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Dr. R. K. Ingle
• The solved examples included in this document are based on a draft code being developed under IITK-GSDMA Project on Building Codes. The draft code is available at http://www.nicee.org/IITK-GSDMA/IITKGSDMA.htm (document number IITK-GSDMA-EQ11-V3.0). • This document has been developed through the IITK-GSDMA Project on Building Codes. • The views and opinions expressed are those of the authors and not necessarily of the GSDMA, the World Bank, IIT Kanpur, or the Bureau of Indian Standards.
Examples on 13920
Sl. No 1. 2. 3. 4. 5. 6. 7. 8. 9. Type of Design Beam Design of an RC Frame Building in Seismic Zone V Beam Design of an RC Frame Building in Seismic Zone II Interior Column Design of an RC Frame Building in Seismic Zone V Exterior Column Design of an RC Frame Building in Seismic Zone V Interior Column-Beam Joint Design for Zone V Exterior Column -Beam Joint Design for Zone V Interior Column-Beam Roof Joint Design for Zone-V Exterior Column-Beam Roof Joint Design for Zone V Shear Wall Design for a Building in Seismic Zone III Page No. 4 15 24 33 42 48 56 62 69
IITK-GSDMA-EQ22-V3.0
However. The sizes of the beams and columns are given in Table 1.8.. It is assumed that there is no parking floor for this building.1: Plan of building (All dimensions in meters) Table 1.g.4 shows beam-loading diagram for dead load and live load. the effect of shear deformation is considered. Seismic analysis is performed using the codal seismic coefficient method.Beam Design of an RC Frame Building in Seismic Zone V 1 Problem Statement:
A ground plus four storey RC office building of plan dimensions 19 m x 10 m located in seismic zone V on medium soil is considered. 1. a dynamic analysis need not be carried out.1. FB1 RB2.1 of IS 1893 (Part 1): 2002. FB2 PB1 PB2 Slab thickness: 125 Note: All dimensions in mm.
1. Since the structure is a regular building with a height less than 16. The effect of finite size of joint width (e.
C C2 1 4
C1 2 4
C1 3 3
C1 5 4
Figure 1.2 and 1. Detailed design of the beams along the grid line ‘2’ as per recommendations of IS 13920:1993 has been carried out. Beam 300 x 600 300 x 500 300 x 400 300 x 350
IITK-GSDMA-EQ22-V3. on an intermediate frame in the transverse direction.1 :Schedule of member sizes Column C1 C2 C3 300 x 500 400 x 400 400 x 500 RB1.0
Example 1 /Page 4
.3.1. Figure 1. rigid offsets at member ends) is not considered in the analysis.1 Preliminary Data
Plan of the building and sectional elevations of different RC frames are shown in Figures 1.50 m. respectively.Examples on 13920
Example 1 . as per Clause 7.
Figure 1. B & C
Figure 1. Dead Load
b. EQX implies earthquake loading in X direction and EQY stands for earthquake loading in Y direction.5 kN/m2 (Nil for earthquake) Live load on floors = 3 kN/m2 (25% for earthquake) Roof finish = 1 kN/m2 Floor finish = 1 kN/m2 Brick wall on peripheral beams = 230 mm thick
Brick wall on internal beams = 150 mm thick Density of concrete = 25 kN/m3 Density of brick wall including plaster = 20 kN/m3
1.2.4: Loading diagram for an intermediate frame 2-5
1. The emphasis here is on showing typical calculations for ductile design and detailing of Example 1 /Page 5
IITK-GSDMA-EQ22-V3.Examples on 13920
Load combinations are considered as per IS 456: 2000 and are given in Table 1.2
Other relevant data are as follows: Grade of concrete: M20 Grade of steel = Fe 415 Live load on roof = 1.5
1.2: Elevation of frame A.3: Elevation of transverse frame 1&6
9DL-1.5 0.1 of IS 1893 (Part 1): 2002.9 0.5 -1. Beams parallel to Y direction are designed for earthquake loading in Y direction only.5 1.5 1.2 1. In practice.5 -1.5LL 1.5* 0.9DL+1.5* 0. even though it is proposed to drop this clause in the new edition of the Code.5 +1.2 1.4
Design of Middle Floor Beam
Beam marked ABC in Figure 1.2 +1.0
Example 1 /Page 6
.5 -1.2(DL+LL*+EQY) 1.5 for frame 2 is considered for design.5 1.2(DL+LL*+EQX) 1.2: Load combinations for earthquake loading S.2 reduce to 7 as shown in Table 1.25/0.2(DL+LL*-EQY) 1.5 EQY DL 1. Torsion effect is not considered in this example.5(DL-EQX) 1.No.9 0.5(DL+EQY) 1. and vice versa.4 which show force resultants for different load combinations.2(DL+LL*-EQX) 1.4
1.5: Beam ABC
1.5 +1. As the beam under consideration is parallel to Y direction.
IITK-GSDMA-EQ22-V3.9 0.
For the beam AB. force resultants for various load cases and load combinations have been obtained from computer analysis and are summarised in Table 1.5 EQY 0.5 two symmetrical spans. 1 2 3 4 5 6 7 8 9 10 11 12 13 Load Combination 1. wind load should also be considered in lieu of earthquake load and the critical of the two load cases should be used for design. (live load of 3 kN/m2) 25% of live load is considered for seismic weight calculations. Table 1.25/0.2 1.Examples on 13920 building elements subjected to earthquakes.5 +1. earthquake loads in Y direction are predominant and hence the 13 load combinations of Table 1.3 : Force resultants in beam AB for various load cases Load Case Left end Shear (kN) DL LL
Centre Shear (kN)
Right end Shear (kN)
-37 -12
-56 -16 -191
79 209 79 11 79 EQY Note: The results are rounded of to the next higher integer value.5* 0.
Figure 1.2 +1.5(DL+EQX) 1.2 1.9 LL 1.5(DL-EQY) 0.5DL+1.3. with the maximum values to be used for design being underlined.5 -1.5 0. Beams parallel to the Y direction are not significantly affected by earthquake force in the X direction (except in case of highly unsymmetrical buildings).2 -1.5 EQX 0.2 -1.9DL+1.3 and Table 1. For the present case.9DL-1.5
*Note: Reduced Live loads are considered as per Clause 7.25/0.5 EQX 0.5* EQ +1. Since the beam consists of Table 1. calculations need to be performed for one span only.5 1.25/0.
L 5.24 × 20 415
Factored axial force = 0.1.5DL+1. = 0.5(DL+EQY) 1.1
= 532 mm.1(b) of IS 13920: 1993) Maximum reinforcement = 2. B = 300 mm > 200 mm.33 > 4.2. centre and right end of the beam AB as per IS 13920:1993. in brief. D = 600 mm
B 300 = = 0.2 of IS 13920: 1993)
Width of beam.3.6.5 > 0. No.2(DL+LL*-EQY) 1.5LL 1.6.5 EQY
-98 29 -160 42 -195 73 -164
8 100 -90 125 -113 122 -115
113 170 -19 207 -30 172 -65
-74 203 -299 258 -369 280 -347
65 55 29 65 32 45 12
-108 -301 157 -371 203 -337 236
* Appropriate fraction of live load has been used
1. Detailed calculations at left end are given in the following sections. Design aid SP: 16 has been used for this purpose.0 MPa < 0.2(DL+LL*+EQY) 1. ok Check for Member Size (Clause 6.3.6
1.1 Design for Hogging Moment
Mu = 369 kN-m
IITK-GSDMA-EQ22-V3. Minimum reinforcement = 0.26%.26 x 300 x 532/100 = 415 mm2 (Clause 6. hence ok D 600
(Clause 6.2 Hence.5 shows.5(DL-EQY) 0. the reinforcement calculations at left end.Examples on 13920 Table 1.1. a spread sheet can be used conveniently.24
f ck fy = 0. (Clause 6.9DL+1.000 = = 8.00 kN Factored axial stress = 0.5 EQY 0. hence ok D 600
1.1.1.4 Force resultants in beam AB for different load combinations S.10 fck Hence. IS 13920:1993)
= 0.5% = 2.2.9DL-1.2. IS 13920:1993) 1.3 Check for Limiting Longitudinal Reinforcement
Effective depth for moderate exposure conditions with 20 mm diameter bars in two layers on an average = 600 – 30 – 8 – 20 – (20/2)
(Clause 6. In actual practice. design as flexural member.1.7. L = 5.5 x 300 x 532 /100 = 3. 1.0
Example 1 /Page 7
. 1 2 3 4 5 6 7 Load Combination Left end Shear (kN) 1. Depth of beam.990 mm2 (Clause 6.4 of IS: 13920-1993) 1. IS 13920:1993) Span.6.
a total of 0. revise to 1. Df = depth of flange = 125 mm xu = depth of neutral axis xu.330 mm2 and 756 m2. At the right end. Bottom reinforcement required is larger of 1.000 6
Referring to Table-50 of SP: 16. at centre. we get Ast at bottom = 1..990 mm2) Asc at bottom = 0. The limiting capacity of the T-beam assuming xu < Df and xu < xu.46/2 = 0.2.487 %) and 3-20Φ extra at bottom (i. provide 2.87 f y Ast d (1 − Ast f y b f d f ck )
(Annex G of IS 456: 2000) Where. But Asc must be at least 50% of Ast hence.633 mm (lowest of the above) (Clause 23. a total of 0.8
M u = 0.2 of IS 456: 2000) Substituting the values and solving the quadratic equation.e.512 mm2.512mm2 > 415 mm2 < 3.2 Design for Sagging Moment
= 1.87 f y Ast 0.Examples on 13920
Mu bd 2 369 × 10 6 = 4.max i.48 x 532 = 255 mm bw = width of rib = 300 mm bf = width of flange =
Lo + bw + 6d f or c/c of beams 6
Top reinforcement required is larger of 2.330 mm2 > Minimum reinforcement (415 mm ) < Maximum reinforcement (3. and at right end of the beam AB.max may be calculated as follows.165 mm2 and 1. provide 1.73 x 300 x 532 /100 = 1.54 % But Asc must be at least 50% of Ast. < 255 mm ok Asc at top = not required. 5-20Φ + 1-16Φ extra at top (i. Asc at bottom = 0.87 × 415 × 1512 = 47.83%) bars are provided.36 f ck b f
0.e.46 x 300 x 532 /100 = 2. Hence. Hence.7. over the central support.512 mm2.e. hence.e.512 /2 = 756 mm2 (Clause 6.35 300 × 532 × 532
0.e.e.3 of IS 13920: 1993) 1.2.000 mm = 1. A total of 3-16Φ straight bars each are provided throughout the length of the beam at both top and bottom. 5-20Φ+1-16Φ extra at top (i.330 mm2.990 mm2 It is necessary to check the design assumptions before finalizing the reinforcement.1.487%) and 1-20Φ + 2-16Φ extra at bottom (i.
1.max = limiting value of neutral axis = 0.633 mm or 4.
0. a total of 1.44 mm 0..36 × 20 × 1633
< xu. we get Ast at top = 1.165 mm2 1.13.. i.7 × 5000 + 300 + 6 × 125 Or 4.
IITK-GSDMA-EQ22-V3. For d’/d = 68 / 532 = 0.73 % (Clause 6.0
Example 1 /Page 8
.7.46 % = 1.3 of IS: 13920-1993) Hence.48 x d = 0. a total of 1.6 shows summary of reinforcement provided at left end. revise to 1.3 Required Reinforcement
Mu = 280 kN-m The beam is designed as T beam.97%) are provided at the left end.
8 Φ = 49 Φ = 980 mm for 20 Φ bar = 784 mm for 16 Φ bar
Ast f y b f d f ck
= 286 kN-m
IITK-GSDMA-EQ22-V3. referring to Table 50 of SP: 16.Examples on 13920 In an external joint.87 f y Ast d (1 −
1.e. pc = 0. from Table 65 of SP: 16.97% and compressive steel pc = 1. the tensile steel pt = 0. MuAh and MuBh) at both ends of beam are calculated on the basis of the actual area of steel provided in the section. 1.487 %) at top and 1. respectively.6.715 MPa
(IS 456:2000 Table 19)
Design shear strength of concrete = τcbd = 0.2.97%) at bottom on the left end of the beam.374 mm2 (i.3 of IS 13920:1993 and is given by Vswaytoright = Vswaytoleft =
± 1 .000 = 114 kN Similarly.715 x 300 x 532 /1.. sagging moment capacity at A for xu < Df and xu < xu.44
Hogging moment capacity at A.
τ c = 0.545 mm2 (i. MuAh = 4.97% whichever gives lowest value in the table). ld = 47 Φ + 10 Φ .9.487% is used. For calculation of MuAs. for Fe415 steel and M20 grade concrete. (for pt = 1.max may be calculated as given below. Referring to Annex G of IS: 456-2000. This method is iterative but gives more appropriate values of Mu.9. The contribution of the compressive steel is ignored while calculating the sagging moment capacity as T-beam. MuBs.4 ( M u ± 1 . For pt = 1.9
1.6: Anchorage of reinforcement bars
in an external joint In this case.44 x 300 x (532)2/(1 x 106) = 377 kN-m The limiting moment carrying capacity of a beam section can also be evaluated from the first principle. MuAs = M u = 0.487% or pc = 0. design shear strength of concrete at center and right end is evaluated as 69 kN and 114 kN.487% Permissible design shear stress of concrete.5 of IS 13920:1993) as shown in Figure 1.3.e.2 Shear Force Due to Plastic Hinge Formation at the ends of the Beam
The additional shear due to formation of plastic hinges at both ends of the beams is evaluated as per clause 6.97%. both the top and bottom bars of the beam shall be provided with an anchorage length beyond the inner face of the column equal to the development length in tension + 10 times bar diameter minus the allowance for 90 degree bend (Clause 6. The beam is provided with a steel area of 2.0
Example 1 /Page 9
. pt =1.487% and pc = 0.1
Tensile steel provided at left end = 1.4 ( M u
+ Mu )
20 # 980 784 16 #
The sagging and hogging moments of resistance (MuAs.
Summary of required reinforcement Top = 1.374 mm2 i.37% < 1.2.2. 0. 1.73 ok.79 % > 1.37% Top = 1.73% (Clause 6.37
1.37/2 = 0.83%
IITK-GSDMA-EQ22-V3.e.e.
Asc at top
0.47/4=0.26% < 1.4 of IS13920: 1993)
236 Ast required = 1264 mm2 = 0.6 Summary of reinforcement for beam AB Longitudinal Reinforcement Left end Center Right end Top 3-16Φ straight + 3-16Φ straight + 5-20Φ +13-16Φ straight reinforcement 5-20Φ +1-16Φ extra 16Φ extra Steel Provided = 603 mm2 Steel Provided = 2.0
Example 1 /Page 10
65 Ast required = 335 mm2 = 0. 0.46% Bottom = 0.3.47/2 > 0.37% > 1.37% ok Hence.735 % ok. 0.47/4=0.47% Bottom = 0. revise to 0.e.47/2 Hence revise to 0.e.e.26% > 0.37 %.378% i. Hence revise to 0.7335% (Clause 6.46%
Top reinforcement Center
Right end -371 4.Examples on 13920 Table 1. Hence.e.37%.21% < 0.79/2 = 0.46/2 i.26% < 1. IS13920: 1993)
0.378% i.47/4=0.185 % < 0.2.545 mm Steel Provided = 603 mm Steel Provided =1.319 mm2 i.487% Bottom 3-16Φ straight + (2-16Φ+13-16Φ straight reinforcement 3-16Φ straight + 3-20Φ extra 20φ) extra 2 2 Steel Provided = 1.165 % 0.5: Flexural design for beam AB
Beam AB Left end Hogging moment (kN-m) -Mu/bd2 Ast at top Asc at bottom -369 4.374 mm2 Steel Provided = 2. 0.2.945% Top = 0.37%. IS13920: 1993)
Sagging moment (kN-m) Ast at bottom
280 Ast required = 1512 mm2 = 0.55% < 1.97% i.e.79%
Table 1.33/2 = 0.37% (Clause 6.46/2 Hence revise to 0.3.487% i.1(b) and 6.35
1.47 /4 = 0.395% 0.37% Bottom = 0. 1.54% < 1.945% > 1. revise to 0. 0.
3) Hence.2 LL + 21.b =
1.3 Design Shear
Referring to the dead and live load diagrams (Figure 1.a =
1. Vu.0
.7 shows the shear force diagram for the beam considering plastic hinge formation at ends.6 kN
+ S.6 kN
+ S.4( M u + M u ) + L 2
Ah Bs
= 1.2( DL + LL) 1.5 x (1 + 5) /2 + 10.6 kN
+ S.4 (286 + 377) / 5 = ±186 kN Vswaytoleft = ±
1.Examples on 13920
61.2( DL + LL ) 1. Vu. due to 1.2 x (103 +36) /2 .2 x (103 + 36) /2 + 175 = 259 kN Shear at right end for sway to right.3 of IS 13920:1993.9.
1.4( M u + M u Vu.F.4( M u + M u ) − L 2
= ±1.F.575 x 5 = 103 kN LL = 12 x (1 + 5) / 2 = 36 kN Figure 1.3.8 kN
21.7: Shear diagram
Similarly.2 DL + 61.175 = 92 kN Figure 1.4(377 + 246)/5 = ±175 kN
1. the design shear force to be resisted shall be the maximum of: i) Calculated factored shear forces as per analysis (Refer Table 1. for the right end of the beam we obtain.4 ( M u
Shear at left end for sway to left. due to 1.2( DL + LL) 1.2( DL + LL) 1. Vu. As per Clause 6.186 = 103 kN
IITK-GSDMA-EQ22-V3.2 x (103 +36) /2 + 186 = 270 kN Shear at right end for sway to left.4( M u + M u + 2 L
= ±1.8 kN
61. Shear is calculated as below: Vswaytoright =
± 1 . due to 1.9. due to 1.4) ii) Shear forces due to formation of plastic hinges at both ends of the beam plus factored gravity load on the span (as calculated in Section 1.6 kN
+ Shear due to sway to right = 270 kN 186 kN
+ Shear due to sway to left = 175 kN
259 kN 92 kN
Figure 1.4( M u
= 1. LL and due to hinge formation at the ends of beam.7 shows the shear force diagram due to DL.2 LL + 21. Shear at left end for sway to right. DL = Trapezoidal dead load + Wall and self load = 16.2 x (103 + 36) /2 .4).F.F. design shear force (Vu) will be 259 kN (maximum of 195 kN from analysis and 259 kN corresponding to hinge formation) for left end of
Example 1 /Page 11
= 1.8 kN
+ S.2 DL + 61.b =
1. MuBh = 377 kN-m and MuBs = 246 kN-m.a = − 2 L
As per Clause 6.064 mm.5 of IS 13920:1993.87 x 415 / (300 x 0. the spacing of links shall be limited to 150 mm centers as per clause 6.
