compression is resisted by concrete
C2=As’fs’ c
εs’
Mu2
Compression is resisted by As’
c-d’
d-c
As2
T2=As2fy
d’
εs
Compression reinforcement is provided to ensure ductile failure ( tension steel must yield) thus the
stress in tension steel must always be equal to fy.
On the other hand the stress in compression
steel may be equal to or less than fy. This stress must always be checked.
Maximum permissible tensile steel area – NSCP states that for members with compression
reinforcement, the portion of pb equalized by compression reinforcement need not be multiplied by
the 0.75 factor thus
As max
,
f
0.75 pb bd As' s
fy
Stress in compression steel
From the strain diagram
also
s'
c d'
0.003
c
'
f
s
s'
200000
then
and
a
c
1
'
600
(
a
d
1 )
f s'
a
Other double reinforced beam formulas ( derived from stress diagrams)
Mu = Mu1 +Mu2
As = As1 + As2
C 1 = T1
0.85fc’ab =As1fy
C 2 = T2
As2 fy = As’fs’ if fs’ = fy
Mu1 =ФO.85fc’ab(d-a/2)
Mu2 =ФAs2fy(d-d’)
As2 =As’
Mu1 = ФAs1fy(d-a/2)
Mu2 =ФAs’fs’(d-d’)
If fs’=fy
Mu2 =ФAs’fy(d-d’)
FLEXURAL ANALYSIS :DOUBLY REINFORCED BEAMS
Case 1 :Compression and Tension steel yields at failure
Given :b,d,d’,As,As’,fc’,fy
Req’d :Mu
1. Assume that compression steel yields at failure
fs’ = fy
As’ = As2
As1 = As - As’
2. Solve for a
a
As1 f y
0.85 fc' b
'
600
(
a
d
'
1 )
f
3. Solve for fs’ s
a
4.
If
fs’ ≥ fy
Mu1 = ФAs1fy(d-a/2)
Mu2 =ФAs’fy (d-d’)
Mu = Mu1 +Mu2
5. Check for yielding of tension steel
As max 0.75 pbbd As'
≥ As
Problem#1:
Determine the permissible ultimate moment capacity of the beam
shown in figure. fc’= 20.7 MPa, fy = 345 MPa.
350 mm
2 of 28 mm
60 mm
600 mm
4 of 36 mm
Case 2 : Compression steel does not yield at failure
tension steel yields at failure
Given :b,d,d’,As,As’,fc’,fy
Req’d :Mu
1. Assume that compression steel yields at failure
fs’ = fy As’ = As2 As1 = As - As2
2. Solve for a a As1 f y
0.85 fc' b
'
600
(
a
d
)
3. Solve for fs’
'
1
fs
a
4. If fs’< fy
From the stress diagrams
∑F = 0 C1 + C2 = T1 + T2
0.85fc’ab + As’fs’ = As1fy + As2fy
0.85fc’ab + As’fs’ = Asfy
600(a 1 d ' )
'
fs
a
5.Using EQ. A and EQ. B solve for a and fs’
As1fy + As2fy = Asfy
EQ.A
EQ.B
6. Solve for Mu
Mu1 = Ф0.85fc’ab(d-a/2)
Mu2 =ФAs’fs’(d-d’)
Mu = Mu1 +Mu2
7. Check for yielding of tension steel
'
As max 0.75 pb bd As
≥
As
Problem#2:
Determine the permissible ultimate moment capacity of
the beam shown in figure. fc’= 27.5 MPa, fy = 345 MPa.
350 mm
As’ = 775mm2
63 mm
600 mm
As =3625 mm2
Problem#1:
Determine the permissible ultimate moment capacity of the beam
shown in figure. fc’= 20.7 MPa, fy = 345 MPa.
350 mm
2 of 28 mm
Solution to Problem #1
60 mm
600 mm
4 of 36 mm
(36) 2
As 4
4071.5mm 2
4
(28) 2
As' 2
1231.5mm 2
4
Assume that compression steel yields at failure
As1 As As ' 4071.5 1231.5 2840mm 2
As1 f y
2840(345)
a
159.1mm
0.85 fc' b 0.85(20.7)350
600(a 1d ' ) 600(159.1 [0.85(60)])
f
407.67 MPa f y
a
159.1
'
s
Compression steel yields at failure
159.1
2840(345)(600
)
a
2
Mu1 As1 f y (d ) 0.9
458.94kN .m
6
2
10
Mu2 As ' f y (d d ' ) 0.9
1231.5(345)(600 60 )
206.48kN .m
6
10
Mu Mu1 Mu2 458.94 206.48 665.42kN .m
Solution to #2
As1 As As ' 3625 775 2850mm 2
As1 f y
2850(345)
a
120.18mm
0.85 fc' b 0.85(27.5)350
600(a 1d ' ) 600(120.18 [0.85(63)])
f
332.65MPa f y
a
120.18
'
s
Compression steel does not yields at failure
∑Fx =0
0.85fc’ab + As’fs’ = Asfy
0.85(27.5)350a + 775fs’ = 3625(345)
10.56a +fs’ = 1613.7
fs’ =1613.7 – 10.56 a
EQ.1
'
600
(
a
d
600(a [0.85(63)])
'
'
1 )
f
fs
s 334.7 MPa f y
a
a
a
Mu1 As1 f y (d )
600(a 53.55)
'
2
fs
EQ.2
121.12
a
0.9( 2850)345)(600
)
EQ.1 EQ.2
600(a 53.55)
1613.7 10.56a
a
1613.7 a 10.56a 2 600a 32130
Mu1
10 6
Mu1 477.36kN .m
2
Mu2 As ' fs ' (d d ' )
0.9(775)334.7(600 63)
10 6
Mu2 125.36kN .m
Mu2
10.56a 2 1013.7 a 32130 0
a 2 96a 3042.6 0
M u Mu1 Mu2 602.72kN .m
0.85( 27.5)334.7(600)
96 (96) 2 4(3042.6)
p
0.0365
a
121.12mm b
(600 345)345
2
' fs '
2
A
0
.
75
p
bd
A
6500
.
6
mm
s max
b
s
600(121.12 53.55)
'
fy
fs
121.12
T – BEAMS
Reinforced concrete floor systems, roof ,decks etc. are almost always
monolithic. Forms are built for beam soffits and sides and for the underside of
the slabs, and the entire construction is poured at once. Beam reinforcement and
stirrups extend up into the slab thus part of the slab will act with the upper part of
the beam to resist longitudinal compression. The resulting shape of the beam is
in the form of a T rather than rectangular. Figure below shows the stress and
strain diagrams of a T- beam.
Stress Diagrams
(bf –bw)
bf
0.85fc’
t
As
bw
Mu
Cf =0.85fc’(bf-bw)
d
d-t/2
Asf
bw
t
Muf
Tf =Asffy
Strain
Diagram
a
d-a/2
Asw
0.003
0.85fc’
Cw=0.85fc’abw
Muw
T w =Aswfy
εs
Compression is resisted by the
Compression is resisted by the
overhanging flange
Web
Where
bf = flange width bw = width of the web
t = thickness of the slab
c
For yielding of tension steel
p w max 0.75( p b p f ) p w
Where
0.85 fc' 1 600
pb
(600 f y ) f y
As
pw
bw d
pf
Asf
bw d
Code requirements of T- beams
1. In T-beam construction, the flange and the web shall be built
integrally or effectively bonded together
2. The width of the flange effective as a T- beam shall not exceed ¼ of
the span , and the effective overhanging flange on each side of the
web shall not exceed :
a) eight times the slab thickness
b) ½ the clear spacing to the next web
3. For beams with slab on one side only, the effective overhanging
flange shall not exceed :
a) 1/12 the span length of the beam
b) 6 times the slab thickness
c) ½ the clear distance to the next web
bf’
bf
t
bw’
S3
bw
S2
bw
S1
For interior beam:
bf is the smallest of
1. bf = L/4
2. bf = bw + 16t
3. bf = S1 /2 + S2 /2 + bw
For End beams:
bf’ is the smallest of
4. bf’ = L/12 + bw’
5. bf’ = bw’ + 6t
6. bf’ = S3/2 + bw’
For Symmetrical interior beam (S1 =S2 = S3)
bf is the smallest of
7. bf = L/4
8. bf = bw + 16t
9. bf = center to center spacing of beams
T- BEAMS FORMULAS
M u M uf M uw
From the stress diagrams
As Asf Asw
C f T f
C w Tw
Asf f y 0.85 fc' (b f bw )t
0.85 fc ' abw Asw fy
Asf
0.85 fc' (b f bw )t
fy
t
M uf 0.85 fc' (b f bw )t (d )
2
t
M uf Asf fy (d )
2
Asw fy
a
0.85 fc ' bw
M uw
M uw
a
0.85 fc' abw (d )
2
a
Asw f y (d )
2
The compression block of a T- beam can fall either within the flange
only or partly in the web. If it falls within the flange, the rectangular
beam formulas apply, if it falls partly on the web the beam should be
considered as a T- beam .
z
NA
t
bf
z
NA
As
As
Criterion for selection of analysis:
0.85fc’bfz = Asfy
z
As f y
0.85 fc ' b f
if z < t
wide rectangular beam
if z > t
T-beam
Flexural analysis of T- beams ( z > t)
Given: bw,t,As,d,center to center spacing of beams
(assuming symmetrical interior beam), L, fc’,fy
Required: MU
1. Determine the effective flange width bf
bf = L/4
bf = bw + 16t
bf = center to center spacing of beams
Use the smallest value of bf
Note: if bf is given omit step 1
2. Determine if it is to be analyzed as T- beam
z
As f y
0.85 fc' b f
z>t
3. Solve for Asf
Asf
0.85 fc' (b f bw )t
fy
4. Check for yielding of tension steel
As
pw
bw d
pf
Asf
bwd
5. Solve for Muf
0.85 fc ' 1 600
pb
(600 f y ) f y
pw max 0.75( pb p f ) pw
t
M uf 0.85 fc' (b f bw )t (d )
2
t
M uf Asf fy (d )
2
6. Solve for Asw
7. Solve for a
8. Solve for Muw
Asw As Asf
a
As f y
0.85 fc' bw
a
M uw 0.85 fc' abw (d )
2
M uw
9. Solve for Mu
a
Asw f y (d )
2
M u M uf M uw
Problem :
A reinforced concrete T- beam spaced at 2.0 m on centers has a
span of 3.0 m with a slab thickness of 100 mm. The effective depth
is 750 mm and the width of the web is 350 mm. The beam is
reinforced with steel of area 5200 mm2. If fc’ = 20.7 MPa and
fy= 345 MPa, calculate the ultimate moment capacity.
Solution
Determine the effective flange width bf
bf = L/4 = 3000/4 = 750 mm
bf = bw + 16t = 350+16(100) = 1950 mm
bf = center to center spacing of beams = 2000 mm
Use bf = 750 mm
Determine if it is to be analyzed as T- beam
As f y
5200(345)
z
135.94mm 100mm
0.85 fc' b f 0.85(20.7)(750)
Analyze as T - beam
Asf
0.85 fc ' (b f bw )t
fy
0.85(20.7)(750 350)100
2040mm 2
345
As
5200
pw
0.0198
bw d 350(750)
Asf
2040
pf
0.0078
b w d 350(750)
0.85 fc ' 1 600 0.85( 20.7)0.85(600)
pb
0.0275
(600 f y ) f y
(600 345)345
pw max 0.75( pb p f )
pw max 0.75(0.0275 0.0078) 0.02649 pw
tension steel yields at failure
t
M uf Asf fy (d )
2
0.9(2040)345(750
10 6
100
)
2
M uf 443.39kN .m
Asw As Asf 5200 2040 3160mm 2
Asw f y
3160(345)
a
177.03mm 100mm
0.85 fc ' bw 0.85( 20.7)350
177.03
0.9(3160)345(750
)
a
2
M uw Asw f y ( d )
649.03kN .m
6
2
10
M u M uf M uw 443.39 649.03 1092.42kN .m
Flexural analysis of T- beams ( z < t)
Given: bw,t,As,d,center to center spacing of beams(assuming
symmetrical interior beam), L, fc’,fy
Required: MU
1. Determine the effective flange width bf
bf = L/4
bf = bw + 16t
bf = center to center spacing of beams
Use the smallest value of bf
Note: if bf is given omit step 1
2. Determine if it is to be analyzed as T- beam
z
As f y
0.85 fc' b f
z<t
z=a
3.
Solve for MU
z
M u As f y (d )
2
4. Check for yielding of tension steel
As
p
bd
0.85 fc ' 1 600
pb
(600 f y ) f y
p max 0.75 pb p
Problem :
A reinforced concrete T- beam has an effective flange width of
1500 mm span, slab thickness of 100 mm, effective depth of 600
mm and the width of the web is 250 mm. The beam is reinforced
with steel of area 4500 mm2. If fc’ = 20.7 MPa and fy= 345 MPa,
calculate the ultimate moment capacity.
solution
As f y
4500(345)
z
58.8mm 100mm
0.85 fc' b f 0.85(20.7)(1500)
Analyze as wide rectangular beam
58.8
0.9(4500)345(600
)
z
2 797.3kN .m
M u As f y (d )
6
2
10
As
4500
p
0.005
bd 1500(600)
0.85 fc' 1 600 0.85( 20.7)0.85(600)
pb
0.0275
(600 f y ) f y
(600 345)345
p max 0.75 pb 0.0206 p
tension steel yields at failure
Plate #2
Doubly Reinforced beams
1.
A rectangular concrete beam has a width of 300 mm and an effective
depth to bottom bars of 450 mm. The beam is reinforced with six
32 mm bars and two 28 mm top bars located 65 mm from the top
of the beam. If fc’=34.5 MPa, fy =345 MPa, calculate the ultimate
moment capacity of the beam.
2.
Calculate the ultimate moment capacity of the beam shown in figure.
fc’=34.5 MPa, fy = 415 MPa.
400 mm
As’ = 1850mm2
80 mm
700 mm
As =4820 mm2
Non- Rectangular Beams
To deal with these beams, the code requirements and principles
of rectangular beams are applied. Stress and strain diagrams plays
an important role in establishing the formulas that will analyze these
beams.
Problem
375 mm
3 of 16 mm
75 mm
375 mm
Compute the ultimate moment capacity
Of the beam shown in figure. Assume
fc’=21 MPa, fy = 345 MPa
Stress diagram
a
Ac
375 mm
x
Strain Diagram
0.003
c
C=0.85fc’Ac
375 - 2a/3
3 of 16 mm
75 mm
T=Asfy
375 - c
s
375 mm
Solution
3 (16) 2
As
603.2mm 2
4
Assume that steel yields at failure( subject to checking)
C=T 0.85fc’Ac=Asfy
0.85(21)Ac =603.2(345)
Ac =11,658.48 mm2
By similar triangles
x
375
x 0.833a EQ.1
a
450
1
1
a 167.3mm
11658
.
48
a (0.833a)
Ac ax
2
2
a 167.3
c
196.82mm
1 0.85
From the strain diagram
s
fs
375 c
s
200000
0.003
c
fs
375 196.82
(200000)0.003
196.82
f s 543.18MPa f y
2a
M u Asf fy (375 )
3
tension steel yields at failure
0.9(603.2)345(375
106
[2]167.3
)
3
49.34kN .m
Problem
Compute the ultimate moment capacity of the beam shown in
figure. Assume fc’=20.7 MPa, fy = 345 MPa
400 mm
3 of 16 mm
75 mm
400 mm
Stress diagram
a
Ac
400 mm
x
Strain Diagram
0.003
C=0.85fc’Ac
400 - 2a/3
3 of 16 mm
75 mm
T=Asfy
c
400 - c
s
400 mm
Solution
3 (16) 2
As
603.2mm 2
4
Assume that steel yields at failure( subject to checking)
C=T 0.85fc’Ac=Asfy
0.85(20.7)Ac =603.2(345)
Ac =11,829.8 mm2
By similar triangles
x
400
x 0.842a EQ.1
a
475
1
1
a 167.63mm
11829
.
8
a (0.842a)
Ac ax
2
2
a 167.63
c
197.21mm
1
0.85
From the strain diagram
s
fs
375 c
s
200000
0.003
c
fs
400 197.21
(200000)0.003
197.21
f s 616.98MPa f y
2a
M u Asf fy (400 )
3
tension steel yields at failure
0.9(603.2)345(400
106
[2]167.63
)
3
53.98kN .m
Beam Deflections
Unless stiffness values are obtained by a more comprehensive
analysis,immediate deflection shall be computed with the
modulus of elasticity of concrete and with an effective moment
of inertia as follows, but not greater than Ig .
M
M cr
cr
Ie
Ig 1
M a
Ma
fr I g
Where
M cr
yt
3
3
I cr
f r 0.7
fc '
fc’ in MPa
fr = modulus of rupture of concrete
Ma = maximum moment in member at stage deflection is computed.
Ig = moment of inertia of gross concrete section about centroidal
axis, neglecting reinforcement.
Icr = moment of inertia of cracked section transformed to concrete
yt = distance from centroidal axis of gross cross section , neglecting
reinforcement, to extreme fiber in tension
TRANSFORMED SECTION
b
b
x
d
d-x
As
n = modular ratio
Es
n
Ec
nAs
To Locate nuetral axis:
Moment of area of Moment of area of
concrete about NA = steel about NA
x2
b
nAs (d x)
2
To determine Icr
3
bx
I cr
nAs (d x) 2
3
yt = distance from centroidal axis of gross section
neglecting reinforcement, to extreme fiber in tension.
Unless stiffness values are obtained by a more comprehensive analysis,
additional long term deflection resulting from creep and shrinkage of
flexural members shall be determined by multiplying the immediate
deflection caused by the sustained load considered, by the factor
1 50 p '
where p’ shall be the value of reinforcement ratio for non prestress
compression reinforcement at midspan for simple and continuous spans,
and at support for cantilever. It is permitted to assume the time-dependent
factor ξ for sustained load to be equal to
5 years or more
12 months
6 months
3 months
…………………………..
…………………………..
…………………………..
…………………………..
2
1.4
1.2
1.0
Problem:
A concrete beam 6 m long is 300 mm wide and 600 mm deep and
carries a dead load of 9 kN/m and live load of 12 kN/m. The beam
is reinforced for tension only with four 25 mm bars with an effective
depth to tension bars of 530 mm. fc’ = 20.7 MPa , fy = 345 MPa ,
fr = 2.832 MPa, Ec = 20,000 MPa, Es = 200,000 MPa.
Covering of bars is 70 mm.
a) Calculate the maximum instantaneous deflection due to dead load
and live load.
b) Calculate the deflection due to the same loads after five years
assuming that 30% of the live load is sustained.
Gross moment of inertia
bh 3 300(600)3
Ig
5400 x106 mm 4
12
12
M cr
fr I g
yt
2.832(5400)10 6
50.97 x10 6 N .mm
300
M cr 50.97 kN .m
Transformed section
b
b=300
x
530-x
4 (25) 2
As
1963.2mm 2
4
d=530
As
nAs
Es
200000
10
Ec
20000
n
x2
b
nAs (d x)
2
x2
b
nAs (d x)
2
x 2 69377 130.9 x
x2
300
10(1963.5)(530 x)
2
x 2 130.9 x 69377 0
130.9 (130.9) 2 4(69377)
x
205.96mm
2
bx 3
300(205.96) 3
2
I cr
nAs (d x)
10(1963.5)(530 205.96) 2
3
3
I cr 2,935.38 x106 mm 4
a)
WT L2 (12 9)(6) 2
Ma
94.5kN .m
8
8
Effective moment of inertia
M
M cr
cr
I cr
Ie
Ig 1
M a
Ma
3
50.97
50.97
6
6
Ie
5400
x
10
1
2935
.
38
x
10
94.5
94.5
6
4
3
3
3
I e 3,322.1x10 mm
a) Instantaneous deflection
5WT L4
5(21)(6000) 4
3.44mm
6
384 Ec I e 384(20000)3322.1(10)
a)
Long term deflection
Since only 30% l of the live load is sustained
WT 12 0.3(9) 14.7kN / m
5WT L4
5(14.7)(6000) 4
'
2.41mm
6
384 Ec I e 384(20000)3322.1(10)
2
2
'
1 50 p 1 50(0)
Long term deflection
L ' 3.44 2( 2.41) 8.26mm
Design for Flexure : Beams Reinforced for tension
Derivation of designing formulas
Stress Diagram
0.85fc’
b
Compression
Zone
a
d
Strain Diagram
0.003
C = 0.85fc’ab
c
d-a/2 Mu
T = Asfy
a
M u As f y (d ) EQ.1
2
s
a
fy
Es
As f y
0.85 fc' b
EQ.2
EQ.2 in EQ.1
M u As f y (d
Mu
As f y
bd
As f y
2[0.85 fc ' b]
bd ( d
)
As f y d
2[0.85 fc' b]d
)
fy
As
As
Mu
bdf y ( d
d)
bd
bd 2[(0.85 fc ' ]
Let
As
p
bd
and
m
fy
0.85 fc '
pm
M u pbdf y ( d
d)
2
pm
2
M u bd pf y (1
)
2
pm
)
Let Ru pf y (1
2
M u bd Ru
2
pm
Ru pf y (1
)
2
p 2m
Ru pf y 1
fy
2
Coeffecient of resistance
For proportioning of section
p 2m
2
{Ru pf y 1
f y}
2
mfy
2
2p
Ru
p2
mf y
m
2p
2
p
Ru 0
m mf y
2
p
2 Ru
2
2 2
( ) 4
m
m
mf y
2
p
p
2mRu
2
2 2
( ) 4 2
m
m
m fy
2
2mRu
2 2
1
m m
fy
2
2mRu
1
p (1 1
)
m
fy
p
p
2mRu
2
2 2
( ) [1
]
m
m
fy
2
2mRu
2
[1 1
]
m
fy
2
Actual tensile steel ratio
NSCP COEFFECIENTS FOR CONTINUOUS BEAMS AND SLABS
NSCP states that in lieu of frame analysis, the following approximate moments
and shear are permitted for design of continuous beams and one way slabs
provided that :
There are two or more spans
Spans are approximately equal,with the large of two adjacent spans not greater
than the shorter by more than 20 %
Loads are uniformly distributed
Unit live load does not exceed three times the unit dead load
The members are prismatic
Positive moment
End spans
Discontinuous end unrestrained
Discontinuous end integral with support
Interior spans
W ULn2/11
WULn2/14
W ULn2/16
Negative moment at exterior face of first interior support
Two spans
W ULn2/9
More than two spans
W ULn2/10
Negative moment at other faces of interior supports W ULn2/11
Negative moment at face of all supports for
Slabs with span not exceeding 3 m; and beams
where ratio of column stiffness to beam stiffness
exceeds eight at each end of the span
W ULn2/12
Negative moment at interior face of exterior supports
for members built integrally with supports
Where support is a spandrel beam
Where support is a column
W ULn2/24
W ULn2/16
Shear in end members at face of first interior support 1.15W ULn/2
Shear at face of all other supports
W ULn/2
Where Ln = the clear span for positive moment or shear and average of
adjacent clear spans for negative moments.
column
column
L1
L2
Shear
1.15wL2
2
wL1
2
2
1
wL
14
1.15wLn
2
wL22
11
Moment
wL12
16
wL2n
9
Ln
L1 L2
2
Shear and moment for continuous beams or slab with two spans
discontinuous edge integral with support,discontinous end unrestrained
column
column
L1
column
L3
Spandrel beam
L2
Shear
wL1
2
wL12
14
1.15wLn
2
wL22
16
1.15wLn
2
wL3
2
2
wL3
14
Moment
wL12
16
wL
10
L L2
Ln 1
2
2
n
wL2n
10
L2 L3
Ln
2
wL23
24
Shear and moment for continuous beams or slab with more than
two spans and discontinuous end integral with support
column
L1
column
L3
L2
Shear
wL1
2
wL12
11
1.15wLn
2
wL22
16
1.15wLn
2
wL3
2
2
wL3
11
Moment
wL
10
L L2
Ln 1
2
2
n
wL2n
10
L2 L3
Ln
2
Shear and moment for continuous beams or slab with more than
two spans and discontinuous end unrestrained
Case 1 : Design for balanced strain condition with given dimensions
Given : b,d, fc’ and fy
Required : Steel area that would produce balance strain condition
General Procedure:
1. Solve for pb
pb
0.85 fc ' 1 600
(600 f y ) f y
2. Solve for Asb
Asb = pbbd
Problem:
A reinforced concrete rectangular beam 300 mm wide has an effective
depth of 460 mm and is reinforced for tension only. If fc’=20 MPa,
fy = 300 MPa, determine the balance steel area in mm2.
