Structural Design Report Rev 02 24.01.2018

DESIGN REPORT-CONTROL BUILDING OF MANNAR TRANSMISSION INFRASTRUCTURE-LOT A-REV-02 CONTROL BUILDING OF MANNAR TRANSMISSION INFRASTRUCTURE

Design Report-Control Building of Mannar Transmission Infrastructure-Lot A-rev-02

1.0

INTRODUCTION

This report is prepared to highlight the Structural Analysis and Design Procedure of the Proposed Control Building of Mannar Transmission Infrastructure-Lot A Project. The Structural Engineering Consultancy Services involves designing of a Three Storey Building with a Basement Floor. The Structural Form of the building is selected as a "Framed Building". The Columns and Beams will resist the gravity loads and Lateral Loads will be resisted by the "Frame Action". Ground Floor is designed as a Part Slab on Grade and Part Part Suspended Slab. All the Other Floors are Designed Designed as Suspended Slabs. Foundations Foundations of the Buildings Buildings is designed as a combination of Beam-Raft Beam-Raft Slab and Individual Footings. Reinforcement detailing of elements were done considering the resistance to moderate earthquakes. This report consists of Load evaluation, structural modeling and Analysis, design calculations procedure for foundations and columns. Designs will be carried out as per relevant British Standards. Structure is analyzed in SAP 2000, version 16, an finite element structural analysis software package.

Prepared by Gihan Chathuranga, BSc.Eng(Hons), PG PG Dip in Strcut.Eng, Strcut.Eng, Design, C.Eng,MIE(SL),MSSE(SL) C.Eng,MIE(SL),MSSE(SL)

Page

1

Design Report-Control Building of Mannar Transmission Infrastructure-Lot A-rev-02

1.0

INTRODUCTION

This report is prepared to highlight the Structural Analysis and Design Procedure of the Proposed Control Building of Mannar Transmission Infrastructure-Lot A Project. The Structural Engineering Consultancy Services involves designing of a Three Storey Building with a Basement Floor. The Structural Form of the building is selected as a "Framed Building". The Columns and Beams will resist the gravity loads and Lateral Loads will be resisted by the "Frame Action". Ground Floor is designed as a Part Slab on Grade and Part Part Suspended Slab. All the Other Floors are Designed Designed as Suspended Slabs. Foundations Foundations of the Buildings Buildings is designed as a combination of Beam-Raft Beam-Raft Slab and Individual Footings. Reinforcement detailing of elements were done considering the resistance to moderate earthquakes. This report consists of Load evaluation, structural modeling and Analysis, design calculations procedure for foundations and columns. Designs will be carried out as per relevant British Standards. Structure is analyzed in SAP 2000, version 16, an finite element structural analysis software package.

Prepared by Gihan Chathuranga, BSc.Eng(Hons), PG PG Dip in Strcut.Eng, Strcut.Eng, Design, C.Eng,MIE(SL),MSSE(SL) C.Eng,MIE(SL),MSSE(SL)

Page

1

Design Report-Control Building of Mannar Transmission Infrastructure-Lot A

2.0

DESIGN INFORMATION AND REFERENCES

Design Information 2.1

Materials Density of Concrete

3

24kN/m

Characteristic Strength of Concrete (f cu ) For All Structural Elements of S ub Structure (Up to Ground

2 35N/m

Floor Level). For All Structural Elements Elements of Super Structure (From

2 25N/m

Ground Floor to Roof Level).

2.2

2

Yield Strength of Tor Steel (f y )

460N/mm

Yield Strength of Mild Steel (f vy )

250N/mm

2

Fire rating 2 hr fire rating

2.3

Exposure Conditions Severe Exposure Conditions

2.4

Cover to reinforcement

2.5

Slabs

25mm

Beams and Columns

30mm

Foundations

50mm

Loads 2.5.1

Dead Loads Dead load due to floor finishes

2 1.25kN/m

Prepared by Gihan Chathuranga, BSc.Eng(Hons), PG Dip in Strcut.Eng, Strcut.Eng, Design, C.Eng,MIE(SL),MSSE(SL) C.Eng,MIE(SL),MSSE(SL)

Page 2


Design Information 2.5.2

2.5.3

Live Loads Cable Basement

2 5kN/m

36kV Switch Gear Room

2 11.5kN/m

220kV Panel Room

2 11.5kN/m

Control Room

2 11.5kN/m

Auxiliary Room

2 11.5kN/m

Office Room

2 3.0kN/m

Roof Slab

2 1.5kN/m

Basic wind speed

47m/s

(This wind load complies to Zone 2 of Sr i Lanka)

2.5.4

2.6

Surcharge Loads for Retaining Walls

2 10kN/m

Soil Properties Bearing Capacity (refer to Annex 1-Extract from Soil Report)

2 175kN/m φ' 28 =



Friction Angle


Page 3


References Code of Practices 1

BS 8110:Part 1:1997 Structural use of Concrete

2

BS 8110:Part 2:1995 Structural use of Concrete

3

BS 8007:1987 Design of Concrete Structures Str uctures for retaining aqueous liquids.