Minimum shear reinforcement as Clause 26. Hence.6 of IS 13920:1993. Hence.Examples on 13920
the beam and 270 kN (maximum of 207 kN and 270 kN) for the right end. respectively.3. In case of splicing of reinforcement. the design shear at centre of the span is taken as 186 kN. The reinforcement Figure1.2.3. provide stirrups at 165 mm centers. However.8.5 of IS13920: 1993 shall be the least of: i) d/4 = 532 /4 = 133 mm ii) 8 times diameter of smallest bar = 8 x 16 = 128 mm However.2 legged links @ 165 mm c/c 3-16Ø straight 300 Cross Section C-C
Figure 1.4) = 300 mm. it need not be less than 100 mm.
Spacing of links over a length of 2d at either end of beam as per Clause 6.4 b) = 2 x 50 x 0.2 legged links @ 125 mm c/c upto 1065mm 3-16Ø straight + 3-20extra 300
3-16Ø straight + 5-20Ø + 1-16Øextra 8Ø . The required capacity of shear reinforcement at the left end of the beam is: Vus = Vu – Vc = 259-114 = 145 kN Similarly the.8: Reinforcement details for the beam ABC
IITK-GSDMA-EQ22-V3. From analysis. Elsewhere. required capacity of shear reinforcement at the right end and at mid-span is 156 and 117 kN. shear due to formulation of plastic hinges at both the ends of the beams has been calculated as 186 kN and 175 kN.87 fy /(0.8 φ stirrups @125mm c/c at left and at right end over a length of 2d = 2 x 532 = 1. Referring to Table 62 of SP:16. the shear at the mid-span of the beam is 125 kN. the spacing of stirrups in the mid-span shall not exceed d/2 = 532/2 = 266 mm.2 legged links @ 125 mm c/c upto 1065mm 3-16Ø straight + 1-20Ø+2-16Ø extra 300
3-16Ø straight 8Ø . provide 2 Legged .1.5. we get the required spacing of 2 legged 8φ stirrups as 145 mm.0
Example 1 /Page 12
. detailing is shown in
3-16Ø straight + 5-20Ø+1-16Øextra A 600
3-16Ø straight
3-16Ø straight+ 5-20Ø +1-16Ø extra
A 3-16Ø 3-16Ø straight + straight 3-20Øextra 1250 5000
B 3-16Ø straight + 1-20Ø+2-16Ø extra 1250
C 8Ø-2 legged links @ 165mm c/c 500 1250 5000
8Ø-2 legged links @ 125mm c/c 500 upto 1090mm
3-16Ø straight + 5-20Ø +1-16Øextra 8Ø . centre and right end. 165 mm and 135 mm respectively at left end.6 of IS 456:2000 is given by: Sv = Asv x 0.
2 legged links @ 230 mm c/c upto 1065mm 2-12Ø+1-16Ø straight + 1-16Ø + 3-20Ø extra 300 Cross Section A-A 300 Cross Section B-B 2-12Ø+1-16Ø str + 5-20Ø +2-16Øextra 8Ø .9: Reinforcement details for the beam ABC as per IS 456:2000 (with R = 5)
IITK-GSDMA-EQ22-V3.
Table 1.0.2 legged links @ 230 mm c/c upto 1065mm 2-16Ø str +1-16Ø 3-20Ø extra 300 Cross Section C-C 2-12Ø+1-16Ø straight 8Ø . The columns is considered.10 Impact of Ductile Detailing on Bill of Quantities
To compare the impact of ductile detailing (as per IS 13920:1993) on the bill of quantities.Examples on 13920
1. only the steel between c/c of i.36 28
Steel required kg
Longitudinal Transverse Longitudinal Transverse Longitudinal Transverse in 95 120 1. For the purpose of seismic forces are the same as computed earlier.e. the beam under consideration has been redesigned as follows:
b) Design and detailing as per IS 456:2000. Table 1. seismic forces increased by a factor of 5/3 to account for R = 3.89 14 Detailing as per IS 456:2000 (Seismic loads as per R = 3) 135 163 1.10.2 legged links @ 300 mm c/c 2-12Ø+1-16Ø straight
Figure 1. with response reduction factor R = 5.0
Example 1 /Page 13
.20Ø 2-12Ø+1-16Ø straight + 5-20Ø +2-16Øextra 8Ø .7 Comparison of bill of quantities for steel in the beam ABC Description Detailing as per IS 13920: 1993 Detailing as per IS 456: 2000 (Seismic loads as per R = 5) 93 107 0.9. The reinforcement details are shown in Figure 1. comparison.0. reinforcement details are shown in Figure 1.0 25
Total steel in kg Ratio
2-12Ø+1-16Ø straight + 5-20Ø + 2-16Øextra A 600 A 2-12Ø+1-16Ø straight + 1-16Ø+3-20Øextra 5000
2-12Ø+1-16Ø B straight
2-12Ø+1-16Ø straight+ 5-20Ø + 2-16Øextra
B 2-12Ø+1-16Ø straight 1250 500
C 8Ø-2 legged links @ 300mm c/c 1250 5000
8Ø-2 legged links @ 230mm c/c 500 upto 1065mm
2-12Ø+1-16Ø + 3.7 compares the quantity of reinforcement a) Design and detailing as per IS 456:2000. for the three cases.
In case.3 and 1. care shall be taken to model the same in the static analysis. respectively.2 legged links @ 120 mm c/c 2-12Ø+1-16Ø str + 4-25Ø+1-16Ø extra 300
2-12Ø+1-16Ø straight + 6-25Ø+1-16Ø extra 8Ø . in the detailed calculations shown in this example. The results without and with finite size corrections can be compared from Tables 1.. The results with finite size joint widths in the analysis are presented in Table 1.3a.
Table 1.e.3a. However. the designer wishes to take advantage of the finite size joint correction. the effect of finite size joint corrections (i.0
Example 1 /Page 14
.10: Reinforcement details for the beam ABC as per IS 456:2000 (with R = 3)
Effect of Finite Size Correction
As mentioned in the problem statement.Examples on 13920
2-12Ø+1-16Ø straight + 6-25Ø+1-20Ø extra 2-12Ø+1-16Ø A straight 600
2-12Ø+1-16Ø straight+ 6-25Ø+1-16Ø extra
2-12Ø+1-16 Ø A B straight 2-12Ø+1-16Ø str + 4-25Ø+1-16Ø extra 1250 1250 5000 2-12Ø+1-16Ø straight+ 4-25Ø+1-12Ø+1-16Ø extra
C 8Ø-2 legged links @ 125mm c/c 500 1250 5000
8Ø-2 legged links @ 120mm c/c
2-12Ø+1-16Ø straight + 6-25Ø+1-20Ø extra 8Ø .
IITK-GSDMA-EQ22-V3.2 legged links @ 120 mm c/c 2-12Ø+1-16Ø straight + 4-25Ø+1-12Ø+1-16Ø extra 300
2-12Ø+1-16Ø straight 8Ø . this correction has been ignored.3a Force resultants in the beam AB for various load cases with Finite Size Correction
Left end Shear (kN) Moment (kN-m)
-29 -10 191
Center Shear (kN)
55 16 83
-45 -13 177
DL LL EQY
-48 -14 83
Note: The results are rounded of to the next integer value. rigid offsets at member ends) has been ignored in the analysis.2 legged links @ 145 mm c/c 2-12Ø+1-16Ø straight 300 Cross Section C-C
Example 2 / Page 15
.2 Force resultants in beam AB for different load cases
Left end V kN
Centre M kN-m
Right end V M kN kN-m
53 15 22 -50 -16 -50
M V kN kN-m
-39 -14 59 2 0 22
-48 -15 22
Note: V = Shear.Examples on 13920
Example2 .
Solution Design of Middle Floor Beam
The beam marked ABC in Figure 2. is explained.2. design as flexural member.2.1
Column C1 C2 C3 230 x 500 350 x 350 300 x 500
Beam RB1.2
2. force resultants for various load cases and load combinations have been obtained from computer analysis and are summarized in Table 2.00 kN Factored axial stress = 0.1 Schedule of member sizes
2. the earthquake loads are much lower for zone-II.1 Beam ABC Table 2.1-1.1
For beam AB.10 fck Hence. as per recommendations of IS13920:1993. (Clause. calculations are performed for one span only. IS 13920:1993)
Slab Thickness: 125
IITK-GSDMA-EQ22-V3.1.
Figure 2. Since the beam consists of two symmetrical spans. M = Moment.Beam Design of an RC Frame Building in Seismic Zone II 2 Problem Statement:
The ground plus four storey RC office building of Example-1 (Refer Figures 1.1.1. The results are rounded of to the next higher integer value. 6. The design of a beam along grid line 2.1 for frame 2 (Figure 1.
2.4) is assumed to be located in seismic zone II on medium soil. Table 2.3 shows force resultants for different load combinations with the maximum values to be used for design being underlined. Hence.0 MPa < 0. reduced member sizes are considered as shown in Table 2.
Table 2. However. FB1 RB2. The dead load and live loads are the same as in Example-1.1 of Example 1) is considered for design. FB2 PB1 PB2 230 x 500 230 x 400 230 x 350 300 x 300
Factored axial force = 0.
the reinforcement calculations at left end.5LL 1.5(DL-EQY) 0.132 x 230 x 438/100 = 1.1.518 mm2)
Example 2 / Page 16
0.3 Check for Limiting Longitudinal Reinforcement
Table 2. L = 5. IS 13920:1993) Depth of beam.4 shows. a spread sheet can be used conveniently.3.2. ok.
2.4 of IS 13920:1993)
2.46 > 0. 1 2 3 4 5 6 7 8
Load Combination Shear (kN) 1. centre and right end as per IS 13920:1993.24 = 0.10 and 0.5 x 230 x 438/100 = 2.14 and interpolating between d’/d of 0.5% = 2. Detailed calculations at left end are given in the following sections.140 mm2 > Minimum reinforcement (262 mm2) < Maximum reinforcement (2.5 EQY
-95 -36 -89 -39 -105 -10 -76
Left end Moment (kN-m)
-80 20 -122 30 -147 53 -124
3 29 -24 36 -30 35 -31
102 95 42 113 47 81 15
59 43 33 50 38 32 20
-99 -125 -5 -150 0 -120 30
* Appropriate fraction of live load has been used 2.2 of IS 13920:1993)
Hence.15.132% = 1.2.3.2 Check for Member Size
= 262 mm
Width of beam. D = 500 mm
B 230 = = 0.2. In actual practice. B = 230 mm > 200 mm Hence.1.Examples on 13920 Table 2.5DL+1.33 230 × 438 × 438
Referring to Table-50 of SP: 16 For d’/d = 62/446 = 0.518 mm2 (Clause 6.9DL+1.2(DL+LL*-EQY) 1.2. in brief.0
f ck fy
147 × 10 6 = 3.2(DL+LL*+EQY) 1. (Clause 6.1.26%. = 0.24 x 20 415
.26 x 230 x 438/100
IITK-GSDMA-EQ22-V3. IS 13920:1993) Span.9DL-1.3 D 500
(Clause 6. ok.5(DL+EQY) 1. ok.000 = = 10 > 4 500 D
2. (Clause 6. we get Ast at top = 1.1(b) of IS13920: 1993) Maximum reinforcement = 2.3 Force resultants in beam AB for different load combinations
S.2.1 Design for Hogging Moment
Mu = 147 kN-m
Effective depth for moderate exposure condition with 16 mm diameter bar in two layers on an average = 500 – 30 – 16 – (16/2) – 8 = 438 mm. Design aid SP: 16 has been used for the purpose. Minimum reinforcement.000 mm
L 5. = 0.5 EQY 0. (Clause 6. No.3
4-16Φ +1-12Φ extra bars at the top and 2-12Φ extra bottom bars are provided. ok. 6
= 339 x 230 x 438 /100 = 0.87 f y Ast d (1 − Ast f y b f d f ck
Asc at top = not required.5 show a summary of reinforcement provided at the left end. Hence.36 × 20 × 1..48 x 438 = 210 mm bw = width of rib = 230 mm bf = width of flange = =
0.3. as per Clause 6.( i )
(Annex G of IS 456: 2000) Where.1.max may be calculated as given below.3 of IS 13920:1993) Hence.3 Required reinforcement
Top reinforcement required is the larger of 1.2.Examples on 13920
Asc at bottom = 0. revise to 1.
Mu = 53 kN-m The beam is designed as T beam.87 f y Ast 0.000 .132 mm2. The limiting capacity of the T-beam assuming xu < Df and xu < xu.3 of IS 13920: 1993)
Ast at bottom = 339 mm2
< df < xu. i. at center and at the right end of the beam AB.87 × 415 × 339 = 10.e.132 / 2 = 0. Hence. 3-12Φ straight bars are provided throughout the length of the beam at the top and 4-12Φ straight bars are provided throughout at the bottom.48 x d = 0. i.000 mm = 1. 210 mm ok.4
Table 2.36 f ck b f
0.518 mm2
IITK-GSDMA-EQ22-V3. both the top and bottom bars of the beam shall be provided with an anchorage length beyond the inner face of the column equal to the development length in tension + 10 times
whichever is less = 1.2.132 mm2 and 170 mm2. 4-16Φ +1-12Φ extra bars at the top and 1-12Φ extra bar at the bottom at the left end are also provided.563 ok. Df = depth of flange = 125 mm xu = depth of neutral axis xu.5 of IS 13920:1993.3. over the central support.19 % But Asc must be at least 50% of Ast.
M u = 0. provide 571 mm2.
But Asc must be at least 50 % of Ast.26 % < 4% Hence. At an external joint.
2.max = limiting value of neutral axis = 0.2 of IS 456: 2000) Substituting the relevant values in (i) and solving the resulting quadratic equation. At the right end.563 mm (lower of the above) (Clause 23.0
Example 2 / Page 17
. revise to 339 /2 = 170 mm2 (Clause 6.563 mm or 4.566 % (Clause 6. Bottom reinforcement required is the larger of 339 mm2 and 571 mm2. hence. Hence.88 mm 0.7 × 5. Asc at bottom = 0.566 x 230 x 438 /100 = 571 mm2
2. provide 1.e.
0.2.2 Design for Sagging Moment
It is necessary to check the design assumptions before finalizing the reinforcement.000 + 230 + 6 × 125 or 4.33 % > 0. we get Ast at bottom = 339 mm2 > 262 mm2 < 2.
e. 0.566% (Clause 6.291% > 0.16 % < 0.566% Hence revise to 0.2. 0.246% 4-12Φ str + 2-10Φ extra Steel Provided = 609 mm2 i. 1.582% Hence revise to 0.6%
Longitudinal reinforcement Center 3-12Φ straight Steel Provided = 339 mm2 i.3. (Clause 6.163 % Bottom = 0.582%
Table 2. revise to 0.5 Summary of reinforcement provided for the beam AB
Beam AB Top reinforcement Left end 3-12Φ straight + 4-16Φ extra Steel Provided = 1.33%
4-12Φ straight Steel Provided = 452 mm2i.e. 0.3.37 % Bottom = 0.44%
Right end 3-12Φ straight + 4-16Φ +1-12Φ) extra Steel Provided = 1.291%.566%
Summary of required reinforcement Top = 0.3.26% <1.165% < 0.
IITK-GSDMA-EQ22-V3.132/2 = 0. 0.26% ok.132%
Top = 1.33%
=0.582% Hence revise to 0.19% < 1.26 > 1. (Refer Figure 2.Examples on 13920
the bar diameter minus the allowance for 90 degree bend.582 /2 = 0.e.33
1. IS13920: 1993)
0.163/4 = 0.291% ok
Ast required = 192 mm2
Ast required = 339 mm
= 0.224% < 1. IS13920: 1993)
0.e.291 % Hence.582% (Clause 6.291% Hence.185% < 0.256 mm2 i.291%.582%.26% < 1.134% 4-12Φ straight + 2-10Φ extra Steel Provided = 609 mm2 i.4 Flexural design for beam AB Beam AB Left end Hogging moment (kN-m) -Mu/bd2 Ast required at top
Asc required at bottom
Right end -150 3.163%
-147 3.0
Example 2 / Page 18
.e.37% > 0. IS 13920:1993)
Ast required = 371 mm2 = 0.163/2 = 0. revise to 0.26% < 1.2.2.2) Table 2.e.
0.33/2 = 0.37 / 2 = 0.163/2 = 0.143 mm2 i.4
1.291% Top = 1.163/4=0.132% Bottom = 0.163/4 = 0.
36 bd 2
Hogging moment capacity at A.4 ( M u
= ±1.. respectively.
Shear is calculated as below:
Vswaytoright =
± 1.4( M u
= ±1. for Fe 415 steel and M20 grade concrete.1
Tensile steel provided at left end = 1.e.e.4(94 + 165) /5 = ±72kN
Vswaytoleft =
MuAh = 3.134% Permissible design shear stress of concrete.
τ c = 0. referring to Table 50 SP: 16(for pt = 1.
Figure 2. The contribution of the compression steel is ignored while calculating the sagging moment capacity as T-beam. For pt = 1.11% and pc = 0. MuAh and MuBh) at both ends of the beam are to be calculated on the basis of the actual area of steel provided in the section. pc = 0.2 Anchorage of beam bars in an external joint
In this case.000 = 66 kN Similarly.66 x 230 x 438 /1.134%) at top and 609 mm2 (i. the tensile steel pt = 0.60%) at bottom on the left end of the beam.2 Shear Force Due to Plastic Hinge Formation at the ends of the Beam
= 94 kN-m Similarly.max may be calculated as given below.60% and compressive steel pc = 1.8 Φ = 49 Φ
Mu = 3.87 f y Ast d (1 − Ast f y b f d f ck
2. Referring to Annex G of IS: 456-2000.60% whichever gives lowest value in the table).134% is used.5
2.. for the right side joint we obtain.36 x 230 x 438 x 438 / 106
= 784 mm for 16 Φ bar = 588 mm for 12 Φ bar
= 149 kN-m For calculation of MuAs.4(149 + 94) /5 = ±68 kN
IITK-GSDMA-EQ22-V3.3 of IS 13920:1993 and is given by
± 1 .Examples on 13920
16 # 784 588 12 #
± 1 . The beam is provided with a steel area of 1.134% or pc = 0.4( M u
The additional shear due to formation of plastic hinges at both ends of the beams is evaluated as per clause 6.0
Example 2 / Page 19
MuAs = M u = 0. from Table 65 of SP: 16.4 ( M u
The sagging and hogging moments of resistance (MuAs . pt = 1.