Solution
pb
0.85 fc ' 1 600 0.85(20)0.85(600)
0.032
(600 f y ) f y
(600 300)300
Asb = pbbd=0.032(300)460 = 4416 mm2
Case 2 : Design of cross section and reinforcement from given loads
and type of beam dimensions
Given : fc’ , fy, Loads, type of beam, Wc
Required : cross section dimension ,size and number of steel bars
General Procedure:
•Assume the weight of the beam ( DL) as 20 to 25% of (DL + LL). add
this to the given dead load.
•By any available method, determine designing moment M u.
•Assume a value of steel ratio p ( 0.3pb to 0.6pb but not less than pmin).
This will provide enough allowance for the rounding off of values of
number of bars to be used.
•Solve for the following design constants :
m
5.
Solve for bd
2
fy
0.85 fc'
Mu
bd
Ru
2
Ru pf y (1
pm
)
2
6. Try a ratio of d/b ( from d= b to d = 2b ) , then solve for b
and d. Alternatively b may be assumed until a reasonable value of d is
attained. Provide concrete cover to get total thickness. Check for
minimum depth if required.
7. Compute actual weight ( W = bDWc) and compare with assume
Weight.
1
2mR
8. Solve for actual p p m (1 1 f u )
y
9.Solve for As:
As = pbd
10.Solve for the number of bars by dividing As by the area of one bar
to be used. Round off this number to the next integer.
Problems :
Design a rectangular beam for a 6 m simple span to support a uniform
dead load of 15 kN/m and uniform live load of 24 kN/m applied along it’s
entire length. fc’=20 MPa, fy = 350 MPa,Wc=23.5 kN/m3.
Design a rectangular beam reinforced for tension only to support a
service dead load moment of 85 kN.m ( including its weight) and service
live load moment of 102 kN.m. Use p =0.6pb, d/b = 1.75 , fc’=28 MPa,
fy = 276 MPa.
Problems :
Design a rectangular beam for a 6 m simple span to support a uniform
dead load of 15 kN/m and uniform live load of 24 kN/m applied along it’s
entire length. fc’=20 MPa, fy = 350 MPa,Wc=23.5 kN/m3
wB 0.20(24 15) 7.8
WD 15 7.8 22.8kN / m
2
2
22.8(6)
24(6)
M u 1.4
1.7
327.4kN .m
8
8
0.85(20)0.85(600)
p 0.4 pb 0.4
0.0104
(600 350)350
fy
350
m
20.59
0.85 fc' 0.85(20)
pm
0.0104(20.59)
Ru pf y (1
) 0.0104(350)[1
] 3.25
2
2
Mu
bd
Ru
2
Mu
327.24(10) 6
d
570mm
Ru b
0.9(3.25)350
Trial section
350 mm x 570mm effective depth, total depth 650 mm
Actual weight
wB 0.35(0.65)23.5 5.35kN / m 7.8kN / m
6
Mu
327.24(10)
Ru
3. 2
2
2
bd
0.9(350)(570)
2mRu
1
p (1 1
)
m
fy
1
2( 20.59)(3.2)
p
(1 1
) 0.102
20.59
350
2
As pbd 0.0102(350)570 2035mm
try 20mmbars
2035
N
7
2
( 20)
4
Plate # 3 Design of beams reinforced for tension
Problems :
Design a rectangular beam for a 5 m simple span to support a uniform
dead load of 12 kN/m and uniform live load of 20 kN/m applied along it’s
entire length. fc’=20 MPa, fy = 400 MPa,Wc=23.5 kN/m3.
Design a rectangular beam reinforced for tension only to support a
service dead load moment of 65 kN.m ( including its weight) and service
live load moment of 80 kN.m. Use p =0.45pb, d/b = 1.5 , fc’=20 MPa,
fy = 300 MPa.
A reinforced concrete T- beam spaced at 3.0 m on centers has a
span of 4.0 m with a slab thickness of 75 mm. The effective depth
is 750 mm and the width of the web is 300 mm. The beam is
reinforced with steel of area 4200 mm2. If fc’ = 20.7 MPa and
fy= 345 MPa, calculate the ultimate moment capacity.
Case 3 : Design of reinforcement of a beam with given moment and
cross sectional dimension
Given : b,d, Mu ,fc’ , fy,
Required : Number of steel bars
General Procedure :
1. Solve for Ru and m
Mu
Ru
bd 2
m
fy
0.85 fc'
2. Solve for p
p
1
(1
m
1
2mRu
)
fy
3. Check for yielding of tension steel and pmin if required.
4. Solve for As : As = pbd
Problem :
Determine the required tension steel area for a rectangular beam with
b =250 mm, d =330mm, fc’ =20.7 MPa, fy = 414 MPa. The beam is
required to support a factored moment of 110 kN.m.
Solution
6
Mu
110 (10)
Ru
4.49
2
2
bd
0.90(250)(330)
fy
414
m
23.53
0.85 fc' 0.85(20.7)
p
1
(1
m
1
1
p
(1
23.53
2mRu
)
fy
Note :
If p > 0.006 no need to
check for pmin
p < 0.02 no need to
check for pmax
As = pbd
As = 0.01276(250)330
As = 1052.7 mm2
2( 23.53) 4.49
1
) 0.01276
414
DESIGN OF CONTINUOUS BEAM
In the design of continuous beam, the cross section is determined
by the maximum moment obtained by any structural analysis
method or by its equivalent NSCP coefficients.
The reinforcements are designed from the moment obtained at the
different sections of maximum positive and negative moments.
Problem:
Figure shows a continuous beam of three spans with the left and right
ends discontinuous and integral with the support . Design the section
and reinforcements at critical sections using the given service uniform
loading. Given dead loads includes the weight of the beam. fc’= 28
MPa, fy =350 MPa . Use NSCP coefficients to determine the moments.
DL = 12 kN/m
LL =16 kN/m
4m
A
B
C
DL = 15 kN/m
LL = 18 kN/m
DL = 20 kN/m
LL = 24 kN/m
5m
6m
D
E
F
G
Factored loads
W1 = 1.4(12)+1.7(16)= 44 W2=1.4(15)+1.7(18)= 51.6 W3 = 1.4(20)+1.7(24)=68.8
Design moments by NSCP coeffecients
w1 L12 44(4) 2
MA
44kN .m
16
16
w1 L12 44(4) 2
MB
50.28kN .m
14
14
w2 L2n1
MC
10
45 2
)
2
104.49kN .m
10
51.6(
w2 L22 51.6(5) 2
MD
80.63kN .m
16
16
w3 L2n 2
ME
10
65 2
68.8(
)
2
208.12kN .m
10
w3 L23 68.8(6) 2
MF
176.91kN .m
14
14
w3 L23 68.8(6) 2
MG
157.9kN .m
16
16
Proportioning of uniform beam size
Note: use the biggest computed design Moment
Mu =208.12 kN.m
0.85 fc' 1 600 0.85(28)0.85(600)
pb
0.0365
(600 f y ) f y
(600 350)350
p 0.5 pb (0.5)0.0365 0.01825
fy
350
m
14.71
0.85 fc' 0.85(28)
pm
Ru pf y (1
)
2
0.01825(14.71)
Ru 0.01825(350)(1
) 5.53MPa
2
Try b = 250 mm
Mu
d
Ru b
pmin
208.12(10) 6
410mm
0.9(5.53)(250)
1.4 1.4
0.004
fy
350
Section A
M u 44kN .m
Mu
44(10) 6
Ru
1.16
2
2
bd
0.9( 250)(410)
1
2mRu
1
2(1.16)14.71
p (1 1
)
(1 1
) 0.0034
m
fy
14.71
350
Use p = 0.004
As pbd 0.004(250)410 410mm 2
Try 20 mm bars
N
410( 4)
2 pcs
2
( 20)
Section B
Top bars
M u 50.28kN .m
Mu
50.28(10) 6
Ru
1.32
2
2
bd
0.9(250)(410)
p
1
2mRu
1
2(1.32)14.71
(1 1
)
(1 1
) 0.0034
m
fy
14.71
350
Use p = 0.004
As pbd 0.004(250)410 410mm 2
Try 20 mm bars
N
410( 4)
2 pcs
2
( 20)
Section C
bottom bars
M u 104.49kN .m
Mu
104.49(10) 6
Ru
2.74
2
2
bd
0.9(250)(410)
p
1
2mRu
1
2(2.74)14.71
(1 1
)
(1 1
) 0.00834
m
fy
14.71
350
As pbd 0.00834( 250)410 854.86mm 2
Try 20 mm bars
N
854.86(4)
3 pcs
2
(20)
Section D
top bars
M u 80.63kN .m
Mu
80.63(10) 6
Ru
2.11
2
2
bd
0.9(250)(410)
2mRu
1
1
2(2.11)14.71
p (1 1
)
(1 1
) 0.00632
m
fy
14.71
350
As pbd 0.00632(250)410 648.06mm 2
Try 20 mm bars
N
648.06(4)
3 pcs
2
(20)
Section E
bottom bars
M u 208.12kN .m
Mu
208.12(10) 6
Ru
5.44
2
2
bd
0.9(250)(410)
2mRu
1
1
2(5.44)14.71
p (1 1
)
(1 1
) 0.018
m
fy
14.71
350
As pbd 0.018(250) 410 1845mm 2
Try 20 mm bars
N
1845(4)
6 pcs
2
(20)
Section F
Top bars
M u 176.91kN .m
Mu
176.91(10) 6
Ru
4.62
2
2
bd
0.9(250)(410)
2mRu
1
1
2(4.62)14.71
p (1 1
)
(1 1
) 0.0148
m
fy
14.71
350
As pbd 0.0148(250) 410 1517 mm 2
Try 20 mm bars
N
1517(4)
5 pcs
2
(20)
Section G
bottom bars
M u 157.9kN .m
Mu
157.9(10) 6
Ru
4.12
2
2
bd
0.9(250)(410)
2mRu
1
1
2(4.12)14.71
p (1 1
)
(1 1
) 0.013
m
fy
14.71
350
As pbd 0.013(250)410 1332.5mm 2
Try 20 mm bars
N
1332.5(4)
5 pcs
2
(20)
top bars
Gross moment of inertia using 100 mm covering
bh 3 300(600)3
Ig
5400 x106 mm 4
12
12
M cr
fr I g
yt
3.1(5400)10 6
55.79 x10 6 N .mm
300
Placement of bars
Requirement
2 of 20
2 of 20
A
B
3 of 20
3 of 20
C
D
6 of 20
E
5 of 20
F
5 of 20
G
Layout
2 of 20
1 of 20
2 of 20
A
B
C
3 of 20
1 of 20
2 of 20
D
3 of 20
3 of 20
E
3 of 20 2 of 20
2 of 20
F
G
Section at A and B
250 mm
Section at C
Section at D
250 mm
250 mm
500 mm
500 mm
Section at F
Section at E
250 mm
Section at G
250 mm
500 mm
500 mm
250 mm
500 mm
500 mm
Plate # 4: Design of continuous beams
DL = 9 kN/m
LL =12 kN/m
DL = 12 kN/m
LL = 14 kN/m
3.6 m
4.0 m
DL = 15kN/m
LL = 18 kN/m
4.5 m
Problem
Design the uniform size and reinforcements at critical section of
the continuous beam shown above. fc’= 20 MPa, fy = 300 MPa.
Given dead loads includes the weight of the beam.
DESIGN FOR FLEXURE :DOUBLY REINFORCED BEAMS
Given :b,d,d’, Mu ,fc’,fy
Req’d : As, As’
General Procedure
1. Solve for Ru
Mu
Ru
bd 2
2. Solve for p
2mRu
1
p (1 1
)
m
fy
3. Check if the beam needs compression reinforcement
0.85 fc ' 1 600
pb
(600 f y ) f y
p max 0.75 pb
if p > pmax then compression reinforcement is necessary
4. Solve for As1
9. Solve for fs’
As1 p max bd
'
600
(
a
d
'
1 )
fs
a
5. Solve for a
a
6. Solve for Mu1
As1 f y
0.85 fc' b
a
M u1 As1 f y (d )
2
7. Solve for Mu2
M u 2 M u M u1
8. Solve for As2
M u2
As 2
f y (d d ' )
Case 1
If fs’ ≥ fy then fs’ = fy
( compression steel yields at failure)
As' As 2
Case 2
If fs’ < fy then use fs’
( compression steel does not yield at failure)
A
'
s
As 2 f y
f s'
10. Solve for As
As As1 As 2
Problem:
Design the reinforcement of a rectangular beam to carry a factored
moment of 272 kN.m. The beam width is 250 mm,effective depth
400mm. Use fc’ = 20.7 MPa, fy = 345 MPa, d’ =60 mm.
Solution
Mu
272(10) 6
Ru
7.55
2
2
bd
0.9(250)(400)
fy
345
m
19.61
0.85 fc ' 0.85( 20.7)
1
2(19.61)7.55
p
(1 1
) 0.0317
19.61
345
0.85 fc ' 1 600 0.85(20.7)0.85(600)
pb
0.0275
(600 f y ) f y
(600 345)345
pmax 0.75(0.0275) 0.0206
p > pmax compression reinforcement is necessary
As1 pmax bd 0.0206(250)400 2060mm 2
As1 f y
2060(345)
a
161.57mm
0.85 fc' b 0.85(20.7)250
161.57
0.9(2060)345(400
)
a
2
M u1 As1 f y (d )
204.18kN .m
6
2
10
M u 2 M u M u1 272 204.18 67.82kN .m
M u2
67.82(10) 6
As 2
642.42mm 2
f y (d d ' ) 0.9(345)(400 60)
600(a 1d ' ) 600(161.57 [0.85(60)])
f
410.6MPa
a
161.57
'
s
As' As 2 642.42mm 2
Compression
Steel yields
at failure
As As1 As 2 2060 642.42 2702.42mm 2
Plate # 5 :
DESIGN FOR FLEXURE :DOUBLY REINFORCED BEAMS
Design the reinforcement of a rectangular beam to resist a dead load
moment of 200 kN.m(including its own weight) and a live load
moment of 300 kN.m . The beam is limited in size to 350 mm by
600mm overall depth. Steel covering ( from centroid of bars to
outermost fiber is 100 mm for both tension and compression
reinforcement. Use fc’ = 27.5 MPa, fy = 414 MPa
DESIGN FOR BENDING : T-BEAMS (z > t)
Given: bw,t,Mu,d,center to center spacing of beams(assuming symmetrical
interior beam), L, fc’,fy
Required: As
1. Determine the effective flange width bf
bf = L/4
bf = bw + 16t
bf = center to center spacing of beams
Use the smallest value of bf
Note: if bf is given omit step 1
2. Solve for Asf
Asf
3. Solve for Muf
M uf
0.85 fc' (b f bw )t
fy
t
Asf fy (d )
2
If Mu > Muf then design as T- beam
4. Solve for Muw
M uw M u M uf
5. Solve for a
a
M uw 0.85 fc' abw (d )
2
6. Solve for Asw
0.85 fc' abw
Asw
fy
7. Solve for As
As Asf Asw
8. Check for yielding of tension steel
As
pw
bw d
pf
Asf
bwd
0.85 fc' 1 600
pb
(600 f y ) f y
pw max 0.75( pb p f ) pw
Design the reinforcement of a T- beam to resist a factored moment of
750 kN.m. Properties of the T- beam are as follows : bf = 550 mm,
bw = 300 mm, d = 600 mm, t = 110 mm, fc’ = 20.7 MPa, fy = 345 MPa.
solution
Asf
M uf
0.85 fc ' (b f bw )t
fy
0.85(20.7)(550 300)110
1402.5mm 2
345
110
0.9(1402.5)345(600
)
t
2 237.33kN .m
Asf fy (d )
2
106
Mu > Muf design as T- beam
M uw M u M uf 750 237.33 512.67kN .m
a
M uw 0.85 fc' abw (d )
2
a
512.67(10) 6 (0.9)0.85( 20.7)300a (600 )
2
a2
107915.8 600a
2
a 2 1200a 215831.6 0
1200 (1200) 2 4( 215831.6)
a
220.3mm 110 mm
2
0.85 fc ' abw 0.85(20.7)220.3(300)
Asw
3370.6mm 2
fy
345
As Asf Asw 1402.5 3370.6 4773.1mm 2
Check for yielding of tension steel
As
4773.1
pw
0.0265
bw d 300(600)
Asf
1402.5
pf
0.008
b w d 300(600)
0.85 fc' 1 600 0.85(20.7)0.85(600)
pb
0.0275
(600 f y ) f y
(600 345)345
pw max 0.75( pb p f ) 0.75(0.0275 0.008) 0.0266 pw
Tension steel yields at failure
DESIGN FOR BENDING : T-BEAMS (z < t)
Given: bw,t,Mu,d,center to center spacing of beams(assuming symmetrical
interior beam), L, fc’,fy
Required: As
1. Determine the effective flange width bf
bf = L/4
bf = bw + 16t
bf = center to center spacing of beams
Use the smallest value of bf
Note: if bf is given omit step 1
2. Solve for Asf
Asf
0.85 fc' (b f bw )t
fy
3. Solve for Muf
M uf
t
Asf fy (d )
2
If Mu < Muf then z < t ,design as wide rectangular beam
4. Solve for a
a
M u 0.85 fc' ab f (d )
2
5. Solve for As
As
0.85 fc ' b f a
fy
6. Check for yielding of tension steel
As
p
bf d
0.85 fc' 1 600
pb
(600 f y ) f y
p max 0.75 pb p
Problem :
A reinforced concrete T-beam with bf = 810 mm, d = 300 mm,
bw = 200 mm , t = 100 mm, fc’=20.7 MPa, fy = 414 MPa is to be
designed to carry an ultimate moment of 221 kN.m Determine the
required steel area.
solution
Asf
M uf
0.85 fc ' (b f bw )t
fy
0.85(20.7)(810 200)100
2683.2mm 2
414
100
0.9(2683.2)414(300
)
t
2 249.9kN .m
Asf fy (d )
2
10 6
Design as wide rectangular beam
a
M u 0.85 fc' ab f (d )
2
a
221(10) 0.9 0.85(20.7)810a(300 )
2
2
a
17229.6 300a
2
2
a 600a 34459.2 0
6
600 (600) 2 4(34459.2)
a
64.33mm
2
0.85 fc ' b f a 0.85( 20.7)(810)64.33
As
2214.5mm 2
fy
414
As
2214.5
p
0.0091
b f d 810(300)
0.85(20.7)0.85 600
pb
0.0213
(600 414)414
pmax 0.75(0.0213) 0.016 0.0091
Tension steel yields at failure
Plate # 6: Flexural design of T-beams
A reinforced concrete T-beam with d = 550 mm, bw = 300 mm , slab
thickness =100 mm is 4.8 m long and spaced 3 m on centers. The beam
support a service dead load moment of 400 kN.m (including its weight) and
service live load moment of 500 kN.m. If fc’=27.5 MPa, fy = 414 MPa ,
determine the required steel area.
Design the reinforcement of a T- beam to support a uniform service
dead load of 25 kN/m and service live load of 30 kN/m on a simple span
of 8 m . Properties of the T- beam are as follows : bf = 1500 mm,
bw = 250 mm, d = 600 mm, t = 100 mm, fc’ = 20.7 MPa, fy = 345 MPa.
ONE WAY SLAB
Reinforced concrete slab are large flat plates that are supported at its
sides by reinforced concrete beams, walls, column, steel beams or by
the ground. If the slabs are supported on opposite sides only they are
called one way slab since bending will occur on one direction only.
A one way slab is considered as a wide & shallow rectangular beam.
Reinforcing steel is usually spaced uniformly over its width. One way
slabs are analyzed by considering a one meter strip, which is assumed
independent of the adjacent strips.
Maximum spacing of reinforcement
Flexural reinforcement shall not be spaced farther apart than 3 times
the slab thickness nor 450 mm.
Minimum size of flexural reinforcement = 12 mm
Shrinkage and temperature reinforcement
The area of shrinkage reinforcement shall be
where Grade 275 deformed bars are used… 0.002bt
where Grade 415 deformed bars are used… 0.0018bt
where reinforcement with fy > 415 MPa measured
at yield strain of 0.35% are used …. 0.0018(400)bt/fy
Shrinkage reinforcement shall not be spaced farther apart than 5
times the slab thickness nor 450 mm.
Minimum size of shrinkage and temperature bars = 10 mm
Design of one way slab
Given: Loads, type of slab, fc’,fy,Wc
Req’d: t, size and spacing of main bars and
temperature bars
General Procedure
Determine the minimum slab thickness t using table for minimum
thickness of non prestressed beams and one way slab. This
thickness should be at least 75 mm
2. Compute the weight of the slab ( this is to be added to the given
dead load)
3. Calculate the design moment Mu
4. Compute the effective depth d
d = t – covering – ½ bar diameter ( minimum of 12 mm)
1.
5. Compute design constants
Mu
Ru
bd 2
m
fy
0.85 fc '
1
2mRu
p (1 1
)
m
fy
p > pmin
5. Solve for As
As = pbd
6. Solve for the spacing of bars
a)S
A1
S 1000
As
where : A1 = area of 1 bar
Use the smallest of the following
b)3t
c) 450 mm
7. Solve for area of temperature bars
At = 0.002bt , At = 0.0018bt, At =0.0018(400)bt/fy
A1
1000
8. Solve for the spacing of temperature bars St
At
where : A1 = area of 1 temperature bar ( minimum of 10mm dia.)
Use the smallest of the following
a) St
b) 5t
c)450 mm
Problem:
Design a one way slab having a simple span of 3.0 m. The slab is to
carry a uniform dead load of 2.5 KPa and uniform live load of 4.2 kPa.
fc’ = 27.6 MPa , fy = 276 MPa for main bars and temperature bars.
Concrete weighs 23.5 kN/m3
Solution
Slab thickness
fy
L
t
(0.4
)
20
700
3000
276
t
(0.4
) 120mm 75mm
20
700
Weight of Slab (assuming 1 m wide strip)
Ws 23.5(1)0.12 2.82kN / m
Total dead load
WD 2.5 2.82 5.32kN / m
Factored uniform load
Wu 1.4WD 1.7WL 1.4(5.32) 1.7(4.2) 14.59kN / m
Design Moment
L2 14.59(3) 2
M u Wu
16.41kN .m
8
8
Effective depth assuming 12 mm bar
1
12
d t 20 120 20 94mm
2
2
Mu
14.59(10) 6
Ru
1.8346 MPa
2
2
bd
0.9(1000)(94)
m
fy
0.85 fc'
276
11.76
0.85(27.6)
p
1
2(11.76)(1.8346)
(1 1
) 0.00693
11.76
276
pmin
1.4 1.4
0.00507
f y 276
As pbd 0.00693(1000)94 651.42mm 2
Using 12 mm bars
(12) 2
A
4 (1000) 173.6 say170mm 3t 360mm
S 1 1000
As
651.42
Temperature bars
At 0.002bt 0.002(1000)120 240mm 2
Using 10 mm bars
(10) 2
A1
4 (1000) 327 say 320mm 5t 600mm
S t 1000
At
240
10 mm temp bars @ 320 mm o.c
12 mm main bars @ 170 mm o.c
120 mm
Problem:
Design a 4 m long one way slab with one end discontinuous as shown
in
the figure . The slab is to carry a uniform dead load of 3.6 KPa
and uniform live load of 4.0 kPa. fc’ = 20.7 MPa , fy = 415 MPa for
main bars and fy = 276 MPa for temperature bars.
