Designproject

INDEX 1.

Problem

2.

Drafting in AUTOCAD

3.

Placing the columns

4.

Modeling the structure in SAP 2000 V12

5.

Defining Sections and Materials

6.

Defining load cases

7.

Dead Load & Live load

8.

Earthquake Loads

9.

Analyzing and Grouping the results

10.

Design of Beams

11.

Design of Columns

Problem: To design a RCC shopping mall with the following dimensions given.

Dimensions of the mall = 101m x 60m x 12m. The boxes marked ‘S’ are the shops and the ones marked in ‘T’ are the toilets. There will be two entrances, one on each side. The parking will be on one side and the other side is for a green area. Four lifts marked ‘L’ are provided for transport of goods and people. There will be an escalator at the central area which is near to both the entrances and two staircases on each wing end. People will move around through the corridors provided all along the length in front of the showrooms (marked with dotted lines). The beams, columns, as well as slabs are made of RCC and walls will be there on left and right faces of the structure. The other faces will be covered by unbreakable glass. The inner dimensions are marked on the diagram.

Fig.1. AUTOCAD drawing showing the building plan and elevation

1.

AUTOCAD DRAFTING The drawing is drafted in AUTOCAD with the desired dimensions. The dimensions comply with the BUILDING BY LAWS as described in the Indian Standard Code. The following drawings describe the design of ground floor and the rest of the floors.

2.

PLACING COLUMNS The structure is designed to carry the dead load (self weight, finishing loads & the loads of the goods), live load and the Earthquake loads. The following diagram explains the placement of columns and the beams joining the columns throughout the structure.

Fig.2. Placing Beams and Columns in the structure

3.

MODELLING IN SAP 2000 (V 12) After we have the exact locations of the beam column joints and the dimensions of each element in the structure, we construct the structure in SAP 2000, assign the loads, and run the analysis to find the resultant moment and axial forces about each of the axes. The snapshots below show the plan of the frames at different floors. The first picture is the position of the foundation, the 2nd one is of first floor, 3rd one is of second floor, and the 4th one is the top floor.

Analysis

Fig. 3 XY PLANE at Z=0

Fig.4. XY PLANE AT Z=4

Fig.5. XY plane at Z=8

Fig.6. XY plane at Z=12

Fig.7. XY plane at Z=16

Fig.8. 3D view of the structure After drawing is completed, the material property is defined and the section dimensions of each of the frames are to be specified. The material of the beam, column and slabs in our case is RCC with different section dimensions. It should be noted that a linear analysis is to be carried out and hence the designs should be carried out with different cross-sections till the suitable one with allowable deflection is obtained.

Material Properties: i.

Beams and Columns( case “M25”) Weight per unit volume, fck = 25 Mpa = 25 x 106 N/m3 Modulus of elasticity, E = 5000sqrt (fck) = 25000000 N/m3 Poisson’s ratio = 0.15

ii. Walls (case “MASONRY”) Weight per unit volume = 20 Mpa

SECTION PROPERTIES I.

Beams ( case identifier “BEAM”) Depth=0.4 m , Width= 0.3 m, Material= M25 RCC(as described above)

II. Columns (identifier “COLUMN”) Depth= 0.4m; Width=0.4m; Material=M25 RCC

III.

Walls ( identifier “masonry”)

Calculation of Earthquake Loads Using Clause 7.53 of IS 1893:2002 Total design lateral Forces = Vb = Ah×W (1) Where Ah = horizontal acceleration spectrum which is given as := (Z×I×Sa)/ (2×R×g) (from clause 6.4.2, IS 1893:2002) (2) Where, Z= Zone factor from table 2 of IS 1893:2002 I = Importance factor depends upon the functional use of the structures R = Response reduction factor I/R < = 1 (From table 7, IS 1893:2002) Sa/g = Avg. response acceleration coefficient for rock or soil W = Seismic weight of the building ( Clause 7.4.2) For Sa/g we require natural period of vibration ( Ta) ) for RC frame building where h = height of the building in ‘m’ Ta = (0.075 ×

(3)

) (4) Design lateral force at floor I ; Qi = Vb × Where Wi = Seismic weight at floor I, Hi = Height of floor ‘i’ measured from base& N = no. of storey in the building is the no. of levels at which the masses are located

1)

For design purpose we assume the following things:Calculation of Vb ; Vb = Ah×W ; Ah = (Z×I×Sa)/ (2×R×g) Where Z= 0.36(table 2 IS 1893:2002), I = 1, R= 5, Sa/g = 1.36 / T (from table 6.4.5) T= 0.6 0.55 T 4.0 Sa/g = 2.27