MuBh = 165 kN-m and MuBs = 94 kN-m.5.143 mm2 (i.MuBs. the design shear strength of concrete at mid-span and at the right end is evaluated as 46 kN and 66 kN.66 MPa
(Table 19 of IS 456:2000)
Design shear strength of concrete = τcbd = 0. sagging moment capacity at A for xu < Df and xu < xu.60%.5.
ld = 47 Φ + 10 Φ .
5.4( M u + M u + 2 L
= 2 x 50 x 0. Vu.2( DL + LL ) 1. Minimum shear reinforcement Clause 26. we get the required spacing of 2 legged 8φ stirrups as 230 mm centers at left and at the right end.575 x 5 = 103 kN LL = 12 x (1 + 5) / 2 = 36 kN Shear at left end for sway to right.a = − 2 L
= 152-66 = 86 kN Similarly.4( M u + M u Vu.
1.6.87 x 415 / (300 x 0.2 DL + 21.
Vu.0
Example 2 / Page 20
. The required capacity of shear reinforcement at the left end.3) ii) Shear force due to formation of plastic hinges at both ends of the beam plus due to factored gravity load on the span (as calculated in 2.4( M u + M u ) Vu.Examples on 13920
= 1.4( M u + M u ) + 2 L
= 1.b = − 2 L
= 1.4 kN 68 kN 21.4 kN The design shear force shall be the maximum of: i) Calculated factored shear force as per analysis (Refer Table 2.
IITK-GSDMA-EQ22-V3. respectively.2 x (103 + 36)/2 + 72
61.1. the spacing of stirrups in the rest of member shall be limited to d/2 = 438/2 = 219 mm. due to 1.2 x (103 + 36) /2 .
< 438 x 0.2( DL + LL) 1. the design shear force (Vu) will be 152 kN (maximum of 105 kN from analysis and 152 kN corresponding to hinge formation) for the left end of beam and 155 kN (maximum of 113 kN from analysis and 155 kN corresponding to hinge formation) for the right end.72 = -11.2 x (103 + 36) /2 . shear due to formation of plastic hinges at both the ends of the beams will be 72 kN.87 fy /(0.2( DL + LL ) 1.4 b)
= 1.4 kN Shear at left end for sway to left.2 LL + 68 kN + Shear due to sway to left = 152 kN + 72 kN 15.75 = 328 mm Hence.F. the required capacity of shear reinforcement at the right end and at mid-span can be calculated as 26 kN and 89 kN. However.8 kN
= 155 kN Shear at right end for sway to left.2 x (103 + 36) /2 + 68 = 152 kN Shear at right end for sway to right.3) Hence.3.6 of IS 456:2000
Sv = Asv x 0.8 kN + S.b =
1.6 kN + S.
Vus = Vu – Vc
Figure 2.3 Shear diagram due to sway to left
2. Referring to Table 62 of SP: 16. due to 1.F.a =
1. ok.5 of IS 13920:1993.2( DL + LL ) 1.5 x (1 + 5)/2 + 10.6 kN 61.68 = 15. Shear at the mid-span from analysis is 36 kN. As per Clause 6.4) = 300 mm.6
Referring to the dead and live load diagrams (Figure 1.
1. DL = Trapezoidal DL+ Brick wall & Self load = 16.4 of Example 1).
6 of IS 13920:1993.4. In case of splicing of main reinforcement.2.5.2 legged links @ 100mm c/c upto 900mm 230 4-12Ø straight + 2-10 extra 500
3-12Ø straight 8Ø . provide stirrups at 215(< 219 mm) centers. provide 2 legged 8 φ stirrups @100 mm c/c at left and at the right end of the member over
3-12Østraight + 4-16Øextra A 500 A B 4-12Østraight 4-12Ø straight + 4-12Ø straight + 2-10Ø extra 2-10Øextra 1250 1250 5000 3-12Østraight
a length of 2d = 2 x 438 = 876 mm at either end of the beam. the beam has been redesigned as follows: a) Design and detailing as per IS 456:2000. detailing is shown in
3-12Ø straight+ 4-16Ø + 1-12Ø extra
C 8Ø-2 legged links @ 215mm c/c 500 1250 5000
8Ø-2 legged links @ 100mm c/c 500 upto 900mm
3-12Ø straight + 4-16Øextra 500 8Ø .6 compares the quantity of reinforcement for the three design cases. Hence. Spacing of links over a length of 2d at either end of the beam as per Clause 6.0. seismic forces are increased by a factor of 5/3 to account for R = 3.5 of IS 13920:1993 shall be least of i) d/4 = 438/4= 109 mm ii) 8 times diameter of smallest bar = 8 x12 = 96 mm However. While calculating the quantities c/c span is considered. Table 2. Elsewhere.. i.0.e.Examples on 13920
The spacing of minimum stirrups is kept at 300 mm. The reinforcement details are shown in Figure 2.7
Impact of Ductile Detailing on Bill of Quantities
factor R = 5. seismic forces are the same as computed earlier. it should not less than 100 mm. The reinforcement Figure 2.3.
To compare the impact of ductile detailing (as per IS 13920:1993) on the bill of quantities. with response reduction
IITK-GSDMA-EQ22-V3.2 legged links @ 100mm c/c upto 900mm 230 4-12Ø straight + 2-10 extra 500
3-12Ø straight + 4-16Ø + 1-12Øextra 8Ø .2 legged links @ 215mm c/c 4-12Ø straight 230 Cross Section C-C
Figure 2.6.4 Reinforcement details for the beam ABC
2. The reinforcement details are shown in Figure 2. the spacing of links shall be limited to 150 mm centers as per Clause 6. b) Design and detailing as per IS 456:2000.0
Example 2 / Page 21
6 Reinforcement detail for the beam ABC as per IS 456:200 (with R = 3.2 legged links @ 300 mm c/c 4-12Ø straight 230 Cross Section C-C
Figure 2.5 Reinforcement detail for the beam ABC as per IS 456:2000 (with R = 5.Examples on 13920
2-12Østraight + (4-16Ø+1-12Ø) extra 2-12Østraight A 500 A 2-12Østraight 2-12Ø straight + + 2-12Ø extra 1-12Ø extra 1250 5000
2-12Ø straight+ (4-16Ø+2-12Ø) extra
B 2-12Ø straight 1250 500 1250
C 8Ø-2 legged links @ 300mm c/c 5000
8Ø-2 legged links @ 230mm c/c 500 upto 900mm
2-12Ø straight + (4-16Ø +1-12Ø)extra 8Ø .2 legged links @ 230 mm c/c upto 900mm 2-12Ø straight + 2-12Ø extra
2-12Ø straight 8Ø .0
Example 2 / Page 22
2-12Østraight
2-12Ø straight+ 7-16Øextra
B A 2-12Ø straight + 2-12Østraight 2-12Østraight+ 1-12Øextra + 2-12Ø extra 3-16Ø extra 1250 5000 1250
C 8Ø-2 legged links @ 300mm c/c 500 1250 5000
8Ø-2 legged links @ 230mm c/c
2-12Ø straight + 7-16Øextra 500 8Ø .2 legged links @ 300 mm c/c 2-12Ø straight +2-12Ø extra
Figure 2.2 legged links @ 230 mm c/c 2-12Ø straight + 3-16Ø extra 230 Cross Section B-B 500
2-12Ø straight 8Ø .2 legged links @ 230 mm c/c 2-12Ø straight + 3-16Ø extra 230 230 500
2-12Ø straight + 7-16Øextra 8Ø .2 legged links @ 230 mm c/c upto 900mm 2-12Ø straight + 1-12Ø extra
2-12Ø straight + (4-16Ø +2-12Ø)extra 8Ø .0)
Table 2.6 Comparison of bill of quantities of steel in the beam ABC
Detailing as per IS 13920: 1993
Detailing as per IS 456:2000 (Seismic loads with R = 5) 46 59 0.79 13
Detailing as per IS 456:200 (Seismic loads with R = 3) 64 77 1.03 13
Longitudinal Transverse Longitudinal Transverse Longitudinal Transverse Steel required (kg) Total steel (kg) Ratio 52 75 1.0 23
Example 2 / Page 23
Example 3 - Interior Column Design of an RC Frame Building in Seismic Zone V 3 Problem Statement:
For the ground plus four storey RC office building of Example-1 (Refer Figures 1.1-1.4 of Example 1), design of an interior column element is explained here. The column marked AB in Figure 3.1 for frame 2 is considered for design.
400 600 300 500
Y Figure 3.1 Column location in elevation Solution:
Load combinations derived from recommendations of Clause 6.3.1.2 of IS 1893(Part 1): 2002 and given in Table 1.4 of Example-1 are considered for analysis.
For column AB, the force resultants for various load cases and load combinations are shown in Tables 3.1 and 3.2. In case of 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 effect. 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 for these moments.
Since the column must be designed for 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 check the column for each combination of loads. Checking the column for all load combinations at all the sections is indeed tedious if carried out by hand. Hence, a computer program is best suited for column design. In the absence of a computer program, one can make a judgment of which two or three load cases out of the thirteen may require the maximum reinforcement and design accordingly. Referring to Table 3.2, it can be observed that out of the various load combination, one design load combination with earthquake in either (X or Y) direction can be identified, which is likely to be critical. These critical design forces are summarised in Table 3.3. Table 3.4 and Table 3.5
Example 3 /Page 24
Pu , f ck bD
M2 f ck b D
give factors such as
M3 f ck bD 2
calculated and summarised in Table 3.6. The detailed calculations are shown in Section 3.4.
Using these factors and the charts given
in SP: 16, the required maximum reinforcement is Table 3.1 Force resultants in column AB for different load cases
Load case Axial (kN) DL LL EQx EQy -961 -241 -22 0 AB M2 (kN-m) 1 0 169 0 M3 (kN-m) 0 0 0 -198 Axial (kN) -764 -185 -11 0 AT M2 (kN-m) -1 0 -169 0 M3 (kN-m) 0 0 0 191 Axial (kN) -749 -185 -11 0 BB M2 (kN-m) 1 0 173 0 M3 (kN-m) 0 0 0 -194 Axial (kN) -556 -131 -4 0 BT M2 (kN-m) -1 1 -148 0 M3 (kN-m) 0 0 0 166
Table 3.2 Force resultants in column AB for different load combinations
Load Combinations 1.5(DL+LL) 1.2(DL+LL+EQX) 1.2(DL+LL-EQX) 1.2(DL+LL+EQY) 1.2(DL+LL-EQY) 1.5(DL+EQX) 1.5(DL-EQX) 1.5(DL+EQY) 1.5(DL-EQY) 0.9DL + 1.5 EQX 0.9DL - 1.5 EQX 0.9DL + 1.5 EQY 0.9DL - 1.5 EQY Axial (kN) -1803 -1252 -1199 -1226 -1226 -1475 -1409 -1442 -1442 -898 -832 -865 -865 AB M2 (kNm) 2 204 -202 1 1 255 -252 2 2 254 -253 1 1 M3 (kN-m) 0 0 0 -238 238 0 0 -297 297 0 0 -297 297 Axial (kN) -1424 -986 -959 -972 -972 -1163 -1130 -1146 -1146 -704 -671 -688 -688 AT M2 (kNm) -2 -204 202 -1 -1 -255 252 -2 -2 -254 253 -1 -1 M3 (kN-m) 0 0 0 229 -229 0 0 287 -287 0 0 287 -287 Axial (kN) -1401 -968 -941 -954 -954 -1140 -1107 -1124 -1124 -691 -658 -674 -674 BB M2 (kNm) 2 209 -206 1 1 261 -258 2 2 260 -259 1 1 M3 (kN-m) 0 0 0 -233 233 0 0 -291 291 0 0 -291 291 Axial (kN) -1031 -711 -702 -707 -707 -840 -828 -834 -834 -506 -494 -500 -500 BT M2 (kNm) 0 -179 177 -1 -1 -224 221 -2 -2 -223 221 -1 -1 M3 (kNm) 0 0 0 199 -199 0 0 249 -249 0 0 249 -249
Factored axial stress = 6,58,000 / (400 x 500) = 3.29 MPa > 0.10 fck Hence, design as a column member. (Clause 7.1.1; IS 13920:1993)
3.3.1 Check for Axial Stress
Lowest factored axial force = 658 kN (Lowest at At or Bb among all load combination is considered)
Example 3 /Page 25
the effective length is taken as 0.4 Design of Column
3.31 < 12.8 x 400 x 500/100 = 1.15 20 × 400 × 500 × 500
(Clause 26. B = 400 mm > 300 mm Hence. IS 13920:1993) Depth of column.600 mm
M u2 f ck bD 2
297 × 10 6 = 0.1.000 mm The effective length of column can be calculated using Annex E of IS 456: 2000.000 mm2 (Clause 26.1 of IS 456: 2000) Maximum reinforcement = 4% (Limited from practical considerations) = 4 x 400 x 500/100 = 8.3.8 % = 2.Examples on 13920 3. we get p/fck = 0.85 times the unsupported length. take p/fck = 0.11
The required steel will be governed by the higher of the above two values and hence. IS 13920:1993) Span..
Design for Earthquake in X-direction
Pu = 1.4.442 × 10 3 = = 0. = 0.36 f ck bD 20 × 400 × 500
Minimum reinforcement.5.1.1. The column is oriented in such a way that depth of column is 400 mm for X direction earthquake and 500 mm for Y direction earthquake force. In this example as per Table 28 of IS 456: 2000. Short Column.14.
(a) Approximate Design
(Clause 7.1 of IS 456: 2000)
Referring to Charts 44 of SP16 For d’/D = (40 + 25 / 2) /500 = 0. 400 D
In this case.2 of proposed draft IS 13920)
3.85 = = 5. = 0.0
Example 3 /Page 26
.14 x 20) % = 2.3.600 mm2
IITK-GSDMA-EQ22-V3.2 Check for member size
3.16 20 × 500 × 400 × 400
Referring to Charts 44 and 45 of SP16 For d’/D = (40 + 25 / 2) / 400 = 0. Required steel = (0.e.37 f ck bD 20 × 400 × 500
i.442 kN Mu2 = 297 kN-m
Pu 1. L = 3.3. we get p/fck = 0.8 %.3.2 of IS 456: 2000) In case of slender column. (Clause 25. the moment about one axis dominates and hence the column is designed as an uniaxially loaded column. hence ok D 500
First approximate design is done and finally it is checked for all force combinations. ok (Clause 7.105.1.
L (3000 − 500) × 0. Minimum dimension of column = 400 mm ≥ 15 times the largest diameter of beam longitudinal reinforcement = 15 x 20 = 300 ok (Clause 7.1 Sample Calculation for Column Reinforcement at AB End
Width of column.3.8 x 400 x 500 /100 = 5. additional moment due to P-δ effect needs to be considered.2. D = 500 mm
B 400 = = 0.5. which is in between that of fixed and hinged case.14
Design for Earthquake in Y-direction
Pu = 1.3 Check for Limiting Longitudinal Reinforcement
255 × 10 6 = 0.8 > 0. Hence ok.475 kN Mu2 = 255 kN-m
Pu 1475 × 10 3 = = 0.13.4.
0) =1.367
= 0.527 = 0.85%.000-600)/500) + (500 / 30) = 21.1 ⎥ ⎦ ⎣
⎡ 31 ⎤ =⎢ ⎥ ⎣ 288 ⎦
1.475 x 0.46 mm > 20 mm
Mu2 = 1.816 <1.1 = 0.00 Hence.6 of IS 456:2000) = 0.6 of IS 456: 2000)
⎡ M u2 ⎤ ⎥ ⎢ ⎢ M u 2.367
⎡ 297 ⎤ +⎢ ⎥ ⎣ 360 ⎦
1.75fy-0.5 kN-m
Eccentricity = clear height of column /500 + lateral dimension / 30 = ((3.7687 = 0.18
M u 2. Moment about other axis may occur due to torsion of building or due to minimum eccentricity of the axial load.02 = 29.714 = 3.1 ⎥ ⎦ ⎣
⎡ 255 ⎤ ⎡ 29.4 of IS 456:2000) = ((3.88 +0.45 x 20 x 400 x 500 + (0.5 ⎤ =⎢ +⎢ ⎥ ⎥ ⎣ 280 ⎦ ⎣ 350 ⎦ = 0.175 × 20 × 400 × 400 × 500 /(1 × 10 6 )
= 280 kN-m
M u 3.92 < 1.45 x 20) x 5.1 = 0.02146 = 31 kN-m Pu = 0.442 x 0.442 kN Mu3 = 297 kN-m
Eccentricity = Clear height of column/500 + lateral dimension / 30 (Clause 25.42 − 0. design eccentricity = 20 mm
Mu3 = 1. 5.e. ok
Checking for Critical Combination with Earthquake in Y Direction (Transverse direction)
Width = 500 mm.
