waa structure dhamaystiran waxaana diyaariyay ardayda waxkabarata jaamacad camuud

1. 2. REINFORCEMENT CONCRETE DESIGN
2.1 DESIGN DATA
Live load 2.5kn/m2 at typical floor
Finishing load 1kn/m2
Water proofing load 1kn/m2
Load of the partition walls 12kn/m2
Location erigavo
Depth of foundation below the ground 0.6m
Type of soil rocky sand
Allowable bearing capacity 450kn/m2
Story height 3m
Floors G+1
Walls 200mm hallow block
Material properties
All components in the design will be used M20 grade
Reinforcement of grade Fe 250 confirming to BS: 8110 is used throughout.

2. 2.2. Slab design
Slabs used in floors and roofsof buildingsmostly integrated with supporting
beams that carry the distributed loadsby bending.
Slabs have differentshapes, and may be rectangular, squire, circular and many
differentirregular shapes dependingof architectural design, so in our design
we userectangular shape.
Slabs can be designed either oneway or two way slabs dependingon
1. If Ly/Lx<2 the slab is two way where
2. If Ly/Lx>=2 the slab is oneway
DATA OF SLAB
Characteristic strength of concrete fcu=25N/mm2
Characteristic strength of main reinforcementfy=460 N/mm2
Characteristic strength of links fyv=250 N/mm2
Size of all columns =200 x400(mm)
Cover of slab =20 mm
Finishing and service = 1.0KN/m2
Live load = 2.5KN/m2
Density of concrete = 24 KN/m3

3. Solution
(a)Loadingcalculation
Dead loads
Self weight of slab = 0.15 x24 =3.6KN/m2
Finishing = 1.0KN/m 2
Total GK= 4.6KN/m2
Imposed load
QK =2.5KN/m2
Design ultimate load (N) = (1.4 x4.6 + 1.6 x2.5)= 10.44KN/m2
𝐿𝑦
𝐿𝑥
=
4
4
=1.0 < 2 (two way slab)
Mid span (bothway):

4. d= 150-20-10-10/2=115mm
Ms= βsnlx2
= 0.024 x10.44x(4)2 = 4.0KNm/m
K= M/fcubd2 =4.0 x106/1000 x1152 x35 = 0.009<0.156
Z= d [0.5 + √ 𝑜. 25 − 𝑘/0.9]= 0.99d > 0.95dthereforeapply 0.95d
As = M/0.87fyZ= 4.0 x106/0.87x460x 0.95 x 115= 92mm2< Asmin
Asmin 0.13%bh0.13/100x1000 x150 =195mm2/m
Therefore provideT10-300 c/c(Asprov= 263mm2/m
Continuous Edge:
d= 150-20-10/2= 125mm
Ms= βsnlx2 = 0.032 x10.44x(4)2 = 5.34KNm/m
K= M/bd2fcu =5.34 x106/1000 x1252 x35 = 0.0097 < 0.156
Z= d [0.5 + √0.25 − 𝑘/0.9]= 0.99d > 0.95dthereforeapply 0.95d
Asmin 0.13%bh0.13/100x1000 x125 =195mm2/m
As = M/0.87fyZ= 5.34 x106/0.87x460x125= 107mm2<Asmin
Therefore provideT10-300 c/c(Asprov= 263mm2/m)
A) Shear check
Vs = βvnlx= 0.33 x10.44x4 = 13.8KN/m
V= V/bd = 13.8 x103/1000x125= 0.11KN/mm2

5. 100As/bd = 100 x263/1000 x125 = 0.12 < 3 (satisfactory)
400/d = 400/125=3.2> 1 satisfactory
Vc =0.79/1.25 (100As/bd)1/3(400/d)1/4(fcu/25)1/3
Vc =0.79/1.25 (0.21)1/3(3.2)¼(1.4)1/3= 0.56N/mm2> v = 0.15N/mm2
Deflectioncheck
(L/d) basic = 26
(L/d)actual =4000/115= 35
M/bd2=5.34 x106/1000x1152 =0.40
fs= 5fy Asreq /8Asprov = 5 x 460 x195 / 8 x 263 = 214
M.F= 0.55+ (477-214/120(0.9+0.40)
= 2.23> 2 thereforeuse 2.0
(L/d) allow = 26 x2 =52
(L/d) allow> (L/d) actual (Satisfactory)
2.3. Beam Design
Beamsare flexural memberswhich are used to transfer the loads from slab to
columnsmay be designed for flexuralmomentin differentfashions depending

