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1 Inspiration Graphics merchandise Tripureswor, Kathmandu Tel: Fax: inspirations@mail.com.np

2 Designer's H a n d b o o k Hama Steel

3 Relevant portions of IS:456-:000, IS , SP-16 &SP-4, NS-191 have been extensively used for preparation of this handbook. Compiled by: Hama Iron and Steel Industries 06

4 Table of Contents Chapter 1 Economy Through Hama 500 RE-BARS Over Plain & CTD/TOR Bars 5 Chapter Material Strength and Stress-Strain Relationships 3 7 Chapter 3 Fiexural Members 75 Chapter 4 Compression Members 95 Chapter 5 Shear and Torsion 113 Chapter 6 Development Length and Anchorage 119 Annexure I Conversion table 13 Annexure II Areas of Reinforcement Bars 14 Annexure III Fixed end moments 16 Annexure IV Substitution tables for Hama Annexure V Guidlines for switching over from CTD/TOR to Hama

5 5 CHAPTER - 1 Economy Through Hama SG 500 RE-BARS Over Plain & CTD/TOR Bars

6 6 1.1 General It is evident that TMT rebars irrespective of grades are technically superior to conventional plain and cold twisted deformed bars. Besides, the replacement of conventional bars by TMT rebars will result in savings in terms of weight and there by the cost. However, the percentage saving will depend upon the type of structural member and the grade of steel. A brief summary of probable range of savings arising due to the use of HAMA 500 quenched rebars is presented for structural members in the following paragraphs with a view to make the user feel about its cost effectiveness. This study is based on various charts and tables corresponding to CTD/TOR and TMT rebars with detailed discussions which will follow subsequently in this handbook. i. Doubly Reinforced Beams Use of TMT 500 rebars will result in the saving of reinforcement of about 44% to 47% over mild steel and 14% to 15% over CTD/TOR bars respectively and in terms of cost will be about 35% to 37% over plain bars and 6% to 8% over CTD/TOR bars. If TMT 550 is used, the savings in weight will be 47% to 51 % over plain bars and 19% to 0% over CTD/TOR bars. If converted in cost, the savings will be 39% to 41% over plain and CTD/TOR bars respectively (refer table 1.). If the cost of bending, binding and transportation are taken into consideration, it will result in further savings. ii. Axially Loaded Columns If TMT rebars are used instead of plain and CTD/TOR bars, the savings by weight will be about 44% respectively. Because, the cost of TMT 500 and CTD/TOR is the same, the cost saving will remain equal to weight-wise saving, i.e. 10 % (refer table 1.5). iii. Columns with Uni-axial Banding The saving by using TMT 500 reinforcing steel over plain bars will be about 39% by weight and 34% cost wise. However, the savings with respect of CTD/TOR bars could be summarised as following:

7 7 If the axial load on a column is comparatively more,tmt rebars have a distinct advantage over CTD/TOR bars between 8% to 10%. When the magnitude of moment is very high as compared to the axial load i.e. the column is subjected to more bending, the saving will be around % to 4%. In the following text, few model examples have been worked out to appraise the users about the economy which can be achieved in terms " of weight and cost by using TMT rebars in lieu of conventional plain/ctd/tor bars-the results of these examples are summarised in tabular form. 1. Doubly Reinforced Beam Example -1 Design a Rectangular Beam with reinforcement grades offy 50, CTD/TOR. TMTSOOand 550. Grade of concrete fck= 15 N/mm and following input data. M =189KN.m u SF =137.5KN Assume, Case-1 B =300mm D =500 d'/d =0.1 M /Dd =(189x10 6)/300x455 y =3.0 f =15N/mrh f = 50N/mm ck y d'/d =0.1 M /bd u = 3.0 FromSP-16,T-45 A t = 1.71x0.3x45.5 = 335cm A sc = 0.403x0.3x45.5 = 5.5 cm Provide 5 - dia dia 18 = cm at bottom Provide 5 - dia 1 = 5.65 cm at top Designing for shear Provide 8 50 c/c L -Vertical stirrup Area of shear reinforcement = 11.6 cm d = kg/cm

8 8 0dia -5x6.0x.466 (CTD) 8-3x159x dia -3x6.0x.0 1dia -5x6.0x0^88 Total steel = 74.0kg = 14.0kg = 36.0kg = 7.0kg = 151 kg Cost of the reinforcement = Rs x = Rs.6655 NB cost of deformed bar assumed to be Rs /ton Class-II Designing for shear f ck =15N/mm d'/d = 0.1 M /bd u =3.0 f = 415N/mm y (CTD) FromSP-16T-49 A t = 1.003x0.3x45.5 = 13.69cm A sc = 0.99x0.3x45.5 = 4.08cm Provide 3 - (CTD) (CTD) 16 = cm at bottom Provide - (CTD) 16 = 4.0 cm at top Provide (CTD) 300 c/c L - vertical stirrup Area of shear reinforcement = 9.6cm Reinforcement requirement by weight (CTD) 16 5x6x1.578 = 47.0kg (CTD) 8 19x1.59x0.395 = 1.0 kg Total steel = 95.0kg Saving over mild steel =(151-95)/151% =37% Cost of the reinforcement = Rs = Rs.44070x0.095 Saving in cost over mild steel = Rs = 37%