Checking for Critical Combination with Earthquake in X Direction (Longitudinal direction)
⎡M ⎤ + ⎢ u3 ⎥ ⎢ M u 3.75fy Asc (Clause 39.1 ⎥ ⎦ ⎣
⎡M ⎤ + ⎢ u3 ⎥ ⎢ M u 3. Hence. p/fck provided = 2. we get For
Mu f ck bD 2 = 0. and referring f ck bD to Charts 44 and 45 of SP: 16.0 − 1.367
Using the interaction formula of clause 39.367
1.85/20 = 0.75 x 415 – 0.143
(b) Checking of Section
αn = 1. Depth = 400 mm
Pu = 1.Examples on 13920
Provide 10-25Φ + 4-16Φ bars with total
Asc provided = 5.527 kN Pu/Puz = 1.714 x100 /(400 x 500) = 2.355 and p/fck= 0.1 = 0.475 /3.00
Example 3 /Page 27
IITK-GSDMA-EQ22-V3.18 × 20 × 400 × 400 × 500 / 1 × 10 6
= 288 kN-m
M u 3.1 = 0.0 +
0.18 × 20 × 400 × 500 × 500 / 1 × 10 6 = 360 kN-m
= 350 kN-m Puz = 0.45fck Ac + 0.45 fck ) Asc = 0. f ck bD
Pu = 0. we get
M u 2.0473 +0.175 × 20 × 400 × 500 × 500 /(1 × 10 6 )
M u 2.714 mm2
The constant αn which depends on factored axial compression resistance Puz is evaluated as
i.527 kN
αn = 1.2 (2.45fck Ag + (0. Depth = 500 mm
Pu = 1.143.000-500) / 500) + (400 / 30) = 18.1 ⎥ ⎦ ⎣
The column should be checked for bi-axial moment.33 mm < 20 mm Hence.367 0.475 kN Mu2 = 255 kN-m
Width = 400 mm.0
Puz = 3..175
Referring to Chart 44 of SP: 16..1 f ck bD 2 = 0.37 and p/fck= 0.35
Using the interaction formula
⎡ M u2 ⎤ ⎥ ⎢ ⎢ M u 2.143.8 − 0.
140 -1.0
Pu f ck bD
Table.29
Table -3.145
0.424 -1.4 Design of column AB for earthquake in X direction
M2 2 f ck b D
Gravity 0. The Tables 3.29
Figure 3.1455
0.8 -0.5 show brief calculations at AB.Examples on 13920
Hence.00 0. i.163 -1.8 0.124
1.4.475 EQX Critical comb with -1.803 Gravity Critical comb with -1.36 2.8 0.124
0 0 -291
-1.3.442 EQY
0 0 -297
-1.8% steel is needed.8 0. BB and BT locations. Similarly.45 Critical comb 0. higher of the reinforcement required at BB and BT.8 2.401 -1.8 2.4 and 3.4% is needed in the column at joint B. AT.146
-1. 2.8 0.14
0.2 0.36 with EQY
1.031 -840 -834
Table. BB and BT is calculated.e.16
0.5 Design of column AB for earthquake in Y direction AB AT BB p p Pu Pu Pu M3 M3 p M3
f ck bD
Critical comb 0.21
-0. The column at joint A should have higher of the
reinforcement required at AB and AT.3.0
Example 3 /Page 28
. the steel required at AT. ok
3.00 0.2 shows the reinforcement in the column along with the steel provided in the transverse and longitudinal beams.1.37 with EQX
0. and hence 2.6
Note: b = 400 mm and D = 500mm
IITK-GSDMA-EQ22-V3.5
Similar to the sample calculations shown in Section 3.4 0.3 Critical forces for design of column AB
AB M2 2 255 2
AT M2 -2 -255 -2
BB M2 2 261 2
BT M2 0 -224 -2
-1.26 2.00 0.
2.1.5(DL+EQY) 186 -1 192 -1 192 -1 171 -1 1.Top steel 3-20 # + 4-16 # .5(DL-EQY) -186 -1 -192 -1 -192 -1 -171 -1 0.9DL + 1.86% Reinforcement 8-25Φ+ 6-16Φ at B Steel provided= 5.5 EQX 0 166 0 170 0 170 0 152 0.5(DL+EQX) 0 -167 0 -171 0 -153 -171 0 1..Bottom steel)
Longitudinal beam 300 x 500 (4-20 # + 5-16 # ..3.7 Shear forces in column AB for different load combinations AB AT BB BT Load Combination EQX EQY EQX EQY EQX EQY EQX EQY (kN) (kN) (kN) (kN) (kN) (kN) (kN) (kN) 1.Bottom steel)
Figure 3.57%
Confining Links: 8 # links @ 85 c/c Nominal Links: 8 # links @ 200 C/C 10-25 # + 4-16 #
8-25 # + 6-16 #
Reinforcement at A
Reinforcement at B
10-25# + 4-16# at A 8-25# + 6-16# at B
Transverse beam 300 x 600 (5-20 # + 4-16 # .5 EQY -186 0 0 -192 0 -171 0 -192
IITK-GSDMA-EQ22-V3.e.5 EQX 0 -167 0 -171 0 -171 0 -153 0.Examples on 13920
Table 3.2(DL+LL-EQY) -149 0 -154 0 -154 0 -136 0 1.Top steel 5-16 # + 1-20 # .9DL .9DL .2(DL+LL+EQX) 0 -133 0 -137 0 -137 0 -122 1.2 Reinforcement details of longitudinal and transverse beam
Table .5(DL-EQX) 0 166 0 170 0 170 0 152 1.134 mm2 i.2(DL+LL-EQX) 0 132 0 136 0 136 0 121 1.5(DL+LL) 0 -1 0 0 0 0 0 0 1.e. 2.5 EQY 186 0 192 0 192 0 171 0 0.1.9DL + 1.0
Example 3 /Page 29
.6 Summary of reinforcement for column AB
Longitudinal Reinforcement Reinforcement 10-25Φ +4-16Φ at A Steel provided = 5.714 mm2 i.2(DL+LL+EQY) 149 0 154 0 154 0 136 0 1.
The hogging and sagging moment capacity is evaluated as 377 kN-m and 246 kN-m.86% / 2 = 1.4 × (377 + 246) 3
= 291 kN
3.7.2 Earthquake in Y-Direction
(Clause 40.3. 0.2. i.6.1
Referring to Figure 3.000 = 187 kN
Details of Transverse Reinforcement
Design of Links in X Direction
Vc = 1.7) and the shear force due to plastic hinge formation in either of the transverse or the longitudinal beam. the maximum factored shear force in X and Y direction is 192 and 171 kN.7.6. the shear force corresponding to plastic hinge formation in the transverse beam is
1.4 (M u
Effective depth in Y direction = 500-40-25/2 = 447.3..70 Mpa (Table 19 of IS 456: 2000) Considering lowest Pu = 658 kN..34%) at bottom.87 × 415 × 347.5 /1.3.
3.3 Column shear due to plastic hinge formation in beams IITK-GSDMA-EQ22-V3.e.043 x 500 x 347.Examples on 13920
3.043 MPa
Effective depth in X direction = 400-40-25/2 = 347.374 mm2.3.5 /1.6. 1.74%) at top and 3-20Φ extra + 4-16Φ str (1.043 x 400 x 447.261 mm2. Ast = 2. the design shear in Y direction is 291 kN which is the higher of 171 kN and 291 kN. 1..
3.70 x 1. Referring to Figure 3.e.2 of IS 456: 2000)
τ c = 0.4
+ M h st
Spacing of 4 Legged 8 Φ Links =
4 × 50 × 0.5 mm
Vc = 1.
= 1 . IS 13920: 1993) The design shear in X direction is 237 kN which is the higher of 192 kN and 237 kN.4 (M u
Assuming 50% steel provided as tensile steel to be on conservative side. Similarly.49 = 1..6. respectively.4 x (288 +221) /3 = 237 kN
3.485%) at top and 3-16Φ str + 3-20Φ extra (1545 mm2. the shear force corresponding to plastic hinge formation in the longitudinal beam is evaluated as:
1.978%) at bottom.000
The design shear force for the column shall be the higher of the calculated factored shear force as per analysis (Table 3. respectively.3. i. 1.1
Vs = 237 – 181 = 56 kN . (Clause7.49 < 1.e.5 Multiplying factor = δ = 1 + Ag f ck
= 1.2 Shear As Per Analysis
As per Table 3.e.000 = 181 kN
The transverse beam of size 300 x 600 is reinforced with 3-16Φ str + 5-20Φ + 1-16Φ extra (2.4 Design Shear
The longitudinal beam of size 300 x 500 is reinforced with 4-20Φ extra + 5-16Φ str (2. i. we get
3Pu =1. The hogging and sagging moment capacities are evaluated as 288 kN-m and 221 kN-m.0
Example 3 /Page 30
.1 Earthquake in X-Direction
1. respectively.6.3 Shear Force Due to Plastic Hinge Formation at Ends of Beam 3.5 = 448 mm 56. i.746 mm2.43% Permissible shear stress τ c = 0.
or ((400 – 40 – 40.4. however.00. Whichever is higher.4.
Table 3.e.8 for various detailing options.5 mm.2 x 40 +2 x 8)
Figure 3.4]
3. Link spacing for confining zone shall not exceed: (a) ¼ of minimum column dimension i.. Ash .5 × 20 ⎛ 2.46. [Refer Figure 3..3
Nominal Links
The spacing of hoops shall not exceed half the least lateral dimension of the column i.7.000 ⎞ ⎜ ⎜ 1.
. (Clause 7.00.
0. i. (Clause 7. IS 13920:1993) Provide 8 Φ links @ 200 c/c in mid-height portion of the column.496 − 1⎟ ⎟ 415 ⎝ ⎠
Substituting we get S = 84 mm. 400 / 4 =100 mm (b) But need not be less than 75 mm nor more than 100 mm. to be used as special confining reinforcement shall not be less than (Clause 7.2 Design of Links in Y Direction
Vs = 287 – 187 = 100 kN
Spacing of 3 legged 8 Φ Links
3 × 50 × 0.4 Reinforcement details for column
The comparisons of steel quantities are shown in Table 3. 400/ 2 = 200 mm.Examples on 13920 3.1 of IS 13920:1993) Provide confining reinforcement for a distance of lo = 500 mm on either side of the joint.4).3. which shall not be less than: i) Larger lateral dimension = 500 mm ii) 1/6 of clear span = (3000 – 500) / 6 = 417 mm iii) 450 mm (Clause 7.4 Confining Links
The area of cross section.18 × S × h × f ck fy ⎛ Ag ⎞ ⎜ ⎟ ⎜ A − 1⎟ ⎝ k ⎠
h = longer dimension of the rectangular link measured to its outer face
= 172 mm.7.5 = 243 mm = 1.18 × S × 188.
Ag = 400 x 500 = 2.496 mm2 Assuming 8Φ stirrup.87 × 415 × 447. of the bar forming rectangular hoop. Links (kg) Main (kg) steel
Detailing as per IS 13920: 1993 (Seismic loads as per R = 5) 25 128
Detailing Detailing as as per per IS 456: IS 456: 2000 2000 (Seismic (Seismic loads as loads as per per R = 5) R = 3) 14 128 Column needs to be redesigned.000 mm2 Ak = (400.e.8 Comparison of bill of quantities of steel in column
= 336 x 436 = 1.000
Provide 8Φ confining links @ 80 c/c for a distance lo (Refer Figure 3.6 of IS 13920:1993).46.25)/2 +(8 x 2) +25) =188.4.7. Ash = 50 mm 50 =
= ((500 – 40 – 40 – 25) /3 + (8 x 2)) + 25)
0.2 x 40 +2 x 8) x (500.5 mm.00. h = 188.8 of IS 13920:1993). that the column designed above has not been checked for requirements related to
Example 3 /Page 31
IITK-GSDMA-EQ22-V3.3.
which are being incorporated in the new edition of IS 13920.
IITK-GSDMA-EQ22-V3. The applications of these provisions are illustrated in Examples 5-8 and may require modifications in column size and /or longitudinal reinforcement.Examples on 13920
the joint region.0
Example 3 /Page 32
300 600 300 500
Y Figure 4. In the absence of a computer program.4 (Example-1) need to be considered. a computer program is best suited for column design. Checking the column for all load combinations at all the sections is indeed tedious if carried out by hand. design of an exterior column element is explained here. the force resultants for various load cases and load combinations are shown in Tables 4. it can be observed that out of the various load combination. one design load combination with earthquake in either (X or Y)
Example 4 /Page 33
Load combinations derived from recommendations of Clause 6.2.2 of IS 1893(Part 1): 2002 and given in Table 1.3.0
.4 of Example-1 are considered for analysis. Similarly earthquake in Y direction causes a large moment in column about X axis and a small moment about Y axis.
4.1 for frame 2 is considered for design. Hence.1 Load Combinations
needs to be designed as a biaxial member for these moments.4 of Example 1). all 13 load combinations as shown in Table 1.2
For column AB.Examples on 13920
Example 4 . Column
IITK-GSDMA-EQ22-V3. one can make a judgment of which two or three load cases out of the thirteen may require the maximum reinforcement and design accordingly.1 and 4.2.Exterior Column Design of an RC Frame Building in Seismic Zone V 4 Problem Statement:
For the ground plus four storey RC office building of Example-1 (Refer Figures 1. The column marked AB in Figure 4. Since the column must be designed for earthquake in both X-direction and Y-direction. It is necessary to check the column for each combination of loads.1 Column location in elevation
Solution: 4. Referring to Table 4.1. column gets a large moment about Y axis and a small moment about X axis due to gravity. In case of earthquake in X direction.1-1. minimum eccentricity and torsional effect.
9DL + 1. These design forces are summarised in Table 4.6.2(DL+LL-EQY) 1.5 EQY
-1143 -867 -747 -483 -1131 -1040 -890 -560 -1370 -654 -504 -174 -984
0 130 -130 0 0 162 -162 0 0 162 -162 0 0
30 20 20 -115 154 23 23 -146 191 14 14 -155 182
-909 -678 -609 -414 -873 -813 -726 -483 -1056 -505 -418 -175 -748
2 -133 136 1 1 -167 170 2 2 -167 169 1 1
-44 -29 -29 88 -145 -33 -33 113 -179 -20 -20 126 -165
-893 -665 -596 -401 -860 -797 -710 -467 -1040 -495 -408 -165 -738
-2 133 -136 -1 -1 167 -170 -2 -2 167 -169 -1 -1
42 27 27 -98 152 32 32 -125 188 19 19 -137 175
-657 -483 -449 -323 -609 -579 -537 -380 -737 -356 -314 -156 -513
2 -115 118 1 1 -144 147 2 2 -145 146 1 1
-41 -26 -26 77 -129 -30 -30 99 -159 -18 -18 111 -147
direction can be identified. The detailed calculations are shown in Section 4.1 Force resultants in column AB for different load cases
DL LL EQx EQy
Table 4. and . which is likely to be critical.4 and Table 4.2(DL+LL+EQY) 1. 2 f ck bD f ck b D f ck bD 2
Load Case AB Axial (kN) -643 -119 -50 270 M2 (kN-m) 0 0 108 0 M3 (kN-m) 15 5 0 -112 Axial (kN) -513 -93 -29 191 AT M2 (kN-m) 1 0 -112 0 M3 (kN-m) -22 -7 0 97
Using these factors and the charts given in SP: 16.
BB Axial (kN) -502 -93 -29 191 M2 (kNm) -1 0 112 0 M3 (kN-m) 21 7 0 -104 Axial (kN) -372 -66 -14 119 BT M2 (kN-m) 1 0 -97 0 M3 (kN-m) -20 -7 0 86
Table 4.1.2(DL+LL-EQX) 1.0
Example 4 /Page 34
.5(DL+EQX) 1. Table 4.1.5 EQX 0.5 EQX 0. .5(DL+EQY) 1.4.2(DL+LL+EQX) 1.9DL + 1.5(DL-EQX) 1.5 EQY 0.2 Force resultants in column AB for different load combinations
Load Combinations P
M2 kN-m M3 P kN-m kN
M2 kN-m M3 kN-m
1. the required maximum reinforcement is calculated the same being summarised in Table 4.9DL .3.5(DL-EQY) 0.9DL .5 give factors Pu M3 M2 such as .5(DL+LL) 1.
Pu = 1.8 x 300 x 500 /100
IITK-GSDMA-EQ22-V3. for the load combination 0.4 4. ok.347 f ck bD 20 × 300 × 500
162 × 10 6 = 0.0
191 × 10 6 = 0. Hence. B = 300 mm ≥ 300 hence. (Clause 7. (Clause 7.
(Clause 7. D = 500 mm
B 300 = = 0.5 EQY the member needs to be checked as flexural member.3 Check for Reinforcement
Pu = 1.85 times the unsupported length.10 fck Hence.1
Factored axial force = 166 kN (Lowest at At or Bb among all load combination is considered) Factored axial stress = 1. the effective length is taken as 0.200 mm2 (Clause 26. IS 13920:1993) Span.2 of IS 456: 2000) In case of slender column. IS 13920:1993)
Design of Earthquake in Y-Direction
Short column. The column is oriented in such a way that depth of column is 300 mm for X direction earthquake and 500 mm for Y direction earthquake force.185.175.Examples on 13920
4.18 20 × 500 × 300 × 300
Referring to Charts 45 and 46 of SP16 For d’/D = (40 +25 /2)/300 = 0.1. the column is designed as an uniaxially loaded column in that direction. the moment about one axis dominates and hence.3
4. ok.127 20 × 300 × 500 × 500
Referring to Charts 44 of SP16 For d’/D = (40 +25 / 2)/500 = 0.3.6 > 0. D 500
In this case. which is in between that of fixed and hinged case. (Clause 25.1. (Clause 7.85 = = 7. we get
Example 4 /Page 35
.000 mm The effective length of column can be calculated using Annex E of IS 456: 2000.1.5.1 of IS 456: 2000) Maximum reinforcement = 4% (Limited from practical considerations) = 4 x 300 x 500 / 100 = 6.4.3. = 0.456 fckbD 20 × 300 × 500
Minimum reinforcement.8 %.000 mm2 (Clause 26..3. we get
p/fck = 0.370 kN Mu2 = 191kN-m
Pu 1370 × 10 3 = = 0.66.2 of proposed draft IS 13920)
4.10 MPa < 0. i.3.1
= 1. Minimum Dimension of Column = 300 mm ≥ 15 times the largest diameter of beam longitudinal reinforcement = 15 x 20 = 300 ok.3. hence ok. L = 3.2 Check for Member Size
Sample Calculation for Column Reinforcement at AB End
First approximate design is done and finally it is checked for all force combinations.9DL + 1.
Width of column.3.5.1. For all other load combinations design is done as a Column member.1.040 kN Mu2 = 162 kN-m
Pu 1040 × 10 3 = = 0. additional moment due to P-δ effect needs to be considered.1.1 < 12.
L (3000 − 500) × 0.105.4.000 /300 x 500 = 1. In this example as per Table 28 of IS 456: 2000. = 0.2.1 of IS 456: 2000)
4.e. IS 13920:1993) Depth of column.
1.92 < 1.e.216 0.0.92/20 = 0.75fy .18
Eccentricity.88 + 0.370 x 0.550 mm2 Provide 12-25Φ bars with total Asc = 5.45 x 20 x 300 x 500 + (0.
