Detailed Design and Analysis: Globe View Residential Building

This document presents a comprehensive structural design for a residential building, the Globe View Residential Building, located near the River Thames. The project includes detailed calculations for slab, beam, and column design, adhering to relevant building codes and standards such as IS 456:2000. The design process encompasses paperwork, layout marking, foundation work, concrete pouring, column and wall construction, roofing, and the installation of windows and doors. The slab design is analyzed as a two-way slab, with calculations for bending moments and reinforcement. Beam design considers loads, shear reinforcement, and stirrup spacing. Column design focuses on load calculations, axial load determination, and eccentricity analysis. The document provides insights into material selection and construction techniques.

DESIGN OF GLOBE VIEW RESIDENTIAL BUILDING
Introduction
The proposed site is located at Queenhithe and near river Thames and close to Millennium
Bridge. The area contains a residential building known as Globe view that provides a clear view
of River Thames. The total area of the site is approximately 235.1 square meters and the building
consists of two bedrooms.
Construction work and sequences
Construction of the residential building would involve the following;
a. Paper work: this would involve drawing preparation, material estimation cost, cost of
labor and drawing approval from the authorities.
b. Layout mark: The approved drawings would then be marked on the ground with claear
and accurate orientation and dimensions
c. Foundation work: It would involve excavation and compaction of the bottom, reinforced
footing would be laid, and concrete shuttering would be done based on the mentioned
dimensions in the drawing.
d. Footing concrete: The foundation levels would be checked before concrete work, the
concrete would then be poured based on the drawing.
e. Column casting: This would be done through shutter fixing formwork then after the
concrete would be poured into the formwork
f. Wall construction: Walls would be constructed either using wooden, brick or precast
concrete.
g. Roofing: The roof slab would first be poured after the masonry work would be completed
h. Fixing windows and doors: Frames would be fixed during masonry work, thereafter the
panel would be fixed using hinges
i. Electrical and plumbing work: This would be done before applying final finishes
Detailed design
Slab design
In reference to IS 456:2000 the slabs would be designed under limit state
Take fck = 15 N/mm2
Take fy =415 N/m2
Span

a. Shorter span:- Lx = 5.8 m
Longer span:-Ly =7.6 m
b. Check Lx/Ly= 7.6/5.8 =1.3<2 (the slab will be designed as two way slab)
c. Take the overall slab depth be = 120mm and cover depth be 15 mm
Effective depth= Depth -15- ∅ /2 =120-15-10/2=100mm
d. The slab is supported on its four sides.
e. Calculation of loads:
Dead load = 25x0.12 m x1m = 3.0KN/m
Live load =2 x 1 = 2.0KN/m
Floor finish = 1x1KN/m
Total load = 6.0 KN/m
f. Calculation of Bending moment : (Based on IS code 456-2000)
Panel types:- Discontinuous two adjacent edges
ax(+) = 0.049 ax(-) = 0.065
ay(+) = 0.035 ay(-) = 0.047
In shorter directions
Positive bending moment at mid-span
Mx(+) = ax(+)wlx2 = 0.049x6x5.82 = 9.9 kN-m
Factored bending moment = 9.9 x 1.5 =14.9 kN-m
Diameter and spacing
Based on sp- 16
Provide a bar of diameter 8 mm at a spacing of 210 mm
Negative bending moment at continuous edge
Mx(-) = ax(-)wlx2 = 0.062x6x5.82 = 13.1 kN-m

