STRUCTURE CALCULATION NOTE

PROJECT: BUILDING WITH TWO LEVELS BY : SOGEIB LTD

STRUCTURE CALCULATION NOTE

November 27

2018

CONSTRUCTION OF A BUILDING WITH

TWO LEVELS (G + TWO ON PLOT NUMBER

502 / UPI 3/03/04/02/502 LOCATED AT

RUBAVU – GISENYI. BUGOYI

OWNER : NDURUMA ELIZABETH CHUCHU

1

PROJECT: Building with two levels BY: Eng. Augustine HABINSHUTI

Contents I. LIST OF SYMBOLS ........................................................................................................................................... 3

1. DESIGN INFORMATION ................................................................................................................................. 5

1.1. Code of practice used ................................................................................................................................... 5

1.2. Units ............................................................................................................................................................. 5

1.3. Values of loads ............................................................................................................................................. 5

1.3.1. Dead loads (Self weight of materials of construction). ............................................................................ 5

1.3.2. Live loads................................................................................................................................................. 5

1.4 Characteristics of used materials .................................................................................................................. 6

1.4.1 The concrete. ................................................................................................................................................ 6

1.4.2 The reinforcement .................................................................................................................................... 6

2. ESTIMATION OF STRUCTURAL ELEMENTS DIMENSIONS. .................................................................... 6

2.1 Estimation of slab dimensions...................................................................................................................... 6

2.2 Estimation of beam dimensions. .................................................................................................................. 6

2.3 Estimation of stair dimensions. .................................................................................................................... 6

3 ANALYSIS AND DESIGN OF SLAB ............................................................................................................... 7

3.1.1 Calculation and design ............................................................................................................................. 7

4 CALCULATION AND DESIGN OF BEAM .................................................................................................. 11

4.1 Calculation and design of inner beam ........................................................................................................ 11

4.1.1 Predesigned flanged section beam(T-beam section) .............................................................................. 11

4.1.2 Computation of uniformly distributed loading on the beam .................................................................. 12

On span3 n3=n1= 49.406KN/m ............................................................................................................................. 12

4.2 Calculation and design of outer beam ........................................................................................................ 19

4.2.1 Predesigned flanged section beam (T-beam section) ............................................................................. 19

4.2.2 Computation of uniformly distributed loading on the beam .................................................................. 20

4.3 Calculation and design of the Ring Beam .................................................................................................. 23

5. CALCULATION AND DESIGN OF COLUMNS .......................................................................................... 24

5.1 Influence area on the column ..................................................................................................................... 24

5.2 Loads take down ........................................................................................................................................ 25

5.3 Calculation and design ............................................................................................................................... 26

5.3.1 Hypotheses ............................................................................................................................................. 26

5.3.2 Reinforcement in the column ................................................................................................................. 26

................................................................................................................................................................................. 31

6. CALCULATION AND DESIGN OF FOUNDATION ................................................................................... 32

6.1 Calculation and design of foundation group1 ............................................................................................ 32

a) Hypothesis .................................................................................................................................................. 32

b) Punching shear ........................................................................................................................................... 33

c) Check for shear force ................................................................................................................................. 34

6.2. Moment Steel ............................................................................................................................................. 34

2


6.3 Design Sketch ............................................................................................ Error! Bookmark not defined.

6.2 Calculation and design of foundation group2 ............................................................................................ 34

d) Hypothesis .................................................................................................. Error! Bookmark not defined.

e) Punching shear ........................................................................................... Error! Bookmark not defined.

f) Check for shear force ................................................................................. Error! Bookmark not defined.

6.3. Moment Steel ............................................................................................. Error! Bookmark not defined.


6.3 Calculation and design of foundation group3 ............................................ Error! Bookmark not defined.

g) Hypothesis .................................................................................................................................................. 34

h) Punching shear ........................................................................................................................................... 35

6.4. Moment Steel ............................................................................................. Error! Bookmark not defined.


