Document No. :: IITK-GSD-EQ-V3.0 Final Report :: - Earthquake Codes IITK-GSD Project on Building Codes Explanatory Examples or Ductile Detailing o RC Buildings by Dr. R. K. Ingle Department o pplied echanics Visvesvaraya National Institute o Technology Nagpur Dr. Sudhir K Jain Department o Civil Engineering Indian Institute o Technology Kanpur Kanpur
The solved examples included in this document are based on a drat code being developed under IITK-GSD Project on Building Codes. The drat code is available at http://www.nicee.org/iitk-gsd/iitk- GSD.htm (document number IITK-GSD-EQ11-V3.0). This document has been developed through the IITK-GSD Project on Building Codes. The views and opinions expressed are those o the authors and not necessarily o the GSD, the World Bank, IIT Kanpur, or the Bureau o Indian Standards. Comments and eedbas may please be orwarded to: Pro. Sudhir K Jain, Dept. o Civil Engineering, IIT Kanpur, Kanpur 08016, email: nicee@iitk.ac.in
Examples on 1390 CONTENTS Sl. No Type o Design Page No. 1. Beam Design o an RC Frame Building in Seismic Zone V 4. Beam Design o an RC Frame Building in Seismic Zone II 15 3. Interior Column Design o an RC Frame Building in Seismic Zone V 4. Exterior Column Design o an RC Frame Building in Seismic Zone V 4 33 5. Interior Column-Beam Joint Design or Zone V 4 6. Exterior Column -Beam Joint Design or Zone V 48 7. Interior Column-Beam Roo Joint Design or Zone-V 56 8. Exterior Column-Beam Roo Joint Design or Zone V 6 9. Shear Wall Design or a Building in Seismic Zone III 69 IITK-GSD-EQ-V3.0 Page 3
Examples on 1390 Example 1 - Beam Design o an RC Frame Building in Seismic Zone V 1 Problem Statement: ground plus our storey RC oice building o plan dimensions 19 m x 10 m located in seismic zone V on medium soil is considered. It is assumed that there is no parking loor or this building. Seismic analysis is perormed using the codal seismic coeicient method. Since the structure is a regular building with a height less than 16.50 m, as per Clause 7.8.1 o IS 1893 (Part 1): 00, a dynamic analysis need not be carried out. The eect o inite size o joint width (e.g., rigid osets at member ends) is not considered in the analysis. However, the eect o shear deormation is considered. Detailed design o the beams along the grid line as per recommendations o IS 1390:1993 has been carried out. Solution: 1.1 Preliminary Data Plan o the building and sectional elevations o dierent RC rames are shown in Figures 1.1, 1. and 1.3. The sizes o the beams and columns are given in Table 1.1. Figure 1.4 shows beam-loading diagram or dead load and live load, respectively, on an intermediate rame in the transverse direction. C C1 C1 C1 C1 C Y 5 5 B C1 C3 C3 C3 C3 C1 X C C C1 C1 C1 C1 C 1 3 4 5 6 4 4 3 4 4 Figure 1.1: Plan o building (ll dimensions in meters) Table 1.1 :Schedule o member sizes Column Beam C1 x RB1, FB1 x 600 C 400 x 400 RB, FB x C3 400 x PB1 x 400 PB x 350 Slab thiness: 15 Note: ll dimensions in mm. IITK-GSD-EQ-V3.0 Example 1 /Page 4
Examples on 1390 Roo RB1 4th FB1 3 3rd FB1 3 nd 1st GL FB1 FB1 PB1 1.5 3 3 3 C 5 C1 5 C Figure 1.: Elevation o rame, B & C Figure 1.3: Elevation o transverse rame 1&6 a. Dead Load b. Live Load Figure 1.4: Loading diagram or an intermediate rame -5 1. General Other relevant data are as ollows: Grade o concrete: 0 Grade o steel = Fe 415 Live load on roo = 1.5 kn/m (Nil or earthquake) Live load on loors = 3 kn/m (5% or earthquake) Roo inish = 1 kn/m Floor inish = 1 kn/m Bri wall on peripheral beams = 30 mm thi Bri wall on internal beams = 150 mm thi Density o concrete = 5 kn/m 3 Density o bri wall including plaster = 0 kn/m 3 1.3 Load Combinations Load combinations are considered as per IS 456: 000 and are given in Table 1.. EQX implies earthquake loading in X direction and EQY stands or earthquake loading in Y direction. The emphasis here is on showing typical calculations or ductile design and detailing o IITK-GSD-EQ-V3.0 Example 1 /Page 5
Examples on 1390 building elements subjected to earthquakes. In practice, wind load should also be considered in lieu o earthquake load and the critical o the two load cases should be used or design. Beams parallel to the Y direction are not signiicantly aected by earthquake orce in the X direction (except in case o highly unsymmetrical buildings), and vice versa. Beams parallel to Y direction are designed or earthquake loading in Y direction only. Torsion eect is not considered in this example. Table 1.: Load combinations or earthquake loading S.No. Load Combination DL LL EQ 1 1.5DL+1.5LL 1.5 1.5-1.(DL+LL * +EQX) 1. 0.5/0.5 * +1. 3 1.(DL+LL * -EQX) 1. 0.5/0.5 * -1. 4 1.(DL+LL * +EQY) 1. 0.5/0.5 * +1. 5 1.(DL+LL * -EQY) 1. 0.5/0.5 * -1. 6 1.5(DL+EQX) 1.5 - +1.5 7 1.5(DL-EQX) 1.5 - -1.5 8 1.5(DL+EQY) 1.5 - +1.5 9 1.5(DL-EQY) 1.5 - -1.5 10 0.9DL+1.5 EQX 0.9 - +1.5 11 0.9DL-1.5 EQX 0.9 - -1.5 1 0.9DL+1.5 EQY 0.9 - +1.5 13 0.9DL-1.5 EQY 0.9 - -1.5 *Note: Reduced Live loads are considered as per Clause 7.3.1 o IS 1893 (Part 1): 00, even though it is proposed to drop this clause in the new edition o the Code. For the present case, (live load o 3 kn/m ) 5% o live load is considered or seismic weight calculations. 1.4 Design o iddle Floor Beam two symmetrical spans, calculations need to be perormed or one span only. Figure 1.5: Beam BC 1.5 ember Forces For the beam B, orce resultants or various load cases and load combinations have been obtained rom computer analysis and are summarised in Table 1.3 and Table 1.4 which show orce resultants or dierent load combinations; with the maximum values to be used or design being underlined. s the beam under consideration is parallel to Y direction, earthquake loads in Y direction are predominant and hence the 13 load combinations o Table 1. reduce to 7 as shown in Table 1.4 Beam marked BC in Figure 1.5 or rame is considered or design. Since the beam consists o Table 1.3 : Force resultants in beam B or various load cases Load Case Shear (kn) Let end Centre Right end oment (kn-m) Shear (kn) oment (kn-m) Shear (kn) oment (kn-m) DL -51-37 4 3 59-56 LL -14-1 1 11 16-16 EQY 79 09 79 11 79-191 Note: The results are rounded o to the next higher integer value. IITK-GSD-EQ-V3.0 Example 1 /Page 6