M u 3. Required steel = 0.1 = 0.2 × 20 × 300 × 300 × 500 /(1 × 10 6 )
= 180 kN-m.130 = 0. ok.1 = 0. Referring to f ck bD Chart 44 of SP: 16.130 kNαn = 1.216
Hence.33 − 0.18 × 20 × 500 × 500 × 300 / 1× 10 6
= 270 kN-m
M u 2.20 > 0. Moment due to other axis may occur due to torsion of building or minimum eccentricity of the axial load.185 Hence.e3 = clear height of column /500 + lateral dimension /30 = ((3.7 % = 3. minimum eccentricity = 20 mm.20.45fck) Asc
(Clause 39. Mu2 = 1.216
1. we get M u 2.1 ⎥ ⎣ ⎦
⎡M ⎤ + ⎢ u3 ⎥ ⎢ M u 3.040 kN Mu2 = 162 kN-m
Width = 300 mm.130 kN
Pu/Puz = 1. Pu = 1.1 ⎥ ⎣ ⎦
⎡162 ⎤ ⎡ 20.6 of IS 456:2000)
IITK-GSDMA-EQ22-V3.1 = 0.040 / 3.1 = 0.28
Example 4 /Page 36
.46 mm > 20 mm Hence.040 x 0.2
Using the interaction formula of Clause 39.2 × 20 × 300 × 500 × 500 /(1 × 10 6 )
M u 3.0) =1.347 and p/fck= 0.8 − 0.92%.370 kN Mu3 = 191 kN-m Eccentricity.45 fckAg + (0.892 x 100 / (300 x 500) = 3.
The column should be checked for bi-axial moment..0 − 1.8 kN-m
Pu = 0.4 kN-m For
Pu = 0. e3 = clear height of column /500 + lateral dimension / 30 = ((3.0 +
Puz = 0.892 = 3.185.2
M u 2.18 × 20 × 300 × 300 × 500 / 1× 10 6 = 162 kN-m
= 300 kN-m.45 x 20) x 5.0. ok.185 x 20% = 3.000 – 600)/500) + (500 / 30) = 21.0
Puz = 3.02 = 20.02146 = 29.7 x 300 x 500 /100 = 5.456 and p/fck = 0.33
The required steel will be governed by the higher of the above two values and hence p/fck = 0.000-500) / 500) + (300 /30) = 15 mm > 20 mm
Mu3 = 1. and referring f ck bD to Charts 44-45 of SP: 16 and we get Mu f ck bD 2 = 0.
⎡ M u2 ⎤ ⎢ ⎥ ⎢ M u 2.8 ⎤ =⎢ +⎢ ⎥ ⎥ ⎣180 ⎦ ⎣ 300 ⎦ = 0.
Width = 500 mm.20.6.1 f ck bD 2 = 0.Examples on 13920
p/fck = 0. IS 456: 2000). Hence p/fck provided = 3.039 = 0. 5.75 x 415 .12
= 0.2 (2.892 mm2 i. Depth = 300 mm
Pu = 1.
αn = 1. Depth = 500 mm.
7% steel is needed.78 < 1.1 ⎥ ⎣ ⎦
⎡ 29.370
0.00 ok.
4.1 ⎥ ⎣ ⎦
⎡M ⎤ + ⎢ u3 ⎥ ⎢ M u 3.2 shows the reinforcement in the column along with steel provided in the transverse and longitudinal beams.8
Table .03
M2 f ck b 2 D
0.3 Critical forces for design of column AB Load Combination Gravity Critical comb with EQX Critical comb with EQY P
0. BB and BT locations.39
0.1.5% is needed in the column at the joint B.66
Example 4 /Page 37
⎡ 191 ⎤ ⎢ 270 ⎥ ⎣ ⎦
1.143 -1.4 ⎤ =⎢ + ⎥ ⎣ 162 ⎦ = 0.7
IITK-GSDMA-EQ22-V3.347
0. Critical comb with EQY
BB M3
0. and hence. the steel required at AB .185
0.5 Design of column AB for earthquake in Y-Direction
Load Comb.5
The Tables 4. 3.4.040
-909 -813
-44 -33
-893 -797
-657 -579
2 -144
-1..03
Gravity Critical comb with EQX
0. AT.040
Table -4.22
3. The column at the joint A should have the higher of the reinforcement required at AB and AT.12 + 0.11
1. i.8
0. Figure 4. BB and BT is calculated.5
⎡ M u2 ⎤ ⎢ ⎥ ⎢ M u 2.216
= 0.5 show brief calculations at AB. 3.13
2.e.46
2. Table -4.4 and 4. Similarly higher of the reinforcement required at BB and BT.056
Similar to the sample calculations shown in Section 4.4.4 Design of column AB for earthquake in X-Direction
Load Comb AB AT p BB p BT p
0. AT.
9DL + 1.9DL + 1.5(DL-EQX) 1.5(DL+EQY) 1..0
Example 4 /Page 38
.1.5 EQX 0.892 mm2 i.5(DL-EQY) 0.e.93%.2(DL+LL+EQY) 1.5(DL+LL) 1.57%
10-25 # + 2-16 #
Table -4.2(DL+LL-EQX) 1. Reinforcement 10-25Φ+ 2-16Φ at B Steel provided= 5.7 Tabulation of shear forces in column AB for different load combinations
AB Load Combination
1.Examples on 13920 Table .6 Summary of reinforcement for column AB
Column AB Longitudinal Reinforcement Reinforcement Details
Confining links: 10 # @ 85 c/c Nominal links: 8 # @ 150 C/C 12-25 #
Reinforcement 12-25Φ Bars at A Asc = 5.2(DL+LL+EQX) 1.5(DL+EQX) 1.5 EQY 0. 3.2(DL+LL-EQY) 1. 3.1.4.e.360 mm2 i.5 EQY
AT EQY (kN)
0 -83 83 0 0 -104 104 0 0 -104 104 0 0
BB EQY (kN)
1 -89 90 0 1 -111 112 0 1 -111 112 0 1
BT EQY (kN)
EQX (kN)
-14 -9 -9 78 -97 -10 -11 99 -120 -6 -7 103 -116
-29 -18 -19 62 -99 -21 -22 80 -123 -12 -14 88 -114
-27 -17 -18 56 -92 -20 -21 72 -113 -11 -13 80 -105
EQY (kN)
2 -79 81 0 2 -98 101 0 2 -99 100 0 2
IITK-GSDMA-EQ22-V3.9DL .9DL .5 EQX 0..
Ast = 3.978%) at bottom.6.7)
4. 0. i.6.167 = 0.Bottom steel)
Figure 4.96% Permissible shear stress τ c = 0. i.2 Reinforcement details of longitudinal and transverse beam 4. 1.5 mm Vc = 0.5 mm Vc = 0.3.e.
kN-m and
kN-m.3.2 of IS 456: 2000)
τ c = 0. The hogging and sagging moment capacities are
IITK-GSDMA-EQ22-V3.2 Shear As Per Analysis
Vu Vu Mu + Mu = 1.e.79 Mpa (Table 19 of IS 456: 2000) Considering lowest Pu = 166 kN.0
The transverse beam of size 300 x 600 is reinforced with 3-16Φ str + 5-20Φ extra + 1-16Φ extra (2..Top steel 2-20 # + 4-16 # . the shear force corresponding to plastic hinge is evaluated as:
(Clause 40.
Example 4 /Page 39
.374 mm2.1
evaluated as respectively.4 x (238 +180) /3 = 195 kN
4. i.Bottom steel)
Longitudinal beam 300 x 500 (4-20 # + 3-16 # .
3Pu =1.5 Ag f ck
Referring to Figure 4.6.. 1.
Assuming 50% steel provided as tensile steel to be on conservative side.43%) at top and 2-20Φ extra + 4-16Φ str (1432 mm2.4 hst
Figure 4.6.1 Earthquake in X-Direction
= 1.92 x 500 x 247.3 Column shear due to plastic hinge formation in longitudinal beams
1.92 MPa
Effective depth in X direction = 300-40-25/2 = 247. (Refer Table 4. The hogging and sagging moment capacities are evaluated as 377 kN-m and 286 kN-m.e.167 < 1.e.10%) at bottom..3 Shear Force Due to Plastic Hinge Formation at Ends of Beam 4.92 x 300 x 447.5 /1.487 %) at top and 3-16Φ str + 3-20Φ extra (1.92% / 2 = 1.79 x 1.3. 1.000 = 114 kN Effective depth in Y direction = 500-40-25/2 = 447.4 (M u
The maximum factored shear force in X and Y direction is 123 and 112 kN respectively. respectively.000 = 123 kN
4.2 Earthquake in Y-Direction
The longitudinal beam of size 300 x 500 is reinforced with 4-20Φ extra +3-16Φ str (1859 mm2..2.Top steel 3-16 # + 3-20 # .545 mm2.6.Examples on 13920
12-25 # at A 10-20 # +2-16# at B
Transverse beam 300 x 600 (5-20 # + 4-16 # . i.5 /1.
40 – 40.7.4 × (377 ) 3
(Clause 7.3 of IS 13920: 1993) Provide 8 Φ links @ 150 c/c in mid-height portion of column.05. Ash = 78.e.4 Column shear due to plastic hinge formation in transverse beams
Vu = =
Substituting we get S = 90 mm.600 mm2 Assuming 10Φ stirrup. 300 / 4 = 75 mm
Vs = 195 – 114 = 81 kN
IITK-GSDMA-EQ22-V3.7) and the shear force due to plastic hinge formation in either of the transverse or longitudinal beams. 300/ 2 = 150 mm.4 Confining Links
Figure 4.6.3.6.4.
4. which is the higher of 123 kN and 176 kN.3.87 x 415 x 447.2
Design of Links in Y Direction
Vs = 176 – 123 = 53 kN Spacing of 2 Legged 8 Φ Links
br Mu
= 2 x 50 x 0. Link spacing for confining zone shall not exceed: ¼ of minimum column dimension i. (Clause 7.7.4.e.0
Example 4 /Page 40
. Ag = 300 x 500 = 1.25)/3 + 25 +10 +10 =177 mm or 300 . (Clause7.7
4. the design shear in X direction is 195 kN which is the higher of 112 kN and 195 kN.54 =
0.54 mm2 78.e.000 mm2 Ak = (300 –2 x 40 +2 x10) x (500-2 x 40 + 2 x 10) = 240 x 440 = 1.18 × S × 240 × 20 ⎛ 1.8 of IS 13920: 1993).4 Mu hst
The spacing of hoops shall not exceed half the least lateral dimension of the column.600 − 1⎟ ⎟ 415 ⎝ ⎠
= 176 kN. Similarly the design shear in Y direction is 176 kN. h = 240.000 ⎞ ⎜ ⎜ 1.3 Nominal Links
br = 1.7. Assuming h = longer dimension of the rectangular link measured to its outer face = ((500.000 = 221 mm
4. whichever is higher.7. IS 13920: 1993) From Section 4.5 /81..Examples on 13920
Referring to Figure 4.000 = 305 mm
4.4.50.4 (M u hst
The area of cross section Ash of the bar forming rectangular hoop to be used as special confining reinforcement shall not be less than Ash =
0.05. the shear force corresponding to plastic hinge formation in the transverse beam is
Spacing of 4 legged 8 Φ Links = 4 x 50 x 0.5 /53.50.4 Design Shear
The design shear force for the column shall be the higher of the calculated factored shear force as per analysis (Table 4. i.3 above.18 × S × h × f ck fy ⎛ Ag ⎞ ⎜ ⎟ ⎜ A − 1⎟ ⎝ k ⎠
1.40 – 40 +10 +10 = 240 mm.87 x 415 x 247..
450 mm (Clause 7.883% = 0.0
. (Clause 7. (Refer Figure 4.
Table 4.5 2 Referring to Table 2 of SP: 16.60 bd 2 300 × 447. Provide confining reinforcement for a distance of Lo = 500 mm on either side of the joint.883 x 300 x 447.8
Check as Flexural Member for Load Comb 0. Provide 10Φ confining links @ 75 c/c for a distance Lo (Refer figure 4. which are being incorporated in the new edition of IS 13920.185 mm2 Ast provided on one face = 3 – 25Φ = 3 x 491 = 1.8 Comparison of bill of quantities of steel in column
Description Detailing Detailing Detailing as per as per as per IS 13920: IS 456: IS 456: 1993 2000 2000 (Seismic (Seismic (Seismic loads as loads as loads as per R = 5) per R = 3) per R = 5) Links (kg) Main steel (kg) 25 132 10 132 Column needs to be redesigned.25/2 = 447. Larger lateral dimension = 500 mm b.4.1 of IS 13920: 1993).185 mm2 Hence.9 DL + 1.5 Reinforcement details for column
4. however.5 EQY
Example 4 /Page 41
Factored moment = 156 kN-m
IITK-GSDMA-EQ22-V3.473 mm2 >1.5)
Effective depth = 500 – 40 .
Note.1/6 of clear span = (3.5 / 100 = 1. which shall not be less than: a.6 of IS 13920: 1993). we get Ast = 0. ok.4.000 –500)/6 = 417 mm c.5mm Mu 156 × 10 6 = = 2. that the column designed above has not been checked for requirements related to the joint region.
Figure 4. These provisions are illustrated in Examples 5-8 and may require modifications in column size and / or longitudinal reinforcement.Examples on 13920
But need not be less than 75 mm nor more than 100 mm.5).
A C2 C1 C1 C1 C1 C2
C C2 1 4 C1 2 4 C1 3 3 C1 4 4 C1 5 4 C2 6
Figure 5. Preliminary Data
The joint of column marked in Figure 5.4)
5.1-1.2 Column location in elevation
IITK-GSDMA-EQ22-V3.2. 1 Plan of building (All dimensions in meters)
Transverse beam 600 500 300 300 Longitudinal beam
Figure 5.1.Examples on 13920
Example 5 – Interior Beam – Column Joint Design for Zone -V
Detailed design as per draft revision of IS 13920:1993 of an interior joint in an intermediate RC frame is explained for the ground plus four storey RC office building of Example-1 (Refer Figures 1.1 and 5.0
500 Longitudinal beam
Example 5 /Page 42
. The plan of the building and the sectional elevation of a typical RC frame is shown in Figures 5.1 for Frame 2 is considered for design.
i.34%) at bottom.319 mm2. i.231 kN = C1
IITK-GSDMA-EQ22-V2. 1.4 × ⎜ ⎟ 3 ⎠ ⎝ = 291 kN
Force Developed in Beam Reinforcement
Figures 5.25 x fy = 2.261 mm2. 0.Examples on 13920 Column Shear The column shear is as explained below.3.2.4
Mh + Ms hst
Figure 5.487%) at top and 1-20Φ + 5-16Φ (1.
5.4 Joint shear
Force developed in the top bars T1 = Ast x 1. 1.83%) at bottom..e.B o tto m s te e l) 1 0 -2 5 # + 4 -1 6 #
L o n g itu d in a l b e a m 3 0 0 x 5 0 0 (4 -2 0 # + 5 -1 6 # .1 Joint Shear
The joint shear equilibrium is shown in Figure 5. Design Data
The details of the column and beam reinforcement meeting at the joint are shown in Figure 5.3 Check for Earthquake in Y-Direction
5. ⎛ M + Mh ⎞ ⎟ Vcol = 1. respectively. respectively.e.6 for sway to right and left conditions respectively). 1.
T ra n s v e rs e b e a m 3 0 0 x 6 0 0 (5 -2 0 # + 4 -1 6 # .74%) at top and 3-20Φ + 4-16Φ (1.5 and 5.0
Example 5 /Page 43
.7 and 5. The hogging and sagging moment capacity is evaluated as 288 kN-m and 221 kN-m.6 Column with sway to left
5.3 Reinforcement details for column and beams.T o p s te e l 3 -2 0 # + 4 -1 6 # .5 Column with sway to right
The transverse beam of size 300 x 600 is reinforced with 5-20Φ + 4-16Φ (2. (Refer Figures 5. VJoint
Figure 5.T o p s te e l 5 -1 6 # + 1 -2 0 # .000 = 1.e. for sway to right and left.374 x 1.e.
Figure 5.374 mm2.4 ⎜ s ⎜ ⎟ hst ⎝ ⎠ ⎛ 377 + 246 ⎞ = 1.25 x 415 /1. respectively.4 Vcol C2 T
For both the above cases.
Vcol Vcol = 1.4.4 Mh + Ms hst
Figure 5.B o tto m s te e l)
Vcol = 1.8 show the development of forces in the joint due to beam reinforcement.. The hogging and sagging moment capacity is evaluated as 377 kN-m and 246 kN-m. i.746 mm2 i. The longitudinal beam of size 300 x 500 is reinforced with 4-20Φ + 5-16Φ (2.
3 Check for Flexural Strength Ratio
20 x 400 x 500 /1.000
= 1.m.9.e.4. respectively.624 kN Maximum value of T1 and minimum value of Vcol are used in the above equation.7 Free body diagram of the joint
P Vcol T = A st x 1.25Φ + 4 .0
The hogging and sagging moment capacities of the transverse beam are as 377 kN-m and 246 kN .Examples on 13920
The factor 1.073 kN < 1.25 fy
strength of the joint is based on the draft revision of IS 13920:1993.25 fy 1 C1 = T 1 Mh
MAB P
Sway to right
Figure 5.25 fy
Sway to left
Figure 5.4.1.4.9 Effective width for joint
bj = bb + h/2
C1 = T1 Ms
= 300 + 500 /2 = 550 mm Or bj = bc = 400 mm Take effective width as 400 mm.8 Free body diagram of the joint
Force developed in the bottom bars T2 = Ast x 1.
5. [Draft revision of IS 13920]
P Vcol C2 = T 2 Ms T = A st x 1.000 = 684 kN = C2 Referring to Figure 5.2
f ck Ac
T = A st x 1.5 x h ii) bj = bc
T = A st x 1.231 + 684-291 = 1.25 fy 2 Mh C2 = T 2 MA T
Figure 5.16Φ bars with total Asc = 5.2 x Hence.319 x 1.25 is to account for the actual ultimate strength being higher than the actual yield strength. as per Clause 8. not Safe.3 of draft revision of IS 13920:1993 = 1. The effective width of the joint is lesser of the i) bj = bb + 0. h = full depth of column = 500 mm Effective shear area of the joint = Ac = bj h Shear strength = 1. The column is reinforced with 10 .714 mm2 i.