Factored bending moment = 13.1 x 1.2 =19.7 kN-m
Longer directions
Positive bending moment at mid-span
My(+) = ay(+)wlx2 = 0.035x6x5.82 = 7.1 kN-m
Factored bending moment = 7.1 x 1.5 =10.7 kN-m
Negative bending moment at continuous edge
My(-) = ay(-)wlx2 = 0.047x6x5.82 = 9.5 kN-m
Factored bending moment = 9.5 x 1.5 =14.2 kN-m
Checking depth
Permissible depth is 100 mm
Mu. limit =0.36 * Xu,max/d * (1 – 0.42* Xu,max/d)fck*bd2
14.86 * 106 = 0.36* Xu,max/d * (1 – 0.42*0.48)*15 *1000*d2
d = 84.7 < 100 mm
Therefore, it’s ok
BEAM DESIGN
Beam member in a building helps in transferring slab load to column and for foundation into the
soil
 Slab span would determine the beam spacing
 Loads that acts on the beam includes; Live load, dead load and wind load
Beam loads
Beam: B1
Shorter span = 5.8 m
Consider the size of the beam = 230 * 405 mm

Height of wall = 3 m
Calculation of loads
Wall load = 0.23 * 19 = 13.1 kN-m
Self-load = 0.23 *0.406 * 25 = 2.3 kN/m
Load of the slab
W = 6 kN
Lx = 5.8
W*Lx/3 = (6 * 5.8)/3 = 11.6 kN/m
Total load = 13.1 + 2.3 + 11.6
= 27.0 kN.m
Design of stirrups
Beam: B1
Va = Vb = (27.0 *5.8)/2
= 78.42 kN
 Normal shear calculation
Tv = Vu/Bd = (1.5 * 71.42 * 103)/(373 *230)
= 1.37
 Permissible shear stress calculations
Tc = percentage of steel
Pt = Ast/Bd * 100
Ast = (2*162 * pi)/4 = 402.12 mm2
Pt = (402.12 * 100)/(230 *373) = 0.60%
Tc = 0.50

Tv > Tc
0.76 > 0.05
Therefore, shear reinforcement would be provided
Shear design
Vs = (Tv – Tc)bd
= (0.76 – 0.5)*373 *230
= 22.3 kN
Vus/D = 22.3/37.3
= 0.59 kN/cm
Therefore, provide a 6 mm diameter at 200 mm center/ center spacing
Space checking
The provided spacing should be minimum for the following;
a. 0.7d = 0.75 * 373 = 279.8 mm
b. (Asvfy)/(0.4b) = (2*62*pi/4)*250)/(0.4*230
= 153.2 mm
c. Space design of 450 mm c/c
Therefore; stirrups of 6 mm diameter @ 150 mm c/c will be provided
Beam loads
Beam: B2
Longer span = 7.62 m
Consider the size of the beam = 230 * 405 mm
Height of wall = 3 m
Calculation of loads
Wall load = 0.23 * 19 = 13.1 kN-m
Self-load = 0.23 *0.406 * 25 = 2.3 kN/m
Load of the slab

W = 6 kN
Ly = 7.62
W*Ly/3 = (6 * 7.62)/3 = 15.24 kN/m
Total load = 13.1 + 2.3 + 15.24
= 30.7 kN.m
Design of stirrups
Beam: B2
Va = Vb = (30.7 *7.62)/2
= 116.9 kN
 Normal shear calculation
Tv = Vu/Bd = (1.5 * 116.9 * 103)/(373 *230)
= 2.04
 Permissible shear stress calculations
Tc = percentage of steel
Pt = Ast/Bd * 100
Ast = (2*162 * pi)/4 = 402.12 mm2
Pt = (402.12 * 100)/(230 *373) = 0.60%
Tc = 0.50
Tv > Tc
0.85 > 0.05
Therefore, shear reinforcement would be provided
Shear design
Vs = (Tv – Tc)bd
= (0.85 – 0.5)*373 *230
= 30.02 kN

Vus/D = 30.02/37.3
= 0.89 kN/cm
Therefore, provide a 6 mm diameter at 150 mm center/ center spacing
Space checking
The provided spacing should be minimum for the following;
d. 0.7d = 0.75 * 373 = 279.8 mm
e. (Asvfy)/(0.4b) = (2*62*pi/4)*250)/(0.4*250)
= 153.2 mm
f. Space design of 450 mm c/c
Therefore; stirrups of 6 mm diameter @ 150 mm c/c will be provided
Beam design
Mu @ left span = 11.6 kN-m
Mu @ Mid span = 19.2 kN-m
Mu @ Right span = 20.4 kN-m
Checking
Mu = 11.6 kN-m
Mu, limit = 0.138 *fck *bd2
= 0.138 * 20 * 230 * 3052
= 59.1 kN-m

