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

Workshop on

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 IslamAssociate Professor

Department of Civil Engineering

Dhaka University of Engineering and Technology (DUET)

Gazipur-1700

Tel: 8932939 Mobile: 01711389584

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|

Three aims of Civil Engineering design

Materials used in Civil Engineering RCC Construction

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

Reinforced concrete is concrete in which reinforcement bars ("rebar") have been incorporated to strengthen the material

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

60 (420) Grade:

Yield Strength = 60 ksi 420 MPa

40 (300) Grade :

Yield Strength = 40 ksi 300 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.

b

nAs

kd

d

T= As fs

c

s

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.

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

b

d

ca

T= As fy

0.85 f ’c

= 1c

u

y

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.

Mn = As fy (d-a/2) where, a= As fy/0.85 f’cb

b = 0.851 f’c/fy (87,000/87,000+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 LWhere 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 = VnWhere Mu and Vu is external factored moment and shear forces Mn and Vn nominal ultimate moment and shear capacity of the member

Factor of Safety =

1

1

6.12.116.12.1

DL

DL

LD

LD

Factor of safety for the various values of and L/D ratio are shown in the following table

0.9 0.75 0.7 0.65 L/D 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 FS 1.33 1.56 1.63 1.67 1.6 1.87 1.96 2.0 1.71 2.00 2.10 2.14 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 gradeFor 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 M=43 k-ft d= 12” 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

9- 16 mm 6-16 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 P=91 k, 10”x12” column 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

6- 16 mm 4-16 mm

40 Grade 60 Grade

8- 16 mm 4-16 mm

Slab Design d = Sqrt (M/Rb) spacing = as*12/As Grade Slab

thickness Moment Grade Slab

thickness Moment % save

40 grade 4” slab10mm spacing

5.2 60 grade 4” slab10mm spacing

8.1 36%

40 grade 4.5” slab10mm spacing

6.1 60 grade 4.5” slab10mm spacing

9.4 36%

40 grade 5” slab10mm spacing

6.9 60 grade 5” slab10mm spacing

10.7 36%

40 grade 4” slab12mm spacing

9.4 60 grade 4” slab10mm spacing

14.6 36%

40 grade 4.5” slab12mm spacing

11.0 60 grade 4.5” slab10mm spacing

17.0 36%

40 grade 5” slab12mm spacing

12.5 60 grade 5” slab10mm spacing

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 one-way slab

l/20 l/24 l/28 l/10

Beams or ribbed one way slab

l/16 l/18.5 l/21 l/8

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 With drop panels Exterior panels Interior

panels Exterior panels Interior

panels

fy (psi) Without

edge beams With edge beams

Without edge beams

With edge beams

40,000 ln/33 ln/36 ln/36 ln/36 ln/40 ln/40 60,000 ln/30 ln/33 ln/33 ln/33 ln/36 ln/36 75,000 ln/28 ln/31 ln/31 ln/31 ln/34 ln/34

ln is length of clear span in long direction

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

Interior Panel Panel Size 18’-0’’x 18’-0”

Beam 12”x20”; Column 24”x24” Panel Size 26’-0’’x 26’-0” Beam 12”x24”; Column 24”x24”

Items 300 Grade 420 grade 500 grade 300 grade 420 grade 500 grade Thickness (in) 5.0 5.5 5.75 7.0 7.5 8.0 Floor finish (psf) 25 25 25 25 25 25 Partition wall (psf) 50 50 50 47.5 47.5 47.5 Live load (psf) 60 60 60 60 60 60 +M (kip-ft/ft) 1.06 1.057 1.071 2.33 2.335 2.398 -M (kip-ft/ft) 2.297 2.297 2.332 5.04 5.039 5.218 + As (in2/ft) 0.15 0.101 0.077 0.22 0.1482 0.1124 - As (in2/ft) 0.32 0.219 0.1672 0.48 0.3212 0.2446 As minimum (in2 /ft) 0.11 0.108 0.088 0.225 0.1512 0.126 Total steel (cft) 1.21 0.81 0.63 3.67 2.448 1.9103 Total steel (kg/sft) 0.83 0.556 0.432 1.21 0.805 0.628 Total Concrete (cft) 120.42 132.46 138.4 364.6 390.6 416.6 Savings in steel (%) 22.3 over 420 grade

47.9 over 300 grade 22.0 over 420 grade 48.0 over 300 grade

Extra concrete (%) 5.0 over 420 grade 14.9 over 300 grade

7.1 over 420 grade 14.3 over 300 grade

USD Design SlabOne way slab

Span length 10’ Span 15’ Items 420 grade 500 grade 420 grade 500 grade Thickness (in) 4.5 5.0 6.5 7.5 Floor finish (psf) 25 25 25 25 Partition wall (psf) 50 50 47.5 47.5 Live load (psf) 60 60 60 60 +M (kip-ft/ft) 1.811 1.865 4.557 4.725 -M (kip-ft/ft) 2.817 2.90 7.088 7.35 + As (in2/ft) 0.124 0.089 0.193 0.136 - As (in2/ft) 0.193 0.138 0.3 0.211 As minimum (in2 /ft) 0.097 0.084 0.14 0.126 Total steel (cft) 0.65 0.43 2.092 1.6 Total steel (kg/sft) 0.58 0.38 0.886 0.677 Total Concrete (cft) 93.75 104.17 284.3 328.1 Savings in steel (%) 33.4 23.6 Extra concrete (%) 11.11 15.38

USD Design Column

Factored Moment = 30 kip-ft Pu = 330 kipSize = 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 500Mn = As500 * fy500 * (d-a/2)As75 = (fy420/fy500) * As420As500 = 0.82 * As420Savings 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.

Product ASTM Specification

Designation Minimum Yield Strength, ksi (MPa)

Minimum Tensile Strength, ksi (MPa)

A615 Grade 40 (300) Grade 60 (420) Grade 75 (520)

40 (300) 60 (420) 75 (520)

60 (420) 90 (620) 100 (690)

Reinforcing bars

A706 Grade 60

60 (420) Max 78 (540)

80 (550)

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.

60 Grade or higher 40 Grade or higher

The marking system of rebars to meet ASTM specifications.

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 Transportation cost

Handling cost

Total

300 (40)

graded

1 ton 69,000/- 1000/- 200/- 70,200/-

420 (60)

Graded

0.67 ton 48,240/- (@72,000/-

per ton)

670/- 135/- 49,045/-

500 Graded

0.55 ton 39,600/- (@72,000/-

per ton)

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