9 9 Case-III (TMT-500) f ck = 15N/mm f y = 500N/mm (TMT) d'/d = 0.1 M u/bd = 3.0 (From Table 9 of handbook) A t = x0.3x45.5 = cm A sc = 0.60x0.3x45.5 = 3.54 cm Provide 6 -(TMT) cm at bottom Provide 3-(TMT) cm at top Designing for shear Provide (CTD) c/c L - vertical stirrup Area of shear reinforcement = cm Reinforcement requirement by weight (TMT) 16-6x6x1.578 (TMT) 1-3x6.0x0.888 (CTD) 8-x1.59x0.395 = kg = 16.0 kg = 14.0 kg Case-IV Total steel = 86.8 kg Saving over mild steel = 4% Saving over CTD/TOR = 8% Cost of the reinforcement Cost saving over mild steel = 45.6% Cost saving over CTD/TOR = 13.6% (TMT-550) fck = 15N/mm f y = 550 N/mm d'/d = 0.1 M /bd u = 3.0 (FromTable 13 of handbook) A t =0.7436x0.3x45.5 = 10.15cm Age =0466x0.3x45.5 = 3.36cm Provide 5-(TMT) 16 = 10.05cm at bottom Provide 3-(TMT) 1 = 3.39 cm at top

10 10 Designing for shear Provide (CTD) 55 c/c L - vertical stirrup Area of shear reinforcement = 11.07cm Reinforcement requirement by weight (TMT) 16 5x6.0x1578 =47.0kg (TMT) 1 3x6.0x0.888 =16.0kg (CTD) 8 x1.59x0395 =14.0 kg Total sheet =77 kg Saving over mild steel =49% Saving over CTD/TOR =11% Cost of the reinforcement = Rs x Example -1 Cost saving over mild steel = 49% Cost saving over mild steel CTD/TOR = 18.9% a) Same as Example -1 with f = 0 N/mm ck Case-1 (Ref.Sp.16)T-46 F = 0 N/mm f =50 N/mm ck y d'/d = 0.1 M y/bd =3.0 A t = 1.77x03x455 =4.16cm A cs = 0.011x03x455 =0.15cm =4.15 cm at bottom =.6cm at top Provide4 - DIA5 +4-DIA1 Provide - DIA1 Designing for shear Provide dia 95 c/c L - vertical stirrup Area of shear reinforcement = 9.6 cm Reinforcement requirement weight Dia5-4x6.0x3.854 = 9.0 kg Dia1-6x6.0x0.888 = 3.0 kg Dia8-19x1.59x0.395 = 1.0 kg Total steel = 136 kg

11 11 Cost of the reinforcement Case-II = Rs.44070x0.136 = Rs (Ref.SP-16)T-50 f ck =0N/mm f =415N/mm y (CTD) d'/d =0.1 M /bd u =3.0 A t =1.09x0.3x45.5 =14.04cm A sc =0.077x0.3x45.5 = 1.05cm Provide 4 - (CTD) 18+-(CTD) 16 =14.19cmat bottom Provide - (CTD) 1 = 6 cm at top Designing for shear Provide (CTD) 310 c/c L - vertical stirrup Area of shear reinforcement = 9.6cm Reinforcement requirement by weight (CTD) 18-4x6.0x.0 = 48.0kg (CTD) 16 -x6.0x1.578 = 19.0kg (CTD) 1 -x6.0x0.888 = 11.0kg (CTD) 18-19x1.59x0.395 = 1.0 kg Total steel = 90 kg Saving over mild steel = 34% Cost of the reinforcement = Rs x 0.09 = Rs Cost saving over f y 50 =33.8% Case-III (Refer Table 10) f ck = 0N/mm f = 500 N/mm y (TMT) d'/d = 0.1 M /bd u =3.0 A t = x0.3x45.5 = 115cm A sc = x0.3x45.5 =1. cm Provide 3 - (TMT) = 11.4 cm at bottom Provide - (TMT) 1 =.6 cm at top Designing for shear Provide (CTD) 75 c/c L - vertical stirrup Area of shear reinforcement = cm

12 1 Reinforcement requirement by weight (TMT) (TMT) 1-3x6.0x.98 -x6.0x0.888 =53.6 kg =10.6 kg (CTD) 8-1x1.59x0.395 =13. kg Total steel =77 kg Saving over mild steel = 43.5% Saving over CTD/TOR = 14.5% Cost of the reinforcement = Rs.44070x0.077 =Rs Cost saving over mild steel =43.3% Cost saving over CTD/TOR = 14.4% Case- IV (Ref.Table-14) F =0N/mm f = 550 N/mm ck y (TMT) d'/d =0.1 M /bd u =3.0 A t =0.759x0.3x45.5 =10.4cm A sc =0.0937x0.3x45.5 =1.8cm Provide - (TMT) (TMT) 16 = 10.3 cm at bottom Provide - (TMT) 1 =.6 cm at top Designing for shear Provide (CTD) 75 c/c L - vertical stirrup Area of shear reinforcement = 10.6 cm Reinforcement requirement by weight (TMT) 0 -x6.0x.466 = 9.6 kg (TMT) 16 -x6.0x1.578 = 18.9 kg (TMT) 1 -x.6.0x0.888 = 10.7 kg (CTD) 8-1x1.59x0.395 = 13. kg Total steel = 7 kg Saving over mild steel = 47% Saving over CTD/TOR = 0% Cost of the reinforcement = Rs.44070x0.07 = Rs Cost saving over mild steel = 47% Cost saving over CTD/TOR = 19.9%