Concrete
weighs 22.56 kN/m3.
column
column
column
4m
4m
C
t
B
A
L
4000
166.67 say170mm
24
24
Weight of Slab (assuming 1 m wide strip)
Ws 22.56(1)0.17 3.83kN / m
Total dead load
WD 3.6 3.83 7.43kN / m
Factored uniform load
Wu 1.4WD 1.7WL 1.4(7.43) 1.7( 4.0) 17.2kN / m
Effective depth assuming 12 mm bar
1
12
d t 20 170 20
144mm
2
2
Design Moments
L2 17.2(4) 2
M A Wu
17.2kN .m
16
16
L2 17.2(4) 2
M B Wu
19.66kN .m
14
14
17.2(10) 6
Ru
0.92 MPa
2
0.9(1000)(144)
19.66(10) 6
Ru
1.05MPa
2
0.9(1000)(144)
30.58(10) 6
L2 17.2(4) 2
1.635MPa
M C Wu
30.58kN .m Ru
2
0.9(1000)(144)
9
9
fy
415
m
23.58
0.85 fc' 0.85(20.7)
pmin
1.4 1.4
0.00337
f y 415
1
2(23.58)(0.92)
pA
(1 1
) 0.00228use0.00337
23.58
415
1
2(23.58)(1.05)
pB
(1 1
) 0.00261use0.00337
23.58
415
pC
1
2(23.58)(1.635)
(1 1
) 0.00415
23.58
415
AsA AsB pmin bd 0.00337(1000)144 485.28mm 2
Using 12 mm bars
(12) 2
A1
4 (1000) 233say 230mm 450mm
S
1000
As
485.28
AsC pc bd 0.00415(1000)144 597.6mm 2
Using 12 mm bars
(12) 2
A
4 (1000) 189 say180mm 450mm
S 1 1000
As
597.6
Temperature bars
At 0.002bt 0.002(1000)170 340mm 2
Using 10 mm bars
(10) 2
A1
4 (1000) 230.9 say 230mm 450mm
S t 1000
At
340
170mm
C
B
A
12 mm extra bars at 230 mm 0.C
10 mm temperature
bars at 230 mm 0.C
12 mm continuous bent bars at 230 mm 0.C
Plate # 7: Design of one way slab
Design a one way cantilever slab of 2.0 m span . The slab is to
carry a uniform dead load of 2.4 KPa and uniform live load of 3.6 kPa.
fc’ = 27.6 MPa , fy = 415 MPa for main bars and temperature bars.
Concrete weighs 22.56 kN/m3. Draw layout of bars.
Design a 4.5 m long one way slab with one end discontinuous as shown
in
the figure . The slab is to carry a uniform dead load of 4.2 KPa
and uniform live load of 4.5 kPa. fc’ = 20.7 MPa , fy = 345 MPa for
main bars and fy = 276 MPa for temperature bars. Concrete weighs
23.5 kN/m3. Draw layout of bars.
column
4.5m
column
4.5m
column
Prelim Exam
A rectangular beam has b =300 mm, d = 500 mm, As = 6 of 32 mm,
fc’ =27.6 MPa,fy =414 MPa. If the beam is simply supported on a span of 6 m,
determine the concentrated live load that could be applied at the third points on
the beam if steel covering is 80 mm and concrete weighs 23.5 kN/m3.
A doubly reinforced rectangular concrete beam has b =350 mm,d =600mm,
fc’=27.5 MPa, fy = 345 MPa, As =3625 mm2, As’ = 775mm2 ,covering for tension
and compression bars 80 mm and 63 mm respectively. If the beam is an interior
span of a three span continuous beam supporting a service dead load of 20 kN/m
(weight included) determine the maximum uniformly distributed live load it can
support on an average clear span of 5.0 m. Use NSCP moment coeffecients.
400 mm
3 of 16 mm
75 mm
400 mm
Determine the ultimate moment capacity
of the triangular beam shown in figure.
fc’ =20.7 MPa, fy = 345 MPa.
Solution to #2
As1 As As ' 3625 775 2850mm 2
As1 f y
2850(345)
a
120.18mm
0.85 fc' b 0.85(27.5)350
600(a 1d ' ) 600(120.18 [0.85(63)])
f
332.65MPa f y
a
120.18
'
s
Compression steel does not yields at failure
∑Fx =0
0.85fc’ab + As’fs’ = Asfy
0.85(27.5)350a + 775fs’ = 3625(345)
10.56a +fs’ = 1613.7
fs’ =1613.7 – 10.56 a
EQ.1
'
600
(
a
d
600(a [0.85(63)])
'
'
1 )
f
fs
s 334.7 MPa f y
a
a
a
Mu1 As1 f y (d )
600(a 53.55)
'
2
fs
EQ.2
121.12
a
0.9( 2850)345)(600
)
EQ.1 EQ.2
600(a 53.55)
1613.7 10.56a
a
1613.7 a 10.56a 2 600a 32130
Mu1
10 6
Mu1 477.36kN .m
2
Mu2 As ' fs ' (d d ' )
0.9(775)334.7(600 63)
10 6
Mu2 125.36kN .m
Mu2
10.56a 2 1013.7 a 32130 0
a 2 96a 3042.6 0
M u Mu1 Mu2 602.72kN .m
0.85( 27.5)334.7(600)
96 (96) 2 4(3042.6)
p
0.0365
a
121.12mm b
(600 345)345
2
' fs '
2
A
0
.
75
p
bd
A
6500
.
6
mm
s max
b
s
600(121.12 53.55)
'
fy
fs
121.12
#1
p=0.03217
pmax = 0.021 tension steel does not yield at failure
a = 260.22 mm
fs =379.92 MPa
Mu = 610.31 kN.m
WD = 4.1 kN/m
P = 171.9 kN
#2
a = 121.12 mm compression steel does not yield at failure
fs’=334.7 MPa
Mu = 602.72 kN.m
WL = 125.34 kN.m
0.003
a
Ac
400mm
x
C=0.85fc’Ac
375 - 2a/3
3 of 16 mm
75 mm
T=Asfy
c
375 - c
s
400 mm
Solution
3 (16) 2
As
603.2mm 2
4
Assume that steel yields at failure( subject to checking)
C=T 0.85fc’Ac=Asfy
0.85(20.7)Ac =603.2(345)
Ac =11,827.45 mm2
By similar triangles
x
400
x 0.842a EQ.1
a
475
1
1
a 167.6mm
11827
.
45
a (0.842a)
Ac ax
2
2
a 167.6
c
197.17 mm
1 0.85
From the strain diagram
s
fs
400 c
s
2000000
0.003
c
fs
400 197.17
(200000)0.003
197.17
f s 617.22MPa f y
2a
M u As f y (375 )
3
tension steel yields at failure
0.9(603.2)345(400
10 6
[2]167.6
)
3
53.91kN .m
Shear and Diagonal Tension
Another type of beam failure other than bending is shear failure.
Shear failure are very dangerous if it happens before flexure failure
because they can occur without warning
BASIC CODE REQUIREMENT
Factored shear strength Vu shall be equal or less than design shear ФVn
Vu Vn
where:
V n Vc V s
Vc = shear carried by concrete
Vs = shear carried by the stirrups
Vu = factored shear strength
Shear Strength provided by concrete
Shear strength provided by concrete subject to shear and
flexure only;
1
Vc
6
fc'bw d
or in more detailed calculation
Vc
Vu d
fc' 120 p w
Mu
where :
7
bw d 0.3 fc'bw d
Vu d
1. 0
Mu
fc' is in MPa and shall not exceed 0.7 MPa
bw = width of the beam web for T-beams,
width of the beam for rectangular beams
d =effective depth of the beam
pw = As/bwd
Spacing limits of shear reinforcement
Spacing S of shear reinforcement placed perpendicular to
the axis of the member shall not exceed d/2 for
nonprestressed members and 3/4 h for prestressed
members, nor 600mm. When V s 0.33 fc 'bw d
maximum spacing given by the above limits shall be reduced by
one half.
Minimum shear reinforcement
Vc , minimum area of shear reinforcement
When
Vu
2
shall be provided in all reinforced concrete flexural members
except in the following conditions:
a)slabs and footings
b)concrete joist construction
c)beams with total depth not greater than 250 mm,2.5 times
flange thickness or half the width of the web whichever is
greatest.
Where shear reinforcement is required, the minimum area of
shear reinforcement shall be computed by:
Where:
bw S
Av
3fy
Av =cross sectional area of the stirrups
taken twice for u-shaped stirrups
Shear strength provided by reinforcement
a) When shear reinforcement perpendicular to
the axis of
the member is used
Av f y d
2
Vs
S
3
fc'bw d
b) When inclined stirrups are used as shear reinforcement
Vs
Av f y d (sin cos )
S
2
3
fc 'bw d
CRITICAL SECTION FOR BEAM SHEAR
Maximum factored shear force Vu maybe computed in accordance
with the following provided that:
a) the support reaction ,in the direction of the applied shear
introduces compression on the end regions of member
no concentrated load occur between the face of the support and
location of the critical section.
b) For non prestressed members,sections located less than a
distance of d from the face of the support maybe designed for the
same shear Vu as that computed at a distance of d
Size of stirrups
Main bars smaller than or equal to 32 mm diameter: 10 mm
Main bars greater than 32 mm diameter
: 12 mm
Shear carried by stirrups
Vs
Vu
Vc
Problems:
Determine the minimum cross section required for a rectangular
beam to satisfy the condition that web reinforcement be neglected
Vu = 72 kN,fc’ =27.6MPa. Assume d = 1.6b.
Vu
Vu
Vu
Vc
2
fc 'bd
( 6) 2
b
72000(12)
0.86 27.6 (1.6)
b 350mm
d (1.6)350 560mm
fc 'b(1.6b)
(6) 2
0.85 27.6 (1.6b 2 )
72(1000)
(6)2
A rectangular beam with b = 270 mm, d = 500 mm is provided with 10
mm vertical stirrups with fy = 276 MPa. Assuming fc’ = 21 MPa;
a) Determine the required spacing if Vu = 40 kN
b) Determine the required spacing if Vu = 92 kN
c) Determine the required spacing if Vu = 236 kN
d) Determine the required spacing if Vu = 473 kN
Solution
Shear carried by concrete
1
Vc
6
fc'bw d
1
Vc
21(270)500 103,108 N
6
Vc 0.85(103108)
43821N
2
2
a)
b)
Vc
Stirrups not necessary
Vu 42000 N
2
Vu 92000 N
Vu
Vs
Vc
92000
Vs
103108 5127.3N
0.85
2 (10) 2
Av
157.1mm 2
4
S
Av f y d
Vs
157.1(276)500
4228.3mm
5127.3
1
1
fc 'bw d
21(270)(500) 206216 N
3
3
1
Vs
fc 'bw d
3
d 500
S
250mm Use S = 250 mm
2
2
c)
Vu 236000 N
Vu
Vs
Vc
236000
Vs
103108 174539 N
0.85
S
Av f y d
Vs
157.1( 276)500
124mmsay120mm
174539
1
fc 'bw d 206216 N
3
1
Vs
fc 'bw d
3
d 500
S
250mm
2
2
2
3
fc 'bw d 412432 N
2
Vs
3
Beam size is inadequate for shear
Use S = 120 mm
d)
fc 'bw d
Vu 473000 N
Vu
Vs
Vc
473000
Vs
103108 453362.5 N
0.85
1
fc 'bw d 206216 N
3
Design of vertical stirrups
Given :bw or b,d,fc’,fy, beam loading & span,
Required: size and spacing of stirrups
General Procedure
1.Calculate factored shear force VU at the critical section.
2.Calculate shear strength of concrete:
1
Vc
6
fc'bw d
Vc
Vu
2
Vc
Vu
2
provide stirrups
stirrups not necessary
Vc
Vu
2
Assuming that
3. Calculate the shear strength provided by the stirrups
Vu
V s Vc
Note: if
2
Vs
3
fc'bw d
adjust the beam size
4. Calculate the required spacing of stirrups
Spacing is the smallest of:
a)
Av f y d
S
Vs
Calculate
1
3
fc 'bw d
b)
S = d/2
when
c)
S = d/4
when
1
fc'bwd
3
1
Vs
fc'bw d
3
Vs
5. Check for minimum required area of stirrups
Av
bw S
3fy
Note: Av must be less than or equal to
the actual area of Stirrups
Problem:
A simply supported reinforced concrete beam 230 mm wide with an effective
depth of 500 mm has a span of 6m. The beam carries a dead load of 9
kN/m ( including its own weight) and live load of 18 kN/m applied throughtout
its entire span. Determine the required spacing of 10 mm stirrups. fc’ = 28
MPa ,fy= 345 MPa.
1
Vc
6
1
fc'bw d
28 (230)500 101420 N
6
Wu =1.4(9)+1.7(18)=43.2 kN/m
wu
6m
R
R =Wu(3)=43.2(3)=129.6kN
43.2 kN/m
Shear force at crtical section
VU
Vu = 129.6-0.5(43.2)=108kN =108000N
0.5
R=129.6kN
Vc
Vc 0.85(101420)
Vu
43103.5 N
2
2
2
Vu
108000
Vs
Vc
101420
0.85
2 (10) 2
2
A
157
.
1
mm
Vs 25639 N
v
4
S
Av f y d
Vs
157.1(345)500
1056mm
25639
1
fc 'bw d 202840 N
3
1
Vs
fc 'bw d
3
d 500
S
250mm
2
2
Use
S 250mm
Minimum required area
bw S
Av
3fy
230(250)
Av
55.56mm 2 157.1mm 2
3(345)
Use 10 mm u shaped stirrups spaced at 250 mm on centers
PLATE # 7 : SHEAR
A simply supported reinforced concrete beam 250 mm wide with an
effective depth of 600 mm has a span of 7.5m. The beam carries a
dead load of 12 kN/m ( including its own weight) and live load of 24
kN/m applied throughtout its entire span. Determine the required
spacing of 10 mm stirrups. fc’ = 28 MPa ,fy= 345 MPa.
A rectangular beam with b = 300 mm, d = 550 mm is provided with 10
mm vertical stirrups with fy = 276 MPa. Assuming fc’ = 21 MPa;
a) Determine the required spacing if Vu = 50 kN
b) Determine the required spacing if Vu = 220 kN
c) Determine the required spacing if Vu = 360 kN
d) Determine the required spacing if Vu = 500kN
Design of beams for bending shear and deflection
General procedure
1.
2.
3.
Design section and reinforcement by bending
Design stirrups by shear
Check adequacy of design by deflection
Problem :
Design a rectangular beam for a 6 m simple span to support a
uniform dead load of 18 kN/m(weight included) and uniform live load
of 12 kN/m applied along it’s entire length. fc’=20 MPa, fy = 345 MPa
for main bars and stirrups,Wc=23.5 kN/m3 ,p=0.6pmax Consider
immediate deflection due to live load only with an allowable of 1/360 of
span length. fr=3.1 MPa n=10,Ec=20000 MPa. Use 100 mm covering.
Solution
Wu 1.4(18) 1.7(12) 45.6kN / m
Wu L2 45.6(6) 2
Mu
205.2kN .m
8
8
0.85 fc ' 1 600 0.85(20)0.85(600)
pb
0.02659
(600 f y ) f y
(600 345)345
p (0.6)(0.75 pb ) 0.6(0.75)0.02659 0.012
fy
345
m
20.29
0.85 fc' 0.85(20)
Ru pf y (1
pm
0.012[20.29]
) 0.012(345)(1
) 3.635MPa
2
2
Try b = 250 mm
Mu
202.5(10) 6
d
500mm
Ru b
0.9(3.635)250
Mu
202.5(10) 6
Ru
3 .6
2
2
bd
0.9( 250)(500)
1
2mRu
1
2(20.29)3.6
p (1 1
)
(1 1
) 0.01186
m
fy
20.29
345
As pbd 0.01186 (250)500 1,482.5mm 2
1482.5
N
5 pcs
2
(25)
4
1
Vc
6
1
fc'bw d
20 (250)500 93169 N
6
R =Wu(3)=45.6(3)=136.8kN
wu
6m
R
45.6 kN/m
Shear force at critical section
VU
0.5
R=136.8
Vu = 136.8- 0.5(45.6)=114kN =114000N
Vc
Vc 0.85(93169)
Vu
39596.8 N
2
2
2
Vu
114000
Vs
Vc
93169
0.85
Vs 40949 N
Using 10 mm u shape stirrups
2 (10) 2
Av
157.1mm 2
4
S
1
3
Av f y d
Vs
157.1(345)500
661mm
40949
fc 'bw d 186338 N
d 500
S
250mm
2
2
Use
S 250mm
Minimum required area
bw S
Av
3fy
230(250)
Av
55.56mm 2 157.1mm 2
3(345)
Gross moment of inertia
bh 3 250(600) 3
Ig
4500 x106 mm 4
12
12
M cr
fr I g
yt
3.1(4500)10 6
46.5 x10 6 N .mm 46.5kN .m
300
Transformed section
b
b=250
x
d=500
500-x
As
nAs
100
5 (20) 2
As
1570mm 2
4
x2
b
nAs (d x)
2
x 2 62800 125.6 x
x2
250
10(1570)(500 x)
2
x 2 125.6 x 62800 0
125.6 (125.6) 2 4(62800)
x
195.6mm
2
3
bx 3
250
(
195
.
6
)
I cr
nAs (d x) 2
10(1570)(500 195.6) 2
3
3
I cr 2,078.3x106 mm 4
a)
WL L2 (12)(6) 2
Ma
54kN .m
8
8
Effective moment of inertia
M
M cr
cr
I cr
Ie
Ig 1
M a
Ma
3
3
46.5
46.5
6
6
Ie
4500
x
10
1
2078
.
3
x
10
54
54
3
3
I e 3,624.6 x106 mm 4
Instantaneous deflection due to live load
5WL L4
5(12)(6000) 4
2.79mm
6
384 Ec I e 384(20000)3624.6(10)
allowable deflection due to live load
allowable
L
6000
16.67mm 2.79mm 3
360 360
Section is adequate
Plate #8:Design of beams for bending shear and deflection
Make a complete design of a rectangular beam reinforced for tension
only for a 7.5 m simple span to support a uniform dead load of
24 kN/m(weight included) and uniform live load of 18 kN/m applied
along it’s entire length. fc’=20 MPa, fy = 300 MPa for main bars and
stirrups,Wc=23.5 kN/m3 ,p=0.18(fc’/fy). Consider immediate deflection
0.7 fcof' span length.
due to total load with an allowablef rof1/360
Es = 200000MPa,Ec=18500 MPa,
. Use 100 mm covering.
BOND and DEVELOPMENT LENGTH
Bond
In reinforced concrete, concrete and steel act as a unit. For this to
happen, there must be absolutely no slippage of the bars in relation
to the surrounding concrete. The steel and concrete must stick or
bond together so that there will be transfer of stress from steel to
concrete and vice-versa. Failure of transfer of stress makes the
concrete an unreinforced member thus it will be subject to collapse.
Development Length
Bar development length is the embedment necessary to assure that
the bar can be stressed to its yield point with some reserved to
ensure member toughness.
Basic concept of development length
2
d b Ld d b f y
4
fy
Ld
db
4
: μ = average bond stress
db
F
Ld
ΣF = 0
F=T
T = Abfy
Where
Ld = minimum development length
The code however provides the basic development length
Ldb for various conditions. The values provided are modified
for different conditions. Thus the minimum development length
provided by the code is;
Ld = Ldb(applicable modification factors) ≥ 300 mm
Basic Development Length of bars in tension
For 32 mm bar & smaller and deformed
wire
Ldb
0.02 Ab f y
fc '
For 36 mm bar
Ldb
25 f y
fc '
For deformed wire
Ldb
3d b f y
8
but not to be taken less than 0.6dbfy
Where
Ldb = basic development length ( mm)
Ab = area of one bar (mm2)
db = diameter of one bar (mm)
fc’,fy in MPa
fc '
Basic Development Length of bars in compression
Ldb
0.24d b f y
fc'
but not to be taken less than 0.04dbfy
Modification Factors for bars in tension
Condition
Modification
Factor,m
a) For bars in beams or column with a minimum cover not less than specified by the code
1.0
b)
1.0
For bars in beams or column with transverse reinforcement satisfying the requirement of the code
c) Bars in beams or column with a clear spacing not less than 3db
1.0
d) Bar in the inner layer of slab or wall reinforcement and with a clear spacing not less than 3db
1.0
e) Any bars with cover of not less than 2db and with a clear
1.0
spacing not less than 3db
f) For bars with cover of db or less with a clear spacing of 2db or less
2.0
g) For bars not included in items a to f
1.4
h) For 32 mm bar and smaller with clear spacing not less than 5db and with cover from face of the
member to edge bar, measured in the plane of the bar, not less than 2.5db, the factors from items
a to g may be multiplied by 0.8
0.8
i) Top reinforcement
1.3
j) Lightweight aggregate concrete
K)Lightweight aggregate concrete when fct is specified
l) For reinforcement enclosed within special reinforcement not less than 6 mm diameter and not
more than 100 mm pitch, within 12 mm or larger circular ties spaced at not more than
100 mm on center or larger ties or stirrups spaced not more than 100 mm on center and arranged
such that alternate bars shall have supported by the corner of a tie hoop with an included angle
not more than 1350,the factors in items a
through g maybe multiplied by 1.8
m) Excess Reinforcement. Development length maybe reduced where reinforcement in a flexural
member is more than required by analysis
fc1.3
'
1.8 f ct
1.8
As required
As provided
Modification Factors for bars in Compression
Condition
Modification
Factor,m
As required
a)
Excess reinforcement. Reinforcement more than
that required by analysis
b) Spiral and Ties. Reinforcement enclosed within
spiral reinforcement not less than 6 mm diameter
and not more than 100 mm pitch or within 10 mm
ties and spaced not more than 100 mm on center.
As provided
0.75
Problems
A rectangular beam 200 mm wide and 400 mm deep is reinforced with 3
of 22 mm tension top bars with fc’ = 20.7 MPa and fy = 275 MPa.
Calculate the required development length.
Solution
Ldb
Ldb
0.02 Ab f y
fc '
0.02
( 22) 2
4
20.7
275
460mm
Ldb 0.06d b f y
Ldb 0.06( 22)(275) 363mm
Use
Modification Factor
Top bar = 1.3
Required development length
Ld 1.3Ldb 1.3(460) 598mm
Ldb 460mm
A rectangular beam 250 mm wide and 500 mm deep is reinforced with 4 of 25 mm
with fc’ = 27 MPa and fy = 345 MPa. If the member is made up of lightweight
aggregate with fct = 2.88 MPa ,Calculate the required development length.
Solution
Ldb
Ldb
Modification Factor
Lightweight concrete
With specified fct
0.02 Ab f y
fc '
0.02
( 25) 2
4
27
345
652mm
Ldb 0.06d b f y
Ldb 0.06( 25)(345) 518mm
Use
Ldb 652mm
fc'
27
1.00
1.8 f ct 1.8(2.88)
Ld 1.0 Ldb
Ld 1.0(652)
Ld 652mm
A cantilever beam 320 mm wide and 500 mm deep is reinforced with 3 of 36 mm
straight top bars with fc’ = 27 MPa and fy = 345 MPa. Calculate required
development length.