5. VJoint = T1 + C2 – Vcol = 1.2 Check for Joint Shear Strength
Shear strength of joint confined on two opposite faces.624kN
The effective width provisions for joints are shown in Figure 5.
Example 5 /Page 44
.25 x 415 /1.25 x fy = 1. The calculation of the effective width of the joint and the design shear
IITK-GSDMA-EQ22-V2.
25 x fy = 1.1
Hence. for p/fck = 0.0
Example 5 /Page 45
.852 / 20 = 0.143 and d’/D = (40 + 25 /2) / 500 = 0.261 x 1. the joint is checked for strong column .3 of draft revision (bc > ¾ bb on two opposite faces ) of IS 13920:1993 Shear strength = 1.
Shear due to formation of plastic hinge in beams
Referring to Figures 5.714 x 100 / (400 x 500) = 2.173 + 905-238 = 1.8.4.
Mu = 0. (Clause 7.weak beam condition
∑Mc = 380+380 = 760 kN-m ∑Mb = 377 + 246 = 623 kN-m The ratio of
∑ ∑M
= 760 /623 = 1.1.weak beam.22 > 1. for both the cases.4.4. In actual Mu practice it is desirable to take minimum f ck bD 2 Pu corresponding to actual obtained from f ck bD different load combinations.25 x 415/1.10 Check for strong column .000 = 905 kN = C2 The joint shear is evaluated maximum T1 and minimum Vcol.19 x 20 x 400 x 500 x 500 / 1x106 = 380 kN-m Referring to Figure 5.
⎛ M + Mh ⎞ ⎟ Vcol = 1.25 x 415/1.4 × ⎜ ⎟ 3 ⎝ ⎠
= 238 kN = 0. VJoint = T1 + C2 – Vcol = 1. we get. requirement of strong column-weak beam condition as per proposed draft IS 13920 is satisfied.852%.000 = 1.19.4 Check for Earthquake in X Direction
Figure 5.840 kN considering
Mc B P
5.6.Examples on 13920
Referring to Figures 5. Referring to chart 44 Pu of SP: 16.1 Joint Shear
The joint shear equilibrium is shown in Figure 5.0 = 1.1426 It is conservative here to calculate the moment capacity of column with zero axial loads.173 kN = C1 T2 = Ast x 1.e.00 at AB f ck bD to be on the conservative side.7 and 5. corresponding to = 0 .10.4 ⎜ s ⎜ ⎟ hst ⎝ ⎠ ⎛ 288 + 221 ⎞ = 1 . bj = 500 mm h = Depth of column or full depth of beam = 400 mm Shear strength of joint not confined as per Clause 8.105. we get
Mu f ck bD 2
5. T1 = Ast x 1.1 of IS 13920 proposed draft)
20 x 500 x 400 /1000
IITK-GSDMA-EQ22-V2.5 and 5. p/fck = 2.2.25 x fy = 2.2
Check for joint shear strength
bj = bb + h/2 = 300 + 400 /2 = 500 or bj = bc = 500 mm Adopt lesser of the two values i.746 x 1.
the column size now requires 4. (Clause 7.00 at AB.178. Hence revise the main longitudinal steel to 8-20φ + 8-16φ bars (4.3. 1.90/20 = 0.13125.11 > 1. The section needs to be checked for flexure for these load combinations. Member forces are taken as calculated earlier without reanalysis of the structure. In such cases the following three alternatives can be tried. The hogging and sagging moment capacity is evaluated as 293 kN-m and 229 kN-m.5
As can be seen from the checks in section 5.4.045.1%. The redesigned longitudinal beam of size 300 x 600 is reinforced with 6-20Φ (1.2.1
Hence.2. As per analysis results. 0.4.0
Example 5 /Page 46
. ii) Increase the size of the beam section.1 fck. increase in beam width can be considered if the difference between the shear strength of joint and joint shear is small. requirement of strong column-weak beam condition is satisfied.9% steel) as main longitudinal reinforcement. If this option is adopted.1 x 522 (i. respectively.884 mm2. 1.77%) at bottom. corresponding to = 0. and 5.e. While redesigning the column few load combinations may give an axial stress less than 0.178 x 20 x 400 x 400 x 500 /1x 106 = 284 kN-m ∑Mc = 284+284 = 568 kN-m ∑Mb = 288+221 = 509 kN-m The ratio of
= 568/509 = 1. 293+223)= 574 kN-m in longitudinal direction.120 mm2.266 mm2.
iii) Increase the grade of concrete. This option will increase the shear strength of joint and also reduce the steel required in columns.Examples on 13920
= 894 kN < 1. (Clause 7.2.11. i. This will reduce the steel required in the beam and hence will reduce the joint shear.1 = 685 kN-m and 1. Hence required moment capacity for column is Mc = 685/2 = 343 kN-m in Y direction and 574 / 2 = 287 kN-m in X direction.
It is conservative here to calculate moment capacity of column with zero axial loads. In actual practice it is desirable to take minimum Pu Mu corresponding to actual obtained 2 f ck bD f ck bD from different load combinations.1426 and d’/D = (40 + 25/2) /400 = Mu 0.e.20φ + 10-16φ bars (3. Referring to Pu chart 44. The above column section will satisfy the flexural strength check. 0.e.
IITK-GSDMA-EQ22-V2.231 mm2 i. (Clause 8. In case of depth restriction in the beam..1 of proposed draft IS13920)
5. i) Increase the column section so that the joint area is increased. it is advisable to increase the depth of the beam.1 of IS 13920 proposed draft) Using SP 16 the steel required to get the above moment capacity of column is calculated as 1.14% steel). respectively. This will also reduce the main longitudinal steel requirement in the column owing to larger column size.3 of IS 13920 proposed draft)
5. the joint is not safe. The column is redesigned to 600 x 600 with 420φ + 10 -16φ bars (3.840 kN Hence not safe. we get = 0.266 mm2. f ck bD for p/fck= 0.9% steel) as main longitudinal steel. It is proposed to increase the size of column from 400 x 500 to 600 x 600 and longitudinal beam size from 300x500 to 300x600. f ck bD 2 Mu= 0. 0.2.3 Check for flexural strength ratio The limiting hogging and sagging moments capacity of the longitudinal beam is 288 kN-m and 221 kN-m.18%) at top and 2-20Φ + 3-16Φ (1. The ∑Mc required in transverse direction is 623 x 1. In practice the structure may be reanalyzed. This design is made based upon the existing forces without reanalysis. The revised reinforcement details are shown in Figure 5. The value of p/fck = 0.
0 Example 5 /Page 47
.000 = 978kN = C1 T2 = Ast x 1.0 = 1.3 of IS 13920:1993 proposed draft)
Shear strength = 1. 3/4 x 600 = 450 mm. we get.25 x fy = 1. ok.231 x 1. VJoint = T1 + C2 – Vcol = 978 + 638-244 = 1. for both the cases.25 x 415/1. ok.0 x Hence.610 kN ≈ 1.372 kN bj = bb + h/2 = 300 + 600 /2 = 600 mm Or bj = bc = 600 mm h = full depth of column = 600 mm Take bj = 600 mm considering
Longitudinal beam 300 x 600 (6-20 # .
= 244 kN
IITK-GSDMA-EQ22-V2. Hence.4 × ⎜ ⎟ 3 ⎝ ⎠
5. full confining reinforcement is required in the joint.Examples on 13920 Force Developed in Beam Reinforcement
T ransverse beam 300 x 600 (5-20 # + 4-16 # .Top steel 2-20 # + 3-16 # .610 kN < 1.624 kN
= 1.6 Confining Links
In this case with the column dimensions revised to 600 x 600.25 x fy = 1. .5 and 5.000
= 1.7 and 5.B ottom steel) -16 # + 8-20 #
Referring to Figures 5.4 ⎜ s ⎜ ⎟ hst ⎝ ⎠ ⎛ 293 + 229 ⎞ = 1 .0 = 1.
20 x 600 x 600 /1.884 x 1.372 kN
Check for Earthquake in X Direction
Referring to Figures 5.11 Revised reinforcement details for column and beams
Check for Earthquake in Y Direction
bj = bb + h /2 = 300 + 600 /2 = 600 mm Or bj = bc = 600 mm h = full depth of column = 600 mm Take bj = 600 mm Shear strength = 1.6.000
20 x 600 x 600 /1.0 x Hence.8. shear due to formation of plastic hinge in beams is
⎛ M + Mh ⎞ ⎟ Vcol = 1.000 = 638 kN = C2 The joint shear is evaluated maximum T1 and minimum Vcol. (Clause 8.e. T1 = Ast x 1. which is less than 3/4 width of column i.25 x 415/1.T op steel 5-16 # + 1-20 # .B ottom steel)
Figure 5. the width of beam is 300 mm.
60.6 of IS 13920:1993) The area of cross section Ash of the bar forming rectangular hoop to be used as special confining reinforcement shall not be less than
S = 80 mm Provide 8Φ confining links @ 80 c/c in the joint.12 Confinement of joint concrete by beams
The spacing of links for the confining zone shall not exceed: i) ¼ of minimum column dimension i.000 ⎞ ⎜ ⎜ 2. 600 / 4 = 150 mm ii) But need not be less than 75 mm nor more than 100 mm.2 x 40 +2 x 8) = 536 x 536 = 2.60.18 × S × 286 × 20 ⎛ 3. (Clause 7. 87.Examples on 13920
0.18 × S × h × f ck fy
⎛ Ag ⎞ ⎜ ⎟ ⎜ A − 1⎟ ⎝ k ⎠
(Clause 7.e.40 .296 mm2 Ash = 50 mm2 50 =
0.8 of IS 13920:1993)
Assuming h = longer dimension of the rectangular confining measured to its outer face
= (600 .296 − 1⎟ ⎟ fy ⎝ ⎠
b y ≥ 3 /4 h y b x ≥ 3 /4 h x
Figure 5.4.40 -20) / 2 +8 x 2 + 20 = 286 mm Ag = 600 x 600 = 3.4.
IITK-GSDMA-EQ22-V2.0
Example 5 /Page 48
.87.000 mm2 Ak = (600-2 x 40 +2 x8) x (600.
6.Examples on 13920
Example 6 — Exterior Beam-Column Joint Design for Zone V
6.1 for Frame 2 is considered for design.2. The plan of the building and the sectional elevation of a typical RC frame are shown in Figures 6.0
Example 6 /Page 49
. Problem Statement:
Detailed design as per draft revision of IS 13920:1993 of an exterior joint in an intermediate RC frame is explained for the ground plus four storey RC office building of Example-1 (Refer Figures 1.
Figure 6. 1 Plan of building (All dimensions in meters)
T r a n s v e rs e beam 600 500 300 300 L o n g itu d in a l beam T ra n s v e rs e beam 500 300 300 P la n
Figure 6-2 Column location in elevation
IITK-GSDMA-EQ22-V2.1-1.1 and 6.1 Preliminary Data
The joint of column marked in Figure 6.
e.3 Reinforcement details for column and beams.Top steel 2 -20 # + 4-1 6 # .4 Joint shear
IITK-GSDMA-EQ22-V2.B o ttom steel)
The column shear is evaluated as explained below. 0.4 ⎜ s ⎟ ⎜h ⎟ ⎝ st ⎠ ⎛ 286 ⎞ = 1.487 %) at top and 3-20Φ + 3-16Φ (1.Examples on 13920
6.e.6 Column with sway to left
⎛M ⎞ Vcol = 1.4 ⎜ h ⎟ ⎜h ⎟ ⎝ st ⎠ ⎛ 377 ⎞ = 1 . i. 1.4 × ⎜ ⎟ ⎝ 3 ⎠ = 133 kN
Figure 6.4 Vcol T
Vcol Ms Vcol = 1. 1.5 and 6. The longitudinal beam of size 300 x 500 is reinforced with 4-20Φ +3-16Φ (1.0
Example 6 /Page 50
. i.545 mm2.
The transverse beam of size 300 x 600 is reinforced with 5-20Φ + 4-16Φ (2.10%) at bottom.6 for sway to
right and left..
Tra nsverse bea m 300 x 60 0 (5-2 0 # + 4-16 # .374 mm2. i.432 mm2. respectively.43%) at top and 2-20Φ + 4-16Φ (1.e.4 hst
Figure 6.859 mm2.1 Joint Shear
The joint shear equilibrium is shown in Figure 6.3 Check for Earthquake in Y-Direction
6. (Refer Figures 6.
1 2-2 5 #
Lo ngitud ina l bea m 300 x 5 00 (4-20 # + 3 -16 # .4 hst
Figure 6. 1.97%) at bottom..3.B ottom stee l)
Vcol Mh Vcol = 1.4 × ⎜ ⎟ ⎝ 3 ⎠ = 176 kN
Figure 6.5 Column with sway to right
⎛M ⎞ Vcol = 1. The hogging and sagging moment capacity is evaluated as 377 kN-m and 286 kN-m.Top steel 3-16 # + 3-20 # .3. i. conditions respectively).
The details of the column and beam reinforcement meeting at the joint are shown in Figure 6.e.. The hogging and sagging moment capacities are evaluated as 247 kN-m and 180 kN-m respectively.
9.25 x 415 /1.4. VJoint = T1 – Vcol = 1.000 = 1.8 shows the development of forces in the joint due to beam reinforcement. For sway to left. Force developed in top bars T1 = Ast x 1.7 and 6.0
. respectively. Referring to Figure 6.2 Check for Joint Shear Strength The effective width provisions for joints are shown in Figure 6.545 x1. The calculation of the effective width of the joint and the design shear strength of the joint is based on the draft revision of IS 13920:1993 The effective width of the joint is the lesser of::
Figures 6.Examples on 13920 Force Developed in Beam Reinforcement
= 801 kN.25 x 415 /1.9 Effective width for joint
C1 = T 1
= 300 + 500 /2 = 550 mm bj = bc = 300 mm Take effective width of joint as 300 mm h = full depth of column = 500 mm Effective area of joint resisting shear = Ac = bj h Shear strength of joint not confined (bc < ¾ bb only on one faces and bc > ¾ bb on other two faces) as per Clause 8.231-176 = 1.25 fy
Figure 6.4.3 of draft revision of IS 13920:1993
Example 6 /Page 51
T = A x 1.7 Free body diagram of the joint
P Vcol MA T
Figure 6.1.374 x1. = 801 -133 = 668 kN
6.25 fy 1 C1 = T 1 Mh
= bb + 0.25 x fy = 2. for sway to right and left.000
IITK-GSDMA-EQ22-V2.055 kN for sway to right.5 x h
ii) bj = bc
Figure 6.25 is to account for the actual ultimate strength being higher than the actual yield strength (Draft revision of IS 13920:1993)
i) bj
T = A st x 1.25 x fy = 1.8 Free body diagram of the joint
Force developed in bottom bars T1 = Ast x 1.231 kN The factor 1.
00 at AB.Examples on 13920
Shear strength = 1.3 Check for Flexural Strength Ratio
The hogging and sagging moment capacities of transverse beam are 377 kN-m and 286
kN-m.11. In actual practice it is desirable to take minimum Pu Mu corresponding to actual obtained 2 f ck bD f ck bD from different load combinations.10 Check for strong column . i.000
= 671 kN < 1. the beam-column joint is checked for strong column-weak beam condition.10.19 It is conservative here to calculate the moment capacity of column with zero axial loads.1
Hence. not safe.11 Joint shear
IITK-GSDMA-EQ22-V2. p/fck = 0. Column is reinforced with 10-25Φ + 4-16Φ bars with total Asc = 5.weak beam condition
= 706 / 377 = 1. p/fck = 3.055 kN Hence.e.235 x 20 x 300 x 500 x 500 / (1x 106) = 353 kN-m ∑Mc = 353 + 353 = 706 kN-m ∑Mb = 377 kN-m (Maximum moment resistance is considered) As shown in Figure 6. (Clause 8.714 mm2. requirement of strong column-weak beam condition as per draft revision of IS 13920:1993 is satisfied
Example 6 /Page 52
.87 > 1.4 Check for Earthquake in X.3 of IS 13920 proposed draft)
6.105.1 Joint Shear
The joint shear equilibrium is shown in Figure 6. respectively. Vcol C2 T1
Shear strength of joint confined on three faces or on two opposite faces.direction 6..4.
M c = 0.714 x100 / (300 x 500) = 3. VJoint
Figure 6.0 x
20 x 300 x 500 /1. as per draft revision IS 13920:1993
= 1.235. we get
Figure 6.4. Referring to Pu chart 44 of SP: 16.8%.8 /20 = 0. corresponding to = f ck bD 0.19 and d’/D = (40 +25 /2) / 500 = 0.
25 x 415 /1.0
Example 6 /Page 53
.13 for sway to right and left.4 ⎜ s ⎜ ⎟ hst ⎝ ⎠ ⎛ 247 + 180 ⎞ = 1 .25 f y
Vcol Vcol = 1.15 show the development of forces in the joint due to beam reinforcement.14 Free body diagram of the joint
Vcol Vcol = 1.4 × ⎜ ⎟ 3 ⎝ ⎠ = 200 kN
MA B P
Figure 6.11.432 x1. respectively.859 x 1.25 x 415 /1. respectively.4 Mh + Ms hst
C2 = T2
T = A x 1.25 f y
T = A st x 1.000 = 964 kN = C2 Referring to Figure 6.25 x 415 /1.859 x1.25 x fy = 1.12 Column with sway to right
T2 = A st x 1.4 Mh + Ms hst
T1 = Ast x 1.14 and 6.25 fy 1 C1 = T 1 Mh
Vcol P
Figure 6.000 = 742 kN = C2
Figure 6.25 f y
Figure 6.000 = 964 kN = C1 T2 = Ast x 1.200 = 1.432 x 1.25 x 415 /1. Refer Figures 6. VJoint = T1 + C2 – Vcol = 964 + 742 .15 Free body diagram of the joint
T1 = Ast x 1.25 x fy = 1.12 and 6.506 kN
Figures 6.Examples on 13920 Shear due to formation of plastic hinges in beams The column shear is evaluated as below.25 x fy = 1.
IITK-GSDMA-EQ22-V2.13 Column with sway to left
⎛ M + Mh ⎞ ⎟ Vcol = 1.000 = 742 kN = C1 T2= Ast x 1. for sway to right and left.25 x fy = 1.
P Vcol C2 = T 2 Ms T = A st x 1.