13 13 Design a Rectangular Beam with f y CTD/TOR, TMT500 and TMT 550 N/mm. Grade of concrete f = 15 N/mm ck. Design by limit state method. Case-I M u = 35 KN.m SF = kn Assume B = 300 D = 600 d'/d = 0.1 M /bd = (35 x 106) / (300 x 550) = 3.9 u (Refer SP-16)T-45 f =15 N/mm f = 50 N/mm ck y d'/d = 0.1 My/bd = 3.9 A t =.1697x0.3x55 = 35.8cm A sc = x0.3x55 = 14.5cm Provide 6 - dia dia 0 = 35.73cmat bottom Pro vide 4- dia dia 1 = cm at top Designing for shear Provide 170 c/c L - vertical stirrup Area of shear reinforcement =.64 cm Reinforcement requirement by weight Dia 5-6 x 8 x = kg Dia0 - x8x.466 = 39.5kg Dia 18-4x8x.0 = 64.0kg Dia 1-4x8x0.888 = 8.5kg Dia 8-45x.1.79x0.395 = 3.0kg Total steel = 349 kg Cost of the reinforcement = Rs x = Rs Case-II (Refer SP-16)T-49 f =15 N/mm f = 415N/mm ck y (CTD) d'/d =0.1 M /bd u = 3.9 A t =1.8x0.3x55 = 1.1cm A sc =0.587x0.3x55 = 9.7cm

14 14 Provide 3 - CTD + - CTD 5 Provide -CTD +1-CTD 16 = 11 cm at bottom = 9.61cm at top Designing for shear Provide (CTD) 60 c/c L - vertical stirrups Area of shear reinforcement = 15.1 cm Reinforcement requirement by weight (CTD) 5 - x 8 x = 61.7 kg (CTD) - 5x8x.98 = 119.kg (CTD) 16-1x8x1.578 = 1.6kg (CTD) 8-30x1.79x0.395 = 1.kg Total steel = 15 kg Saving of steel over f y 50 = 38.3% Cost of the reinforcement = Rs x 0.15 = Rs Cost Saving over f y 50 = 38.3% Case-III (Refer Table 9) f =15N/mm f = 500 N/mm ck y (TMT) d'/d = 0.1 M /bd u = 3.9 A t = x0.3x55 = 17.8cm A sc = 0.494x0.3x55 = 8.15cm Provide 4 - (TMT) + - (TMT) 1 = cm at bottom Provide 4 - (TMT) 16 = 8.04 cm at top Designing for shear Provide (CTD) 30 c/c L - vertical stirrups Area of shear reinforcement = 17.1 cm Reinforcement requirement by weight (TMT) - 4x8x.98 = 95.4 kg (TMT) 16-4x8x1.578 = 50.5 kg (TMT) 1 - x8x0.888 = 14.kg (CTD) 8-34x1.79x0.395 = 4.1 kg Total steel = 184 kg Saving over mild steel = 47.3% Saving over CTD/TOR = 14.5% Cost of the reinforcement Rs x = Rs Cost saving over mild steel = 47.% Cost saving over CTD/TOR = 14.4%

15 15 Case-IV (Refer Table 13) f ck = 15 N/mm f = 550 N/mm y (TMT) d'/d = 0.1 M /bd u = 3.9 A t = 0.956x0.3x55 = 15.7cm A sc = x0.3x55 = 756cm Provide 5 - (TMT) 0 = 15.7 cm at bottom Provide 3 - (TMT) l0 = 7.63 cm at top Designing for shear Provide (CTD) 5 c/c L - vertical stirrup Area of shear reinforcements 17.1 cm Reinforcement requirement by weight (TMT) 0-5x8x.466 = 98.6 kg (TMT) 18-3x8x.0 = 48kg (CTD) 8-34x1.79x0395 = 4.1 kg Total steel = kg Saving of steel over mild steel = 51% Saving of steel over CTD/TOR = 1% Cost of the reinforcement = Rs.44070x0.170 = Rs Cost saving over mild steel = 51.% Cost saving over CTD/TOR = 0.9% Example - a) Same as Example- with fck = 0 N/mm Case-I (Refer SP-16)T-4 f ck = 0 N/mm f y = 50 N/mm d'/d = 0.10 M u/bd = 3.9 A t = 3x0.3x55 = 39.8 cm A sc = x0.3x55 = 8.1 cm Provide 6 - dia dia 1 = 363cm Provide 4 - dia 16 = 8.04 cm at bottom Designing for shear Provide dia 0 c/c L - vertical stirrup Area of shear reinforcement = 17.6 cm

