as built train up a child structural intergrity -report

K.C.C.A Structural integrity report.

[M.T.L] Page 1

MACRO TECHNICS LIMITED

AS BUILT

TRAIN UP A CHILD PRIMARY SCHOOL ON PLOT NO. 33,

LUTHULI DRIVE

M.T.L

AUGUST 2013

Prepared by karija Paul.

Analysis and design calculations for key structural elements pertaining to the development of this project

are presented with in this report.


[M.T.L] Page 2

TABLE OF CONTENTS

1.0 INTRODUCTION .................................................................................................................. 5

1.1Basic details of Structure, Materials, and Loading ............................................................................... 6

1.2 Structural summary ............................................................................................................................. 7

2.0 ANALYSIS AND DESIGN OF FLOOR SLABS ......................................................................... 8

2.1Design of maxspan/ribbed slabs .......................................................................................................... 8

2.1.1 Maxspan slab case 2 ...................................................................................................................... 9

General Analysis .................................................................................................................................... 9

3.0 ANALYSIS AND DESIGN OF FLOOR BEAMS ..................................................................... 13

3.1Critical floor beams ............................................................................................................................ 13

3.1.1 FBM 112 ..................................................................................................................................... 13

4.0 ANALYSIS AND DESIGN OF COLUMNS ............................................................................. 21

4.1 Briefing .............................................................................................................................................. 21

4.2 Column C1 ........................................................................................................................................ 21

4.3Column C2 ......................................................................................................................................... 25

4.4Column C3 ......................................................................................................................................... 29

5.0 ANALYSIS AND DESIGN OF FOUNDATIONS. ................................................................... 33

5.1 Briefing .............................................................................................................................................. 33

5.2 Footing F1 ......................................................................................................................................... 33

5.3 Footing F2 ......................................................................................................................................... 39

6.0 ANALYSIS AND DESIGN OF STAIRCASE ........................................................................... 46

6.1 Staircase type 1. ................................................................................................................................. 46

7.0 ANALYSIS AND DESIGN OF RETAINING WALL................................................................ 50

Indicative retaining wall reinforcement diagram .................................................................................... 59

8.0 APPENDIX .......................................................................................................................... 60

8.1 RING BEAM 01 ................................................................................................................................ 60

8.2 RING BEAM 02 ................................................................................................................................ 62

8.3 Maxspan slab case 1 ........................................................................................................................... 64

8.4 FBM 103 ........................................................................................................................................... 66

8.5 FBM 110 ........................................................................................................................................... 68

8.6 FBM 114 ........................................................................................................................................... 70


[M.T.L] Page 3

9.0 CHECKS FOR ADDITIONAL SECOND FLOOR. ................................................................. 74

9.1 COLUMN CHECKS ........................................................................................................................ 74

9.1.1 Column C1 ................................................................................................................................. 74

9.1.2 Column C2 ................................................................................................................................. 86

9.1.3 Column C3 ................................................................................................................................. 92

9.2 FOUNDATION CHECKS ............................................................................................................... 96

9.2.1 Footing F1 .................................................................................................................................. 96

9.2.2 Footing F2 ................................................................................................................................ 103

9.3 SYNOPSIS ...................................................................................................................................... 109

END... ..................................................................................................................................... 109


[M.T.L] Page 4


[M.T.L] Page 5

1.0 INTRODUCTION

Design calculations for the main structural elements are provided in this report with reference to

Reinforced concrete design codes Euro code 2 and BS8110. Computer aided analysis and design by

PROKON, CSC Tedds, and LIN PRO is also used as detailed and incorporated in this report.

The design calculations have concentrated on structural members which are of critical structural integrity

to the development. These include:

Ribbed Slabs-

Critical maxspan slab case 2.

Critical Floor beams

FBM103, FBM112, FBM 106 and FBM110

Columns C1, C2 & C3.

Foundation Footings F1 & F2.

Design for the means of access and egress (R.C) Stair case

Critical stair case type 1

Retaining wall.


[M.T.L] Page 6

1.1Basic details of Structure, Materials, and Loading

Design information

Client Peter Mutebi & Rebecca Mutebi.

Architect Arch. Geoffrey Muhanguzi

Eng. Responsible Eng. Mubiru Frederick.

Building Regulations Authority K.C.C.A

Relevant Building Regulations and Design codes

1.0 BS8110- Structural use of concrete Part 1& Part 2 2.0 Euro code 2- Design of concrete structures 3.0 BS6399- Loading for buildings Part1Code of practice for dead and imposed loads, Part 2 Code of practice for wind loads, and Part 3 Code of practice for imposed wind loads

Intended use of structure Primary school (Educational institution)

Fire resistance required 1 hour for all elements

General loading conditions Roof imposed loading (pitch


[M.T.L] Page 7

1.2 Structural summary

Notes

Slabs; 350mm thick double maxspan slab & 250mm thick solid slab.

Beams; 500 X 230mm

Columns; 400X 230mm & 230 X 230mm

Column bases; F1& f2 pad footing distributing loading from column C1 & C2 respectively. Check

structural drawings for other varying pad sizes.

Wind resistance- Provided by 350mm thick slab acting as a horizontal beam, lateral bracing by masonry

infill of the blocks and concrete staircase.

Figure showing the typical 1st floor slab layout showing the critical elements whose analysis and design are to be

presented in this structural integrity report.


[M.T.L]

2.0 ANALYSIS AND DESIGN

The design of the slabs was based on the ultimate limit states. The analysis of the slabs are done and

shown in the proceeding pages. The bending moment, shear force diagrams and the deflection diagrams

were plotted by the use of PROKON, a computer aided program. Reference was made to the BS8110

&Euro code in the design and analysis.

2.1Design of maxspan/ribbed slab

Design Data:

Steel Reinforcement bars (fy)

Concrete grade (fcu)

Concrete cover

Loading Data:

Slab thickness

Max-pan depth

Concrete Topping

Finishes (under-plaster & screed)

Mass of max-pan

Unit weight of concrete

Unit weight of finishes

Single max-pan dimensions

Figure showing design rib and the facing dimensions of a maxspan unit


DESIGN OF FLOOR SLABS


the proceeding pages. The bending moment, shear force diagrams and the deflection diagrams


&Euro code in the design and analysis.

2.1Design of maxspan/ribbed slabs

Steel Reinforcement bars (fy) - 460 N/mm2

Concrete grade (fcu) - 25 N/mm2

Concrete cover - 25 mm

= 350mm

= 300mm

= 50mm

plaster & screed) = 50mm

pan = 9.0 kg/unit

Unit weight of concrete = 24 Kn/m3

Unit weight of finishes = 24Kn/m3

pan dimensions = (380x300x150)mm

howing design rib and the facing dimensions of a maxspan unit

Page 8


the proceeding pages. The bending moment, shear force diagrams and the deflection diagrams



[M.T.L] Page 9

2.1.1 Maxspan slab case 2

General Analysis

Loading per rib/ 0.45m length

Permanent loading, Gk

Total dead weight, Gk [1.62 + 0.594 +0.9] = 3.11 Kn/m

Variable load (Qk), for classroom loading = 3.0Kn/m2

Qk = 3.0x0.45 = 1.35 Kn/m

W = (1.35 Gk + 1.5Qk)

W = (1.35 x 3.115) + (1.5 x 1.35) = 6.23 KN/0.45m strip

Using Prokon design program to analyze the rib, the dead and live loads, moments and shear reactions to

be taken up by the beams determined.

Figure .showing longitudinal and cross section through rib.

Results from Prokon,

Long term deflection drawing (Max 27.32mm)


[M.T.L] Page 10

Max shear force, vED= 29.49Kn

Maximum span moment = 32.13Knm, Maximum support moment = 37.03Knm,

Reactions on the beams

Support Dead Load(kN) Live Load(kN) Ultimate(kN)

1 9.29 3.91 18.43

2 14.41 6.06 38.31

3 20.56 8.64 48.20 (onto beam FBM 24)

4 10.19 4.29 20.22

---------------------------------------------------------------------

SHEAR: X-X V max = 29.49kN @ 9.18m

-25.0-20.0-15.0-10.0-5.00

5.0010.015.020.025.030.0

.50

0

1.00

1.50

2.00

2.50

3.00

3.50

4.00

4.50

5.00

5.50

6.00

6.50

7.00

7.50

8.00

8.50

9.00

9.50

10.0

10.5

11.0

11.5

12.0

12.5

13.0

13.5

14.0

14.5

15.0

15.5

16.0

16.5

MOMENTS: X-X M max = -37.03kNm @ 9.18m

30.025.020.015.010.05.00

-5.00-10.0-15.0-20.0-25.0-30.0-35.0

.50

0

1.00

1.50

2.00

2.50

3.00

3.50

4.00

4.50

5.00

5.50

6.00

6.50

7.00

7.50

8.00

8.50

9.00

9.50

10.0

10.5

11.0

11.5

12.0

12.5

13.0

13.5

14.0

14.5

15.0

15.5

16.0

16.5

REFERENCE HOLLOW SLAB OUTPUT

EC2

2.5.3.5.5(2)

a)Longitudinal Reinforcement,

Span Reinforcement,

Maximum span moment ,Med=32.13Knm

Effective depth d= h-cover-0.5

d= 185 -20- 6

d = 319 mm

Reinforcement per rib

k= M/ (bd2fck)

k = 32.13x106 / (450x3192x25)


[M.T.L] Page 11

EC2

4.3.2.3

Eqn 4.18

Table 4.8

4.4.3.2(5)

NAD table 7

k = 0.028 < 0.167 Therefore section is singly reinforced

z = d (0.5+ (0.25-k/1.134))

z = 319 (0.5+ (0.25-0.028 /1.134))

z = 310.89mm

As = M/(0.87fyz)

As =32.13x106 / (0.87x460x310.89)

Asreq =258.23 mm2,

Provide 2Y16 As = 402.1 mm2

Support reinforcement,

Maximum support moment = 37.03Knm,

k= M/ (bd2fck)

k = 37.03x106 / (150x3192x25)

k = 0.097 < 0.167 Therefore section is singly reinforced

z = d (0.5+ (0.25-k/1.134))

z = 319 (0.5+ (0.25-0.097 /1.134))

z = 288.9mm

As = M/(0.87fyz)

As =37.03x106 / (0.87x460x288.9)

Asreq =320.3mm2

Provide 2Y16 As = 402.1 mm2

b)Shear check

VRd1 = [Rd k (1.2+40) bwd]

From table 4.8

Rd = 0.3 for grade 25 concrete

k= 1.6 -0.319= 1.6 0.319= 1.281

= As/bwd

= 402.1/ (150x319)

= 0.0084

VRd1 = [0.3x1.281 (1.2+40x0.0084) x450x319] x10-3

= 84.73kN

VRd1 = 84.73kN > VED= 29.49kN

Therefore shear reinforcement is not required.

c) Deflection check

For an end span of continuous beam or one-way continuous slab or two-way spanning slab continuous over one long side

Provide 2Y16

Provide 2Y16


[M.T.L] Page 12

Check appendix for maxspan slab case 1 design results.

member 100As/bwd Basic span/d

highly stressed 1.5 20

considered 0.84 25.2

lightly stressed 0.5 27.83

100As/bwd = 100x0.0084= 0.84

Basic span/d ratio from table 7 = 32 for lightly stressed member

with fy= 400N/mm2

for grade 460 steel,

Basic span/d ratio = 32x400/460 = 27.83 for lightly stressed

members

Basic span/d ratio from table 7 = 23 for highly stressed member

with fy= 400N/mm2


Basic span/d ratio = 23x400/460 = 20 for highly stressed

members.