6. on the magnitudeof the momentand the dimensions. Accordingly the beam
can be singly reinforced, doubly reinforced Tor Lsection. The proceduresand
why each is used for specific design type are donebelow.
Given data
Live load = 2.5 KN/m2
Finishing& service = 1.0 KN/m2
Thickness of slab = 150 mm
Fcu = 25 N/mm2
Fy = 460 N/mm2
Fyv = 250 N/mm2
Density of concrete = 24 KN/m3
Size if beam = 250 x450 mm
Solution
Ly/Lx= 4.2/4.2 = 1.0 < 2 (Two way slab)
(a) LoadingCalculation
Dead loads

7. Self-weight of slab = 0.15 x24 = 3.6 kN/m2
Finishing + service= = 1.0 kN/m2
Total dead loads, Gk = 4.6 kN/m2
Imposed Load
Live load, Qk= 2.5 kN/m2
Thus dead load, Gk = (4.6 x2.5)+ [0.25x(0.45-0.15)x24]
= 13.3kN/m
Live load, Qk= 2.5 x 2.5 = 6.25 kN/m
Design load, w = 1.4Gk+ 1.6Qk
= 1.4(13.3)+1.6(6.25)
= 28.62 kN/m
Load, F = W x L = 28.62 x4.2
= 120 KN

8. Design for main reinforcement
Assumeфt = 20mm, фlink = 10mm, cover = 25mm
d= h – c - фlink – ½фt
= 450 – 25 – 10 – 20/2 = 405mm
Mid span of 1-2 and 3-4

9. K= M/fcubd2 =45.4 x106/250 x4052 x25 = 0.044< 0.156(singlyreinforcement)
Z= d {0.5 + √0. 25 − 𝑘/0.9]= 0.96d > 0.95d therefore use0.95d
As = M/0.87fyZ= 45.4x106/0.87x460x0.95 x405= 295mm2>Asmin
Asmin= 0.13%bh= 0.13/100 x450 x250 =146mm2
Asmax = 4%bh = 4/100 x250 x450 = 4500mm2
Use 2T20 (629mm2)
Mid span of 2-3
K= M/fcubd2 =35.3 x106/250 x4052 x25 =0.034 < 0.156(singlyreinforcement)
Z= d {0.5 + √0. 25 − 𝑘/0.9]= 0.96d > 0.95d therefore use0.95d
As = M/0.87fyZ= 35.3x106/0.87x460x0.95 x405= 229mm2>Asmin
Asmin= 0.13%bh= 0.13/100 x450 x250 =146mm2
Asmax = 4%bh = 4/100 x250 x450 = 4500mm2
Provide2T20 (629mm2)
Support at2-3
K= M/fcubd2 =55.4 x106/250 x4052 x25 =0.053 < 0.156(singlyreinforcement)
Z= d {0.5 + √0. 25 − 0.053/0.9]= 0.94d < 0.95d (OK)
As = M/0.87fyZ= 55.4x106/0.87x460x0.95 x405= 360mm2>Asmin
Asmin= 0.13%bh= 0.13/100 x450 x250 =146mm2
Asmax = 4%bh = 4/100 x250 x450 = 4500mm2
Provide2T16 + 2T16 (As= 804mm)
Design for Shear reinforcement

