Diagrid Structural System: Comparative Study to Reduse the Earthquake Forces on High Rise Building

Construction of multi-storey building is rapidly increasing throughout the world. But when the height of structure increases then the consideration of lateral load is very much important. For that the lateral load resisting system becomes more important than the structural system that resists the gravitational loads. The lateral load resisting systems that are widely used are rigid frame, shear wall, wall frame, braced tube system, outrigger system and tubular system. Recently the diagrid –diagonal grid structural system is widely used for tall buildings due to its structural efficiency and aesthetic potential provided by the unique geometric configuration of the system. Generally, for tall buildings diagrid structure steel is used. In present work concrete diagrid structure is analysed and compared with convensional concrete building. DIAGRID is combination of words ―diagonal‖ and ―grid‖. It is a system of triangulated beams -straight or curved and horizontal rings that together make up a Diagrid structural system. Therefore the diagonal members in diagrid structures act both as inclined columns and as bracing elements, and carry gravity loads as well as lateral forces due to their triangulated configuration. There are engineering based reasons that would suggest the use of diagrid, Some of them are: ● Increased the stability due to Triangulation. ● Combination of the gravity and lateral load bearing system potentially provide more efficiency. ● Provide alternate load path (redundancy) in the event of structural failure. ● Reduced weight of superstructure can translate into a reduced load on foundation. Load distribution in diagrid structure is as follows-

Fig. 01: Load transfer system of Diagrid Structure The load path can be divided into two main parts, vertical load and horizontal shear as shown in fig. vertical load will be transferred in the form of an axial load from the diagrid members above the node to the gusset plate and stiffners, then to diagrid members below the nodes as shown. The horizontal shear will be in the form of an axial load in the diagrid members above the nodes with one in compression and one in tension to the gusset plate and stiffners. The force will be then transferred as shear force in the gusset plate and then to the other pair of tensile and compressive forces on the diagrid member below the node. Famous examples of diagrid structure all around the world are the Swiss Re London, Hearst Tower in New York, Capital Gate tower in Abu Dhabi, IBM Building in Pittsburgh, Cyclone Tower in Asan (Korea) and new World Trade Centre in New York.

In present work concrete diagrid structure is analysed and compared with conventional concrete building. A regular 4 and 12 storey building with plan size 20m x 20m located in seismic zone III is considered for analysis. All the structural members are design as per IS-800 (2000).STAAD-PRO software is used for the analysis of structural members. All load combinations of seismic forces are considered as per IS-1893:2002. Comparison of analysis results in terms of mode shape, lateral displacement, storey drift.

(A) Slab Design- Slabmark- S-1,S-2 & S-3 Type- Two way continuous with corners restrained Spans- Short span (Lx) in metre=5,in mm=5000 Long span (Ly) in metre= 5,in mm=5000 Aspect ratio= Ly ,so =1 Lx Live Load (in KN/m²)= 4 Floor Finish (in KN/m²) = 2.5 Trial Depth- In case of two-way slab, (L/d) ratio for deflection criteria is related to short span. In case of two-way slab, the design moments are small compared to those in one-way slabs, percentage of steel required in two way slabs in general is very low (Between 0.20 to 0.30% for M-20 and fe-415) so assumed 0.28% steel. For, Pt=0.28% …..so Pt= 0.28 corresponding to fs= 240 N/mm²

for continuous slab Basic (L/d) ratio (rb= 26 ) …… As per IS 456:2000, clause no. 23.2.1 Required (d) in mm= Lx/(α*rb)= 113.12217 , say (d) in mm ≈ 120 Assuming (d′) in mm= 20 Required (D) in mm= 140,in m0.14 Loads- consider 1 metre width of slab Wu= 1.5 [(25 x D)+ LL+ FF ] So Wu in KN/m²= 15 Design moments- Boundary condition no.=4 …….. As per IS 456:2000, table 26,pg. no.91 Wu .(Lx)²= 375 Ly / Lx is =1 ….. As per IS 456:2000, table 26,pg. no.91 Check for concrete depth- for slabs (b) in mm= 1000 Mur.max= Rumax.b.(d)² …..Rumax= 0.138xfck here fck= 20 for outer bars (KN.m)=39.744……safe for inner bars (KN.m)=33.396…..safeRequired Ast =

For , M 20 , so fck =20 For , fe 415,so fy= 415 0.5fck/fy=0.02409639 fckxbx(d)² =288000000 b.d =120000

Spacing= 299.047619 ≈290 Check for deflection - Required (Pt) at mid-span of short span= Ast x100 bx dRequired (Pt) in % = 0.2733333 ……………….Safe Check for shear - β = Ly/Lx ,so = 1 (a) Long edge – continuous