! ! ! ! ! ! ! % ' ' ' ' ' '

!"!# !"!# !"!# !"!# !"!# !"!# !"!# !" !"!$ !"!$ !"!$ !"!$ !"!$ !"!$

#" $ ! "$ '! "$ &"'%

%"$ %"$ %"$ %"$ %"$ %"$ %"$ "$ "!# "!# "!# "!# "!# "!#

&"&$ &"&$ &"&$ &"&$ &"&$ &"&$ &"&$ "' %"'# %"'# %"'# %"'# %"'# %"'#

$ "' $ "' $ "' $ "' $ "' $ "' $ "' '#" %&" %&" %&" %&" %&" %&"

!"' !"' !"' !"' !"' !"' !"' $" $"& $"& $"& $"& $"& $"&

"!# "!# " " " " "% "% "% "% " " "

" " " " " " '" '" '" '" " " "

% "' % "' #"% #"% #"% #"% &&" &&" &&" &&" #"% #"% #"%

!" !" #$" $ #$" $ #$" $ #$" $ #" #" #" #" #$" $ #$" $ #$" $

( & & # # # #

# # #

!"!$ !"!$ !"!% !"!% !"!% !"!% !" !" !" !" !"!% !"!% !"!%

#" $ ! "$ '! "$ &"'%

Dead loads 1)

For roof Wt. of slabs = 12675 KN Wt. of beams = 2708 KN

2)

For 2nd floor Wt. of slabs = 9378.5 KN Wt. of beams = 2170 KN

3)

For 3rd floor Wt. of slabs = 9378.5 KN Wt. of beams = 2170 KN

4)

For 4th floor Wt. of ground floor = 12675 KN Wt. of beams = 2728 KN

Distribution of Lateral Loads

Wt. of columns = 7744 KN Total Weight of the structure = 60876.3 KN = W Vb = 0.08 × 60876.3 = 4870.104 KN W1 = 33770.8 KN ( Seismic wt. of ground floor ) W2 = 15582.35 ( seismic weight of 1st floor) W3 = 15582.35KN ( seismic weight of 2nd floor) W4 = 32520 ( Seismic weight of the roof ) H1 = 4 ( height of ground floor from the base ) H2 = 9 ( height of 1st floor from base) H3 = 12 ( height of the 2nd floor from base) H4 = 16 ( height of roof from base) Using the Eq. 4 we can calculate the equivalent earthquake loads on each floor Q1 = 217 KN Q2= 402KN Q3 = 902KN Q4= 3349KN

Dimension of beam: 300 mm × 400 mm , Dimension of column: 400 mm × 400 mm The loads are tested for the following load combinations as per code IS 1893 which are as follows: 1) 1.5 ( DL + LL ) 2) 1.2 ( DL + LL + Elx/y ) 3) 1.2 ( DL + LL – Elx/y ) 4) 1.5 ( DL +Elx/y ) 5) 1.5 ( DL – Elx/y ) 6) 0.9 DL + 1.5 Elx/y 7) 0.9 DL- 1.5 Elx/y One of the load combination data (identified as “1.2 (DL + LL + Elx)”) is shown below. The other load combinations are done similarly.

Fig. Showing the different load combinations

Modal Analysis is performed on the structure and the maximum time period comes out to be 0.77 seconds. By the formula in IS 1893, T= 0.075 h0.75 Where h= Height of the building in m T= 0.7sec (as per code formula) This is quite close. The excess time period is due to the negation of the lateral earth pressures on the basement storey which SAP takes as another storey.

Analyses is then run for all the load combinations and from the moment and shear force obtained, beams and columns are grouped together as per a nearby values for convenience in design and construction.

Fig. The 3D view of the deformed shape in one of the load combinations

Grouping Force Resultants in Beams

.

At Floor Height = 4m

At Floor Height = 8m

At Floor Height = 12 m

At floor height = 16 m

Force Resultants in Columns !" # $ (

% )

%& '

&

'

&

'

!" # $

%& '

&

'

*

+

,

-

.

