Design Example of a Six Storey Building, Sample of Reports

18 pages, 8567 words

Document No. :: IITK-GSDMA-EQ26-V3.0 Final Report :: A – Earthquake Codes IITK-GSDMA Project on Building Codes

Design Example of a Six Storey Building

Dr. H. J. Shah

Department of Applied Mechanics M. S. University of Baroda Vadodara

Dr. Sudhir K Jain

Department of Civil Engineering Indian Institute of Technology Kanpur Kanpur

•

This document has been developed under the project on Building Codes sponsored by Gujarat State Disaster Management Authority, Gandhinagar at Indian Institute of Technology Kanpur. 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. Comments and feedbacks may please be forwarded to: Prof. Sudhir K Jain, Dept. of Civil Engineering, IIT Kanpur, Kanpur 208016, email: [email protected]

•

Design Example of a Building

Example — Seismic Analysis and Design of a Six Storey Building

Problem Statement:

A six storey building for a commercial complex has plan dimensions as shown in Figure 1. The building is located in seismic zone III on a site with medium soil. Design the building for seismic loads as per IS 1893 (Part 1): 2002.

General

1. The example building consists of the main block and a service block connected by expansion joint and is therefore structurally separated (Figure 1).

Analysis and design for main block is to be performed. 2 The building will be used for exhibitions, as an art gallery or show room, etc., so that there are no walls inside the building. Only external walls 230 mm thick with 12 mm plaster on both sides are considered. For simplicity in analysis, no balconies are used in the building. 7. Sizes of all columns in upper floors are kept the same; however, for columns up to plinth, sizes are increased. 8. The floor diaphragms are assumed to be rigid. 9. Centre-line dimensions are followed for analysis and design. In practice, it is advisable to consider finite size joint width. 10. Preliminary sizes of structural components are assumed by experience. 11. For analysis purpose, the beams are assumed to be rectangular so as to distribute slightly larger moment in columns. In practice a beam that fulfils requirement of flanged section in design, behaves in between a rectangular and a flanged section for moment distribution. 12. In Figure 1(b), tie is shown connecting the footings. This is optional in zones II and III; however, it is mandatory in zones IV and V. 13. Seismic loads will be considered acting in the horizontal direction (along either of the two principal directions) and not along the vertical direction, since it is not considered to be significant. 14. All dimensions are in mm, unless specified otherwise.

4 pages, 1550 words

The Term Paper on Design and Analysis of a Laminated Composite Tube

A composite material is a material made from two or more constituents with significantly different physical or chemical properties. There are two main categories of constituent materials, matrix and reinforcement. At least one material of each type is required. (Composite Material). The phase distribution and geometry of the two materials have been ordered to optimise its properties, this has led ...

3. At ground floor, slabs are not provided and the floor will directly rest on ground. Therefore, only ground beams passing through columns are provided as tie beams. The floor beams are thus absent in the ground floor. 4. Secondary floor beams are so arranged that they act as simply supported beams and that maximum number of main beams get flanged beam effect. 5. The main beams rest centrally on columns to avoid local eccentricity. 6. For all structural elements, M25 grade concrete will be used. However, higher M30 grade concrete is used for central columns up to plinth, in ground floor and in the first floor.

IITK-GSDMA-EQ26-V3.0

Page 3

Design Example of a Building

(7.5,0) C1 (0,0) A F.B. 7.5 m B 15 B 18 B1 C2 B2

(15,0) C3 B3 F.B. B 21 C4 (22.5,0) X Main block B 24 A

3 pages, 1212 words

The Essay on Design of a Hotel Building

... multi storey building is a building that has multiple floors above ground in the building. Multi-storey buildings aim to increase the floor area ... IS 875 Part -2 (Design loads for buildings and structures). Overall depth of beam or slab d Effective depth ... and overnight accommodation for meeting attendees. They also provide High quality audiovisual equipments, business services, flexible seating ...

F.B.

F.B. B4

F.B.

F.B. C7 B6

4 (22.5,7.5) C8 A B 23

B 7.5 m

C5 (0,7.5) B 14 B 17 F.B. F.B. A

(7.5, 7.5) F.B. B 20 F.B. C10 B8

(15, 7.5) F.B. F.B.

Service block

Expansion joint

C 7.5 m

C9 (0,15) B 13

B7 F.B. B 16 F.B. B10

C11

(7.5,15) B 19 F.B. F.B.

(15, 15) F.B. F.B. B12 C15 (15,22.5) B 22

C12

C x z D

(22.5,15)

B11 C14 (7.5,22.5)

D C13 (0,22.5) Z C16 (22.5,22.5)

7.5 m

7.5 m (a) Typical floor plan

7.5 m

300 × 600 5 m 500 × 500

+ 31.5 m + 30.5 m Terrace 7 + 25.5 m Fifth Floor 5m

+ 30.2 m M25 + 25.2 m M25 + 20.2 m M25 + 15.2 m M25 + 10.2 m M25 + 5.2 m M25 + 1.1 m M25

5m + 20.5 m Fourth Floor 5m + 15.5 m Third Floor 5m + 10.5 m Second Floor 5m + 5.5 m First Floor 4m

0.10 0.60 0.80 0.90 0.10

5 y 4 3

5m x 5m 4.1 m 1.1 m + 0.0 m

300 × 600

2.5

+ 2.1 m Ground Floor Plinth + 0.0 Tie

2 1 Storey numbers

600 × 600

(b) Part section A-A

Figure 1 General lay-out of the Building.

IITK-GSDMA-EQ26-V3.0

Page 4

Design Example of a Building

1.1.

Data of the Example

The design data shall be as follows: Live load Floor finish Water proofing Terrace finish Location Wind load : 4.0 kN/m2 at typical floor : 1.5 kN/m2 on terrace : 1.0 kN/m2 : 2.0 kN/m2 : 1.0 kN/m2 : Vadodara city : As per IS: 875-Not designed for wind load, since earthquake loads exceed the wind loads. : As per IS-1893 (Part 1) – 2002 : 2.5 m : Type II, Medium as per IS:1893 : 200 kN/m2 : 0.9 m, assume isolated footings : Typical floor: 5 m, GF: 3.4 m : G.F. + 5 upper floors. : To be provided at 100 mm below G.L. : 0.6 m : 230 mm thick brick masonry walls only at periphery.

Earthquake load Depth of foundation below ground Type of soil Allowable bearing pressure Average thickness of footing Storey height Floors Ground beams Plinth level Walls Material Properties Concrete

All components unless specified in design: M25 grade all Ec = 5 000

7 pages, 3267 words

The Term Paper on Ground Floor Design

Third Edition Concrete industrial ground floors A guide to design and construction Report of a Concrete Society Working Party Concrete Society Report TR34- Concrete industrial ground floors Third Edition 2003 IMPORTANT Errata Notification Would you please amend your copy of TR34 to correct the following:On page 50 - symbols and page 63 - Clause 9. 11. 3, change the word "percentage" to "ratio" in ...

f ck N/mm2 = 5 000

f ck MN/m2

= 25 000 N/mm 2 = 25 000 MN/m 2 . For central columns up to plinth, ground floor and first floor: M30 grade Ec = 5 000

f ck N/mm2 = 5 000

f ck MN/m2

= 27 386 N/mm 2 = 27 386 MN/m 2 . Steel HYSD reinforcement of grade Fe 415 confirming to IS: 1786 is used throughout.

1.2.

Geometry of the Building

The general layout of the building is shown in Figure 1. At ground level, the floor beams FB are IITK-GSDMA-EQ26-V3.0

not provided, since the floor directly rests on ground (earth filling and 1:4:8 c.c. at plinth level) and no slab is provided. The ground beams are

Page 5

Design Example of a Building provided at 100 mm below ground level. The numbering of the members is explained as below. 1.2.1. Storey number from upper to the lower part of the plan. Giving 90o clockwise rotation to the plan similarly marks the beams in the perpendicular direction. To floor-wise differentiate beams similar in plan (say beam B5 connecting columns C6 and C7) in various floors, beams are numbered as 1005, 2005, 3005, and so on. The first digit indicates the storey top of the beam grid and the last three digits indicate the beam number as shown in general layout of Figure 1. Thus, beam 4007 is the beam located at the top of 4th storey whose number is B7 as per the general layout.

Storey numbers are given to the portion of the building between two successive grids of beams. For the example building, the storey numbers are defined as follows: Portion of the building Foundation top – Ground floor Ground beams – First floor First Floor – Second floor Second floor – Third floor Third floor – Fourth floor Fourth floor – Fifth floor Fifth floor – Terrace 1.2.2. Column number Storey no. 1 2 3 4 5 6 7

1.3.

Gravity Load calculations

1.3.1. Unit load calculations Assumed sizes of beam and column sections are: Columns: 500 x 500 at all typical floors Area, A = 0.25 m2, I = 0.005208 m4 Columns: 600 x 600 below ground level Area, A = 0.36 m2, I = 0.0108 m4 Main beams: 300 x 600 at all floors Area, A = 0.18 m2, I = 0.0054 m4 Ground beams: 300 x 600 Area, A = 0.18 m2, I = 0.0054 m4 Secondary beams: 200 x 600

In the general plan of Figure 1, the columns from C1 to C16 are numbered in a convenient way from left to right and from upper to the lower part of the plan. Column C5 is known as column C5 from top of the footing to the terrace level. However, to differentiate the column lengths in different stories, the column lengths are known as 105, 205, 305, 405, 505, 605 and 705 [Refer to Figure 2(b)]. The first digit indicates the storey number while the last two digits indicate column number. Thus, column length 605 means column length in sixth storey for column numbered C5. The columns may also be specified by using grid lines. 1.2.3. Floor beams (Secondary beams)

2 pages, 637 words

The Essay on End plate connection Steel beam to steel column

End plate connection (Steel beam to steel column) Introduction An end plate connection (Steel beam to steel column) is a type of connection used in steel structures. In this type of steel connection, a piece of steel known as the end plate is fixed onto a column and a steel beam is bolted to it. This type of connection is mainly used in purely steel structures. The end plate can either be ...

Member self- weights: Columns (500 x 500) 0.50 x 0.50 x 25 = 6.3 kN/m Columns (600 x 600) 0.60 x 0.60 x 25 = 9.0 kN/m Ground beam (300 x 600) 0.30 x 0.60 x 25 = 4.5 kN/m Secondary beams rib (200 x 500) 0.20 x 0.50 x 25 = 2.5 kN/m Main beams (300 x 600) 0.30 x 0.60 x 25 = 4.5 kN/m Slab (100 mm thick) 0.1 x 25 = 2.5 kN/m2 Brick wall (230 mm thick) 0.23 x 19 (wall) +2 x 0.012 x 20 (plaster) = 4.9 kN/m2 Page 6

All floor beams that are capable of free rotation at supports are designated as FB in Figure 1. The reactions of the floor beams are calculated manually, which act as point loads on the main beams. Thus, the floor beams are not considered as the part of the space frame modelling. 1.2.4. Main beams number

Beams, which are passing through columns, are termed as main beams and these together with the columns form the space frame. The general layout of Figure 1 numbers the main beams as beam B1 to B12 in a convenient way from left to right and IITK-GSDMA-EQ26-V3.0

Design Example of a Building Floor wall (height 4.4 m) 4.4 x 4.9 = 21.6 kN/m Ground floor wall (height 3.5 m) 3.5 x 4.9 = 17.2 kN/m Ground floor wall (height 0.7 m) 0.7 x 4.9 = 3.5 kN/m Terrace parapet (height 1.0 m) 1.0 x 4.9 = 4.9 kN/m 1.3.2. Slab load calculations Main beams B1–B2–B3 and B10–B11–B12 Component From Slab 0.5 x 2.5 (5.5 +1.5) Parapet Total 6.9 + 1.9 4.9 + 0 11.8 + 1.9 kN/m Typical (DL + LL) 2.5 + 0.0 0.0 + 0.0 1.0 + 0.0 0.0 + 4.0 3.5 + 4.0 kN/m2 0+0 4.9 + 0 4.9 + 0 kN/m B1-B3 B2

Component

Terrace (DL + LL)

Two point loads on one-third span points for beams B2 and B11 of (61.1 + 14.3) kN from the secondary beams. Main beams B4–B5–B6, B7–B8–B9, B16– B17– B18 and B19–B20–B21 From slab 0.5 x 2.5 x (5.5 + 1.5) = 6.9 + 1.9 kN/m Total = 6.9 + 1.9 kN/m Two point loads on one-third span points for all the main beams (61.1 + 14.3) kN from the secondary beams. Main beams B13–B14–B15 and B22–B23–B24 Component B13 – B15 B22 – B24 B14 B23 6.9 + 1.9

3 pages, 1305 words

The Essay on Load Cell System Data Cells

Biologics and Contract Manufacturing Company ensures batch control system data accuracy Precision Pharma Services, Inc. is a new business a biological products facility originally started in 1979 as part of the Greater NYC Blood Program. The FDA approved facility has been operational in the production of blood products for the past decade and has been a privately held company since 2001. With ...