4

BS 6399: PART 1: 1984 Imposed loads on Buildings

5

CP3 : Chapter V : Part II, 1972 Basic data for design of buildings – Wind loads

6

BS 8002:1994 Earth Retaining Structures

Other References 1

Reinforced Concrete Designer’s Hand Book (Eleventh Edition) C. E. Reynolds & James C. Steedman

2

Manual for the Design of Reinforced Concrete Building Structures Published by the Institute of Structural Engineers (UK)

3

Standard Method of detailing structural concrete Published by the Institute of Structural Engineers (UK)-August 1985

4

Standard Method of detailing structural concrete-Third edition Published by the Institute of Structural Engineers (UK

5

Reinforced Concrete Design to BS 8110 Simply Explained, by A.H. Allen.

6

Reinforced concrete design-Fifth edition W.H. Mosley,J.H. Bungey and R.Hussle


Page 4


References Other References cont’d. 7

Graded Example in Reinforced Concrete Design Prof.W.P.S Dias

8

Design of Buildings for High Winds in Sri Lanka Published by Ministry of local Government Housing and Construction

9

Reinforcement Detailing to mitigate seismic effects Published by Society of Structural Engineers, Sri Lanka

10

Examples of the design of Reinforced Concrete Buildings to BS 8110-Fourth Edition C. E. Reynolds & James C. Steedman

11

Reinforced and Prestressed Concrete Design-The Complete Process Eugene J.O’Brien & Andre S.Dixion

Prepared by Gihan Chathuranga, BSc.Eng(Hons), PG Dip in Strcut.Eng, Design, C.Eng,MIE(SL),MSSE(SL)

Page 5


3.0

STRUCTURAL MODELING IN SAP 2000

3.1

INTRODUCTION

Super Structure is modeled in SAP 2000, a Finite Element Structural Analysis package popular among Structural Engineers. Main steps in the modeling procedure are as follows. (1) Definition of Grids In this step building grid lines are defined in the model. (2) Definition of Materials Following material properties of the Concrete is defined as per given in Table 3.1 & Table 3.2

Table 3.1: Material Properties of Grade 25 Concrete

Property

Value

Density

24kN/m3

Young Modulus

6 2 25 × 10 kN/m

Coefficient of Thermal Expansion

10 × 10

Poisson’s Ration

0.2

Characteristic strength

2 25N/mm

-6

Table 3.2: Material Properties of Grade 35 Concrete

Property

Value

Density

24kN/m3

Young Modulus

6 2 27 × 10 kN/m

Coefficient of Thermal Expansion

10 × 10

Poisson’s Ration

0.2

Characteristic strength

2 35N/mm

-6


Page 6


(3) Definition of Frames and Area Sections Beams and columns were modeled as frame elements. Initial sizes of the members were determined carrying out a preliminary member size design calculations. Raft Slabs and Basement Walls are modeled as Shell Elements. (4) Assignment of Loading Vertical and lateral loads is assigned to model (5) Analysis of Structure Structure is analyzed as a rigid frame structure. 3D model of the Structure is shown in Figure 3.1.

Figure 3.1: 3D Model of the Super Structure


Page 7


3.2

SPECIAL CONSIDERATIONS IN STRUCTURAL MODELING

3.2.1

BEAM RAFT SLAB MODELLING

Foundation is modeled as a Beam-Raft Slab with Soil Springs. The Basement Slab and Basement Walls are modeled as Thin Shells of approximately 0.75mx0.75m and thicknesses are 350mm and 225 mm respectively. Basement Slab and the Basement Walls are manually meshed to ensure the proper connectivity between elements. The Ground Beams are modeled as Frame Elements. In order to simulate the propped cantilever action due to presence of Ground Floor Slab, the Ground Floor Slab is also modeled.

Modulus of Sub Grade Reaction (K)

The Modulus of Sub grade Reaction is given by following equation. K = SF × 40 × Q allowable

SF is the Factor of Safety and usually taken as 2. According to the Soil Report, the allowable bearing capacity is Q allowable Hence K = SF × 40 × Q allowable

=

2 × 40 × 175 = 14000kN/m

=

2 175kN/m

3

Hence Corresponding Spring Constant at Joints is, K = 14000 × 0.75 × 0.75 = 7875kN/m


Page 8


3.3

LOAD EVALUATION

3.3.1

GRAVITY LOADS

3.3.1.1 DETERMINATION OF LOADS TRANSFERRED ON TO BEAMS FROM SLABS

Slab loads were assigned to supporting Beams as per Shear Force Coefficients given in Table 3.16 of BS 8110 -1:1997. Slab loads were assigned to SAP 2000 model under following load cases as given in Table 3.2 Consider "Unit_Slab FF" Load case assigned to First Floor L evel is shown in Figure 3.2. The Loads on the First Floor Slab are as follows. Self Weight of 175mm thick slabs

=

0.175 × 24 = 4.2kN/m2

Finishes

=

1.25kN/m2

Live Loads on first floor slab

=

11.5kN/m2

Hence "Unit_Slab FF" Load case was scale multiplied as given in Table 3.3 to represent actual loads. Similar Procedure was carried to model the floor loads of other floors as well.