Allowable basic span/d = 25.2mm

Actual span/d ratio = 7780/319= 24.4< 25.2

Actual < Allowable deflection.

d)Check for cracking

Bar spacing 3d or 500mm

3d=3 x319= 957mm>rib thickness

e)Check Anchorage & lap length

From the table of lapping lengths in the appendix, KA = KL For apr. 25% bars lapped at section, KA=40 Bar lapping length L= KA x bar dai = 40 x16 =640mm Therefore take lapping length of 650mm.

Deflections are

acceptable

Safe against

cracking

La =650mm


[M.T.L] Page 13

3.0 ANALYSIS AND DESIGN OF FLOOR BEAMS

Beams are structural elements spanning between supports to carry vertical loads from floors, walls and the

any other superimposed loading.

The design of beams was based on the ultimate limit states. The run down analysis and design of the

beams is shown in the proceeding pages. The bending moment, shear force diagrams and the deflection

diagrams were plotted by the use of PROKON, a computer aided program. Reference was made to the

BS8110 &Euro code in the design and analysis.

3.1Critical floor beams

General Design data for beams,

Concrete cover=30mm, Assumed diameter of bars=16 mm, Concrete density=24kN/m3, Wall density=5kN/m2 for a 230mm thick wall, Concrete grade=25N/mm2, Steel grade=450N/mm2, Weight of finishes is negligible, Beam = 500X230mm, Maxspan slab loading, Gk= [3.11 /0.45] = 6.911Kn/m2

3.1.1 FBM 112

Beam 24 designed as a rectangular beam


Total dead weight, Gk [34.59 +12.5+2.76] = 60.95 Kn/m

Variable load (Qk),

Qk = [8.64/0.45] = 19.2 Kn/m

W = (1.35 Gk + 1.5Qk)

W = (1.35 x 60.95) + (1.5 x 19.2) = 111.083Kn/m,

Using Prokon to analyze and design the beam, span and support reinforcements are designed for and other limit states are checked as shown below.

From critical maxspan slab [20.56/0.45] = 45.69 Kn/m From walling [5kN/m2 (3-0.5)] = 12.5 Kn/m Beam self weight [24kN/m3x 0.23 x0.5] = 2.76 Kn/m

111.083Kn/m


[M.T.L] Page 14

Results

Long term deflection drawing (Max 11.83mm)


[M.T.L] Page 15

Column moments (Ultimate limit state)

Columns below

Support Bending Moment(kNm) Shear Force (kN)

No. Min. Max. Min. Max.

1 0.00 0.00 0.00 0.00

2 0.98 3.98 0.49 1.99

3 -7.60 -2.61 -3.80 -1.30

4 -2.12 6.27 -1.06 3.13

5 -10.27 -0.72 -5.13 -0.36

6 0.00 0.00 0.00 0.00

Columns above

Support Bending Moment(kNm) Shear Force (kN)

No. Min. Max. Min. Max.

1 0.00 0.00 0.00 0.00

2 0.98 3.98 0.49 1.99

3 -7.60 -2.61 -3.80 -1.30

4 -2.12 6.27 -1.06 3.13

5 -10.27 -0.72 -5.13 -0.36

6 0.00 0.00 0.00 0.00

--------------------------------------------------------------------- -----------------------------------------------------------------

REFERENCE FLOOR BEAMS OUTPUT

EC2 2.5.3.5.5(2)

a)Longitudinal Reinforcement,

Span Reinforcement,

Maximum span moment, Med= 226.6Knm

Effective depth d= h-cover-link -

main bar

d= 500 -30- 10- 16/2

d = 452 mm

Reinforcement per rib

k= M/ (bd2fck)


[M.T.L] Page 16

Table 15.5 of the

concise EC2

k = 226.6x106 / (230x4522x25)

k = 0.193 > 0.167 Therefore section is doubly reinforced

Area of compression steel

;

where

Therefore provide3 Y16 having a total area of 603.2mm2

Area of tension steel

Where;

Provide 3Y16

( )( )'87.0

2'

ddfbdfKKA

yk

ckbals

=

barofxdiametererd 21cov' +=

2

2

2

1533.43

181.43 452 82. 0 45087.0

452230 25 167.0

'

87.0

mm Axxx

xxxA

AZf bd f KA

s

s

s

balyk

ckbals

=

+=

+=

( )( )

2

2

181.4 ' 22.0 45245087.0

452230 25 167 . 0 193 . 0 '

mmAxx

x xxA

s

s

=

=

steel ncompressioofarea A mm d mm b

mmNfmmNf

K K

s

ck

yk

bal

=

=

=

=

=

=

=

'

452230

2 /25

2 /450193.0

167 .0

mmx d .022 162 130 ' =+=


[M.T.L] Page 17

Provide 8Y16 As = 1608.5mm2

Support reinforcement,

Maximum support moment, Med= 308.2Knm

k= M/ (bd2fck)

k = 308.2x106 / (230x4522x25)

k = 0.262 > 0.167 Therefore section is doubly reinforced

Area of compression steel

;

where

Therefore provide4 Y16 having a total area of 804.2mm2

Area of tension steel

Provide 8Y16

Provide 4Y16

( )( )'87.0

2'

ddfbdfKKA

yk

ckbals

=

barofxdiametererd 21cov' +=

( )( )

2

2

662.93 ' 22.0 45245087.0

452230 25 167 . 0 262 . 0 '

mm A x x

x x x A

s

s

=

=

steel n compressio of area A mm d mm b

mm N f mm N f

K K

s

ck

yk

bal

=

=

=

=

=

=

=

'

452230

2 /25

2/450262.0

167.0

mm x d .022 162 130 ' =+=

steel n compressio of area A d Z

mm d mm b

f K

s

bal

ck

bal

=

=

=

=

=

=

'

82. 0 452230

25167 .0


[M.T.L] Page 18

Where;

Provide 12Y16 As = 2412.74mm2

Check for minimum tension reinforcement;

Asmin> 0.0015bd , b = 230mm, d = 452mm.

Asmin > 0.0015x300x457

Asmin > 155.94mm2

Asmin = 206 mm2< As prov = 2412.74 mm2 ok

Check for maximum reinforcement.

As not greater than 0.04bd

0.04bd = 0.04x230x452 = 6000mm2

As = 2412.74 mm2< 4158.4mm2 ok

b) Design for shear.

Maximum shear = 339.5kN,

links = 10mm

Relating to shear force diagram, Shear at face of support, Vsd =

324.5KN

Using the standard method.

VRd2 = ( fcdbw0.9d)

= 0.7-25/200 = 0.575 > 0.5 ok

VRd2 = 1/2x0.575x25/1.5x230x0.9x452

= 448.33kN

VRd2 = 448.33kN > Vsd = 324.5KN ok

Provide 12Y16

Section is

steel n compressio of area A d Z

mm d mm b

f K

s

bal

ck

bal

=

=

=

=

=

=

'

82. 0 452230

25167 .0

2

2

2

2014.93

662.93 452 82. 0 45087.0

452230 25 167.0

'

87.0

mm A x x x

x x x A

A Z f bd f K A

s

s

s bal yk

ck bal s

=

+=

+=


[M.T.L] Page 19

EC2

4.3.2.3

Eqn 4.18

Table 4.8

EC2

Table 5.5

Table NA5 of the

national annex to

EC2.

Concrete shear resistance, VRd1

VRd1 = (Rdk(1.2+401)+0.15cp)bwd

For fck = 25, Rd=0.3

K = 1.6-d > 1

= 1.6 0.452

= 1.148>1 ok

1 = Asprov/(bwd) At section , Asprov= 14Y16 bars= 2814.86

1 =2814.86/(230x452)

= 0.0213

VRd1 = (0.3x1.148(1.2+40x0.0282))x230x452

= 94669.3N

= 94.66kN

VRd1 = 94.66kN > Vsd =324.5KN

VRd3 = Vsd = VRd1 +Vwd

Vwd = Vsd-VRd1

= 324.5-94.66

= 229.85 kN

Vwd = Asw/s(0.9dfywd) = 229.85

fywd = 0.87fyk

Asw/s = 229.85 x103/(0.9x452x.87x250)

Asw/s = 2.59, Asw= 2(102/4)=157.1

s = 157/(2.59) = 60.62 mm, s = 50 mm

Provide R10 @ 50mm

Check minimum reinforcement

w = Asw/(sbw)

As /s= wbw

Assume S220 = 250 N/mm2

w = 0.0024

s= As/(bww) = 157/(0.0024x230)

= 284.4 mm

Provide R10 @100 mm at low shear zones as shown in the

curtailed drawing

c) Deflection check

For an end span of continuous beam or one-way continuous slab or two-way spanning slab continuous over one long side

member 100As/bwd Basic span/d

considered 2.13

highly stressed 1.5 23

lightly stressed 0.5 32

adequate

Provide R10 @

50mm for Main

shear

Provide R10 @

100mm for

nominal shear


[M.T.L] Page 20

Check appendix for other critical floor beam design results. The critical beams include; FBM 103, 114,

110.

Table A.6 of the

Mosley

11Y16 provided at midspan for tension reinforcement.

100As/bwd =

100x1211.68/(230x452)= 2.13>1.5 Hence highly stressed

Basic span/d ratio from table 7 = 23 for highly stressed member

with fy= 400N/mm2


Basic span/d ratio = 23x400/250 = 36.8 for highly stressed

members.

Allowable basic span/d = 36.8mm

Actual span/d ratio = 5000/452= 11.06 < 36.8

Actual < Allowable deflection.

d)Check lap length

From the table of lapping lengths in the appendix, For apr. 25% bars lapped at section, KA=40 Bar lapping length L= KA x bar dai = 40 x16 =640mm Therefore take lapping length of 650mm.

Safe against

deflection

LL=650mm


[M.T.L] Page 21

4.0 ANALYSIS AND DESIGN OF COLUMNS

4.1 Briefing

Columns are vertical compression structural members that carry loads from beams and slabs to foundations. Columns may be classified as;

Short columns

Slender columns

The analysis & design of the example in this report was done based on BS8110

Note; The columns C1, C2 & C3 analysis and design is provided in this chapter.

4.2 Column C1

From analysis, Column C1 was found to be the critical (most loaded column).