10. At support 2: V= 72.0 KN
d= 450 – 25 – 10 – 16 - 2/3(20)/2 = 392mm
v = V/bd
v = 72.0 x103/250 x392 = 0.73<0.8√25= 4 kN/mm2 (Ok)
V at d = 72 - 0.392(28.62)
= 60.7 kN
v = V/bd
v = 60.7 x103/250 x392 = 0.62N/mm2
100As /bd = 100(804)/250 x392 = 0.82 < 3 (ok)
400/d = 400/392= 1.02 > 1 (ok)
vc= 0.79/1.25(0.82)1/3 (1.02)1/4 = 0.59N/mm2
0.5vc= 0.5 x0.59 = 0.29 N/mm2
vc+ 0.4 = 0.99 N/mm2
Therefore 0.5vc< v < (vc+ 0.4)
: - Require minimum shear reinforcement
: - provideR10 – 275 c/c
At support 1: V= 54.0 KN
d= 450 – 25 – 10 – 16 - (20)/2 = 405mm
v = V/bd

11. v = 54.0 x103/250 x392 = 0.53<0.8√25= 4 kN/mm2 (Ok)
V at d = 54 – 0.405(28.62)
= 52.7 kN
v = V/bd
v = 52.7 x103/250 x405 = 0.52N/mm2
100As /bd = 100(629)/250 x405 = 0.62 < 3 (ok)
400/d = 400/405= 0.99 > 1 (Use1)
vc= 0.79/1.25(0.62)1/3 (1.0)1/4 = 0.54N/mm2
0.5vc= 0.5 x0.54 = 0.27 N/mm2
vc+ 0.4 = 0.94 N/mm2
Therefore 0.5vc< v < (vc+ 0.4)
: - Require minimum shear reinforcement
: - provideR10 – 275 c/c
Deflectionchecking
Checking to be doneat the most critical span, 1-2 and 3-4
M/bd2=45.4 x106/250 x4052= 1.1
fs= 5fy Asreq /8Asprov = 5 x 460 x295 / 8 x 629 = 135

12. Modification factor M.F= 0.55+ (477-135/120(0.9+1.1)= 1.9< 2.0(satisfactory)
(L/d) basic = 26
(L/d) allow = 26 x 1.9 = 49.4
(L/d) actual = 4200/405 = 10.4< (L/d) allow
The deflection is satisfactory
2.4. Column design
Columnsare compression membersthat transfer loads of slabs and any
structureabove them to foundation.
There are two types of columnswhich are short column and slender column.
The followingcondition is determined whether the column is short or slender.

13. If Lex/D <12 and
Ley/b <12
The column is short column otherwise it is slender or long column.
And also weconsidered that all the loads are axially loaded and there is no
eccentricity.
Given data
fcu=25N/mm2
fy =460N/mm2
Cover = 25mm

14. Solution
d = 400 – 25 – 10 – 20/2 = 355
N/bd = 1320 x103/200x400 = 15.37N/mm2
M/bd2= 50 x 106/200 x400 = 1.56N/mm2
From chart
fc = 25N/mm2
fy =460N/mm2
100As/bh = 0.7
d/h = 355/400= 0.9
As = (0.7 /100)x200 x35 = 497mm2
Detailing:

15. Use 4T10 (As = 505)SV not morethan 10 ɸ, R6 @ 175 c/c
2.5. Foundation design
Foundation issub structurebelow the ground leveland all elements of supper
structurestransform all loads and momentsto foundation structureand
foundation transformsallloadsto the under layer soil or rock.
Thus the foundation structureseffectively supportthesuper structure

16. Given data
Column size = 400 x200 mm
Dead load = 1000KN
Live load = 350KN
Soil bearing Capacity = 200
fc = 35 N/mm2
fy = 460N/mm2
Solution
1) Determiningslab thickness, h
Use cover 50mm and ɸ of stump = 25mm deformed type2
From table 3.29 la = 27ɸ27x25 = 675 mm
Assumethat starting bar is bent 200 mm
l = 675 – 200 = 475 mm
h = 475 + 50 + 2(25)= 575mm
Take h 600
2) Determinefooting area ( serviceability limit state )
GK + QK = 1350 KN
Self weight = 10% total = 1350 + 135 = 1485KN
Footing area = 1485/200 = 7.84m2
3) Reinforcementdesign ( ultimate limit state )
Design load = 1.4GK+ 1.6 QK =
1.4(1000)+ 1.6 (350) = 1960 KN
Soil ultimate pressure = 1960/7.84= 250KN/m2