Examples on 1390 Table 1.4 Force resultants in beam B or dierent load combinations S. No. Load Combination Let end Centre Right end Shear (kn) oment (kn-m) Shear (kn) oment (kn-m) Shear (kn) oment (kn-m) 1 1.5DL+1.5LL -98-74 8 65 113-108 1.(DL+LL*+EQY) 9 03 100 55 170-301 3 1.(DL+LL*-EQY) -160-99 -90 9-19 157 4 1.5(DL+EQY) 4 58 15 65 07-371 5 1.5(DL-EQY) -195-369 -113 3-30 03 6 0.9DL+1.5 EQY 73 80 1 45 17-337 7 0.9DL-1.5 EQY -164-347 -115 1-65 36 * ppropriate raction o live load has been used 1.6 Various Ches 1.6.1 Che or xial Stress Factored axial orce = 0.00 kn Factored axial stress = 0.0 Pa < 0.10 Hence, design as lexural member. (Clause 6.1.1; IS 1390:1993) 1.6. Che or ember Size Width o beam, B = mm > 00 mm, Hence, ok (Clause 6.1.3; IS 1390:1993) Depth o beam, D = 600 mm B = = D 600 0.5 > 0.3, hence ok (Clause 6.1.; IS 1390:1993) = 53 mm. inimum reinorcement = 0.4 y 0.4 0 = 415 = 0.6%. = 0.6 x x 53/100 = 415 mm (Clause 6..1(b) o IS 1390: 1993) aximum reinorcement =.5% =.5 x x 53 /100 = 3,990 mm (Clause 6.. o IS 1390: 1993) Span, L = 5,000 mm L D 5,000 = = 8.33 > 4, hence ok 600 (Clause 6.1.4 o IS: 1390-1993) 1.6.3 Che or Limiting Longitudinal Reinorcement Eective depth or moderate exposure conditions with 0 mm diameter bars in two layers on an average = 600 30 8 0 (0/) 1.7 Design or Flexure Table 1.5 shows, in brie, the reinorcement calculations at let end, centre and right end o the beam B as per IS 1390:1993. Design aid SP: 16 has been used or this purpose. Detailed calculations at let end are given in the ollowing sections. In actual practice, a spread sheet can be used conveniently. 1.7.1 Design or Hogging oment u = 369 kn-m IITK-GSD-EQ-V3.0 Example 1 /Page 7
Examples on 1390 u 6 369 10 = = 4.35 bd 53 53 Reerring to Table-50 o SP: 16, For d /d = 68 / 53 = 0.13, we get st at top = 1.46 % = 1.46 x x 53 /100 =,330 mm > inimum reinorcement (415 mm ) < aximum reinorcement (3,990 mm ) sc at bottom = 0.54 % But sc must be at least 50% o st, hence, revise to 1.46/ = 0.73 % (Clause 6..3 o IS: 1390-1993) Hence, sc at bottom = 0.73 x x 53 /100 = 1,165 mm 1.7. Design or Sagging oment u = 80 kn-m The beam is designed as T beam. The limiting capacity o the T-beam assuming x u < D and x u < x u,max may be calculated as ollows. u = 0.87 y st st d(1 b d (nnex G o IS 456: 000) Where, D = depth o lange = 15 mm x u = depth o neutral axis x u,max = limiting value o neutral axis = 0.48 x d = 0.48 x 53 = 55 mm b w = width o rib = mm b = width o lange L o = bw + 6d 6 y + or c/c o beams ) 0.7 0 = + + 6 15 Or 4,000 6 = 1,633 mm or 4,000 mm = 1,633 mm (lowest o the above) (Clause 3.1. o IS 456: 000) Substituting the values and solving the quadratic equation, we get st at bottom = 1,51mm > 415 mm < 3,990 mm It is necessary to che the design assumptions beore inalizing the reinorcement. x u 0.87 y = 0.36 b st 0.87 415 151 = = 47.44 mm 0.36 0 1633 < d ok < x u,max i.e. < 55 mm ok sc at top = not required. But sc must be at least 50% o st hence, revise to 1,51 / = 756 mm (Clause 6..3 o IS 1390: 1993) 1.7.3 Required Reinorcement Top reinorcement required is larger o,330 mm and 756 m. Hence, provide,330 mm. Bottom reinorcement required is larger o 1,165 mm and 1,51 mm. Hence, provide 1,51 mm. 1.8 Details o Reinorcement Table 1.6 shows summary o reinorcement provided at let end, at centre, and at right end o the beam B. total o 3-16Φ straight bars each are provided throughout the length o the beam at both top and bottom. 5-0Φ+1-16Φ extra at top (i.e., a total o 1.487 %) and 3-0Φ extra at bottom (i.e., a total o 0.97%) are provided at the let end. t the right end, i.e., over the central support, 5-0Φ + 1-16Φ extra at top (i.e. a total o 1.487%) and 1-0Φ + -16Φ extra at bottom (i.e. a total o 0.83%) bars are provided. IITK-GSD-EQ-V3.0 Example 1 /Page 8
Examples on 1390 In an external joint, both the top and bottom bars o the beam shall be provided with an anchorage length beyond the inner ace o the column equal to the development length in tension + 10 times bar diameter minus the allowance or 90 degree bend (Clause 6..5 o IS 1390:1993) as shown in Figure 1.6. 1.9. Shear Force Due to Plastic Hinge Formation at the ends o the Beam The additional shear due to ormation o plastic hinges at both ends o the beams is evaluated as per clause 6.3.3 o IS 1390:1993 and is given by V swaytoright = s Bh u u ) ± 1.4 ( + L 980 784 0 # 16 # h ± 1.4 ( u + u ) V swaytolet = L The sagging and hogging moments o resistance s ( u, Bs h u, u and u Bh ) at both ends o beam are calculated on the basis o the actual area o steel provided in the section. The beam is provided with a steel area o,374 mm (i.e., p t =1.487 %) at top and 1,545 mm (i.e. p c = 0.97%) at bottom on the let end o the beam. For p t = 1.487% and p c = 0.97%, reerring to Table 50 o SP: 16, (or p t = 1.487% or p c = 0.97% whichever gives lowest value in the table), Bs Figure 1.6: nchorage o reinorcement bars in an external joint In this case, or Fe415 steel and 0 grade concrete, rom Table 65 o SP: 16, l d = 47 Φ + 10 Φ - 8 Φ = 49 Φ = 980 mm or 0 Φ bar = 784 mm or 16 Φ bar 1.9 Design or Shear 1.9.1 Design Shear Strength o Concrete Tensile steel provided at let end = 1.487% Permissible design shear stress o concrete, τ c = 0.715 Pa (IS 456:000 Table 19) Design shear strength o concrete = τ c b d = 0.715 x x 53 /1,000 = 114 kn Similarly, design shear strength o concrete at center and right end is evaluated as 69 kn and 114 kn, respectively. h u = 4.44 bd Hogging moment capacity at, h u = 4.44 x x (53) /(1 x 10 6 ) = 377 kn-m The limiting moment carrying capacity o a beam section can also be evaluated rom the irst principle. This method is iterative but gives more appropriate values o u. For calculation o s u, the tensile steel p t = 0.97% and compressive steel p c = 1.487% is used. The contribution o the compressive steel is ignored while calculating the sagging moment capacity as T-beam. Reerring to nnex G o IS: 456-000, sagging moment capacity at or x u < D and x u < x u,max may be calculated as given below. s st y u = u = 0.87 y st d(1 ) b d = 86 kn-m IITK-GSD-EQ-V3.0 Example 1 /Page 9