Loading onto column from beams and slabs obtained from analysis of FBM 112 & 103 as shown in the table below;

Loading,

Floor to floor height = 3m

Column self weight = 0.4x 0.23x24x3 Gk = 6.63KN

Gk Qk My Mx

1st floor slab & beams 433.75KN 108.42KN 6.4 Knm 10.27 Knm

Column self weight 6.63KN 0KN

Ring beams 100.26KN 11.34KN 3.82knm 3.23knm

i) At Ground level

Gk = 547.27KN Qk=119.76KN

Ultimate load =1.4Gk+ 1.6Qk =957.8KN

TEDDS calculation version 2.0.01

RC COLUMN DESIGN (BS8110:PART1:1997) TEDDS calculation version 2.0.01

Column definition Column depth (larger column dim); h = 400 mm Nominal cover to all reinforcement (longer dim); ch = 30 mm Depth to tension steel; h' = h - ch Ldia Dcol/2 = 354 mm


[M.T.L] Page 22

Column width (smaller column dim); b = 250 mm Nominal cover to all reinforcement (shorter dim); cb = 30 mm Depth to tension steel; b' = b - cb - Ldia Dcol/2 = 204 mm Characteristic strength of reinforcement; fy = 450 N/mm2 Characteristic strength of concrete; fcu = 25 N/mm2

Braced Column Design to cl 3.8.4

Check on overall column dimensions Column OK - h < 4b

Braced column slenderness check Column clear height; lo = 2500 mm Slenderness limit; llimit = 60 b = 15000 mm

Column slenderness limit OK

Short column check for braced columns Column clear height; lo = 2500 mm Effect. height factor for braced columns - maj axis; x = 0.75

BS8110:Table 3.19 Effective height major axis; lex = x lo = 1.875 m; Slenderness check; lex/h = 4.69

The braced column is short (major axis) Effect height factor for braced columns - minor axis; y = 0.75

BS8110:Table 3.19 Effective height minor axis; ley = y lo = 1.875 m Slenderness check; ley/b = 7.50

The braced column is short (minor axis) Short column - bi-axial bending

Define column reinforcement Main reinforcement in column Assumed diameter of main reinforcement; Dcol = 16 mm Assumed no. of bars in one face (assumed sym); Lncol = 3 Area of "tension" steel; Ast = Lncol pi Dcol2 / 4 = 603 mm2 Area of compression steel; Asc = Ast = 603 mm2 Total area of steel; Ascol = Ast + Asc = 1206.4 mm2 Percentage of steel; (Ast + Asc) / (b h) = 1.2 %


[M.T.L] Page 23

Design ultimate loading Design ultimate axial load; N = 958 kN Design ultimate moment (major axis); Mx = 6 kNm Design ultimate moment (minor axis); My = 10 kNm Minimum design moments Min design moment (major axis); Mxmin = min(0.05 h, 20 mm) N = 19.2 kNm Min design moment (minor axis); Mymin = min(0.05 b, 20 mm) N = 12.0 kNm Design moments Design moment (major axis); Mxdes = max (abs(Mx), Mxmin) = 19.2 kNm Design moment (minor axis); Mydes = max (abs(My) , Mymin) = 12.0 kNm Simplified method for dealing with bi-axial bending:- h' = 354 mm; b' = 204 mm Approx uniaxial design moment (Cl 3.8.4.5) = 1 - 1.165 min(0.6, N/(bhfcu)) = 0.55 Design moment;

Mdesign = if(Mxdes/h' < Mydes/b' , Mydes + b'/h' Mxdes , Mxdes + h'/b' Mydes ) = 18.1 kNm Set up section dimensions for design:- Section depth; D = if(Mxdes/h' < Mydes/b' , b, h) = 250.0 mm Depth to "tension" steel; d = if(Mxdes/h' < Mydes/b' , b', h') = 204.0 mm Section width; B = if(Mxdes/h' < Mydes/b' , h, b) = 400.0 mm Check of design forces - symmetrically reinforced section

NOTE Note:- the section dimensions used in the following calculation are:- Section width (parallel to axis of bending); B = 400 mm Section depth perpendicular to axis of bending); D = 250 mm Depth to "tension" steel (symmetrical); d = 204 mm

Compression steel yields (0.9x


[M.T.L] Page 24

Column depth; h = 400 mm Effective depth to steel; h' = 354 mm Area of concrete; Aconc = b h = 100000 mm2 Design ultimate shear force (major axis); Vx = 340 kN Characteristic strength of concrete; fcu = 25 N/mm2

Is a check required? (3.8.4.6) Axial load; N = 957.9 kN Major axis moment; Mx = 6.4 kNm Eccentricity; e = Mx / N = 6.7 mm Limit to eccentricity; elimit = 0.6 h = 240.0 mm Actual shear stress; vx = Vx / (b h') = 3.8 N/mm2 Allowable stress; vallowable = min ((0.8 N1/2/mm) (fcu ), 5 N/mm2 ) = 4.000 N/mm2

No shear check required Define Containment Links Provided Link spacing; sv = 150 mm; Link diameter; Ldia = 8 mm; No of links in each group; Ln = 1

Minimum Containment Steel (Cl 3.12.7) Shear steel Link spacing; sv = 150 mm Link diameter; Ldia = 8 mm

Column steel Diameter; Dcol = 16 mm Min diameter; Llimit = max( (6 mm), Dcol/4) = 6.0 mm

Link diameter OK Max spacing; slimit = 12 Dcol = 192.0 mm

Link spacing OK Reinforcement Summary;

Ascrequired = 408 mm2

PROVIDED 4Y25mm + 4Y16mm bars (Ascprovided) = 2768 mm2) & Links, provide R8 bars @150 c/c

At the First floor level PROVIDED - 8Y16mm bars (Ascprovided) = 1608.5 mm2) & Links, provide R8 bars @150 c/c


[M.T.L] Page 25

4.3Column C2

Loading onto column from beams and slabs obtained from analysis of FBM 1110 as shown in the table below;

Loading,



Gk Qk My Mx

1st floor slab & beams 381.08KN 99.0KN 0 Knm 6.4 Knm



i) At Ground level

Gk = 495KN Qk=110.34KN

Ultimate load =1.4Gk+ 1.6Qk =869.6KN


Column definition Column depth (larger column dim); h = 250 mm Nominal cover to all reinforcement (longer dim); ch = 30 mm Depth to tension steel; h' = h - ch Ldia Dcol/2 = 204 mm Column width (smaller column dim); b = 400 mm Nominal cover to all reinforcement (shorter dim); cb = 30 mm Depth to tension steel; b' = b - cb - Ldia Dcol/2 = 354 mm Characteristic strength of reinforcement; fy = 450 N/mm2 Characteristic strength of concrete; fcu = 25 N/mm2


[M.T.L] Page 26










Define column reinforcement Main reinforcement in column Assumed diameter of main reinforcement; Dcol = 16 mm Assumed no. of bars in one face (assumed sym); Lncol = 3 Area of "tension" steel; Ast = Lncol pi Dcol2 / 4 = 603 mm2 Area of compression steel; Asc = Ast = 603 mm2 Total area of steel; Ascol = Ast + Asc = 1206.4 mm2 Percentage of steel; (Ast + Asc) / (b h) = 1.2 % Design ultimate loading Design ultimate axial load; N = 870 kN Design ultimate moment (major axis); Mx = 0 kNm Design ultimate moment (minor axis); My = 6 kNm Minimum design moments Min design moment (major axis); Mxmin = min(0.05 h, 20 mm) N = 10.9 kNm Min design moment (minor axis); Mymin = min(0.05 b, 20 mm) N = 17.4 kNm Design moments Design moment (major axis); Mxdes = max (abs(Mx), Mxmin) = 10.9 kNm Design moment (minor axis); Mydes = max (abs(My) , Mymin) = 17.4 kNm Simplified method for dealing with bi-axial bending:- h' = 204 mm; b' = 354 mm Approx uniaxial design moment (Cl 3.8.4.5) = 1 - 1.165 min(0.6, N/(bhfcu)) = 0.59 Design moment;

Mdesign = if(Mxdes/h' < Mydes/b' , Mydes + b'/h' Mxdes , Mxdes + h'/b' Mydes ) = 16.8 kNm


[M.T.L] Page 27

Set up section dimensions for design:- Section depth; D = if(Mxdes/h' < Mydes/b' , b, h) = 250.0 mm Depth to "tension" steel; d = if(Mxdes/h' < Mydes/b' , b', h') = 204.0 mm Section width; B = if(Mxdes/h' < Mydes/b' , h, b) = 400.0 mm Check of design forces - symmetrically reinforced section




[M.T.L] Page 28




Link spacing OK Reinforcement summary;


PROVIDED 4Y25mm + 4Y16mm bars (Ascprovided) = 2768 mm2) & Links, provide R8 bars @150 c/c

At the First floor level PROVIDED - 8Y16mm bars (Ascprovided) = 1608.5 mm2) & Links, provide R8 bars @150 c/c


[M.T.L] Page 29

4.4Column C3

Loading onto column from beams and slabs obtained from analysis of FBM 112 & 114 as shown in the table below;

Loading,



Gk Qk My Mx

1st floor slab & beams 245KN 34.94KN 13.98 Knm 2.03 Knm



i) At Ground level

Gk = 358.52KN Qk=46.3KN

Ultimate load =1.4Gk+ 1.6Qk =576KN


Column definition Column depth (larger column dim); h = 230 mm Nominal cover to all reinforcement (longer dim); ch = 30 mm Depth to tension steel; h' = h - ch Ldia Dcol/2 = 184 mm Column width (smaller column dim); b = 400 mm Nominal cover to all reinforcement (shorter dim); cb = 30 mm Depth to tension steel; b' = b - cb - Ldia Dcol/2 = 354 mm Characteristic strength of reinforcement; fy = 450 N/mm2 Characteristic strength of concrete; fcu = 25 N/mm2

184

230

30


[M.T.L] Page 30










Define column reinforcement Main reinforcement in column Assumed diameter of main reinforcement; Dcol = 16 mm Assumed no. of bars in one face (assumed sym); Lncol = 2 Area of "tension" steel; Ast = Lncol pi Dcol2 / 4 = 402 mm2 Area of compression steel; Asc = Ast = 402 mm2 Total area of steel; Ascol = Ast + Asc = 804.2 mm2 Percentage of steel; (Ast + Asc) / (b h) = 0.9 % Design ultimate loading Design ultimate axial load; N = 576 kN Design ultimate moment (major axis); Mx = 2 kNm Design ultimate moment (minor axis); My = 14 kNm Minimum design moments Min design moment (major axis); Mxmin = min(0.05 h, 20 mm) N = 6.6 kNm Min design moment (minor axis); Mymin = min(0.05 b, 20 mm) N = 11.5 kNm Design moments Design moment (major axis); Mxdes = max (abs(Mx), Mxmin) = 6.6 kNm Design moment (minor axis); Mydes = max (abs(My) , Mymin) = 14.0 kNm Simplified method for dealing with bi-axial bending:- h' = 184 mm; b' = 354 mm Approx uniaxial design moment (Cl 3.8.4.5) = 1 - 1.165 min(0.6, N/(bhfcu)) = 0.71 Design moment;

Mdesign = if(Mxdes/h' < Mydes/b' , Mydes + b'/h' Mxdes , Mxdes + h'/b' Mydes ) = 23.0 kNm


[M.T.L] Page 31

Set up section dimensions for design:- Section depth; D = if(Mxdes/h' < Mydes/b' , b, h) = 400.0 mm Depth to "tension" steel; d = if(Mxdes/h' < Mydes/b' , b', h') = 354.0 mm Section width; B = if(Mxdes/h' < Mydes/b' , h, b) = 230.0 mm Check of design forces - symmetrically reinforced section




[M.T.L] Page 32




Link spacing OK

Reinforcement summary;


PROVIDED 4Y16mm bars (Ascprovided) = 804.2 mm2) & Links, provide R8 bars @150 c/c

At the First floor level

PROVIDED - 4Y16mm bars (Ascprovided) = 804.2 mm2) & Links, provide R8 bars @150 c/c


[M.T.L] Page 33

5.0 ANALYSIS AND DESIGN OF FOUNDATIONS.

5.1 Briefing

Foundations transfer and spread loads from columns and walls onto the ground. In order to achieve this,

the safe bearing capacity of the soil must not be exceeded.