Solution
Ldb
Ldb
25 f y
fc '
25(345)
27
1660mm
Modification Factor
Top bar = 1.3
Required development length
Ld 1.3Ldb 1.3(1660) 2158mm
A rectangular beam 250 mm wide and 410 mm deep is reinforced with
3 of 20 mm compression bars with fc’ = 20.7 MPa and fy = 275 MPa.
Calculate the required development length.
Ldb
0.24d b f y
fc'
0.24(20)275
Ldb
254mm
20.7
Ldb 0.04d b f y
Ldb 0.04( 20)(275) 220mm
Use
Ldb 254mm
No applicable modification factor
Ld 254mm
Plate # 9 Development Length
A rectangular beam 300 mm wide and 450 mm deep is reinforced with 4 of 25 mm
tension top bars with fc’ = 20.7 MPa and fy = 345 MPa. Calculate the required
development length.
A rectangular beam 200 mm wide and 350 mm deep is reinforced with 3 of 20 mm
with fc’ = 30 MPa and fy = 415 MPa. If the member is made up of lightweight
aggregate with fct = 2.4 MPa ,Calculate the required development length.
A cantilever beam 300 mm wide and 450 mm deep is reinforced with 6 of 22 mm
straight top bars with fc’ = 20.7 MPa and fy = 415 MPa. Calculate required
development length.
A rectangular beam 350 mm wide and 600 mm deep is reinforced with 4 of 32 mm
compression bars with fc’ = 20.7 MPa and fy = 275 MPa. Calculate the required
development length.
Development Length of Flexural Reinforcement
Tension reinforcement in flexural members maybe developed by :
a) bending across the web to be anchored
b) made continues with reinforcement on opposite face ofthe member
Critical points for development length in flexural members are at points
of maximum stress and at points where the adjacent reinforcements
terminates or is bent. Reinforcement shall extend beyond the point at
which it isno longer required to resist flexure for a distance equal to the
effective depth of member or 12db whichever is greater, except at
supports of simple beam and free end of cantilevers. Continuing
reinforcement shall have an embedment length not less than the
development length beyond the point where bent or terminated tension
reinforcement is no longer required to resist flexure
Development of positive moment bars
The code stipulates that at least one third the positive
reinforcement in simple members and one fourth the positive
reinforcement in continuous members shall extend along the
same face in the member into the supports. In beams such
reinforcement shall extend into the support at least 150mm.
At simple supports and at points of inflection,positive
moment tension reinforcement shall be limited to a diameter
such that Ld computed by
Mn
Ld
la
Vu
Where:
Mn = nominal moment strength assuming that all reinforcement
at section are stressed to specified yield strength f y
Vu = factored shear force at supports for simple beams and at
points of inflection for continuous beams
la = end anchorage ;at support shall be the embedment length beyond
the center of support; at point of inflection limited to the effective depth
or 12db whichever is greater.
Value of Mn/Vu maybe increased by 30% when the ends of the
reinforcement are confined by a compressive reaction such as a
column below but not when the beam frame into the girder.
Mn
Ld 1.3
l a
Vu
When
Where:
Mn
Ld 1.3
la
Vu
or
a
M n As f y (d )
2
Mn
Ld
la
Vu
a
As f y
0.85 fc' b
Use smaller bar
Size or increase
End anchorage la
Max Ld
la
1.3Mn/Vu
at least 1/3 of positive reinforcement
VU
Max Ld
tension bars
Mn/Vu
la
Max Ld
la is the larger
Value of d or 12db
la
Mn/Vu
tension bars
150 mm min
at least ¼ of positive moment reinforcement
.
CL
Point of inflection
Development length for Negative moment Reinforcement
Negative moment reinforcement should have an embedment length into the span to
develop the calculated tension in the bar,or a length equal to the effective depth of the
member or 12db whichever is the greatest. At least one third of the total negative
reinforcement should have an embedment length beyond the point of inflection not less
than the effective depth of the member or 12 db or 1/16 of the clear span whichever is
greatest.
Ld
`
Development length for Negative moment
the larger value of 12db,d or Ln/16
at least 1/3 of the total Negative moment reinforcement
Clear span =Ln
Point of inflection
Recommended bar details for continuous beams
Ln1/4
0.3 Ln1
Ln1/8
150 mm min
0.3 Ln2
Ln1/8
Ln1
150 mm min
Ln2/8
Ln2
Hooks
If sufficient space is not available to anchor tension bars by running
them straight for the required development length as required by the
code, hooks maybe used.
12db
r
db
r
90 hook
D
0
bend diameter,D =2r
D =6db for 10 mm through 25 mm bars
D =8db for 28 mm through 32 mm bars
D =10db for 36 mm bars
4db
1800 hook
STANDARD HOOKS
Ldb
65 mm min
12db
4db
4db for 10 mm through 25 mm bars
5db for 28 mm through 32 mm bars
6db for 36 mm bar
Development of standard hooks
Basic Development Length of standard hooks
l hb
100d b
fc '
Actual Development Length of standard hooks
ldh = lhb(applicable modification factors) ≥8db nor 150 mm
Modification Factors
1. If the reinforcing bar has an fy other 415 MPa, lhb is to be multiplied by fy/415
2. When 900 hooks and 32 mm or smaller bar are used and when 60 mm or
more of side cover normal to the hook is present, together with at least 50 mm
cover for the bar extension, lhb is to be multiplied by 0.7.
3. When hooks made of 32 mm and smaller are enclosed vertically and
horizontally within ties or stirrups ties spaced no farther apart than 3db, lhb is to be
multiplied by 0.8
4. Where the amount of flexural reinforcement exceeds the theoretical amount
required and where the specifications being used do not specifically require that
development lengths be based on fy the value of lhb is to be multiplied by
Asrequired/As provided.
5. When light weight concrete are used, apply a modification factor of 1.3
6. For bars being developed by standard hook at discontinuous end with side
cover and top or bottom cover over hook less than 60 mm,hooked bar shall be
enclosed within ties or stirrups spaced along the full development length ldh not
greater than 3db where db is the diameter of hooked bar. For this case, the
factor mentioned in item 3 shall not apply.
Splices of Reinforcement
Splicing maybe done by welding, by mechanical connections or most
frequently by lapping bars. Lapped bars are usually tied in contact.
Lap splice must not be used for bars larger than 32 mm.
Splices in tension
The minimum length of lap for tension lap splice shall be as required for
class A or class B, but shall not be less than 300 mm, where
Class A splice
1.0Ld
Class B splice
1.3Ld
Lap splices of deformed bars and wires in tension shall be class B splice
except that class A splice are allowed when
•the area of the reinforcement provided is at least twice than that
required by analysis.
•One half or less of the total reinforcement is spliced within the required
lap strength.
Splices of deformed bars in compression
Compression bars maybe spliced by lapping,end bearing, welding or
mechanical devices. The minimum length of such bars should be the
development length Ld but may not be less than 0.07d bfy for fy of 415
MPa or less, or (0.13fy -24)db for fy greater than 415 MPa.
Problems
A simply supported beam is reinforced with three of 28 mm
bars with fc’= 27.6 MPa and fy = 275 MPa. Assuming that side, bottom and top
cover to be greater than 60 mm, determine the following:
a) the required development length if a 900 hook is used
b) the required development length if a 1800 hook is used
Solution
Using a 900 hook
l hb
lhb
ldh
ldh
100d b
fc'
100(28)
533mm
27.6
Modification factor for fy other than 415 MPa
m
fy
415
275
0.6626
415
Modification factor for 900 hook = 0.7
Required development length
Ldh =0.6626(533)0.7=247.2 say 250 mm
ldh
Solution
Using a 1800 hook
l hb
lhb
100d b
fc'
100(28)
533mm
27.6
Modification factor for fy other than 415 MPa
m
fy
415
275
0.6626
415
Required development length
Ldh =0.6626(533)=353.17 say 355 mm
Problem
For the simply supported beam shown in figure below, investigate whether the
bars size is satisfactory for the required development length. The beam is
reinforced with 4 of 25 mm bars . fc’ = 20.7 MPa and fy = 414 MPa, Vu = 270 kN.
The beam is made up of normal sand concrete and the reaction produces
compression on concrete.
300mm
600 mm
Ldb 0.06d b f y
Solution
175 mm
Basic development length
Ldb
Ldb
Ldb 0.06( 25)(414) 621mm
Use
0.02 Ab f y
0.02
Since there is no applicable
modification factor
fc '
( 25) 2
4
20.7
Ldb 894mm
414
894mm
Ld 894mm
a
As f y
425.14 x106
1.3
175 2221.2mm
270(1000)
0.85 fc' b
(25) 2
4
414
4
a
154mm
0.85(20.7)300
a
M n As f y (d )
2
(25) 2
154
Mn 4
414(600
)
4
2
M n 425.14 x106 N .mm
Mn
1.3
la
Vu
894mm 2221.2mm
Mn
Ld 1.3
la
Vu
bars are adequate
A rectangular beam has b = 380 mm and d = 500 mm. The beam is simply
supported over a length of 6 m measured from the center of the support and is
reinforced for tension with 6 of 25 mm diameter bars. Assuming fc’ = 21 MPa,
fy = 276 MPa, draw the details of the bar showing the cut off points of each bar.
The beam carries a uniformly distributed load throughout its length and is made up
of normal sand- gravel concrete. The reactions at the ends produces compression
on concrete. Maximum value of la is 150 mm. Vu = 210 kN at support.
Ldb 0.06d b f y
Solution
Basic development length
Ldb
Ldb
Ldb 0.06( 25)(276) 414mm
Use
0.02 Ab f y
0.02
Since there is no applicable
modification factor
fc '
( 25) 2
4
21
Ldb 591mm
276
591mm
Ld 591mm
BAR LAYOUT & CUT OFF POINTS
L3
L1
2 of 25
L2
CL
x2
x1
3m
y
2y 3
3
y
Parabolic
Moment
diagram
Location of cut off points
By square property of parabola
2y
x12
3
32
y
x1 2.45m
y
x22 3
2
3
y
x2 1.73m
Available development length
Reinforcement shall extend beyond the point at which it is no longer required to resist
flexure for a distance equal to the effective depth of member or 12db whichever is greater
12(25) 300mm
d 500mm
Use extension = 500 mm
L1 x1 500 2450 500 2950mm Ld
L2 x2 500 1730 500 2230mm Ld
No need for hook
The code stipulates that at least one third the positive reinforcement (2 bars) in
simple members shall extend along the same face in the member into the supports
For the last two bars the code requires that the be bars be limited to a diameter so that
Ld 1.3
Mn
l a
Vu
2 (25) 2
AS
981.75mm 2
4
As f y
a
0.85 fc ' b
981.75(276)
a
40mm
0.85(21)380
a
M n As f y (d )
2
130.1x106
1.3
150 955mm Ld
210(1000)
L3 3000 150 3150mm Ld
40
M n 981.75(276)(500 )
2
M n 130.1x106 N .mm
2230
2950
2230
2950
3150
3150
DETAILS
Problem
Figure shows the first interior support of a continuous beam reinforced with 6 of 25
mm bars to resist a negative moment of 496 kN.m for which the calculated steel area
required is 2600 mm2. Determine and draw the details of the lengths of the bars
required if normal gravel-sand concrete is used. fy =414 MPa ,fc’ = 20.7 MPa,
b = 350mm,d= 600 mm.
7m
10 m
2.4 m
2.1 m
350mm 320mm
Mu for 4 bars
point of inflection
496 kN.m
point of inflection
7m
10 m
Bar 3
Bar 2 Bar 1
Ld1
L1
Ld2
L2
L3
L4
2.4 m
2.1 m
Mu for 2 bars
350mm 320mm
Mu for 4 bars
point of inflection
496 kN.m
(Mu for 6 bars)
point of inflection
Solution
Basic development length
Ldb
Ldb
0.02 Ab f y
0.02
Ldb 0.06d b f y
fc'
( 25) 2
4
20.7
For bar 1
Ldb 0.06( 25)(414) 621mm
414
893mm Use Ldb 893mm
(As at section = 2 of 25 mm)
At least one third of the total negative reinforcement should have an embedment
length beyond the point of inflection not less than the effective depth of the
member or 12 db or 1/16 of the clear span whichever is greatest.
1/3 of 6 = 2 bars
12db = 12(25)=300 mm
d=600 mm
1/16 of 7000=438 mm
1/16 of 10000=625mm
7m span
L3 2400 600 3000mm
10m span
L4 2100 625 2725mm
for 7 m span use 600 mm
for 10 m span use 625 mm
For bar 2
(As at section = 4 of 25 mm)
Negative moment reinforcement should have an embedment length into the
span to develop the calculated tension in the bar,or a length equal to the
effective depth of the member or 12db whichever is the greatest
12db = 12(25)=300 mm d=600 mm use 600 mm
L1 350 600 950mm
For bar 3
As provided
L2 320 600 920mm
(As at section = 6 of 25 mm)
Modification factor
6 ( 25) 2
4
2945.24
Asrequired 2600mm 2
m
Asrequired
As provided
2600
0.88
2945.24
Ld1 0.88(893) 785mm
Ld 2 0.88(893) 785mm
Plate # 10. BAR DETAILING
A rectangular beam has b = 350 mm and d = 450 mm. The beam is simply
supported over a length of 5 m measured from the center of the support and is
reinforced for tension with 7 of 20 mm diameter bars. Assuming fc’ = 27 MPa,
fy = 345 MPa, draw the details of the bar showing the cut off points of each bar.
The beam carries a uniformly distributed load throughout its length and is made up
of normal sand- gravel concrete. The reactions at the ends produces compression
on concrete. Maximum value of la is 150 mm. Vu = 150 kN at support.
350 mm
450 mm
Hint : extend 3 bars beyond the support
Figure shows the first interior support of a continuous beam reinforced with 6 of 25
mm bars to resist a negative moment of 450 kN.m for which the calculated steel area
required is 2400 mm2. Determine and draw the details of the lengths of the bars
required if normal gravel-sand concrete is used. fy =414 MPa ,fc’ = 20.7 MPa,
b = 350mm,d= 600 mm.
6m
9m
2.5 m
2.0 m
330mm 300mm
point of inflection
450 kN.m
point of inflection
Axially Loaded Columns
Columns are classified as pedestal, short column and long
column. Pedestal is a column whose height is less than
three times its least lateral dimension. They maybe
designed without reinforcement with a maximum permissible
compressive strength of Φ0.85fc’ where Φ =0.70.
If the column does not qualify as pedestal then it maybe
classified as a short column. They fail due to initial material
failure. The load on the column depends on the dimension
and the strength of the material it is made.
If the length of the column is increased, chances of lateral
buckling increases. Column that fails due to lateral bucking
are classified as long columns.
AXIALLY LOADED SHORT TIED COLUMNS
The axial load capacity of tied columns is given by:
Pu =Ф0.80{0.85fc’(Ag-Ast) + Astfy}
Ф = 0.70
Ag = gross area of the column
Ast = area of reinforcement
Ast
Limits of reinforcement for tied columns
pg
1. Pg ranges from 0.01 to 0.06
Ag
2. The minimum number of longitudinal bars is 4 for bars within
rectangular or circular ties, 3 for bars within triangular bars.
Sizes and spacing of main bars and ties
1. Clear distance between longitudinal bars shall not be less than
1.5dbnor 40 mm.
2. Use 10 mm diameter ties for 32 mm bars or smaller and at least
10 mm in size for 36 mm and bundled longitudinal bars.
Vertical spacing of ties shall be the smallest of the following:
16 times longitudinal bar diameter
48 times tie diameter
least dimension of the column
Axially loaded short spiral columns
The axial load capacity of spiral columns is given by:
Pu =Ф0.85{0.85fc’(Ag-Ast) + Astfy}
Ф = 0.75
Limits of reinforcement for spiral columns
1. Pg ranges from 0.01 to 0.06
2. The minimum number of longitudinal bars is 6
Sizes and spacing of main bars and ties
1. Clear spacing between spirals shall not exceed 75 mm,
nor less than 25 mm
2. For cast in place construction, size of spirals shall not be less
than 10 mm.
3. The percentage of spiral reinforcement is computed by
4a s ( Dc d b )
ps
SDc2
Where:
Dc = diameter of the concrete core
as = cross sectional area of the spiral
S = spacing of the spirals
db = diameter of main bars
4. The minimum spiral percentage is given by
p s min
Ag
0.45 fc' ( 1)
Ac
fy
Where: Ac = area of the concrete core
D
A
4
2
c
c
Clear cover ( min. of 40 mm)
DC D
Clear cover ( min. of 40 mm)
Dc = D – 2 ( clear cover )
Problems
A square tied column 350 mm by 350 mm is reinforced with 6 of 25 mm
bars with fc’ = 20.7 MPa and fy = 345 MPa. Determine the following :
a) Ultimate axial load capacity of the column.
b) spacing of 10 mm lateral ties
Solution
(25) 2
Pu =Ф0.80{0.85fc’(Ag-Ast) + Astfy} Ast 6
2945.25mm 2
4
0.7(0.80){0.85(20.7)[(350)(350) 2945.25] 2945.25(345)}
Pu
1747 kN
1000
Spacing of 10 mm ties
S 16(25) 400mm
S 48(10) 480mm
S 350mm
Use S = 350 mm
Problems
A circular spiral column 400 mm in diameter is reinforced with 8 of
25 mm bars with fc’ = 20.7 MPa and fy = 345 MPa. Determine the
following :
a) ultimate axial load capacity of the column.
b) the spacing of 10 mm spirals assuming clear covering of 40 mm
Solution
Pu =Ф0.85{0.85fc’(Ag-Ast) + Astfy}
(25) 2
Ast 8
3927 mm 2
4
( 400) 2
Ag
125664mm 2
4
0.75(0.85){0.85(20.7)[(125664 3927] 3927(345)}
Pu
2229.2kN
1000
Diameter ot the concrete core
Dc = D – 2 ( clear cover )=400-2(40) = 320 mm
Dc2 (320) 2
Ac
80425mm 2
4
4
0.45 fc' (
ps
fy
Ag
Ac
1)
125664
0.45(20.7)(
1)
80425
0.0152
345
Ds2 (10) 2
As
78.54mm 2
4
4
4as ( Dc d b ) 4(78.54)(320 25)
S
60mm 25mm 75mm
2
2
Ps Dc
0.0152(320)
Use s = 60 mm
Short Composite column
Design strength of composite members
Pu 0.85 0.85 fc ' Ac Ast f y Ass Fy
Where:
Ф= 0.75 for composite members with spiral reinforcement
Ф= 0.70 for other reinforcement
Ast = area of reinforcing bars of strength fy
Ass = area of structural shape of strength Fy
Ac = net concrete area
For evaluation of slenderness effect, radius of gyration of composite
members should not exceed
Ec I g
r
5
E c Ag
5
Es I t
E s At
Structural steel encased concrete core
Steel pipe filled with concrete
t
Dc D
Concrete
Core
t
t min D
fy
8Es
Steel tubing filled with concrete
t2
Concrete
Core
b2
t1
b1
t1min b1
fy
3E s
and
t 2 min b2
fy
3E s
Calculate the maximum axial load that the concrete-filled pipe shown can
resist. fc’ = 25 MPa, Fy =248 MPa for steel pipe. Check if the section
complies with the code.
10 mm
Concrete
core
300 mm
[(300) 2 (280) 2 ]
Ap
9111mm 2
4
[(280) 2 ]
Ac
61575mm 2
4
Pu 0.85 0.85 fc' Ac Ap Fy
Pu (0.7)0.85 0.85(25)(61575) 9111(248) 2122958 N
Minimum required thickness of pipe
t min
fy
248
D
300
3.74mm 10mm
8Es
8(200000)
Calculate the ultimate axial load capacity of the composite column shown
below. fc’ = 21 MPa, Fy = 248 MPa
500 mm
Properties of W 14 x 210
Area = 40000 mm2
Depth = 400 mm
500 mm
Ac 500(500) 40000 210000mm 2
W14 x 210
Pu 0.85 0.85 fc' Ac AWF Fy
Pu (0.7)0.85 0.85(21)(210000) 40000(248) 8132757 N
Calculate the ultimate axial load capacity of the composite column shown
below .fc’=21 MPa, Fy=248 MPa,fy =276MPa.
Reinforcing bars consist of 8 of 20 mm diameter bars.
500 mm
500 mm
Properties of W 14 x 210
Area = 40000 mm2
Depth = 400 mm
8 [(20) 2 ]
As
2513mm 2
4
W14 x 210
Ac 500(500) 40000 2513 207487 mm 2
Pu 0.85 0.85 fc ' Ac As f y AWF Fy
Pu (0.7)0.85 0.85(21)(207487) 2513(276) 40000(248)
Pu 8518752 N
Design of axially Loaded Tied Columns
Given :axial load, fc’, fy
Required : Column size, size of main bars, size and spacing
of lateral ties
General Procedure
1.Solve for design axial load Pu
2.Assume pg
0.01 to 0.06
3. Solve for required gross area to get column size
Pu
Ag
(0.7)0.8 0.85 fc ' (1 p g ) p g f y
4. Solve for Ast
Ast = pgAg
5. Using adjusted values of column size and Ast, check
column capacity (optional)
Pu =Ф0.80{0.85fc’(Ag-Ast) + Astfy}
6. Assume size ot ties and solve for the spacing
Use the smallest value from the ff:
• 16 times longitudinal bar diameter
•
48 times tie diameter
c) least dimension of the column
Problem:
Design a square tied column to support an axial dead load
of 600 kN and axial live load of 750 kN. fc’ =20.7 MPa, fy = 345 MPa.
Pu 1.4 PD 1.7 PL 1.4(600) 1.7(750) 2115 kN
Assume pg = 0.02
Pu
Ag
(0.7)0.8 0.85 fc ' (1 p g ) p g f y
2115(1000)
Ag
156435mm 2
(0.7)0.8 0.85( 20.7)(1 0.02) 0.02(345)
t
Ag 156435 395mm
Say 400 mmx 400 mm square column
As p g Ag 0.02(156435) 3128mm 2
Using 20 mm bars
No.
3128( 4)
10 pcs
2
( 20)
Note: No need to check column capacity
Spacing of 10 mm ties
S 16( 20) 320mm
S 48(10) 480mm
S 400mm
Use S = 320 mm
Design of axially Loaded Short Spiral Columns
Given :axial load, fc’, fy
Required : Column size, size of main bars, size and spacing
of spirals
General Procedure
1.Solve for design axial load Pu
2. Assume pg
0.01 to 0.06
3. Solve for required gross area to get column size
Pu
Ag
(0.75)0.85 0.85 fc' (1 p g ) p g f y
4. Solve for Ast
Ast = pgAg
5. Using adjusted values of column size and Ast, check
column capacity (optional)
Pu =Ф0.85{0.85fc’(Ag-Ast) + Astfy}
6. Solve for the diameter of the concrete core
Dc = D – 2 ( clear cover)
7. Solve for the spiral percentage
0.45 fc ' ( Ag Ac )
ps
fy
8. Assume size of spirals and solve for required spacing
4as ( Dc d b )
S
Ps Dc2
where: S ≥ 25 mm, S≤ 75 mm
Problem:
Design a circular spiral column to support an axial dead load of 600 kN and axial
live load of 750 kN. fc’ =20.7 MPa, fy = 345 MPa.