Vumax= 30KN Ast= Provided Ast of short span of support (in mm²)=449 Pt = Ast x100 , Pt (in%)=0.374167 bx d K=1.3…….As per IS 456:2000, clause no. 40.2.1.1 ζuc= ?…….. As per IS 456:2000, table 19,pg. no.73 Check for Development length- (a) Long edge – continuous Required (Ld)=ɸ . Ϭs 4. ζbd

ζbd =1.2*1.6 =1.92 Required (Ld) in mm=470.117188≈471 Available (Ld)=0.3 * Lx

(b) Long edge - discontinuous Assuming 50% bars bent up, M1=Mu, midspan of longspan (Mu) 2 So M1=6.5625 KN.m Now, K*M1*(1000/Vumax)+Lo (c) Short edge - continuous Available (Ld)=0.4 * Ly so = 2000……………..safe (d) Short edge - discontinuous Assuming 50% bars bent up, M1=Mu, midspan of longspan (Mu) 2 So M1=6.5625 KN.m Now, K*M1*(1000/Vumax)+Lo =599.16667……………….safe So provide slab thickness (D) in mm= 140 ,0.14 m

(B) BEAM DESIGN

(i) INTERNAL TRANSVERSE BEAM:-(ROOF AND SLAB) (1) End Condition: Simply Supported (2) Span L (meter):5 (3) Section : Assumed b (mm)= 300in merer0.3 Depth (mm) =500in meter0.5 d" (mm)=40 d(mm)=460 Wu=1.5 x(Self weight of Beam+2*RoofSlab'sTriangular Load) Roof Slab Load =(self-weight + Floor Finish ) Floor finish =2.5 Live load=1.5 Roof Slab Load = 6 i.e. Self-weight of Beam =2.7 Slab's Triangular Load = 20 Wu =34.05Kn/m Wus =15.3Kn/m slab=30 Total dead load=32.7 slab=7.5 Total working load=40.2 maximum load Wmax=60.3≈61 minimum load Wmin=32.7≈33 Ratio LL/Total Load=0.19 (5) Design Moment: Mu = (Wu. L²)/8 Mu=106.40625Kn.m (6) Maximum Ultimate Moment of Resistance of Rectengular Section = For M 25 ,fck=20 fck 500,fy=415 Mur.max= Ru.max b.d²i.e583.04961 Ru.max =0.36 fck.Kumax (1-0.42 Kumax)

=2.75543292

Kumax=700/1100+0.87fy =0.4791075 If Murmax > Mu……………….Singly Reinforced (7) Main steel :

b.d=138000 {1-(4.6*Mu.10^6/fck*b*d²)}^1/2 = Required Ast=3325.301205mm² assumed Diameter of stirrups ɸ (mm)=6 assumed Diameter of Bar ɸ (mm)=25 Number of Bars=6.777683984≈ 7 (8) check for Width= Required b=n.ɸ+(n+1)x25+2.ɸst Required b (mm)=287……………….ok (9) Check for Effective cover d : For Moderate Environment (M 25) nominal cover in mm=20 d'=Nominal cover + Diameter of stirrups

+ (ɸ/2)

d'=38.5…………..ok (10) Design for Shear : Vu.max = Wu*L/2so,85.125 Ast1=3434.375 Pt =100.Ast1/b.d = 2.488677536….in % ζuc= ?…….. As per IS 456:2000, table 19,pg. no.73 ζuc (N/mm²)= 0.815 ζuc=0.815 Vuc= ζuc.bd , so, Vuc=112.47Kn Vur.min=167.67 ……………..min. stirrups are sufficient (11) Check for Development Length : Required Ld= 47 ɸ = 1175in mm

1.3 M1/V+L0

275.8408984 Kn.m………. M1 L0=Ld/3-bs/2 = 276.6666667

1.3 M1/V+L0 =4489.214894

Required Ld=1175 mm (12) Check for Deflection : Actual L/d = 10.86956522 Basic L/d = 20 …………………. safe (13) Load on column : Vu.max = Wu*L/2 Load on column = so, in KN = 85.13 ………………. safe (ii)INTERNAL&EXTERNAL LONGITUDINAL ROOF BEAM & INTERNAL LONGITUDINAL FLOOR BEAM:- (1) End Condition: Simply Supported (2) Span L (meter): 5 (3) Section : Assumed b (mm)= 300 in merer 0.3 Depth (mm) = 300 in meter 0.3 d" (mm)= 40 d(mm)= 260 Slab Thickness (mm)= 140 in merer 0.14 (4) Loads: Wu =1.5x(Selfweight of Beam + 2*Roof Slab's TriangularLoad) Roof Slab Load =(self weight + Floor Finish )

i.e. Self weight of Beam =1.2 Slab's Triangular Load =20 Wu =31.8Kn/m Wus =13.05Kn/m (5) Design Moment :Mu = (Wu. L²)/8 Mu=99.375 Kn.m (6) Maximum Ultimate Moment of Resistance of Rectengular Section = For M 25 , fck= 20 fck 500, fy= 415 Mur.max= Ru.max b.d² i.e 186.26727 Ru.max = 0.36 fck.Kumax (1-0.42 Kumax)