In this particular design,

Pad foundations and combined footings are provided for columns. Combined footings are

provided were columns are close together in that their individual pads have to be joined so

that they do a combined action of transferring load.

Unreinforced strip foundations are provided for the strip walls.

Footings F1 & F 2 are the critical pads to be designed in this report.

5.2 Footing F1

Pad footing F1 carries loading from column C1;

Total column loading at Ground level , Gk = 547.3KN, Qk=119. 8Kn

Ultimate loading =1.4Gk+1.6Qk = 957.8Kn

Design information;

Safe bearing capacity of soil = 250kN/m2

Strength of concrete = 24kN/m2

Column 400x230mm

Grade 25 concrete

500mm overall depth

40mm cover

16mm bars

a) Determining of pad size

Service load (Ns)

Ns = Gk + Qk + Wb where;

Gk = characteristic permanent load from column

Qk = characteristic variable load from column

Wb = weight of base

Wb= 10% of (Gk + Qk)

Gk + Qk= 667.1KN each column C1

Ns= 733.81 each column C1

Safe bearing pressure= 250kN/m2


[M.T.L] Page 34

Area of footing = service load/ Bearing pressure

= 733.81 /250

= 2.94m2 required

Considering a square footing of 1.8m x 1.8m. (3.24 m2)

PAD FOOTING ANALYSIS AND DESIGN (BS8110-1:1997) TEDDS calculation version 2.0.03.00

Pad footing details Length of pad footing; L = 1800 mm Width of pad footing; B = 1800 mm Area of pad footing; A = L B = 3.240 m2 Depth of pad footing; h = 350 mm Depth of soil over pad footing; hsoil = 200 mm Density of concrete; conc = 24.0 kN/m3

Column details Column base length; lA = 250 mm Column base width; bA = 400 mm Column eccentricity in x; ePxA = 0 mm Column eccentricity in y; ePyA = 0 mm

Soil details Density of soil; soil = 20.0 kN/m3 Design shear strength; = 25.0 deg Design base friction; = 19.3 deg Allowable bearing pressure; Pbearing = 250 kN/m2

Axial loading on column Dead axial load on column; PGA = 547.3 kN Imposed axial load on column; PQA = 119.8 kN Wind axial load on column; PWA = 0.0 kN Total axial load on column; PA = 667.1 kN


[M.T.L] Page 35

Foundation loads Dead surcharge load; FGsur = 0.000 kN/m2 Imposed surcharge load; FQsur = 0.000 kN/m2 Pad footing self weight; Fswt = h conc = 8.400 kN/m2 Soil self weight; Fsoil = hsoil soil = 4.000 kN/m2 Total foundation load; F = A (FGsur + FQsur + Fswt + Fsoil) = 40.2 kN Calculate pad base reaction Total base reaction; T = F + PA = 707.3 kN Eccentricity of base reaction in x; eTx = (PA ePxA + MxA + HxA h) / T = 0 mm Eccentricity of base reaction in y; eTy = (PA ePyA + MyA + HyA h) / T = 0 mm Check pad base reaction eccentricity abs(eTx) / L + abs(eTy) / B = 0.000

Base reaction acts within middle third of base

Calculate pad base pressures q1 = T / A - 6 T eTx / (L A) - 6 T eTy / (B A) = 218.295 kN/m2 q2 = T / A - 6 T eTx / (L A) + 6 T eTy / (B A) = 218.295 kN/m2 q3 = T / A + 6 T eTx / (L A) - 6 T eTy / (B A) = 218.295 kN/m2 q4 = T / A + 6 T eTx / (L A) + 6 T eTy / (B A) = 218.295 kN/m2 Minimum base pressure; qmin = min(q1, q2, q3, q4) = 218.295 kN/m2 Maximum base pressure; qmax = max(q1, q2, q3, q4) = 218.295 kN/m2

PASS - Maximum base pressure is less than allowable bearing pressure

Partial safety factors for loads Partial safety factor for dead loads; fG = 1.40 Partial safety factor for imposed loads; fQ = 1.60 Partial safety factor for wind loads; fW = 0.00


[M.T.L] Page 36

Ultimate axial loading on column Ultimate axial load on column; PuA = PGA fG + PQA fQ + PWA fW = 957.9 kN

Ultimate foundation loads Ultimate foundation load; Fu = A [(FGsur + Fswt + Fsoil) fG + FQsur fQ] = 56.2 kN Ultimate horizontal loading on column Ultimate horizontal load in x direction; HxuA = HGxA fG + HQxA fQ + HWxA fW = 0.0 kN Ultimate horizontal load in y direction; HyuA = HGyA fG + HQyA fQ + HWyA fW = 0.0 kN

Ultimate moment on column Ultimate moment on column in x direction; MxuA = MGxA fG + MQxA fQ + MWxA fW = 0.000 kNm Ultimate moment on column in y direction; MyuA = MGyA fG + MQyA fQ + MWyA fW = 0.000 kNm

Calculate ultimate pad base reaction Ultimate base reaction; Tu = Fu + PuA = 1014.1 kN Eccentricity of ultimate base reaction in x; eTxu = (PuA ePxA + MxuA + HxuA h) / Tu = 0 mm Eccentricity of ultimate base reaction in y; eTyu = (PuA ePyA + MyuA + HyuA h) / Tu = 0 mm Calculate ultimate pad base pressures q1u = Tu/A - 6TueTxu/(LA) - 6TueTyu/(BA) = 313.008 kN/m2 q2u = Tu/A - 6TueTxu/(LA) + 6Tu eTyu/(BA) = 313.008 kN/m2 q3u = Tu/A + 6TueTxu/(LA) - 6TueTyu/(BA) = 313.008 kN/m2 q4u = Tu/A + 6TueTxu/(LA) + 6TueTyu/(BA) = 313.008 kN/m2 Minimum ultimate base pressure; qminu = min(q1u, q2u, q3u, q4u) = 313.008 kN/m2 Maximum ultimate base pressure; qmaxu = max(q1u, q2u, q3u, q4u) = 313.008 kN/m2 Calculate rate of change of base pressure in x direction Left hand base reaction; fuL = (q1u + q2u) B / 2 = 563.415 kN/m Right hand base reaction; fuR = (q3u + q4u) B / 2 = 563.415 kN/m Length of base reaction; Lx = L = 1800 mm Rate of change of base pressure; Cx = (fuR - fuL) / Lx = 0.000 kN/m/m Calculate pad lengths in x direction Left hand length; LL = L / 2 + ePxA = 900 mm Right hand length; LR = L / 2 - ePxA = 900 mm

Calculate ultimate moments in x direction Ultimate moment in x direction; Mx = fuLLL2/2+CxLL3/6-FuLL2/(2L) = 215.528 kNm Calculate rate of change of base pressure in y direction Top edge base reaction; fuT = (q2u + q4u) L / 2 = 563.415 kN/m Bottom edge base reaction; fuB = (q1u + q3u) L / 2 = 563.415 kN/m Length of base reaction; Ly = B = 1800 mm Rate of change of base pressure; Cy = (fuB - fuT) / Ly = 0.000 kN/m/m Calculate pad lengths in y direction Top length; LT = B / 2 - ePyA = 900 mm Bottom length; LB = B / 2 + ePyA = 900 mm


[M.T.L] Page 37

Calculate ultimate moments in y direction Ultimate moment in y direction; My = fuTLT2/2+CyLT3/6-FuLT2/(2B) = 215.528 kNm Material details Characteristic strength of concrete; fcu = 25 N/mm2 Characteristic strength of reinforcement; fy = 500 N/mm2 Characteristic strength of shear reinforcement; fyv = 250 N/mm2 Nominal cover to reinforcement; cnom = 30 mm

Moment design in x direction Diameter of tension reinforcement; xB = 12 mm Depth of tension reinforcement; dx = h - cnom - xB / 2 = 314 mm Design formula for rectangular beams (cl 3.4.4.4) Kx = Mx / (B dx2 fcu) = 0.049 Kx = 0.156

Kx < Kx' compression reinforcement is not required Lever arm; zx = dx min([0.5 + (0.25 - Kx / 0.9)], 0.95) = 296 mm Area of tension reinforcement required; As_x_req = Mx / (0.87 fy zx) = 1674 mm2 Minimum area of tension reinforcement; As_x_min = 0.0013 B h = 819 mm2 Tension reinforcement provided; 17 No. 12 dia. bars bottom (100 centres) Area of tension reinforcement provided; As_xB_prov = NxB pi xB2 / 4 = 1923 mm2

PASS - Tension reinforcement provided exceeds tension reinforcement required

Moment design in y direction Diameter of tension reinforcement; yB = 12 mm Depth of tension reinforcement; dy = h - cnom - xB - yB / 2 = 302 mm Design formula for rectangular beams (cl 3.4.4.4) Ky = My / (L dy2 fcu) = 0.053 Ky = 0.156

Ky < Ky' compression reinforcement is not required Lever arm; zy = dy min([0.5 + (0.25 - Ky / 0.9)], 0.95) = 283 mm Area of tension reinforcement required; As_y_req = My / (0.87 fy zy) = 1749 mm2 Minimum area of tension reinforcement; As_y_min = 0.0013 L h = 819 mm2 Tension reinforcement provided; 17 No. 12 dia. bars bottom (100 centres) Area of tension reinforcement provided; As_yB_prov = NyB pi yB2 / 4 = 1923 mm2


Calculate ultimate shear force at d from right face of column Ultimate pressure for shear; qsu = (q1u + Cx (L / 2 + ePxA + lA / 2 + dx) / B + q4u) / 2 qsu = 313.008 kN/m2 Area loaded for shear; As = B min(3 (L / 2 - eTx), L / 2 - ePxA - lA / 2 - dx) = 0.830 m2