Self (100 mm thick) Water proofing Floor finish Live load Total

2.5 + 0.0 2.0 + 0.0 1.0 + 0.0 0.0 + 1.5 5.5 + 1.5 kN/m2

1.3.3.

Beam and frame load calculations:

From Slab 0.5 x 2.5 (5.5 +1.5)

—-

(1) Terrace level: Floor beams: From slab 2.5 x (5.5 + 1.5) Self weight Total Reaction on main beam 0.5 x 7.5 x (16.3 + 3.8) = 61.1 + 14.3 kN. = = 13.8 + 3.8 kN/m 2.5 + 0 kN/m

Parapet Total

4.9 + 0 4.9 + 0 kN/m

4.9 + 0 11.8 + 1.9 kN/m

= 16.3 + 3.8 kN/m

Two point loads on one-third span points for beams B13, B15, B22 and B24 of (61.1+14.3) kN from the secondary beams. (2) Floor Level: Floor Beams: From slab 2.5 x (3.5 + 4.0) Self weight Total Reaction on main beam 0.5 x 7.5 x (11.25 + 10.0)

Note: Self-weights of main beams and columns will not be considered, as the analysis software will directly add them. However, in calculation of design earthquake loads (section 1.5), these will be considered in the seismic weight.

= = = =

8.75 + 10 kN/m 2.5 + 0 kN/m 11.25 + 10 kN/m 42.2 + 37.5 kN.

IITK-GSDMA-EQ26-V3.0

Page 7

Design Example of a Building

Main beams B1–B2–B3 and B10–B11–B12

Component From Slab 0.5 x 2.5 (3.5 + 4.0) Wall Total 4.4 + 5.0 21.6 + 0 0+0 21.6 + 0 B1 – B3 B2 Two point loads on one-third span points for beams B13, B15, B22 and B24 of (42.2 +7.5) kN from the secondary beams. (3) Ground level: 26.0 + 5.0 21.6 + 0 kN/m kN/m Outer beams: B1-B2-B3; B10-B11-B12; B13B14-B15 and B22-B23-B24 Walls: 3.5 m high 17.2 + 0 kN/m Inner beams: B4-B5-B6; B7-B8-B9; B16B17-B18 and B19-B20-B21 Walls: 0.7 m high 3.5 + 0 kN/m Loading frames The loading frames using the above-calculated beam loads are shown in the figures 2 (a), (b), (c) and (d).

There are total eight frames in the building. However, because of symmetry, frames A-A, B-B, 1-1 and 2-2 only are shown. It may also be noted that since LL

Two point loads on one-third span points for beams B2 and B11 (42.2 + 37.5) kN from the secondary beams.

Main beams B4–B5–B6, B7–B8–B9, B16– B17–B18 and B19–B20–B21

1 page, 366 words

The Essay on Truss Construction Fire Beams Building

On September 11, 2001 the United States was brought to its knees as group of terrorists successfully took down the World Trade Center. There were many factors that contributed to the collapse. When the World Trade Center was built, a protective fire coating was sprayed on the beams used to construct the building. This coating was used so that in case of a fire, the beams would not be affected. ...

From slab 0.5 x 2.5 (3.5 + 4.0) = 4.4 + 5.0 kN/m Total = 4.4 + 5.0 kN/m Two point loads on one-third span points for all the main beams (42.2 + 37.5) kN from the secondary beams. Main beams B13–B14–B15 and B22–B23–B24

Component

B13 – B15 B22 – B24

B14 B23

From Slab 0.5 x 2.5 (3.5 + 4.0) Wall Total —21.6 + 0 21.6 + kN/m 4.4 + 5.0 21.6 + 0 0 26.0 + 5.0 kN/m

IITK-GSDMA-EQ26-V3.0

Page 8

Design Example of a Building

61.1 + 14.3

61.1 + 14.3 kN

(11.8 + 1.9) kN/m 7001 5m 702 701 (4.9 + 0) kN/m 7002

42.2+37.5 42.2+37.5 kN

(11.8 + 1.9) kN/m 7003 703 704 (26 + 5) kN/m (21.6 + 0) kN/m 6002 602

42.2+37.5 42.2+37.5 kN

(26 + 5) kN/m 6001 5m 601

6003 603 604 (26 + 5) kN/m 5003 503 504 (26 + 5) kN/m 4003 403 404 (26 + 5) kN/m 3003 303 304 (26 + 5) kN/m 2003 203 204 (17.2 + 0) kN/m 103 1003 B3 7.5 m 104 C4

(26 + 5) kN/m 5001 5m 501 502 (26 + 5) kN/m 4001 5m 401 402 (26 + 5) kN/m 3001 5m 301 302 (26 + 5) kN/m 2001 201 4.1 m 202 (17.2 + 0) kN/m 101 1001 B1 7.5 m 102 (21.6 + 0) kN/m 2002 (17.2 + 0) kN/m 1002 B2 7.5 m C2 (21.6 + 0) kN/m 3002

42.2+37.5 42.2+37.5 kN

(21.6 + 0) kN/m 5002

42.2+37.5 42.2+37.5 kN

(21.6 + 0) kN/m 4002

42.2+37.5 42.2+37.5 kN

1.1 m

Figure 2 (a) Gravity Loads: Frame AA

IITK-GSDMA-EQ26-V3.0

Page 9

Design Example of a Building

61.1+14.3

61.1+14.3 kN

61.1+14.3

61.1+14.3 kN

61.1+14.3

61.1+14.3 kN

705

706

707

42.2+37.5 42.2+37.5 kN

605

606

607

42.2+37.5 42.2+37.5 kN

505

506

507

42.2+37.5 42.2+37.5 kN

405

406

407

42.2+37.5 42.2+37.5 kN

305

306

307

42.2+37.5 42.2+37.5 kN

205

206

207

4.1 m

(3.5 + 0) kN/m 105 106 1004 B4 7.5 m

(3.5 + 0) kN/m 107 1005 B5 7.5 m

(3.5 + 0) kN/m 1006 B6 7.5 m 108 C8

1.1 m

Figure 2(b) Gravity Loads: Frame BB

IITK-GSDMA-EQ26-V3.0

208

(4.4 + 5) kN/m 2004

(4.4 + 5) kN/m 2005

(4.4 + 5) kN/m 2006

308

(4.4 + 5) kN/m 3004

(4.4 + 5) kN/m 3005

(4.4 + 5) kN/m 3006

408

(4.4 + 5) kN/m 4004

(4.4 + 5) kN/m 4005

(4.4 + 5) kN/m 4006

508

(4.4 + 5) kN/m 5004

(4.4 + 5) kN/m 5005

(4.4 + 5) kN/m 5006

608

(4.4 + 5) kN/m 6004

(4.4 + 5) kN/m 6005

(4.4 + 5) kN/m 6006

708

(6.9+1.9) kN/m 7004

(6.9+1.9) kN/m 7005

(6.9+1.9) kN/m 7006

Page 10

Design Example of a Building

61.1 + 14.3 61.1 + 14.3 kN

(4.9 + 0) kN/m 7013 709 713 5m

42.2+37.5 42.2+37.5 kN

(11.8 + 1.9) kN/m 7014 705

(4.9 + 0) kN/m 7015

42.2+37.5 42.2+37.5 kN

(21.6 + 0) kN/m 6013 609 613 5m

42.2+37.5 42.2+37.5 kN

(26 + 5) kN/m 6014 605

(21.6 + 0) kN/m 6015

42.2+37.5 42.2+37.5 kN

(21.6 + 0) kN/m 5013 509 513 5m

42.2+37.5 42.2+37.5 kN

(26 + 5) kN/m 5014 505

(21.6 + 0) kN/m 5015

42.2+37.5 42.2+37.5 kN

(21.6 + 0) kN/m 4013 409 413 5m

42.2+37.5 42.2+37.5 kN

(26 + 5) kN/m 4014 405

(21.6 + 0) kN/m 4015

42.2+37.5 42.2+37.5 kN

(21.6 + 0) kN/m 3013 309 313 5m

42.2+37.5 42.2+37.5 kN

(26 + 5) kN/m 3014 305

(21.6 + 0) kN/m 3015

42.2+37.5 42.2+37.5 kN

209

213

4.1 m

(17.2 + 0) kN/m 1.1 m 109 113 1013 B 13 7.5 m

(17.2+ 0) kN/m 1014 B 14 7.5 m 105

205

(17.2 + 0) kN/m 1015 B 15 7.5 m 101 C1

C 13

Figure 2(c) Gravity Loads: Frame 1-1

IITK-GSDMA-EQ26-V3.0

201

(21.6 + 0) kN/m 2013

(26 + 5) kN/m 2014

(21.6 + 0) kN/m 2015

301

401

501

601

701

Page 11

Design Example of a Building

61.1 + 14.3 61.1 + 14.3 kN

61.1 + 14.3

61.1 + 14.3 kN

61.1 + 14.3

61.1 + 14.3 kN

(6.9+1.9) kN/m 7016 5m 714 710

42.2+37.5 42.2+37.5 kN

(6.9+1.9) kN/m 7017 706

42.2+37.5 42.2+37.5 kN

(6.9+1.9) kN/m 7018

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 6016 5m 614 610

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 6017 606

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 6018

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 5016 5m 514 510

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 5017 506

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 5018

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 4016 5m 414 410

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 4017 406

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 4018

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 3016 5m 314 310

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 3017 306

42.2+37.5 42.2+37.5 kN

(4.4+5) kN/m 3018

42.2+37.5 42.2+37.5 kN

214

210

206

4.1 m

(3.5 + 0) kN/m 114 110 1016 B 16 7.5 m

(3.5 + 0) kN/m 106 1017 B 17 7.5 m

(3.5 + 0) kN/m 1018 B 18 7.5 m 102 C2

1.1 m

C 14

C 10

Figure 2(d) Gravity Loads: Frame 2-2

IITK-GSDMA-EQ26-V3.0

202

(4.4+5) kN/m 2016

(4.4+5) kN/m 2017

(4.4+5) kN/m 2018

302

402

502

602

702

Page 12

Design Example of a Building Columns Total (4) Storey 1 (plinth): Walls Walls Main beams Column 0.5 x 4 x 22.5 (17.2 + 0) 0.5 x 4 x 22.5 x (3.5 + 0) 8 x 22.5 x (4.5 + 0) 16 x 0.5 x 4.1 x (6.3 + 0) 16 x 0.5 x 1.1 x (9.0 + 0) DL + LL 774 + 0 158 + 0 810 + 0 206 + 0 79 + 0 2 027 + 0 = 2 027 kN 16 x 0.5 x (5 + 4.1) x (6.3 + 0) 459 + 0 5 125 +1 013 = 6 138 kN

1.4.

