Transcript Environmental Management

Workshop on
Cost Saving in Steel by Using
500-Graded Rebar in Place of
Ordinary Steel As Reinforcement
Barisal
21 July 2008
Organized by
BSRM Steels Limited
Design Theory on Building Design
and Cost Saving in Steel by Using
500-Graded Rebar in Place of
Ordinary Steel As Reinforcement
Dr. S.M. Atiqul Islam
Associate Professor
Department of Civil Engineering
Dhaka University of Engineering and Technology (DUET)
Gazipur-1700
Tel: 8932939 Mobile: 01711389584
Three aims of Civil Engineering design
 Safety and durability of the structure
 Beauty of the structure
 Economy for constructions
The responsibility of a design engineer is to fulfill these aims
 use of appropriate and quality materials
 Use of modern technology with the advancement of technology
 Proper design and method of construction
Proper maintenance|
Materials used in Civil Engineering RCC Construction
Reinforced concrete is concrete in which reinforcement bars
("rebar") have been incorporated to strengthen the material
 Reinforcement as Ms. bar
 Cement
 Brick
 Stone
 Lime, etc.
Proper design and appropriate use of these materials ensures the safety and
durability of the structure
For the economic consideration the use materials should be such that ensure
the durability and safety within the affordability
Structural Design consists the following works
 Calculation and analysis of loads.
 Realize of effects due to loads.
 Findings way to prevent them by selecting and providing
proper materials in proper places.
In order to understand the structural design it is necessary to
understand the failure mechanism of a structure
Flexural failure mechanism of
RCC Structure
40 (300) Grade :
Yield Strength = 40 ksi
300 MPa
60 (420) Grade:
Yield Strength = 60 ksi
420 MPa
500 (72.5) Grade :
Yield Strength = 72.5 ksi
or 500 MPa
Typical reinforced concrete beam in walk-though within transparent reinforced
concrete
Structural Design
It is required for a Structural Engineer to design the structure to
satisfy four major criteria
•Appropriateness- The arrangement of spaces, spans, ceiling
heights, access and traffic flow must competent to intended use.
•Economy- The overall cost of the structure should not exceed
the Client Budget
•Structural adequacy- A structure must be sufficiently strong to
support safety, without collapse, all anticipated loadings and a
structure must not deflect, tilt, vibrate, or crack in a manner that
impairs its usefulness
•Maintainability- A structure should be designed to require a
minimum of maintenance and/or to be able to be maintain in a
simple fashion
Structural Design
The structural design process is a sequential and iterative
decision making process consists of three major phases
•Definition of the client’s need and priorities
•Development of concept of project
•Design of individual system
Design Methods
Usual design methods are based on limit states of the
structural elements. Limit states design is a design process
that involve the followings
•Identification of all potential modes of failure
•Determination of acceptable levels of safety against
occurrence of each limit state
•Consideration by the designer of the significant limit states
Design Methods
Generally there are two types of design method to
select the concrete and steel reinforcement.
First design concept is Working Stress Design
and second one is Ultimate Strength Design.
In Working Stress Design, members are designed so that stresses in
the steel and concrete resulting from normal service loads are
within specified limits. These limits, known as allowable stresses,
are only fractions of the failure stresses of the materials. Concrete
responds reasonably elastically up to compression stresses not
exceeding about half its strength, while steel remains elastic
practically up to the yield stress. Hence members may be designed
on an elastic basis as long as the stresses under service loads
remain below these limits.
c
kd
d
nAs
b
s
T= As fs
Beam Design
Load
• Dead Load: Self wt and furniture, equipment, etc
• Live load
• Lateral Load; Wind, Earthquake
• Impact load (if any)
Column
Lateral Load
Earthquake
Wind/
Earthquake
From Load Analysis by software
or manual
Moment and shear can be
obtained
Flexural Moment M= 0.5 fc kjbd2
M=Asfsjd
Ultimate Strength Design reinforced concrete structures at loads
close to and at failure, one or both of the materials, concrete
and steel, are invariably in their nonlinear inelastic range. That
is, concrete in a structural member reaches its maximum
strength and subsequent fracture at stresses and strains far
beyond the initial elastic range in which stresses and strains are
fairly proportional.
b
0.85 f ’c
u
a =  1c
c
d
y
Mn = As fy (d-a/2) where, a= As fy/0.85 f’cb
b = 0.851 f’c/fy (87,000/87,000+fy)
T= As fy
The ACI code (2002 and 2005 version) establishes the value of the required
strength called U not less than
U = 1.2 D + 1.6 L
Where D is effect of dead load and L is effect of live load.
