Bsrm Steels Limited · DSEBD:BSRMSTEEL
Bsrm Steels — Technical Presentation (July 2008)
43 slides from Bsrm Steels's technical presentation, Jul 21, 2008
Slide 1: Workshop on Cost Saving in Steel by Using 500-Graded Rebar
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
Slide 2: Design Theory on Building Design and Cost Saving in Steel
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: 0171138958
Slide 3: Three aims of Civil Engineering design
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 technol
Slide 4: Materials used in Civil Engineering RCC Construction
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 t
Slide 5: Structural Design consists the following works
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 m
Slide 6: Flexural failure mechanism of RCC Structure
Flexural failure mechanism of RCC Structure
Slide 7: Rebar grades and yield strength
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
Slide 8: Typical reinforced concrete beam
Typical reinforced concrete beam in walk-though within transparent reinforced concrete
Slide 9: Structural Design
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
Slide 10: Structural Design
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
Slide 11: Design Methods
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 •Consid
Slide 12: Design Methods
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.
Slide 13: Working Stress 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
Slide 14: Beam Design Load
Beam Design Load • Dead Load: Self wt and furniture, equipment, etc • Live load • Lateral Load; Wind, Earthquake • Impact load (if any) Column
Slide 15: Lateral Load Earthquake Wind
Lateral Load Earthquake Wind/ Earthquake
Slide 16: Load analysis for moment and shear
From Load Analysis by software or manual Moment and shear can be obtained
Slide 17: Flexural Moment
Flexural Moment M= 0.5 fc kjbd2 M=Asfsjd
Slide 18: Ultimate Strength Design
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
Slide 19: ACI code required strength
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, earthqu
Slide 20: Factor of safety table
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 Factor of Safety = L 1.2 D 1.6 L 1 1.2 1.6 D 1 DL 1 L D Factor of safety for the various values of
Slide 21: Higher strength steel requires less material
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
Slide 22: Beam and footing steel savings
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%
Slide 23: Beam example: 40 grade vs 60 grade
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
Slide 24: Column design
For Column Design P= 0.85 x Ag x (0.25xfc’+ fsg) 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%
Slide 25: Column examples: 40 grade vs 60 grade
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.
Slide 26: Slab Design
Slab Design d = Sqrt (M/Rb) spacing = as*12/As Grade Slab Moment Grade Slab Moment % save thickness thickness 40 grade 4” 5.2 60 grade 4” 8.1 36% slab10mm slab10mm spacing spacing 40 grade 4.5” 6.1 60 grade 4.5” 9.4 36% slab10mm slab10mm spacing spacing 40 grade 5” 6.9 60 grade 5” 10.7 36% slab10mm
Slide 27: USD Design Slab: minimum thickness
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 supporte
Slide 28: USD Design Slab: minimum thickness without beams
USD Design Slab 9.5 (c) Minimum thickness of slabs without interior beams Without drop panels With drop panels Exterior panels Interior Exterior panels Interior panels panels fy (psi) Without With edge Without With edge edge beams beams edge beams beams 40,000 ln/33 ln/36 ln/36 ln/36 ln /40 ln /40 6
Slide 29: USD Design Slab: minimum reinforcement
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 rei
Slide 30: USD Design Slab: interior panel comparison
USD Design Slab Interior Panel Panel Size 18’-0’’x 18’-0” Panel Size 26’-0’’x 26’-0” Beam 12”x20”; Column 24”x24” 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
Slide 31: USD Design Slab: one way slab
USD Design Slab One 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
Slide 32: USD Design Column
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 #
Slide 33: USD Design Beam
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%
Slide 34: Cost Saving in Steel by Using 500-Graded Rebar
Cost Saving in Steel by Using 500-Graded Rebar in Place of Ordinary Steel As Reinforcement
Slide 35: Trend toward higher strength materials
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 strengt
Slide 36: Available reinforcing steel grades and ASTM specifications
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 Designation Minimum Yield Minimum Specification Strength, ksi Tensile Strength, (MPa) ksi (MPa) Reinforcing bars A615 Grade 40 (300)
Slide 37: Rebar marking system
The marking system of rebars to meet ASTM specifications. 60 Grade or higher 40 Grade or higher
Slide 38: Stress-strain diagram for different graded steel
Stress-strain diagram for differernt graded steel
Slide 39: Advantages of 500 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 ha
Slide 40: Advantages of 500 graded steel (bonding and safety)
•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)
Slide 41: Lower steel area requirement for 500 graded steel
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 42
Slide 42: Example of cost savings by using 500 graded steel
Example of cost savings by using 500 graded steel Rebars Steel area Steel price Transportation Handling Total cost cost 300 1 ton 69,000/- 1000/- 200/- 70,200/- (40) graded 420 0.67 ton 48,240/- 670/- 135/- 49,045/- (60) (@72,000/- Graded per ton) 500 0.55 ton 39,600/- 550/- 110/- 40,260/- Graded (@
Slide 43: Cost savings by using 500 graded steel
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)