7 STAIRS DESIGN .............................................................................................. Error! Bookmark not defined.

8 RESULTS PRESENTATION ........................................................................................................................... 37

3


I. LIST OF SYMBOLS

Ac : net area of the column section

As : area of steel in tension

Asc : total area of steel in a column cross section

Asmin : minimum area of steel in tension

Aspov : provided area steel in tension

Asreqd : required area of steel in tension

Asv : area of steel in vertical links

b : width of reinforced concrete section

bw : width of web in a beam

c : cover

dx : effective depth in x direction

dy : effective depth in y direction

E : modulus elasticity

fcu : characteristic yield strength of concrete at 28 days

fs : design service stress

Fv : vertical forces

fy : characteristic yield strength of steel

GK : permanent load

h : overall depth of a concrete section

hf : thickness of flange in a T-beam

I : Inertial moment

K : coefficient

L : span length

l0 : clear height

le : effective height

Lx : short side of the span panel

Ly : long side of the span panel

M : bending moment

Md : design bending moment

4


Mmax : maximum bending moment

Msx : bending moment in x direction

Msy : bending moment in y direction

Mu : ultimate bending moment

Nd : axial load on column

Nd : axial load on the column

Nu : ultimate design load

Nuz : design ultimate capacity of a section subjected to axial load only

Ø : diameter of reinforcement

P : pressure of pad foundation

Pcr : critical perimeter

Px, Py : percentage of reinforcement

Qadm : admissible bearing pressure of soil

Qk : imposed load

S : spacing

Scr : critical surface

Sv : spacing of vertical links

V : shear force in concrete section

νc : design concrete shear strength

νsx, vsy : design shear capacity of shear reinforcement

x : depth of neutral axis from compression face

y : diameter of steel

z : depth of lever arm

βb : ration of redistributed moment over elastic analysis moment

βɣ : factor governing moment of resistance of concrete T-section

βsx , βsy : bending moment coefficient

βvx, βvy : shear force coefficient

5


1. DESIGN INFORMATION

1.1. Code of practice used

In order to design the building, the sections of reinforcements of different elements of the

structure will be determined according to British Standards BS 8110.

1.2. Units

Unit Weight: kN/m3

Uniformly distributed load: kN/m2

Linear distributed load: kN/m

Concentrated load: kN

1.3. Values of loads

1.3.1. Dead loads (Self weight of materials of construction).

-Roof load: 1kN/m2

-Floor finishes: 1kN/m2

-Tiling: 0.38kN/m2

-Masonry in bricks: 22kN/m3

-Concrete: 25kN/m3

-Bearing capacity of the soil 350kN/m2

1.3.2. Live loads

The building being residential building the BS ‘Code of practice’ stipulates that live load to be

used should be equal to 2.5KN/m2 for normal slabs and 3KN/m

2 for stair slabs. We are going to

use 3KN/m2 for all slabs in our design.

6


1.4 Characteristics of used materials

1.4.1 The concrete.

The concrete to be used will have a resistance in compression fcu = 25N/mm2 for slabs and beams

and fcu = 30 N/mm2 for columns and foundations.

1.4.2 The reinforcement

The reinforcements to be used will be of the grade of 460N/mm2.

2. ESTIMATION OF STRUCTURAL ELEMENTS DIMENSIONS.

2.1 Estimation of slab dimensions.

Since , the slab will be reinforced in two ways.

The estimated thickness h of the slab is between

, let us take h=15 cm

The thickness h of the slab is 15cm made of reinforced concrete.

2.2 Estimation of beam dimensions.

For the continuous beam:

with l: the length of the considered span and h: the overall

depth of the beam.

We adopt h = 550mm including the depth of the slab.

For this project we will design the overloaded continuous beams (inner beams) and side beams.

2.3 Estimation of stair dimensions.

R: Riser of the stairs

G: Going

C: Slab waist

7


The Blonder formula: 2R + G = 60 to 66 cm.