Pu 1.4 PD 1.7 PL 1.4(600) 1.7(750) 2115 kN
Assume pg=0.02
Pu
Ag
(0.75)0.85 0.85 fc ' (1 p g ) p g f y
2115(1000)
Ag
137427 mm 2
(0.75)0.85 0.85(20.7)(1 0.02) 0.02(345)
D 2
137427 mm 2
4
D 420mm
Ast p g Ag 0.02(137427) 2749mm 2
25 mm main vertical bars
Using 25 mm bars
N
2749(4)
6
2
(25)
40
Diameter ot the concrete core
Dc = D – 2 ( clear cover )=420-2(40) = 340 mm
Dc2 (340) 2
Ac
90792mm 2
4
4
0.45 fc' (
ps
fy
Ag
Ac
1)
420
40
10 mm spirals at
65 mm pitch
137427
0.45(20.7)(
1)
90792
0.014
345
Using 10 mm spirals
Ds2 (10) 2
As
78.54mm 2
4
4
S
4as ( Dc d b ) 4(78.54)(340 25)
65mm 25mm 75mm
2
2
Ps Dc
0.014(340)
Use 10 mm spirals at 65 mm pitch
Plate # 11 : Axially loaded short columns
A square tied column 400 mm by 400 mm is reinforced with 8 of 25 mm bars
with fc’ = 20.7 MPa and fy = 345 MPa. Determine the following :
a) Ultimate axial load capacity of the column.
b) spacing of 10 mm lateral ties
A circular spiral column 450 mm in diameter is reinforced with 10 of 25 mm
bars with fc’ = 27 MPa and fy = 415 MPa. Determine the following :
a) ultimate axial load capacity of the column.
b) the spacing of 10 mm spirals assuming clear covering of 40 mm
Calculate the maximum axial load that the concrete-filled pipe shown can resist.
fc’ = 20 MPa, Fy =250 MPa for steel pipe. Check if the section complies with
the code.
12mm
Concrete
core
350 mm
12mm
Calculate the ultimate axial load capacity of the composite column shown
below .fc’=25 MPa, Fy=248 MPa,fy =345MPa. Reinforcing bars consist of 8 of
20 mm diameter bars.
450 mm
450 mm
Properties of W 12 x 150
Area = 30000 mm2
Depth = 300 mm
W12 x 150
Design a circular spiral column to support an axial dead load of 450 kN and axial live
load of 600 kN. fc’ =20.7 MPa, fy = 345 MPa. Use pg =0.025 and 10 mm spirals
Design a square tied column to support an axial dead load of 450 kN and axial
live load of 600 kN. fc’ =20.7 MPa, fy = 345 MPa. Use pg =0.03 and 10 mm ties.
SHORT COLUMNS SUBJECT TO AXIAL LOAD AND
BENDING
( Eccentrically loaded columns)
All columns are subject to axial force and bending and they must be
proportioned to resist these forces. Eccentricities of 0.10h for tied
column and 0.05h for spiral columns can be permitted for axially
loaded columns. Beyond this, analysis for the effect of axial load
and bending must be undertaken.
Columns will tend to bend under the action of moment and produces
compression on one side and tension on the other side. The
following failures are possible under combined axial load and
bending.
Pn
1.
Plastic centroid
e
Pn
Large axial load and negligible
moment. Failure under this condition
occur by sudden crushing of concrete
with all bars reaching their
yeild stress simultaneuosly
2. Large axial load and small moment.
with entire cross section in
compression . Failure under this
Plastic centroid
condition occur by sudden crushing of
concrete with all bars are in
compression
Pn
e
Plastic centroid
3. Large axial load with moment bigger in
2. Bars in the far side in tension has not
yielded. Failure occur by crushing of
concrete
e
Pn
Plastic centroid
e
4. Balance loading condition – bars on the
tensile side yield at the same time that
concrete crushes at 0.85fc’
Pn
Plastic centroid
5. Large moment, small axial load
Failure initiated by tensile yielding
of tension bars
M
Plastic centroid
6. Moment, no axial load - Fails as a
beam
Plastic centroid – the point in the column through which the resultant of
the column load passes through to produce uniform strain at failure. It is
the location of the resultant force of concrete and steel.
In locating this point, all concrete is assumed to be stressed in
compression to 0.85fc’ and all steel bars in compression at (fy – 0.85fc’).
For symmetrical cross sections the plastic centroid coincides with the
centroid of the cross section.
Eccentricity of a column is the distance from the load to the plastic
centroid of the cross section.
eccentricity
Pn
Plastic centroid
centroid of cross section
Plastic centroid
Centroid of cross section
Problem:
The T shaped cross section shown below is reinforced with 4 of 32 mm bars
with fc’= 20.7 MPa and fy = 345 MPa. Determine the location of the plastic
centroid measured from the 450 mm side.
150 mm 200 mm
100 mm
450mm
250 mm
100 mm
75mm
75mm
150 mm 200 mm
100 mm
1
450mm
250 mm
2
100 mm
75mm
75
75mm
P1 P
2
P4
75
275
P
x
P2 0.85 fc' A2 0.85(20.7)200(250) 879.75kN
(32) 2
P3 ( fy 0.85 fc' ) As {345 0.85(20.7)}2
526.63kN
4
(32) 2
P4 ( fy 0.85 fc' ) As {345 0.85(20.7)}2
526.63kN
4
P P1 P2 P3 P4 3120.7kN
P x Px
P x P1 x1 P2 x2 P3 x3 P4 x4
250
P3
P1 0.85 fc' A1 0.85(20.7)150(450) 1187 .63kN
3120.7 x 1187 .63(75) 879.75(250) 526.63(75) 526.63 275
x 158mm
Balanced Loading condition
Balanced loading occur when the tension steel just reaches its yield
strain as concrete is strained to 0.003.
Every column has always
have a loading situation where an ultimate load Pnb placed at eccentricity
eb will produce a moment Mnb. If the eccentricity of the column is less
than eb (e < eb) , the column fails in compression( compression
controlled column); if e > eb the column fails in tension( tension
controlled column)
Balanced loading condition
d-d’
2
Pnb
eb
Determination of Pnb
From stress Diagram
ΣFy =0
Pnb - C1 - C2+ T = 0
d
b
Axis of bending
d – d’
T = Asfy
s
Es
0.85fc’
C1= 0.85fc’abb
εs’
EQ. 1
From strain Diagram
C2 = As’fy
ab
cb
If As = As’
C2 = T
Pnb =C1
Pnb =0.85fc’abb
d’
d-ab/2
fy
Stress Diagram
Strain Diagram
0.003
600 1 d
ab
600 f y
EQ. 2
To determine eb
ΣM @ center of tension steel = 0
Pnb( eb + ½{d-d’}) –C1( d-ab/2) – C2(d –d’)= 0
Pnb( eb + ½{d-d’}) –0.85fc’abb( d-ab/2) – As’fy(d –d’)= 0
'
'
ab As f y (d d ) (d d ' )
eb ( d )
2
Pnb
2
Balanced Moment
EQ. 3
Mnb = Pnbeb
Determination of eb, Pnb and Mb of column with given section and reinforcement.
1. Solve for ab
ab
600 1 d
600 f y
2. Solve for Pnb
Pnb =0.85fc’abb
3. Solve for eb
'
'
ab As f y (d d ) (d d ' )
eb ( d )
2
Pnb
2
4. Solve for the balanced Moment
Mnb = Pnbeb
Problem:
The column shown is reinforced with 6 of 25 mm diameter
bars. If fc’ = 21 MPa, fy = 345 MPa determine P nb, eb and Mnb.
65 mm
500 mm
Axis of bending
65 mm
400 mm
3 (25) 2
As As '
1472.6mm 2 d 500 65 435mm
4
ab
600 1d 600(0.85)435
234.76mm
600 f y
600 345
Pnb =0.85fc’abb
Pnb =0.85(21)234.76(400)= 1,676,186 N
'
'
A
f
(
d
d
) (d d ' )
ab
s y
eb ( d )
2
Pnb
2
234.76 1472.6(345)(435 65) (435 65)
eb (435
)
2
1676186
2
eb 244.76mm
M nb Pnb eb (1676186)(244.76) 410.26 x10 6 N .mm
M nb 410.26kN .m
ANALYSIS OF ECCENTRICALLY LOADED COLUMNS: ANALYTICAL METHOD
e
e
eb
tension controlled columns
Compression controlled columns
Plastic centroid
Analysis of tension controlled columns
e
Plastic centroid
Large moment, small axial load
Failure initiated by tensile yielding
of tension bars
eb e
fs f y
fs ' f y
c 0.003
Tension steel yields at failure
Compression steel yields at failure
d-d’
2
Pn
e
1. From stress Diagram
ΣFy =0
Pn - C1 - C2+ T = 0
d
b
Axis of bending
d – d’
T = Asfy
Stress Diagram
Pn =0.85fc’ab
d’
C2 = As’fy
a
0.85fc’
s
fy
Es
C1= 0.85fc’ab
c
εs’
If As = As’
C2 = T
Strain Diagram
0.003
EQ. 1
2. ΣM @ center of tension steel = 0
Pn( e + ½{d-d’}) –C1( d-ab/2) – C2(d –d’)= 0
Pn( e + ½{d-d’}) –0.85fc’ab( d-a/2) – As’fy(d –d’)= 0
0.85fc’ab( e + ½{d-d’}) –0.85fc’ab( d-a/2) – As’fy(d –d’)= 0
Solve for a
3. Check for yielding of compression steel
c
a
1
s'
cd
0.003
c
4. Solve for Pn
Pn =0.85fc’ab
5. Solve for Pu
Pu Pn
6. Solve for Mu
M u Pu e
'
'
(
c
d
)
f s' 600
fy
c
Plate #12: Eccentrically loaded columns
The column shown is reinforced with 6 of 32 mm diameter
bars. If fc’ = 27 MPa, fy = 345 MPa , e = 370 mm, determine
Pu, and Mu .
65 mm
600 mm
65 mm
450 mm
Axis of bending
Plate #12: Eccentrically loaded columns
The column shown is reinforced with 6 of 28 mm diameter
bars. If fc’ = 27 MPa, fy = 345 MPa , e = 150 mm, determine
Pu, and Mu .
65 mm
500 mm
65 mm
400 mm
Axis of bending
Problem:
The column shown is reinforced with 6 of 25 mm diameter
bars. If fc’ = 21 MPa, fy = 345 MPa , e = 200 mm, determine
Pu, and Mu .
65 mm
500 mm
Axis of bending
65 mm
400 mm
3 (25) 2
As As '
1472.6mm 2 d 500 65 435mm
4
From the preceding problem
eb 244.76mm 260mm
Tension controls
Pn =0.85fc’ab
Pn=0.85(21)400a=7140a
EQ.1
ΣM @ center of tension steel = 0
0.85fc’ab( e + ½{d-d’}) –0.85fc’ab( d-a/2) – As’fy(d –d’)= 0
1
a
7140a{260 (435 65)} 7140a (435 ) 1472.6(345)(435 65) 0
2
2
2
a
445a 435a
26327.4 0
2
a 2 20a 52654.8 0
20 (20) 2 4(52654.8)
a
219.7 mm
2
a 219.7
c
258.47 mm
1 0.85
compression steel
(258.47 65)
f 600
449.1MPa f y yields at failure
258.47
'
s
Pn 7140a 7140(219.7) 1568658 N 1568.65kN
Pu (0.7)1568.65 1098kN
M u Pu e 1098 0.26 285.48kN .m
Verify yeilding of tension steel( not required)
s
d c
0.003
c
(d c)
f s 600
fy
c
(435 258.47)
tension steel
f s 600
409.8MPa f y yields at failure
258.47
Analysis of compression controlled columns
e
Plastic centroid
2. Large axial load and small moment.
with entire cross section in compression.
Failure under this condition occur by
sudden crushing of concrete with all bars
are in compression
3. Large axial load with moment bigger in
2. Bars in the far side in tension has not
yielded. Failure occur by crushing of
concrete
eb e
fs f y
fs ' f y
c 0.003
Tension steel does not yield at failure
Compression steel yields at failure
d-d’
2
Pn
e
d
b
Axis of bending
d – d’
T = Asfs
d
s
fy
Es
a
2
Stress Diagram
d’
Pn 0.85 fc' ab As' f y As f s 0
C2 = As’fy
a
0.85fc’
εs’
Pn 0.85 fc' ab As' f y As f s EQ.1
Note: unknowns are Pn , a &
C1= 0.85fc’ab
c
1. From stress Diagram
ΣFy =0
Pn - C1 - C2+ T = 0
Strain Diagram
0.003
fs
1
a
Pn {e (d d ' )} C 1(d ) C2 (d d ' ) 0
2
2
1
a
Pn {e (d d ' )} 0.85 fc' ab(d ) As' f y (d d ' ) 0
2
2
As' f y (d d ' )
0.85 fc' ab(d 0.5a )
Pn
EQ.2
{e 0.5(d d ' )}
{e 0.5(d d ' )}
M centerofT .S 0
EQ.1 EQ.2
Pn 0.85 fc' ab As' f y As f s EQ.1
As' f y (d d ' )
0.85 fc' ab(d 0.5a)
0.85 fc' ab A f y As f s
EQ.3
{e 0.5(d d ' )}
{e 0.5(d d ' )}
'
s
a
600(d )
(d c)
1
f s 600
a
c
1
From the strain diagram
( 1d a )
f s 600
EQ.4
a
0.85 fc' ab As' f y As 600
As' f y (d d ' )
( 1d a ) 0.85 fc ' ab(d 0.5a)
a
{e 0.5(d d ' )}
{e 0.5(d d ' )}
Solve for a
Check for yielding of compression steel
a
c
1
Solve for fs
Solve for Pn
Solve for Pu
Solve for Mu
s'
cd
0.003
c
'
(c d ' )
f 600
fy
c
'
s
( 1d a )
f s 600
a
Pn 0.85 fc' ab As' f y As f s EQ.1
Pu Pn
M u Pu e
Problem:
The column shown is reinforced with 6 of 25 mm diameter
bars. If fc’ = 21 MPa, fy = 345 MPa , e = 200 mm, determine
Pu, and Mu .
65 mm
500 mm
Axis of bending
65 mm
400 mm
3 (25) 2
As As '
1472.6mm 2
4
d 500 65 435mm
From the preceding problem
eb 244.76mm 200mm
Compressions controls
0.85(21)400a 1472.6(345) 1472.6(600)
{0.85 435 a} 0.85(21) 400a (435 0.5a ) 1472.6(345)(435 65)
a
{200 0.5( 435 65)}
{200 0.5(435 65)}
{369.75 a} a(435 0.5a)
a 71.16 123.75
68.38
a
385
385a 2 27396.6a 17616276.5 47643.75a 435a 2 0.5a 3 26326.3a
a 3 100a 2 97428.1a 35232553 0
a 256.28mm
Can be solved by
Newton’s Method
of approximation
or by trial and error
(256.28) 3 100(256.28) 2 97428.1(256.28) 35232553 0
703.1
0k
( 1d a )
{ 0.85 435 256.28}
f s 600
600
214.73MPa
a
256.28
c
256.28
301.5mm
0.85
(301.5 65 )
f 600
470.65MPa f y
301.5
'
s
Compression steel yields
Pn 0.85 fc' ab As' f y As f s EQ.1
Pn 0.85(21)256.28 400 1472.6(345) 1472.6(214.73)
Pn 2021682 N 2021.68kN
Pu (0.7)2021.68 1415.2kN
M u 1415.2(0.2) 283.04kN .m
COLUMN INTERACTION DIAGRAM
It was observed that the use of analytical method is very complicated and requires accuracy of
manipulation of algebraic equations. An option is to analyze the column by using the column
interaction diagram.
Interaction diagram is the graph of the axial load capacity and the moment capacity of the column of
the M-P axes. The diagram is very useful for studying the strength of the column with varying
proportions of the loads and moments.
Pn
Pn
e
A
Pure axial load
B
Compression Failure
Balanced Loading
Compression
zone
C
Pn
Mn
Tension failure
Tension
zone
Pure bending
Mn
D
The diagram is made by plotting the axial load capacity when the moment is zero( point A), then the
balanced loading( point C) and the moment capacity when the axial load is zero( point D). In between
points A and D , the column fail due to axial and bending. Point C the balanced point. In reference to
Point B, the vertical and horizontal dotted line represents the particular load combination of axial load
and moment.
How to use the column interaction Diagram
Process
Given
Required
calculate
Design
Loads, Pu and e
section dimension
Amount of reinforcement
Pn
Ag
and
M n
Ag h
Intersection of
Locate
Pn
Ag
and
M n
Ag h
and determine pg
obtain
Ast
Analysis
Section properties
eccentricity
axial load and
moment capacity
Ast
pg
Ag
e/h
Intersection of pg & e/h
project horizontally to
The left to get Pn
Ag
project vertically
M n
down to get
Ag h
Pn or Pu
Mn or Mu
Problem
Calculate the ultimate axial load capacity of the column section
shown if the eccentricity is 200 mm. fc’ = 20.7 MPa, fy = 413.7 MPa.
Solve using the column interaction diagram.
400 mm
75 mm
4 of 25 mm
Axis of
bending
450 mm
4 of 25 mm
75 mm
Solution
1. Solve for
,pg, and e/h
(25) 2
8
Ast
4
pg
0.016
Ag
(400)(600)
h 450
600 450
0.75
e 200
0.33
h 600
2.
Locate the intersection of pg and e/h. Project horizontally to the left
Pn
to obtain Pn
Ag
Ag
8.52MPa
Pu Pn
Pu =8.52Ag = 8.52(400)600 = 2044.8 kN
Pn
(MPa)
Ag
50
Interaction Diagram
0.1
0
45
e/
h=
0.
20
35
30 .33
.
=0 h =0
h
e/
e/
.40
0
=
e/h
. 50
0
=
e/h
30
20
15
0.016
8.2 MPa
M n
(MPa)
Ag h
10
0.75
e/h
=
40
25
fc’= 3 ksi ( 20.7 MPa)
fy = 60 ksi ( 413.7 MPa)
pg=0.08
0.07
0.06
0.05
04
0.03
0.02
0.01
h
h
Pn
e
0
e/h=1.
5
0
1
2
3
4
5
6
7 8
9 10 11 12 13 14 15
Problem
Calculate the nominal axial load capacity of the colum shown applied at
an eccentricity of 400 mm. fc’=20.7 MPa,fy=413.7 MPa.
Reinforcement consist of 16 of 28 mm bars.
500 mm
70 mm
Axis of
bending
360 mm
70 mm
(28) 2
10
Ast
4
pg
0.0394
Ag
(500)(500)
h 360
500 360
0.72
Since there is no available interaction diagram for 0.72
Use the interaction diagrams for 0.60 and 0.75
With fy = 413.7 MPa and fc’=20.7 MPa and interpolate the values
obtained
e/h = 400/500 = 0.80
From column interaction diagram for
fc’=20.7 MPa
0.60
Pn
4.6 MPa
Ag
f y = 413.7 MPa and
From column interaction diagram for
f y = 413.7 MPa and fc’=20.7 MPa
0.75
Pn
5.8MPa
Ag
Interpolating values
x 5.8
0.72 .075
4.6 5.8 0.60 0.75
X = 5.56 MPa
Pn
5.56 MPa
Ag
0.60
0.72
0.75
0.7 Pn
5.56
(500)(500)
Pn = 1,985,714 N = 1,985.7 kN
Pn
Ag
4.6
x
5.8
50
Interaction Diagram
0.1
0
45
0.039
20
0.08
0.07
0.06
0.05
0.04
15 0.03
10
4.6 MPa
5
0
0.02
0.01
e/
h=
0.
20
35
25
0.60
e/h
=
40
30
fc’= 3 ksi ( 20.7 MPa)
fy = 60 ksi ( 413.7 MPa)
0
.3
0
h=
/
e
40
.
0
=
e/h
. 50
0
=
e/h
h
h
Pn
e
0.8
e/h =
0
e/h=1.
50
Interaction Diagram
0.1
0
45
30
25
0.039
15
10
5.8 MPa
5
0
0.0
4
e/
h=
0.
20
35 0.0
8
0.75
e/h
=
40
20
fc’= 3 ksi ( 20.7 MPa)
fy = 60 ksi ( 413.7 MPa)
30
.
=0
h
e/
40
.
0
=
e/h
50
.
0
=
e/h
h
h
Pn
e
0.
e/h =
0
e/h=1.
8
Plate # 13: Analysis of eccentric columns using interaction diagram
Calculate the ultimate axial load capacity of the column section
shown if the eccentricity is 300 mm. fc’ = 27.57 MPa(4 ksi), fy= 413.MPa(60ksi).
Solve using the column interaction diagram.
350 mm
75 mm
1ksi =6.8928 MPa
4 of 25 mm
Axis of
bending
400 mm
4 of 25 mm
75 mm
Calculate the nominal axial load capacity of the colum shown applied at
an eccentricity of 400 mm. fc’=27.57 MPa,fy=413.7 MPa.
Reinforcement consist of 16 of 22 mm bars.
450 mm
70 mm
Axis of
bending
330 mm
70 mm
Problem
CE Board May 2000
For the spiral column shown, determine the maximum nominal axial load
capacity if the eccentricity is 150 mm. fc’= 27.6 MPa, fy = 413.7 MPa.
Use the column interaction diagram. 1 ksi = 6.8928 MPa
8 of
22 mm
375 mm 500 mm
(22) 2
8
Ast
4
pg
0.0155
2
(500)
Ag
4
h 375
500 375
0.75
e 150
0.3
h 500
Pn
1.3ksi 1.3 x6.8928 8.961MPa
Ag
6
ΦPn/Ag(ksi)
0.08
0.75
0.
20
5
INTERACTION DIAGRAM
fc’ = 4ksi(27.6 MPa)
fy = 60 ksi ( 413.7 MPa)
0.10
e/h=0.
05
7
0.07
h
e/
0.06
4
0.05
0.04
30
.
=0
Pn
40
0.
0.03
3
h
h
0.02
e
0
0.5
0.0155
0.01
2
1.0
1.3 ksi
1
0
0.2
0.4
0.6
0.8
1.0
1.2
ΦMn/Agh(ksi)
1.4
1.6
1.8
2.0
Determination of steel area
A rectangular column is as shown in the
figure. It is acted upon by the following loads:
axial load due to dead load = 450 kN
axial load due to live load = 500 kN
Moment due to dead load = 134 kN.m
Moment due to live load
= 150 kN.m
The column is to be analyzed and designed
by ultimate strength design method.
Use fc’ = 20.7MPa, fy = 413.7 MPa.
Determine
a)
Eccentricity of the column
b)
Area of steel bars required
400 mm
75 mm
Axis of
bending
450 mm
75 mm
Solution:
Pu = 1.4PD + 1.7PL
Pu =1.4 (450) + 1.7(500) = 1480 kN
Pn = Pu/Φ
Pn = 1480/0.7
Pn = 2114.28 kN
h 450
600 450
0.75
e
300
0.50
h
600
Pn 0.7(2114 .28)(1000)
6.17 MPa
Ag
(400)(600)
6
M
0
.
7
(
632
.
28
)(
10
)
n
Mu =1.4MD + 1.7ML
3.07 MPa
(600)(400)(600)
Mu =1.4(134)+1.7(150) = 442.6 kN.m Ag h
Mn =Mu/Φ = 442.6/0.7 = 632.28 kN.m
Pn
M n
Intersection of
and
Eccentricity
Ag
Ag h
e= M /P = 632.28/ 2114.28
n
n
e = 0.3 m = 300 mm
pg =0.017
Ast =pgAg= 0.017(400)(600)
Ast = 4080 mm2
50
Interaction Diagram
0.1
0
45
e/
h=
0.
20
35
30
25
15
Pn
e
0
e/h=1.
01
0.
Pn
6.17 MPa
Ag
10
h
30
.
=0
h
e/
40
.
0
=
e/h
. 50
0
=
e/h
0.
0. 08
0. 07
0. 06
0.0 05
0.0 4
0. 3
02
20
h
0.60
e/h
=
40
pg=0.017
fc’= 3 ksi ( 20.7 MPa)
fy = 60 ksi ( 413.7 MPa)
5
0
1
2
3
4
5
6
M n
3.07 MPa
Ag h
7
8
9
10
11 12
13
14
15
Design of Eccentric columns( section & reinforcement)
The size of column can be obtained by assuming an average
compressive stress of concrete from 0.5fc’ to 0.6 fc’ under the load
PU. Design of reinforcement follows the procedure of the preceding
topic. If value of pg is not reasonable, section maybe revised.