Mu < Mu, limit
Designed of simply reinforced beam
Mu/bd2 = (11.6 *106)/(230 *3052)
= 1.39
Taking, fy 415 N/mm2 and fck = 20 N/mm2
Mu/bd2 = 1.39
Using interpolation
1.35 ……. 0.409
1.40 …….. 0.426
1.39 …… ?
Pt = 0.42
Pt = 0.42%
Area of reinforcement
Pt = (Ast*100)/Bd
= 0.42*305*230/100
= 393.1 mm2
Provide 3 bars of diameter 12 mm
Hence; provide area = 400 mm2
Mid-span reinforcement
Limiting moment of resistance
Mu = 19.18 kN-m
Mu,limit = 0.138 * 20 *230 *3052
= 59.1 kN-m
Mu < Mu,limit
Therefore; it would be designed as a simply reinforcement.

Mu/bd2 = (19.18 *106)/(230 *3052)
= 0.66
Taking, fy 415 N/mm2 and fck = 20 N/mm2
Mu/bd2 = 0.66
Using interpolation
0.65 ……. 0.187
0.70 …….. 0.203
0.66 …… ?
Pt = 0.190%
Pt = 0.42%
Area of reinforcement
Pt = (Ast*100)/Bd
= 0.19*305*230/100
= 133.29 mm2
Provide 2 bars of diameter 12 mm
Hence; provided area = 155.2 mm2
COLUMNS DESIGN
 Column are regarded as compression members
 Columns transmits load from the slab to the foundations
 Considerations are always placed in place to avoid use of large span beams to control
cracking and deflection
COLUMNS:
The loads which are taken to column include; (a) Beam loads (b) wall loads (c) Slab loads (d)
Self-weight of column
Load type Roof load Floor load

Wall load (7.6 + 5.8)/2 * 0.115 *0.91
*19
= 12.1 kN
(7.6 + 5.8)/2 * 0.23*3 *19
= 29.3 kN
Self-weight of beam 0.23 *0.406 * (7.6 *5.5)/2 *
25
= 25 kN
0.23 *0.406 * (7.6 *5.5)/2 *
25
= 25 kN
Slab load (7.6 + 5.8)/2 * 6
= 40.3 kN
(7.6 + 5.8)/2 * 6
= 40.3 kN
Total load 77.4 kN 94.6 kN
Total load
Load of roof = 77.4 kN
Load of floor = 94.6 kN
Self-weight = 0.23 *0.23 * 3*25
= 34.5 kN
Total load = 77.4 + 94.6 + 34.5 = 167 kN
Axial load
Pu = 167 kN
Area = 230 m * 230 mm
Pu/(fck*b*d) = (167 *103)/(20 *230 *230)
= 0.15
Eccentricity calculation
e = 1/500 + b/30
= 4640/500 + 230/30 = 16.9 mm
e < 20 mm
Mu*e = Pu*e
= 167 * 0.02
= 3.34 kN-m
Mue/(fck*bd2) = (3.34*106)/(20 *230 *2302)

= 0.0112
d’/D = 0.2
P/fck = 0.02
P = 0.02*fck
= 0.02*20
= 0.4%
Area of steel = 0.8Bd/100 = (0.8*20*230)/100
= 423.2 mm
Number of bars for diameter 12 mm
= 432.2/(pi/4*122)
= 4 bars
Conclusion
The slabs were designed based on the relationship of the ratio between longer to shorter span of
panel. Therefore, the slab in this case was designed as two way slab based on the end condition,
that would correspond to bending moment. The coefficient calculated corresponded to lx/ly ratio.
The calculation for loads on columns and beams in addition to designed frame using moment
distribution were done.
Appendix
Plan

Beam