6.5. This will reduce the steel required in the beam and hence will reduce the joint shear.
IITK-GSDMA-EQ22-V2. it is advisable to increase depth of the beam. i) Increase the column section so that the joint area is increased.0 x Hence. respectively.2.2 Check for Joint Shear Strength
The effective width calculations for the joint are based on Figure 6.15.3.0
Example 6 /Page 54
.1.503 kN
The hogging and sagging moment capacities are evaluated as 247 kN-m and 188 kN-m.000
= 604 kN << 1. 6.5.5. not safe. In case of depth restriction on the beam. bj = bb + h /2 = 300 + 300 /2 = 450 mm bj = bc = 500 mm Take bj = 450 mm h = full depth of column = 300 mm
Mc B P Ms A
Figure 6. If this option is adopted.16 Effective width of joint
Shear strength of joint not confined (bc > ¾ bb on both opposite faces) as per Clause 8. the joint is not safe. iii) Increase the grade of concrete. (Clause 7. The limiting moment capacity of the column calculated using SP: 16 is 212 kN-m As shown in Figure 6.99 < 1.2 and 6. the beam-column joint is checked for strong column-weak beam condition.0 = 1.
6. This will also reduce the main longitudinal steel requirement in the column owing to larger column size. In such cases the following three alternatives can be tried separately or in combination. Member forces are taken as calculated earlier
20 x 450 x 300 /1.17.3 of draft revision of IS 13920:1993 Shear strength = 1. increase in beam width can be considered if the difference between the shear strength of joint and joint shear is small.4. This option will increase the shear strength of joint and also reduce the steel required in columns. not ok.5 Revision
As can be seen from the checks in sections 6.5.1 of IS 13920 proposed draft)
Strong column-weak beam condition is not satisfied Hence. ii) Increase the size of the beam section.Examples on 13920 6.2.17 Check for strong column-weak beam condition ∑Mc = 212 + 212 = 424 kN-m ∑Mb = 247 + 180 = 427 kN-m
= 424/427 = 0. It is proposed to increase the column size from 300 x 500 to 400 x 500 and the beam depth from 600 mm to 750 mm and 500 mm to 600 mm for the transverse and longitudinal beams respectively.
231 mm2.17.859 x 1. 0.4 × ⎜ ⎟ ⎝ 3 ⎠ = 173 kN Max. 0.206 mm2. 0. T1 = Ast x 1. As per analysis results. T1.
⎛ M + Ms Vcol = 1. Hence. 1. In practice the structure may be reanalyzed. force developed in the top bars. revise the main longitudinal steel to 1420φ bars (4396 mm2. Using SP: 16. ok.59% steel) and 3-16φ + 1-16φ at bottom (804 mm2.T o p ste el 4 -1 6 # .0 = 1.12φ at top (1. The revised reinforcement details are shown in Figure 6.25 x fy = 1. Hence. required moment capacity for column is Mc = 415/2 = 208 kN-m in transverse direction and 449/ 2 = 225 kN-m in longitudinal direction.91% steel) and 3-16φ + 2-20φ at bottom (1.18 Revised reinforcement details for column and beams. 2.2. (Clause 7. the moment capacity of the beam section is calculated as Mh = 265 kN-m and Ms = 184 kN-m.4 ⎜ s ⎟ ⎜ h ⎟ ⎝ st ⎠ ⎛ 371 ⎞ = 1.0 x
20 x 400 x 500 /1. the column size now requires 14-16φ bars (2.4 ⎜ h ⎜ hst ⎝ ⎞ ⎟ ⎟ ⎠
Figure 6.000 = 964 kN Joint shear is calculated as VJoint = T1 – Vcol = 964-173 = 791 kN bJ = bb +h/2 = 300 + 500 /2 = 550 mm or bJ = bc = 400 mm Take bJ = 400 mm h = full depth of column = 500 mm Shear strength = 1.2% steel).41% steel) as main longitudinal steel. respectively.0
Example 6 /Page 55
. ⎛M ⎞ V col = 1 . 0.39% steel).B o ttom ste e l)
Check for Earthquake in Y Direction Column sway to right
The column shear is evaluated as below. respectively.000
1 4 -2 0 #
= 894 kN > 791 kN Hence.07.25 x 415 /1.859 mm2. Similarly. Using SP: 16.
IITK-GSDMA-EQ22-V2. The transverse beam is redesigned to 300 x 750 with 4-16φ + 3-20 φ +1.Examples on 13920
without reanalysis of the structure.
L o ng itud ina l be a m 30 0 x 7 50 (6-16 # . The ∑Mc required in transverse direction is 371 x 1.60% steel). The value of p/fck = 1.1 of IS 13920 proposed draft) Using SP-16.1 (226 + 208) = 477 kN-m in longitudinal direction. the longitudinal beam is redesigned to 300 x 750 with 3-16φ + 3-16φ at top (1.
T ran s ve rs e b e a m 3 0 0 x 75 0 (3 -2 0 # + 4 -1 6 # + 1 -1 2# T o p s te e l 3 -1 6 # + 2 -2 0 # .41/20 = 0. the moment capacity of the beam is calculated as Mh = 371 kN-m and Ms = 297 kN-m.B o tto m ste e l)
Check for Earthquake in X direction Column sway to right
The column shear is evaluated as below.1 = 408 kN-m and 1.814 mm2. the steel required to get the above moment capacity of column is calculated as 2%.
25 x 415 /1.000 ⎞ ⎜ ⎜ 1.600 mm2 Assuming 10 diameter links. ok.Examples on 13920
⎛ 265 + 184 ⎞ = 1. 3/4 x 500 = 375 mm) in one direction.25 x fy = 804 x1.4.00.206 x1.4.18 × S × 240 × 20 ⎛ 2.600 − 1⎟ ⎟ fy ⎝ ⎠
20 x 500 x 400 /1.000
= 894 kN > 714 kN Hence.25 x 415 /1.18 x S × h × f ck fy
⎛ Ag ⎞ ⎜ − 1⎟ ⎜A ⎟ ⎝ k ⎠
h = longer dimension of the rectangular confining measured to its outer face = (500 – 40 – 40 – 20) / 2 + 10 x 2 + 20 = 240 mm Ag = 500 x 400 = 2.54 =
S = 112 mm Provide 10Φ confining links @ 100 c/c in the joint
Example 6 /Page 56
.314 = 714 kN bJ = bb + h /2 = 300 + 500 /2 = 550 mm or bJ = bc = 500 mm Take bJ = 500 mm h = full depth of column = 400 mm Shear strength = 1. (Clause 7. Force developed in the top bars T1 = Ast x 1.0 = 1.25 x fy = 1.6 Confining Links
The column dimensions have been revised to 400 x 500.2 of IS 13920:1993.49.4 × ⎜ ⎟ 3 ⎠ ⎝ = 314 kN
IITK-GSDMA-EQ22-V2.000 = 403 kN = C2 VJoint = T1 + C2 – Vcol = 626 + 403 . which is less than ¾ of column width (i. 400 / 4 = 100 mm But need not be less than 75 mm nor more than 100 mm.300)/2 = 100 mm of concrete is exposed on either side of beam As per Clause 8.50. 6. .8 of IS 13920:1993) Ash =
0.6 of IS 13920:1993) The area of cross section Ash of the bar forming rectangular hoop to be used as special confining reinforcement shall not be less than (Clause 7.49.000 mm2 Ak = (500-2 x 40 +2 x 10) x (400.0 x
b y ≥ 3/4 hy
Figure 6.000 = 626 kN = C1 T2 = Ast x 1. and the width of beam is 300 mm. since the joint is not confined by beams framing into its two vertical faces and also since the width of the longitudinal beam is less than ¾ of the column width.e. Ash = 78.19 Confinement of joint concrete by beams
The spacing of links used as special confining reinforcement shall not exceed: (i) (ii) ¼ of minimum column dimension i. An offset of (500 . special confining reinforcement is required in the joint.54 mm2 78.2 x 40 +2 x 10) = 440 x 340 = 1.
IITK-GSDMA-EQ22-V2.3.1-1.2 Column location in elevation
The details of the column and beam reinforcement meeting at the joint are shown in Figure 7.4)
The joint of column marked in Figure 7.1 for Frame 2 is considered for design.1 and 7. 1 Plan of building (All dimensions in meters)
Section Transverse beam 500
Longitudinal beam Plan
Example 7 /Page 57
Figure 7. The plan of the building and the sectional elevation of a typical RC frame are shown in Figures 7. Problem Statement:
Detailed design as per draft revision of IS 13920:1993 of an interior roof joint in an intermediate RC frame is explained here as per IS 13920 (proposed draft) for the ground plus four storey RC office building of Example-1 (Refer Figures 1.Examples on 13920
Example 7 – Interior Beam-Column Roof Joint Design for Zone-V
4 Mh + Ms hst 2
Figure 7.e.25 is to account for the actual ultimate strength being higher than the actual yield strength (Draft revision IS 13920:1993)
The column shear is evaluated as explained below.25 x 415 /1. 0.B o tto m s te e l)
Mh + Ms hst 2
Figure 7.7 and 7.510%) at top and 4-12Φ (452 mm2 i. i. 0..6 Column with sway to left
For both the above cases.8 show the development of forces in the joint due to beam reinforcement.4 shows the joint shear equilibrium. for sway to right and to left respectively. The longitudinal beam of size 300 x 500 is reinforced with 6-12Φ (678 mm2 i. 0.Examples on 13920
T ra n s v e rs e b e a m 3 0 0 x 6 0 0 (7 -1 2 # . (Refer Figures 7.5 Column with sway to right
The transverse beam of size 300 x 600 is reinforced with 7-12Φ (791 mm2 .
Figure 7. respectively.0
Example 7 /Page 58
Vcol Vcol = 1.T o p s te e l 4 -1 2 # .3 Reinforcement details for column and beams.4 × ⎜ ⎟ ⎝ 3/ 2 ⎠
= 207 kN
Figures 7.T o p s te e l 4 -1 2 # .4
L o n g itu d in a l b e a m 3 0 0 x 5 0 0 (6 -1 2 # .48%) at top and 4-12Φ (452 mm2 . The hogging and sagging moment capacity is 105 kN-m and 66 kN-m. i.000 = 410 kN = C1 The factor 1.e.
IITK-GSDMA-EQ22-V2.34%) at bottom.4 Joint shear
T1 = Ast x 1.25 x fy = 791 x 1.e.27%) at bottom.4 ⎜ s ⎜ h /2 ⎟ ⎟ st ⎝ ⎠ ⎛ 139 + 83 ⎞ = 1 . respectively.5 and 7.6 for sway to right and left condition respectively). The hogging and sagging moment capacity is evaluated as 139 kN-m and 83 kN-m.B o tto m s te e l) 1 0 -1 2 # + 4 -1 6 #
Vcol = 1. Vcol C2 T1
⎛ M + Mh ⎞ Vcol = 1. 0.e.. Force developed in the top bars VJoint
4.25 fy 1 C1 = T 1 Mh
Figure 7.8 Free body diagram of the joint
Force developed in the bottom bars T2 = Ast x 1.25 fy
Figure 7.e.25 x 415 /1.2 x Hence.000 = 235 kN = C2 Referring to Figure 7.048 It is conservative here to calculate the moment capacity of the column with zero axial loads.2 = 1.
T = A x 1.9.967 / 20 = 0.3 of draft revision of IS 13920:1993 Shear strength = 1.3.25 x fy = 452 x 1. ok.000
= 1. as per Clause 8.3 Check for flexural strength ratio
The hogging and sagging moment capacity of the transverse beam is evaluated as 139 kN-m and 83 kN-m. The column is reinforced with 10 .25 fy 2 Mh C2 = T 2
C1 = T 1 Ms
Figure 7. The calculation of the effective width and the design shear strength of the joint is based on the draft revision of IS 13920:1993 The effective width of the joint is the lesser of: i) bj = bb + 0.1.073 kN > 645 kN
7.934 mm2 i.967%.9 Effective widths for joint
bj = bb + h /2 = 300 + 500 /2 = 550 mm h = full depth of column = 500 mm bj = bc = 400 mm Take effective width of joint as 400 mm Effective area of joint resisting shear = Ac = bj h Shear strength of joint confined on two opposite faces. respectively.5 x h ii) bj = bc
20 x 400 x 500 /1. p/fck = 0.12Φ + 4 . 1.Examples on 13920
C2 = T 2 Ms T = A st x 1.934 x 100 / (400 x 500) = 0.7 Free body diagram of the joint
T = A st x 1.16Φ bars with total Asc = 1.3.25 fy
T = A st x 1. In actual practice it is desirable to take minimum
IITK-GSDMA-EQ22-V2. VJoint = T1 + C2 – Vcol = 410 + 235-207 = 438 kN
7.2 Check for joint shear strength
The effective width provisions for joints are shown in Figure 7.0
Example 7 /Page 59
corresponding to Pu = 0 . p/fck = 0. requirement of strong column-weak beam condition as per draft revision of IS 13920:1993 is not satisfied.Examples on 13920
Pu obtained f ck bD f ck bD from different load combinations.4.0 x Hence.05 x 20 x 400 x 500 x 500 / 1x10 = 100 kN-m
The joint shear is evaluated maximum T1 and minimum Vcol.3 of IS 13920 proposed draft)
7.3 Check for Joint Shear Strength
The effective width of the joint is evaluated as: bj = bb + h /2 = 300 + 400 /2
= 500 mm h = full depth of column = 400 mm bj = bc = 500 mm Take bj = 500 mm Shear strength of joint not confined as per Clause 8.4 ⎜ s ⎜ ⎟ hst ⎝ ⎠ ⎛ 105 + 66 ⎞ = 1 .1
20 x 500 x 400 /1.25 x 415 /1. It is conservative here to calculate moment capacity of column with zero axial loads.weak beam condition.25 x fy = 678 x1.10 Check for strong column .6.0 = 1.
= 100/222 = 0.05.4 × ⎜ ⎟ ⎝ 3/ 2 ⎠ = 160 kN
The hogging and sagging moment capacity of the longitudinal beam is 105 kN-m and 66 kN-m.10.
⎛ M + Mh ⎞ ⎟ Vcol = 1.0
Example 7 /Page 60
.1. we get
corresponding to actual
Referring to Figures 7.000
= 894 kN > 587 kN
The joint equilibrium is shown in Figure 7.25 x fy = 452 x1.4 Check for Flexural Strength Ratio
Figure 7. VJoint = T1 + C2 – Vcol = 352 + 235 -160 = 427 kN
As per Figure 7.000 = 352 kN = C1 T2 = Ast x 1. (Clause 8.105.000 = 235 kN = C2
Mu f ck bD
= 0.4. we get.weak beam condition ∑Mc = 100 kN-m ∑Mb = 139 + 83 = 222 kN-m The ratio of
Hence.3 of draft revision of IS 13920:1993 Shear strength = 1. ok. for both the cases.4.45 < 1.25 x 415 /1. respectively.00 at AB.7 and 7.
Mu = 0.05 and f ck bD d’/D = (40 + 25 /2) / 500 = 0.
7.5 and 7. Referring to chart 44 of SP: 16. In actual practice it is desirable to take minimum
IITK-GSDMA-EQ22-V2. T1 = Ast x 1.
Shear Due to Plastic Hinge in Beam
Referring to Figure 7.8. the joint is checked for strong column .
Example 7 /Page 61
.1 fck.3 and 7.300) / 2 = 100 mm.718 mm2.05
Mu = 0.Bottom steel)
= 80/171 = 0. the structure may be reanalyzed. The revised reinforcement details are shown in Figure 7. Hence revise the main longitudinal steel to 8-20φ+6-16φ bars (3. Hence.1
Figure 7.Top steel 4-12 # . (Clause 7.Top steel 4-12 # .
Transverse beam 300 x 600 (7-12 # .4. 1.4.5.1 of IS 13920 proposed draft) Using SP: 16.1 = 244 kN-m and 1. Hence required moment capacity for the column is Mc = 244 kN-m in the transverse direction and 188 kN-m in the longitudinal direction.e.6 and 7.4.13125. In practice. corresponding to = 0. p/fck = 0.e.12).8 of IS 13920: 1993 is required.4.00 at f ck bD AB. The redesigned column section is expected to satisfy the flexural strength check. which is less than 3/4 width of column (i.
IITK-GSDMA-EQ22-V2. In such cases it is recommended to either increase the column section or the reinforcement or both so that ∑Mc is increased It is proposed to increase the reinforcement in the column.11 Revised reinforcement details for column and beams
Hence.4. The spacing of hoops used as special confining reinforcement shall not exceed: (i) (ii) ¼ of minimum column dimension i.11.6 of IS 13920:1993)
As can be seen from the checks in section 7. The maximum column offset on either side of the framing beam is (500 . The ∑Mc required in the transverse direction is 222 x 1.05 x 20 x 400 x 400 x 500 /1x106 = 80 kN-m ∑Mc = 80 kN-m ∑Mb = 105 + 66 = 171 kN-m The ratio of
Longitudinal beam 300 x 500 (6-12 # . (Clause 7. adequate confinement can be assumed. 3/4 x 500 = 375 mm).05 and d’/D = (40 + 25/2) /400 = 0.. a few load combinations may give axial stresses less than 0.Bottom steel) 8-20 # + 6-16 #
= 0. the steel required to get the above moment capacity of the column is calculated as 1.1 x 171 = 188 kN-m in the longitudinal direction. strong column weak beam condition is not satisfied. (Draft revision of IS13920: 1993) In this case the column dimensions are 400 x 500.47 < 1. the joint is not safe.2.4.8%. The width of the beam is 300 mm. if the beam width is at least 3/4 of the column width and if no more than 100 mm of column offset is exposed on either side of the beams. special confining reinforcement as per provisions of Clause 7. 400 / 4 =100 mm But spacing not be less than 75 mm nor more than 100 mm. (Refer Figure 7. Member forces are taken as calculated earlier without reanalysis of the structure. we get
While redesigning the column. (Clause 7.86% steel).Examples on 13920
Pu obtained f ck bD f ck bD from different load combinations.2. The section then needs to be checked for flexure load combinations.1 of IS 13920 proposed draft)
7.5 Re-design of Column
Confining Links
In case of an internal joint like the one being designed where beams frame into all vertical faces of the joint. Referring Chart Pu 44 of SP: 16.