For this project we adopt 2R+G = 64 cm.

Private Common

Riser R

Going G

Slope

No of steps in flight

< 220mm

>220mm

<42o

-

<190mm

>230mm

<38o

<16

Adopt R = 160 mm

Breakaway = 4200mm/3 = 1400 mm

First breakaway @ 1400mm

Second breakaway @ 2800mm

Number of risers = (1400/160)-1 = 8 risers

(2* 160) + G = 640(mm)

Adopt G = 300 mm

We adopt slab waist C = 20cm.

3 ANALYSIS AND DESIGN OF SLAB

In this project, we shall use slab made of reinforced concrete. The figure below shows how our

slab is divided into panels.

3.1.1 Calculation and design

Given that the imposed loads are the same on all panels; our calculations are limited on a panel

which is more loaded (P3)

3.1.2 Hypotheses

Characteristic material strengths:

Concrete 2/30 mmNf

Steel 2/460 mmNf y

Self-weight of concrete 3/25 mKN

8


Thickness of slab (h) = 150 mm

Self-weight of slab 2/75.32515.0 mKN

Finishes 2/1 mKN

Total dead load (GK) 2/75.4)175.3( mKN

Imposed load (Qk) 2/3 mKN

Ultimate design load (n)

2/45.118.465.66.14.1 mKNQkGk

Cover (c) = 25mm (Fire resistance required: 2 hour)

Assumed diameter ( ) of main reinforcement = 10mm

Effective depth mmchd x 1202

1025150

2

Effective depth mmchd y 1102

101025150

2

3.1.3 Bending reinforcement

Results given by PROKON:

9


Interpretation of results:

In the direction of short span(x) we provide:

Y10@200mm in the span

And Y12@200mm on the support

In the direction of long span(y) we provide:

Y10@200mm in the span

And Y12@200mm on the support

5.3

m

Y12@180

Y1

0 @

20

0

A

Y1

2@

18

00

Y10@200 Y

12

@1

80

A

Y12@1800

4.0m

A A

4000 mm

150 mm

Y12@180 mm Y12@180 mm

Y10 @ 180 mm

150 mm

Y

10


Section A-A

a) Check for deflection

For deflection we use the maximum span moment

M = 13.1KNm

6.41120

102.4 3

dx

Lx

Basic span/ effective depth ratio = 26

bprovAs

reqdAsffs y

1

8

2.4

2/8.27218.252

88.239460

8

5mmNfs

84.0

1251000

101.132

6

2

bd

M Modification factor =1.5 (from BS8110)

Modified span/effective depth ratio = 26X1.5 = 39 > 32.8

The slab is adequate, and there is no deflection.

b) Check for crack

BS8110 states that the maximum spacing of bars in tension is given by 3d = 3X120 = 360mm

Since all the spacing is less than 360mm, there is no cracking required.

Since Vc>Vx and Vy no shear reinforcement required (from PROKON).

11


4 CALCULATION AND DESIGN OF BEAM

4.1 Calculation and design of inner beam

To design a beam, we considered a continuous beam which is highly loaded as it is shown by a

figure below: To see beam 4.

Beam 4-4

Influence area of slab on the beam

A1= 12.125m2

A3= 17.500m2

A5= 1.020m2

4.1.1 Predesigned flanged section beam (T-beam section)

The overall depth (h) is given by 1012

Lh

L

L=5.3, then 10

5.5

12

5.5 h

53.044.0 h

We take h=450 mm

And bw= 200 mm

For flanged beam: b = bw + 0.2Lz, where Lz=0.7Lx

Lz = 0.7x6000 = 4200mm

b =200+0.2x4200= 1040mm

hf=150mm (thickness of the slab)

b

h

bw

hf

b

h

bw

hf

12


4.1.2 Computation of uniformly distributed loading on the beam

a) Dead load

Self-weight of the slab =25x0.15=3.75KN/ 2m

Finishes =1KN/ 2m

Uniformly distributed load on span1 of the beam is given by mKN /31.85.3

125.675.4

Uniformly distributed load on span2 of the beam is given by mKN /85.136

5.1775.4

Uniformly distributed load on span3 is equal to 8.31KN/m


175.4

Self-weight of the beam =25x0.2x0.40= 2KN/m

Self-weight of the wall = 22x0.2x4.2=18.48KN/m

On span1 we have:

Total dead load (Gk) =8.31+2+18.48

Gk=29.29KN/m

Imposed load (Qk) = 5.3

6.1253

Qk=5.25 mKN /

mKNxxQkGkn /406.49)25.56.1()29.294.1(6.14.11

On span2 we have:

Total dead load (Gk)=13.85+2+18.48=34.83KN/m

Imposed load (Qk)= mKN /75.86

5.173

mKNxxQkGkn /762.62)75.86.1()83.344.1(6.14.12

On span3 n3=n1= 49.406KN/m

On span4 we have:

Total dead load (Gk)=2.375+2+18.48= 23.355KN/m

13



13

mKNxxQkGkn /097.35)5.16.1()355.234.1(6.14.14

b) Calculation of bending moment and shear force

Hypotheses:


Concrete 2/30 mmNf

Steel 2/460 mmNf y (for main reinforcement)

2/250 mmNf yv (for shear reinforcement)

Cover (c) = 25mm (Fire resistance required: 2 hours)

Effective width b=1040mm

Thickness of the slab hf=150mm

Overall depth h = 450mm

Thickness of the beam bw = 200mm

Proposed size of reinforcement:

Y16 for bending moment and

Y8 for shear reinforcement

Effective depth mmchd 509162555082

16

14


c) Shear reinforcement

Maximum shear V=214KN computed by PROKON

The maximum shear stress is given by:

23

/7.1509200

10214mmN

bwd

V

The beam section is well designed if fcu8.0

4.4308.0 OK

84.0509200

780100100

bwd

As

From BS8110, we find =0.65N/mm2

( ) OK

( )

( )

Assume mm8 for rings with two legs

22

5.1004

828 mmAsv

mmAsv

SvmmS

As28.61

64.1

5.100

64.164.1

We take Sv=100mm

We provide R8@100mm

01.0509200

5.942

bwd

As

015.0460

3023.023.0

fy

fb

15


Since b , the beam section is adequate

009.0509200

5.942

bwd

As

015.0460

3023.023.0

fy

fb

Since b , the beam section is adequate

16


17


(Section A-A, C-C, E-E, G-G) (Section B-B, D-D, F-F)

Reinforcement in the beam

18


200 200

35

0

15

0

26 R8 @ 100mm

15

0

35

0

R8@100mm

2Y1

2Y14

SECTION B-B SECTION A-A

2Y12

3Y16 A

2Y14

R8@100

B

3Y16 A

3Y16

3Y16

2Y16

2Y16

B

1

50

R8 @ 100 mm 26 R8 @ 100mm

15

0

6m

R8@180

2Y16

SECTION B-B SECTION A-A

3 A

R8@100

B

A

B

19


4.2 Calculation and design of outer beam

Beam F-F

Influence area of slab on the beam, to see beam 5

A1= 9.688m2

A2= 5.688m2

A3= 5.688m2

A4= 5.688m2

4.2.1 Predesigned flanged section beam (T-beam section)

The overall depth (h) is given by 1012

Lh

L

L=5.3, then 10

5.5

12

5.5 h

53.044.0 h

We take h=550mm

And bw= 200mm

For flanged beam: b = bw + 0.2Lz, where Lz=0.7Lx

Lz = 0.7x5000 = 3500mm

b =250+0.2x3500= 900mm

hf=150mm (thickness of the slab)

b

h

bw

hf

b

h

bw

hf

20


4.2.2 Computation of uniformly distributed loading on the beam

b) Dead load

Self-weight of the slab =25x0.15=3.75KN/ 2m

Finishes =1KN/ 2m


688.575.4

Uniformly distributed load on span2 and 4 is equal to 5.40KN/m


688.975.4

Self-weight of the beam =25x0.2x0.4= 2KN/m

Self-weight of the wall = 22x0.2x4.2=18.48KN/m

On span1 we have:

Total dead load (Gk) =5.40+2+18.48

Gk=25.75KN/m

Imposed load (Qk) = 5

5.6883

Qk=3.41 mKN /

mKNxxQkGkn /506.41)41.36.1()75.254.1(6.14.11

On span2 and 4 we have: n2=n4=n1=41.506KN/m

On span3 we have:

Total dead load (Gk)=9.20+1.87+18.48= 29.55KN/m


688.93

mKNxxQkGkn /66.50)81.56.1()55.294.1(6.14.14

d) Calculation of bending moment and shear force

Hypotheses:


21


Concrete 2/30 mmNf

Steel 2/460 mmNf y (for main reinforcement)

2/250 mmNf yv (for shear reinforcement)

Cover (c) = 25mm (Fire resistance required: 2 hours)

Effective width b=900mm

Thickness of the slab hf=150mm

Overall depth h=550mm

Thickness of the beam bw=200mm

Proposed size of reinforcement:

Y16 for bending moment and

Y8 for shear reinforcement

Effective depth mmchd 509162555082

16

22


(Section A-A, C-C, E-E, G-G) (Section B-B, D-D, F-F)

23


4.3 Calculation and design of the Ring Beam

The maximum span of the Ring Beam is 6m, n= 62.762KN/m

Let take h= 400mm and b= 400mmd

24


Section A-A

5. CALCULATION AND DESIGN OF COLUMNS

The design is done on a highly loaded column as it is shown on the figure below:

5.1 Influence area on the column

2Y12

25


5.2 Loads take down

Story

ni Load case

Unity load

(KN/m2) or

(KN/m3)

rea or

volume in

m2 or m

3

(C1)

Area or

volume in

m2 or m

3

(C2)

Total load in KN

(C1)

n4

Roof 5 23.75 8.75 118.75

Top beam 25 0.85 0.45 21.25

Dead load(G4)

140

Design load n4=1.4*G4

196

n3

From n4 196

Column 25 0.512 0.512 12.8

Brick masonry 25 6.24 3.2 156

concrete slab 25 3.56 0.9 89

Finishes 1 23.75 8.75 23.75

Beam 25 0.85 0.45 21.25

Dead load(G3)

302.8

Imposed

load(Q3) 3 23.75 8.75 71.25

Design load n3=n4+1.4*G3+1.6*Q3

733.92

n2

From n3

733.92

Column 25 0.512 0.512 12.8



Finishes 1 23.75 8.75 23.75

Beam 25 0.85 0.45 21.25

Dead load(G2)

302.8

Imposed

loads(Q2) 3 23.75 8.75 71.25


1271.84

n1

From n2 1271.84

Column 25 0.512 0.512 12.8



Finishes 1 23.75 8.75 23.75

Beam 25 0.85 0.45 21.25

Dead load(G1)

302.8

Imposed loads (Q1) 3 23.75 8.75 71.25


1809.76

26


5.3 Calculation and design

5.3.1 Hypotheses

The design load from load take down is KNNd 76.1809

Characteristic material strength:

Concrete 2/30 mmNf

Steel: 2/460 mmNf y (for longitudinal reinforcement)

2/250 mmNf yv (For ring)

Cover (c) = 25mm

Size: h= 400mm b= 400mm

L0 = 4.20m for the Ground Floor Column

L0 = 3.20 for the remaining Columns

5.3.2 Reinforcement in the column

5.3.2.1 Ground floor column

a) Design Column Section

27


28


b) Design Sketch

Figure 1

29


c) Check slenderness of column

The effective height of the column is given by:

lole

Where le -effective height

-coefficient used to determine effective height of the column: 0.75< <1

lo -clear height

mle 96.27.38.0

1586.93.0

96.2

h

le

158.142.0

96.2

b

le

Since the ratios above are all less than 15, we have the short column, therefore there is no

buckling.