16 16 Reinforcement requirement by weight Dia5-6x8x3.854 = kg Dia16-4x8x1.578 = 50.5 kg Dia1-6x8x0.888 = 4.6kg Dia8-35 x1.79x0.395 = 4.8 kg Total steel = 303 kg Cost of the reinforcement = Rs x = Rs Case-II (Refer SP-16)T-50 F = 0N/mm f = 415N/mm ck y (CTD) d'/d = 0.1 M /bd u = 3.9 A t = 1.306x0.3x55 = 155cm A sc = 0.369x0.3x55 = 6.08 cm Provide 4-(CTD) 5+-(CTD) 1 = 1.9 cm at bottom Provide 3 - (CTD) 16 = 6.03 cm at top Designing for shear Provide (CTD) 65 c/c L - vertical stirrup Area of shear reinforcement = 14.6 cm Reinforcement requirement by weight (CTD) 5-4x8x3.854 = 13.3kg (CTD) 16-3x8x1.578 = 37.9kg (CTD) 1 -x8x0.888 =14.kg (CTD) 8-9 x 1.79 x = 0.5 kg Total steel =196 kg Saving of steel over f y 50 = 35.3% Cost of the reinforcement = Rs x = Rs Case-III Cost of saving over f 50 = 35.3% y (Refer table 10) f ck = 0N/mm f = 500 N/mm y (TMT) d'/d = 0.1 Mu/bd = 3.9 A t = x0.3x55 = 17.7cm A sc = 0.33x0.3x55 = 5.33 cm Provide 4 - (TMT) 0+ - (TMT) 18 = cm at bottom

17 5 Provide - (TMT) 18 = 5.1 cm at top Provide (CTD) 40 c/c L - vertical stirrup Area of shear reinforcement by = 16.1 cm Reinforcement requirement by weight (TMT) 0-4x8x.466 = 79.0 kg (TMT) 18-4x8x.0 = 64.0kg (CTD) 8-3x1.79x0.395 =.7 kg Total steel = 166 kg Saving over mild steel = 45.% Saving over CTD/TOR = 15.3 % Cost of the reinforcement Rs.44070x0.166 =Rs Cost saving over mild steel = 45.% Cost saving over CTD/TOR = 15.3% Case-IV (Refer Table-14) f ck = 0N/mm f = 550 N/mm y (TMT) d'/d = 0.1 Mu/bd = 3.9 A t = 0.968x0.3x55 = 16.0 cm A sc = x0.3x55 = 5.06cm Provide - (TMT) 5+ - (TMT) 0 = cm at bottom Provide - (TMT) 18 = 5.10 cm at top Designing for shear Provide (CTD) 30 c/c L - vertical stirrup Area of shear reinforcement =17.1 cm Reinforcement requirement by weight (TMT) 5 ' - x8x3.854 = 61.7kg (TMT) 0 - x8x.466 = 39.5 kg (TMT) 18 - x8x = 3.0kg (CTD) 8-34x1.79x0.395 = 4.0 kg Total steel = 157 kg Saving over mild steel = 48. kg Saving over CTD/TOR = 0% Cost of the reinforcement Rs x = Rs Cost saving over mild steel = 48% Cost saving over CTD/TOR = 19.8% The above information are presented in tables 1.1 and 1.

18 18 Tabel Economy in using TMT 500 & TMT 500 in Place of fy 50 and CTD/TOR For Doubly Reinforced Sections. Dimesnion of the Section (MM) Mu (KN-m) fck N/ mm Mu/bd =3.0 (EXAMPLE 1) 15 Grade of Steel fy-50 CTD-415 TMT-500 TMT-500 Area of steel in Sq. cm At Asc Share No. and Dia. of Bars Tension Compression 5-dia 30+3 dia 18 3 (CDT 18+CDT 16) 6-TMT TMT 16 5-dia 1 -CTD 16 3-TMT 1 3-TMT 1 Shear Dia 8@50 TOR 8@300 TOR 8@65 TOR 8@55 Total Wt. in Kg Total Cost Rs Rs Rs Rs % Saving (Wt.) with respect to Sy -50 CTD % 4% 49% 8% 18.9% % Saving ( Cost with repspect to fy-50-37% 4.6% 49% CTD % 18.9% 0 fy-50 CTD-415 TMT-500 TMT dia 5+4 dia 1 4-CDT 18+ CDT 16 3-TMT -TMT 0+-TMT16 -dia 1 -CTD 1 -TMT 1 -TMT 1 Dia 8@95 TOR 8@310 TOR 8@75 TOR 8@ Rs Rs Rs Rs % 43.5% 47% 14.5% 0% % 43.3% 47% 14.4% 19.9% Dimesnion of the Section (MM) Mu (KN-m) Mu/bd 35 Mu/bd=3.9 (EXAMPLE ) fck N/ mm 15 Grade of Steel fy-50 CTD-415 TMT-500 TMT-500 Area of steel in Sq. cm At Asc Share No. and Dia. of Bars Tension Compression 6-dia 5+ dia 0 -CDT 5+3+CDT 4-TMT +-TMT 1 5- TMT 0 Shear 4-dia18+4-dia1 Dia 8@70 4 TMT 16 TOR 8@60 -TMT0+-TMT TOR 8@30 3-TMT 18 TOR 8@55 Total Wt. in Kg Total Cost Rs Rs Rs Rs % Saving (Wt.) with respect to Sy -50 CTD % 47.% 51% 8% 14% 18.9% 1% % Saving ( Cost with repspect to fy-50 CTD % 47.% 51.% 14.4% 0.9% 0 fy-50 CTD-415 TMT-500 TMT dia 5+6 dia 1 4-CDT 5++CDT 1 4-TMT 0+-TMT 18 - TMT 5+-TMT0 4-dia15 3 TMT 16 -TMT18 -TMT 18 Dia 8@0 TOR 8@56 TOR 8@40 TOR 8@ Rs Rs Rs Rs % 45.% 48.% 15.3% 0% % 4.% 48% 15.3% 19.8%