Examples on 1390 Beam B Hogging moment (kn-m) Table 1.5: Flexural design or beam B Top reinorcement Let end Center Right end -369 - -371 - u /bd 4.35-4.37 st at top 1.46% - 1.47% sc at bottom Sagging moment (kn-m) st at bottom 0.54% < 1.46/ Hence revise to 0.73% (Clause 6..3; IS1390: 1993) - Bottom reinorcement 0.55% < 1.47/ Hence revise to 0.7335% (Clause 6..3; IS1390: 1993) 80 65 36 st required = 151 mm = 0.945% > 1.46/ i.e. 0.73 ok. sc at top 0.33/ = 0.165 % < 0.6% < 1.47/4=0.37% Hence, revise to 0.37%. Top = 1.46% Bottom = 0.945% st required = 335 mm = 0.1% < 0.6% < 1.47 /4 = 0.37 %, Hence revise to 0.37% (Clause 6..1(b) and 6..4 o IS1390: 1993) 0.37/ = 0.185 % < 1.47/4=0.37% Hence, revise to 0.37%. Summary o required reinorcement Top = 0.37% Bottom = 0.37% st required = 164 mm = 0.79 % > 1.47/ > 0.735 % ok. 0.79/ = 0.395% > 0.6% > 1.47/4=0.37% ok Top = 1.47% Bottom = 0.79% Beam B Top reinorcement Bottom reinorcement Table 1.6 Summary o reinorcement or beam B Longitudinal Reinorcement Let end Center Right end 3-16Φ straight + 5-0Φ +1-16Φ extra Steel Provided =,374 mm i.e. 1.487% 3-16Φ straight + 3-0Φ extra Steel Provided = 1,545 mm i.e. 0.97% 3-16Φ straight Steel Provided = 603 mm i.e. 0.378% 3-16Φ straight Steel Provided = 603 mm i.e. 0.378% 3-16Φ straight + 5-0Φ +1-16Φ extra Steel Provided =,374 mm i.e. 1.487% 3-16Φ straight + (-16Φ+1-0φ) extra Steel Provided =1,319 mm i.e. 0.83% IITK-GSD-EQ-V3.0 Example 1 /Page 10
Examples on 1390 61.8 kn + S.F. due to 1. DL + - 61.8 kn 61.8 kn + S.F. due to 1. DL + - 61.8 kn 1.6 kn + S.F. due to 1. LL + - 1.6 kn 1.6 kn + S.F. due to 1. LL + - 1.6 kn 186 kn + 175 kn + - 186 kn - 175 kn Shear due to sway to right = Shear due to sway to let = 103 kn - 70 kn 59 kn - 9 kn Figure 1.7: Shear diagram Similarly, or the right end o the beam we obtain, Shear at let end or sway to let, Bh u = 377 kn-m and Bs u = 46 kn-m, h 1.( DL + LL) 1.4( + Shear is calculated as below: V u, a = + L s Bh ±1.4( u + u ) V swaytoright = = 1. x (103 + 36) / + 175 L = 59 kn = ±1.4 (86 + 377) / 5 Shear at right end or sway to right, = ±186 kn s 1.( DL + LL) 1.4( + h Bs V 1.4( u + u ) u, b = + L V swaytolet = ± L = 1. x (103 +36) / + 186 = ±1.4(377 + 46)/5 = 70 kn = ±175 kn Shear at right end or sway to let, 1.9.3 Design Shear Reerring to the dead and live load diagrams (Figure 1.4), DL = Trapezoidal dead load + Wall and sel load = 16.5 x (1 + 5) / + 10.575 x 5 = 103 kn LL = 1 x (1 + 5) / = 36 kn Figure 1.7 shows the shear orce diagram due to DL, LL and due to hinge ormation at the ends o beam. Shear at let end or sway to right, s Bh u u ) 1.( DL + LL) 1.4( + V u, a = L = 1. x (103 + 36) / - 186 = 103 kn Bs u u ) Bh u u ) 1.( DL + LL) 1.4( u + u ) V u, b = L = 1. x (103 +36) / - 175 = 9 kn Figure 1.7 shows the shear orce diagram or the beam considering plastic hinge ormation at ends. s per Clause 6.3.3 o IS 1390:1993, the design shear orce to be resisted shall be the maximum o: i) Calculated actored shear orces as per analysis (Reer Table 1.4) ii) Shear orces due to ormation o plastic hinges at both ends o the beam plus actored gravity load on the span (as calculated in Section 1.9.3) Hence, design shear orce (V u ) will be 59 kn (maximum o 195 kn rom analysis and 59 kn corresponding to hinge ormation) or let end o h Bs IITK-GSD-EQ-V3.0 Example 1 /Page 11
Examples on 1390 the beam and 70 kn (maximum o 07 kn and 70 kn) or the right end. From analysis, the shear at the mid-span o the beam is 15 kn. However, shear due to ormulation o plastic hinges at both the ends o the beams has been calculated as 186 kn and 175 kn. Hence, the design shear at centre o the span is taken as 186 kn. The required capacity o shear reinorcement at the let end o the beam is: V us = V u V c = 59-114 = 145 kn Similarly the, required capacity o shear reinorcement at the right end and at mid-span is 156 and 117 kn, respectively. Reerring to Table 6 o SP:16, we get the required spacing o legged 8φ stirrups as 145 mm, 165 mm and 135 mm respectively at let end, centre and right end. s per Clause 6.3.5 o IS 1390:1993, the spacing o stirrups in the mid-span shall not exceed d/ = 53/ = 66 mm. inimum shear reinorcement as per Clause 6.5.1.6 o IS 456:000 is given by: S v = sv x 0.87 y /(0.4 b) = x 50 x 0.87 x 415 / ( x 0.4) = mm. Spacing o links over a length o d at either end o beam as per Clause 6.3.5 o IS1390: 1993 shall be the least o: i) d/4 = 53 /4 = 133 mm ii) 8 times diameter o smallest bar = 8 x 16 = 18 mm However, it need not be less than 100 mm. Hence, provide Legged - 8 φ stirrups @15mm c/c at let and at right end over a length o d = x 53 = 1,064 mm. Elsewhere, provide stirrups at 165 mm centers. In case o splicing o reinorcement, the spacing o links shall be limited to 150 mm centers as per clause 6..6 o IS 1390:1993. The reinorcement detailing is shown in Figure1.8. 3-16Ø straight + 5-0Ø+1-16Øextra 3-16Ø straight B 3-16Ø straight+ 5-0Ø +1-16Ø extra C 600 150 3-16Ø straight + 3-0Øextra 3-16Ø straight 0 B 3-16Ø straight + 1-0Ø+-16Ø extra 150 150 C 8Ø- legged links @ 165mm c/c 0 8Ø- legged links @ 15mm c/c upto 1090mm 600 Cross Section - 3-16Ø straight + 5-0Ø +1-16Øextra 8Ø - legged links @ 15 mm c/c upto 1065mm 600 3-16Ø straight + 3-0extra 3-16Ø straight + 5-0Ø + 1-16Øextra 8Ø - legged links @ 15 mm c/c upto 1065mm 600 3-16Ø straight + 1-0Ø+-16Ø extra Cross Section B-B Cross Section C-C 3-16Ø straight 8Ø - legged links @ 165 mm c/c 3-16Ø straight Figure 1.8: Reinorcement details or the beam BC IITK-GSD-EQ-V3.0 Example 1 /Page 1
Examples on 1390 1.10 Impact o Ductile Detailing on Bill o Quantities To compare the impact o ductile detailing (as per IS 1390:1993) on the bill o quantities, the beam under consideration has been redesigned as ollows: a) Design and detailing as per IS 456:000; seismic orces are the same as computed earlier, i.e, with response reduction actor R = 5.0. The reinorcement details are shown in Figure 1.9. Table 1.7 Comparison o bill o quantities or steel in the beam BC Description Steel required kg in Total steel in kg Detailing as per IS 1390: 1993 Longitudinal Transverse Longitudinal b) Design and detailing as per IS 456:000; seismic orces increased by a actor o 5/3 to account or R = 3.0. The reinorcement details are shown in Figure 1.10. Table 1.7 compares the quantity o reinorcement or the three cases. For the purpose o comparison, only the steel between c/c o columns is considered. Detailing as per IS 456: 000 (Seismic loads as per R = 5) Detailing as per IS 456:000 (Seismic loads as per R = 3) Transverse Longitudinal Transverse 95 5 93 14 135 8 10 107 163 Ratio 1.0 0.89 1.36-1Ø+1-16Ø straight + 5-0Ø + -16Øextra -1Ø+1-16Ø straight B -1Ø+1-16Ø straight+ 5-0Ø + -16Øextra C 600-1Ø+1-16Ø straight + 1-16Ø+3-0Øextra 150 0 B -1Ø+1-16Ø straight 150 150 C 8Ø- legged links @ mm c/c 0 8Ø- legged links @ 30mm c/c upto 1065mm -1Ø+1-16Ø + 3-0Ø 600 Cross Section - -1Ø+1-16Ø straight + 5-0Ø +-16Øextra 8Ø - legged links @ 30 mm c/c upto 1065mm 600-1Ø+1-16Ø straight + 1-16Ø + 3-0Ø extra -1Ø+1-16Ø str + 5-0Ø +-16Øextra 8Ø - legged links @ 30 mm c/c upto 1065mm -16Ø str +1-16Ø 3-0Ø extra Cross Section B-B 600 Cross Section C-C -1Ø+1-16Ø straight 8Ø - legged links @ mm c/c -1Ø+1-16Ø straight Figure 1.9: Reinorcement details or the beam BC as per IS 456:000 (with R = 5) IITK-GSD-EQ-V3.0 Example 1 /Page 13