Other adjustment factors are provided when design conditions involve
consideration of the effect of wind, earthquake, differential settlement, creep,
shrinkage, temperature change. Design strength of structure is determined by
the application of assumptions and requirements given in the code and is
further modified by the use of a strength reduction factor  as follows:
 = 0.9 for flexure, axial tension, and flexure + tension
 = 0.7 for columns with spirals
 = 0.65 for columns with ties
 = 0.75 for shear and torsion
 = 0.65 for compressive bearing
Mu =  Mn
and Vu =  Vn
Where Mu and Vu is external factored moment and shear forces
Mn and Vn nominal ultimate moment and shear capacity of the member
  1 
    
L
1.2 D  1.6L  1  1.2  1.6 D
  
Factor of Safety =
D  L  
1 L
D
Factor of safety for the various values of  and L/D ratio are shown
in the following table

L/D
FS
0.9
0 1
2
3
1.33 1.56 1.63 1.67
0.75
0 1
2
3
1.6 1.87 1.96 2.0
0.7
0 1
2
3
1.71 2.00 2.10 2.14
0.65
0 1
2
3
1.85 2.15 2.26 2.31
 What ever is the design method and what ever is the
element for design, strength of concrete and steel are the main
factors
If the strength is higher the required amount
materials will be less
 High strength steel will require less amount of
steel
For 60 grade steel required amount is 2/3 of 40
grade
For 500 grade required amount is 20% less than
60 grade
For beam and footing
As= M/fsjd
n= Es/Ec k=nfc/(nfc+fs)
j=1-k/3
for 40 grade j= 0.87
for 60 grade j=0.9
As (40 grade) = M/ (0.45x40x0.87xd)
As (60 grade) = M/ (0.45x60x0.9xd)
As (60 grade)/As (40 grade)= 0.64 = 64%
Net save in steel = 34%
Example 1
40 Grade
60 Grade
M=43 k-ft d= 12”
9- 16 mm
6-16 mm
As (40 grade) = 43/ (0.45x40x0.87x1)= 2.7 in2 9 – 16 mm
As (60 grade) = 43/ (0.45x40x0.9x1)= 1.76 in2 6 – 16 mm
M=28 k-ft d= 12”
As (40 grade) = 28/ (0.45x40x0.87x1)= 1.79 in2 6 – 16 mm
As (60 grade) = 28/ (0.45x40x0.9x1)= 1.15 in2 4 – 16 mm
40 Grade
60 Grade
6- 16 mm
4-16 mm
For Column Design
P= 0.85 x Ag x (0.25xfc’+ fsg)
As (40 grade)= (P/0.85Ag - 0.25xfc’)/(0.45x40xAg)
As (60 grade)= (P/0.85Ag - 0.25xfc’)/(0.45x60xAg)
As (60 grade)/ As (40 grade) =0.66 = 66%
Net Save = 34%
Example 1
40 Grade
60 Grade
P=91 k, 10”x12” column
6- 16 mm
4-16 mm
As (40 grade) = (91/0.85*120-0.25*2.5)/0.45*40*120= 1.78 in2 = 6-16 mm
As (60 grade) = (91/0.85*120-0.25*2.5)/0.45*60*120 = 1.18 in2 = 4-16 mm
Example 2
P=117 k, 10”x15” column
As (40 grade) = (117/0.85*150-0.25*2.5)/0.45*40*150= 2.43 in2 = 8-16 mm
As (60 grade) = (117/0.85*150-0.25*2.5)/0.45*60*150 = 1.62 in2 = 4-16 mm
40 Grade
60 Grade
8- 16 mm
4-16 mm
Slab Design
d = Sqrt (M/Rb)
spacing = as*12/As
Grade
40 grade
40 grade
40 grade
40 grade
40 grade
40 grade
Slab
thickness
4”
slab10mm
spacing
4.5”
slab10mm
spacing
5”
slab10mm
spacing
4”
slab12mm
spacing
4.5”
slab12mm
spacing
5”
slab12mm
spacing
Moment
Grade
5.2
60 grade
6.1
60 grade
6.9
60 grade
9.4
60 grade
11.0
60 grade
12.5
60 grade
Slab
thickness
4”
slab10mm
spacing
4.5”
slab10mm
spacing
5”
slab10mm
spacing
4”
slab10mm
spacing
4.5”
slab10mm
spacing
5”
slab10mm
spacing
Moment
% save
8.1
36%
9.4
36%
10.7
36%
14.6
36%
17.0
36%
19.4
36%
USD Design Slab
ACI code provision for minimum slab thickness (9.5.2.1 table 9.5
(a))
9.5 (a) Minimum thickness of nonprestressed beams or one
way slabs unless defection are calculated
Minimum thickness, h