Ultimate shear force; Vsu = As (qsu - Fu / A) = 245.329 kN Shear stresses at d from right face of column (cl 3.5.5.2) Design shear stress; vsu = Vsu / (B dx) = 0.434 N/mm2 From BS 8110:Part 1:1997 - Table 3.8 Design concrete shear stress; vc = 0.469 N/mm2 Allowable design shear stress; vmax = min(0.8N/mm2 (fcu / 1 N/mm2), 5 N/mm2) = 4.000 N/mm2


[M.T.L] Page 38

PASS - vsu < vc - No shear reinforcement required

Calculate ultimate punching shear force at face of column Ultimate pressure for punching shear; qpuA = q1u+[(L/2+ePxA-lA/2)+(lA)/2]Cx/B-[(B/2+ePyA-bA/2)+(bA)/2]Cy/L qpuA = 313.008 kN/m2 Average effective depth of reinforcement; d = (dx + dy) / 2 = 308 mm Area loaded for punching shear at column; ApA = (lA)(bA) = 0.100 m2 Length of punching shear perimeter; upA = 2(lA)+2(bA) = 1300 mm Ultimate shear force at shear perimeter; VpuA = PuA + (Fu / A - qpuA) ApA = 928.335 kN Effective shear force at shear perimeter; VpuAeff = VpuA = 928.335 kN

Punching shear stresses at face of column (cl 3.7.7.2) Design shear stress; vpuA = VpuAeff / (upA d) = 2.319 N/mm2 Allowable design shear stress; vmax = min(0.8N/mm2 (fcu / 1 N/mm2), 5 N/mm2) = 4.000 N/mm2

PASS - Design shear stress is less than allowable design shear stress

Calculate ultimate punching shear force at perimeter of 1.5 d from face of column Ultimate pressure for punching shear; qpuA1.5d = q1u+[(L/2+ePxA-lA/2-1.5d)+(lA+21.5d)/2]Cx/B-[B/2]Cy/L qpuA1.5d = 313.008 kN/m2 Average effective depth of reinforcement; d = (dx + dy) / 2 = 308 mm Area loaded for punching shear at column; ApA1.5d = (lA+21.5d)B = 2.113 m2 Length of punching shear perimeter; upA1.5d = 2B = 3600 mm Ultimate shear force at shear perimeter; VpuA1.5d = PuA + (Fu / A - qpuA1.5d) ApA1.5d = 333.136 kN Effective shear force at shear perimeter; VpuA1.5deff = VpuA1.5d 1.25 = 416.420 kN

Punching shear stresses at perimeter of 1.5 d from face of column (cl 3.7.7.2) Design shear stress; vpuA1.5d = VpuA1.5deff / (upA1.5d d) = 0.376 N/mm2 From BS 8110:Part 1:1997 - Table 3.8 Design concrete shear stress; vc = 0.474 N/mm2 Allowable design shear stress; vmax = min(0.8N/mm2 (fcu / 1 N/mm2), 5 N/mm2) = 4.000 N/mm2

PASS - vpuA1.5d < vc - No shear reinforcement required

REQUIRED - Square footing of 1.8m X 1.8m, depth 350mm


[M.T.L] Page 39

PROVIDED - Combined footing of 4.5m X 2.5m, depth 500mm

5.3 Footing F2

Pad footing F2 carries loading from column C2;

Total column loading at Ground level , Gk = 495KN, Qk=110.34Kn

Ultimate loading =1.4Gk+1.6Qk = 869.6Kn

Design information;

Safe bearing capacity of soil = 250kN/m2

Strength of concrete = 24kN/m2

Column 400x230mm

Grade 25 concrete

500mm overall depth

40mm cover

16mm bars

a) Determining of pad size

Service load (Ns)

Ns = Gk + Qk + Wb where;

Gk = characteristic permanent load from column

Qk = characteristic variable load from column

Wb = weight of base

Wb= 10% of (Gk + Qk)

Gk + Qk= 665.8KN


[M.T.L] Page 40

Ns= 665.87Kn

Safe bearing pressure= 250kN/m2

Area of footing = service load/ Bearing pressure

= = 665.87/250

= 2.66m2 required

Considering a square footing of 1.8m x 1.8m. (3.24 m2)

PAD FOOTING ANALYSIS AND DESIGN (BS8110-1:1997) TEDDS calculation version 2.0.03.00

Pad footing details Length of pad footing; L = 1800 mm Width of pad footing; B = 1800 mm Area of pad footing; A = L B = 3.240 m2 Depth of pad footing; h = 350 mm Depth of soil over pad footing; hsoil = 200 mm Density of concrete; conc = 24.0 kN/m3

Column details Column base length; lA = 250 mm Column base width; bA = 400 mm Column eccentricity in x; ePxA = 0 mm Column eccentricity in y; ePyA = 0 mm

Soil details Density of soil; soil = 20.0 kN/m3 Design shear strength; = 25.0 deg

1800

700

700


[M.T.L] Page 41

Design base friction; = 19.3 deg Allowable bearing pressure; Pbearing = 250 kN/m2

Axial loading on column Dead axial load on column; PGA = 495.0 kN Imposed axial load on column; PQA = 110.3 kN Wind axial load on column; PWA = 0.0 kN Total axial load on column; PA = 605.3 kN

Foundation loads Dead surcharge load; FGsur = 0.000 kN/m2 Imposed surcharge load; FQsur = 0.000 kN/m2 Pad footing self weight; Fswt = h conc = 8.400 kN/m2 Soil self weight; Fsoil = hsoil soil = 4.000 kN/m2 Total foundation load; F = A (FGsur + FQsur + Fswt + Fsoil) = 40.2 kN Calculate pad base reaction Total base reaction; T = F + PA = 645.5 kN Eccentricity of base reaction in x; eTx = (PA ePxA + MxA + HxA h) / T = 0 mm Eccentricity of base reaction in y; eTy = (PA ePyA + MyA + HyA h) / T = 0 mm Check pad base reaction eccentricity abs(eTx) / L + abs(eTy) / B = 0.000

Base reaction acts within middle third of base

Calculate pad base pressures q1 = T / A - 6 T eTx / (L A) - 6 T eTy / (B A) = 199.233 kN/m2 q2 = T / A - 6 T eTx / (L A) + 6 T eTy / (B A) = 199.233 kN/m2 q3 = T / A + 6 T eTx / (L A) - 6 T eTy / (B A) = 199.233 kN/m2 q4 = T / A + 6 T eTx / (L A) + 6 T eTy / (B A) = 199.233 kN/m2 Minimum base pressure; qmin = min(q1, q2, q3, q4) = 199.233 kN/m2 Maximum base pressure; qmax = max(q1, q2, q3, q4) = 199.233 kN/m2

PASS - Maximum base pressure is less than allowable bearing pressure


[M.T.L] Page 42

Partial safety factors for loads Partial safety factor for dead loads; fG = 1.40 Partial safety factor for imposed loads; fQ = 1.60 Partial safety factor for wind loads; fW = 0.00

Ultimate axial loading on column Ultimate axial load on column; PuA = PGA fG + PQA fQ + PWA fW = 869.5 kN

Ultimate foundation loads Ultimate foundation load; Fu = A [(FGsur + Fswt + Fsoil) fG + FQsur fQ] = 56.2 kN Ultimate horizontal loading on column Ultimate horizontal load in x direction; HxuA = HGxA fG + HQxA fQ + HWxA fW = 0.0 kN Ultimate horizontal load in y direction; HyuA = HGyA fG + HQyA fQ + HWyA fW = 0.0 kN

Ultimate moment on column Ultimate moment on column in x direction; MxuA = MGxA fG + MQxA fQ + MWxA fW = 0.000 kNm Ultimate moment on column in y direction; MyuA = MGyA fG + MQyA fQ + MWyA fW = 0.000 kNm

Calculate ultimate pad base reaction Ultimate base reaction; Tu = Fu + PuA = 925.8 kN Eccentricity of ultimate base reaction in x; eTxu = (PuA ePxA + MxuA + HxuA h) / Tu = 0 mm Eccentricity of ultimate base reaction in y; eTyu = (PuA ePyA + MyuA + HyuA h) / Tu = 0 mm Calculate ultimate pad base pressures q1u = Tu/A - 6TueTxu/(LA) - 6TueTyu/(BA) = 285.738 kN/m2 q2u = Tu/A - 6TueTxu/(LA) + 6Tu eTyu/(BA) = 285.738 kN/m2 q3u = Tu/A + 6TueTxu/(LA) - 6TueTyu/(BA) = 285.738 kN/m2 q4u = Tu/A + 6TueTxu/(LA) + 6TueTyu/(BA) = 285.738 kN/m2 Minimum ultimate base pressure; qminu = min(q1u, q2u, q3u, q4u) = 285.738 kN/m2 Maximum ultimate base pressure; qmaxu = max(q1u, q2u, q3u, q4u) = 285.738 kN/m2 Calculate rate of change of base pressure in x direction Left hand base reaction; fuL = (q1u + q2u) B / 2 = 514.328 kN/m Right hand base reaction; fuR = (q3u + q4u) B / 2 = 514.328 kN/m Length of base reaction; Lx = L = 1800 mm Rate of change of base pressure; Cx = (fuR - fuL) / Lx = 0.000 kN/m/m Calculate pad lengths in x direction Left hand length; LL = L / 2 + ePxA = 900 mm Right hand length; LR = L / 2 - ePxA = 900 mm

Calculate ultimate moments in x direction Ultimate moment in x direction; Mx = fuLLL2/2+CxLL3/6-FuLL2/(2L) = 195.647 kNm Calculate rate of change of base pressure in y direction Top edge base reaction; fuT = (q2u + q4u) L / 2 = 514.328 kN/m Bottom edge base reaction; fuB = (q1u + q3u) L / 2 = 514.328 kN/m Length of base reaction; Ly = B = 1800 mm


[M.T.L] Page 43

Rate of change of base pressure; Cy = (fuB - fuT) / Ly = 0.000 kN/m/m Calculate pad lengths in y direction Top length; LT = B / 2 - ePyA = 900 mm Bottom length; LB = B / 2 + ePyA = 900 mm

Calculate ultimate moments in y direction Ultimate moment in y direction; My = fuTLT2/2+CyLT3/6-FuLT2/(2B) = 195.647 kNm Material details Characteristic strength of concrete; fcu = 25 N/mm2 Characteristic strength of reinforcement; fy = 500 N/mm2 Characteristic strength of shear reinforcement; fyv = 250 N/mm2 Nominal cover to reinforcement; cnom = 30 mm

Moment design in x direction Diameter of tension reinforcement; xB = 16 mm Depth of tension reinforcement; dx = h - cnom - xB / 2 = 312 mm Design formula for rectangular beams (cl 3.4.4.4) Kx = Mx / (B dx2 fcu) = 0.045 Kx = 0.156

Kx < Kx' compression reinforcement is not required Lever arm; zx = dx min([0.5 + (0.25 - Kx / 0.9)], 0.95) = 296 mm Area of tension reinforcement required; As_x_req = Mx / (0.87 fy zx) = 1521 mm2 Minimum area of tension reinforcement; As_x_min = 0.0013 B h = 819 mm2 Tension reinforcement provided; 10 No. 16 dia. bars bottom (200 centres) Area of tension reinforcement provided; As_xB_prov = NxB pi xB2 / 4 = 2011 mm2