Examples on 1390-1Ø+1-16Ø straight + 6-5Ø+1-0Ø extra -1Ø+1-16Ø straight B -1Ø+1-16Ø straight+ 6-5Ø+1-16Ø extra C 600-1Ø+1-16 Ø B straight -1Ø+1-16Ø str + 4-5Ø+1-16Ø extra 150 150 0-1Ø+1-16Ø straight+ 4-5Ø+1-1Ø+1-16Ø extra 150 C 8Ø- legged links @ 15mm c/c 0 8Ø- legged links @ 10mm c/c 600 Cross Section - -1Ø+1-16Ø straight + 6-5Ø+1-0Ø extra 8Ø - legged links @ 10 mm c/c -1Ø+1-16Ø str + 4-5Ø+1-16Ø extra 600 Cross Section B-B -1Ø+1-16Ø straight + 6-5Ø+1-16Ø extra 8Ø - legged links @ 10 mm c/c -1Ø+1-16Ø straight + 4-5Ø+1-1Ø+1-16Ø extra 600 Cross Section C-C -1Ø+1-16Ø straight 8Ø - legged links @ 145 mm c/c -1Ø+1-16Ø straight Figure 1.10: Reinorcement details or the beam BC as per IS 456:000 (with R = 3) Eect o Finite Size Correction s mentioned in the problem statement, the eect o inite size joint corrections (i.e., rigid osets at member ends) has been ignored in the analysis. In case, the designer wishes to take advantage o the inite size joint correction, care shall be taken to model the same in the static analysis. The results with inite size joint widths in the analysis are presented in Table 1.3a. The results without and with inite size corrections can be compared rom Tables 1.3 and 1.3a, respectively. However, in the detailed calculations shown in this example, this correction has been ignored. Table 1.3a Force resultants in the beam B or various load cases with Finite Size Correction Load Case Shear (kn) Let end Center Right end oment (kn-m) Shear (kn) oment (kn-m) Note: The results are rounded o to the next integer value. Shear (kn) oment (kn-m) DL -48-9 4 8 55-45 LL -14-10 0 10 16-13 EQY 83 191 83 8 83 177 IITK-GSD-EQ-V3.0 Example 1 /Page 14
Examples on 1390 Example - Beam Design o an RC Frame Building in Seismic Zone II Problem Statement: The ground plus our storey RC oice building o Example-1 (Reer Figures 1.1-1.4) is assumed to be located in seismic zone II on medium soil. The dead load and live loads are the same as in Example-1. However, the earthquake loads are much lower or zone-ii. Hence, reduced member sizes are considered as shown in Table.1. The design o a beam along grid line, as per recommendations o IS1390:1993, is explained. Solution Design o iddle Floor Beam The beam marked BC in Figure.1 or rame (Figure 1.1 o Example 1) is considered or design. Since the beam consists o two symmetrical spans, calculations are perormed or one span only. Note: ll dimensions in mm.1 ember Forces For beam B, orce resultants or various load cases and load combinations have been obtained rom computer analysis and are summarized in Table.. Table.3 shows orce resultants or dierent load combinations with the maximum values to be used or design being underlined. Table. Force resultants in beam B or dierent load cases Load Case Let end Centre Right end V kn kn-m V kn kn-m V kn kn-m DL -48-39 9 53-50 LL -15-14 0 10 15-16 EQY 59 4-50 Figure.1 Beam BC Table.1 Schedule o member sizes Column Beam C1 30 x RB1, FB1 30 x C 350 x 350 RB, FB 30 x 400 C3 x PB1 30 x 350 Note: V = Shear; = oment, The results are rounded o to the next higher integer value.. Various Ches..1 Che or xial Stress Factored axial orce = 0.00 kn Factored axial stress = 0.0 Pa < 0.10 Hence, design as lexural member. (Clause, 6.1.1; IS 1390:1993) PB x Slab Thiness: 15 IITK-GSD-EQ-V3.0 Example / Page 15
Examples on 1390 Table.3 Force resultants in beam B or dierent load combinations S. No. Load Combination Let end Centre Right end 1 Shear (kn) oment (kn-m) Shear (kn) oment (kn-m) Shear (kn) oment (kn-m) 1.5DL+1.5LL -95-80 3 59 10-99 3 4 1.(DL+LL*+EQY) -36 0 9 43 95-15 1.(DL+LL*-EQY) -89-1 -4 33 4-5 5 1.5(DL+EQY) 6 1.5(DL-EQY) 7 0.9DL+1.5 EQY 8 0.9DL-1.5 EQY * ppropriate raction o live load has been used -39 30 36 50 113-150 -105-147 -30 38 47 0-10 53 35 3 81-10 -76-14 -31 0 15 30 = 6 mm.. Che or ember Size Width o beam, B = 30 mm > 00 mm Hence, ok. (Clause 6.1.3; IS 1390:1993) Depth o beam, D = mm B 30 = = 0.46 > 0.3 D Hence, ok. (Clause 6.1.; IS 1390:1993) Span, L = 5,000 mm L 5,000 = = 10 > 4 D Hence, ok. (Clause 6.1.4 o IS 1390:1993)..3 Che or Limiting Longitudinal Reinorcement Eective depth or moderate exposure condition with 16 mm diameter bar in two layers on an average = 30 16 (16/) 8 = 438 mm. inimum reinorcement, = 0.4 y = 0.4x 0 415 = 0.6%. = 0.6 x 30 x 438/100 (Clause 6..1(b) o IS1390: 1993) aximum reinorcement =.5% =.5 x 30 x 438/100 =,518 mm (Clause 6.. o IS 1390:1993).3 Design or Flexure Table.4 shows, in brie, the reinorcement calculations at let end, centre and right end as per IS 1390:1993. Design aid SP: 16 has been used or the purpose. Detailed calculations at let end are given in the ollowing sections. In actual practice, a spread sheet can be used conveniently..3.1 Design or Hogging oment u = 147 kn-m u 147 10 = = 3.33 bd 30 438 438 Reerring to Table-50 o SP: 16 6 For d /d = 6/446 = 0.14 and interpolating between d /d o 0.10 and 0.15, we get st at top = 1.13% = 1.13 x 30 x 438/100 = 1,140 mm > inimum reinorcement (6 mm ) < aximum reinorcement (,518 mm ) IITK-GSD-EQ-V3.0 Example / Page 16