19 19 Tabel Summary of Quantity and cost of Different Grade of Concrete and Sell Section Example Grade of Concrete Grade of Steel QTY. in Kg. % saving in wt. wi th respect to fy-50 CTD-415 Cost in Rs. % saving in cost with respect to fy-50 CTD Doubly 1.(a) fy 50 CTD 415 TMT 500 TMT % 4% 49% 6% 19% Rs Rs RS Rs % 45.6% 49% % 18.9% Reinforced fy 50 CTD 415 TMT 500 TMT % 43.5% 47% 145% 0% Rs Rs.3966 RS Rs % 45.6% 49% % 19.9% 15 Beam.(a) fy 50 CTD 415 TMT 500 TMT % 47.3% 51.% 14.55% 1% Rs Rs RS Rs % 47.% 51.% % 0.9% fy 50 CTD 415 TMT 500 TMT % 45.% 48.% 15.3% 0% Rs Rs RS Rs % 45.% 48.% % 19.8%

20 0 Fig 1.1 gives the percent saving in usingtmt415 rebars instead of CTD/TOR bars of Mu/bd different values of d7d and also the average savings. It can be seen that there is maximum saving of about 3.8% and average saving of about 1.5%. These curves are plotted for M0 concrete. The percentage saving is more if we usetmt 500 rebars and TMT 550 rebars instead of CTD/TOR bars. The percentage saving is indicated in Fig 1. & 13.The maximum saving is about 19.7% in case of TMT 500 rebars and about 7.0% in case of TMT 550 rebars. The average saving is about 17.5% and 4.5% respectively. The above figures are in tonnage only. In terms ofcosuhese figures will be slightly more since TMT 500 and TMT 550 rebars are the same if not less costlier than CTD/TOR bars.

21 Fig1.1 - % Saving in Using TMT 415 Over CTD/TOR in Doubly Reinforced Beam d'/d=0.05 d'/d=0.10 d'/d=0.15 d'/d=0.0 Average Only main reinforcement has been taken into consideration. % Saving (wt)

22 Fig1. - % Saving in Using TMT 500 Over CTD/TOR in Doubly Reinforced Beam d'/d=0.05 d'/d=0.10 d'/d=0.15 d'/d=0.0 Average Only main reinforcement has been taken into consideration. % Saving (wt)

23 Fig1.3 - % Saving in Using TMT 550 Over CTD/TOR in Doubly Reinforced Beam d'/d=0.05 d'/d=0.10 d'/d=0.15 d'/d=0.0 Average Only main reinforcement has been taken into consideration. % Saving (wt)

24 4.3 Axially Loaded Columns Example 3 Axially Loaded Column Design Given: Factored loaded P u Grade of concrete M0 Unsupported length of column = 5000KN = 3.0 m Reqd: Case-I: Cross-section of column and reinforcement Assuming: Column size 800 mm x 500 mm Column with reinforcement off y=50 N/mm Area of concrete = A 5 g = 800 x 500 = 4 x 10 mm For given values of P u,a g, and f ck = 0 Corresponding to Chart 1 (or Ch. 4 of SP-16) P =.9% Required Area of reinforcement Age =.9x(80x50)/100 =116cm Provide (1-dia 3 +4-dia 5) Eq. area = cm Case-II: Column with CTD/TOR reinforcement. Case-III: As per chart 5 of SP -16: for given loading. Column area & grade of concrete: P = 1.67% Required area of reinforcement = A sc = 1.67 x (80x50)7100 = 66.8cm Provide (1-dia CTD 10) Eq. area = 6630 cm Column with TMT 415 reinforcement From chart 1 of handbook P =1.45% Therefore area of reinforcement required A sc = 1.45 x (50x80)/100 = 58.0cm Provide (8-TMT TMT 18) Eq. area = 59.6cm This information is summarized in Table 1.3 and 1.5

25 5.4 Uniaxial Bending Example-4 Rectangular Column with Uniaxial Bending Given: Factored Load P u= 500 KN Factored Moment My = 50 KNm (along longer direction) Concrete grade M0 Unsupported length of column = 3.0 M Required: Cross-section of column & reinforcement Assumptions: 1) Columnsize = 50mmx500 ) Dia.ofbar = 0 mm 3) Clear cover to main reinforcement = 40 mm d' =40+0/ = 50mm d'/d =50/500 = 0.1 P u/fckbd =500x103/(0x50x500) = 0. M /f bd u ck =50x103x103/0x50x500x500=0. Case-I: Column with reinforcement f y= 501 N/mm Referring Ch.8 of SP-16 p/f ck =0.17 P =0.17x0 Area of steel reqd. Ag = 3.44 x 5 x 50 /100 = 3.44% = 43 cm Provide (4 - dia dia 5) A s = 44.6 cm Case-II: Column with CTD/TOR reinforcement Asper Ch.3 of SP-16 p/f ck =0.11 P =0.11x0 =.% Area of steel A =x5x50/100 = 7.5cm s Provide (4 - CTD +4 - CTD 0) A s = 7.76 cm