17. Moment= 250 x1000 x1.2 x 1.2/2 = 225 KNm
Becauseof it is a square footing, design for critical d only
Assume ɸ= 20mm
d = h-c-ɸ-ɸ/2 = 600-50 -20 -20/2 = 520mm
NOTE: Take the lesser `d` on onedirection as critical
d = 500-50-12-12/2= 432mm
K = M/bd2fcu = 225 x106/1000x5202 x35 = 0.023 < 0.156
Z = d [0.5√0.25− 0.023/0.9] = 0.97d > 0.95d∴ 𝑢𝑠𝑒 0.95𝑑
Asreq = 504 x106/0.87 x460 x0.95520= 2549mm2
Check Asmin = 0.13/100 x1000x600 = 2166mm2< Asreq(Ok)
So use Asreq.

18. Provide9T20 (Asreq=2826mm2)
Check cracking
Allowable distance between bar = 3d @ 375mm (lesser)
3d = 3 x 520 = 1560mm > 375mm
∴Allowable distance = 750mm
Actual distance between bar
= {1000-2(50)-9(20)}/8 = 90mm < 375mm(ok).
h>200mm ∴need further checking
Check 100A /bd = 0.14 < 0.3 ∴no furtherchecking required
Detailing
ly = lx = 1000mm
1.5 (ly+ 3d)= 1.5(lx+3d)= 1.5 (400 + 3 x520)= 2940mm
L = 1000 > 1.5 (c+3d)so reinforcementsdistribution isuniform.
2.6. Staircase design
Staircase is an importantcomponentof buildingprovidingaccess to different
floorsand roofs of the building. it consist of a flight of steps(stairs) and one or
moreintermediate landingslabs between the floor levels.
Given data :
Fcu = 30 N/mm2
Fy= 460 N/mm2
Riser(R) = 175mm
Going(G) = 250mm
Live load =5kn/m2

19. Finishing = 0.4kn/m2
Solution
a) Assumingthe thickness of waist
Becausespan is simply supported chooseL/d = 24
(No additionalfactor for L/d because stairs span is less than 60% of total
span)
d= 6350/24= 265 , assumeф = 16 and cover = 20mm
:- therefore h = 263+(16/2)+20= 293 mm
Hence take h = 300mm and d = 300-(16/2)-20 = 272mm
b) Loading( on plan view):
i. Stairs:
Dead load : waist = 0.3 x 1.22 x24 = 8.8 KN/m2

20. Stair = (0.175/2)x24 = 2.1 kN/m2
Finishing = 0.4 kN/m2
Total Gk = 11.3kN/m2
Imposed load Qk = 5.0 kN/m2
:- Designload = 1.4(11.3)+ 1.6(5.0)= 23.8kN/m2
ii. Landing:
Dead load: waist = 0.3 x 24 = 7.2 kN.m2
Finishing = 0.4 kN/m2
Total Gk = 7.6 kN/m2
:- Design load = 1.4(7.6)+ 1.6(5.0)= 18.6kN/m2
∑ M cutting = 0 ;
M max –(18.6 x1.5 x2.425) – (23.8 x1.6752 x/2 ) + 67.8(3.175)= 0
:- M max= 114 kN/m width

21. K= M/fcubd2 = 114 x106 / 1000 x2722 x30 = 0.051 < 0.156
Z = d [0.5 + √0.25 − 0.051/0.9] = 0.94d < 0.95d (ok)
As = 114 x106 / 0.87 x460 x0.94 x272 = 1114 mm2/m
ProvideT16 – 175 c/c (Asprovide = 1150 mm2/m)
Reinforcementdistribution, useAsmin = 0.13%bh
= 0.13/100 x (1000)(300)
=390 mm2/m.
Deflectionchecking
M/bd2 = 114 x106/1000 x2722 = 2.40
fs= 5 x 460 x1114 /8 x1150 = 278
M.F= 0.55 + (477-278)/120(0.9 + 2.40)= 1.7
(l/d) allowable = 20 x .07 = 34.00
(l/d) actual = 6350/272 = 23 < 34 ( Ok )