=2.75543292

Kumax= 700/1100+0.87fy = 0.4791075 If Murmax > Mu……….Singly Reinforced (7) Main steel : 0.5* fck/fy= 0.024096386 b.d= 78000 {1-(4.6*Mu.10^6/fck*b*d²)}^1/2 = Required Ast= 1879.518072 mm² assumed Diameter of stirrups ɸ (mm)= 6 assumed Diameter of Bar ɸ (mm)= 25 Number of Bars=3.830864861≈ 4 (8) Check for Width= Required b=n.ɸ+(n+1)x25+2.ɸst Required b (mm)=212……………….ok (9) Check for Effective cover d : For Moderate Environment (M 25) d'= 38.5…………..ok (10) Design for Shear : Vu.max = Wu*L/2 so,79.5 Ast1= 1962.5 Pt = 100.Ast1/b.d = 2.52 ….in % ζuc= ? …….. As per IS 456:2000, table 19,pg. no.73 ζuc (N/mm²)= 0.8108 ζuc= 0.8108 Vuc= ζuc.bd , so, Vuc=63.2424Kn Vusv.min=0.4 bd, so, Vusv.min=31.2 Vur.min= 94.4424…..min. stirrups are sufficient (11) Check for Development Length : Required Ld= 47 ɸ = 574 in mm 61.82785102 Kn.m………. M1 L0=Ld/3-bs/2 = 73

1.3 M1/V+L0= 1701.903485 MM

Required Ld=564 mm…………………. safe (12) Check for Deflection : Actual L/d =12.19512 Basic L/d =20……………….safe (13) Load on column : Vu.max = Wu*L/2 so, in KN49.34 Load on column =98.69

Column Height=3 assume,column width=300 =0.3 METER assume,column depth=500 = 0.5 METER Shear from longitudinal beam=171.5 Maximum Shear from transverse beam=182.75 self weight of beam=16.875 Load= 3340.125 Shear from longitudinal beam=98.69 Maximum Shear from transverse beam=91.38 self weight of beam=16.875 Load=2483.34 Shear from longitudinal beam=98.69 self weight of beam=16.875 Load=462.26 Total load on column=251.429≈252 Pu (Kn)=987 Assume Asc=1%Ag Asc=0.01Ag Ac=0.9999Ag Assumed Column is short column Slenderness Ratio=Leff / Least lateral dimension 12 =2700/Least lateral dimension Least lateral dimension (b) =225≈300 Calculation of Ag Pu= 0.4 fck.Ac + 0.67 fy.Asc fck= 20 fy=415 2.7805 Ag 987000 10.7797 Ag Ag =91560.98964 Column is Rectangular Other Dimension =Ag /b Other Dimension =305.2033≈500 So provide 300 X 500 mm column

3.PROBLEM STATEMENT

Type- G+3 Storeyed R.C. Framed Structure Floor to Floor Height- 3 meter Loads- Live Load-Roof-1.5 KN/m2 -Floors-4KN/m2 Dead Load-Floor Finish-Roof-2.5 KN/m2 Floor-1KN/m2 Slab Thickness-140 mm Beam Section (a)Roof level- Longitudinal External and Internal-300 x 300 mm Transverse-300 x 500 mm Beam Section (b) Floor level- Longitudinal External-300 x 450 mm Longitudinal Internal-300 x 300 mm Transverse-300 x 500 mm Column Section-300 x 500 mm 4. G+3 STAAD_Pro Model

DETERMINATION OF BASE SHEAR MANUALLY

• FRAME STRUCTURE

(1) Mass of column- [(0.300x0.500x25)x25]x4 = 375 KN (2) Mass of Beam- [(0.300x0.300x25)x(20x14)]=630 KN [0.300x0.450x25)x(20x6)]= 405 KN [0.300x0.500x25)x(20x20)]=1500 KN So total mass of beam = 2535 KN (3) Mass of Slab- Roof Slab- [(20x20)x(0.140x25+2.5)]