Seismic Weight Calculations

The seismic weights are calculated in a manner similar to gravity loads. The weight of columns and walls in any storey shall be equally distributed to the floors above and below the storey. Following reduced live loads are used for analysis: Zero on terrace, and 50% on other floors [IS: 1893 (Part 1): 2002, Clause 7.4) (1) Storey 7 (Terrace): From slab Parapet Walls 22.5 x 22.5 (5.5+0) 4 x 22.5 (4.9 + 0) DL + LL 2 784 + 0 441 + 0 972 + 0 338 + 0 810 + 0 252 + 0 5 597 + 0 = 5 597 kN DL + LL 1 772 + 1 013 1 944 + 0 338 + 0 810 + 0 504+0 5 368 +1 013 = 6 381 kN DL + LL 1 772 + 1 013 972 + 0 774 + 0 338 + 0 810 + 0

0.5 x 4 x 22.5 x (21.6 + 0) Secondary 18 x 7.5 x (2.5 + 0) beams Main 8 x 22.5 x (4.5 + 0) beams Columns 0.5 x 5 x 16 x (6.3 + 0) Total (2) Storey 6, 5, 4, 3: From slab Walls Secondary beams Main beams Columns Total (3) Storey 2: From slab Walls Walls Secondary beams Main beams 22.5 x 22.5 x (3.5 + 0.5 x 4) 0.5 x 4 x 22.5 x (21.6 + 0) 0.5 x 4 x 22.5 x (17.2 + 0) 18 x 7.5 x (2.5 + 0) 8 x 22.5 x (4.5 + 0) 22.5 x 22.5 x (3.5 + 0.5 x 4) 4 x 22.5 x (21.6 + 0) 18 x 7.5 x (2.5 + 0) 8 x 22.5 x (4.5 + 0) 16 x 5 x (6.3 + 0)

Total

Seismic weight of the entire building = 5 597 + 4 x 6 381 + 6 138 + 2 027 = 39 286 kN The seismic weight of the floor is the lumped weight, which acts at the respective floor level at the centre of mass of the floor. 1.5. Design Seismic Load

The infill walls in upper floors may contain large openings, although the solid walls are considered in load calculations. Therefore, fundamental time period T is obtained by using the following formula:

Ta = 0.075 h0.75 [IS 1893 (Part 1):2002, Clause 7.6.1]

= 0.075 x (30.5)0.75 = 0.97 sec.

Zone factor, Z = 0.16 for Zone III IS: 1893 (Part 1):2002, Table 2

Importance factor, I = 1.5 (public building) Medium soil site and 5% damping

S a 1.36 = = 1.402 g 0.97 IS: 1893 (Part 1): 2002, Figure 2.

IITK-GSDMA-EQ26-V3.0

Page 13

Design Example of a Building Table1. Distribution of Total Horizontal Load to Different Floor Levels 1.5.1. Accidental eccentricity:

Design eccentricity is given by edi = 1.5 esi + 0.05 bi or esi – 0.05 bi

Storey Wi (kN)

hi (m)

Wihi2 x10-3

Qi =

Wi h i2 ∑ Wi h i2

Vi (kN)

7 6 5 4 3 2 1 Total

5 597 6 381 6 381 6 381 6 381 6 138 2 027

30.2 25.2 20.2 15.2 10.2 5.2 1.1

5 105 4 052 2 604 1 474 664 166 3 14 068

x VB (kN) 480 380 244 138 62 16 0 1 320

IS 1893 (Part 1): 2002, Clause 7.9.2.

480 860 1 104 1 242 1 304 1 320 1 320

For the present case, since the building is symmetric, static eccentricity, esi = 0.

0.05 bi = 0.05 x 22.5 = 1.125 m. Thus the load is eccentric by 1.125 m from mass centre. For the purpose of our calculations, eccentricity from centre of stiffness shall be calculated. Since the centre of mass and the centre of stiffness coincide in the present case, the eccentricity from the centre of stiffness is also 1.125 m. Accidental eccentricity can be on either side (that is, plus or minus).

Hence, one must consider lateral force Qi acting at the centre of stiffness accompanied by a clockwise or an anticlockwise torsion moment (i.e., +1.125 Qi kNm or -1.125 Qi kNm).

Forces Qi acting at the centres of stiffness and respective torsion moments at various levels for the example building are shown in Figure 3. Note that the building structure is identical along the X- and Z- directions, and hence, the fundamental time period and the earthquake forces are the same in the two directions.

S a 1.36 = = 1.402 g 0.97 IS: 1893 (Part 1): 2002, Figure 2.

Ductile detailing is assumed for the structure. Hence, Response Reduction Factor, R, is taken equal to 5.0. It may be noted however, that ductile detailing is mandatory in Zones III, IV and V. Hence, Ah =

Z 2 × S × a R g I

0.16 1.5 × × 1.402 = 0.0336 2 5

Base shear, VB = Ah W = 0.0336 x 39 286 = 1 320 kN. The total horizontal load of 1 320 kN is now distributed along the height of the building as per clause 7.7.1 of IS1893 (Part 1): 2002. This distribution is shown in Table 1.

IITK-GSDMA-EQ26-V3.0

Page 14

Design Example of a Building

Mass centre ( Centre of stiffness) 480 kN 5m 380 kN

540 kNm

428 kNm

5m 244 kN 275 kNm

5m 138 kN 155 kNm

5m 62 kN 70 kNm

5m 16 kN 18 kNm

4.1 m

22 .5

0 kN 1.1 m

0 kNm

22.5 m

All columns not shown for clarity Figure not to the scale

Figure 3

Accidental Eccentricity Inducing Torsion in the Building

IITK-GSDMA-EQ26-V3.0

Page 15

Design Example of a Building

1.6.

Analysis by Space Frames

The space frame is modelled using standard software. The gravity loads are taken from Figure 2, while the earthquake loads are taken from Figure 3. The basic load cases are shown in Table 2, where X and Z are lateral orthogonal directions.

Table 2 Basic Load Cases Used for Analysis

No. 1 2 3 4 5 6

Load case DL

IL(Imposed/Live load)

Directions Downwards Downwards +X; Clockwise torsion due to EQ

+X; Anti-Clockwise torsion due to EQ +Z; Clockwise torsion due to EQ +Z; Anti-Clockwise torsion due to EQ

For design of various building elements (beams or columns), the design data may be collected from computer output. Important design forces for selected beams will be tabulated and shown diagrammatically where needed. . In load combinations involving Imposed Loads (IL), IS 1893 (Part 1): 2002 recommends 50% of the imposed load to be considered for seismic weight calculations. However, the authors are of the opinion that the relaxation in the imposed load is unconservative. This example therefore, considers 100% imposed loads in load combinations. For above load combinations, analysis is performed and results of deflections in each storey and forces in various elements are obtained.

Table 3 Load Combinations Used for Design

EXTP (+Torsion) EXTN (-Torsion) EZTP (+Torsion) EZTN (-Torsion)

No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Load combination 1.5 (DL + IL) 1.2 (DL + IL + EXTP) 1.2 (DL + IL + EXTN) 1.2 (DL + IL – EXTP) 1.2 (DL + IL – EXTN) 1.2 (DL + IL + EZTP) 1.2 (DL + IL + EZTN) 1.2 (DL + IL – EZTP) 1.2 (DL + IL – EZTN) 1.5 (DL + EXTP) 1.5 (DL + EXTN) 1.5 (DL – EXTP) 1.5 (DL – EXTN) 1.5 (DL + EZTP) 1.5 (DL + EZTN) 1.5 (DL – EZTP) 1.5 (DL – EZTN)

EXTP: EQ load in X direction with torsion positive EXTN: EQ load in X direction with torsion negative EZTP: EQ load in Z direction with torsion positive EZTN: EQ load in Z direction with torsion negative.

1.7.

Load Combinations

As per IS 1893 (Part 1): 2002 Clause no. 6.3.1.2, the following load cases have to be considered for analysis: 1.5 (DL + IL) 1.2 (DL + IL ± EL) 1.5 (DL ± EL) 0.9 DL ± 1.5 EL Earthquake load must be considered for +X, -X, +Z and –Z directions. Moreover, accidental eccentricity can be such that it causes clockwise or anticlockwise moments. Thus, ±EL above implies 8 cases, and in all, 25 cases as per Table 3 must be considered. It is possible to reduce the load combinations to 13 instead of 25 by not using negative torsion considering the symmetry of the building. Since large amount of data is difficult to handle manually, all 25-load combinations are analysed using software.

IITK-GSDMA-EQ26-V3.0

Page 16

Design Example of a Building

18 19 20 21 22 23 24 25

0.9 DL + 1.5 EXTP 0.9 DL + 1.5 EXTN 0.9 DL – 1.5 EXTP 0.9 DL – 1.5 EXTN 0.9 DL + 1.5 EZTP 0.9 DL + 1.5 EZTN 0.9 DL – 1.5 EZTP 0.9 DL – 1.5 EZTN

Maximum drift is for fourth storey = 17.58 mm. Maximum drift permitted = 0.004 x 5000 = 20 mm. Hence, ok. Sometimes it may so happen that the requirement of storey drift is not satisfied. However, as per Clause 7.11.1, IS: 1893 (Part 1): 2002; “For the purpose of displacement requirements only, it is permissible to use seismic force obtained from the computed fundamental period (T ) of the building without the lower bound limit on design seismic force.” In such cases one may check storey drifts by using the relatively lower magnitude seismic forces obtained from a dynamic analysis.

1.8.

Storey Drift

1.9.

Stability Indices

As per Clause no. 7.11.1 of IS 1893 (Part 1): 2002, the storey drift in any storey due to specified design lateral force with partial load factor of 1.0, shall not exceed 0.004 times the storey height. From the frame analysis the displacements of the mass centres of various floors are obtained and are shown in Table 4 along with storey drift.

It is necessary to check the stability indices as per Annex E of IS 456:2000 for all storeys to classify the columns in a given storey as non-sway or sway columns. Using data from Table 1 and Table 4, the stability indices are evaluated as shown in Table 5. The stability index Qsi of a storey is given by

Qsi =

Since the building configuration is same in both the directions, the displacement values are same in either direction.

Table 4 Storey Drift Calculations

∑P Δ

H u hs

Where

Qsi = stability index of ith storey

Storey

Displacement (mm) 79.43 72.20 60.01 44.33 26.75 9.49 0.41 0

Storey drift (mm) 7.23 12.19 15.68

∑P

= sum of axial loads on all columns in the ith storey = elastically computed first order lateral deflection = total lateral force acting within the storey = height of the storey.

7 (Fifth floor) 6 (Fourth floor) 5 (Third floor) 4 (Second floor) 3 (First floor) 2 (Ground floor) 1 (Below plinth) 0 (Footing top)

Hu hs

17.58 17.26 9.08 0.41 0 As per IS 456:2000, the column is classified as non-sway if Qsi ≤ 0.04, otherwise, it is a sway column. It may be noted that both sway and nonsway columns are unbraced columns. For braced columns, Q = 0.

IITK-GSDMA-EQ26-V3.0

Page 17

Design Example of a Building

Table 5 Stability Indices of Different Storeys

Storey

Storey seismic weight Wi (kN)

Axial load ΣPu=ΣWi, (kN) 5 597 11 978 18 359 24 740 31 121 37 259 39 286

(mm)

Lateral load

Hu = Vi (kN)

Qsi

Classification

H u hs

(mm)

∑ Pu Δ u

7 6 5 4 3 2 1

5 597 6 381 6 381 6 381 6 381 6 138 2 027

7.23 12.19 15.68 17.58 17.26 9.08 0.41

480 860 1 104 1 242 1 304 1 320 1 320

5 000 5 000 5 000 5 000 5 000 4 100 1 100

0.0169 0.0340 0.0521 0.0700 0.0824 0.0625 0.0111

No-sway No-sway Sway Sway Sway Sway No-sway

1.10.

Design of Selected Beams

Here,

Lc = 7500 – 500 = 7000 mm D = 600 mm

The design of one of the exterior beam B2001-B2002-B2003 at level 2 along Xdirection is illustrated here.