Examples on 1390 sc at bottom = 0.19 % But sc must be at least 50% o st. Hence, revise to 1.13 / = 0.566 % (Clause 6..3 o IS 1390:1993) Hence, sc at bottom = 0.566 x 30 x 438 /100 = 571 mm.3. Design or Sagging oment u = 53 kn-m The beam is designed as T beam. The limiting capacity o the T-beam assuming x u < D and x u < x u,max may be calculated as given below. u st y = 0.87 y st d(1 ) ------- ( i ) b d (nnex G o IS 456: 000) Where, D = depth o lange = 15 mm x u = depth o neutral axis x u,max = limiting value o neutral axis = 0.48 x d = 0.48 x 438 = 10 mm b w = width o rib = 30 mm b = width o lange L o = bw + 6d 6 + or c/c o beams 0.7 5,000 = + 30 + 6 15 or 4,000, 6 whichever is less = 1,563 mm or 4,000 mm = 1,563 mm (lower o the above) (Clause 3.1. o IS 456: 000) Substituting the relevant values in (i) and solving the resulting quadratic equation, we get st at bottom = 339 mm > 6 mm <,518 mm It is necessary to che the design assumptions beore inalizing the reinorcement. x u 0.87 y = 0.36 b st 0.87 415 339 = = 10.88 mm 0.36 0 1,563 < d ok. < x u,max i.e. 10 mm ok. sc at top = not required. But sc must be at least 50 % o st, hence, revise to 339 / = 170 mm (Clause 6..3 o IS 1390: 1993) st at bottom = 339 mm = 339 x 30 x 438 /100 = 0.33 % > 0.6 % < 4% Hence, ok..3.3 Required reinorcement Top reinorcement required is the larger o 1,13 mm and 170 mm. Hence, provide 1,13 mm. Bottom reinorcement required is the larger o 339 mm and 571 mm. Hence, provide 571 mm..4 Details o Reinorcement Table.5 show a summary o reinorcement provided at the let end, at center and at the right end o the beam B. 3-1Φ straight bars are provided throughout the length o the beam at the top and 4-1Φ straight bars are provided throughout at the bottom. 4-16Φ +1-1Φ extra bars at the top and 1-1Φ extra bar at the bottom at the let end are also provided. t the right end, i.e., over the central support, 4-16Φ +1-1Φ extra bars at the top and -1Φ extra bottom bars are provided. t an external joint, as per Clause 6..5 o IS 1390:1993, both the top and bottom bars o the beam shall be provided with an anchorage length beyond the inner ace o the column equal to the development length in tension + 10 times IITK-GSD-EQ-V3.0 Example / Page 17
Examples on 1390 the bar diameter minus the allowance or 90 degree bend. (Reer Figure.) Table.4 Flexural design or beam B Beam B Top reinorcement Let end Center Right end Hogging moment -147 - -150 (kn-m) - u /bd 3.33-3.4 st required at top 1.13% - 1.163% sc required at bottom 0.19% < 1.13/ = 0.566% Hence revise to 0.566% (Clause 6..3; IS1390: 1993) Sagging moment (kn-m) st at bottom st required = 339 mm Bottom reinorcement - 0.4% < 1.163/ = 0.58% Hence revise to 0.58% (Clause 6..3; IS 1390:1993) 53 58 30 = 0.33% sc at top 0.33/ = 0.165% < 0.6% < 1.163/4=0.91% Hence, revise to 0.91%. st required = 371 mm = 0.37% > 0.6 > 1.163/4 = 0.91% ok 0.37 / = 0.185% < 0.6% < 1.163/4 = 0.91 % Hence, revise to 0.91%. st required = 19 mm =0.16 % < 0.6% <1.163/ = 0.58% Hence revise to 0.58%. (Clause 6..3; IS1390: 1993) 0.58 / = 0.91% > 0.6% ok. Top = 1.13% Bottom = 0.566% Summary o required reinorcement Top = 0.91% Top = 1.163 % Bottom = 0.37 % Bottom = 0.58% Beam B Top reinorcement Bottom reinorcement Table.5 Summary o reinorcement provided or the beam B Longitudinal reinorcement Let end Center Right end 3-1Φ straight Steel Provided = 339 mm i.e. 0.33% 3-1Φ straight + 4-16Φ extra Steel Provided = 1,143 mm i.e. 1.134% 4-1Φ straight + -10Φ extra Steel Provided = 609 mm i.e. 0.6% 4-1Φ straight Steel Provided = 45 mm i.e. 0.44% 3-1Φ straight + 4-16Φ +1-1Φ) extra Steel Provided = 1,56 mm i.e. 1.46% 4-1Φ str + -10Φ extra Steel Provided = 609 mm i.e. 0.6% IITK-GSD-EQ-V3.0 Example / Page 18
Examples on 1390 784 588 16 # 1 # Figure. nchorage o beam bars in an external joint In this case, or Fe 415 steel and 0 grade concrete, rom Table 65 o SP: 16, l d = 47 Φ + 10 Φ - 8 Φ = 49 Φ = 784 mm or 16 Φ bar = 588 mm or 1 Φ bar.5 Design or Shear.5.1 Design Shear Strength o Concrete Tensile steel provided at let end = 1.134% Permissible design shear stress o concrete, τ c = 0.66 Pa (Table 19 o IS 456:000) Design shear strength o concrete = τ c b d = 0.66 x 30 x 438 /1,000 = 66 kn Similarly, the design shear strength o concrete at mid-span and at the right end is evaluated as 46 kn and 66 kn, respectively..5. Shear Force Due to Plastic Hinge Formation at the ends o the Beam The additional shear due to ormation o plastic hinges at both ends o the beams is evaluated as per clause 6.3 o IS 1390:1993 and is given by V swaytoright = s Bh u u ) ± 1.4( + L h ± 1.4( u + u ) V swaytolet = L The sagging and hogging moments o resistance s ( u, Bs h u, u and Bh u ) at both ends o the beam are to be calculated on the basis o the actual area o steel provided in the section. The beam is provided with a steel area o 1,143 mm (i.e., p t = 1.134%) at top and 609 mm (i.e., p c = 0.60%) at bottom on the let end o the beam. For p t = 1.11% and p c = 0.60%, reerring to Table 50 SP: 16(or p t = 1.134% or p c = 0.60% whichever gives lowest value in the table), h u = 3.36 bd Hogging moment capacity at, h u = 3.36 x 30 x 438 x 438 / 10 6 = 149 kn-m For calculation o s u, the tensile steel p t = 0.60% and compressive steel p c = 1.134% is used. The contribution o the compression steel is ignored while calculating the sagging moment capacity as T-beam. Reerring to nnex G o IS: 456-000, sagging moment capacity at or x u < D and x u < x u,max may be calculated as given below. s st y u = u = 0.87 y st d(1 ) b d = 94 kn-m Similarly, or the right side joint we obtain, = 165 kn-m and Bs u = 94 kn-m. Shear is calculated as below: u Bh ± 1.4( + u V swaytoright = L = ±1.4(94 + 165) /5 = ±7kN Bs s Bh u ) h Bs u ) ± 1.4( + u V swaytolet = L = ±1.4(149 + 94) /5 = ±68 kn IITK-GSD-EQ-V3.0 Example / Page 19
Examples on 1390 61.8 kn 1.6 kn + 68 kn + 15 kn + + S.F. due to 1. DL + S.F. due to 1. LL + - Shear due to sway to let = 7 kn - - 61.8 kn 1.6 kn 68 kn 15.4 kn Figure.3 Shear diagram due to sway to let.6 Design Shear Reerring to the dead and live load diagrams (Figure 1.4 o Example 1), DL = Trapezoidal DL+ Bri wall & Sel load = 16.5 x (1 + 5)/ + 10.575 x 5 = 103 kn LL = 1 x (1 + 5) / = 36 kn Shear at let end or sway to right, s Bh u u ) 1.( DL + LL) 1.4( + V u, a = L = 1. x (103 + 36) / - 7 = -11.4 kn Shear at let end or sway to let, h Bs u u ) 1.( DL + LL) 1.4( + V u, a = + L = 1. x (103 + 36) / + 68 = 15 kn Shear at right end or sway to right, s Bh u u ) 1.( DL + LL) 1.4( + V u, b = + L = 1. x (103 + 36)/ + 7 = 155 kn Shear at right end or sway to let, 1.( DL + LL) 1.4( u + u ) V u, b = L = 1. x (103 + 36) / - 68 = 15.4 kn The design shear orce shall be the maximum o: i) Calculated actored shear orce as per analysis (Reer Table.3) ii) Shear orce due to ormation o plastic hinges at both ends o the beam plus due to actored gravity load on the span (as calculated in.6.3) Hence, the design shear orce (V u ) will be 15 kn (maximum o 105 kn rom analysis and 15 kn corresponding to hinge ormation) or the let end o beam and 155 kn (maximum o 113 kn rom analysis and 155 kn corresponding to hinge ormation) or the right end. Shear at the mid-span rom analysis is 36 kn. However, shear due to ormation o plastic hinges at both the ends o the beams will be 7 kn. The required capacity o shear reinorcement at the let end, V us = V u V c = 15-66 = 86 kn Similarly, the required capacity o shear reinorcement at the right end and at mid-span can be calculated as 6 kn and 89 kn, respectively. Reerring to Table 6 o SP: 16, we get the required spacing o legged 8φ stirrups as 30 mm centers at let and at the right end. s per Clause 6.3.5 o IS 1390:1993, the spacing o stirrups in the rest o member shall be limited to d/ = 438/ = 19 mm. inimum shear reinorcement as per Clause 6.5.1.6 o IS 456:000 S v = sv x 0.87 y /(0.4 b) = x 50 x 0.87 x 415 / ( x 0.4) = mm. < 438 x 0.75 = 38 mm Hence, ok. h Bs IITK-GSD-EQ-V3.0 Example / Page 0