d) Check for crushing

The crushing is caused by:

scy AfAcfNuz 87.045.0

3573546087.0600003045.0 Nuz

KNNKNNuz d 46.16841.2240 no crushing required.

e) Check for cracking

BS8110: part 1: 1985 states that if the design load on the column (Nd) is greater than Acf2.0 ,

there is no cracking of the column.

KNNAcf 360360000460302.02.0

3602.046.1684 AcfSinN d , there is no cracking required.

f) Shear reinforcement

BS8110 states that the diameter of ring ring is equal to ¼ of the bigger longitudinal bar, and

this must be greater or equal to 6mm.

30


mm25.6254

1ring , we take ring=8mm

The maximum spacing S is equal to 12 times the diameter of the smaller longitudinal bar.

S=12X20=240mm, we take S=200mm

g) Check the percentage of reinforcement

The maximum area of reinforcement should not exceed 6%:

69.560000

3.3573100100

Ac

Asc

The percentage of reinforcement is adequate

5.3.2.2 First floor column

31


5.3.2.3 Second floor Column

5.3.2.4 Fourth Floor Column

32


6. CALCULATION AND DESIGN OF FOUNDATION

The bearing pressure of the soil 2/90.180 mKNadm and we adopt the square isolated pad

foundation.

6.1 Calculation and design of foundation group1- inner foundations

a) Hypothesis

Bearing pressure of soil 2/90.180 mKNadm

Gk = 1048.4KN and Qk=213.75KN

The self-weight of the isolated pad foundation (w) is between 10% and 20% of the design load of

the column.

KNW 181100

1076.1809

For Serviceability limit states, the axial design load is given by:

QkWGkNd 0.1)(0.1

KNNd 15.144375.2130.1)1814.1048(0.1

The area of square pad foundation is given by:

2197.790.180

15.1443m

Qadm

Nd

We make the base of 2.2mx2.2m; B=L=2.20m

For ultimate serviceability limit states, the ultimate design load on the column is given by:

QkGkNu 6.14.1

KNNu 46.168475.2136.176.18094.1

The pressure of pad foundation on the soil is given by:

2/51.2692.22.2

46.1684mKN

LB

NuP

P<Qadm OK

Assume:

Overall depth of pad foundation h=400mm

Concrete 2/30 mmNf

33


2/460 mmNf y (For longitudinal reinforcement)

Diameter of reinforcement = 16mm

Cover (c) = 40mm

The effective depth is given by 2

-c-h=d

d=350-40-8=302mm

The shear stress at the column c is less than the minimum value of f8.0 and 5KN

2

3

/4.48.006.33024004002

1046.1684mmNf

dPc

Nuc

Where Pc-perimeter of the column.

b) Punching shear

The critical perimeter of pad foundation is given by:

Pcr=Pc+8x1.5d

Pcr=2x(400+400)=8x1.5x302=7600mm

The critical surface is given by :

Scr=(400+2x1.5d)x(400+2x1.5d)=3607500mm2

Scr=3607500 22 61.3 mmm

The punching shear force is given by:

V=P(BxL-Scr) where P-pressure of foundation

V=269.51(2.2x2.2-3.61)=711.51KN

The punching shear stress is given by:

23

/17.03027600

1051.711mmN

dPcr

Vv

We provide Y16@150mm with Approve.=2243

Check the percentage of reinforcement

418.03502200

224310010013.0

bh

As

The percentage of reinforcement is adequate.