Table 3.3: Load Cases defined in SAP 2000

"Load Case" Name

Represent

Unit_Slab GF

Slab Load due to Unit Load Intensity acting on Ground Floor Slabs

Unit_Slab FF

Slab Load due to Unit Load Intensity acting on First Floor Slabs

Unit_Slab RF

Slab Load due to Unit Load Intensity acting on Roof Floor Slabs

Mass-Concrete Fill

Mass Concrete Filling on First Floor

Water_Tank

Slab Load due to Unit Load Intensity acting on Water Tank Level Slabs

Wall

Wall Loads


Page 9


Figure 3.2: "Unit_Slab FF" Load Case assigned on First Floor Beams

Table 3.4: Scale Multiplier for "Load Cases"

"Load Case" Name

Scale Multiplier for Load Cases

Self Weight

Finishes

Live Loads

Unit_Slab GF

4.2

1.25

11.5

Unit_Slab FF

4.2

1.25

115

Unit_Slab RF

3.6

1.25

1.5

Mass-Concrete Fill

10

-

-

Water_Tank

3.0

-

2

Wall

-

-

-

Prepared by Gihan Chathuranga, BSc.Eng(Hons), PG Dip in Strcut.Eng, Design, C.Eng,MIE(SL),MSSE(SL) Page 10


3.3.1.2 WALL LOADS

Walls are basically 225mm thick brick walls. Weight of walls on elevation is 5kN/m2 . Floor to floor height of the structure is 3900mm For example assume beam height is 75 0mm. Hence height of the wall is 3150mm. = 15.75kN/m . Hence load on a beam due to wall is 5 × 3.150

Wall loads assigned to first floor level are shown in Figure 3.3.

Figure 3.3 : Wall Loads assigned to First Floor Beams



3.3.1.3 ROOF LOADINGS

Roof will be a double pitch steel portal frame roof. The portals will be span between Columns on Grid A and Columns on Grid C. Hence Roof Loads are assigned to Modal as Point Loads to Columns. (Please refer to Figure 3.4). The Wind Loads usually will generate uplifts, hence it will have beneficiary effects for the axial loads in columns, as such roof loads due to dead and live loads are considered as follows. Weight of Asbestos Roofing Sheets with Insulations 0.25kN/m Weight of Purlins 0.04kN/m

2

2

2 Imposed load on roof is 0.5kN/m Hence Serviceability Load Intensity of the Roof is n sls Hence Ultimate Load Intensity of the Roof is n uls

=

=

2 0.25 + 0.04 + 0.5 = 0.79kN/m

1.4[0.25 + 0.04 ] + 1.6 × 0.5 = 1.206kN/m

2

The Plan area of the Roof is 19.8mx15.6m. There will be Total 12 Columns to support Roof Loads. (Column grid is approximately equal and roof reactions are compared to other loads).

0.79 × 19.8 × 15.5

Hence Serviceability Point Load

=

Hence Ultimate Point Load

=

12

=

20kN

1.206 × 19.8 × 15.5 12

=

30kN

Figure 3.4 : Roof Loads assigned to Columns



3.3.2

LATERAL LOADS

3.3.2.1 WIND LOADS

3.3.2.1.1

INTRODUCTION

A wind force on Structure is calculated as per CP3: Chapter V: Part 2: 1972 .According to “Design of Buildings for High Winds in Sri Lanka” Country is divided in to t hree wind zones. (See Figure 3 .5).

Figure 3.5: Wind Zones in Sri Lanka



Basic wind speeds for each zone is given in Table 3.5 below

Table 3.5: Basic Wind Speeds in Sri Lanka

Zone

Basic Wind Speed Post Disaster Structure

Normal Structure

1

53.5 m/s

49 m/s

2

47 m/s

42.5 m/s

3

38 m/s

33.5 m/s

Structure is located in Mannar and is designed as a “Post Disaster Structure”. Hence corresponding Basic Wind Speed (V) is 47m/s. Wind force calculation can be done as follows.

Figure 3.6: Wind Loads directions on Structure



Reference

Calculations

Out Put

Basic Wind Speed V = 47m/s (Corresponds to Post Disaster Structure located in Wind Zone 2 of Sri Lanka). Clause 4.3

Design Wind Speed (V ) s

CP3: Chapter V

V s

Clause 5.4

Topography Factor (S ) 1

CP3: Chapter V

S 1

Clause 5.5

Ground roughness, building size and height above ground

CP3: Chapter V

factor (S ) 2

Clause 5.5.2

Building is located in a Open Country with No Obstructions.