Examples on 1390 The spacing o minimum stirrups is kept at mm. Spacing o links over a length o d at either end o the beam as per Clause 6.3.5 o IS 1390:1993 shall be least o i) d/4 = 438/4= 109 mm ii) 8 times diameter o smallest bar = 8 x1 = 96 mm However, it should not less than 100 mm. Hence, provide legged 8 φ stirrups @100 mm c/c at let and at the right end o the member over a length o d = x 438 = 876 mm at either end o the beam. Elsewhere, provide stirrups at 15(< 19 mm) centers. In case o splicing o main reinorcement, the spacing o links shall be limited to 150 mm centers as per Clause 6..6 o IS 1390:1993. The reinorcement detailing is shown in Figure.4. 3-1Østraight + 4-16Øextra 3-1Østraight B 3-1Ø straight+ 4-16Ø + 1-1Ø extra C 4-1Østraight 4-1Ø straight + -10Øextra 150 0 B 4-1Ø straight + -10Ø extra 150 150 C 8Ø- legged links @ 15mm c/c 0 8Ø- legged links @ 100mm c/c upto 900mm 30 Cross Section - 3-1Ø straight + 4-16Øextra 8Ø - legged links @ 100mm c/c upto 900mm 4-1Ø straight + -10 extra 3-1Ø straight + 4-16Ø + 1-1Øextra 8Ø - legged links @ 100mm c/c upto 900mm 4-1Ø straight + -10 extra 30 Cross Section B-B 30 Cross Section C-C 3-1Ø straight 8Ø - legged links @ 15mm c/c 4-1Ø straight Figure.4 Reinorcement details or the beam BC.7 Impact o Ductile Detailing on Bill o Quantities To compare the impact o ductile detailing (as per IS 1390:1993) on the bill o quantities, the beam has been redesigned as ollows: a) Design and detailing as per IS 456:000; seismic orces are the same as computed earlier, i.e., with response reduction actor R = 5.0. The reinorcement details are shown in Figure.5. b) Design and detailing as per IS 456:000; seismic orces are increased by a actor o 5/3 to account or R = 3.0. The reinorcement details are shown in Figure.6. Table.6 compares the quantity o reinorcement or the three design cases. While calculating the quantities c/c span is considered. IITK-GSD-EQ-V3.0 Example / Page 1
Examples on 1390-1Østraight + (4-16Ø+1-1Ø) extra -1Østraight B -1Ø straight+ (4-16Ø+-1Ø) extra C 150-1Ø straight + 1-1Ø extra -1Østraight + -1Ø extra 0 B -1Ø straight 150 150 C 8Ø- legged links @ mm c/c 0 8Ø- legged links @ 30mm c/c upto 900mm -1Ø straight + (4-16Ø +1-1Ø)extra 8Ø - legged links @ 30 mm c/c upto 900mm -1Ø straight + (4-16Ø +-1Ø)extra 8Ø - legged links @ 30 mm c/c upto 900mm -1Ø straight 8Ø - legged links @ mm c/c 30 Cross Section - -1Ø straight + 1-1Ø extra 30 Cross Section B-B -1Ø straight + -1Ø extra 30 Cross Section C-C 4-1Ø straight Figure.5 Reinorcement detail or the beam BC as per IS 456:000 (with R = 5.0) -1Østraight B -1Ø straight+ 7-16Øextra C 150-1Ø straight + 1-1Øextra -1Østraight + -1Ø extra 0 B -1Østraight+ 3-16Ø extra 150 150 C 8Ø- legged links @ mm c/c 0 8Ø- legged links @ 30mm c/c 30-1Ø straight + 7-16Øextra 8Ø - legged links @ 30 mm c/c -1Ø straight + 3-16Ø extra -1Ø straight + 7-16Øextra 8Ø - legged links @ 30 mm c/c -1Ø straight + 3-16Ø extra 30 Cross Section B-B 30 Cross Section C-C -1Ø straight 8Ø - legged links @ mm c/c -1Ø straight +-1Ø extra Figure.6 Reinorcement detail or the beam BC as per IS 456:00 (with R = 3.0) IITK-GSD-EQ-V3.0 Example / Page
Examples on 1390 Description Table.6 Comparison o bill o quantities o steel in the beam BC Detailing as per IS 1390: 1993 Detailing as per IS 456:000 (Seismic loads with R = 5) Detailing as per IS 456:00 (Seismic loads with R = 3) Longitudinal Transverse Longitudinal Transverse Longitudinal Transverse Steel required (kg) 5 3 46 13 64 13 Total steel (kg) 75 59 77 Ratio 1.0 0.79 1.03 IITK-GSD-EQ-V3.0 Example / Page 3
Examples on 1390 Example 3 - Interior Column Design o an RC Frame Building in Seismic Zone V 3 Problem Statement: For the ground plus our storey RC oice building o Example-1 (Reer Figures 1.1-1.4 o Example 1), design o an interior column element is explained here. The column marked B in Figure 3.1 or rame is considered or design. B B T B B 600 T B 600 Z 400 Y Solution: Figure 3.1 Column location in elevation 3.1 Load Combinations Load combinations derived rom recommendations o Clause 6.3.1. o IS 1893(Part 1): 00 and given in Table 1.4 o Example-1 are considered or analysis. 3. Force Data For column B, the orce resultants or various load cases and load combinations are shown in Tables 3.1 and 3.. In case o earthquake in X direction, column gets a large moment about Y axis and a small moment about X axis due to gravity, minimum eccentricity and torsional eect. Similarly earthquake in Y direction causes a large moment in column about X axis and a small moment about Y axis. Column needs to be designed as a biaxial member or these moments. Since the column must be designed or earthquake in both X direction and Y direction, all 13 load combinations as shown in Table 1.4 (Example-1) need to be considered. It is necessary to che the column or each combination o loads. Cheing the column or all load combinations at all the sections is indeed tedious i carried out by hand. Hence, a computer program is best suited or column design. In the absence o a computer program, one can make a judgment o which two or three load cases out o the thirteen may require the maximum reinorcement and design accordingly. Reerring to Table 3., it can be observed that out o the various load combination, one design load combination with earthquake in either (X or Y) direction can be identiied, which is likely to be critical. These critical design orces are summarised in Table 3.3. Table 3.4 and Table 3.5 IITK-GSD-EQ-V3.0 Example 3 /Page 4