26 6 Case-III: Column with TMT 415 reinforcement As per Ch. 3 of handbook p/f ck = 0.10 P = 0.10x0 =.0% Area of steel A s = x5x50/100 = 5 cm Provide (8 -TMT 0):A s = 5.13 cm This information is presented in Table 1.4 & 1.5 In Fig 1.5 to 1.7 the interaction diagrams of CTD/TOR bars are superimposed on the TMT rebars of different grades and d'/d =0.10. Also, the interaction curves for CTD/TOR bars are superimposed on TMT 500 for comparison Fig 1.6a. It is explicit that in case of TMT rebars, as the reinforcement percentage increases the curves tend to bulge out For the same percentage of steel and P u/fckbd, the section can resist a higher moment In other words, for a particular P u/fckbd and M /fckbd u, p/fck values are less in case of TMT rebars. For same concrete grade, reinforcement percentage will be less fortmt rebars which are used and this results in savings-the percentage savings will vary depending on the P /f bd and M /f bd u ck u ck values.

27 7 Tabel Economy in using TMT 415 in Place of fy 50 and CTD/TOR For Axially Loded Colums Dimesnion of the Section (MM) Mu (KN) fck N/ sqmi Grade of Steel Area of steel As Links No. and Dia. of Bars Compression Shear Total Wt. in Kg Total cc of R/F % Saving with respect to fy - 50 & fy in wt. % pf Saving with respect to fy - 50 & fy in Rs. fy dia 3+4-dia 5 dia 8@50 c/c 30 Rs Pu=5000KN 0 CTD CTD 8+6-CTD 10 CTD 8@50 c/c 188 Rs % over ty % over ty-50 (Example) TMT T0+8-TMT 18 CTD-8@50 c/c 169 Rs % over ty-50 44% over ty-50 10% over CTD % over ty-50 Tabel Economy in using TMT in Place of fu 50 and CTD/TOR For Uniaxial bending with Compression Dimesnion of the Section (MM) Pu (KN) Mu (KN-m) fck N/ mm Grade of Steel Area of steel sq. cm As Links No. & Dta. of bars Compression Share Total Wt. in Kg Total cc of R/F % Saving with respect % to pf Saving with respect to fy-50 & CTD-415 in wt. fy-50 & CTD-415 in Rs. fy dia8+4-dia5 dia-8@50c/c 115 Rs Pu=500 Mu=50 0 CTD CTD+4-CTD0CTD-8@50c/c75.9 Rs % over fy % over fy-50 (Example-4) TMT TMT 0 CTD-8@50c/c69.7 Rs % over fy % over fy-50 8% over CTD % overctd-415

28 8 Tabel Summary of Quantity and cost of Different Grade of Concrete and Steel Section Grade of Example Concrete Grade of Steel QTY. in Kg. % of Seving in fy-50 CTD-415 Cost in Rs. % of Seving incost over fy-50 CTD-415 AXIALLY 0 fy Rs LOADED 3 0 CTD % - Rs % - COLUMNS 0 TMT % 10.1% Rs % 10.1 UNIAXIAL BENDING 0 0 fy-50 CDT % - - Rs Rs % - - WITH 4 0 TMT % 8% Rs % 8% COMPRESSION

29 9 Fig % Saving in Using TMT 415 CTD 415 for Axially Loded Coluns Grade of Concrete, fck (N/sq. mm) Only Main reinforcement is taken into consideration

30 fy = 415N/mm d'/d=0.10 D TMT b As=pbD/100 d Axis of banding 1.1 CTD Pu / fck bd P Lc 1= Mu / fck bd Fig Comparative Chart for Compression with Bending for TMT 415 and CTD/TOR (Rectangular Section-Reinforcement Distributed equally on two sides)

31 fy = 415N/mm d'/d=0.10 D TMT b As=pbD/100 d Axis of banding 1.1 CTD Pu / fck bd P Lc 1= Mu / fck bd Fig Comparative Chart for Compression with Bending for TMT 500 and CTD/TOR (Rectangular Section-Reinforcement Distributed equally on two sides)

32 fy = 415N/mm d'/d=0.10 D TMT b As=pbD/100 d Axis of banding 1.1 CTD Pu / fck bd P Lc 1= Mu / fck bd Fig Comparative Chart for Compression with Bending for TMT 500 and CTD/TOR (Rectangular Section-Reinforcement Distributed equally on two sides)

33 TMT fy = 415N/mm d'/d=0.10 D 1.3 b As=pbD/ CTD d Axis of banding Pu / fck bd P Lc 1= Mu / fck bd Fig Comparative Chart for Compression with Bending for TMT 550 and CTD/TOR (Rectangular Section-Reinforcement Distributed equally on two sides