=

=

V×S ×S ×S 1 2 3

1

CP3: Chapter V Consider the Wind along Y direction.

b-dimension of the Building normal to wind d-dimension of the building measured in the direction perpendicular to wind b = 19.8m and d = 15.5m

(See Figure 3.6) Hence greatest horizontal dimension of the building is 19.8m. Height of the building above ground is 8.375m Hence neither the greatest horizontal dimension nor the vertical dimension of the building exceeds 50m. Clause 5.5.2

Hence Building Class-Class B

CP3: Chapter V

Consider the Second floor level. Height above ground is 8.375m



Reference Table 3

Calculations S

0.91

=

2

Out Put

CP3: Chapter V Clause 5.6 CP3: Chapter V

Probability Factor (S ) 3 S

=

3

1

Hence design wind speed is, V s

=

47 × 1 × 0.91 × 1 = 42.77m/s

Clause 4.3

Dynamic Pressure (q)

CP3: Chapter V

q = kVs

2

k = 0.613

2 q = 0.613 × 42.77 Table 3.10

Clause 4.3

l

b

CP3: Chapter

H Table 10.

w

d

b

=

=

=

19.8 15.5 19.8 15.5

1.27 ≈ 1.5

=

1.27 ≈ 1.5

=

15.5 f

2 1121N/mm

=

8.375

Hence C

=

=

0.54 ≈ 1/2

0.95

Floor to Floor Height from First to Second is 3.475m Hence effective area (A ) corresponds to second floor level is e A

e

=

19.8 × 3.475 ×

1 2

=

2 34.40m

Clause 7.3

Hence Wind force (F) on Second Floor level is

CP3: Chapter

F = C qA e f

=

0.95 × 1121 × 34.40 = 36.63kN

Similarly wind load is calculated for other floors and results are given in Table 3.6 & 3.7.



Table 3.6 : Summary of Wind Load Calculations-X Direction Level

F.F

Elevation

E.Height

Ae

2.45

38.22

S2

Vs

Q

Fx

0.78

36.66

823.8448

25.82

Height

1

1

3.9

4.9

3.6875

57.525

0.8

37.6

866.6349

40.88

3.475

8.375

1.7375

27.105

0.91

42.77

1121.344

24.92

Vs

Q

0.78

36.66

792.9338

36.54

Ground First Second

Table 3.7 : Summary of Wind Load Calculations-Y Direction Level

F.F

Elevation

E.Height

Ae

2.45

48.51

S2

Fy

Height Ground First Second

1

1

3.9

4.9

3.6875

73.0125

0.8

37.6

834.1184

57.86

3.475

8.375

1.7375

34.4025

0.91

42.77

1121.34

36.63



3.3.2.1 EARTH PRESSURE

Figure 3.7 : Lateral Loads on Basement Walls

Active Pressure Coefficient- Ka

Ka

=

1 − sinφ 1 + sinφ

=

1 − sin28 1 + sin28

=

0.36

Hence Active Soil Pressure at the Bottom of the Wall is ,

σ

A

=

2 K γh = 0.36 × 18 × 2 = 19.44kN/m a

The Pore Water Pressure at the Bottom of the Wall is,

σ

A

=

δ

2 h = 9.81 × 3 = 29.43kN/m water



2

As per Clause 3.3.4.1 of BS 8002:1994, the minimum surcharge load should be 10kN/m .

σ

A

4.0

=

2 Ka × Surcharge = 0.36 × 10 = 3.63kN/m

LOAD COMBINATIONS

Following ULS and SLS Combinations were defined in SAP 2000 as per Table 2.1 of BS 8110-1:1997. (Please refer to Figure 3.8). ULS Com 1

-1.4 Dead Loads + 1.6 Live Loads+1.2 Earth Pressure

ULS Com 2

-1.0 Dead Loads + 1.4 Wind Loads in X direction+1.2 Earth Pressure

ULS Com 3

-1.0 Dead Loads + 1.4 Wind Loads in Y direction+1.2 Earth Pressure

ULS Com 4

-1.2 Dead Loads + 1.2 Live Loads + 1.2 Wind Loads in X direction+1.2 Earth Pressure

ULS Com 5

-1.2 Dead Loads + 1.2 Live Loads + 1.2 Wind Loads in Y direction+1.2 Earth Pressure

ULS Com 6


ULS Com 7


SLS Com 1

-1.0 Dead Loads + 1.0 Live Loads

SLS Com 2


SLS Com 3


SLS Com 4

-1.0 Dead Loads + 1.0 Live Loads + 1.0 Wind Loads in X direction+1.0 Earth Pressure

SLS Com 5

-1.0 Dead Loads + 1.0 Live Loads + 1.0 Wind Loads in Y direction+1.0 Earth Pressure

Figure 3.8 : Extract of Table 2.1 of BS 8110-1:1997.



