STUCTURAL ANALYSIS AND DESIGN REPORT

REPUBLIC OF RWANDA

PROJECT: CONSTRUCTION OF RESIDENTIAL BUILDING

Structure 3D View

STUCTURAL ANALYSIS AND DESIGN REPORT

PROJECT OWNER: ABBAS MUKAMA

Eng BIRASA ALEXIS

Design code: BS 8110-1997

Software: ROBOT

APRIL 2018

2

1. GENERAL

1.1 SCOPE

The numerical values of actions on buildings and civil Engineering works

to be taken into account in the design are applicable to the various types of

construction.

1.2 The purpose of the building is, residential building. the materials used

are R.C structures of framed type, Solid two way slabs and cement blocks for wall

elevation, whereas the roof is made up with wood trusses covered with metal

sheets.

1.3 The execution of construction of this building is covered by various

code of designs to the extent that is necessary to indicate the quality of

construction materials and products which should be used and the standard of

workmanship on site needed to be supervised by qualified and experienced

Engineer .Some lab test like compressive strength test for concrete should be done

as far as the importance of building structure is concerned.

1.4 The method of design is Limit state design method accordingly to BS

8110-1997, and the use of ROBOT ® Structural software for all members except

to slab design where we use Prokon structure software.

3

DESIGN INFORMATION

UWAMAHORO Faustin Client

The structural use of Concrete

British Standard (BS 8110-1997)

Relevant Building

Regulations and

Design Code

Public Building. Intended use of the

building

following codes: BS 648, BS 649, and BS 6399.

Self-weight of Bac Autoportant Sheet = 0.11 KN/m2, Self-weight of

truss: 0.15 KN/m2, Ceiling: 0.1KN/m2, Services: 0.2 KN/m2

and Finishes: 0.01 KN/m2. Dead load (Gk) = 0.11+0.1+0.2+0.01 =

0.42 KN/m2

Floor –Imposed (3 )and partitions(1) 1.5 kN/m2 (BS 6399)

- Finishes 1.0 kN/m2

Stairs –Imposed 4 kN/m2

- Finishes 1.0 kN/m2

General loading

conditions

Severe (external ) and Mild (internal) Exposure

conditions

Compact sandy-gravel

Allowable bearing pressure Calculated to 200kN/ m2

Subsoil conditions

Reinforced Concrete footing to columns. Foundation type

4

Concrete: grade fck=25MPa (with 20mm max. aggregates.

Mix ratio 350 kg/ m3

Reinforcement :-Characteristic strength:

ƒy = 460 N/mm High yield steel

for stirrups ƒy = 250 N/mm2 Mild steel

Material data

Self weight of Reinforced concrete = 25 KN/ m3

Self weight of masonry = 18 KN/ m3

Other relevant

information

For dead load 1.4

For live load 1.6

-Concrete is assumed to carry zero tensile stresses.

-Plane sections of structural member before bending

remain plane after bending.

-At the ultimate limit state, the strain in concrete is 0.0035

-The strains in the concrete and in the reinforcing steel are

directly proportional to the distances from the neutral axis

at which the strain is zero.

-Partial safety factors for material strength according to

BS8110 part1,

γm is 1.15 for steel and

1.5 for concrete

Partial safety

factor

5

2. ANALYSIS AND DIMENSION OF SLAB

2.1. Analysis of slab

During our calculations the following notations are going to be used:

hs = thickness of the slab

d = efficient thickness of the slab

Lx = short span

ly = long span

Ly/lx > 2 one way span

Ly/lx <2 two way span

Figure 1.1 Plan view show slab , beam and Column, foundation

6

2.1.1. Dimension of slab

2.1.1.1. Preliminary dimension

Efficient thickness of slab = Po rtee(lx )

2 0*2.0

The value of the small range (lx) is lx = 4.63m

hs 3.20 *10 3

1115.75mm; 20 * 2.0

Let take a slab of thickness hs = 120mm for the whole slab

Let assume h (the overall depth of slab) to be 120mm so that construction reinforcement bar

being easy on site. We will only consider the fire resistance of 1.5 hour to get the cover. From

table 3.4 given in the BS 8110, the cover for 1.5 hour fire resistance is 20mm.

2.2 Panel on floor

2.2.1Checking slab

The ratio of = 5.21/4.63=1.125 < 2.0

Slab is designed as two ways

We assumed 10Ø bar

Cover =20mm

2.2.2 Load distribution

-Thickness of slab 120mm

-Self weight of slab = 25×0.12 = 3 kN/m2

Finishes 1.5 kN/m2

Dead load = 3+1.5 = 4.5 kN/m2

Imposed load = 1.5 kN/m2

Design load = 1.4 Gk+1.6Qk= 1.4×4.5+1.6x 1.5 = 8.7 kN/m

7

2.2.3 Moment result:

Max Moment at mid-span is 12.7 KNm/m

Max Moment at support is 33.5 KNm/m

2.2. 4 Deflection

Deflection is 2.8mm<Max deflection=l/500=10.4mm

8

Reinforcement area

Maximum Reinforcement areas:

At support is 555.6 mm2/m and Reinforcement is Y10@150 mm (T1 and T2)