Examples on 1390 give actors such as Pu, bd, and b D calculated and summarised in Table 3.6. The detailed calculations are shown in Section 3.4. 3 Using these actors and the charts given bd in SP: 16, the required maximum reinorcement is Table 3.1 Force resultants in column B or dierent load cases Load case xial (kn) B T B B B T (kn-m) 3 (kn-m) xial (kn) (kn-m) 3 (kn-m) xial (kn) (kn-m) 3 (kn-m) xial (kn) (kn-m) 3 (kn-m) DL LL EQx EQy -961 1 0-764 -1 0-749 1 0-556 -1 0-41 0 0-185 0 0-185 0 0-131 1 0-169 0-11 -169 0-11 173 0-4 -148 0 0 0-198 0 0 191 0 0-194 0 0 166 Table 3. Force resultants in column B or dierent load combinations Load Combinations xial (kn) B T B B B T 3 3 3 (kn-m) (kn-m) (kn-m) (knm) xial (kn) (knm) 1.5(DL+LL) -1803 0-144 - 0-1401 0-1031 0 0 1.(DL+LL+EQX) -15 04 0-986 -04 0-968 09 0-711 -179 0 1.(DL+LL-EQX) -1199-0 0-959 0 0-941 -06 0-70 177 0 1.(DL+LL+EQY) -16 1-38 -97-1 9-954 1-33 -707-1 199 1.(DL+LL-EQY) -16 1 38-97 -1-9 -954 1 33-707 -1-199 1.5(DL+EQX) -1475 55 0-1163 -55 0-1140 61 0-840 -4 0 1.5(DL-EQX) -1409-5 0-1130 5 0-1107 -58 0-88 1 0 1.5(DL+EQY) -144-97 -1146-87 -114-91 -834-49 1.5(DL-EQY) -144 97-1146 - -87-114 91-834 - -49 0.9DL + 1.5 EQX -898 54 0-704 -54 0-691 60 0-506 -3 0 0.9DL - 1.5 EQX -83-53 0-671 53 0-658 -59 0-494 1 0 0.9DL + 1.5 EQY -865 1-97 -688-1 87-674 1-91 - -1 49 0.9DL - 1.5 EQY -865 1 97-688 -1-87 -674 1 91 - -1-49 3.3 Design Ches xial (kn) (knm) xial (kn) (knm) Factored axial stress = 6,58,000 / (400 x ) 3 (knm) 3.3.1 Che or xial Stress Lowest actored axial orce = 658 kn (Lowest at t or B b among all load combination is considered) = 3.9 Pa > 0.10 Hence, design as a column member. (Clause 7.1.1; IS 1390:1993) IITK-GSD-EQ-V3.0 Example 3 /Page 5
Examples on 1390 3.3. Che or member size Width o column, B = 400 mm > mm Hence, ok (Clause 7.1.; IS 1390:1993) Depth o column, D = mm B 400 = = 0.8 > 0.4, hence ok D (Clause 7.1.3; IS 1390:1993) Span, L = 3,000 mm The eective length o column can be calculated using nnex E o IS 456: 000. In this example as per Table 8 o IS 456: 000, the eective length is taken as 0.85 times the unsupported length, which is in between that o ixed and hinged case. L (0 ) 0.85 = = 5.31 < 1, D 400 i.e., Short Column. Hence ok. (Clause 5.1. o IS 456: 000) In case o slender column, additional moment due to P-δ eect needs to be considered. inimum dimension o column = 400 mm 15 times the largest diameter o beam longitudinal reinorcement = 15 x 0 = ok (Clause 7.1. o proposed drat IS 1390) 3.3.3 Che or Limiting Longitudinal Reinorcement inimum reinorcement, = 0.8 %. = 0.8 x 400 x /100 = 1,600 mm (Clause 6.5.3.1 o IS 456: 000) aximum reinorcement = 4% (Limited rom practical considerations) = 4 x 400 x /100 = 8,000 mm (Clause 6.5.3.1 o IS 456: 000) 3.4 Design o Column 3.4.1 Sample Calculation or Column Reinorcement at B End First approximate design is done and inally it is cheed or all orce combinations. (a) pproximate Design In this case, the moment about one axis dominates and hence the column is designed as an uniaxially loaded column. The column is oriented in such a way that depth o column is 400 mm or X direction earthquake and mm or Y direction earthquake orce. Design or Earthquake in X-direction P u = 1,475 kn u = 55 kn-m 3 Pu 1475 10 = = 0.37 bd 0 400 u bd 55 10 = = 0.16 0 400 400 6 Reerring to Charts 44 and 45 o SP16 For d /D = (40 + 5 / ) / 400 = 0.13, we get p/ = 0.14 Design or Earthquake in Y-direction P u = 1,44 kn u = 97 kn-m 3 Pu 1,44 10 = = 0.36 bd 0 400 u bd 97 10 = = 0.15 0 400 6 Reerring to Charts 44 o SP16 For d /D = (40 + 5 / ) / = 0.105, we get p/ = 0.11 Longitudinal Steel The required steel will be governed by the higher o the above two values and hence, take p/ = 0.14. Required steel = (0.14 x 0) % =.8 % =.8 x 400 x /100 = 5,600 mm IITK-GSD-EQ-V3.0 Example 3 /Page 6
Examples on 1390 Provide 10-5Φ + 4-16Φ bars with total sc provided = 5,714 mm i.e., 5,714 x100 /(400 x ) =.85%. Hence, p/ provided =.85/0 = 0.143 (b) Cheing o Section The column should be cheed or bi-axial moment. oment about other axis may occur due to torsion o building or due to minimum eccentricity o the axial load. Cheing or Critical Combination with Earthquake in X Direction (Longitudinal direction) Width = mm; Depth = 400 mm P u = 1,475 kn u = 55 kn-m Eccentricity = Clear height o column/ + lateral dimension / 30 (Clause 5.4 o IS 456:000) = ((3,000-) / ) + (400 / 30) = 18.33 mm < 0 mm Hence, design eccentricity = 0 mm u3 = 1,475 x 0.0 = 9.5 kn-m Pu For = 0. 37 and p/ = 0.143, and reerring bd to Charts 44 and 45 o SP: 16, we get u = 0.175 bd u u,1 = 0.175 0 400 400 /(1 10 = 80 kn-m 3,1 = 0.175 0 400 /(1 10 = 350 kn-m P uz = 0.45 c + 0.75 y sc (Clause 39.6 o IS 456:000) = 0.45 g + (0.75 y -0.45 ) sc = 0.45 x 0 x 400 x + (0.75 x 415 0.45 x 0) x 5,714 = 3,57 kn P u /P uz = 1,475 /3,57 = 0.4 6 6 ) ) The constant α n which depends on actored axial compression resistance P uz is evaluated as 0.4 0. α n = 1.0 + (.0 1.0) =1.367 0.8 0. Using the interaction ormula o clause 39.6 o IS 456: 000) αn αn 1.367 1.367 u u3 55 9.5 + = +,1 3,1 80 350 u u = 0.88 +0.04 = 0.9 < 1.00 Hence, ok Cheing or Critical Combination with Earthquake in Y Direction (Transverse direction) Width = 400 mm; Depth = mm P u = 1,44 kn u3 = 97 kn-m Eccentricity = clear height o column / + lateral dimension / 30 = ((3,000-600)/) + ( / 30) = 1.46 mm > 0 mm u = 1,44 x 0.0146 = 31 kn-m Pu For = 0. 355 and p/ = 0.143, bd Reerring to Chart 44 o SP: 16, we get u,1 bd = 0.18 6 u,1 = 0.18 0 400 400 /1 10 = 88 kn-m 6 u3,1 = 0.18 0 400 /1 10 = 360 kn-m P uz = 3,57 kn α n = 1.35 Using the interaction ormula u u,1 αn + u3 u3,1 αn = 31 88 1..367 97 + 360 = 0.0473 +0.7687 = 0.816 <1.00 1.367 IITK-GSD-EQ-V3.0 Example 3 /Page 7
Examples on 1390 Hence, ok 3.5 Details o Longitudinal Reinorcement Similar to the sample calculations shown in Section 3.4.1, the steel required at T, B B and B T is calculated. The Tables 3.4 and 3.5 show brie calculations at B, T, B B and B T locations. The column at joint should have higher o the reinorcement required at B and T, and hence.8% steel is needed. Similarly, higher o the reinorcement required at B B and B T, i.e..4% is needed in the column at joint B. Figure 3. shows the reinorcement in the column along with the steel provided in the transverse and longitudinal beams. Table -3.3 Critical orces or design o column B Load B T B B B T Combination P 3 P 3 P 3 P 3 Gravity -1,803 0-1,44-0 -1,401 0-1,031 0 0 Critical comb with -1,475 55 0-1,163-55 0-1,140 61 0-840 -4 0 EQX Critical comb with -1,44-97 -1,146-87 -1,14-91 -834-49 EQY Table- 3.4 Design o column B or earthquake in X direction Load Comb P u bd B T B B B T P P p P p b D u bd b p D u bd b D u bd b D p Gravity 0.45 0.00 0.8-0.36 0.00 0.8-0.35 0.00 0.8 0.6 0.00 0.8 Critical comb with EQX 0.37 0.16.8 0.9 0.16.4 0.9 0.16.4 0.1 0.14.0 Load Comb P u bd Table- 3.5 Design o column B or earthquake in Y direction B T B B B T 3 p Pu 3 p Pu 3 p Pu 3 bd bd bd bd bd bd bd p Critical comb with EQY 0.36 0.145.8 0.9 0.133 1.8 0.8 0.1455. 0.1 0.14 1.6 Note: b = 400 mm and D = mm IITK-GSD-EQ-V3.0 Example 3 /Page 8