5.0

STRUCTURAL DESIGN OF BASEMENT WALLS-SPECIMEN CALCULATION

References

Calculation

Out Put

Consider a Long Wall Panel on Grid A. The Serviceability Vertical Bending Envelop (In the Vertical Direction) is 28 kNm/m. (See Figure 5.1). Cover to reinforcement

= 50 mm

Diameter of main bar (φ )

= 12 mm

(The Reinforcement Diameter was measured from the c ore cutter sample). The Spacing of the Main Reinforcement = 150mm

Assessment of Flexural Crack Width

2 A s = 754 mm /m d = 250 –50 – 12/2 = 194mm

A 754 = 3.887 × 10 − 3 ρ= s = bd 1000 × 194

X- Depth to the Neutral Axis α

e

X d

-Modular Ratio take as 15.

 = α eρ 1 + 2

X 194



α eρ

 

− 1



2



− 15 × 3.887 × 10 3

= 15 × 3.887 × 10 − 3 ×  1 +



− 1 



X = 55.90 mm

Z- Lever arm

Z = d−

X 3

= 194 −

55.90 3

= 175.36mm



Figure 5.1 : Serviceability Bending Moments (Vertical) in Long Wall Panel



References

Calculation

Out Put

Ms- Serviceability Moment of the Wall Base is 28kNm/m

f s -Steel Stress f s =

M A Z s

=

28 × 106 754 × 175.36

= 211.77N/mm2

B:2

Limiting Steel Stress = 0.8f y = 0.8 × 460 = 368N/mm

BS 8007:1987

Hence Steel stress is within acceptable limits.

2

f -concrete Stress cb 2M

2 × 28 × 106

2 f = 5.71N/mm = = cb Zbx 175.36 × 1000 × 55.90 B:2

Limiting Concrete Stress = 0.45f cu = 0.45 × 35 = 15.75N/mm

BS 8007:1987

Hence Concrete stress is within acceptable limits.

2

f (h − x) (250 − 55.90) 211.77 ε1 = s × = × = 1.488 × 10 − 3 3 (194 − 55.90) E s d − x 200 × 10 For a Limiting design crack width of 0.2mm,

b (h − x)(a′ − x) ε = t 2 3Es A s (d − x) ε = 2

1000 × (250 − 55.90)(250 − 55.90) −4 = 6 × 10 3 3 × 200 × 10 × 754 × (194 − 55.90)

εm = ε1 − ε2 = 1.488 × 10

− 3 − 6 × 10 − 4 = 8.88 × 10 − 4

2  150 2  12   12 acr =   +  50 +   − = 87.60mm 2   2  2    4 3acrεm 3 × 87.60 × 8.88 × 10 − = = 0.17mm < 0.2mm w=  87.60 − 50   acr − Cmin  1 + 2 1+2   h − x   250 − 55.90  Hence Crack with is Ok.



References

Calculation

Out Put

Assessment of Crack Width Due to Thermal and Moisture Effects

ρ -Reinforcement Ratio for a Surface Zone Clause 2.2.3.3 BS 8007:1987

ρ=

As bh

=

754 1000 × 0.5x250

= 0.006032

The minimum Reinforcement ratio is 0.0035 Hence minimum reinforcement ratio is satisfactory.

S-Crack Spacing

f φ smax = ct × f 2ρ b Table A.1 BS 8007:1987

f ct = 0.67 f b smax = 0.67 ×

12 2 × 0.006032

= 666.45mm

W-Crack Width

Wmax = Smax R α(T + T ) 1 2

R-Restraint Factor is taken as 0.5.

T1 = 30 C and T2 = 10 C for Sri Lankan context. 

α = 10 × 10



−6

Wmax = 666.45 × 0.5 × 10 × 10

− 6 (30 + 10) = 0.13mm < 0.2mm

Hence Crack width is satisfactory.



6.0

STRUCTURAL DESIGN OF GROUND BEAMS-SPECIMEN CALCULATION

Reference

Calculations

Out Put

Consider the Ground Beam on Grid 3. The Bending Moments and Shear Forces for ULS ENVELOP are shown in Figure 6.1 and Figure 6.2.

Checks for Bending

Consider the B/3 Support, the support moment is 315kNm Assume T20 Main Reinforcement, T10 Links and 50mm cover .

d = 600 − 50 − 10 − Cl.3.4.4.4 BS 8110:1-1997

k=

M f b d2 cu f

=

20

35 × 450 × 5302

 

 

As

= 530mm

315 × 10 6

z = d 0.5 + 0.25 −

z = d 0.5 +

2

0.25 −

= 0.070

  0.9  K

0.070  0.9

 = 0.92d < 0.95d 

M

=

0.95f y z

=

6 315 × 10 0.95 × 460 × 0.92 × 530

2

= 1478mm

3T20+3T20 has been provided at bottom. A

s,provided

= 1884mm2

(See the Reinforcement Detailing in Annex 2)



Figure 6.1 : Ground Beam Bending Moment

Figure 6.2 : Ground Beam Shear Force



Reference

Table 3.25 BS 8110:1-1997

Calculations

Out Put

Checks for Minimum area of reinforcement

100A

s

bh

100 × 1884

=

=

450 × 600

0.70 > 0.13

Hence minimum steel requirement is satisfied. Clause 3.12.6

Checks for Maximum area of reinforcement

BS 8110:1-1997

Neither the area of tension reinforcement nor compression

Minimum r/f Ok.