34 34 Characteristics Mild Steel CTD 415 HAMA TMT 500 HAMA TMT 550 Design Strength 50N/mm 415N/mm 500N/mm 550N/mm Quantity Required M. T M.T M.T M.T. % Saving in Weight Over fy-50 % Saving in Weight Cost of Reinforcement Per M.T. Rs % 45.5% 48.5% 14.60% 19.15% Rs Rs Rs Cost of Reinforcement Excluding Fabrication Saving in cost With Respect to Fy -50 Per 1 M.T. Reinforcement % of Cost Saving with Respect to Fy - 50 Saving in Cost with Respect to CTD/TOR Per 1M.T. Reinforcement % of Cost Saving With Respect to CTD-415 Rs Rs. 807 Rs Rs. 696 Rs Rs Rs % 54.6% 48.5% Rs Rs % 19.1% Note: 1. Cost towards cutting, bending, binding, fixing and placing not included. If this cost is staken into consideration then There will be further savings.. Price of steel may increase, but percentage savings will be same

35 35 Table 1.6a - Additional Cost Saving by Using TMT bars With Welden Lap Joints Type of Lap Lap Le n gth reqd Material Requirement Cost Percentage cost saving per joint of single bar Conventional based on concrete-steel bonding Av. 50 dia=1000mm 1w=6.3dia.74 Kg. of Rs per tonne Rs For welded Lap joints 6.3 dia=16mm bw 1w=6.3dia dw=0.dia or 4mm bw=0.7dia or 10mm 0.31 Kg. of Rs per tonne.5 electrodes of 4 mm dia & 450 mm long Rs Rs.5 Rs.11 Rs.7% Recommended welding length For TMT 415 1w = 6.3dia TMT 500 1w = 7.0 dia TMT 550 1w = 7.0 dia Including cost of electrodes. 0, Da & others

36 6 CHAPTER - Material Strength and Stress-Strain Relationships

37 38.1 Grade of Concrete The grade of concrete used in this hand book M15,M0,5&M30. Types and Grades of Steel The reinforcement steel used in this handbook conform to 15:43 in case of mild steel bus and IS: 1786/NS191 in case of re-bars..3 STRESS-STRAIN RELATIONSHIP FOR CONCRETE The stress-strain relationship for concrete is shown in Fig.1.4 STRESS-STRAIN RELATIONSHIP FOR STEEL The stress-strain relationship for CTD/TOR and TMT Rebars is shown in Figs.. to.5.5 Typical stress-strain relationship for 1mm, 16mm, 0mm and 5mm TMT bars as shown in Figs.6 and.7..6 The actual stress-strain relationship for 8mm to3mm HAMA 500 rebars along with theirtypical values of mechanical properties are indicated in Figs..8 to.15. Figs. I and II are copies of test certificates from SAIL indicating typical chemical properties. Fig. Ill shows a typical test certificate issued with each dispatch.

38 fck Stress Parabolic Curve Strain Fig.1 - Stress-Strain Curve for Concrete

39 Characteristic Curve 400 CTD 415 (0.06-fy for stress 0.00) Stress,10mm Es=.10 N/mm Strain Fig. - Stress-Strain Curve for CTD 415

40 Characteristic Curve CTD Design Curve TMT Stress,10mm Es=.10 N/mm Strain Fig.3 - Stress-Strain Curve for TMT 415

41 Characteristic Curve TMT Design Curve TMT Stress,10mm Es=.10 N/mm Fig.4 - Stress-Strain Curve for TMT 500

42 Characteristic Curve TMT Design Curve TMT Strain,N/mm Es=.10 N/Sq. mm Strain Fig.4 - Stress-Strain Curve for TMT 550

43 44 Fig Stress (Kg/mm ) Strain

44 45 Fig Stress (Kg/mm ) Strain

45 46 Fig..8a

46 47 Fig..8b

47 48 Fig..8c

48 49 Fig..9a

49 50 Fig..9b

50 51 Fig..9c

51 5 Fig..10a

52 53 Fig...10b

53 54 Fig...10c

54 55 Fig...11a

55 56 Fig...11b

56 57 Fig...10cc

57 58 Fig...1a

58 59 Fig...1b

59 60 Fig...1c

60 61 Fig...13a

61 6 Fig...13b

62 63 Fig...13c

63 64 Fig...14a

64 65 Fig...14b

65 66 Fig...14c

66 67 Fig...15a

67 68 Fig...15b

68 69 Fig...15c

69 70

70 71

71 7 Fig. I

72 73 Fig. II

73 74 Fig. III

74 75 CHAPTER - 3 Flexural Members

75 Assumptions: The basic assumptions in design offlexural members for limit state of collapse are as given below (in accordance with Clause 37.1 of Code. a) Plane sections normal to the axis remain plane after bending. b) The maximum strain in concrete at the outermost compression fiber is taken as c) The design stress-strain relationship for concrete is as given in Fig.1 d) The tensile strength of concrete is ignored. e) The stresses for reinforcement are derived from the curves given in Fig.3,.4, &.5 (for bars oftmt415,tmt 500 and TMT 550) f) Maximum strain in tension reinforcement is less than (fy/1.15es)+0.00 to ensure ductile failure. 3. Under Reinforced Sections: 3..1 Under reinforced sections are those single reinforcement sections in which the amount of reinforcement percentage does not exceed the values given in Table E of SP-16. Here steel fails first. At collapse, the strain in steel will be more than that given in code i.e. (0.00+fy/I.ISEs) so the depth of neutral axis is less than Xu.max. and it could be calculated by equating the forces in tension and compression. The expression for M u/bd can be written as: M y/bd 0.87 f ycpt/100) x [ (f y/f ck) (p t/100)] 3.. One worksheet was developed and the same was used to prepare tables by only varying different variable parameters. Tables show the variation of P t.i.e percentage reinforcement with the change in M u/bd and for different values of fy. In worksheet the formula was fed for Pt for the corresponding value of M u/bd i.e. P = f (M u/bd ). The limiting condition is also introduced so that it returns for the values of Pt above the balanced section. Thus after preparing one worksheet only value of fck had to be changed to obtain different worksheets Tables 1 to 4 give the values percentage of steel for different values of M u/ bd and grade of concrete.