Simply supported One end continuous Both end continuous cantilever
Member
Members not supported to partitions or other construction likely to be damaged
by large deflection
Solid
onel/20
l/24
l/28
l/10
way slab
Beams
or
l/16
l/18.5
l/21
l/8
ribbed
one
way slab
Notes:
Values given shall be used directly for members with normal weight concrete and 60 grade steel
b) for fy other than 60 ksi values shall be multiplied by (0.4+fy/100,000)
USD Design Slab
9.5 (c) Minimum thickness of slabs without interior beams
Without drop panels
Exterior panels
fy (psi)
40,000
60,000
75,000
Without
edge beams
ln /33
ln /30
ln /28
Interior
panels
With drop panels
Exterior panels
With edge
Without
With edge
beams
edge beams beams
ln /36
ln /36
ln /36
ln /40
ln /33
ln /33
ln /33
ln /36
ln /31
ln /31
ln /31
ln /34
ln is length of clear span in long direction
Interior
panels
ln /40
ln /36
ln /34
USD Design Slab
ACI code minimum reinforcement
7.12.2.1
(b) slab where 60 grade deformed bar or welded wire
reinforcement are used ….0.0018
(c) slab where reinforcement with yield stress exceeding 60,000
psi measured at a yield strain of 0.35 percent is used …..7.12.2.2
Shrinkage and temperature reinforcement shall be spaced not
farther apart than 5 times the slab thickness, nor farther apart than
18 in.
USD Design Slab
Items
Thickness (in)
Floor finish (psf)
Partition wall (psf)
Live load (psf)
+M (kip- ft/ft)
-M (kip- ft/ft)
+ As (in2 /ft)
- As (in2 /ft)
As minimum (in2 /ft)
Total steel (cft)
Total steel (kg/sft)
Total Concrete (cft)
Savings in steel (%)
Extra concrete (%)
Interior Panel
Panel Size 18’-0’’x 18’-0”
Beam 12”x20”; Column 24”x24”
300 Grade 420 grade 500 grade
5.0
5.5
5.75
25
25
25
50
50
50
60
60
60
1.06
1.057
1.071
2.297
2.297
2.332
0.15
0.101
0.077
0.32
0.219
0.1672
0.11
0.108
0.088
1.21
0.81
0.63
0.83
0.556
0.432
120.42
132.46
138.4
22.3 over 420 grade
47.9 over 300 grade
5.0 over 420 grade
14.9 over 300 grade
Panel Size 26’-0’’x 26’-0”
Beam 12”x24”; Column 24”x24”
300 grade 420 grade 500 grade
7.0
7.5
8.0
25
25
25
47.5
47.5
47.5
60
60
60
2.33
2.335
2.398
5.04
5.039
5.218
0.22
0.1482
0.1124
0.48
0.3212
0.2446
0.225
0.1512
0.126
3.67
2.448
1.9103
1.21
0.805
0.628
364.6
390.6
416.6
22.0 over 420 grade
48.0 over 300 grade
7.1 over 420 grade
14.3 over 300 grade
USD Design Slab
One way slab
Items
Thickness (in)
Floor finish (psf)
Partition wall (psf)
Live load (psf)
+M (kip- ft/ft)
-M (kip- ft/ft)
+ As (in2 /ft)
- As (in2 /ft)
As minimum (in2 /ft)
Total steel (cft)
Total steel (kg/sft)
Total Concrete (cft)
Savings in steel (%)
Extra concrete (%)
Span length 10’
420 grade
500 grade
4.5
5.0
25
25
50
50
60
60
1.811
1.865
2.817
2.90
0.124
0.089
0.193
0.138
0.097
0.084
0.65
0.43
0.58
0.38
93.75
104.17
33.4
11.11
Span 15’
420 grade
6.5
25
47.5
60
4.557
7.088
0.193
0.3
0.14
2.092
0.886
284.3
500 grade
7.5
25
47.5
60
4.725
7.35
0.136
0.211
0.126
1.6
0.677
328.1
23.6
15.38
USD Design Column
Factored Moment = 30 kip-ft Pu = 330 kip
Size = 12” x 12”
Designed by 420 grade steel required reinforcement = 4 # 8 bars
Designed by 500 grade steel required reinforcement = 4 # 7 bars
Reinforcement ratio 0f 420 grade to 500 grade = 2.4/3.16 = 0.76 =
76%
Savings in steel = 26%
4 # 8 (grade 420)
4 #7 (grade 500)
USD Design Beam
For a beam it is known that, Mn = As * fy * (d-a/2)
Beam with Grade 420 (60)
Mn = As60 * fy60 * (d-a/2)
Beam with Grade 500