Maximum r/f Ok

reinforcement should exceed 4% of cross sectional area of the concrete.

Checks for Shear at RHS Support

Maximum shear at support =297 kN (From SAP Model) 3 297× 10

Cl.3.4.5.2 BS 8110:1-1997

Cl.3.4.5.2 BS 8110:1-1997

vmax =

2

= 1.25N/mm

(450 × 530)

Maximum permissible shear stress is lesser of 2

0.8 f cu or 5N/mm

2 2 0.8 35 = 4.73N/mm > 1.25N/mm

Hence maximum shear is O.K

Cl. 3.4.5.4 BS 8110:1-1997

Maximum Shear Ok

Concrete shear Stress

The area of tension reinforcement at a distance “d” from the

2

face of the support is 1884mm

Table 3.8

100As

BS 8110:1-1997

bd

 400   d 

=

100 × 1570 450 × 530

14

= 0.78 < 3

1

 400  4 = 0.93 < 1 =  530 



Reference

Calculations

Out Put

Hence design concrete shear stress is,

1

1

1

 100As  3 ×  400  4 × 1 ×  f cu  3 = 0.79 ×       δm  25   d   bd 

vc

1

 35  3 = 0.79 × (0.78 ) 3 × 0.93 × ×  1.25  25  1

1

= 0.60N/mm2 Table 3.7 BS 8110:1-1997

Spacing of the shear links 2 vc + 0.4 = 0.60 + 0.4 = 1.0N/mm vc + 0.4 < v < v d max

We have provided 10 Links at supports. Number of shear legs is 2.

 π × 102  2 Asv = 2 ×   = 157mm  4  Hence maximum possible shear link spacing at the support is,

Sv ≤

A sv × 0.95 × f yv

(

bv × v − v

c

)

=

157 × 0.95 × 460 = 235mm 450 × (1.25 - 0.6)

We have provided T10 @ 200mm. Cl. 3.4.5.5

Maximum possible spacing of links is 0.75d = 397.5mm

BS 8110:1-1997

Hence maximum spacing is satisfied.

Cl.3.4.6

Checks for Deflection

Table 3.9

The Span of the Beam in 4575 mm and beam is continuous

T10@ 200mm Links

over supports. BS 8110:1-1997 Basic

  Span   = 26 Effective Depth   

 4750 = 8.96 = Effective Depth 530  

Actual 

Span

Hence Deflection of the Beam is Satisfactory.



7.0

STRUCTURAL DESIGN OF BASEMENT SLAB-SPECIMEN CALCULATION

7.1

CHECKS FOR BEARING

Maximum Serviceability Joint Reaction (Corresponds to SLS Combination 01) is 73kN. This is corresponds to a area of 0.75mx0.75m. Hence Maximum Serviceability Pressure is

73

Q = applied

7.2

= 130kN/m2 < 175kN/m2

2 0.75

CHECKS FOR BENDING

Maximum Moment for any direction is 90kNm/m. (See Figure 7.1 and 7.2). Assume T16 is to be used. Effective depth (d)

d = 300 − 50 − 16 −

K=

M 2 f cubd

=

16 2

= 226mm

6 90 × 10 2 35 × 1000 × 226

= 0.050 < 0.156

Hence section can be designed as a singly reinforced section.

 

 d 0.9 



0.050 

z = 0.5 + 0.25 −

z = 0.5 + 0.25 −



K

0.9

 d = 0.94d < 0.95d 

Hence Area of reinforcement required , A s,req

=

M 0.95f y Z

=

90 × 10

6

0.95 × 460 × 0.95 × 226

=

969

2 mm m

Hence provide T16 @ 150mm C/C.

A

s,provided

= 1340

2 mm m



Figure 7.1 : M11 Bending

Figure 7.1 : M22 Bending



7.3

CHECKS FOR SHEAR

Maximum possible shear , lesser of

0.8 f cu or 5N/mm2 2 0.8f cu = 0.8 × 35 = 4.73N/mm

1

1

1

 100As  3  400  4 1  f cu  3 vc =0.79×  ×  × δ × 25  b d d    v  m 

100As   100× 1340   =  = 0.59 < 3 b d 1000 226 ×   v  1 400   4

 d 

1 400   4

=

 226 

= 1.15 > 0.67

δm = 1.25 1 1 1 35 3  2 vc = 0.79 × [0.59] 3 × 1.15 × × = 0.68N/mm   1.25  25 

Maximum Shear Stress as per SAP Model is 0.53N/mm^2 (See Figure 7.3 and Figure 7.4).

v < vc



Figure 7.3 : S13 Shear

Figure 7.4 : S23 Shear



8.0

SPECIMEN CALCULATION FOR A INDIVIDUAL FOOTING

The Specimen Calculations for the Column Footing at Grid C/4 presented here. The SLS and ULS reactions of the Column are given in Figure 8.1 and Figure 8 .2.