Moment design in y direction Diameter of tension reinforcement; yB = 16 mm Depth of tension reinforcement; dy = h - cnom - xB - yB / 2 = 296 mm Design formula for rectangular beams (cl 3.4.4.4) Ky = My / (L dy2 fcu) = 0.050 Ky = 0.156

Ky < Ky' compression reinforcement is not required Lever arm; zy = dy min([0.5 + (0.25 - Ky / 0.9)], 0.95) = 279 mm Area of tension reinforcement required; As_y_req = My / (0.87 fy zy) = 1614 mm2 Minimum area of tension reinforcement; As_y_min = 0.0013 L h = 819 mm2 Tension reinforcement provided; 10 No. 16 dia. bars bottom (200 centres) Area of tension reinforcement provided; As_yB_prov = NyB pi yB2 / 4 = 2011 mm2


Calculate ultimate shear force at d from right face of column Ultimate pressure for shear; qsu = (q1u + Cx (L / 2 + ePxA + lA / 2 + dx) / B + q4u) / 2 qsu = 285.738 kN/m2 Area loaded for shear; As = B min(3 (L / 2 - eTx), L / 2 - ePxA - lA / 2 - dx) = 0.833 m2

Ultimate shear force; Vsu = As (qsu - Fu / A) = 223.666 kN Shear stresses at d from right face of column (cl 3.5.5.2) Design shear stress; vsu = Vsu / (B dx) = 0.398 N/mm2


[M.T.L] Page 44

From BS 8110:Part 1:1997 - Table 3.8 Design concrete shear stress; vc = 0.478 N/mm2 Allowable design shear stress; vmax = min(0.8N/mm2 (fcu / 1 N/mm2), 5 N/mm2) = 4.000 N/mm2

PASS - vsu < vc - No shear reinforcement required

Calculate ultimate punching shear force at face of column Ultimate pressure for punching shear; qpuA = q1u+[(L/2+ePxA-lA/2)+(lA)/2]Cx/B-[(B/2+ePyA-bA/2)+(bA)/2]Cy/L qpuA = 285.738 kN/m2 Average effective depth of reinforcement; d = (dx + dy) / 2 = 304 mm Area loaded for punching shear at column; ApA = (lA)(bA) = 0.100 m2 Length of punching shear perimeter; upA = 2(lA)+2(bA) = 1300 mm Ultimate shear force at shear perimeter; VpuA = PuA + (Fu / A - qpuA) ApA = 842.706 kN Effective shear force at shear perimeter; VpuAeff = VpuA = 842.706 kN

Punching shear stresses at face of column (cl 3.7.7.2) Design shear stress; vpuA = VpuAeff / (upA d) = 2.132 N/mm2 Allowable design shear stress; vmax = min(0.8N/mm2 (fcu / 1 N/mm2), 5 N/mm2) = 4.000 N/mm2

PASS - Design shear stress is less than allowable design shear stress

Calculate ultimate punching shear force at perimeter of 1.5 d from face of column Ultimate pressure for punching shear; qpuA1.5d = q1u+[(L/2+ePxA-lA/2-1.5d)+(lA+21.5d)/2]Cx/B-[B/2]Cy/L qpuA1.5d = 285.738 kN/m2 Average effective depth of reinforcement; d = (dx + dy) / 2 = 304 mm Area loaded for punching shear at column; ApA1.5d = (lA+21.5d)B = 2.092 m2 Length of punching shear perimeter; upA1.5d = 2B = 3600 mm Ultimate shear force at shear perimeter; VpuA1.5d = PuA + (Fu / A - qpuA1.5d) ApA1.5d = 308.205 kN Effective shear force at shear perimeter; VpuA1.5deff = VpuA1.5d 1.25 = 385.256 kN

Punching shear stresses at perimeter of 1.5 d from face of column (cl 3.7.7.2) Design shear stress; vpuA1.5d = VpuA1.5deff / (upA1.5d d) = 0.352 N/mm2 From BS 8110:Part 1:1997 - Table 3.8 Design concrete shear stress; vc = 0.485 N/mm2 Allowable design shear stress; vmax = min(0.8N/mm2 (fcu / 1 N/mm2), 5 N/mm2) = 4.000 N/mm2

PASS - vpuA1.5d < vc - No shear reinforcement required


[M.T.L] Page 45

REQUIRED - Square footing of 1.8m X 1.8m, depth 350mm

PROVIDED - Square footing of 2.3m X 2.3m, depth 500mm

Shear at d from column face

Punching shear perimeter at 1.5 d from column face

10 No. 16 dia. bars btm (200 c/c)


Shear at d from column face

Punching shear perimeter at 1.5 d from column face




[M.T.L] Page 46

6.0 ANALYSIS AND DESIGN OF STAIRCASE

Design done here is for the staircase type 1.

The staircase is longitudinally spanned (spans in the direction of the flight) and is supported by beams

that are also supported by columns.

6.1 Staircase type 1.

Design information

QK= 3kN/m2

Risers= 150mm

Goings = 250mm

50mm granolithic finish on treads and landings

50mm plaster finish on underside of stair flight and landing

fCK= 20 N/mm

175mm waist thickness

12mm bars with 30 mm cover

Unit weight of finishes and concrete = 24kN/m3

RC STAIR DESIGN (BS8110-1:1997) TEDDS calculation version 1.0.03

Stair geometry Number of steps; Nsteps = 10 Waist depth for stair flight; hspan = 175 mm Going of each step; Going = 250 mm


[M.T.L] Page 47

Rise of each step; Rise = 150 mm Angle of stairs; Rake = atan(Rise / Going) = 30.96 deg Upper landing geometry Support condition; Continuous interior support Landing length; Lupper = 250 mm Depth of landing; hupper = 250 mm

Lower landing geometry Support condition; Continuous first interior support Landing length; Llower = 1215 mm Depth of landing; hlower = 175 mm

Material details Characteristic strength of concrete; fcu = 20 N/mm2 Characteristic strength of reinforcement; fy = 500 N/mm2 Nominal cover to reinforcement; cnom = 30 mm Density of concrete; conc = 24.0 kN/m3

Partial safety factors Partial safety factor for imposed loading; fq = 1.60 Partial safety factor for dead loading; fg = 1.40

Loading details Characteristic imposed loading; qk = 3.000 kN/m2 Characteristic loading from finishes; gk_fin = 1.200 kN/m2 Average stair self weight; gk_swt = (hspan / Cos(Rake) + Rise / 2) conc = 6.698 kN/m2 Design load; F = (gk_swt + gk_fin) fg + qk fq = 15.857 kN/m2 Mid span design Midspan moment per metre width; Mspan = 0.063 F L2 = 13.787 kNm/m Diameter of tension reinforcement; span = 12 mm Depth of reinforcement; dspan = hspan - cnom - span / 2 = 139 mm Design formula for rectangular beams (cl 3.4.4.4) Moment redistribution ratio; b = 1.25 Kspan = Mspan / (dspan2 fcu) = 0.036 Kspan = 0.402 (b - 0.4) - 0.18 (b - 0.4)2 = 0.212

Kspan < K'span compression reinforcement is not required Lever arm; zspan = dspan min([0.5 + (0.25 - Kspan / 0.9)], 0.95) = 132

mm

Area of tension reinforcement required; As_span_req = Mspan / (0.87 fy zspan) = 240 mm2/m Minimum area of tension reinforcement; As_span_min = 0.13 hspan / 100 = 228 mm2/m Tension reinforcement provided; 12 dia.bars @ 200 centres Area of tension reinforcement provided; As_span_prov = 565 mm2/m


Basic span/effective depth ratio (cl 3.4.6.3) From BS8110 : Part 1 : 1997 Table 3.9 Basic span/effective depth ratio; ratiobasic = 26.0


[M.T.L] Page 48

Modification of span/effective depth ratio for staircases without stringer beams (cl 3.10.2.2) Modification factor for stairs without stringers; factorflight = 1.15

Modification of span/effective depth ratio for tension reinforcement (cl 3.4.6.5) From BS8110 : Part 1 : 1997 Table 3.10 Design service stress; fs = 2 fy As_span_req / (3 As_span_prov b) = 113.189

N/mm2 Modification factor for tension reinforcement; factortens = 0.55 + (477 N/mm2- fs)/(120 (0.9 N/mm2+ (Mspan

/ dspan2))) factortens = 2.429

Check span/effective depth ratio (cl 3.4.6.1) Allowable span/effective depth ratio; ratioadm = ratiobasic factorflight factortens = 72.623 Actual span/effective depth ratio; ratioact = L / dspan = 26.727

PASS - Span/effective depth ratio is adequate

Upper landing support design Upper support moment per metre width; Mupper = 0.063 F L2 = 13.787 kNm/m Diameter of tension reinforcement; upper = 12 mm Depth of reinforcement; dupper = hupper - cnom - upper / 2 = 214 mm Design formula for rectangular beams (cl 3.4.4.4) Moment redistribution ratio; b = 0.80 Kupper = Mupper / (dupper 2 fcu) = 0.015 Kupper = 0.402 (b - 0.4) - 0.18 (b - 0.4)2 = 0.132

Kupper < K'upper compression reinforcement is not required Lever arm; zupper = dupper min([0.5 + (0.25 - Kupper / 0.9)], 0.95) = 203

mm

Area of tension reinforcement required; As_upper_req = Mupper / (0.87 fy zupper) = 156 mm2/m Minimum area of tension reinforcement; As_upper_min = 0.13 hupper / 100 = 325 mm2/m Tension reinforcement provided; 12 dia.bars @ 200 centres Area of tension reinforcement provided; As_upper_prov = 565 mm2/m


Shear stress in beam (cl 3.4.5.2) Design shear force; Vupper = 0.500 F L = 29.455 kN/m Design shear stress; vupper = Vupper / dupper = 0.138 N/mm2 Allowable design shear stress; vmax = min(0.8N/mm2 (fcu / 1 N/mm2), 5 N/mm2) = 3.578

N/mm2 PASS - Design shear stress does not exceed allowable shear stress

From BS 8110:Part 1:1997 - Table 3.8 Design concrete shear stress; vc_upper = 0.440 N/mm2

PASS - Design shear stress does not exceed design concrete shear stress

Lower landing support design Lower support moment per metre width; Mlower = 0.086 F L2 = 18.821 kNm/m Diameter of tension reinforcement; lower = 12 mm Depth of reinforcement; dlower = hlower - cnom - getvar(lower, 12 mm) / 2 = 139 mm


[M.T.L] Page 49

Design formula for rectangular beams (cl 3.4.4.4) Moment redistribution ratio; b = 0.80 Klower = Mlower / (dlower 2 fcu) = 0.049 Klower = 0.402 (b - 0.4) - 0.18 (b - 0.4)2 = 0.132

Klower < K'lower compression reinforcement is not required Lever arm; zlower = dlower min([0.5 + (0.25 - Klower / 0.9)], 0.95) = 131

mm

Area of tension reinforcement required; As_lower_req = Mlower / (0.87 fy zlower) = 330 mm2/m Minimum area of tension reinforcement; As_lower_min = 0.13 hlower / 100 = 228 mm2/m Tension reinforcement provided; 12 dia.bars @ 200 centres Area of tension reinforcement provided; As_lower_prov = 565 mm2/m


Shear stress in beam (cl 3.4.5.2) Design shear force; Vlower = 0.600 F L = 35.346 kN/m Design shear stress; vlower = Vlower / dlower = 0.254 N/mm2 Allowable design shear stress; vmax = min(0.8N/mm2 (fcu / 1 N/mm2), 5 N/mm2) = 3.578

N/mm2 PASS - Design shear stress does not exceed allowable shear stress

From BS 8110:Part 1:1997 - Table 3.8 Design concrete shear stress; vc_lower = 0.566 N/mm2

PASS - Design shear stress does not exceed design concrete shear stress

Distribution steel;

Greater of 20%Main steel or nominal steel;

20%Main steel =20%Asprovided

=20%565=113mm2

OR

0.13%Ac =0.13%1000X175=227.5mm2

8 dia.bars @ 200 centres AS PROVIDED


[M.T.L] Page 50

8.0 ANALYSIS AND DESIGN OF RETAINING WALL.

Briefing;

Retaining wall provided for the need to retain earth. Loading on wall is due to self weight, wall on the floor above and the retained earth.