Final check for punching

34


16.03022200

2243100100

bd

As

From BS8110, and after interpolation, we find 20.36N/mm=Vc

The punching shear stress v = 0.17 2N/mm

Since v = 0.17< Vc = 0.36; the overall depth (h=600mm) is adequate, therefore there is no

punching of pad foundation.

c) Check for shear force

The shear force is given by

V=269.51X2X0.35= 188.7KN

The shear stress is given by

223

/36.0/25.05502200

107.188mmNVcmmN

bd

Vv

Since v = 0.25< Vc = 0.36, the section is adequate for shear force.

6.2. Moment Steel.

Reinforcement in pad foundation

6.2 Calculation and design of foundation group2-outer foundation

d) Hypothesis

Bearing pressure of soil 2/80.190 mKNadm

Gk = 538.9KN and Qk=78.75KN

The self-weight of the isolated pad foundation (w) is between 10% and 20% of the design load of

the column.

KNW 1.80100

1086.800

For Serviceability limit states, the axial design load is given by:

QkWGkNd 0.1)(0.1

KNNd 75.69775.780.1)1.809.538(0.1

The area of square pad foundation is given by:

35


29.390.180

75.697m

Qadm

N d

We make the base of 1.8mx1.8m; B=L=1.80m

For ultimate serviceability limit states, the ultimate design load on the column is given by:

QkGkNu 6.14.1

KNNu 96.80075.785.19.5384.1

The pressure of pad foundation on the soil is given by:

2/3145.15.1

86.800mKN

LB

NuP

P<Qadm OK

Assume:

Overall depth of pad foundation h=350mm

Concrete 2/30 mmNf

2/460 mmNf y (For longitudinal reinforcement)

Diameter of reinforcement = 16mm

Cover (c) = 40mm

The effective depth is given by 2

-c-h=d

d=300-40-8=252mm

The shear stress at the column c is less than the minimum value of f8.0 and 5KN

2

3

/4.48.008.32524004002

1046.1684mmNf

dPc

Nuc

Where Pc-perimeter of the column.

e) Punching shear

The critical perimeter of pad foundation is given by:

Pcr = Pc+8x1.5d

Pcr = 2x(400+400)=8x1.5x452=7600mm

The critical surface is given by :

Scr = (400+2x1.5d)x(400+2x1.5d)=3607500mm2

36


Scr = 3607500 22 61.3 mmm

The punching shear force is given by:

V=P(BxL-Scr) where P-pressure of foundation

V=269.51(1.5x1.5-3.61)=711.51KN

The punching shear stress is given by:

23

/18.02527600

1051.711mmN

dPcr

Vv

37


7.RESULTS PRESENTATION

The results obtained are given in a table below:

Project: CONSTRUCTION OF A BUILDING WITH TWO LEVELS (G + 2 ).

CALCULATIONS BY: CHECKED BY: Eng. A.H

SOGEIB Ltd

Table Reinforcement for each structural member

Elements

Materials

Concrete Steel

Longitudinal steel Transversal Steel

1 Slab fcu=30N/mm2 Y10@180

Supports: Y12@180

Y10@180

Supports: Y12@180

2 Inner Beams

fcu=30N/mm2

Spans: 3Y16

Supports: 3Y16 + 2Y14 Y8@180

3 Outer Beams

fcu=30N/mm2

Spans: 2Y16 +2Y14

Supports: 3Y16

Y8@180

4 Ring Beam

fcu=30N/mm2

4Y16 + 2Y12 Y8@180

5 Column ground

floor fcu=30N/mm

2 8Y16 Y8@180

6 Column all floors fcu=30N/mm

2

8Y16 Y8@180

9 Inner Foundation all

groups

fcu=30N/m

m2

Y16@180 Y16@180

10 Outer Foundation all

groups

fcu=30N/m

m2

Y16@180 Y16@180

10 Stairs fcu=30N/mm2 Y12@180 Y10@180

Done at Kigali, on November 28, 2018.

By SOGEIB LTD

STRUCTURE CALCULATION NOTE

Documents

Transcript of STRUCTURE CALCULATION NOTE