Reference

Calculation

Out Put

General Data

Columns are pinned at the base. Figure 4.1

Axial load at SLS

= 598kN

Figure 4.2

Axial load at ULS

= 863kN

Soil Report

Bearing Capacity at 2.0 m below EGL

= 175kN/m2

Cover to reinforcement

= 50mm

Grade of Concrete f cu

=

2 35N/mm

Sizing of the footing

Assume an initial thickness of 300mm.

2 Hence Allowable bearing capacity = 175 − 0.30 × 24 = 167.8kN/m Hence Required Width (B),

B=

598 167.8

= 1.89m

Hence Provide a 2.0mx2.0m Square Footing.

Design for Flexure

Ultimate pressure

=n =

863 = 215.8kN/m2 22

Dimension of the Stub Column 500mmx500mm Maximum moment at Column face (For a Unit Width)

M = 215.8 ×

2 (2 − 0.50) 8

= 60.70kNm/m



Figure 8.1 : SLS Joint Reactions

Figure 8.2 : ULS Joint Reactions



Reference

Calculation

Out Put

Diameter of the main reinforcement to be used =12mm d = 300 − 50 − 16 −

16 2

= 226mm

Consider a Unit Width of the footing

k=

M f bd2 cu

=

60.70 × 106 35 × 1000 × 2262

 

  0.9 



0.034 

z = d0.5 + 0.25 −

z = d0.5 + 0.25 −

k



As =

M

=

0.95f y z

= 0.034 < 0.156

0.9

 = 0.96d ,but should not be greater than 0.95d 

60.70 × 10

6

0.95 × 460 × 0.95 × 226

= 647mm2 /m

Provide T16@ 150mm C/C A

s,provided

T16 @ 150mm

= 1340mm2 /m

Checks for minimum reinforcement

100A bh Table 3.25

3.11.4.5

s

=

100 × 1340 1000 × 300

=

0.45 > 0.13

Hence minimum steel requirement is o.k.

Checks for Maximum shear

Maximum shear stress at column face

=

863 × 10

3

2(500 + 500) × 226

= 1.91N/mm2

Maximum possible shear 0.8

2

f cu or 5N / mm which is lesser

Hence v

max

= 0.8 f cu = 0.8 35 = 4.73N/mm2

Hence Maximum Shear is OK.



Reference

Calculation

Out Put

Shear at 1.0 d from the face of the Column

215.8 × v= Table 3.8

 (2 − 0.50) − 0.226 × 1000  2  = 0.50N/mm2 1000 × 226

Design concrete shear strength 100As

=

bd

100 × 1340 (1000 × 226)

1

= 0.59 < 3

1

 400  4 =  400  4 = 1.15  d   226  Hence design concrete shear strength is,

1

vc

1

 100As  3 ×  400  4 × 1 = 0.79 ×     δm  bd   d  1

 35  3 = 0.79 × (0.59 ) 3 × 1.15 × ×   = 0.68N/mm2 1.25  25  1

1

2

vc=0.68N/mm

Hence shear at 1.0 d is OK. Checks for Punching shear at 1.5 d

Critical Perimeter = 4(0.50 + 3d) = 4(0.50 + 3 × 0.226) = 4 × 1.178 = 4.712m Force outside the critical perimeter

= 215.8 × (22 − 1.1782 ) = 563.74kN Hence Punching shear stress

=

563.74 × 1000 4712 × 226

= 0.53N/mm2

Hence Punching Shear is OK.



8.0

COLUMN DESIGN-SPECIMEN CALCULATIONS

Column layout is determined as per the Client's requirements and initial member sizes were estimated carrying out a preliminary design. Then structure was modeled in SAP 2000 and each column is designed to withstand the bending and axial loads for ULS combination. Specimen Calculation is provided for Column at grid B/3 (Type C2).Columns are designed as unbraced by assuming all the lateral loads are taken by the columns.



Reference

Calculations

Output

Specimen Calculation for Column Type "C2"

The Column size is 450x450 Consider a Column Segment from Ground to First Floor Level. Determination of Short or Slender Column

Bending about X-X Axis Dimension of column about XX axis Bending h = 450mm Depth of the Beam along Y direction is 750mm. Cl.3.8.1.6.2

Hence Top Condition is 1.

BS 8110-1:1997

Slab Thickness at Ground Level is 175mm. Hence Bottom Condition -2

Table 3.2

Hence β = 1.3

BS 8110-1:1997

L ex L ex h

= βL ox = 1.3 × (3900 − 750) = 4095mm =

4095 450

= 9 < 10

Hence Column is Short for X-X axis Bending.

Bending about Y-Y Axis Dimension of column about YY axis Bending h = 450mm Depth of the Beam along X direction is 500mm. Cl.3.8.1.6.2

Hence Top Condition is 1.