1.10.1. General requirements

The flexural members shall fulfil the following general requirements. (IS 13920; Clause 6.1.2)

b ≥ 0.3 D

Here

b 300 = = 0.5 > 0.3 D 600

Hence, ok. (IS 13920; Clause 6.1.3)

b ≥ 200 mm

Here b = 300 mm ≥ 200 mm

1.10.2. Bending Moments and Shear Forces The end moments and end shears for six basic load cases obtained from computer analysis are given in Tables 6 and 7. Since earthquake load along Z-direction (EZTP and EZTN) induces very small moments and shears in these beams oriented along the X-direction, the same can be neglected from load combinations. Load combinations 6 to 9, 14 to 17, and 22 to 25 are thus not considered for these beams. Also, the effect of positive torsion (due to accidental eccentricity) for these beams will be more than that of negative torsion. Hence, the combinations 3, 5, 11, 13, 19 and 21 will not be considered in design. Thus, the combinations to be used for the design of these beams are 1, 2, 4, 10, 12, 18 and 20.

Hence, ok. (IS 13920; Clause 6.1.4) D≤ Lc 4

The software employed for analysis will however, check all the combinations for the design moments and shears. The end moments and end shears for these seven load combinations are given in Tables 8 and 9. Highlighted numbers in these tables indicate maximum values.

IITK-GSDMA-EQ26-V3.0

Page 18

Design Example of a Building

To get an overall idea of design moments in From the results of computer analysis, moment beams at various floors, the design moments and envelopes for B2001 and B2002 are drawn in shears for all beams in frame A-A are given in Figures 4 (a) and 4 (b) for various load combinations, viz., the combinations 1, 2, Tables 11 and 12. It may be noted that values of 4,10,12,18 and 20. Design moments and shears at level 2 in Tables 11 and 12 are given in table 10. various locations for beams B2001-B2002–B2003 are given in Table 10. Table 6 End Moments (kNm) for Six Basic Load Cases S.No. Load case Left 1 2 3 4 5 6 (DL) (IL/LL) (EXTP) (EXTN) (EZTP) (EZTN) 117.95 18.18 -239.75 -200.03 -18.28 19.39 B2001 Right -157.95 -29.85 -215.88 -180.19 -17.25 16.61 Left 188.96 58.81 -197.41 -164.83 -16.32 14.58 B2002 Right -188.96 -58.81 -197.40 -164.83 -16.20 14.70 Left 157.95 29.85 -215.90 -180.20 -18.38 15.47 B2003 Right -117.95 -18.18 -239.78 -200.05 -21.37 16.31

Sign convention: Anti-clockwise moment (+); Clockwise moment (-)

Table 7 End Shears (kN) For Six Basic Load Cases

S.No. Load case Left 1 2 3 4 5 6 (DL) (IL/LL) (EXTP) (EXTN) (EZTP) (EZTN) 109.04 17.19 -60.75 -50.70 -4.74 4.80

B2001 Right 119.71 20.31 60.75 50.70 4.74 -4.80 Left 140.07 37.5 -52.64 -43.95 -4.34 3.90

B2002 Right 140.07 37.5 52.64 43.95 4.34 -3.90 Left

B2003 Right 109.04 17.19 60.76 50.70 5.30 -4.24

119.71 20.31 -60.76 -50.70 -5.30 4.24

Sign convention: (+) = Upward force; (–) = Downward force

IITK-GSDMA-EQ26-V3.0 Page 19

Design Example of a Building

Table 8

Factored End Moments (kNm) for Load Combinations

Combn Load combination No: Left 1 2 4 10 12 18 20 [1.5(DL+IL)] [1.2(DL+IL+EXTP)] [1.2(DL+IL-EXTP)] [1.5(DL+EXTP)] [1.5(DL-EXTP)] [0.9DL+1.5EXTP] [0.9DL-1.5EXTP]

B2001 Right -281.71 -484.43 33.69

-560.76

B2002 Left 371.66 60.44

534.21

B2003 Left 281.71 -33.71 484.45 -86.91

560.78 -181.69

Right -371.66 -534.21 -60.44

-579.55

Right -204.21 -451.10 124.37

-536.60

204.21 -124.34 451.07 -182.69

536.56 -253.47

-12.66 579.55

-126.04

86.90 -465.99

181.67

12.66 -466.18

126.04

182.73 -465.82

253.51

465.79

466.18

466.00

Sign convention: (+) = Anti-clockwise moment; (–) = Clockwise moment

Table 9 Factored End Shears (kN) for Load Combinations

Combn No:

Load combination Left

B2001 Right 210.02 240.92 95.12

270.69

B2002 Left 266.36 149.92 276.26 131.15

289.07

B2003 Left 210.02 95.11 240.93 88.43

270.70

Right 266.36 276.26 149.92

289.07

Right 189.35 224.39 78.57

254.70

1 2 4 10 12 18 20

[1.5(DL+IL)] [1.2(DL+IL+EXTP)] [1.2(DL+IL-EXTP)] [1.5(DL+EXTP)] [1.5(DL-EXTP)] [0.9DL+1.5EXTP] [0.9DL-1.5EXTP]

189.35 78.58 224.38 72.44

254.69

88.44 198.86 16.61

131.15 205.03 47.11

72.43 189.27 7.00

7.01 189.26

47.11 205.03

16.60 198.87

Sign convention: (+) = Upward force; (–) = Downward force

IITK-GSDMA-EQ26-V3.0

Page 20

Design Example of a Building

300 200 100 0

Sagging Moment Envelope

20 12

7000 8000

0 1000 2000 3000 4000 5000 6000

M o m e n ts in K N m

-100 -200 -300 -400

Distance in mm

1 2

Hogging Moment Envelope

-500

Note: 1, 2, 4,10,12,18 and 20 denote the moment envelopes for respective load combinations.

Figure 4(a) Moments Envelopes for Beam 2001

300 200

Sagging Moment Envelope

100 0 0 -100 -200 -300 -400 Hogging Moment Envelope

Note: 1, 2, 4,10,12,18 and 20 denote the moment envelopes for respective load combinations

Figure 4(b) Moment Envelopes for Beam 2002 IITK-GSDMA-EQ26-V3.0 Page 21

1000

2000

20 3000

4000

5000

6000 1

7000

Distance in mm

Design Example of a Building Table 10 Design Moments and Shears at Various Locations

Beam Distance from left end (mm) 0 Moment (kNm)

-537 253

B2001 Shear (kN)

255

B2002 Moment (kNm)

-580 126

B2003 Shear (kN)

289

Moment (kNm)

-561 182

Shear (kN)

271

625

-386 252

226

-407 151

265

-401 188

242

1250

-254 241

198

-249 167

240

-258 181

214

1875

-159 238

169

-123 190

218

-141 172

185

2500

-78 221

140

-27 218

198

-55 165

156

3125

-8 186

112

0 195

103

0 140

128

3750

0 130

-99

0 202

0 130

4375

0 140

-128

0 195

-103

-8 186

-112

5000

-55 165

-156

-27 218

-128

-78 221

-140

5625

-141 172

-185

-123 190

-218

-159 238

-169

6250

-258 181

-214

-249 167

-240

-254 241

-198

6875

-401 187

-242

-407 151

-265

-386 253

-226

7500

-561 182

-271

-580 126

-290

-537 254

-255

IITK-GSDMA-EQ26-V3.0

Page 22

Design Example of a Building Table 11 Design Factored Moments (kNm) for Beams in Frame AA

Level 0 7 (-) (+) 6 (-) (+) 5 (-) (+) 4 (-) (+) 3 (-) (+) 2 (-) (+) 1 (-) (+) 190 47 411 101 512 207 574 274 596 303

537 253

External Span (Beam B1) 1250 71 69 167 137 237 209 279 255 294 274

254 241

Internal Span (B2) 6250 86 33 162 106 226 164 267 202 285 215

259 181

2500 11 87 29 164 67 202 90 227 99 238

78 221

3750 0 67 0 133 0 132 0 131 0 132

0 130

5000 3 54 12 134 41 159 60 176 68 182

55 165

7500 221 2 414 65 512 155 575 213 602 234

561 182

0 290 0 479 25 559 107 611 159 629 175

580 126

1250 91 39 182 99 235 154 270 189 281 199

249 167

2500 0 145 0 190 20 213 37 230 43 235

27 218

3750 0 149 0 203 0 204 0 200 0 202

0 202

250 24

90 63

3 94

0 81

4 87

98 55

264 13

259 10

97 55

5 86

0 76

Table 12

Design Factored Shears (kN) for Beams in Frame AA

Level External Span (Beam B1 ) 0 7-7 6-6 5-5 4-4 3-3 2-2 1-1 110 223 249 264 270

255

Internal Span (B2) 3750 -31 52 77 93 98

-99

1250 79 166 191 207 213

198

2500 49 109 134 150 155

140

5000 -61 -116 -143 -160 -168

-156

6250 -92 -173 -200 -218 -225

-214

7500 -123 -230 -257 -275 -282

-271

0 168 266 284 298 302

289

1250 150 216 235 247 253

240

2500 133 177 194 205 208

198

3750 -23 52 74 88 92

149

108

-31

-72

-112

-153

150

110

-28

IITK-GSDMA-EQ26-V3.0

Page 23

Design Example of a Building

1.10.3. Longitudinal Reinforcement Consider mild exposure and maximum 10 mm diameter two-legged hoops. Then clear cover to main reinforcement is 20 +10 = 30 mm. Assume 25 mm diameter bars at top face and 20 mm diameter bars at bottom face. Then, d = 532 mm for two layers and 557 mm for one layer at top; d = 540 mm for two layers and 560 mm for one layer at bottom. Also consider d’/d = 0.1 for all doubly reinforced sections.

at the face of the support, i.e., 250 mm from the centre of the support are calculated by linear interpolation between moment at centre and the moment at 625 mm from the centre from the table 10. The values of pc and pt have been obtained from SP: 16. By symmetry, design of beam B2003 is same as that of B2001. Design bending moments and required areas of reinforcement are shown in Tables 15 and 16. The underlined steel areas are due to the minimum steel requirements as per the code. Table 17 gives the longitudinal reinforcement provided in the beams B2001, B 2002 and B2003.