Examples on 1390 Column B Reinorcement at Reinorcement at B Table 3.6 Summary o reinorcement or column B Longitudinal Reinorcement 10-5Φ +4-16Φ Steel provided = 5,714 mm i.e.,.86% 8-5Φ+ 6-16Φ Steel provided= 5,134 mm i.e.,.57% Conining Links: 8 # links @ 85 c/c Nominal Links: 8 # links @ 00 C/C 400 Reinorcement at 10-5 # + 4-16 # 400 Reinorcement at B 8-5 # + 6-16 # 10-5# + 4-16# at 8-5# + 6-16# at B Transverse beam x 600 (5-0 # + 4-16 # - Top steel 5-16 # + 1-0 # - Bottom steel) 400 Longitudinal beam x (4-0 # + 5-16 # - Top steel 3-0 # + 4-16 # - Bottom steel) Figure 3. Reinorcement details o longitudinal and transverse beam Table - 3.7 Shear orces in column B or dierent load combinations B T B B B T Load Combination EQX (kn) EQY (kn) EQX (kn) EQY (kn) EQX (kn) EQY (kn) EQX (kn) EQY (kn) 1.5(DL+LL) 0-1 0 0 0 0 0 0 1.(DL+LL+EQX) 0-133 0-137 0-137 0-1 1.(DL+LL-EQX) 0 13 0 136 0 136 0 11 1.(DL+LL+EQY) 149 0 154 0 154 0 136 0 1.(DL+LL-EQY) -149 0-154 0-154 0-136 0 1.5(DL+EQX) 0-167 0-171 0-171 0-153 1.5(DL-EQX) 0 166 0 170 0 170 0 15 1.5(DL+EQY) 186-1 19-1 19-1 171-1 1.5(DL-EQY) -186-1 -19-1 -19-1 -171-1 0.9DL + 1.5 EQX 0-167 0-171 0-171 0-153 0.9DL - 1.5 EQX 0 166 0 170 0 170 0 15 0.9DL + 1.5 EQY 186 0 19 0 19 0 171 0 0.9DL - 1.5 EQY -186 0-19 0-19 0-171 0 IITK-GSD-EQ-V3.0 Example 3 /Page 9
Examples on 1390 3.6 Design or Shear 3.6.1 Shear Capacity o Column ssuming 50% steel provided as tensile steel to be on conservative side, st =.86% / = 1.43% Permissible shear stress τ c = 0.70 pa (Table 19 o IS 456: 000) Considering lowest P u = 658 kn, we get 3Pu ultiplying actor = δ = 1+ =1.49 < 1.5 (Clause 40.. o IS 456: 000) τ c = 0.70 x 1.49 = 1.043 Pa Eective depth in X direction = 400-40-5/ = 347.5 mm V c = 1.043 x x 347.5 /1,000 = 181 kn Eective depth in Y direction = -40-5/ = 447.5 mm V c = 1.043 x 400 x 447.5 /1,000 = 187 kn 3.6. Shear s Per nalysis s per Table 3.7, the maximum actored shear orce in X and Y direction is 19 and 171 kn, respectively. 3.6.3 Shear Force Due to Plastic Hinge Formation at Ends o Beam 3.6.3.1 Earthquake in X-Direction The longitudinal beam o size x is reinorced with 4-0Φ extra + 5-16Φ str (,61 mm, i.e., 1.74%) at top and 3-0Φ extra + 4-16Φ str (1,746 mm, i.e., 1.34%) at bottom. The hogging and sagging moment capacities are evaluated as 88 kn-m and 1 kn-m, respectively. h st bl u V u V u = 1.4 V u g br u h st + bl u br u Figure 3.3 Column shear due to plastic hinge ormation in beams Reerring to Figure 3.3, the shear orce corresponding to plastic hinge ormation in the longitudinal beam is evaluated as: bl 1.4 ( u + V u = h st br u = 1.4 x (88 +1) /3 = 37 kn 3.6.3. Earthquake in Y-Direction ) The transverse beam o size x 600 is reinorced with 3-16Φ str + 5-0Φ + 1-16Φ extra (,374 mm, i.e., 1.485%) at top and 3-16Φ str + 3-0Φ extra (1545 mm, i.e., 0.978%) at bottom. The hogging and sagging moment capacity is evaluated as 377 kn-m and 46 kn-m, respectively. Reerring to Figure 3.3, the shear orce corresponding to plastic hinge ormation in the transverse beam is V u = bl u 1.4 ( + h st br u 1.4 (377 + 46) = 3 = 91 kn 3.6.4 Design Shear ) The design shear orce or the column shall be the higher o the calculated actored shear orce as per analysis (Table 3.7) and the shear orce due to plastic hinge ormation in either o the transverse or the longitudinal beam. (Clause7.3.4; IS 1390: 1993) The design shear in X direction is 37 kn which is the higher o 19 kn and 37 kn. Similarly, the design shear in Y direction is 91 kn which is the higher o 171 kn and 91 kn. 3.7 Details o Transverse Reinorcement 3.7.1 Design o Links in X Direction V s = 37 181 = 56 kn. Spacing o 4 Legged 8 Φ Links = 4 50 0.87 415 347.5 = 448 mm 56,000 IITK-GSD-EQ-V3.0 Example 3 /Page 30
Examples on 1390 3.7. Design o Links in Y Direction V s = 87 187 = 100 kn Spacing o 3 legged 8 Φ Links = 3 50 0.87 415 447.5 = 43 mm 1,00,000 3.7.3 Nominal Links The spacing o hoops shall not exceed hal the least lateral dimension o the column i.e., 400/ = 00 mm. (Clause 7.3.3; IS 1390:1993) Provide 8 Φ links @ 00 c/c in mid-height portion o the column. 3.7.4 Conining Links The area o cross section, sh, o the bar orming rectangular hoop, to be used as special conining reinorcement shall not be less than (Clause 7.4.8 o IS 1390:1993). 0.18 S h g sh = 1 y k h = longer dimension o the rectangular link measured to its outer ace = (( 40 40 5) /3 + (8 x )) + 5) = 17 mm, or ((400 40 40-5)/ +(8 x ) +5) =188.5 mm, Whichever is higher, i.e,. h = 188.5 mm. g = 400 x =,00,000 mm k = (400- x 40 + x 8) x (- x 40 + x 8) = 336 x 436 = 1,46,496 mm ssuming 8Φ stirrup, sh = 50 mm 0.18 S 188.5 0,00,000 50 = 1 415 1,46,496 Substituting we get S = 84 mm. Link spacing or conining zone shall not exceed: (a) ¼ o minimum column dimension i.e, 400 / 4 =100 mm (b) But need not be less than 75 mm nor more than 100 mm. (Clause 7.4.6 o IS 1390:1993). Provide 8Φ conining links @ 80 c/c or a distance l o (Reer Figure 3.4), which shall not be less than: i) Larger lateral dimension = mm ii) 1/6 o clear span = (0 ) / 6 = 417 mm iii) 450 mm (Clause 7.4.1 o IS 1390:1993) Provide conining reinorcement or a distance o l o = mm on either side o the joint. [Reer Figure 3.4] Figure 3.4 Reinorcement details or column The comparisons o steel quantities are shown in Table 3.8 or various detailing options. Table 3.8 Comparison o bill o quantities o steel in column Description Detailing as per IS 1390: 1993 (Seismic loads as per R = 5) Detailing as per IS 456: 000 (Seismic loads as per R = 5) Links (kg) 5 14 ain (kg) steel CUTION 18 18 Detailing as per IS 456: 000 (Seismic loads as per R = 3) Column needs to be redesigned. Note, however, that the column designed above has not been cheed or requirements related to IITK-GSD-EQ-V3.0 Example 3 /Page 31