Design a square tied column to carry a factored axial load of 2,000 kN
And a factored moment of 190 kN.m. Use 25 mm bars to be placed
uniformly along the faces of the column. Assume fc’=20.7MPa and
fy =413.7 MPa.
Solution
Ag = Pu/0.5fc’
Ag = (2000)(1000) /0.5(20.7)
Ag = 193,237 mm2
Using a square column
h = 439 mm say 450 mm
Proposed Layout
450 mm
62.5 mm
Axis of
bending
325 mm
62.5 mm
e = Mu/Pu
e =190/2000 = 0.095 m = 95 mm
e/h = 95/450 = 0.211
h 325
450 325
0.722
Interpolate between 0.60 and 0.75
Pn =Pu/Ф= 2000/0.7=2,857 kN
ФPn/Ag = 0.7(2857)1000/(450)(450)
ФPn/Ag =9.876 MPa/6.8928 = 1.43 ksi
0.60
Intersection of e/h and ФPn/Ag
0.75
Intersection of e/h and ФPn/Ag
0.60
0.722
0.75
pg =0.018
pg =0.016
pg
0.018
x
0.016
x 0.016
0.722 0.75
0.018 0.016
0.6 0.75
x = pg = 0.01637
As =PgAg
As = 0.01637(450)(450)
As = 3315 mm2
7
0.10
ΦPn/Ag(ksi)
5
3
1.43 ksi
0.0
0. 8
0
0.0 7
0.0 6
0.0 5
0. 4
03
h
e/
h
30
.
=0
h
Pn
e
40
.
0
0
0.5
02
0.
2
0.75
0.
20
e/
h=
0.
21
1
6
4
INTERACTION DIAGRAM
fc’ = 3 ksi(20.7 MPa)
fy = 60 ksi ( 413.7 MPa)
1.0
0.0
1
1
0
0.2
0.4
0.6
pg=0.016
0.8
1.0
1.2
1.4
ΦMn/Agh(ksi)
1.6
1.8
2.0
7
0.10
6
0.60
0.
2
e/
h= 0
0.
21
1
5
4
e/
0
0.
3
0
h
Pn
e
0
0.5
2
0. 0 0 1
0.
2
.3
0
h=
h
40
0.
8
ΦPn/Ag(ksi)
INTERACTION DIAGRAM
fc’ = 3ksi(20.7 MPa)
fy = 60 ksi ( 413.7 MPa)
1.0
1.43 ksi
1
0
0.2
0.4
0.6
pg=0.018
0.8
1.0
1.2
ΦMn/Agh(ksi)
1.4
1.6
1.8
2.0
Plate # 14 : Design of eccentric columns by using interaction diagram
For the spiral column shown, determine the maximum nominal axial load
capacity if the eccentricity is 200 mm. fc’= 27.6 MPa, fy = 413.7 MPa.
Use the column interaction diagram. 1 ksi = 6.8928 MPa
8 of
22 mm
400 mm 530 mm
Design a square tied column to carry a factored axial load of 2,500 kN
And a factored moment of 300 kN.m. Use 28mm bars to be placed
uniformly along the faces of the column. Assume fc’=20.7MPa and
fy =413.7 MPa.
A rectangular column is as shown in the
figure. It is acted upon by the following loads:
axial load due to dead load = 500 kN
axial load due to live load = 600 kN
Moment due to dead load = 140 kN.m
Moment due to live load
= 180 kN.m
The column is to be analyzed and designed
by ultimate strength design method.
Use fc’ = 20.7MPa, fy = 413.7 MPa.
Determine
a)
Eccentricity of the column
b)
Area of steel bars required
450 mm
75 mm
Axis of
bending
500 mm
75 mm
Axial Load and Biaxial Bending
In some cases, columns are subject to bending on both axes. This
is the usual case of corner columns where beams or girder frames
into the column from both directions .
ex
ey
My
Mx
Bresler Equation
Biaxially loaded columns can be analyzed by Bresler’s equation.
1
1
1
1
Pn Pnx Pny Pno
Where:
Pnx = axial load capacity of the column for eccentricity ex with ey =0
Pny = axial load capacity of the column for eccentricity ey with ex =0
Pno = axial load capacity of the column when ex = 0 and ey = 0
Pno = 0.85fc’Ag + Asfy
Problem
Calculate the permissible axial load capacity Pu of the short tied
column shown shown if ex= 200 mm and ey = 100 mm.
Use fc’ =20.7 MPa, fy = 413.7 MPa
75 mm
8 of 28 mm
225 mm 375 mm
75 mm
75 mm
450 mm
600 mm
75 mm
Solution
(28) 2
8
Ast
4
pg
0.022
Ag
(375)(600)
For eccentricity ex ( bending about the y axis)
h 450
600 450
0.75
Pn
1.28ksi
Ag
e
200
0.333
h
600
Pn
8.823MPa
Ag
Pnx
8.823(600)(375)
0.7(1000)
Pnx =2836 kN
For eccentricity ey ( bending about the x axis)
h 225
375 225
0.60
e 100
0.267
h 375
Pn
1.35ksi
Ag
Pn
9.305MPa
Ag
9.305(600)(375)
Pny
0.7(1000)
Pnx =2991 kN
Pno = 0.85fc’Ag + Asfy = 5,997kN
1
1
1
1
Pn Pnx Pny Pno
1
1
1
1
Pn 2,836 2,991 5,997
Pn = 1,922 kN
7
0.10
6
ΦPn/Ag(ksi)
0.75
0.
20
5
h
e/
h
30
.
=0
h
33
3
.
=0
h
/
e
4
02
0.
0.0
0. 8
0
0.0 7
p3
0.0 6
g=
0.0 5
0.0
22 0.0 4
3
2
0.0
1
1.28ksi
INTERACTION DIAGRAM
fc’ = 3 ksi(20.7 MPa)
fy = 60 ksi ( 413.7 MPa)
Pn
e
40
.
0
0
0.5
1.0
1
0
ΦMn/Agh(ksi)
7
0.10
6
0.60
0.
20
5
0
0.
4
h
e/
8
e/
3
=
h
67
2
0.
h
30
.
0
h=
=0
pg
ΦPn/Ag(ksi)
INTERACTION DIAGRAM
fc’ = 3ksi(20.7 MPa)
fy = 60 ksi ( 413.7 MPa)
Pn
e
0
0.4
0
0.5
0
22
.0 .02
0.
03
2
1.35 ksi
1.0
1
0
0.2
0.4
0.6
0.8
1.0
1.2
ΦMn/Agh(ksi)
1.4
1.6
1.8
2.0
Using the data of the preceding problem, determine Mux and Muy .
Solution
Pu Pn
Pu 0.7(1922) 1345.4kN
M ux Pu e y 1345.4(0.10) 134.54kN .m
M uy Pu ex 1345.4(0.20) 269.08kN .m
Plate # 14 : Columns subject to Biaxial Bending
Calculate the permissible axial load capacity Pu of the short tied
column shown shown if ex= 250 mm and ey = 150 mm. Solve by
analytical method. Use fc’ =20.7 MPa, fy = 413.7 MPa
75 mm
10 of 28 mm
250 mm 400 mm
75 mm
75 mm
500 mm
650 mm
75 mm
Long Columns
CONSIDERATION OF SLENDERNESS EFFECTS
For compression members braced against sideways,effects
of slenderness maybe neglected when:
kLu
M 1b
34 12
M 2b
r
For compression members not braced against sideways,
effects of slenderness maybe neglected when
kLu
22
r
Where:
M2b = value of the larger factored end moment on a compression
member due to loads that result in no appreciable
sidesway, calculated by conventional elastic frame analysis
M1b = value of the smaller factored end moment on a compression
member due to loads that result in no appreciable sidesway,
calculated by conventional elastic frame analysis positive if
member is bent in single curvature, negative if member is
bent in double curvature
M1b
Lu
M2b
Effective Length Factor
Effective length factor is equal to 1 when the member is braced
against sideway ; for members not braced against sideway
effects of cracking and reinforcement on relative stiffness must
be considered and should be greater than 1.
Radius of Gyration
r = 0.25 D for circular columns ,D the column diameter
r = 0.30 h for rectangular columns ,h the column dimension in
the direction stability is being considered
Unsupported Length of Compression members
Unsupported length Lu a of compression member is taken as the
clear distance between floor slabs, beams or other members
capable of providing lateral support for that compression
member. Where column capitals or haunches are present it
shall be measured to the lower extremity of capital or haunch in
the place considered.
Braced and Unbraced Frame
A frame maybe considered braced if the bracing elements such as shear
walls, shear trusses, or other means resisting lateral movement of a storey,
have a total stiffness at least, six times the sum of the stiffness of all the
columns resisting lateral movement in that storey.
PROBLEM
A square column having an unsupported length of 5 m is used in a
braced frame against sidesway. The column is bent in single
curvature and subject to factored end moments of 58 kN/m at the
top and 42 kN.m at the bottom. Determine its minimum dimension
so that slenderness effects may not be considered in the analysis.
M 2b 58kN .m
Lu 5m 5000mm
M 1b 42kN .m
k 1 .0
r 0.3h
M 1b
kLu
34 12
M 2b
r
1(5000)
(42)
34 12
0.3h
58
h 660mm
+ since the
column is bent
in single
curvature
column
is braced
against
sidesway
MOMENT MAGNIFIER METHOD
According to NSCP 1992 edition, compression members shall be designed using the
factored axial load Pu from a conventional frame analysis and a magnified factored
moment Mc defined by :
M c s M 2 s b M 2b
EQ. 1
Where:
M2b = value of the larger factored end moment on a compression
member due to loads that result in no appreciable sidesway,
calculated by conventional elastic frame analysis. For use in
EQ.1 M2b ≥ Pu(15 + 0.03 h) 15 + 0.03h in mm
M1b = value of the smaller factored end moment on a compression
member due to loads that result in no appreciable sidesway,
calculated by conventional elastic frame analysis positive if member
is bent in single curvature, negative if member is bent in double
curvature
M2s = value of the larger factored end moment on a
compression member due to loads that result in appreciable
sidesway ( wind, earthquake and other lateral loads) ,
calculated by conventional elastic frame analysis .
For use in EQ.1 M2b ≥ Pu(15 + 0.03 h) 15 + 0.03h in mm
b =
moment magnification factor that takes into account
the effect of member curvature in a frame braced
against sideway.
Cm
b
1 .0
Pu
1
Pc
M 1b
Cm 0.6 0.4
M 2b
but not less than 0.4,
For all other cases Cm = 1.
s = moment magnification factor that takes into account
the lateral drift of a column caused by lateral or
gravity loading the frame not braced against sidesway
1
s
1.0
Pu
1
Pc
s 1.0
Pu Pc
2 EI
Pc
( kLu ) 2
for columns braced against sidesway
summation for all columns in a
storey.
EQ 2
For frames not braced against sidesway, both band s should be
computed. For frames braced against sidesway, s should be taken
as 1.0. In calculating Pc ,EI should be determined either by
Ec I g
Ec I g
Es I se
EI 5
1 d
or
EI 2.5
1 d
Where
βd = the ratio of factored dead load moment to maximum total
factored moment ( always positive), where the load is due to
gravity only in the calculation of Pc in computing bor
the ratio of the maximum sustained lateral load to the
maximum total factored lateral load in that story in the
calculation of Pc in computing s
Ig = moment of inertia of gross section about centroidal axis
neglecting reinforcement
Ise = moment of inertia of reinforcement about centroidal axis of
member cross section
Ec =modulus of elasticity of concrete E c 4700 fc '
Es =modulus of elasticity of reinforcement
Moment Magnifier for biaxial bending
For compression members subject to moment on both axes, moment
about each principal axis shall be magnified by computed from the
corresponding restraint about that axis.
Alignment Charts
Effective Length Factor k are obtained using the Jackson and
Moreland alignment chart. To use this chart a parameter ψ A for end A
of column AB and ψB for end B are computed. A straightedge is
placed between A and B. The point where the straightedge cross the
middle monograph is k. The parameter ψ at one end of the column
equals the sum of the stiffnesses of the columns (including the column
under consideration) meeting at the joint divided by the sum of all the
stiffnesses of the beams meeting at that joint.
∑EI/L of Columns
Ψ=
∑EI/L of Beams
ALIGNMENT CHARTS
ψA
∞
50
K
1.0
10
5
3.0
2.0
1.0
0.8
0.6
0.5
10
0.9
0.8
0.7
0.4
0.3
0.2
5
3.0
2.0
1.0
0.8
0.6
0.5
0.4
0.6
0.1
0
ψB
5∞
0
0.3
0.2
0.1
0.5
BRACED FRAME
0
ψA
∞
100.0
K
50.0
30.0
20.0
10.0
8.0
∞
20.0
10.0
5.0
4.0
3.0
ψB
∞
100.0
50.0
30.0
20.0
10.0
8.0
6.0
5.0
4.0
2.0
3.0
6.0
5.0
4.0
3.0
2.0
1.5
1.0
2.0
1.0
0
1.0
0
UNBRACED FRAME
Note: Ψ ═ ∞ for pinned ends and 1.0 for fixed end
PROBLEMS
1. A rectangular concrete column has a dimension of 300 mm by
380 mm. The column length is 4.5 m and is to be used in a
frame braced against sidesway ( k =1.0, EI = 34 MN.m2). The
column is bent in single curvature by ultimate factored moment of
110 kN.m at the top and 120 kN.m at the bottom. The column
carries factored axial dead load of 150 kN and factored axial live
load of 350 kN. Considering the longer dimension of the column,
is slenderness effects be considered in the analysis ? If so,
what is the magnified moment capacity of this column ?
2. Solve the previous problem considering the effect of wind
loads producing an unfactored axial load of 130 kN an unfactored
moment of 100 kN.m
Solution to #1
M 2b 120kN .m
M 1b 110 kN .m
M 2s 0
r 0.3h 0.3(380) 114 mm
Lu 4.5m 4500mm
kLu 1(4500)
39.47
r
114
M 1B
110
34 12
34 12
23
M 2B
120
kLu
M 1B
34 12
r
M 2B
Column is Long
+ since the
column is bent
in single
curvature
M 1b
110
Cm 0.6 0.4
0.6 0.4
0.9668
M 2b
120
Pu 150 350 500kN
2 EI 2 34(10) 6
Pc
16,571,188 N 16,571.18kN
2
2
(kLu ) {(1)(4.5)}
Cm
0.9668
b
1.0103
Pu
500
1
1
0.7(16571.18)
Pc
M c b M 2b c M 2 s 1.0103(120) s (0) 121.24kN .m
Solution to #2
M 2 s 0.75[1.4M D 1.7 M L 1.7 M W ]
M 2 s 0.75[1.4(0) 1.7(0) 1.7(100)] 127.5kN .m
[1.4 PD 1.7 PL ] 150 350 500kN
Specified in the previous
problem
Pu 0.75[1.4 PD 1.7 PL 1.7 PW ] 0.75[500 1.7(130)] 540.7 kN
Cm
0.9668
b
1.014
Pu
540.75
1
1
0.7(16571.18)
Pc
s 1.0 Frame is braced against sidesway
M c b M 2b s M 2 s 1.014(120) 1.0(127.5) 249.18kN .m
Plate No.15
Long Columns ( Part I)
1.A square column having an unsupported length of 4.5 m is used in a
frame braced against sidesway. The column is bent in single curvature and
subject to factored end moments of 52 kN/m at the top and 36 kN.m at the
bottom. Determine its minimum dimension so that slenderness effects may
not be considered in the analysis.
2. A square concrete column has a dimension of 450 mm by 450 mm.
The column length is 5 m and is to be used in a frame braced against
sidesway ( k =1.0, EI = 44 MN.m2). The column is bent in single
curvature by ultimate factored moment of 130 kN.m at the top and 150
kN.m at the bottom. The column carries factored axial dead load of 180
kN and factored axial live load of 390 kN. Is slenderness effects be
considered in the analysis ? If so, what is the magnified moment
capacity of this column ?
3. Solve the previous problem considering the effect of wind loads
producing an unfactored axial load of 150 kN an unfactored moment of
120 kN.m
For the column shown in figure, PD = 266.88 kN, PL = 444.8 kN, MD =
162.72kN.m, and ML = 189.84 kN.m. The column belongs to a braced
frame and is bent in single curvature about the x axis by equal moments
at each end of the member . The unsupported length is 6 m. fc’ = 20.7
MPa, fy = 413.7 MPa . Determine the following :
a) Critical load Pc
b) Magnified moment
375 mm
c) the steel ratio required to
8 of 32 mm
375 mm
500 mm
sustain the magnified moment
Solution:
8 (32) 2
As
6433.98mm 2
4
Pu=1.4PD + 1.7PL = 1129.79 kN
M1b= M2b =1.4MD + 1.7ML = 550.54 kN.m
Ec 4700 fc ' 4700 20.7 21383.7 MPa
375(500)3
Ig
3906.25 x106 mm 4
12
I se As z 2 6433.98(187.5) 2 226.194 x106 mm 4
Es 200000 MPa
1.4M D 1.4(162.72)
d
0.41
Mu
550.54
Ec I g
EI
E s I se
5
1 d
21383.71(3906.35)10 6
200000(226.194)10 6
5
EI
1 0.41
EI 4.4 x1013 N .mm 2
2 EI
Pc
(kLu ) 2
2 4.4 x1013
Pc
12062.85kN
2
(6000 )
M 1b
550.54
Cm 0.6 0.4
0.6 0.4
1.0
M 2b
550.54
Cm
1
b
1.13
Pu
1129 .79
1
1
Pc
0.7(12062.85)
M c b M 2b 1.13(550.54) 622.11kN .m
Pn Pu
1129 .79
6.03MPa
Ag
Ag 375(500)
M n M c
(622.11)(10) 6
6.64 MPa
Ag h Ag h (500)(375)(500)
Using the column interaction diagram
Project 6.03 MPa(0.875ksi) from
P
n
axis horizontally to the right
Ag
M n
Project 6.64 MPa(0.963ksi) from
axis vertically up
Ag h
Locate intersection of projected lines to get pg = 0.05
0.10
INTERACTION DIAGRAM
fc’ = 3 ksi(20.7 MPa)
fy = 60 ksi ( 413.7 MPa)
6
0.
20
5
0.75
e/
4
h
Pn
e
40
.
0
0
0.5
0
0.
2
30
.
0
h=
3
3
3
0.
=
e/h
0.0
0. 8
0
0.0 7
0.0 6
0.0 5
0. 4
03
3
h
1.0
2
0.0
1
1
0.875 ksi
0
0.2
2.0
0.4
0.6
0.8
1.0
1.2 1.4 1.6
ΦMn/Agh(ksi)
0.963 ksi
1.8
Problem:
Design the steel reinforcement for a 7.5 m 600 mm by 600 mm square tied
Column bent in single curvature in a braced frame with ψ equal 1 at one end
and 2.0 at the other end. The column is used to carry the following loads
axial dead load = 800 kN
axial live load = 880 kN
dead load moment = 135 kN. m
live load moment = 175 kN. m
Use fc’ = 20.7 MPa, fy = 413.7 MPa, and place the bars at two end faces
With d’ = 75 mm.
600 mm
450 mm
Axis of
bending
75mm
75 mm
Solution
Pu =1.4 PD + 1.7PL
Pu= 1.4 (800) + 1.7(880) = 2,616 kN
MU = M2b = M1b = 1.4 MD + 1.7 ML
MU = M2b = M1b = 1.4(135) + 1.7 (175)
MU = M2b = M1b = 486.5 kN.m
M1b / M2b = 1.0
r = 03 h = 0.3(600) = 180
from Alignment chart k = 0.82
KLu / r = 0.82( 7500)/180 = 34.17
34 – 12M1b / M2b =34 -12(1) = 24
Check:
ψA
∞
50
M2b > 2616 (15 + 0.03(600))/1000 = 86.33 kN.m
1.0
10
5
3.0
2.0
1.0
0.8
0.6
0.5
0.3
0.9
0.82
0.8
0.7
5
3.0
2.0
1.0
0.8
0.6
0.5
0.4
0.6
0.1
0
ψB
5∞
0
10
0.4
0.2
M2b > Pu (15 + 0.03h)
K
0.3
0.2
0.1
0.5
BRACED FRAME
0
M 1b
Cm 0.6 0.4
0.6 0.4(1) 1.0
M 2b
E c 4700 fc' 4700 20.7 21383.7 MPa
600( 600 )3
Ig
10.8 x10 9 mm 4
12
1.4( 135 )
d
0.3885
486.5
Ec I g
21,373.7( 10.8 )( 10 ) 9
2.5
EI 2.5
66.532 x10 12 N .mm 2
1 d
1 0.3885
EI
( 66.532 )( 10 )
Pc
17 ,361kN
2
2
( kLu )
0.82( 7500 ) ( 1000 )
2
2
9
Cm
1
b
1.274
M 2s 0
Pu
2 ,616
1
1
Pc
0.7( 17 ,361 )
M c s M 2 s b M 2 b 0 ( 1.274 )( 486.5 ) 619.8 kN .m
M u 619.8
e
0.237 m 237 mm
Pu
2 ,616
h 450
600 450
0.75
e 237
0.395
h 600
Pn PU 2 ,616( 1000 )
7.27 MPa 1.0547 ksi
Ag
Ag
600( 600 )
Pg=0.016
As =0.016(600)600 = 5,760 mm2
Using 25 mm bars
5760
No
12 pcs
2
( 25 )
4
0.10
INTERACTION DIAGRAM
fc’ = 3 ksi(20.7 MPa)
fy = 60 ksi ( 413.7 MPa)
6
0.
20
5
0.75
e/
4
0.0
0. 8
0
0.0 7
0.0 6
0.0 5
0. 4
03
3
0
0.
0.016
2
40
.
0
0
0.5
h
30
.
0
h=
h
Pn
5
9
3
0.
=
e/h
e
1.0
2
0.0
1
1.0547 ksi
1
0
0.2
2.0
0.4
0.6
0.8 1.0 1.2
ΦMn/Agh(ksi)
1.4
1.6
1.8
7
0.10
6
0.75
0.
20
5
ΦPn/Ag(ksi)
INTERACTION DIAGRAM
fc’ = 4ksi(27.6 MPa)
fy = 60 ksi ( 413.7 MPa)
e/
4
h
30
.
0
h=
h
40
0.
3
Pn
e
0
0.5
2
1.0
1
0
0.2
0.4
0.6
0.8 1.0 1.2
ΦMn/Agh(ksi)
1.4
1.6
1.8
2.0
Plate # 17: Design of Long columns
Design the steel reinforcement for a 6.0 m 500 mm by 500 mm square tied
Column bent in single curvature in a braced frame with ψ equal 2.0 at one end and
3.0 at the other end. The column is used to carry the following loads
axial dead load = 600 kN
axial live load = 650 kN
dead load moment = 105 kN. m
live load moment = 125 kN. m
Use fc’ = 20.7 MPa, fy = 413.7 MPa, and place the bars at two end faces
With d’ = 75 mm.
500 mm
350 mm
Axis of
bending
75mm
75 mm
For the column shown in figure, PD = 200 kN, PL = 400 kN, MD =
122kN.m, and ML = 148 kN.m. The column belongs to a braced frame
and is bent in single curvature about the x axis by equal moments at
each end of the member . The unsupported length is 6 m. fc’ = 20.7
MPa, fy = 413.7 MPa . Determine the following :
a) Critical load Pc
b) Magnified moment
350 mm
c) the steel ratio required to
8 of 28 mm
300 mm
450 mm
sustain the magnified moment
Two Way Slabs
When a rectangular reinforced concrete slab is supported on four
sides,reinforcement placed perpendicular to the side may be assumed
to be effective in the two directions. These slabs are known as two
way slabs. Bending occurs on both directions. However if the slab is
supported on all sides but the ratio of the long side to short is two or
more times, the slab act as a one way slab with bending occuring in
the short direction.