Design calculations at specific sections for flexure reinforcement for the member B2001 are shown in Table 13 and that for B2002 are tabulated in Table 14. In tables 13 and 14, the design moments

Table 13 Flexure Design for B2001

Location from left support 250

(kNm)

(mm)

Mu bd 2

Type

Ast

(mm )

Asc (mm2)

(N/mm2) -477 +253 300 300 300 300 300 300 300 300 300 300 300 300 300 300 532 540 532 540 557 540 557 560 557 540 532 540 532 540 5.62 2.89 2.99 2.75 0.84 2.53 0 1.38 0.59 1.89 3.04 2.07 5.85 2.08 D S S S S S S S S S S S D S 1.86 0.96 1.00 0.90 0.25 0.81 0 0.42 0.18 0.58 1.02 0.65 1.933 0.65 0.71 0.782 2 969 1 555 1 596 1 458 418 1 312 0 706 301 940 1 628 1 053 3 085 1 053 1 133 1 248 –

1 250

-254 +241

2 500

-78 +221

3 750

0 +130

5 000

-55 +165

6 250

-258 +181

7 250

-497 +182

D = Doubly reinforced section; S = Singly reinforced section

IITK-GSDMA-EQ26-V3.0

Page 24

Design Example of a Building

Table 14 Flexure Design for B2002

Location from left support 250

Mu, (kNm)

(mm)

Mu , 2 bd ( kNm)

Type

Ast

(mm )

Asc (mm2)

-511 +136

300 300 300 300 300 300 300 300 300 300 300 300 300 300

532 540 532 540 557 540 557 560 557 540 532 540 532 540

6.02 1.55 2.93 1.91 0.29 2.49 0 2.15 0.29 2.49 2.93 1.91 6.02 1.55

D S S S S S S S S S S S D S

1.99 0.466 0.97 0.59 0.09 0.80 0 0.67 0.09 0.80 0.97 0.59 1.99 0.466

0.84 0.84 –

3 176 755 1 548 956 150 1 296 0 1 126 150 1 296 1 548 956 3 176 755

744 ,744 ,-

1 250

-249 +167

2 500

-27 +218

3 750

0 +202

5 000

-27 +218

6 250

-249 +167

7 250

-511 +136

D = Doubly reinforced section; S = Singly reinforced section

Table 15 Summary of Flexure Design for B2001 and B2003

B2001 Distance from left (mm)

M (-) at top (kNm)

A 250 477 532 2969 1133 253 540 1555 1250 254 532 1596 241 540 1458 2500 78 557 486 221 540 1312 3750 0 557 486 130 560 706 5000 55 557 486 165 540 940 6250 258 532 1628 181 540 1053

B 7250 497 532 3085 1248 182 540 1053

Effective depth d (mm)

Ast, top bars (mm2) Asc, bottom bars (mm2) M (+) at bottom (kNm)

Effective depth d (mm)

Ast, (bottom bars) (mm2)

IITK-GSDMA-EQ26-V3.0

Page 25

Design Example of a Building Table 16 Summary of Flexure Design for B2002

B2002 Distance from left (mm)

M (-), at top (kNm)

B 250 511 532 3176 744 136 540 755 1250 249 532 1548 167 540 956 2500 27 557 486 218 540 1296 3750 0 557 486 202 560 1126 5000 27 557 486 218 540 1296 6250 249 532 1548 167 540 956

C 7250 511 532 3176 744 136 540 755

Effective depth d, (mm)

Ast, top bars (mm2) Asc, bottom bars (mm2) M (+) at bottom (kNm)

Effective depth d, (mm)

Ast, (bottom bars) (mm2)

IITK-GSDMA-EQ26-V3.0

Page 26

Design Example of a Building

A 2500

H 2500 B 2001

B 2500

K ‘ 2500

C 2500

H ‘

F’ 2500

B 2002 L o c a tio n s fo r c u rta ilm e n t

B 2003

Figure 5 Critical Sections for the Beams

Table 17: Summary of longitudinal reinforcement provided in beams

B2001 and B2003 At A and D (External supports) Bottom bars At Centre Top bars Bottom bars At B and C (Internal supports) B2002 At Centre Top bars Bottom bars 2- 25 #, Ast = 982 mm2 5 – 20 #, Ast = 1570 mm2 Top bars Bottom bars Top bars 7 – 25 #, Ast = 3437 mm2, with 250 mm (=10 db) internal radius at bend, where db is the diameter of the bar 6 – 20 #, Ast = 1884 mm2, with 200 mm (=10 db) internal radius at bend 2- 25 #, Ast = 982 mm2 5 – 20 #, Ast = 1570 mm2 7- 25 # , Ast = 3437 mm2 6 – 20 #, Ast = 1884 mm2

At A and D, as per requirement of Table 14, 5-20 # bars are sufficient as bottom bars, though the area of the compression reinforcement then will not be equal to 50% of the tension steel as required by Clause 6.2.3 of IS 13920:1993. Therefore, at A and D, 6-20 # are provided at bottom. The designed section is detailed in Figure.6. The top bars at supports are extended in the spans for a distance of (l /3) = 2500 mm.

IITK-GSDMA-EQ26-V3.0

Page 27

Design Example of a Building

250 250

1260 A

2500 1 3 4 2 6-20 # 7500 c/c 1010 Dia No 12 # 9

2500 2-25 # + 5-25 # extra 2-25 # 2-25 # + 5-25 # extra

2500

2-25 # 100 500

5-20 # B2001 (300 × 600)

6-20 #

6-20 # 7500 c/c B2002 (300 × 600)

5-20 # 300 100

Section A – A

12 # 8 160 Column bars assume 25 # Maximum 10 # hoops

12 # Rest 200

12 # 8 160

12 #

12 # Stirrups 22 110

12 # Rest 130 100 500

130 SPA 2 1 3/4

130

Elevation

r = 250 mm central r = 262.5

r = 200 central r = 210 300 100

25 40 20 25

Section B- B

275 20 25 (3/4) 25

25 135

140

90 280 (d) Bar bending details in raw1 (Top bars)

140200 (d) Bar bending details in raw 2 (Bottom bars)

Details of beams B2001 – B2002 – B2003

Figure 6

1.10.3.1. Check for reinforcement

Details of Beams B2001, B2002 and B2003

(IS 13920; Clause 6.2.1)

1.10.3.2. (a) Minimum two bars should be

The positive steel at a joint face must be at least equal to half the negative steel at that face. Joint A Half the negative steel =

continuous at top and bottom.

Here, 2–25 mm # (982 mm ) are continuous throughout at top; and 5–20 mm # (1 570 mm2) are continuous throughout at bottom. Hence, ok. (b) p t , min =

Positive steel = 1884 mm2 > 1718 mm2 Hence, ok. Joint B Half the negative steel =

3437 = 1718 mm2 2

0.24 f ck

0.24 25 415

=0.00289, i.e., 0.289%.

Ast , min =

0.289 × 300 × 560 = 486 mm 2 100

Positive steel = 1 884 mm2 > 1 718 mm2 Hence, ok.

3437 = 1718 mm2 2

Provided reinforcement is more. Hence, ok.

(IS 13920; Clause 6.2.2) Maximum steel ratio on any face at any section should not exceed 2.5, i.e.,

(IS 13920; Clause 6.2.4)

Along the length of the beam,

Ast at top or bottom ≥ 0.25 Ast at top at joint A or B Ast at top or bottom ≥ 0.25 × 3 437

p max = 2.5%. Ast ,max = 2.5 × 300 × 532 = 3990 mm 2 100

Provided reinforcement is less. Hence ok.

(IS 13920; Clause 6.2.3)

≥ 859 mm2 Hence, ok.

IITK-GSDMA-EQ26-V3.0

Page 28

Design Example of a Building (IS 13920; Clause 6.2.5)

As Mu = 321 kNm

Ah Mu = 568 kNm

At external joint, anchorage of top and bottom bars = Ld in tension + 10 db.

Ld of Fe 415 steel in M25 concrete = 40.3 db

Bs Mu = 321 kNm

Bh M u = 568 kNm

Here, minimum anchorage = 40.3 db + 10 db = 50.3 db. The bars must extend 50.3 db (i.e. 50.3 x 25 = 1258 mm, say 1260 mm for 25 mm diameter bars and 50.3 x 20 = 1006 mm, say 1010 mm for 20 mm diameter bars) into the column. At internal joint, both face bars of the beam shall be taken continuously through the column.

The moment capacities as calculated in Table 18 at the supports for beam B2002 are:

As Mu = 321 kNm Ah Mu = 585 kNm

Bs Mu = 321 kNm Bh Mu = 585 kNm

1.10.4. Web reinforcements Vertical hoops (IS: 13920:1993, Clause 3.4 and Clause 6.3.1) shall be used as shear reinforcement.

1.2 (DL+LL) for U.D.L. load on beam B2001 and B2003. = 1.2 (30.5 + 5) = 42.6 kN/m. 1.2 (DL+LL) for U.D.L. load on beam B2002 = 1.2 (26.1 + 0) = 31.3 kN/m. 1.2 (DL+LL) for two point loads at third points on beam B2002 = 1.2 (42.2+37.5) = 95.6 kN. The loads are inclusive of self-weights. For beam B2001 and B2003: VaD + L = VbD + L = 0.5 × 7.5 × 42.6 = 159.7 kN. For beam 2002: VaD + L = VbD + L = 0.5 × 7.5 × 31.3 + 95.6 = 213 kN.

Hoop diameter ≥ 6 mm ≥ 8 mm if clear span exceeds 5 m. (IS 13920:1993; Clause 6.3.2) Here, clear span = 7.5 – 0.5 = 7.0 m. Use 8 mm (or more) diameter two-legged hoops.

The moment capacities as calculated in Table 18 at the supports for beam B2001 and B2003 are:

IITK-GSDMA-EQ26-V3.0

Page 29

Design Example of a Building

Beam B2001 and B2003: Sway to right

D+L − 1.4 ⎢ Vu , a = V a

42.6 kN/m B

⎡ M As ⎢ ⎣

Bh u ,lim + M u ,lim L AB

⎤ ⎥ ⎥ ⎦

159.7 kN

7.5 m Loding

159.7 kN

⎡ 321 + 568 ⎤ D+L = Va − 1 .4 ⎢ ⎣ 7 .5 ⎥ ⎦

= 159.7 − 166 = −6.3 kN

Vu ,b = 159.7 + 166 = 325.7 kN .

159.7 kN

– S.F.diagram (i) 1.2 (D + L) 159.7 kN

Sway to left

⎡ M Ah + M Bs u ,lim D + L – 1.4 ⎢ u ,lim Vu ,a = Va ⎢ L AB ⎣ ⎡ 568 + 321 ⎤ = 159.7 − 1.4 ⎢ ⎥ 7.5 ⎣ ⎦ ⎤ ⎥ ⎥ ⎦

– 169.1 kN S.F.diagram (ii) Sway to right

166 kN S.F.diagram (iii) Sway to left 325.7 kN 272.4 219.2

= 159.7 + 166 = 325.7 kN Vu ,b = 159.7 − 166 = −6.3 kN

Maximum design shear at A and B = 325.7 kN, say 326 kN

166

219.2

272.4

325.7 kN

(iv) Design S.F.diagram Beam B2001 and B2003

Figure 7 Beam Shears due to Plastic Hinge Formation for Beams B2001 and B2003

IITK-GSDMA-EQ26-V3.0

Page 30

Design Example of a Building

Beam 2002

95.6 kN 95.6 kN B

Sway to right

D+L − 1.4 ⎢ Vu , a = V a

31.3 kN/m

⎡ M As ⎢ ⎣

Bh u ,lim + M u ,lim L AB

⎤ ⎥ ⎥ ⎦

213 kN

2.5 m

213 kN

2.5 m 7.5 m Loding 134.7 kN 39.1 39.1 134.7 kN S.F.diagram (i) 1.2 (D + L)

213 kN 2.5 m

⎡ 321 + 568 ⎤ D+L = Va − 1 .4 ⎢ ⎣ 7 .5 ⎥ ⎦

= 213 − 166 = 47 kN

–

213 kN

Vu ,b = 213 + 166 = 379 kN .

–

166 kN S.F.diagram

Sway to left

Vu ,a = 213 + 166 = 379 kN Vu ,b = 213 − 166 = 47 kN

379 kN 340

(ii) Sway to right

166 kN S.F.diagram (iii) Sway to left 301 166 127 31.4 – 208.3 301 (iv) Design S.F.diagram Beam 2002 340 379

Maximum design shear at A = 379 kN. Maximum design shear at B = 379 kN.

208.3

31.4 127 166

Figure 8 Beam Shears due to Plastic Hinge Formation for Beam B 2002

IITK-GSDMA-EQ26-V3.0

Page 31

Design Example of a Building

Maximum shear forces for various cases from analysis are shown in Table 19(a).

The shear force to be resisted by vertical hoops shall be greater of: i) Calculated factored shear force as per analysis. ii) Shear force due to formation of plastic hinges at both ends of the beam plus the factored gravity load on the span. The design shears for the beams B2001 and B2002 are summarized in Table 19. As per Clause 6.3.5 of IS 13920:1993,the first stirrup shall be within 50 mm from the joint face. Spacing, s, of hoops within 2 d (2 x 532 = 1064 mm) from the support shall not exceed: (a) d/4 = 133 mm (b) 8 times diameter of the smallest longitudinal bar = 8 x 20 = 160 mm

Hence, spacing of 133 mm c/c governs. Elsewhere

s≤ d

the

span,

spacing,

532 2

= 266 mm.