00.496 mm2 Ash = 78.54 =
0.40 – 40 .46.54 mm2 Substituting we get 78.496 − 1⎟ ⎟ fy ⎝ ⎠
b y ≥ 3/4 hy b x ≥ 3/4 hx
Figure 7.18 × S × 236 × 20 ⎛ 2.12 ) / 2 + 10 x 2 +12 = 186 mm Ag = 400 x 500 = 2.0
Example 7 /Page 62
.00.000 mm2 Ak = (400 -2 x 40 + 2 x 10) x (500.46.18 × S × h × f ck fy
S = 116 mm Provide 10Φ confining links @ 100 c/c in the joint.2 x 40 + 2 x 10)
= 340 x 440 = 1.000 ⎞ ⎜ ⎜ 1.4.
(Clause 7.12 Confinement of joint concrete by beams The area of cross section Ash of the bar forming rectangular hoop to be used as special confining reinforcement shall not be less than Ash =
0.8 of IS 13920:1993) h = longer dimension of the rectangular confining stirrup measured to its outer face
IITK-GSDMA-EQ22-V2.Examples on 13920
= (500 – 40 – 40 + 12 ) /2 + 10 x 2 +12 = 236 mm
or = (400 .
The joint of column marked in Figure 8.
Figure 8.2 Column location in elevation
IITK-GSDMA-EQ22-V2.1 and 8.2. The plan of the building and the sectional elevation of a typical RC frame are shown in Figures 8.1-1.1 for Frame 2 is considered for design. Problem Statement:
Detailed design as per draft revision of IS 13920:1993 of an exterior roof joint in an intermediate RC frame is explained for the ground plus four storey RC office building of Example-1 (Refer Figures 1.0
Example 8 / Page63
. 1 Plan of building (All dimensions in meters)
600 500 300 300 Longitudinal beam Transverse beam 500 300 300 Plan Longitudinal beam
Figure 8.Examples on 13920
Example 8 — Exterior Beam-Column Roof Joint Design for Zone V
4 × ⎜ ⎟ ⎝ 3/ 2⎠
= 113 kN
Figures 8.Bottom steel)
4-16 # + 4-12 #
M Vcol = 1.Bottom steel)
The transverse beam of size 300x600 is reinforced with 6-12Φ (678 mm2 i.7 and 8. respectively.26%) at bottom.Top steel 4-12 # . 0.3.
Column Shear The column shear is evaluated as explained below. The longitudinal beam of size 300 x 500 is reinforced with 4-12Φ (452 mm2 i. for sway to right and to left. 0. VJoint
Figure 8.e.6 for sway to left
and right conditions respectively).Examples on 13920
8.1 Joint Shear
The joint shear equilibrium is shown in Figure 8.4 ⎜ h ⎜h ⎝ st ⎞ ⎟ ⎟ ⎠
8. (Refer Figures 8.e. The hogging and sagging moment capacity is evaluated as 67 kN-m and 52 kN-m. respectively.e.3 Check for Earthquake in Y Direction
8. 0.e. The hogging and sagging moment capacity is evaluated as 121 kN-m and 83 kN-m.4 ⎜ ⎜ h /2⎟ ⎟ ⎝ st ⎠ ⎛ 83 ⎞ = 1 .5 Column with sway to left
⎛ Ms ⎞ Vcol = 1.
Vcol Mh Vcol = 1.5 and 8.27%) at bottom.4 Joint shear IITK-GSDMA-EQ22-V2.3.4 Vcol T1
⎛ 121 ⎞ = 1 .
Transverse beam 300 x 600 (6-12# .4 × ⎜ ⎟ ⎝ 3/ 2 ⎠ = 77 kN
Longitudinal beam 300 x 500 (4-12 # .Top steel 3-12 # .4 hst 2
Figure 8. 0.2 Design Data
The details of the column and beam reinforcement meeting at the joint are shown in Figure 8.8 shows the development of forces in the joint due to beam reinforcement. respectively.3 Reinforcement details for column and beams.4 h s st 2
Figure 8.34%) at top and 3-12 (339 mm2 i.0 Example 8 / Page 64
.41 %) at top and 412Φ (452 mm2 i.6 Column with sway to right
⎛M Vcol = 1.
Figure 8.83%.3 of draft revision of IS 13920:1993 Shear strength = 1.Examples on 13920
T = Ast x 1. VJoint = T1 – Vcol = 352-77 = 275 kN
= 670 kN > 352 kN
8.0 = 1.0 x Hence.256 mm2 i. respectively.4. (Clause 8.3 Check for Flexural Strength Ratio
The hogging and sagging moment capacity of the transverse beam is evaluated as 121 kN-m and 83 kN-m.25 fy
T = A x 1. In actual practice it is desirable to take minimum Pu Mu corresponding to actual obtained 2 f ck bD f ck bD
8.25 x 415 /1.25 x fy = 678 x1.7 Free body diagram of the joint
Force developed in the top bars T1 = Ast x 1. The column is reinforced with 4-16Φ + 4-12Φ bars with total Asc = 1.8 Free body diagram of the joint
Max developed force in the bottom bars T1 = Ast x 1.2 Check for joint shear strength
The effective width provisions for joints are shown in Figure 8.1.3 of IS 13920 proposed draft)
f ck Ac 20 x 300 x 500 /1. Referring to Figure 8.25 fy 1 C1 = T 1 Mh
= bb + 0.9 The calculation of the effective width and the design shear strength of the joint is based on the draft revision of IS 13920:1993 The effective width of the joint is the lesser of:
IITK-GSDMA-EQ22-V2.25 x fy = 452 x 1.256 x 100 / (300 x 500) = 0.0
Example 8 / Page 65
.000 = 235 kN.e.83 / 20 = 0.9 Effective widths for joint
bj = bb +h /2 = 300 + 500 /2 = 550 bj = bc
Figure 8. p/fck = 0.25 x 415 /1.000 = 352 kN
Take effective width of joint as 300 mm h = full depth of column = 500 mm Effective area of joint resisting shear = Ac = bj h Shear strength of joint not confined as per Clause 8. ok. 1.3.3.042 It is conservative here to calculate the moment capacity of the column with zero axial loads.
13 Column with sway to left Joint Shear
8.06 x 20 x 300 x 500 x 500 /1x 106 = 90 kN-m ∑Mc = 90 kN-m ∑Mb = 121 kN-m (Maximum moment of resistance of the beam is considered) As per Figure 8.12.25 x fy = 452 x 1.11 Joint shear
Referring to figure 8.15 show the development of forces in the joint due to beam reinforcement. Referring to chart 44 of SP: 16.000
IITK-GSDMA-EQ22-V2.105.4 ⎜ s ⎜ h /2 ⎟ ⎟ st ⎝ ⎠ ⎛ 67 + 52 ⎞ = 1 .
M = 0.8..
⎛ M + Mh ⎞ Vcol = 1.14 and 8.4 Mh + Ms hst 2
Figure 8.10. corresponding to Pu = 0. the joint is checked for strong column .042 and f ck bD d’/D = (40 +25 /2) / 500 = 0.
= 111 kN
Vcol Vcol = 1.4 × ⎜ ⎟ ⎝ 3/ 2 ⎠
= 0.11 and 8.06. Hence.00 at AB.10 Check for strong column .4 Mh + Ms hst 2
∑M The ratio of ∑M
= 90 / 121 = 0. we get
Figure 8. Force developed in top bars.12 Column with sway to right
Figure 8. requirement of strong column-weak beam condition as per draft revision of IS 13920:1993 is not satisfied.25 x fy = 339 x 1.74 < 1.25 x 415 /1. T2 = Ast x 1. p/fck = 0. T1 = Ast x 1. not ok.25 x 415 /1.1 Shear Due to Plastic Hinge in Beam Vcol
Figures 8.weak beam condition. for sway to right and to left respectively.1
Figure 8.000 = 235 kN = C1 Force developed in bottom bars.4 Check for Earthquake in X Direction
4.0 Example 8 / Page 66
.weak beam condition
Vcol Vcol = 1.Examples on 13920
from different load combinations.
Referring to charts 45/46 of SP: 16.16 Effective width of joint
8.3 of IS: 13920 proposed draft)
Figure 8.256 mm2 i.25 fy
MB A P
Figure 8. ok.25 fy 2 Mh C2 = T 2
C =T 1 1
T = Ast x 1.0
.Examples on 13920
= 176 kN = C2
Shear strength of joint not confined (bc > ¾ bb on two opposite faces) as per Clause 8.4.e.11. respectively.055 x 20 x 300 x 300 x 500 /1x 106 = 50 kN-m ∑Mc = 50 kN-m
Example 8 / Page 67
IITK-GSDMA-EQ22-V2. 1.83%. p/fck = 0.16. we get
Figure 8.256 x100 / (300 x 500) = 0.3 of draft revision of IS 13920:1993
T = Ast x 1. In actual practice it is desirable to take minimum Pu Mu corresponding to actual obtained 2 f ck bD f ck bD from different load combinations.042 It is conservative here to calculate moment capacity of column with zero axial loads.111 = 300 kN
8.25 fy 2 MB A P Sway to right V col
Shear strength = 1. p/fck = 0. corresponding to Pu = 0.
M = 0.000
= 603 kN > 300 kN (Clause 8.042 and f ck bD d’/D = (40 +25 /2) / 300 = 0.
f ck Ac 20 x 450 x 300 /1.055. The effective width of the joint is evaluated as: bj = bb + h/2 = 300 + 300 /2 = 450 mm bj = bc = 500 mm Take bj = 450 mm h = full depth of column or full depth of beam = 300 mm.0 = 1.00 at AB. VJoint = T1 + C2 – Vcol = 235 + 176 . The column is reinforced with 4-16Φ + 4-12Φ bars with total Asc = 1.15 Free body diagram of the joint
Referring to the Figure 8.4.14 Free body diagram of the joint
T = Ast x 1.25 fy 1 C =T 1 1 M h
C =T 2 2 M s T =Ast x 1.83 / 20 = 0.175.3 Check for Flexural Strength Ratio
The hogging and sagging moment capacity of the longitudinal beam is evaluated as 67 kN-m and 52 kN-m.2 Check for joint shear strength
The effective width calculations for joint are explained in Figure 8.0 x Hence.
the required moment capacity for the column is Mc = 133 kN-m in the transverse direction and 131kN-m in the longitudinal direction.Examples on 13920
∑Mb = 67 + 52 = 119 kN-m (Maximum moment resistance is considered) As per Figure 8. The column offset on either side of beam is equal to (500-300) /2 = 100 mm. 2. revise the main longitudinal steel to 8-20φ + 6-16φ bars (3.19 Confinement of joint concrete by beams
The spacing of hoops used as special confining reinforcement shall not exceed: (i) ¼ of minimum column dimension i.3 and 8.6 and 7. Hence.. Member forces are taken as earlier without reanalysis of the structure. the joint is not safe.
8. 3/4 x 500 = 375 mm) in the longitudinal direction. the joint is checked for strong column . The ∑Mc required in the transverse direction is 121 x 1. a few load combinations may give axial stresses less than 0. (Clause 7. The revised reinforcement details are shown in Figure 8.Top steel 4-12 # .e. In practice the structure may be reanalyzed.
IITK-GSDMA-EQ22-V2.18.weak beam condition.718 mm2.1 fck.Bottom steel)
8-20 # + 6-16 #
Figure 8.8 of IS 13920: 1993 is required. The section needs to be checked for flexure for these load combinations.e.1 x 119 = 131 kN-m in longitudinal direction.17 Check for strong column .5 Re-design
As can be seen from the checks in sections 8.4.4% steel.5.weak beam condition
= 50 / 119 = 0. the steel required to get the above moment capacity of the column is calculated as 2.4.6 Confining Links
In this case.. Hence.e.Top steel 3-12 # . requirement of strong column-weak beam condition as per draft revision of IS 13920:1993 is not satisfied.2.4. Using SP: 16. Hence. 3/4 x 300 = 225 mm) in the transverse direction but less than ¾ of column width (i. In such cases it is recommended to either increase the column section or the reinforcement or both so that ∑Mc is increased. It is proposed to increase the reinforcement in the column.18 Revised reinforcement details for column and beams. Also.47% steel). The width of the beam is 300 mm which is more than 3/4 width of the column (i.
Figure 8. While redesigning the column.42 < 1.1 of IS: 13920 proposed draft)
Figure 8.3.0
Example 8 / Page 68
.1 = 133 kN-m and 1. special confining reinforcement as per Clauses 7. not ok.
8. since the column is confined by beams framing into its vertical faces from three sides only.
The redesigned column section satisfies the flexural strength check.17..Bottom steel)
Hence.. the column dimensions have been revised to 300 x 500.1
Longitudinal beam 300 x 500 (4-12 # .
Transverse beam 300 x 600 (6-12# .
IITK-GSDMA-EQ22-V2.6 of IS 13920:1993) The area of cross section Ash of the bar forming the rectangular hoop to be used as special confining reinforcement shall not be less than (Clause 7.8 of IS 13920:1993) Ash =
0.Examples on 13920
300 / 4 = 75 mm (ii) But spacing not be less than 75mm nor more than 100 mm.0
Example 8 / Page 69
.600 − 1⎟ ⎟ fy ⎝ ⎠
Solving we get.05.000 ⎞ ⎜ ⎜ 1.50.4. 50.400 mm2 78.18 × S × 240 × 20 ⎛ 1.40 +10 +10) = 240 mm Ag = 300 x 500 = 1.54 =
0. (Clause 7.40. S = 90 mm.18 × S × h × f ck fy ⎛ Ag ⎞ ⎜ − 1⎟ ⎜A ⎟ ⎝ k ⎠
h = longer dimension of the rectangular confining measured to its outer face = (300 .000 mm2 Ak = (300-2 x 40 +2 x 10) x (500. Adopt 10 mm diameter bar for special confining reinforcement at a spacing of 90 c/c.2 x 40 + 2 x 10) = 240 x 440 = 92.
Example 8 / Page 70
.No 1.1
The problem and the solution have been adopted from Medhekar M S and Jain S K. In practice all other combinations should also be considered. Problem Statement:
Design a shear wall for a two-storey building shown in (Figure 9. “Seismic Behavior and Detailing of R C Shear Walls. 2.Shear Wall Design for a Building in Zone III 9. “The Indian Concrete Journal”. S.5 4830. Roof
tw=230
Figure 9.1– Shear wall details for example
IITK-GSDMA-EQ22-V2.8. Vol. 451-457”.Examples on 13920
Example 9 . Load Case (DL+LL) Earthquake Bending Moment k -577.9 Axial Force 1922.7 699. The unfactored forces in the panel between the ground level and first floor are obtained by analysis as. No.7 Shear Force k 19. part II: Design and Detailing. The example shows design for load combination 1. September 1993.9 255. 67.2(DL + LL +EL) only.The materials are M20 concrete and Fe415 steel.1).
Vu =1.
9.2 × 255.1. Therefore. Thus. shear to be resisted by horizontal reinforcement is Vus = 552 kN.7) -849] = -587 kN. v =1. The arrangement of reinforcement in the boundary element as per Figure 9.
9. Assuming short column action.466 N/mm2. Thus. As xu/lw is less than x*u/lw. the reinforcement shall be in 2 layers. Mu = 1.574 × 1845 = 1059 kN.1. one bar of 16 mm diameter should be provided per layer of reinforcement on each side of the opening.660.
v = 0.1.25%.3 Boundary elements:
The axial compression at the extreme fiber due to combined axial load and bending on the section is 6.1) = 863 kN.1.2 requires 10 mm diameter rectangular hoops to be provided at 95 mm c/c as special confining reinforcement. φ = 0. shear to be resisted by reinforcement on each side of opening is.805 N/mm2.2 × (1922. Therefore. Figure 9. horizontal reinforcement of 8mm diameter bars at 175 mm c/c in 2 layers will suffice.25%) requires this ratio to be 0. Therefore. the maximum factored compression on the boundary element is [849 + 0. The maximum factored shear force. xu/lw = 0.8 × 1922.25%) be provided in the web.233. Vuc
= 311 kN. (3). the moment of resistance of the web is
IITK-GSDMA-EQ22-V2. As tw > 200 mm.8 × 1922.9) = 6490 kNm. Vu = 863 kN.7 = 1845 kN.8% is 2953 kN. provision of minimum horizontal reinforcement (0.213 × (0. the factored axial force should be taken as Pu = 0. Therefore.
9.213 × 1. Thus.76 = 849 kN. Let minimum Therefore. i. 12 bars of 16 mm diameter will be adequate to take the compression as well as tension. as per Table 13 of IS: 456-1978. xu*/lw = 0.1 Solution:
The maximum factored bending moment on the section.2 illustrates the reinforcement details.2fck.9 +1. from equations (2). Cw. the axial load capacity of the boundary element with minimum reinforcement of 0.25%) that is interrupted by it is 690 mm2.516. (MuMuv) = 3194 kN shall be resisted by reinforcement in the boundary elements. load on web = 0. Taking depth of wall on each side of opening that is resisting shear as 1280 mm. Shear carried by concrete.5 + 4830.
τ v = 0. lw.2 × 255.e.2 × (19. Hence. obtained from equation (4) as. is 230 mm. Assuming this axial load to be uniformly distributed. The factored tension on the boundary element is [0. Let the effective depth in resisting shear be 3760 mm (=3380+380).7 + 699. The horizontal bar should be provided with development length in tension beyond the sides of the opening.4 Reinforcement around opening:
The opening is of size 1200 mm by 1200 mm. However. Axial compression will increase the moment capacity of the wall. The vertical bar should extend for the full storey height.9 + 255.9 . is 3. is 4140 mm and its web thickness. Muv = 3296 kNm.Examples on 13920
9. vertical reinforcement (0. Therefore. and (5).1.056.. provide 8 mm diameter 2-legged stirrups at 140 mm c/c on each side of opening. The remaining moment. provision of boundary elements along the wall edges is mandatory. tw. This requires the ratio Ah/Sv to be 0. The length of wall.998 N/mm2.2 Flexural strength of web:
The vertical reinforcement in the web is 0.407. Vus= 326 kN. we get λ = 0. An opening is present at section B-B. Thus. The center to center distance between the boundary elements. The area of vertical and horizontal reinforcement in the web (0.36 N/mm2. As this is greater than 0.0
Example 9 /Page 71
.2 × (577.7)] = 1406 kN. the design shear force is given by. The axial force on the boundary element due to earthquake loading is (Mu-Muv)/Cw = 3194/3.045.76 m.575. and the value of β is 0.1 Shear Design:
At section A-A.
Figure 9.2 Reinforcement details for example
Example 9 /Page 72
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