=2400 KN

Floor slab- [(20x20)x(0.140x25+1+2)]x3

= 7800 KN

Seismic weight of building-

= 375+2535+2400+7800 =13110 KN

Step 2- Determination of Fundamental Natural Period Ta = 0.075 x h ^ 0.75 = 0.075x 12 ^0.75 = 0.48 seconds Step 3- Determination of Design Base Shesr Vb= Ah x W Ah= (Z/2)x(I/R)x(Sa/g) Z= Zone Factor= III=0.16 I= Importance Factor = 1 R= Response Reduction Factor

= SMRF= 1

Sa/g = Average response acceleration coefficient for medium soil Vb = 0.04 x 13110 KN

= 524.40 KN DESIGN SEISMIC BASE SHEAR OF FRAME = 524.40 KN • DIAGRID STRUCTURE

Step 1- calculation of lumped masses at various floor level (1) Mass of column- [(0.300x0.500x25)x4]x4 = 60 KN [(0.300x0.500x25x3.91)/2]x85 = 620 KN So total mass of column = 680 KN (2) Mass of Beam- [(0.300x0.300x25)x(20x16)]=720 KN [0.300x0.450x25)x(20x6)]= 405 KN [0.300x0.500x25)x(20x20)]=1500 KN Deduct releases= (0.300x0.450x25x15x12) +(0.300x0.500x25x2.5x12) = 720 So total mass of beam = 1905 KN (3) Mass of Slab- Roof Slab- [(20x20)x(0.140x25+2.5)]

=2400 KN

Floor slab- [(20x20)x(0.140x25+1+2)]x3

= 7800 KN

Seismic weight of building-

= 680+1905+2400+7800 =12785 KN

Step 2- Determination of Fundamental Natural Period

= 0.075x 12 ^0.75 = 0.48 seconds Step 3- Determination of Design Base Shear Vb= Ah x W Ah= (Z/2)x(I/R)x(Sa/g) Z= Zone Factor= III=0.16 I= Importance Factor = 1 R= Response Reduction Factor

= SMRF= 1

Sa/g = Average response acceleration coefficient for medium soil

= 2.5

SO, Ah= (0.16/2)x(1/5)x2.5 = 0.04 Vb = 0.04 x 12785 KN

= 511.40 KN

Design Seismic Base Shear Of Diagrid= 511.40 Kn

5. DETERMINATION OF BASE SHEAR BY STAAD-PRO

Design Seismic Base Shear Of Frame = 0.0400x 13110 Kn =524.40 Kn Design Seismic Base Shear Of Diagrid = 0.0400x 12785 Kn = 511.40 Kn

BASE SHEAR

………………..Validation Satisfied 7. G+11 STAAD_Pro Model

(1) MODE SHAPE-

(2) DISPLACEMENT IN (MM)-

(a) COLUMN ―A‖

COLUMN ―B‖

COLUMN ―C‖

OLUMN ―D‖

GRAPH OF DISPLACEMENT –

DRIFT IN (MM)-

(a) COLUMN ―A‖

COLUMN ―B‖

(c) COLUMN ―C‖

(d)COLUMN ―D‖

GRAPH OF DRIFT –

In this paper, comparative analysis & design of 4-storey & 20-storey diagrid structural system building and simple frame building is presented here. A regular floor plan of 20m x 20m size is considered. STAAD-PRO software is used for modelling and analysis of structure. Analysis results like mode shape, displacement, storey drift are presented here. Also design of both structures is done and optimum member sizes are decided to satisfy the code criteria. structural system.  As the lateral loads are resisted by diagonal columns, the storey displacement is very much less in diagrid structure as compared to the simple frame building.  The storey drift is very much less for diagrid structural system.  Diagrid provide more resistance in the building which makes system more effective.  The design of both structures is done by using same member size. The higher sizes of members are selected to prevent the failure criteria.

Mr. S.R. Suryavanshi for their guidance and also very much thankful to Civil Engineering Department & Structural Engineering Department, Faculty of JSPM‘s Imperial College Of Engineering, Wagholi, Pune for giving such a good facilities and platform to complete the work and also my dear friends who have support me to complete this work.

Ali, M. M. & Moon, K. (2007). Structural Developments in Tall Buildings: Current Trends and Future Pros-pects. Architectural Science Review, Vol. 50.3, pp. 205-223. Connor, J. J. (2003). Introduction to Structural Motion Control. New York: Prentice Hall. IS: 1893(Part-I)-2002, Criteria for Earthquake Resistant Design of Structures, Bureau of Indian Standard, New Delhi. IS: 456-2000. Plain and Reinforced Concrete- Code of Practice (Fourth Revision), Bureau of Indian Standard, New Delhi. Moon, K. (2008). Optimal Grid Geometry of Diagrid Structures for Tall Buildings.

pp. 205-230.

ME Students, JSPM‘s ICOER, Wagholi