Maximum nominal shear stress in the beam

379 × 10 3 τc = = 2.37 N/mm 2

(τc,max, for M25 mix) The proposed provision of two-legged hoops and corresponding shear capacities of the sections are presented in Table 20.

IITK-GSDMA-EQ26-V3.0

Page 32

Design Example of a Building

Table 18 Calculations of Moment Capacities at Supports

All sections are rectangular. For all sections: b = 300 mm, d = 532 mm, d’=60 mm, d’/d = 0.113 fsc = 353 N/mm2, xu,max = 0.48d = 255.3 mm.

As Mu (kNm) Ah Mu (kNm) Bs Mu (kN-m) Bh Mu (kN-m)

Top bars Bottom bars Ast (mm2) Asc (mm2) C1= 0.36 fck b xu = A xu C2 = Asc fsc (kN) T = 0.87 fy Ast (kN) xu= (T-C2) /A Muc1 = (0.36fck b xu) × (d-0.42xu) Muc2 = Asc fsc (d-d’) Mu = 0.87fyAst × (d-d’) Mu = Mu1+ Mu2, (kNm)

7-25 # = 3 437 7-25 # = 3 437 7-25 # = 3 437 7-25 # = 3 437 mm2 mm2 mm2 mm2 6-20 # = 1 884 6-20 # = 1 884 6-20 # = 1 884 6-20 # = 1 884 mm2 mm2 mm2 mm2 1 884 3 437 1 884 3 437 3 437 1 884 3 437 1 884 2 700 xu 2 700 xu 2 700 xu 2 700 xu 1 213.2 680.2 Negative i.e. xu 20mm 500 30

i.e., p = 0.11 x 25 = 2.75, and

Asc =

pbD 2.75 × 500 × 500 = = 6875 mm 2 . 100 100

ex,min = ez,min = 23.7 mm. Similarly, for all the columns in first and second storey, ex,min = ey,min = 25 mm. For upper height of column, L = 5,000 – 600 = 4,400 mm.

ex ,min = ez,min = 4,400 500 + = 25.46mm > 20mm 500 30

rd th

Trial steel areas required for column lengths C102, C202, C302, etc., can be determined in a similar manner. The trial steel areas required at various locations are shown in Figure 10(a).

As described in Section 1.12. the trial reinforcements are subsequently selected and provided as shown in figure 11 (b) and figure 11 (c).

Calculations shown in Tables 25 and 27 for checking the trial sections are based on provided steel areas. For example, for column C202 (mid-height of second storey to the mid-height of third storey), provide 8-25 # + 8-22 # = 6968 mm2, equally distributed on all faces. Asc = 6968 mm2, p = 2.787,

For all columns in 3 to 7 storey. ex,min = ez,min = 25.46 mm. For column C2 in all floors, i.e., columns C102, C202, C302, C402, C502, C602 and C702, fck = 25 N/mm2, fy = 415 N/mm2, and

d ‘ 50 = = 0.1. d 500

p = 0.111 . f ck

Puz = [0.45 x 25(500 x 500 – 6968) + 0.75 x 415 x 6968] x 10-3 = 4902 kN. Calculations given in Tables 24 to 27 are selfexplanatory.

Calculations of Table 25 and 27 are based on uniaxial moment considering steel on two opposite faces and hence, Chart 32 of SP: 16 is used for determining the trial areas. Reinforcement obtained for the trial section is equally distributed on all four sides. Then, Chart

IITK-GSDMA-EQ26-V3.0

Page 39

Design Example of a Building

402 5230 mm2 6278 mm2 302 8-25 mm # + 8-22 mm # = 6968 mm2

6278 mm 2

302

302 6875 mm2 6562 mm

202 C 6875 mm 2

C 202

202 C

8-25 mm # + 8-22 mm # = 6968 mm 2

7762 mm2 B 3780 mm2 102 A 5400 mm2 C2 (a) Required trial areas in mm 2 at various locations

102

B A C2

7762 mm 2 5400 mm

102

B A

16-25 mm # = 7856 mm2

(b) Proposed reinforcement areas at various joints

C2 (c) Areas to be used for detailing

Figure 11 Required Area of Steel at Various Sections in Column

IITK-GSDMA-EQ26-V3.0

Page 40

Design Example of a Building

TABLE 24 TRIAL SECTION BELOW JOINT C

Pu, Comb. kN No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 4002 3253 3225 3151 3179 2833 2805 3571 3598 3155 3120 3027 3063 2630 2596 3552 3587 1919 1883 1791 1826 1394 1359 2316 2351

Centreline moment Mux, Muz, kNm kNm 107 89 83 82 88 17 23 189 195 65 58 57 65 68 75 190 198 41 33 33 40 92 100 166 173 36 179 145 238 203 12 45 46 13 242 199 279 236 3 38 40 1 249 206 272 229 10 31 32 9

Moment at face Mux, Muz, kNm kNm 93.946 78.14 72.87 72.00 77.26 14.93 20.19 165.94 171.21 57.07 50.92 50.05 57.07 59.70 65.85 166.82 173.84 36.00 28.97 28.97 35.12 80.78 87.80 145.75 151.89 31.608 157.16 127.31 208.96 178.23 10.54 39.51 40.39 11.41 212.48 174.72 244.96 207.21 2.63 33.36 35.12 0.88 218.62 180.87 238.82 201.06 8.78 27.22 28.10 7.90

Cal. Ecc.,mm ex ez 23.47 24.02 22.60 22.85 24.30 5.27 7.20 46.47 47.58 18.09 16.32 16.53 18.63 22.70 25.37 46.97 48.47 18.76 15.39 16.18 19.23 57.95 64.61 62.93 64.61 7.90 48.31 39.48 66.32 56.07 3.72 14.09 11.31 3.17 67.35 56.00 80.93 67.65 1.00 12.85 9.89 0.24 113.92 96.05 133.34 110.11 6.30 20.03 12.13 3.36

Des. Ecc.,mm edx edz 25.00 25.00 25.00 25.00 25.00 25.00 25.00 46.47 47.58 25.00 25.00 25.00 25.00 25.00 25.37 46.97 48.47 25.00 25.00 25.00 25.00 57.95 64.61 62.93 64.61 25.00 48.31 39.48 66.32 56.07 25.00 25.00 25.00 25.00 67.35 56.00 80.93 67.65 25.00 25.00 25.00 25.00 113.92 96.05 133.34 110.11 25.00 25.00 25.00 25.00

Mux, kNm

Muz, kNm

P’u M’uz

Pu’ f ck bD

‘ Mu f ck bD 2

p f ck

100 81 81 79 79 71 70 166 171 79 78 76 77 66 66 167 174 48 47 45 46 81 88 146 152

100 157 127 209 178 71 70 89 90 212 175 245 207 66 65 89 90 219 181 239 201 35 34 58 59

4002 3253 3225 3151 3179 2833 2805 3571 3598 3155 3120 3027 3063 2630 2596 3552 3587 1919 1883 1791 1826 1394 1359 2316 2351

200 238 208 288 258 142 140 255 261 291 253 321 284 132 131 256 264 267 228 284 247 116 122 204 211

0.64 0.52 0.52 0.50 0.51 0.45 0.45 0.57 0.58 0.50 0.50 0.48 0.49 0.42 0.42 0.57 0.57 0.31 0.30 0.29 0.29 0.22 0.22 0.37 0.38

0.06 0.08 0.07 0.09 0.08 0.05 0.04 0.08 0.08 0.09 0.08 0.10 0.09 0.04 0.04 0.08 0.08 0.09 0.07 0.09 0.08 0.04 0.04 0.07 0.07

0.105 0.083 0.078 0.083 0.08 0.042 0.038 0.096 0.1 0.083 0.079 0.097 0.082 0.024 0.024 0.1 0.1 0.04 0.023 0.038 0.03 negative negative 0.038 0.04

IITK-GSDMA-EQ26-V3.0

Page 41

Design Example of a Building

TABLE 25 CHECKING THE DESIGN OF TABLE 24 Pu Comb. No.

P u P uz

0.82 0.66 0.66 0.64 0.65 0.58 0.57 0.73 0.73 0.64 0.64 0.62 0.62 0.54 0.53 0.72 0.73 0.39 0.38 0.37 0.37 0.28 0.28 0.47 0.48

αn

2.03 1.77 1.76 1.74 1.75 1.63 1.62 1.88 1.89 1.74 1.73 1.70 1.71 1.56 1.55 1.87 1.89 1.32 1.31 1.28 1.29 1.14 1.13 1.45 1.47

Pu fckbD

0.64 0.52 0.52 0.50 0.51 0.45 0.45 0.57 0.58 0.50 0.50 0.48 0.49 0.42 0.42 0.57 0.57 0.31 0.30 0.29 0.29 0.22 0.22 0.37 0.38

Mux, kNm

100 81 81 79 79 71 70 166 171 79 78 76 77 66 66 167 174 48 47 45 46 81 88 146 152

Muz, kNm

100 157 127 209 178 71 70 89 90 212 175 245 207 66 65 89 90 219 181 239 201 35 34 58 59

M u1 f ck bd 2

0.09 0.13 0.13 0.13 0.13 0.135 0.135 0.105 0.105 0.13 0.13 0.135 0.135 0.145 0.145 0.105 0.105 0.17 0.18 0.18 0.18 0.175 0.175 0.16 0.16

Mu1

281 406 406 406 406 422 422 328 328 406 406 422 422 453 453 328 328 531 563 563 563 547 547 500 500

⎡ Mux ⎤ ⎢ ⎥ ⎣ Mu1 ⎦

αn

⎡ Muz ⎤ ⎢M ⎥ ⎣ u1 ⎦

0.123 0.186 0.129 0.315 0.237 0.055 0.055 0.086 0.087 0.324 0.233 0.398 0.297 0.049 0.049 0.086 0.087 0.310 0.227 0.335 0.266 0.043 0.043 0.043 0.043

αn

Check

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

4002 3253 3225 3151 3179 2833 2805 3571 3598 3155 3120 3027 3063 2630 2596 3552 3587 1919 1883 1791 1826 1394 1359 2316 2351

0.123 0.058 0.058 0.058 0.058 0.055 0.055 0.277 0.292 0.058 0.058 0.054 0.054 0.049 0.050 0.281 0.302 0.042 0.039 0.040 0.039 0.113 0.127 0.166 0.174

0.246 0.243 0.187 0.373 0.295 0.109 0.109 0.364 0.379 0.382 0.291 0.452 0.351 0.098 0.100 0.368 0.388 0.352 0.266 0.375 0.305 0.156 0.170 0.210 0.218

IITK-GSDMA-EQ26-V3.0

Page 42

Design Example of a Building

Pu, kN Centreline moment Mux, Muz, kNm kNm Moment at face Mux, Muz, kNm kNm TABLE 26 TRIAL SECTION ABOVE JOINT C Mux, Cal. Ecc.,mm Des. Ecc.,mm kNm ex ez edx edz Muz, kNm P’u

Comb. No.