BS 8110-1:1997

Ground Beam Depth is 450mm. Hence Bottom Condition -1

Table 3.2

Hence β

BS 8110-1:1997

L ey L ey b

=

1.2

= βL oy = 1.2 × (3900 − 500) = 4080mm

=

408 450

= 9 < 10

Hence Column is Slender for YY axis Bending.

Short Column

Hence "C2" Column is a short Column.



Figure 9.1: ULS Axial Loads on Column at B/3

Figure 9.2: ULS XX Bending Moment of the Column



Figure 9.3: ULS YY Bending Moment of Column at B/3



Reference

Calculations

Output

Design Loads from SAP 2000 Model

Column is designed for ULS 1 Load Combination.(Critical)

N

=

2223kN

Mxx

= 165kNm

Myy

=

57kNm

Cl.3.8.2.4

Minimum Moments

BS 8110-1:1997

e xx

=

0.05 × 450

=

22.5mm > 20mm

= 2223 × 0.020 = 45kNm < 165kNm M xx, min

e yy

=

0.05 × 450 = 22.5mm

>

20mm

= 1800 × 0.020 = 36kNm < 57kNm M yy,min

Design Moments

Cl.3.8.4.5 BS 8110-1:1997

M xx

= 165kNm

M

= 57kNm

YY

Biaxial Bending

25

b' = 450 − 30 − 10 −

2 25

h' = 450 − 30 − 10 −

β =1−

M xx h′ M yy b′ M xx h′

=

=

>

7N 6bhf cu

165 397.5 57 397.5

=1−

2

=

397.5mm

=

397.5mm

7 × 2223 × 103 6 × 450 × 450 × 25

= 0.49

= 0.42

= 0.14

M YY b′

Hence Column is bending about XX axis.



Reference

Calculations

M′x Column Design

=M

h′

M b′ y

+β

x

= 165 + 0.49 ×

397.5 397.5

Output

× 57 = 193kNm

Design of Main Reinforcement

Charts by IstructE d

397.5

=

h

N

=

bhf cu

M 2 bh f cu Hence

ρ=

2223 × 10

3

450 × 450 × 25

=

193 × 10

= 0.44

6

2 450 × 450 × 25

ρf y

=

f cu

= 0.08

0.2

0.2 × 25 460

A

sc bh

=

A

sc bh

=

A

sc 2

0.90

≈

450

=

0.2 × 25 460 0.2 × 25 460

×

2 450 × 450 = 2201mm

2 1100mm

Provided Reinforcement per Face is 5T25. A

sc, provided

=

5T25 per Face

2 2455mm

Hence provided Reinforcement is Satisfactory.



Reference

Calculations

Table 3.25

Checks for Minimum Area of Reinforcement

BS 8110-

100A sc

1:1997

bh A sc

=

=

Output

0.4

0.4 × 450 × 450 100

=

2 810mm

Hence provided reinforcement satisfied the minimum reinforcement requirement. Checks for Maximum Area of Reinforcement

Cl.3.12.6.1 BS 8110-

100Asc bh

=

6

1:1997 A sc

=

6 × 450 × 450 100

2

= 12150mm

>

2 4910mm

Hence Maximum Area of Reinforcement is Ok. At Laps At a Lap 6T25 bars are lapped. A sc

=

A sc

=

2 6 × 491 = 2946mm (4910 + 2946) 450 × 450

× 100 = 3.87 < 10%

Cl.3.12.7.1

Design of Links

BS 8110-

Largest Compression bar

1:1997

Hence Diameter of the shear link should be larger than 25/4=6mm.

=25mm

Hence T10 Links were provided. Shear Link spacing should be less than 12 times smallest compression bar. (12x25=300mm)

T10-150

Links are provided at 150mm Spacing.

Links



ANNEX 1-LOADINGS


LEGENDS 5 kN/m2

LEGENDS

11.5kN/m2

LEGENDS

11.5kN/m2

LEGENDS

11.5 kN/m2

3kN/m2

LEGENDS

11.5 kN/m2

3kN/m2

1.5kN/m2

1.5kN/m2



ANNEX 2-EXTRACT FROM SOIL REPORT


6.4 Recommendations

Giving due regard to the above factors and inferring similar subsurface conditions throughout the site, it is recommended that individual pad footings be used at each tower leg. The footings can be placed at a depth of 2 m, and designed for an allowable bearing capacity of 175kPa. The above shallow foundation option is based on the presumption that there would be no significant interference between the zones of influence of the designed footings. This condition shall be met if the clear distance between the foundations should be more than the width of the larger footing. Further, it recommended that the stability of the foundation shall checked against possible uplifting and sliding. For the purpose the following design p arameters shall be used: shear strength parameters of the soils above footing level can be taken as c’ = 8 kN/m 2, ф’ = 28. The frictional resistance at the base of the footing can be taken as ca = 5 kN/m, δ= 19.5

16thSeptember 2015

12

Structural Design Report Rev 02 24.01.2018

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