Total dead weight, Gk = 20.2 Kn/m

RETAINING WALL ANALYSIS (BS 8002:1994) TEDDS calculation version 1.2.01.02

Wall details Retaining wall type; Unpropped cantilever Height of retaining wall stem; hstem = 3000 mm Thickness of wall stem; twall = 300 mm Length of toe; ltoe = 900 mm Length of heel; lheel = 1100 mm Overall length of base; lbase = ltoe + lheel + twall = 2300 mm Thickness of base; tbase = 350 mm Depth of downstand; dds = 0 mm Position of downstand; lds = 1550 mm Thickness of downstand; tds = 350 mm

Self weight [0.2x24x1.0] = 7.2 Kn/m From walling [5kN/m2 (3-0.5)] = 13 Kn/m


[M.T.L] Page 51

Height of retaining wall; hwall = hstem + tbase + dds = 3350 mm Depth of cover in front of wall; dcover = 0 mm Depth of unplanned excavation; dexc = 0 mm Height of ground water behind wall; hwater = 0 mm Height of saturated fill above base; hsat = max(hwater - tbase - dds, 0 mm) = 0 mm Density of wall construction; wall = 24.0 kN/m3 Density of base construction; base = 24.0 kN/m3 Angle of rear face of wall; = 90.0 deg Angle of soil surface behind wall; = 0.0 deg Effective height at virtual back of wall; heff = hwall + lheel tan() = 3350 mm Retained material details Mobilisation factor; M = 1.5 Moist density of retained material; m = 18.0 kN/m3 Saturated density of retained material; s = 21.0 kN/m3 Design shear strength; ' = 24.2 deg Angle of wall friction; = 18.6 deg

Base material details Moist density; mb = 18.0 kN/m3 Design shear strength; 'b = 24.2 deg Design base friction; b = 18.6 deg Allowable bearing pressure; Pbearing = 100 kN/m2

Using Coulomb theory Active pressure coefficient for retained material

Ka = sin( + ')2 / (sin()2 sin( - ) [1 + (sin(' + ) sin(' - ) / (sin( - ) sin( + )))]2) = 0.369 Passive pressure coefficient for base material

Kp = sin(90 - 'b)2 / (sin(90 - b) [1 - (sin('b + b) sin('b) / (sin(90 + b)))]2) = 4.187 At-rest pressure At-rest pressure for retained material; K0 = 1 sin() = 0.590 Loading details Surcharge load on plan; Surcharge = 10.0 kN/m2 Applied vertical dead load on wall; Wdead = 20.2 kN/m Applied vertical live load on wall; Wlive = 0.0 kN/m Position of applied vertical load on wall; lload = 1050 mm Applied horizontal dead load on wall; Fdead = 0.0 kN/m Applied horizontal live load on wall; Flive = 0.0 kN/m Height of applied horizontal load on wall; hload = 0 mm


[M.T.L] Page 52

Loads shown in kN/m, pressures shown in kN/m2

Vertical forces on wall Wall stem; wwall = hstem twall wall = 21.6 kN/m Wall base; wbase = lbase tbase base = 19.3 kN/m Surcharge; wsur = Surcharge lheel = 11 kN/m Moist backfill to top of wall; wm_w = lheel (hstem - hsat) m = 59.4 kN/m Applied vertical load; Wv = Wdead + Wlive = 20.2 kN/m Total vertical load; Wtotal = wwall + wbase + wsur + wm_w + Wv = 131.5 kN/m

Horizontal forces on wall Surcharge; Fsur = Ka cos(90 - + ) Surcharge heff = 11.7 kN/m Moist backfill above water table; Fm_a = 0.5 Ka cos(90 - + ) m (heff - hwater)2 = 35.3 kN/m Total horizontal load; Ftotal = Fsur + Fm_a = 47.1 kN/m

Calculate stability against sliding Passive resistance of soil in front of wall; Fp = 0.5 Kp cos(b) (dcover + tbase + dds - dexc)2 mb = 4.4 kN/m Resistance to sliding; Fres = Fp + Wtotal tan(b) = 48.6 kN/m

PASS - Resistance force is greater than sliding force

Overturning moments Surcharge; Msur = Fsur (heff - 2 dds) / 2 = 19.6 kNm/m Moist backfill above water table; Mm_a = Fm_a (heff + 2 hwater - 3 dds) / 3 = 39.5 kNm/m Total overturning moment; Mot = Msur + Mm_a = 59.1 kNm/m

Restoring moments Wall stem; Mwall = wwall (ltoe + twall / 2) = 22.7 kNm/m Wall base; Mbase = wbase lbase / 2 = 22.2 kNm/m Surcharge; Msur_r = wsur (lbase - lheel / 2) = 19.2 kNm/m


[M.T.L] Page 53

Moist backfill; Mm_r = (wm_w (lbase - lheel / 2) + wm_s (lbase - lheel / 3)) = 104 kNm/m Design vertical load; Mv = Wv lload = 21.2 kNm/m Total restoring moment; Mrest = Mwall + Mbase + Msur_r + Mm_r + Mv = 189.3 kNm/m

Check stability against overturning Total overturning moment; Mot = 59.1 kNm/m Total restoring moment; Mrest = 189.3 kNm/m

PASS - Restoring moment is greater than overturning moment

Check bearing pressure Total moment for bearing; Mtotal = Mrest - Mot = 130.2 kNm/m Total vertical reaction; R = Wtotal = 131.5 kN/m Distance to reaction; xbar = Mtotal / R = 990 mm Eccentricity of reaction; e = abs((lbase / 2) - xbar) = 160 mm

Reaction acts within middle third of base Bearing pressure at toe; ptoe = (R / lbase) + (6 R e / lbase2) = 81 kN/m2 Bearing pressure at heel; pheel = (R / lbase) - (6 R e / lbase2) = 33.3 kN/m2

PASS - Maximum bearing pressure is less than allowable bearing pressure


[M.T.L] Page 54

RETAINING WALL DESIGN (BS 8002:1994) TEDDS calculation version 1.2.01.02

Ultimate limit state load factors Dead load factor; f_d = 1.4 Live load factor; f_l = 1.6 Earth and water pressure factor; f_e = 1.4

Factored vertical forces on wall Wall stem; wwall_f = f_d hstem twall wall = 30.2 kN/m Wall base; wbase_f = f_d lbase tbase base = 27 kN/m Surcharge; wsur_f = f_l Surcharge lheel = 17.6 kN/m Moist backfill to top of wall; wm_w_f = f_d lheel (hstem - hsat) m = 83.2 kN/m Applied vertical load; Wv_f = f_d Wdead + f_l Wlive = 28.3 kN/m Total vertical load; Wtotal_f = wwall_f + wbase_f + wsur_f + wm_w_f + Wv_f = 186.3 kN/m

Factored horizontal at-rest forces on wall Surcharge; Fsur_f = f_l K0 Surcharge heff = 31.6 kN/m Moist backfill above water table; Fm_a_f = f_e 0.5 K0 m (heff - hwater)2 = 83.4 kN/m Total horizontal load; Ftotal_f = Fsur_f + Fm_a_f = 115.1 kN/m Passive resistance of soil in front of wall; Fp_f = f_e 0.5 Kp cos(b) (dcover + tbase + dds - dexc)2 mb = 6.1 kN/m

Factored overturning moments Surcharge; Msur_f = Fsur_f (heff - 2 dds) / 2 = 53 kNm/m Moist backfill above water table; Mm_a_f = Fm_a_f (heff + 2 hwater - 3 dds) / 3 = 93.2 kNm/m Total overturning moment; Mot_f = Msur_f + Mm_a_f = 146.2 kNm/m

Restoring moments Wall stem; Mwall_f = wwall_f (ltoe + twall / 2) = 31.8 kNm/m Wall base; Mbase_f = wbase_f lbase / 2 = 31.1 kNm/m Surcharge; Msur_r_f = wsur_f (lbase - lheel / 2) = 30.8 kNm/m Moist backfill; Mm_r_f = (wm_w_f (lbase - lheel / 2) + wm_s_f (lbase - lheel / 3)) = 145.5 kNm/m Design vertical load; Mv_f = Wv_f lload = 29.7 kNm/m Total restoring moment; Mrest_f = Mwall_f + Mbase_f + Msur_r_f + Mm_r_f + Mv_f = 268.9 kNm/m

Check stability against overturning Total overturning moment; Mot = 59.1 kNm/m Total restoring moment; Mrest = 189.3 kNm/m

PASS - Restoring moment is greater than overturning moment

Factored bearing pressure Total moment for bearing; Mtotal_f = Mrest_f - Mot_f = 122.7 kNm/m Total vertical reaction; Rf = Wtotal_f = 186.3 kN/m Distance to reaction; xbar_f = Mtotal_f / Rf = 659 mm Eccentricity of reaction; ef = abs((lbase / 2) - xbar_f) = 491 mm

Reaction acts outside middle third of base Bearing pressure at toe; ptoe_f = Rf / (1.5 xbar_f) = 188.6 kN/m2 Bearing pressure at heel; pheel_f = 0 kN/m2 = 0 kN/m2 Rate of change of base reaction; rate = ptoe_f / (3 xbar_f) = 95.44 kN/m2/m


[M.T.L] Page 55

Bearing pressure at stem / toe; pstem_toe_f = max(ptoe_f - (rate ltoe), 0 kN/m2) = 102.7 kN/m2 Bearing pressure at mid stem; pstem_mid_f = max(ptoe_f - (rate (ltoe + twall / 2)), 0 kN/m2) = 88.4 kN/m2 Bearing pressure at stem / heel; pstem_heel_f = max(ptoe_f - (rate (ltoe + twall)), 0 kN/m2) = 74.1 kN/m2