M’uz

P’ u fckbD

0.53 0.43 0.43 0.42 0.42 0.38 0.38 0.47 0.48 0.42 0.42 0.41 0.41 0.36 0.35 0.47 0.48 0.26 0.25 0.24 0.25 0.19 0.19 0.31 0.31

‘ Mu fckbD2

p fck

0.075 0.095 0.075 0.1 0.09 0.018 0.022 0.095 0.096 0.1 0.082 0.11 0.096 0.038 0.037 0.095 0.102 0.062 0.046 0.07 0.056 0.016 0.016 negative negative

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

3339 2710 2687 2632 2654 2377 2355 2965 2987 2643 2616 2547 2548 2228 2201 2963 2990 1605 1577 1509 1537 1189 1162 1925 1952

131 111 99 98 110 87 98 296 307 78 64 63 77 169 183 310 324 50 36 35 49 197 211 281 295

47 293 238 368 313 11 63 65 13 389 321 437 368 10 55 58 7 399 330 427 358 20 45 48 17

117.9 99.9 89.1 88.2 99 78.3 88.2 266.4 276.3 70.2 57.6 56.7 69.3 152.1 164.7 279 291.6 45 32.4 31.5 44.1 177.3 189.9 252.9 265.5

42.3 263.7 214.2 331.2 281.7 9.9 56.7 58.5 11.7 350.1 288.9 393.3 331.2 9 49.5 52.2 6.3 359.1 297 384.3 322.2 18 40.5 43.2 15.3

35.31 36.86 33.16 33.51 37.30 32.94 37.45 89.85 92.50 26.56 22.02 22.26 27.20 68.27 74.83 94.16 97.53 28.04 20.55 20.87 28.69 149.12 163.43 131.38 136.01

12.67 97.31 79.72 125.84 106.14 4.16 24.08 19.73 3.92 132.46 110.44 154.42 129.98 4.04 22.49 17.62 2.11 223.74 188.33 254.67 209.63 15.14 34.85 22.44 7.84

35.31 36.86 33.16 33.51 37.30 32.94 37.45 89.85 92.50 26.56 25.00 25.00 27.20 68.27 74.83 94.16 97.53 28.04 25.00 25.00 28.69 149.12 163.43 131.38 136.01

25.00 97.31 79.72 125.84 106.14 25.00 25.00 25.00 25.00 132.46 110.44 154.42 129.98 25.00 25.00 25.00 25.00 223.74 188.33 254.67 209.63 25.00 34.85 25.00 25.00

118 100 89 88 99 78 88 266 276 70 65 64 69 152 165 279 292 45 39 38 44 177 190 253 266

83 264 214 331 282 59 59 74 75 350 289 393 331 56 55 74 75 359 297 384 322 30 41 48 49

3339 2710 2687 2632 2654 2377 2355 2965 2987 2643 2616 2547 2548 2228 2201 2963 2990 1605 1577 1509 1537 1189 1162 1925 1952

201 364 303 419 381 138 147 341 351 420 354 457 401 208 220 353 366 404 336 422 366 207 230 301 314

0.06 0.12 0.10 0.13 0.12 0.04 0.05 0.11 0.11 0.13 0.11 0.15 0.13 0.07 0.07 0.11 0.12 0.13 0.11 0.14 0.12 0.07 0.07 0.10 0.10

IITK-GSDMA-EQ26-V3.0

Page 43

Design Example of a Building

TABLE 27

Design Check on Trial Section of Table 26 above Joint C

Comb. No.

P u P uz

0.68 0.55 0.55 0.54 0.54 0.48 0.48 0.60 0.61 0.54 0.53 0.52 0.52 0.45 0.45 0.60 0.61 0.33 0.32 0.31 0.31 0.24 0.24 0.39 0.40

αn

1.80 1.59 1.58 1.56 1.57 1.48 1.47 1.68 1.68 1.57 1.56 1.53 1.53 1.42 1.42 1.67 1.68 1.21 1.20 1.18 1.19 1.07 1.06 1.32 1.33

P u fckbD

0.53 0.43 0.43 0.42 0.42 0.38 0.38 0.47 0.48 0.42 0.42 0.41 0.41 0.36 0.35 0.47 0.48 0.26 0.25 0.24 0.25 0.19 0.19 0.31 0.31

Mux, kNm

118 100 89 88 99 78 88 266 276 70 65 64 69 152 165 279 292 45 39 38 44 177 190 253 266

Muz, kNm

83 264 214 331 282 59 59 74 75 350 289 393 331 56 55 74 75 359 297 384 322 30 41 48 49

M u1 f ck bd 2

0.12 0.145 0.145 0.145 0.145 0.155 0.155 0.13 0.13 0.145 0.14 0.14 0.14 0.17 0.17 0.13 0.13 0.17 0.17 0.17 0.17 0.18 0.18 0.17 0.17

Mu1

375 453 453 453 453 484 484 406 406 453 438 438 438 531 531 406 406 531 531 531 531 563 563 531 531

⎡ M ux ⎤ ⎢ ⎥ ⎣ M u1 ⎦

αn

⎡ M uz ⎤ ⎢M ⎥ ⎣ u1 ⎦

αn

Check

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

3339 2710 2687 2632 2654 2377 2355 2965 2987 2643 2616 2547 2548 2228 2201 2963 2990 1605 1577 1509 1537 1189 1162 1925 1952

0.124 0.091 0.076 0.078 0.092 0.068 0.082 0.493 0.523 0.054 0.052 0.052 0.059 0.168 0.191 0.533 0.572 0.050 0.044 0.044 0.052 0.290 0.316 0.375 0.397

0.067 0.423 0.306 0.613 0.474 0.045 0.045 0.058 0.058 0.668 0.524 0.849 0.653 0.040 0.040 0.058 0.058 0.622 0.497 0.682 0.552 0.043 0.061 0.042 0.042

0.191 0.514 0.382 0.691 0.566 0.113 0.127 0.551 0.581 0.722 0.576 0.901 0.712 0.209 0.231 0.591 0.630 0.672 0.541 0.727 0.603 0.333 0.377 0.417 0.439

IITK-GSDMA-EQ26-V3.0

Page 44

Design Example of a Building

1.11.3. Design of Transverse reinforcement

The spacing should not exceed (i)

Three types of transverse reinforcement (hoops or ties) will be used. These are: i) General hoops: These are designed for shear as per recommendations of IS 456:2000 and IS 13920:1993. ii) Special confining hoops, as per IS 13920:1993 with spacing smaller than that of the general hoops iii) Hoops at lap: Column bars shall be lapped only in central half portion of the column. Hoops with reduced spacing as per IS 13920:1993 shall be used at regions of lap splicing.

1.11.3.1. Design of general hoops

0.87 f y ASV

0.4b reinforcement) =

(requirement for minimum shear

0.87 × 415 × 250 = 451.3 mm 0.4 × 500

(2)

(ii) 0.75 d = 0.75 X 450 = 337.5 mm (iii) 300 mm; i.e., 300 mm … As per IS 13920:1993, Clause 7.3.3, Spacing of hoops ≤ b/2 of column = 500 / 2 = 250 mm …

(3)

(A) Diameter and no. of legs

From (1), (2) and (3), maximum spacing of stirrups is 250 mm c/c.

1.11.3.2. Design Shear

Rectangular hoops may be used in rectangular column. Here, rectangular hoops of 8 mm diameter are used. Here h = 500 – 2 x 40 + 8 (using 8# ties) = 428 mm > 300 mm 13920:1993) (Clause 7.3.1, IS

As per IS 13920:1993, Clause 7.3.4, design shear for columns shall be greater of the followings: (a) Design shear as obtained from analysis For C202, lower height, Vu = 161.2 kN, for load combination 12. For C202, upper height, Vu = 170.0 kN, for load combination 12. (b) Vu = 1.4 ⎢

⎡ M bLlim + M bR ⎤ u, u, lim ⎥. h st ⎢ ⎥ ⎣ ⎦

The spacing of bars is (395/4) = 98.75 mm, which is more than 75 mm. Thus crossties on all bars are required (IS 456:2000, Clause 26.5.3.2.b-1) Provide 3 no open crossties along X and 3 no open crossties along Z direction. Then total legs of stirrups (hoops) in any direction = 2 +3 = 5.

(B) Spacing of hoops

For C202, lower height, using sections of B2001 and B2002

bL M u ,lim

bR M u ,lim

= 568 kNm = 568 kNm,

(Table 18) (Table 18)

As per IS 456:2000, Clause 26.5.3.2.(c), the pitch of ties shall not exceed: (i) b of the column = 500 mm (ii) 16 φmin (smallest diameter) = 16 x 20 = 320 mm (iii) 300 mm ….

(1)

hst = 4.1 m. Hence,

⎡ M bLlim + M bRlim ⎤ ⎡ 568 + 568 ⎤ u, u, Vu = 1.4 ⎢ ⎥ = 1.4⎢ h st ⎣ 4.1 ⎥ ⎦ ⎢ ⎥ ⎣ ⎦

= 387.9 kN say 390 kN. For C202, upper height, assuming same design as sections of B2001 and B2002

bL M u ,lim (Table 18) = 585 kNm bR M u ,lim (Table 18) = 585 kNm, and

The spacing of hoops is also checked in terms of maximum permissible spacing of shear reinforcement given in IS 456:2000, Clause 26.5.1.5 b x d = 500 x 450 mm. Using 8# hoops, Asv = 5 x 50 = 250 mm2.

hst = 5.0 m.

IITK-GSDMA-EQ26-V3.0

Page 45

Design Example of a Building

Then

l0 shall not be less than

⎡ M bLlim + M bRlim ⎤ u, u, Vu = 1.4 ⎢ ⎥ h st ⎢ ⎥ ⎣ ⎦ ⎡ 585 + 585 ⎤ = 1.4⎢ = 327.6 kN. ⎣ 5.0 ⎥ ⎦

Design shear is maximum of (a) and (b).

Then, design shear Vu = 390 kN. Consider the column as a doubly reinforced beam, b = 500 mm and d = 450 mm. As = 0.5 Asc = 0.5 x 6 968 = 3 484 mm2. For load combination 12, Pu = 3,027 kN for lower length and Pu = 2,547 kN for upper length. Then,

(i) D of member, i.e., 500 mm (ii) i.e., Lc , 6

(4100 – 600) = 583 mm for column C202 6 (5000 – 600) =733 mm for column C302. and, 6

Provide confining reinforcement over a length of 600 mm in C202 and 800 mm in C302 from top and bottom ends of the column towards mid height. As per Clause 7.4.2 of IS 13920:1993, special confining reinforcement shall extend for minimum 300 mm into the footing. It is extended for 300 mm as shown in Figure 12. As per Clause 7.4.6 of IS 13920:1993, the spacing, s, of special confining reinforcement is governed by: s ≤ 0.25 D = 0.25 x 500 = 125 mm ≥ 75 mm

≤ 100mm

δ = 1+

= 1+

3 Pu Ag fck

(IS456: 2000, Clause 40.2.2)

3 ×3027×1000 = 2.45, for lower length, and 500× 500× 25 3× 2547×1000 = 2.22, for upper length. = 1+ 500× 500× 25 ≤ 1.5 Take δ = 1.5.

100As 100× 3484 = = 1.58 500× 450 bd τ c = 0.753 N/mm2 andδτc = 1.5 × 0.753 = 1.13 N/mm2 Vuc = δτc bd = 1.13× 500× 450×10-3 = 254.5 kN Vus = 390 − 254.5 = 135.5 kN Asv = 250 mm2 , using 8 mm # 5 legged stirrups. Then sv = 0.87 f y Asvd Vus = 0.87 × 415× 250× 450 = 299.8 mm 135.5 ×1000

i.e. Spacing = 75 mm to 100 mm c/c…… (1) As per Clause 7.4.8 of IS 13920:1993, the area of special confining reinforcement, Ash, is given by:

Ash = 0.18 s ≤ h

f ck fy

⎡ Ag ⎤ – 1.0⎥ ⎢ ⎣ Ak ⎦

Here average h referring to fig 12 is

100 + 130 + 98 + 100 = 107 mm 4

Ash = 50.26 mm2 Ak = 428 mm x 428 mm 50.26 = 0.18 x s x 107 x 50.26 = 0.4232 s s = 118.7 mm

25 ⎡ 500 × 500 ⎤ -1 415 ⎢ 428 × 428 ⎥ ⎣ ⎦

Use 200 mm spacing for general ties.

1.11.3.3. Design of Special Confining Hoops:

As per Clause 7.4.1 of IS 13920:1993, special confining reinforcement shall be provided over a length l0, where flexural yielding may occur.

IITK-GSDMA-EQ26-V3.0

≤ 100 mm

…

(2)

Provide 8 mm # 5 legged confining hoops in both the directions @ 100 mm c/c.