Two methods of designing will be discussed. The direct design
method and the ACI moment coeffecient method. Other methods
available are the strip method and the equivalent frame method.
panel
h =slab thickness
Beam or wall
panel
Beam
Beam
Beam
Beam
Drop panel
Column
capital
Two way edge supported slab
Two way column supported slab
( Flat Slab)
Column and Middle Strip
When the design moments are computed, the moments are distributed
to the column and middle strip. Column strip is a design strip with a width on
each side of a column centerline equal to 0.25L1 or 0.25L2 whichever is less.
Column strip includes beam if any. The middle strip is a design strip bounded
by two column strips.
L2
Column
Strip
Middle
Strip
Smaller of
0.25L1 or 0.25L2
CL of column
Column
Strip
L1
Minimum thickness
Minimum thickness of slabs without interior beams spanning between the
supports shall be in accordance with Table 1 and shall not be less than the
following values:
a) slabs without drop panels
125 mm
b) slabs with drop panels
100 mm
Table 1
Yield
Stress
Fy,MPa(1)
275
415
With drop panels(2)
Without drop panels(2)
Exterior Panels
Without
edge
beam
With
edge
beam(3)
Ln /33
Ln /30
Ln /36
Ln /33
Interior
Panels
Ln /36
Ln /33
Exterior Panels
Without
edge
beam
With
edge
beam(3)
Ln /36
Ln /33
Ln /40
Ln /36
Interior
Panels
Ln /40
Ln /36
(1) For values of reinforcement yield stress between 275 MPa and 415 MPa minimum
thickness shall be obtained by linear interpolation
(2) See definition of drop panel in Sections 5.13.4.7.1, 5.13.4.7.2,NSCP 5th Edition
(3) Slabs with beam between columns along exterior edges. The value of α for the
edge beam shall not be less than 0.8.
DROP PANEL
Drop panels and column capitals are placed around the vicinity of the column
to reduce the effect of negative moment and shear
Not less than 1/6L
h
Not less than 1/4h
Column capital
L
Sections 5.13.4.7.1 Drop panel shall extend in each direction from centerline
of support a distance not less than one sixth the span length measured center
to center of supports in that direction.
5.13.4.7.2 Projection of drop panel below the slab shall be at least one fourth
of the slab thickness beyond the drop.
The minimum thickness of slabs with or without beams
spanning between the supports on all sides and having a
ratio of long span to short span not exceeding 2 shall be
h
1
36000 5000 m 0.12( 1
but not less than
h
EQ 10.1
Ln ( 800 0.73 f y )
Ln ( 800 0.73 f y )
)
EQ 10.2
36000 9000
and need not be more than
h
Ln ( 800 0.73 f y )
36000
EQ 10.3
The values of h obtained from EQS.10.1,10.2 and 10.3 maybe shall be
modified as required by Sections 5.9.5.3.4 and 5.9.5.3.5 but in no case
shall the thickness be less than
m 2.0
m 2.0
------------------------- 125 mm
------------------------- 90 mm
Section 5.9.5.3.4 For slabs without beams, but with drop panels extending
In each direction from centerline of support a distance not less than one sixth
the span length in that direction measured center to center of supports and
projection of slab at least one fourth of slab thickness beyond the thickness
required by EQS.10.1,10.2 and 10.3 shall be reduced by 10%.
Section 5.9.5.3.5 At discontinuous edge, an edge beam shall be provided
With a stiffness ratio σ not less than 0.8; or the minimum thickness
required by EQS.10.1,10.2 and 10.3 shall be increased by 10% in the
panel with a discontinuous edge.
Where
Ln = length of clear span in the long direction of a two- way construction
measured face to face of supports in slabs without beams and face
to face of beams or other supports in other cases.
αm = average value of α for all beams on edges of a panel.
α = ratio of flexural stiffness of beam section to flexural stiffness a
width of the slab bounded laterally by centerline of adjacent
panel ( if any) in each side of the beam.
E cb I b
E cs I s
β = ratio of clear spans in long to short direction of two way slabs.
Ecb = modulus of elasticity of beam concrete
Ecs = modulus of elasticity of slab concrete
Ib =moment of inertia about the centroidal axis of gross section of
section defined in section 5.13.2.4
Section 5.13.2.4 For monolithic or fully composite construction, a beam includes that
portion of the slab on each side of the beam extending a distance equal to the projection
of the beam above or below the slab, whichever is greater, but not greater than four times
the slab thickness
Is =moment of inertia about the centroidal axis of gross section of slab
Is =H3 /12 times width of slab defined in notation α and βt .
Direct Design Method
Limitations of Direct Design Method
There shall be a minimum of three continuous span in each direction
Panel shall be rectangular with a ratio of longer to shorter span center to center of support within a
panel not greater than 2
Successive span lengths center to center of supports in each direction shall not differ by more than
1/3 the longer support
Columns be offset a maximum of 10% of span ( in the direction of the offset) from either axis
between centerlines of successive columns.
All loads shall be due to gravity only and uniformly distributed over an entire panel. Live load shall
not exceed three times the dead load.
For a panel between supports on all sides , the relative stiffness of beams in the two perpendicular
direction,
must be between 0.2 and 0.5.
Where :L1 = length of the span in the direction that moments are being
2
determined, measured center to
center
of supports
L
1
1
L2 = length of span transverse to L I, measured
center to center of
2
L
2
2
supports.
Moments in slabs
The total moment resisted by the slab equals absolute sum of the positive
and negative factored moments in each direction and shall not be less than
( wu L2 )Ln
Mo
8
2
EQ. A
Where: wu is the factored
load
Clear span Ln shall extend face to face of the columns,capitals,bracket or
walls Ln in EQ. A shall not be less than 0.65L1. Circular or polygon supports
shall be treated as square supports with the same area
If the transverse span of panels on either side of the centerline of supports
varies L2 in EQ. A shall be taken as the average of adjacent transverse
spans. When the span adjacent and parallel to an edge is considered, the
distance from edge to panel centerline shall be substituted for L2 in EQ. A.
Negative and Positive Factored Moments
Negative factored moments shall be located at face of rectangular supports.
Circular or regular polygon supports shall be treated as square supports with
the same area
A. In an interior span, total static moment Mo shall be distributed as follows:
Negative factored moments
0.65
Positive factored moments
0.35
B. In an end span, total factored static moment Mo shall be distributed according
to the table below
1
2
3
4
Slabs without beams
Exterior edge Slabs with
b/w interior supports
unrestrained beams b/w
With
Without
all supports
edge beam
edge beam
Interior Negative
Factored Moment
5
Exterior edge
Fully restrained
0.75
0.70
0.70
0.70
0.65
Positive
Factored Moment
0.63
0.57
0.52
0.50
0.35
Exterior Negative
Factored Moment
0
0.16
0.26
0.30
0.65
Factored Moment in Column strips
Column strips shall be proportioned to resist the following portion in percent
Of interior factored negative moments.
TABLE A
l2
l1
l2
0
l1
l2
1.0
l1
0.5
1.0
2.0
75
75
75
90
75
45
Linear interpolation be made for the values shown
Column strips shall be proportioned to resist the following portion in percent
Of exterior factored negative moments.
TABLE B
l2
l1
l
2 0
l1
l
2 1.0
l1
t 0
t 2.5
t 0
t 2.5
0.5
1.0
100
100
100
75
75
75
100
100
100
90
75
45
2.0
Linear interpolation be made for the values shown
Where supports consists of columns or walls extending for a distance equal to
or greater than three – quarter other span length L2 used to compute Mo,
negative moments shall be considered to be uniformly distributed across Lo.
E cb C
t
2 E cs I s
Where :
C = torsional rigidity of the effective transverse beam
Column strips shall be proportioned to resist the following in percent of
positive factored moments.
TABLE C
l2
l1
0.5
1.0
2.0
l2
0
l1
60
60
60
l2
1.0
l1
90
75
45
Linear interpolation be made for the values shown
For slabs with beams between supports, the slab portion of the column strip
shall be proportioned to resist that portion of the column strip moments not
resisted by the beams.
Factored moment in beams
Beams between supports shall be proportioned to resist 85% of column strip
l
moments If l 2 1.0
. If
is between 1.0 and zero
2
,proportion of l1
l1
of column strip moments resisted by beams shall be obtained by linear
interpolation between 85 and zero percent.
Factored moment in middle strip
That portion of negative and positive factored moments not resisted by
column strips shall be proportionately assigned to corresponding half middle
strip. Each middles strip shall be proportioned to resist the sum of moment
assigned to its two half middle strip. A middle strip assigned to and parallel
with one edge supported by a wall shall be proportioned to resist twice the
moment assigned to the half middle strip corresponding to the first row of
Interior supports.
Modification of Factored Moments
Negative and positive factored moments maybe modified by 10 % provided
the total static moment for a panel in the direction considered is not less
than that required by EQ. A.
Factored shear in slab system with beams
l
2 1.0
Beams with l1
shall be proportioned to resist shear caused by factored
loads on tributary areas bounded by 450 lines drawn from the corners of
the panel and the centerlines of adjacent panels parallel to the long side.
Beams with l 2 1.0 maybe proportioned to resist shear obtained by linear
l1
Interpolation, assuming that the beam carries no load at α = 0
C.S for wide beam shear
bw
d
450
x
L2/2
L2
1
450
L1
Where
d =effective depth of the slab
Problem
Using NSCP specifications, determine the minimum slab thickness of the panel
shown. The slab has beam between all supports which are poured
monolothically with the slab. fc’ =21MPa, fy = 414 MPa. Assume Ec be the
same for beam, slab and column.
B
5m
5m
6m
A
Panel
for design
A
6m
6m
B
h
500 mm 500 mm
Section B-B
350 mm
350 mm
Solution
1. Assume
h
h
Section A-A
350 mm
350 mm
Ln ( 800 0.73 f y )
36000 9000
LongClearSpan 6000 350
1.22
ShortClearSpan 5000 350
Ln =6000 -350 =5650 mm
5650( 800 0.73 414 )
h
36000 9000( 1.22 )
h = 133 mm say 150 mm
2. Effective flange projection of beams
Section 5.13.2.4 For monolithic or fully composite construction, a beam includes that portion of the
slab on each side of the beam extending a distance equal to the projection of the beam above or
below the slab, whichever is greater, but not greater than four times the slab thickness
a)
b)
3.
500 – 150 = 350 mm
4(150)
= 600 mm
use 350 mm
Moment of Inertias of beams
Approximate moment of inertias
Edge beams 1.4 to 1.6 of bh3/12
Interior beams 1.5 to 2 of bh3/12
a. Edge beam
b. Interior beam
350( 500 )3
I 1.4
5 ,104 x10 6 mm 4
12
350( 500 )3
I 1 .6
5 ,833 x10 6 mm 4
12
E cb I b
E cs I s
4. Values of α
Note : Ecb =Ecs
a) For edge beam with 3 m wide slab
3000( 150 ) 3
Is
843.75 x10 6 mm 4
12
5104
1
6.05
843.75
b) For interior beam with 5 m wide slab ( 2 beams)
5000( 150 )3
Is
1406.25 x10 6 mm 4
12
2
5833
4.15
1406.25
c) For interior beam with 6 m wide slab
6000( 150 )3
Is
1687.5 x10 6 mm 4
12
3
5833
3.46
1687.5
d) αm = average value of α for all beams on edges of a panel.
1 2 2 3 6.05 2( 4.15 ) 3.46
m
4.45
4
4
Check thickness
a) Minimum thickness
5.
h
Ln ( 800 0.73 f y )
1
36000 5000 m 0.12( 1
)
5650( 800 0.73 414 )
h
1
36000 5000( 1.22 ) 4.45 0.12( 1
)
1.22
h = 101 mm
< 133 mm o.k
The values of h obtained from EQS.10.1,10.2 and 10.3 maybe shall be modified as
required by Sections 5.9.5.3.4 and 5.9.5.3.5 but in no case shall the thickness be
less than
m 2.0
m 2.0
Since
m 4.45
------------------------- 125 mm
------------------------- 90 mm
hmin= 90 mm < 133 mm ok
b) thickness need not be more than
5650( 800 0.73 414 )
h
173mm133mm
36000
o.k
Design the slab given in the preceding problem using auniform dead load
of 1.3 kPa excluding the weight of the slab and a uniform live load of
5.7 kPa. Use 12 mm bars , h = 150 mm and C = 4282 x 106 mm4.
Solution
1. Slab weight :
Ws = Wct(1)
Ws =23.5(0.15) = 3.53 kPa
Total uniform dead load
WD = 1.3 + 3.53 = 4.83 kPa
Total Factored uniform Load
Wu = 1.4WD+1.7WL = 1.4(4.83) +1.7(5.7) = 16.452 kPa
2. Effective depth
d = h - 1/2 bar diameter – cover
d = 150 - 1/2(12) – 20 = 124 mm
3. check
depth for shear
l2
Beams with 1.0 shall be proportioned to resist shear caused by factored loads on
l
tributary areas 1 bounded by 450 lines drawn from the corners of the panel and the centerlines
of adjacent panels parallel to the long side.
5m
l2
5
( 4.15 ) 3.451.0
l1
6
1
450
x = 2.5-0.35/2 -0.124 = 2.201 m
x
0.124 m
450
C.S for wide beam shear
2.5 m
0.35 m
6m
taking b = 1 m
Shear force at critical section
Vu = Wu (shaded area)
Vu =Wu(x)(1)
Vu =16.452(2.201)
Vu = 36.21 kN
Shear carried by concrete
1
Vc
fc' bd
6
1
Vc
21( 1000 )( 124 )
6
Vc 94 ,706 N 94.7 kN
Vc 0.85( 94.7 ) 80.5 kN 36.21kN
4. Moment along the short span ( 5 m interior span)
The total moment resisted by the slab equals absolute sum of the positive
and negative factored moments in each direction and shall not be less than
( wu L2 )Ln
Mo
8
2
EQ. A
Where: wu is the factored
load
Clear span Ln shall extend face to face of the columns,capitals,bracket or
walls Ln in EQ. A shall not be less than 0.65L 1
L1 =5 m
0.35 m
0.35 m
Ln = 5 – 0.35 = 4.65 m
( 16.452( 6 )( 4.65 ) 2
Mo
8
Ln
L2 = 6 m
M o 266.8 kN .m
Ok for shear
In an interior span, total static moment Mo shall be distributed as follows:
Negative factored moments
0.65
Positive factored moments
0.35
Negative factored moment=0.65(266.8) = -173.42 kN.m
Positive factored moment =0.35(266.8) = 93.38 kN.m
Distribute the above negative and positive moment to the
column strip, beam and middle strip
l2
l2 6
3.46 (in the direction of L1) l 3.46( 1.2 ) 4.15 1.0
1.2
1
l 5
1
Factored Moment in Column strips
Column strips shall be proportioned to resist the following portion in percent of interior factored
negative moments.
Table A
l2
l1
0.5
1.0
2.0
l2
1.0
l1
90
75
45
Linear interpolation
be made for the values shown
l2
l1
1.0 - 2.0
1.0 – 1.2
1.0
1.2
2.0
Percentage
75
x
45
75 - x
- 45
1.0 - 2.0
75
75 - 45
75 - x
1.0 - 1.2
x= 69
Interior Negative moment resisted by column strip =0.69(-173.42)= -119.66kN.m
Factored moment in beams
Beams between supports shall be proportioned to resist 85% of column strip moments If
l2
1.0
l1
0.85(-119.66) = -101.71kN.m will be resisted by the beam
0.15(-119.66) = -17.95 kN.m will be resisted by the slab
The remaining –(173.42 – 119.66) = - 53.76 kN.m is allotted to the middle strip
Column strips shall be proportioned to resist the following in percent of positive factored moments.
TABLE C
l2
l1
0.5
1.0
2.0
l2
1.0
l1
90
75
45
Linear interpolation
be made for the values shown
l2
l1
1.0 - 2.0
1.0 – 1.2
1.0
1.2
2.0
Percentage
75
x
45
75 - x
- 45
1.0 - 2.0
75 - x
1.0 - 1.2
75
75 - 45
x= 69
Positive moment resisted by column strip =0.69(93.38)= 64.43kN.m
0.85(64.43) = 54.77kN.m will be resisted by the beam
0.15(64.43) = 9.66 kN.m will be resisted by the slab
The remaining (93.38 – 64.43) =28.95 kN.m is allotted to the middle strip
5. Moment along the short span (along edge beam)
L1 =5 m
Ln = 5 – 0.35 = 4.65 m
If the transverse span of panels on either side of the centerline of
supports varies L2 in EQ. A shall be taken as the average of
adjacent transverse spans. When the span adjacent and
parallel to an edge is considered, the distance from edge to panel
centerline shall be substituted for L 2 in EQ. A.
L2 =6/2 + 0.35/2 = 3.175 m
L2
Ln
0.35 m
0.35 m
( 16.452( 3.175 )( 4.65 ) 2
Mo
8
M o 141.18 kN .m
Negative factored moment=0.65(141.18) = -91.78 kN.m
Positive factored moment =0.35(141.18) = 49.9 kN.m
l2 6
1.2
l1 5
6.05 (for edge beam)
l2
6.05( 1.2 ) 7.26 1.0
l1
3000( 150 ) 3
Is
843.75 x10 6 mm 4
12
E cb C
4282 x10 6
t
2.54
6
2 E cs I s
2( 843.75 )( 10 )
Note: Ecb =Ecs
Column strips shall be proportioned to resist the following portion in percent
Of exterior factored negative moments.
TABLE B
l2
l1
t 0
t 2.5
l2
1.0
l1
0.5
1.0
100
90
100
75
2.0
100
45
Linear interpolation be made for the values shown
l2
l1
1.0 - 2.0
1.0 – 1.2
1.0
1.2
2.0
Percentage
75
x
45
75 - x
- 45
1.0 - 2.0
75
75 - 45
Exterior Negative Moment
Column strip =0.69(-91.78) = 63.33 kN.m
Beam =0.85(-63.33) = - 53.83 kN.m
slab = 0.15(-63.33) = - 9.5 kN.m
Middle strip =-(91.78 -63.33) = - 28.45 kN.m
x= 69
75 - x
1.0 - 1.2
Positive Moment
From Table C
l2
l1
Percentage
1.0
75
75 - x
1.0 - 2.0
1.2
x
75 - 45
2.0
45
Column strip =0.69(49.4) = 34.1 kN.m
Beam =0.85(34.1) = 28.99 kN.m
slab = 0.15(34.1) = 5.12 kN.m
Middle strip = (49.4 -34.1) = 15.30 kN.m
1.0 – 1.2
75 - x
1.0 - 1.2
x= 69
-28.45 -9.5
-26.88
1.25 m
-8.975 -8.975
+15.3
+14.475
+5.12
2.5 m
+4.83 +4.83
-28.45 -9.5
-26.88
-8.975 -8.975
1.25
3.5
1.25
1.25 m
- 45
1.0 - 2.0
75
Exterior Column strip
-17.95/2 =-8.975
+9.66/2= +4.83
-53.76/2=-26.88
+28.95/2 = 14.475
-(26.88+28.45)= -55.33
+(14.475+15.3)=+29.78
interior Column strip
6. Moment along the long span (6m end span)
L2= 5 m
Ln = 6 -0.35 = 5.65 m
( 16.452( 5 )( 5.65 ) 2
Mo
8
M o 328.24 kN .m
Ln
L1 = 6 m
In an end span, total factored static moment Mo shall
be distributed according to the table below
1
2
3
4
Slabs without beams
Exterior edge Slabs with
b/w interior supports
unrestrained beams b/w
With
Without
all supports
edge beam
edge beam
Interior Negative
Factored Moment
5
Exterior edge
Fully restrained
0.75
0.70
0.70
0.70
0.65
Positive
Factored Moment
0.63
0.57
0.52
0.50
0.35
Exterior Negative
Factored Moment
0
0.16
0.26
0.30
0.65
Factored Interior Negative Moment = -0.7(328.24)= -229.77 kN.m
Factored Positive Moment = -0.57(328.24)= -187.1 kN.m
Factored Exterior Negative Moment = -0.16(328.24)= -52.52 kN.m
l2 5
0.83
l1 6
4.15
l2
4.15( 00.83 ) 3.44 1.0
l1
Interior Negative Moment ( Use Table A to get percentage)
l2
l1
0.5 - 1.0 0.5 – 0.83
0.5
0.83
1.0
Percentage
90
x
75
90 - x
90 - 75
Column strip =0.801(-229.77) = -184 kN.m
Beam =0.85(-184) = - 156.4 kN.m
slab = 0.15(-184) = - 27.6 kN.m
Middle strip =-(229.77- 184) = - 45.77 kN.m
90 x 0.5 0.83
90 75
0.5 1.0
x 80.1
Exterior Negative Moment ( Use Table B to get percentage)
Percentage
l2
l1
0.5
90
0.5 – 0.83
90 - x 90 - 75
0.5 - 1.0
0.83
x
1.0
75
Column strip =0.801(-52.52) = -42 kN.m
Beam =0.85(-42) = - 37.5 kN.m
slab = 0.15(-42) = - 6.3 kN.m
Middle strip =-(52.52- 42) = - 10.52 kN.m
90 x 0.5 0.83
90 75
0.5 1.0
x 80.1
Positive Moment ( Use Table C to get percentage)
l2
l1
Percentage
0.5
90
90 - x
0.83
x
90 - 75
1.0
75
Column strip =0.801(187.1) = 150 kN.m
Beam =0.85(150) = 127.5 kN.m
slab = 0.15(150) = 22.5 kN.m
Middle strip =(187.1- 150) = 37.1 kN.m
0.5 - 1.0 0.5 – 0.83
90 x 0.5 0.83
90 75
0.5 1.0
x 80.1
Edge Beam
-10.52
-3.15
-3.15
22.5/2
11.25
11.25
37.1
-27.6/2
-13.8
1.25 m
11.25
-13.8
-45.77
-13.8
1.25 m
2.5 m
1.25 m
11.25
Column strip
Column strip
-3.15
-6.3/2
-3.15
-13.8
1.25 m
Note: Since the column strip exist on both sides of the beam, the value
Of the moment alloted to the slab must be divided by 2. In the middle
Strip it is not divided by 2 since the same moment is also alloted by the
Opposite middle strip.
+37.1
-13.8
-45.77
1.25 m
B
2.5 m
-13.8
C
1.25 m
D
-8.975
+11.25
+4.83
A
F
+11.25
E
-3.15
-9.5
-10.5
Edge Beam
-53.33
-3.15
+5.12
C
+28.97
-9.5
-53.33
B
-8.975
F
E
D
1.25 m
3.5 m
1.25 m
A
Row A & C
Row B
Across F Across E Across D Across F Across E Across D
Mu
b
11.25
-13.8
-10.5
37.1
-45.77
1075
1075
1075
2500
2500
2500
112
124
112
124
124
124
-3.15
d
Ru
0.25955
0.75624
1.13708 0.30350
1.07238
1.32299
ρ
0.00063
0.00187
0.00284 0.00074
0.00267
0.00332
Use ρ
0.00338
0.00338
0.00338
0.00338
0.00338
0.00338
As
407.15
450.77
407.15
1048.31
1048.31
1048.31
3.6
3.99
3.6
9.27
9.27
9.27
s
Use s
298.61
269.71
298.61
269.71
Position
Top
N
pmin
290
260
Bottom
1.4
0.00338
414
290
Top
269.71
269.71
260
260
260
Top
Bottom
Top
Row D & F
Row E
Across A Across B Across C Across A Across B Across C
5.12
-9.5
-55.33
29.78
-55.33
1075
1075
1075
3500
3500
3500
124
124
124
124
124
124
0.34417
0.6386
1.14237
0.61485
1.14237
0.00157
0.00084
0.00157 0.00285
0.00151
0.00285
Use ρ
0.00338
0.00338
0.00338
0.00338
0.00338
0.00338
As
450.77
450.77
450.77
1467.63
1467.33
1467.33
3.99
3.99
3.99
12.98
12.98
12.98
s
Use s
267.71
269.71
269.71
269.71
269.71
269.71
Position
Top
Mu
b
-9.5
d
Ru
0.6386
ρ
N
260
260
Bottom
260
Top
260
260
260
Top
Bottom
Top
Temperature bars:
At =0.0018(1000)(150) = 270 mm2 Using 10 mm S = 418 mm
Maximum spacing = 2t = 300 mm say 260 mm
ACI Moment Coeffecient Method
The complexity of the Direct Design Method particularly in meeting
its requirements paved way to the ACI Moment Coeffecient Method.