76 Assumptions: The basic assumptions in design offlexural members for limit state of collapse are as given below (in accordance with Clause 37.1 of Code. a) Plane sections normal to the axis remain plane after bending. b) The maximum strain in concrete at the outermost compression fiber is taken as c) The design stress-strain relationship for concrete is as given in Fig.1 d) The tensile strength of concrete is ignored. e) The stresses for reinforcement are derived from the curves given in Fig.3,.4, &.5 (for bars oftmt415,tmt 500 and TMT 550) f) Maximum strain in tension reinforcement is less than (fy/1.15es)+0.00 to ensure ductile failure. 3. Under Reinforced Sections: 3..1 Under reinforced sections are those single reinforcement sections in which the amount of reinforcement percentage does not exceed the values given in Table E of SP-16. Here steel fails first. At collapse, the strain in steel will be more than that given in code i.e. (0.00+fy/I.ISEs) so the depth of neutral axis is less than Xu.max. and it could be calculated by equating the forces in tension and compression. The expression for M u/bd can be written as: M y/bd 0.87 f ycpt/100) x [ (f y/f ck) (p t/100)] 3.. One worksheet was developed and the same was used to prepare tables by only varying different variable parameters. Tables show the variation of P t.i.e percentage reinforcement with the change in M u/bd and for different values of fy. In worksheet the formula was fed for Pt for the corresponding value of M u/bd i.e. P = f (M u/bd ). The limiting condition is also introduced so that it returns for the values of Pt above the balanced section. Thus after preparing one worksheet only value of fck had to be changed to obtain different worksheets Tables 1 to 4 give the values percentage of steel for different values of M u/ bd and grade of concrete.

77 Doubly Reinforced Section 33.1 Doubly reinforced section is generally adopted when the dimensions of the beam is predetermined form various considerations and the design moment exceeds the moment of resistance of the singly reinforced sections. The additional moment of resistance needed is obtained by providing compression reinforcement and additional tensile reinforcement. The moment of resistance of a doubly reinforced section is thus the sum of the limiting moment of resistance M u. ulm of singly reinforced section and the additional moment Mu. Given the values of M which is greater than M lim. the values of Mu u u can be calculated. Mu = Mu-M u.lim 3.3. The equations for the moment of resistance may be written as follows: M u = M u, lim + Ptd. 87(d-d') 100 M u=-m u,lim+p t (0-87f y) d-d') bd bd 100 d Where P t is the additional percentage of tensil reinforcement P t =Pt. lim+pt P c =Ptt(0.87f y)/(fsc-t cc)] The values of P t and P tc for four values of d'/d form 0.05 to 0.0 have been tabulated in the tables 45 to 56.The tabulation is done by changing the values of the variables like f y and the starting values of the tables with correspond to the balanced singly reinforced section.

78 78 Table - 1 Flexure Reinforcement Percentage Pt for Singly Reinforced Sections fy (N/mm ) fy (N/mm ) Mu/bd N/mm * Mu/b d N/mm Note: "-"indicates inadmissible reinforcement percentage

79 79 Table - Flexure Reinforcement Percentage Pt for Singly Reinforced Sections fy (N/mm ) fy (N/mm ) Mu/bd N/mm * Mu/b d N/mm Note: "-"indicates inadmissible reinforcement percentage

80 80 Table - 3 Flexure Reinforcement Percentage Pt for Singly Reinforced Sections fy (N/mm ) fy (N/mm ) Mu/bd N/mm * Mu/b d N/mm Note: "-"indicates inadmissible reinforcement percentage * Mu/b d N/mm

81 81 Table - 4 Flexure Reinforcement Percentage Pt for Singly Reinforced Sections fy (N/mm ) fy (N/mm ) Mu/bd N/mm * Mu/b d N/mm Note: "-"indicates inadmissible reinforcement percentage

82 8 Table 5 Flexure - Reinforcement Percentage For Doubly Reinforced Sections d /d = 0.05 Mu/bd ) (N/mm Pt Pc d /d = 0.1 Pt Pc d /d = 0.15 Pt Pc d /d = 0. Pt Pc

83 83 Table 6 Flexure - Reinforcement Percentage For Doubly Reinforced Sections d /d = 0.05 Mu/bd ) (N/mm Pt Pc d /d = 0.1 Pt Pc d /d = 0.15 Pt Pc d /d = 0. Pt Pc

84 84 Table 7 Flexure - Reinforcement Percentage For Doubly Reinforced Sections d /d = 0.05 Mu/bd ) (N/mm Pt Pc d /d = 0.1 Pt Pc d /d = 0.15 Pt Pc d /d = 0. Pt Pc