Design of reinforced concrete retaining wall toe (BS 8002:1994) Material properties Characteristic strength of concrete; fcu = 25 N/mm2 Characteristic strength of reinforcement; fy = 450 N/mm2

Base details Minimum area of reinforcement; k = 0.13 % Cover to reinforcement in toe; ctoe = 30 mm

Calculate shear for toe design Shear from bearing pressure; Vtoe_bear = (ptoe_f + pstem_toe_f) ltoe / 2 = 131.1 kN/m Shear from weight of base; Vtoe_wt_base = f_d base ltoe tbase = 10.6 kN/m Total shear for toe design; Vtoe = Vtoe_bear - Vtoe_wt_base = 120.5 kN/m

Calculate moment for toe design Moment from bearing pressure; Mtoe_bear = (2 ptoe_f + pstem_mid_f) (ltoe + twall / 2)2 / 6 = 85.5 kNm/m Moment from weight of base; Mtoe_wt_base = (f_d base tbase (ltoe + twall / 2)2 / 2) = 6.5 kNm/m Total moment for toe design; Mtoe = Mtoe_bear - Mtoe_wt_base = 79.1 kNm/m

Check toe in bending Width of toe; b = 1000 mm/m Depth of reinforcement; dtoe = tbase ctoe (toe / 2) = 314.0 mm Constant; Ktoe = Mtoe / (b dtoe2 fcu) = 0.032

Compression reinforcement is not required Lever arm; ztoe = min(0.5 + (0.25 - (min(Ktoe, 0.225) / 0.9)),0.95) dtoe ztoe = 298 mm Area of tension reinforcement required; As_toe_des = Mtoe / (0.87 fy ztoe) = 677 mm2/m Minimum area of tension reinforcement; As_toe_min = k b tbase = 455 mm2/m Area of tension reinforcement required; As_toe_req = Max(As_toe_des, As_toe_min) = 677 mm2/m Reinforcement provided; 12 mm dia.bars @ 100 mm centres Area of reinforcement provided; As_toe_prov = 1131 mm2/m

PASS - Reinforcement provided at the retaining wall toe is adequate


[M.T.L] Page 56

Check shear resistance at toe Design shear stress; vtoe = Vtoe / (b dtoe) = 0.384 N/mm2 Allowable shear stress; vadm = min(0.8 (fcu / 1 N/mm2), 5) 1 N/mm2 = 4.000 N/mm2

PASS - Design shear stress is less than maximum shear stress From BS8110:Part 1:1997 Table 3.8 Design concrete shear stress; vc_toe = 0.478 N/mm2

vtoe < vc_toe - No shear reinforcement required

Design of reinforced concrete retaining wall heel (BS 8002:1994) Material properties Characteristic strength of concrete; fcu = 25 N/mm2 Characteristic strength of reinforcement; fy = 450 N/mm2

Base details Minimum area of reinforcement; k = 0.13 % Cover to reinforcement in heel; cheel = 30 mm

Calculate shear for heel design Shear from bearing pressure; Vheel_bear = pstem_heel_f ((3 xbar_f) - ltoe - twall) / 2 = 28.7 kN/m Shear from weight of base; Vheel_wt_base = f_d base lheel tbase = 12.9 kN/m Shear from weight of moist backfill; Vheel_wt_m = wm_w_f = 83.2 kN/m Shear from surcharge; Vheel_sur = wsur_f = 17.6 kN/m Total shear for heel design; Vheel = - Vheel_bear + Vheel_wt_base + Vheel_wt_m + Vheel_sur = 85 kN/m

Calculate moment for heel design Moment from bearing pressure; Mheel_bear = pstem_mid_f ((3 xbar_f) - ltoe - twall / 2)2 / 6 = 12.6 kNm/m Moment from weight of base; Mheel_wt_base = (f_d base tbase (lheel + twall / 2)2 / 2) = 9.2 kNm/m Moment from weight of moist backfill; Mheel_wt_m = wm_w_f (lheel + twall) / 2 = 58.2 kNm/m Moment from surcharge; Mheel_sur = wsur_f (lheel + twall) / 2 = 12.3 kNm/m Total moment for heel design; Mheel = - Mheel_bear + Mheel_wt_base + Mheel_wt_m + Mheel_sur = 67.1 kNm/m

Check heel in bending Width of heel; b = 1000 mm/m Depth of reinforcement; dheel = tbase cheel (heel / 2) = 314.0 mm Constant; Kheel = Mheel / (b dheel2 fcu) = 0.027

Compression reinforcement is not required


[M.T.L] Page 57

Lever arm; zheel = min(0.5 + (0.25 - (min(Kheel, 0.225) / 0.9)),0.95) dheel zheel = 298 mm Area of tension reinforcement required; As_heel_des = Mheel / (0.87 fy zheel) = 574 mm2/m Minimum area of tension reinforcement; As_heel_min = k b tbase = 455 mm2/m Area of tension reinforcement required; As_heel_req = Max(As_heel_des, As_heel_min) = 574 mm2/m Reinforcement provided; 12 mm dia.bars @ 100 mm centres Area of reinforcement provided; As_heel_prov = 1131 mm2/m

PASS - Reinforcement provided at the retaining wall heel is adequate

Check shear resistance at heel Design shear stress; vheel = Vheel / (b dheel) = 0.271 N/mm2 Allowable shear stress; vadm = min(0.8 (fcu / 1 N/mm2), 5) 1 N/mm2 = 4.000 N/mm2

PASS - Design shear stress is less than maximum shear stress

From BS8110:Part 1:1997 Table 3.8 Design concrete shear stress; vc_heel = 0.478 N/mm2

vheel < vc_heel - No shear reinforcement required

Design of reinforced concrete retaining wall stem (BS 8002:1994) Material properties Characteristic strength of concrete; fcu = 25 N/mm2 Characteristic strength of reinforcement; fy = 450 N/mm2

Wall details Minimum area of reinforcement; k = 0.13 % Cover to reinforcement in stem; cstem = 30 mm Cover to reinforcement in wall; cwall = 30 mm

Factored horizontal at-rest forces on stem Surcharge; Fs_sur_f = f_l K0 Surcharge (heff - tbase - dds) = 28.3 kN/m Moist backfill above water table; Fs_m_a_f = 0.5 f_e K0 m (heff - tbase - dds - hsat)2 = 66.9 kN/m

Calculate shear for stem design Shear at base of stem; Vstem = Fs_sur_f + Fs_m_a_f = 95.2 kN/m

Calculate moment for stem design Surcharge; Ms_sur = Fs_sur_f (hstem + tbase) / 2 = 47.4 kNm/m Moist backfill above water table; Ms_m_a = Fs_m_a_f (2 hsat + heff - dds + tbase / 2) / 3 = 78.6 kNm/m Total moment for stem design; Mstem = Ms_sur + Ms_m_a = 126.1 kNm/m

Check wall stem in bending Width of wall stem; b = 1000 mm/m


[M.T.L] Page 58

Depth of reinforcement; dstem = twall cstem (stem / 2) = 260.0 mm Constant; Kstem = Mstem / (b dstem2 fcu) = 0.075

Compression reinforcement is not required Lever arm; zstem = min(0.5 + (0.25 - (min(Kstem, 0.225) / 0.9)),0.95) dstem zstem = 236 mm Area of tension reinforcement required; As_stem_des = Mstem / (0.87 fy zstem) = 1363 mm2/m Minimum area of tension reinforcement; As_stem_min = k b twall = 390 mm2/m Area of tension reinforcement required; As_stem_req = Max(As_stem_des, As_stem_min) = 1363 mm2/m Reinforcement provided; 20 mm dia.bars @ 200 mm centres Area of reinforcement provided; As_stem_prov = 1571 mm2/m

PASS - Reinforcement provided at the retaining wall stem is adequate

Check shear resistance at wall stem Design shear stress; vstem = Vstem / (b dstem) = 0.366 N/mm2 Allowable shear stress; vadm = min(0.8 (fcu / 1 N/mm2), 5) 1 N/mm2 = 4.000 N/mm2

PASS - Design shear stress is less than maximum shear stress From BS8110:Part 1:1997 Table 3.8 Design concrete shear stress; vc_stem = 0.595 N/mm2

vstem < vc_stem - No shear reinforcement required


[M.T.L] Page 59

Indicative retaining wall reinforcement diagram

Toe bars - 12 mm dia.@ 100 mm centres - (1131 mm2/m) Heel bars - 12 mm dia.@ 100 mm centres - (1131 mm2/m) Stem bars - 20 mm dia.@ 200 mm centres - (1571 mm2/m) - AS PROVIDED.


[M.T.L] Page 60

8.0 APPENDIX

8.1 RING BEAM 01

Gk= 9.63Kn/m, Qk = 1.64 Kn/m

Results;

1 2 3 4 5

3.000

SPAN 1

Dead+

Own W

Live

9.63

1.46

5.000

SPAN 2

5.000

SPAN 3

5.000

SPAN 4

230

350

Long-term Deflections (All spans loaded - using entered reinforcement) Def max = 8.885mm @ 15.8m

9.008.007.006.005.004.003.002.001.00

.50

0

1.00

1.50

2.00

2.50

3.00

3.50

4.00

4.50

5.00

5.50

6.00

6.50

7.00

7.50

8.00

8.50

9.00

9.50

10.0

10.5

11.0

11.5

12.0

12.5

13.0

13.5

14.0

14.5

15.0

15.5

16.0

16.5

17.0

17.5

18.0

SHEAR: X-X V max = 46.63kN @ 13.0m

-40.0

-30.0

-20.0

-10.0

10.0

20.0

30.0

40.0

50.0

.50

0

1.00

1.50

2.00

2.50

3.00

3.50

4.00

4.50

5.00

5.50

6.00

6.50

7.00

7.50

8.00

8.50

9.00

9.50

10.0

10.5

11.0

11.5

12.0

12.5

13.0

13.5

14.0

14.5

15.0

15.5

16.0

16.5

17.0

17.5

18.0


[M.T.L] Page 61

Reinforcement diagrams

COLUMN REACTIONS

Support Dead Load(kN) Live Load(kN) Dead + Live(kN) Ultimate(kN)

1 10.11 1.53 11.65 16.96

2 42.23 6.40 48.63 67.96

3 47.95 7.27 55.22 78.19

4 54.40 8.25 62.65 87.09

MOMENTS: X-X M max = -43.24kNm @ 13.0m

25.020.015.010.05.00

-5.00-10.0-15.0-20.0-25.0-30.0-35.0-40.0-45.0

.50

0

1.00

1.50

2.00

2.50

3.00

3.50

4.00

4.50

as built train up a child structural intergrity -report

Documents

design calculations

design of floor beams

design of floor slabs

design of columns

design of foundations

design of staircase

general analysis

design of maxspanribbed