The method makes use of tables of moment coeffecients for a
variety of conditions. These coeffecients are based on elastic
analysis but also takes into account inelastic distribution.
Moments at column strip and middle strip are computed by
M a Ca wl
2
a
M b Cb wl
2
b
Ca , Cb=Tabulated moment coeffecients
la , lb
= Clear span length in short and long directions respectively
w=uniform load in psf,Pa or KPa
la
la/4
lb/4
Column strip
lb
lb/4
Middle
strip
Column strip
Width of middle strip – one-half of the panel
Width of edge or column strip – one-fourth of the panel
Column strip
Column strip
panel
la/4
lb
Table 1: Coeffecients for Negative Moments in Slabs
M a ,neg C aNeg wl a 2
Ratio
m
1.0
la
Case 1 Case 2
Case 3
Case 4
w= total uniform dead plus live load
Case 5 Case 6 Case 7 Case 8 Case 9
lb
Ca,Neg
Cb,Neg
0.95 Ca,Neg
Cb,Neg
0.90 Ca,Neg
Cb,Neg
0.85 Ca,Neg
Cb,Neg
0.80 Ca,Neg
Cb,Neg
0.75
Ca,Neg
Cb,Neg
0.70
Ca,Neg
Cb,Neg
0.65
Ca,Neg
Cb,Neg
0.60
Ca,Neg
Cb,Neg
0.55 Ca,Neg
Cb,Neg
0.50
M b.neg CbNeg wlb
2
Ca,Neg
Cb,Neg
0.045
0.045
0.050
0.041
0.055
0.037
0.060
0.031
0.065
0.027
0.069
0.022
0.074
0.017
0.077
0.014
0.081
0.010
0.084
0.007
0.086.
0.006
0.076
0.072
0.070
0.065
0.061
0.056
0.050
0.043
0.035
0.028
0.022
0.050
0.050
0.055
0.045
0.060
0.040
0.066
0.034
0.071
0.029
0.076
0.024
0.081
0.019
0.085
0.015
0.089
0.011
0.092
0.008
0.094.
0.006
0.075
0.079
0.071
0.075
0.080
0.079
0.082
0.083
0.083
0.086
0.085
0.088
0.071
0.067
0.062
0.057
0.051
0.044
0.086
0.091
0.087
0.093
0.088
0.095
0.089
0.096
0.090
0.097
0.038
0.031
0.024
0.019
0.014
0.033
0.061
0.038
0.056
0.043
0.052
0.049
0.046
0.055
0.041
0.061
0.036
0.068
0.029
0.074
0.024
0.080
0.018
0.085
0.015
0.089.
0.010
0.061
0.033
0.065
0.029
0.068
0.025
0.072
0.021
0.075
0.017
0.078
0.014
0.081
0.011
0.083
0.008
0.085
0.006
0.086
0.005.
0.088
0.033
Table 2: Coeffecients for Dead Load Positive Moments in Slabs
2
w= total uniform dead load
M apos ,dl C a ,dl wl a2
bpos , dl
b.dl
b
M
Ratio
m
1.0
la
Case 1 Case 2
Case 3
C wl
Case 4
Case 5 Case 6 Case 7 Case 8 Case 9
lb
Ca,dl
Cb,dl
0.95 Ca,dl
Cb,dl
0.90 Ca,dl
Cb,dl
0.85 Ca,dl
Cb,dl
0.80 Ca,dl
Cb,dl
0.75 Ca,dl
Cb,dl
0.70
Ca,dl
Cb,dl
0.65
Ca,dl
Cb,dl
0.60
Ca,dl
Cb,dl
0.55 Ca,dl
Cb,dl
0.50 Ca, dl
Cb, dl
0.036
0.036
0.040
0.033
0.045
0.029
0.050
0.026
0.056
0.023
0.061
0.019
0.068
0.016
0.074
0.013
0.081
0.010
0.088
0.008
0.095
0.006
0.018
0.018
0.020
0.016
0.022
0.014
0.024
0.012
0.026
0.011
0.028
0.009
0.030
0.007
0.032
0.006
0.034
0.004
0.035
0.003
0.037
0.002
0.018
0.027
0.021
0.025
0.025
0.024
0.029
0.022
0.034
0.020
0.040
0.018
0.046
0.016
0.054
0.014
0.062
0.011
0.071
0.009
0.080
0.006
0.027
0.027
0.030
0.024
0.033
0.022
0.036
0.019
0.039
0.016
0.043
0.013
0.046
0.011
0.050
0.009
0.053
0.007
0.056
0.005
0.059
0.004
0.027
0.018
0.028
0.015
0.029
0.013
0.031
0.011
0.032
0.009
0.033
0.007
0.035
0.005
0.036
0.004
0.037
0.003
0.038
0.002
0.039
0.001
0.033
0.027
0.036
0.024
0.039
0.021
0.042
0.017
0.045
0.015
0.048
0.012
0.051
0.009
0.054
0.007
0.056
0.006
0.058
0.004
0.061
0.003
0.027
0.033
0.031
0.031
0.035
0.028
0.040
0.025
0.045
0.022
0.051
0.020
0.058
0.017
0.065
0.014
0.073
0.012
0.081
0.009
0.089
0.007
0.020
0.023
0.022
0.021
0.025
0.019
0.029
0.017
0.032
0.015
0.036
0.013
0.040
0.011
0.044
0.009
0.048
0.007
0.052
0.005
0.056
0.004
0.023
0.020
0.024
0.017
0.026
0.015
0.028
0.013
0.029
0.010
0.031
0.007
0.033
0.006
0.034
0.005
0.036
0.004
0.037
0.003
0.038
0.002
Table 3: Coeffecients for Live Load Positive Moments in Slabs
2
w= total uniform live load
M apos ,ll C a ,ll wl a2
bPos ,ll
b ,ll
b
M
Ratio
m
1.0
la
Case 1 Case 2
Case 3
C wl
Case 4
Case 5 Case 6 Case 7 Case 8 Case 9
lb
Ca,ll
Cb,ll
0.95 Ca,ll
Cb,ll
0.90 Ca,ll
Cb,ll
0.85 Ca,ll
Cb,ll
0.80 Ca,ll
Cb,ll
0.75 Ca,ll
Cb,ll
0.70
Ca,ll
Cb,ll
0.65
Ca,ll
Cb,ll
0.60
Ca,ll
Cb,ll
0.55 Ca,ll
Cb,ll
C ,ll
0.50 Ca,ll
b
0.036
0.036
0.040
0.033
0.045
0.029
0.050
0.026
0.056
0.023
0.061
0.019
0.068
0.016
0.074
0.013
0.081
0.010
0.088
0.008
0.095
0.006
0.027
0.027
0.030
0.025
0.034
0.022
0.037
0.019
0.041
0.017
0.045
0.014
0.049
0.012
0.053
0.010
0.058
0.007
0.062
0.006
0.066
0.004
0.032
0.032
0.031
0.029
0.035
0.027
0.040
0.024
0.045
0.022
0.051
0.019
0.057
0.016
0.064
0.014
0.071
0.011
0.080
0.009
0.088
0.007
0.032
0.032
0.035
0.029
0.039
0.026
0.040
0.023
0.048
0.020
0.052
0.016
0.057
0.014
0.062
0.011
0.067
0.009
0.072
0.007
0.077
0.005
0.032
0.027
0.034
0.024
0.037
0.021
0.041
0.019
0.044
0.016
0.047
0.013
0.051
0.011
0.055
0.009
0.059
0.007
0.063
0.005
0.067
0.004
0.035
0.032
0.038
0.029
0.042
0.025
0.046
0.022
0.051
0.019
0.055
0.016
0.060
0.013
0.064
0.010
0.068
0.008
0.073
0.006
0.078
0.005
0.032
0.035
0.036
0.032
0.040
0.029
0.045
0.026
0.051
0.023
0.056
0.020
0.063
0.017
0.070
0.014
0.077
0.011
0.085
0.009
0.092
0.007
0.028
0.030
0.031
0.027
0.035
0.024
0.040
0.022
0.044
0.019
0.049
0.016
0.054
0.014
0.059
0.011
0.065
0.009
0.070
0.007
0.076
0.005
0.030
0.028
0.032
0.025
0.036
0.022
0.039
0.020
0.042
0.017
0.046
0.013
0.050
0.011
0.054
0.009
0.059
0.007
0.063
0.006
0.067
0.004
Table 4: Ratio of load w in la and lb directions for shear and slabs and load
0n supports
Ratio
m
la
Case 3
Case 4
Case 5 Case 6 Case 7 Case 8 Case 9
lb
Wa
Wb
0.95 Wa
Wb
0.90 Wa
Wb
0.85 Wa
Wb
0.80 Wa
Wb
0.75 Wa
Wb
lW
a
0.70
Wb
0.65 Wa
Wb
0.60 Wa
Wb
0.55 Wa
Wb
0.50 Wa
Wb
1.0
Case 1 Case 2
0.50
0.50
0.55
0.45
0.60
0.40
0.66
0.34
0.71
0.29
0.76
0.24
0.81
0.19
0.85
0.15
0.89
0.11
0.92
0.08
0.94
0.06
0.50
0.50
0.55
0.45
0.60
0.40
0.66
0.34
0.71
0.29
0.76
0.24
0.81
0.19
0.85
0.15
0.89
0.11
0.92
0.08
0.94
0.06
0.17
0.83
0.20
0.80
0.23
0.77
0.28
0.72
0.33
0.77
0.39
0.61
0.45
0.55
0.53
0.47
0.61
0.39
0.69
0.31
0.76
0.24
0.50
0.50
0.55
0.45
0.60
0.40
0.66
0.34
0.71
0.29
0.76
0.24
0.81
0.19
0.85
0.15
0.89
0.11
0.92
0.08
0.94
0.06
0.83
0.17
0.86
0.14
088
0.12
0.90
0.10
0.92
0.08
0.94
0.06
0.95
0.05
0.96
0.04
0.97
0.08
0.98
0.02
0.99
0.01
0.71
0.29
0.75
0.25
0.79
0.21
0.83
0.17
0.86
0.14
0.88
0.12
0.91
0.09
0.93
0.07
0.95
0.05
0.96
0.04
0.97
0.03
0.29
0.71
0.33
0.67
0.38
0.62
0.43
0.57
0.49
0.51
0.56
0.44
0.62
0.38
0.69
0.31
0.76
0.24
0.81
0.19
0.86
0.14
0.028
0.030
0.38
0.62
0.43
0.57
0.49
0.51
0.55
0.45
0.61
0.39
0.68
0.32
0.74
0.26
0.80
0.20
0.85
0.25
0.89
0.11
0.030
0.028
0.71
0.29
0.75
0.25
0.79
0.21
0.83
0.17
0.86
0.14
0.89
0.11
0.92
0.08
0.94
0.06
0.95
0.05
0.97
0.03
Discontinuous Edge
Continuous edge
Cases Description
Case 1 :
Case 2 :
Case 3 :
Case 4 :
Case 5 :
Case 6 :
Case 7:
Case 8:
Case 9:
Simply supported
Interior Panel
Edges of the Long span continuous
One edge of Long span, one edge
of short span continuous (corner panel)
Edges of the short span continuous
One edge of short span continuous
One edge of long span continuous
Two edge of long span, one edge
of short span continuous
Two edge of short span, one edge
of long span continuous
Uses of Tables of coeffecients
Table 1 - to compute the total negative moment at the continuous edge(column strip)
Table 2 - to compute the positive dead load moment at the midspan(middle strip)
Table 3- to compute the positive live load moment at the midspan(middle strip)
Table 4 - to compute shear at the long and short span
Total Positive moment at midspan
Sum of positive dead load moment and positive live load moment
Negative Moment at Discontinuous egde =1/3 of Positive moment at the midspan
Minimum thickness
1/180 times the perimeter of the panel but not to be less than 75 mm
hmin
2(la lb )
180
Problem
Design the thickness and reinforcement of the cornel panel shown usin the
ACI moment coeffecient method. The slab has beam between all supports
which are poured monolothically with the slab. fc’ =20 MPa, fy = 350 MPa
The slab is to support a uniform dead load of 4.5 kPa and uniform live load of
3.9 kPa. Concrete weighs 23.5 kN/m3.
B
5.3 m
500 mm
h
Section A-A
300 mm
300 mm
h
Section B-B
300 mm
4.3 m
Panel
for design
A
A
5.3 m
500 mm
5.3 m
300 mm
B
Solution
Clear span length
short span = la = 4.3 – 0.3 = 4.0m
long span = lb = 5.3 – 0.3 = 5.0m
Slab thickness:
hmin
2(la lb )
180
2(4000 5000)
h
100mm 75mm
180
Weight of slab
Ws = 0.10(23.5) =2.35 kN/m
Total Dead load
WD = 4.5 + 2.35 = 6.85 kN/m
Factored loads
Dead load = 1.4 (6.85) = 9.59 kN/m
Live load = 1.7 (3.9) = 6.63 kN/m
Total factored load = 16.22 kN/m
Effective depth ( using 12 mm bars)
d = 100 -1/2(12) – 20 =74 mm
m
la 4
0.80
lb 5
Check for Shear
From Table 4 Wa = 0.71, Wb = 0.21, showing that the shear in the short
direction is more critical than the long direction.
uniform load per m width Wu = 0.71( 16.22) = 11.52 kN/m
11.52 kN/m
c.s for shear
0.074 m
4m
R=11.52(4)/2=23.04 kN
Shear force at critical section
Vu =23.04 -0.074(11.52) = 22.19 kN=22190 N
Allowable shear force
0.85 20 (1000)74
Vu
46883 N
6
Design Moments
Short Direction
Negative Moment at continuous edge:
From Table 1 CaNeg =0.071, w = 16.22
Ma,Neg = 0.071(16.22)(4)2=18.43 kN/m
Positive Moment at the midspan
Dead Load
M apos ,dl C a ,dl wl a2
From Table 2 ,Ca,dl =0.039 w= 9.59
MaPos,dl=0.039(9.59)(4)2 =5.98 kN.m
Live Load
M apos ,ll C a ,ll wl a2
From Table 3 ,Ca,ll =0.048 w= 6.63
MaPos,ll=0.048(6.63)(4)2 =5.09 kN.m
Total Positive Moment at Midspan
MaPos = 5.98 + 5.09 = 11.07 kN.m
Negative Moment at discontinuous edge
MaNeg = -1/3MaPos
MaNeg = -1/3(11.07) = -3.69 kN.m
Long Direction
Negative Moment at continuous edge:
From Table 1 CbNeg =0.029, w = 16.22
M b.neg CbNeg wlb
2
Mb,Neg = 0.029(16.22)(5)2=11.76 kN/m
Positive Moment at the midspan
Dead Load
M bpos ,dl Cb.dl wlb2
From Table 2 ,Cbdl =0.016 w= 9.59
MbPos,dl=0.016(9.59)(5)2 =3.84 kN.m
Live Load
M bPos ,ll Cb,ll wlb2
From Table 3 ,Cb,ll =0.020 w= 6.63
MbPos,ll=0.020(6.63)(5)2 =3.32 kN.m
Total Positive Moment at Midspan
MbPos = 3.84 + 3.32 = 7.16 kN.m
Negative Moment at discontinuous edge
MbNeg = -1/3MaPos
MbNeg = -1/3(7.16) = -2.39kN.m
D
D
1.25 m
-11.76
A
B
C
1.0 m
2.0m
1.0 m
E
B
-3.69
+11.07
F
F
E
-18.43
1.25 m
2.5 m
A
C
-2.39
+7.16
Row A
Row B
Row C
Row D
Row E
-18.43
11.07
-3.69
-11.76
7.16
1000
1000
1000
1000
1000
1000
74
74
74
74
68
74
1.72
Across D,
E&F
Mu
b
Across D, Across D, Across A,
E&F
E&F
B&C
Across A,
B&C
Row F
Across A,
B&C
-2.39
d
Ru
3.74
2.25
2.39
ρ
0.0122
0.00692
0.0074
0.0052
Use ρ
0.0122
0.00692
0.0074
0.0052
As
902.8
512.08
547.6
384.8
s
125
220
206
293.9
618
150
300
300
12 mm
12 mm
12 mm
12 mm
Top
Top
Use s
110
Bar size
12 mm
Position
Top
Smax = 5t = 500 mm
220
12 mm
Bottom
660
220
Bottom
Top
Temperature bars At = 0.0020(1000)100 = 200
S = 392 mm use s = 300 mm Bar size 10 mm
Concrete Design
Semi- Final Exam
1.A square tied column 300 mm by 300 mm is reinforced with 6 of 20 mm bars
with fc’ = 20.7 MPa and fy = 345 MPa. Determine the following :
a) Ultimate axial load capacity of the column.
b) spacing of 10 mm lateral ties
(20) 2
Pu =Ф0.80{0.85fc’(Ag-Ast) + Astfy} Ast 6
1884.95mm 2
4
0.7(0.80){0.85(20.7)[(300)(300) 1884.95] 1884.95(345)}
Pu
1232.38kN
1000
Spacing of 10 mm ties
S 16(20) 400mm
S 48(10) 480mm
S 300mm
Use S = 300 mm
2. Calculate the ultimate axial load capacity of the composite column shown
below .fc’=21 MPa, Fy=250 MPa,fy =350MPa. Reinforcing bars consist of 8 of
20 mm diameter bars.
400 mm
400 mm
W10 x 150
Properties of W 10 x 150
Area = 27000 mm2
Depth = 250 mm
8 [(20) 2 ]
As
2513mm 2
4
Ac 400(400) 27000 2513 130487 mm 2
Pu 0.85 0.85 fc ' Ac As f y AWF Fy
Pu (0.7)0.85 0.85(21)(130487) 2513(350) 27000(250)
Pu 5925452 N 5,925.52kN
3. A cantilever beam 300 mm wide and 450 mm deep is reinforced with
6 of 22 mm straight top bars with fc’ = 20.7 MPa and fy = 415 MPa.
Calculate required development length.
Solution
Ldb
Ldb
25 f y
fc '
25(415)
20.7
2280mm
Modification Factor
Top bar = 1.3
Required development length
Ld 1.3Ldb 1.3( 2280) 2964mm
4. A rectangular beam 350 mm wide and 600 mm deep is reinforced
with 4 of 32 mm compression bars with fc’ = 20.7 MPa and fy = 275
MPa. Calculate the required development length.
0.24d b f y
Ldb
fc'
0.24(32)275
Ldb
464.2mm
20.7
Ldb 0.04d b f y
Ldb 0.04(32)(275) 352mm
Use L
db
464.2mm
No applicable modification factor
Ld 464.2mm
5. A rectangular beam with b = 300 mm, d = 500 mm is provided with 10 mm
vertical stirrups with fy = 276 MPa. Assuming fc’ = 21 MPa;
a) Determine the required spacing if Vu = 42 kN
b) Determine the required spacing if Vu = 95 kN
Solution
Shear carried by concrete
1
Vc
6
fc'bw d
1
Vc
21(300)500 114,564 N
6
Vc 0.85(114564 )
48689.7 N
2
2
a)
b)
Vc
Stirrups not necessary
Vu 42000 N
2
Vu 95000 N
Vu
Vs
Vc
95000
Vs
114564 2799 N
0.85
2 (10) 2
Av
157.1mm 2
4
S
Av f y d
Vs
157.1( 276)500
7745mm
2799
1
3
1
fc 'bw d
21(300)(500) 229129 N
3
1
Vs
fc 'bw d
3
d 500
S
250mm
2
2
Use S = 250 mm
Semi-Final Exam
1. A square concrete column has a dimension of 400 mm by 400 mm.
The column length is 4.5 m and is to be used in a frame braced against
sidesway ( k =1.0, EI = 40MN.m2). The column is bent in single
curvature by ultimate factored moment of 100 kN.m at the top and 120
kN.m at the bottom. The column carries factored axial dead load of 150
kN and factored axial live load of 320 kN. Is slenderness effects be
considered in the analysis ? If so, what is the magnified moment
capacity of this column ?
2. Solve the previous problem considering the effect of wind loads
producing an unfactored axial load of 125 kN an unfactored moment of
90 kN.m.
8 of 28 mm
350 mm
500 mm
3. For the column shown in figure, PD = 250 kN, PL = 410kN, MD = 140
kN.m, and ML = 152 kN.m. The column belongs to a braced frame and
is bent in single curvature about the x axis by equal moments at each
end of the member . The unsupported length is 6 m. fc’ = 20.7 MPa, f y =
413.7 MPa . Determine the following :
a) Critical load Pc
b) Magnified moment
350 mm
Solution to #1
M 2b 120kN .m
M 1b 100kN .m
M 2s 0
r 0.3h 0.3(400) 120mm
Lu 4.5m 4500mm
kLu 1(4500)
37.5
r
120
M 1B
100
34 12
34 12
24
M 2B
120
kLu
M 1B
34 12
r
M 2B
Column is Long
+ since the
column is bent
in single
curvature
M 1b
100
Cm 0.6 0.4
0.6 0.4
0.933
M 2b
120
Pu 150 320 470kN
2 EI 2 40(10) 6
Pc
19,495,515 N 19,495.5kN
2
2
(kLu ) {(1)(4.5)}
Cm
0.933
b
0.967
Pu
470
1
1
0.7(19495.5)
Pc
M c b M 2b c M 2 s 0.967(120) s (0) 116 .04kN .m
Solution to #2
M 2 s 0.75[1.4M D 1.7 M L 1.7 M W ]
M 2 s 0.75[1.4(0) 1.7(0) 1.7(90)] 114 .75kN .m
[1.4 PD 1.7 PL ] 150 320 470kN
Specified in the previous
problem
Pu 0.75[1.4 PD 1.7 PL 1.7 PW ] 0.75[470 1.7(125)] 511.87 kN
Cm
0.933
b
0.97
Pu
511.87
1
1
0.7(19495.5)
Pc
s 1.0 Frame is braced against sidesway
M c b M 2b s M 2 s 0.97(120) 1.0(114 .75) 231.15kN .m
Solution to # 3
8 (28) 2
As
4926mm 2
4
Pu=1.4PD + 1.7PL = 1047 kN
M1b= M2b =1.4MD + 1.7ML = 454.4 kN.m
Ec 4700 fc' 4700 20.7 21383.7 MPa
350(500)3
Ig
3645.83 x106 mm 4
12
I se As z 2 4926(175) 2 150.86 x106 mm 4
Es 200000 MPa
1.4 M D 1.4(140)
d
0.431
Mu
454.4
Ec I g
EI
E s I se
5
1 d
21383.71(3645.83)106
200000(150.86)10 6
5
EI
1 0.431
EI 4.576 x1013 N .mm 2
2 EI
Pc
(kLu ) 2
2 4.576 x1013
Pc
12545.36kN
2
(6000 )
M 1b
550.54
Cm 0.6 0.4
0.6 0.4
1.0
M 2b
550.54
Cm
1
b
1.135
Pu
1047
1
1
Pc
0.7(12545.36)
M c b M 2b 1.135(454.4) 515.78kN .m