/

0

1

1

1

1

1

Here, P = Axial force in the members in KN M2 = Moment about the 2-2 axis (Y axis in the figure) M3= Moment about the 3-3 axis (Z axis in the figure)

&

'

Calculations Beams at Roof ( z=16m) Group I : 3m Moment= 270 KN-m Shear = 130 KN Material and Section Properties: Steel Fe 415 => fy = 415 Mpa Concrete M 25 => fck = 25 Mpa Width of beam, b = 300 >200 mm ok (IS 13920 6.1.3) Depth of beam, D = 500 b/D = 0.6 > 0.3 ok (IS 13920 6.1.2) Assumed d= 440 mm Assuming Moderate Exposure, Nominal Cover = 50 mm(Table 16)

Design for Flexure Mu, lim = 0.36bd2 fck(1-0.42xu,max/d)(xu,max/d) = 201 (Annex G-1.1) Where, xu/d = 0.87 fy Ast/0.36 fck bd Xu, max/d = 0.48 for Fe 415 If Moments of all beams designed are less than, Mu, lim=> Singly Reinforced Compression Steel = Min. Reinforcement = 0.85 bd / fy Mu =270> Mu, lim Hence Doubly Reinforced Mu/bd2=4.64 Pt= 1.56 , Pc= 0.4 ( as per table 51 SP16)

Tensile Steel Minimum no of tensile bars provided = 3 >2 ok (IS 13920 6.2.1) Minimum Steel Ratio

= 0.24sqrt (fck)/fy ok (IS 13920 6.2.1)

Spacing of bars: Spacing between bars = 70mm Diameter of Larger Bar> nominal max. size of coarse agg + 5mm =25 mm

Compression Steel No. of compression bars provided =4 (2-16 , 2-12 ) Spacing between bars =50mm Diameter of Larger Bar > nominal max. size of coarse agg + 5mm = 25 mm

Note: • The positive steel at joint >= 0.5 negative steel at that joint (IS 13920 6.2.3) • Steel on a side of any face >= 0.25 X negative steel at the joint face (IS 13920 6.2.4) Ductility Check: Pc,lim= 0.87fy(Pt-Pt,lim)/(fsc-0.447fck)=0.43 Pc,provided=0.464>Pc,lim Hence OK

Deflection Checks: Percentage of Tensile Steel, Pt = (Ast/bd)*100=1.6 Percentage of Compressive Steel, Pc = (Asc/bd)*100=0.465 From 23.2.1 Fst = 0.58fyAst,req./Ast=234.68 Now, from table: Modification factor for tensile RF, Kt =0.86 Modification factor for compressive RF, Kc =1.11 And from clause 23.2.1 Max. Span to effective depth ratio, (l/d)max = (l/d) basic * Kt * Kc=24.81 For Continuous Spans, (l/d) basic = 26 (Since all spans are = 10* diameter of bar = 100mm

(IS 13920 6.3.1)

Minimum diameter of bar forming the hoop =10mm > 6mm (IS 13920 6.3.2) Hence OK

Anchorage: In external joints, Anchorage length of bottom and top bars = Ld + 10*diameter (IS 13920 6.2.5) = 50*diameter =500mm Anchorage for 30Anchorage for 16-

bars = 1500mm bars = 800mm

Extend the bars to a length of 300 mm in the column then remaining is provided after the bend: For 30For 16-

bars =>300+1200mm bars =>300+500mm

Design of Columns (z=12 – 16m) Group E (500*500) 2nd floor ( z=12 – 16m) Axial Load, Pu = 700KN Moment about X-axis, Mx=275 KN-m Moment about Y-axis, My=700 KN-m Material Used: Concrete, M25 Steel, Fe 415 Unsupported Length = 4m (clause 25.1.3.b) Effective length, “l” = 0.85*4=3.4m Max of (I/Dx , I/Dy) = 5.66 20 (clause 25.4)

Properties of column: Width of column, b =500mm Length of column, D= 500mm Total moment, Mu= 1.15 of resultant (Mux, Muy) = 864.8 KN-m Axial load, Pu= 700KN Therefore, Pu/fck *b*D = 0..112 Mu/fck *b*b*D=0.28 Assume d’= 50mm , for (d’/D) =0.1 , fy=415KN P/fck=0.225 Bars provided: 12-36 and 4-32

, chart 44 used

Uniaxial Moment Capacity Check: From Chart 44: Pu/fck*b*D=0.112 P/fck=0.232 d’/Dy=0.1 Mux1/(fck*bD2)=0.29 Mux1=Muy1= 906KN-m >Mu

Safety Check under Biaxial Bending: Calculation of Puz and

(clause 39.6)

Puz = 0.45*fck*b*D+(0.75+fy-0.45*fck)*Asc Puz = 7541 Pu/Puz =0.08 = 1 (Mux/Mux1)

(clause 39.6) + (Muy/Muy1)

= 0.89 fy = 415 MPa Concrete M 25 => fck = 25 Mpa Width of beam, b = 300 >200 mm ok (IS 13920 6.1.3) Depth of beam, D = 400 b/D = 0.75 > 0.3 ok (IS 13920 6.1.2) Assumed d= 350 mm Assuming Moderate Exposure, Nominal Cover = 50 mm(Table 16)