Page 46

Design Example of a Building

600

8 mm # 5 leg @ 100 mm c/c 500 8 – 25 mm # + 8 – 22 mm # 500 8 mm # 5 leg @ 200 mm c/c (4 no.) 100 130 98 100 100 100 130 98

8 mm # 5 leg @ 150 mm c/c (8 no.) 4400

8 mm # 5 leg @ 200 mm c/c (4 no.)

8 mm # 5 leg @ 100 mm c/c (20 no.)

600

8 – 25 mm # + 8 – 22 mm # 8 mm # 5 leg @ 200 mm c/c ( 2no.)

8 mm # 5 leg @ 150 mm c/c (8 no.)

3500

8 mm # 5 leg @ 200 mm c/c (3 no.) 16 – 25 mm # 8 mm # 5 leg @ 100 mm c/c (25 no.) 600 800 × 800 × 800 Pedestal M25 M20 Concrete 450 28-16 # both ways M10 Grade 4200 n 102 – 202 – 302 re – 9 150

* Beam reinforcements not shown for clarity * Not more than 50 % of the bars be lapped at the section

800

900 100

Figure 12 Reinforcement Details

IITK-GSDMA-EQ26-V3.0

Page 47

Design Example of a Building

1.11.3.4. Design of hoops at lap

As per Clause 7.2.1 of IS 13920:1993, hoops shall be provided over the entire splice length at a spacing not exceeding 150 mm centres Moreover, not more than 50 percent of the bars shall be spliced at any one section. Splice length = Ld in tension = 40.3 db. Consider splicing the bars at the centre (central half ) of column 302. Splice length = 40.3 x 25 = 1008 mm, say 1100 mm. For splice length of 40.3 db, the spacing of hoops is reduced to 150 mm. Refer to Figure 12.

1.11.3.5. Column Details

Mx = 12 kNm, Mz = 6 kNm. At the base of the footing P = 2899 kN P’ = 2899 + 435 (self-weight) = 3334 kN, assuming self-weight of footing to be 15% of the column axial loads (DL + LL).

Mx1 = Mx + Hy × D = 12 + 16 × 0.9 = 26.4 kNm Mz1 = Mz +Hy × D = 6 + 12 × 0.9 = 18.8 kNm. For the square column, the square footing shall be adopted. Consider 4.2 m × 4.2 m size. A = 4.2 × 4.2 = 17.64 m2 Z=

The designed column lengths are detailed in Figure 12. Columns below plinth require smaller areas of reinforcement; however, the bars that are designed in ground floor (storey 1) are extended below plinth and into the footings. While detailing the shear reinforcements, the lengths of the columns for which these hoops are provided, are slightly altered to provide the exact number of hoops. Footings also may be cast in M25 grade concrete.

1 × 4.2 × 4.22 = 12.348 m3. 6

P 3344 = = 189 kN/m2 A 17.64

M x1 26.4 = = 2.14 kN/m2 Zx 12.348

M z1 18.8 = = 1.52 kN/m2 Zz 12.348

Maximum soil pressure = 189 + 2.14 + 1.52 = 192.66 kN/m2 0 kN/m2.

Case 2:

1.12.

Design of footing: (M20 Concrete):

It can be observed from table 24 and table 26 that load combinations 1 and 12 are governing for the design of column. These are now tried for the design of footings also. The footings are subjected to biaxial moments due to dead and live loads and uniaxial moment due to earthquake loads. While the combinations are considered, the footing is subjected to biaxial moments. Since this building is very symmetrical, moment about minor axis is just negligible. However, the design calculations are performed for biaxial moment case. An isolated pad footing is designed for column C2. Since there is no limit state method for soil design, the characteristic loads will be considered for soil design. These loads are taken from the computer output of the example building. Assume thickness of the footing pad D = 900 mm. (a) Size of footing:

Case 1:

Combination 12, i.e., (DL – EXTP) Permissible soil pressure is increased by 25%. i.e., allowable bearing pressure = 200 × 1.25 = 250 kN/m2. P = (2291 – 44) = 2247 kN Hx = 92 kN, Hz = 13 kN Mx = 3 kNm, Mz = 216 kNm.

Page 48

Combination 1, i.e., (DL + LL) P = (2291 + 608) = 2899 kN Hx = 12 kN, Hz = 16 kN

IITK-GSDMA-EQ26-V3.0

Design Example of a Building

At the base of the footing P = 2247 kN P’ = 2247 + 435 (self-weight) = 2682 kN. Mx1 = Mx + Hy × D = 3 + 13 × 0.9 = 14.7 kNm Mz1 = Mz +Hy × D = 216 + 92 × 0.9 = 298.8 kNm.

The same design will be followed for the other direction also. Net upward forces acting on the footing are shown in fig. 13.

1700

800

1700 Z Z2 Z1

P 2682 = = 152.04 kN/m2 A 17.64

M x1 14.7 = = 1.19 kN/m2 12.348 Zx M z1 298.8 = = 24.20 kN/m2 Zz 12.348

Maximum soil pressure = 152.04 + 1.19 + 24.2 = 177.43 kN/m2 0 kN/m2.

Case 1 governs.

167 kN/m2 1700 417 800

‘

826

1700

Z Z2 Z1 417 1283 826 874

(a) Flexure and one way shear

250 kN/m2

In fact all combinations may be checked for maximum and minimum pressures and design the footing for the worst combination. Design the footing for combination 1, i.e., DL + LL.

216.4 224.6 232.7 (b) Upward pressure 4200

P 2899 = = 164.34 kN/mm 2 A 17.64

Factored upward pressures for design of the footing with biaxial moment are as follows. For Mx pup = 164.34 + 2.14 = 166.48 kN/m2 pu,up = 1.5 × 166.48 For Mz pup = 164.34 + 1.52 = 165.86 kN/m2 pu,up = 1.5 × 165.86 = 248.8 kN/m

D 4200

A 1634

= 249.72 kN/m2

417 1283 (c) Plan

Since there is no much difference in the values, the footing shall be designed for Mz for an upward pressure of 250 kN/m2 on one edge and 167 kN/m2 on the opposite edge of the footing.

IITK-GSDMA-EQ26-V3.0

Figure 13

Page 49

Design Example of a Building

(b) Size of pedestal: A pedestal of size 800 mm × 800 mm is used. For a pedestal A = 800 × 800 = 640000 mm2

1449 × 10 6 = 354 mm 2.76 × 4200

Try a depth of 900 mm overall. Larger depth may be required for shear design. Assume 16 mm diameter bars. dx = 900 – 50 – 8 = 842 mm dz = 842 – 16 = 826 mm. Average depth = 0.5(842+826) = 834 mm. Design for z direction.

M uz bd 2 = 1449 × 10 6 = 0.506 4200 × 826 × 826

1 × 800 × 8002 = 85333333 mm3 6

For case 1

2899 × 1000 (26.4 + 18.8) × 106 q01 = + 800 × 800 85333333

= 4.53 + 0.53 = 5.06 N/mm … For case 2 q02 =

(1)

pt = 0.145, from table 2, SP : 16

2247 × 1000 (14.7 + 298.8) × 106 + 800 × 800 85333333

Ast =

0.145 × 4200 × 900 = 5481 mm 2 100 0.12 × 4200 × 900 = 4536 mm 2 100

= 3.51 + 3.67 = 7.18 N/mm2 Since 33.33 % increase in stresses is permitted due to the presence of EQ loads, equivalent stress due to DL + LL is

Ast , min =

(Clause 34.5, IS: 456) Provide 28 no. 16 mm diameter bars. Ast = 5628 mm2.

Spacing = 4200 − 100 − 16 = 151.26 mm 27

7.18 ÷ 1.33 = 5.4 N/mm .

…

(2)

From (1) and (2) consider q0 = 5.4 N/mm2. For the pedestal

tan α ≥ 0.9

This gives

100 × 5.4 +1 20

(d) Development length: HYSD bars are provided without anchorage. Development length = 47 × 16 = 752 mm Anchorage length available = 1700 – 50 (cover) = 1650 mm … (o.k.) (e) One-way shear: About z1-z1 At d = 826 mm from the face of the pedestal

V u= 0.874 × 232.7 + 250 × 4.2 = 886 kN 2

tan α ≥ 4.762 , i.e., α ≥ 78.14 0

Projection of the pedestal = 150 mm Depth of pedestal = 150 × 4.762 = 714.3 mm. Provide 800 mm deep pedestal. (c) Moment steel: Net cantilever on x-x or z-z = 0.5(4.2-0.8) = 1.7 m. Refer to fig. 13.

1 1 2 ⎡1 ⎤ M uz = ⎢ × 216.4 × 1.7 × × 1.7 + × 250 × 1.7 × × 1.7 ⎥ × 4.2 3 2 3 ⎣2 ⎦

b = 4200 mm, d = 826 mm

τv =

Vu 886 × 1000 = = 0.255 N/mm 2 bd 4200 × 826

= 1449 kNm For the pad footing, width b = 4200 mm For M20 grade concrete, Qbal = 2.76. Balanced depth required

100 Ast 100 × 5628 = = 0.162 bd 4200 × 826

τc = 0.289 N/mm2 τv

… … (o.k.)

IITK-GSDMA-EQ26-V3.0

Page 50

Design Example of a Building

(f) Two-way shear: This is checked at d/2, where d is an average depth, i.e., at 417 mm from the face of the pedestal. Refer to fig. 13 (c).

Width of punching square = 800 + 2 × 417 = 1634 mm. Two-way shear along linr AB

⎛ 224.6 + 250 ⎞⎛ 1.634 + 4.2 ⎞ =⎜ ⎟⎜ ⎟ ×1.283 = 883 kN. 2 2 ⎝ ⎠⎝ ⎠

= 1.2 × q02= 1.2 × 7.18 = 8.62 N/mm2. Thus dowels are not required. Minimum dowel area = (0.5/100) × 800 × 800 = 3200 mm2. Area of column bars = 7856 mm2 It is usual to take all the bars in the footing to act as dowel bars in such cases. Minimum Length of dowels in column = Ld of column bars = 28 × 25 = 700 mm. Length of dowels in pedestal = 800 mm. Length of dowels in footing = D + 450 = 900 + 450 = 1350 mm. This includes bend and ell of the bars at the end. The Dowels are lapped with column bars in central half length of columns in ground floors. Here the bars are lapped at mid height of the column width 1100 mm lapped length. Total length of dowel (Refer to fig. 12) = 1350 + 800 + 600 + 1750 + 550 = 5050 mm. Note that 1100 mm lap is given about the midheight of the column. (h) Weight of the footing: = 4.2 × 4.2 × 0.9 × 25 = 396.9 kN

Acknowledgement

τv =

Vu 883 × 1000 = = 0.648 N/mm 2 bd 1634 × 834

Design shear strength = ksτc, where ks= 0.5 + τc and τc = (bc/ l c ) = 500/500 = 1 ks= 0.5 +1 = 1.5 ≤ 1, i.e., ks = 1 Also,

τ c = 0.25 f ck = 0.25 20 = 1.118 N/mm2

Then ksτc = 1.118 = 1.118 N/mm2. Here

τv

…

(o.k.)`

(g) Transfer of load from pedestal to footing: Design bearing pressure at the base of pedestal = 0.45 f ck = 0.45 × 25 = 11.25 N/mm2 Design bearing pressure at the top of the footing

A1 × 0.45 f ck = 2 × 0.45 × 20 = 18 N/mm 2 A2

Thus design bearing pressure = 11.25 N/mm2. Actual bearing pressure for case 1 = 1.5 × q01= 1.5 × 5.06 = 7.59 N/mm2. Actual bearing pressure for case 2 .

The authors thank Dr R.K.Ingle and Dr. O.R. Jaiswal of VNIT Nagpur and Dr. Bhupinder Singh of NIT Jalandhar for their review and assistance in the development of this example problem.

IITK-GSDMA-EQ26-V3.0

Page 51

General

Similar Papers