Mn = As500 * fy500 * (d-a/2)
As75 = (fy420/fy500) * As420
As500 = 0.82 * As420
Savings in steel = 20%
Cost Saving in Steel by
Using 500-Graded Rebar in
Place of Ordinary Steel As
Reinforcement
In reinforced concrete a long trend is evident towards the use of
higher strength materials, both steel and concrete. Reinforcing
bars with 40 ksi (40 grade) yield stress, almost standard 20 years
ago, has largely been replaced by 60 ksi (60 grade) yield stress.
Now a days the demand of high strength steel is increasing. The
use of steel 70-100 ksi is increasing day by day. This is because
they are more economical and their use tends to reduce
congestion of steel in the forms.
Table 1 lists presently available reinforcing steel, their grade
designation, ASTM specifications. The yield strength of 500
grade is 500 MPa or 72.5 ksi.
Product
ASTM
Specification
Designation
Reinforcing bars
A615
Grade 40 (300)
Grade 60 (420)
Grade 75 (520)
Grade 60
A706
Minimum Yield
Strength,
ksi
(MPa)
40 (300)
60 (420)
75 (520)
60 (420)
Max 78 (540)
Minimum
Tensile Strength,
ksi (MPa)
60 (420)
90 (620)
100 (690)
80 (550)
The marking system of rebars to meet ASTM specifications.
60 Grade or higher
40 Grade or higher
Stress-strain diagram for differernt graded steel
The major advantages of using 500 graded steel over any other steel can be
summarized:
•As the steel requirements is lower for 500 graded steel, it is very economical;
•The steel requirement is lower, the steel congestion relief in the form is lower in
the form as shown in Figure 9, Therefore the handling of steel in form is easier
•As the steel congestion in the form is lower, better concrete placement can be
achieved and medium to large course aggregate can be used and higher strength
of concrete can be achieved
(a)
(b)
•Better bonding is achieved for 500 graded steel
•As higher yield strength is achieved in 500 Graded Steel than
specified, then it is more safer to use 500 Graded steel in
reinforcement.
•For heavy construction, the code recommendation is to use 420
(60) Grade or higher grade steel.
(a)
(b)
Example of lower steel area requirement for 500 graded
steel
The following table shows the steel requirement for 500, 420 (60) and 300 (40)
graded steel and preferable bar steel. The area of the steel can be calculated by
the following:
Area required by 500 graded steel = 420/500*Area required by 420 grade
= 300/500*Area required by 300 grade
Cost savings by using 500 graded steel considering 1 ton of steel of 300
grade
Example of cost savings by using 500 graded steel
Rebars
Steel area
Steel price
300
(40)
graded
420
(60)
Graded
500
Graded
1 ton
69,000/-
0.67 ton
48,240/(@72,000/per ton)
39,600/(@72,000/per ton)
0.55 ton
Transportation
cost
1000/-
Handling
cost
200/-
Total
70,200/-
670/-
135/-
49,045/-
550/-
110/-
40,260/-
Example of cost savings by using 500 graded steel
Cost saving by using 500 graded over 300 grade steel = (70,200-40,260)*
100/70,200= 42.6%
Cost saving by using 500 graded over 420 grade steel = (49,045-40,260)*
100/49,045= 18.0%
Cost saving by using 420 graded over 300 grade steel = (70,200-49,045)*
100/70,200= 30.1%
However, using of 500 grade steel will increase the 10-15% of concrete volume
from 300 grade and 5-10% concrete volume from 420 grade, which will reduce
the savings and fall to around 12-15% over 420 grade and 30-35% over 300 grade
steel