Design for Flexure Mu, lim = 0.36bd2 fck(1-0.42xu,max/d)(xu,max/d) = 165.55 KN-m (Annex G-1.1) Where, xu/d = 0.87 fy Ast/0.36 fck bd Xu, max/d = 0.48 for Fe 415 If Moments of all beams designed are less than, Mu, lim=> Singly Reinforced Compression Steel = Min. Reinforcement = 0.85 bd / fy Mu =270 KN-m > Mu, lim Hence Doubly Reinforced Mu/bd2= 7.347 Pt= 2.447, Pc= 1.610 ( as per table 51 SP16)

Tensile Steel Minimum no of tensile bars provided = 4 >2 ok (IS 13920 6.2.1) ( 2- 28 dia + 2- 32 dia ) Minimum Steel Ratio

= 0.24sqrt (fck)/fy ok (IS 13920 6.2.1)

Spacing of bars: Spacing between bars = 70mm Diameter of Larger Bar> nominal max. size of coarse agg + 5mm =25 mm

Compression Steel No. of compression bars provided =4 (4-24 ) Spacing between bars =50mm Diameter of Larger Bar > nominal max. size of coarse agg + 5mm = 25 mm

Note: • The positive steel at joint >= 0.5 negative steel at that joint (IS 13920 6.2.3) • Steel on a side of any face >= 0.25 X negative steel at the joint face (IS 13920 6.2.4) Ductility Check: Pc,lim= 0.87fy(Pt-Pt,lim)/(fsc-0.447fck)=1.514 Pc,provided=1.74>Pc,lim Hence OK

Deflection Checks: Percentage of Tensile Steel, Pt = (Ast/bd)*100=2.59 Percentage of Compressive Steel, Pc = (Asc/bd)*100=1.74 From 23.2.1 Fst = 0.58fyAst,req./Ast=217.88 Now, from table: Modification factor for tensile RF, Kt =0.85 Modification factor for compressive RF, Kc =1.32 And from clause 23.2.1 Max. Span to effective depth ratio, (l/d)max = (l/d) basic * Kt * Kc=29.172 For Continuous Spans, (l/d) basic = 26 (Since all spans are = 10* diameter of bar = 100mm Length of extended 8 mm dia bar > = 10* diameter of bar = 80mm

(IS 13920 6.3.1)

Minimum diameter of bar forming the hoop =8 mm > 6mm (IS 13920 6.3.2) Hence OK

Anchorage: In external joints, Anchorage length of bottom and top bars = Ld + 10*diameter (IS 13920 6.2.5) = 50*diameter =500mm (considering 10 mm dia bars) Anchorage for 32Anchorage for 28-

bars = 1600mm bars = 1400mm

Extend the bars to a length of 300 mm in the column then remaining is provided after the bend: For 32For 28-

bars =>300+1600mm bars =>300+1400mm

Design of Columns (z=4-12 m) Group J (600*600) Axial Load, Pu = 1200KN Moment about X-axis, Mx=325 KN-m Moment about Y-axis, My=700 KN-m Material Used: Concrete, M25 Steel, Fe 415 Unsupported Length = 4m (clause 25.1.3.b) Effective length, “l” = 0.85*4=3.4m Max of (I/Dx , I/Dy) = 5.66 20 (clause 25.4)

Properties of column: Width of column, b =600mm Length of column, D= 600mm Total moment, Mu= 1.15 of resultant (Mux, Muy) = 887.53 Kn-m Axial load, Pu= 1200KN Therefore, Pu/fck *b*D = 0.083 Mu/fck *b*b*D=0.164 Assume d’= 90mm , for (d’/D) =0.15 , fy=415KN

, chart 44 used

Uniaxial Moment Capacity Check: d’=(nominal cover + ties + diameter/2)=50+8+32/2=74 mm Pu/fck*b*D=0.083 ,p/fck=0.164 Bars provided: 16 – 36 mm dia From Chart 44: Mux1/(fck*bD2)=0.099 Mux1=Muy1= 1823.95 KN-m >Mu Pu/fck*b*D=0.083 P/fck=0.0633 d’/Dy=0.15

Safety Check under Biaxial Bending: Calculation of Puz and

(clause 39.6)

Puz = 0.45*fck*b*D+(0.75+fy-0.45*fck)*Asc Puz = 13841.113 KN Pu/Puz =0.0867 = 1 (Mux/Mux1)

(clause 39.6) + (Muy/Muy1)

= 0.925 fy = 415 MPa Concrete M 25 => fck = 25 Mpa Width of beam, b = 400 >200 mm ok (IS 13920 6.1.3) Depth of beam, D = 500 b/D = 0.8 > 0.3 ok (IS 13920 6.1.2) Assumed d= 450 mm Assuming Moderate Exposure, Nominal Cover = 50 mm(Table 16)

Design for Flexure Mu, lim = 0.36bd2 fck(1-0.42xu,max/d)(xu,max/d) = 281 (Annex G-1.1) Where, xu/d = 0.87 fy Ast/0.36 fck bd Xu, max/d = 0.48 for Fe 415 If Moments of all beams designed are less than, Mu, lim=> Singly Reinforced Compression Steel = Min. Reinforcement = 0.85 bd / fy Mu =420> Mu, lim Hence Doubly Reinforced Mu/bd2=5.18 Pt= 1.73 , Pc= 0.55 ( as per table 51 SP16)

Tensile Steel Minimum no of tensile bars provided = 4(20 Minimum Steel Ratio

) >2 ok (IS 13920 6.2.1)

= 0.24sqrt (fck)/fy ok (IS 13920 6.2.1)

Spacing of bars: Spacing between bars = 110mm Diameter of Larger Bar> nominal max. size of coarse agg + 5mm =25 mm

Compression Steel No. of compression bars provided =4 (2-16 , 2-12 ) Spacing between bars =120mm Diameter of Larger Bar > nominal max. size of coarse agg + 5mm = 25 mm

Note: • The positive steel at joint >= 0.5 negative steel at that joint (IS 13920 6.2.3) • Steel on a side of any face >= 0.25 X negative steel at the joint face (IS 13920 6.2.4) Ductility Check: Pc,lim= 0.87fy(Pt-Pt,lim)/(fsc-0.447fck)=0.58 Pc,provided=0.58>Pc,lim Hence OK

Deflection Checks: Percentage of Tensile Steel, Pt = (Ast/bd)*100=1.786 Percentage of Compressive Steel, Pc = (Asc/bd)*100=0.69 From 23.2.1 Fst = 0.58fyAst,req./Ast=222.5 Now, from table: Modification factor for tensile RF, Kt =0.97 Modification factor for compressive RF, Kc =1.18 And from clause 23.2.1 Max. Span to effective depth ratio, (l/d)max = (l/d) basic * Kt * Kc=27.612 For Continuous Spans, (l/d) basic = 26 (Since all spans are = 10* diameter of bar = 80mm

(IS 13920 6.3.1)

Minimum diameter of bar forming the hoop =20mm > 6mm (IS 13920 6.3.2) Hence OK

Anchorage: In external joints, Anchorage length of bottom and top bars = Ld + 10*diameter (IS 13920 6.2.5) = 50*diameter =400mm Anchorage for 32Anchorage for 20-

bars = 1600mm bars = 1000mm

Extend the bars to a length of 300 mm in the column then remaining is provided after the bend: For 32For 20-

bars =>300+1300mm bars =>300+700mm

Design of Columns (Z = 0-4 mm) Group E (600*600) Axial Load, Pu = 700KN Moment about X-axis, Mx=275 KN-m Moment about Y-axis, My=700 KN-m Material Used: Concrete, M25 Steel, Fe 415 Unsupported Length = 4m (clause 25.1.3.b) Effective length, “l” = 0.85*4=3.4m Max of (I/Dx , I/Dy) = 5.66 20 (clause 25.4)

Properties of column: Width of column, b =600mm Length of column, D= 600mm Total moment, Mu= 1.15 of resultant (Mux, Muy) = 864.8 Kn-m Axial load, Pu= 700KN Therefore, Pu/fck *b*D = 0.077 Mu/fck *b*b*D=0.16 Assume d’= 50mm , for (d’/D) =0.1 , fy=415KN

, chart 44 used

Uniaxial Moment Capacity Check: d’=(nominal cover + ties + diameter/2)=50+8+25/2=70.5 Pu/fck*b*D=0.07 ,p/fck=0.16 Bars provided: 12-32 and 4-28 From Chart 44: Mux1/(fck*bD2)=0.17 Mux1=Muy1= 918KN-m >Mu Pu/fck*b*D=0.07 P/fck=0.16 d’/Dy=0.1

Safety Check under Biaxial Bending: Calculation of Puz and

(clause 39.6)

Puz = 0.45*fck*b*D+(0.75+fy-0.45*fck)*Asc Puz = 8469 Pu/Puz =0.08 = 1 (Mux/Mux1)

(clause 39.6) + (Muy/Muy1)

= 0.95