At mid span is 284.2 mm2/m and Reinforcement is Y 10@150 mm (B1 and B2)

9

3. ANALYSIS AND DESIGN OF BEAM

3.0 BEAM 1

Fire rating : 1.5 (h)

Maximum cracking: 0.30 (mm)

Environment class : moderate Concrete creep coefficient : p = 2.00

3.1 Material properties:

Concrete: CONCR fcu = 25 (KN)

Longitudinal reinforcement : T fy = 460 (N/mm2)

Transversal reinforcement : R fy = 250 (N/mm2)

3.2 Geometry:

2.2.1 Span Position L.supp. L R.supp. (m) (m) (m) P1 Span 0.20 4.80 0.20

Span length: Lo = 5.00 (m) Section from 0.00 to 4.80 (m)

20 x 35 (cm) without left slab without right slab

3.3 Calculation options:

Regulation of combinations : BS5950 Calculations according to : BS 8110

Precast beam : no Cover : bottom c = 3 (cm)

: side c1 = 3 (cm) : top c2 = 3 (cm)

3.4 Calculation results:

3.4.1 Internal forces in ULS

Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN) P1 34.75 -0.00 -43.12 -5.23 60.61 -28.20

10

-60

-50

-40

-30

-20

-10

0

10

20

30

40

50

[kN*m]

[m]

0 1 2 3 4 5

Bending Moment ULS: M Mr Mc

150

100

[kN]

50

0

-50

-100

-150

[m]

0 1 2 3 4 5

Shear Force ULS: V Vr Vc(stirrups) Vc(total)

3.4.2 Internal forces in SLS

Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN) P1 0.00 0.00 0.00 0.00 0.00 0.00

3.4.3 Required reinforcement area

500

400

300

200

100

0

100

200

300

400

[mm2]

Span Span (mm2) Left support (mm2) Right support (mm2) bottom top bottom top bottom top

P1 292 0 0 374 11 42

[m]

0 1 2 3 4 5

Reinforcement Area for Bending: Abt Abr Abmin

1000

800

600

400

200

0

200

400

600

800

1000

[mm2/m]

[m]

0 1 2 3 4 5

Reinforcement Area for Shear: Ast Asr AsHang

11

Abscissa

ULS M max.

M min.

SLS M max.

M min.

A bottom

A top

(m) (kN*m) (kN*m) (kN*m) (kN*m) (mm2) (mm2) 0.20 0.00 -43.12 0.00 0.00 0 374 0.60 0.00 -19.59 0.00 0.00 0 157 1.10 4.01 -0.00 0.00 0.00 32 0 1.60 19.83 -0.00 0.00 0.00 159 0 2.10 29.51 -0.00 0.00 0.00 244 0 2.60 34.75 -0.00 0.00 0.00 292 0 3.10 33.29 -0.00 0.00 0.00 279 0 3.60 27.97 -0.00 0.00 0.00 230 0 4.10 17.88 -0.00 0.00 0.00 143 0 4.60 6.66 -0.00 0.00 0.00 53 0 5.00 1.33 -5.23 0.00 0.00 11 42

Abscissa

ULS Q max.

SLS Q max.

Cw

(m) 0.20

(kN) 60.61

(kN) 0.00

(mm) 0.00

0.60 55.81 0.00 0.00 1.10 34.64 0.00 0.00 1.60 28.64 0.00 0.00 2.10 13.55 0.00 0.00 2.60 0.07 0.00 0.00 3.10 -5.93 0.00 0.00 3.60 -17.17 0.00 0.00 4.10 -23.18 0.00 0.00

3.4.4 Deflection and cracking

at(s-t) - initial deflection due to total load ap(s-t) - initial deflection due to long-term load ap(l-t) - long-term deflection due to long-term load a - total deflection aall - allowable deflection

Cw - width of perpendicular cracks

-20

-15

-10

[mm]

Span at(s-t) ap(s-t) ap(l-t) a aall Cw (mm) (mm) (mm) (mm) (mm) (mm)

P1 0 0 0 0=(L0/--) -20 0.00

-5

0

5

10

15 [m]

20 0 1 2 3 4 5

Deflections: ap(s-t) ap(l-t) at(s-t) a a adm

0.3

0.2

[mm]

0.1

0

0.1

0.2

0.3

[m]

0 1 2 3 4 5

Cracking: Cw Cw adm

3.5 Theoretical results - detailed results:

3.5.1 P1 : Span from 0.20 to 5.00 (m)

12

(mm) (m) (kG) (No.) (kG) 8 0.79 0.31 44 13.77

4.60 -23.40 0.00 0.00 5.00 -28.20 0.00 0.00

3.6 Reinforcement:

2.6.1 P1 : Span from 0.20 to 5.00 (m) Longitudinal reinforcement: bottom

3 T 12 l = 5.68 from 0.03 to 5.17

assembling (top) 3 T 12 l = 4.51 from 0.60 to 5.10

support (T) 3 T 12 l = 1.73 from 0.03 to 1.41 1 T 12 l = 1.88 from 0.08 to 1.51 3 T 12 l = 1.23 from 4.28 to 5.17

Transversal reinforcement: main

stirrups 44 R 8 l = 0.79 e = 1*0.09 + 21*0.22 (m)

pins 44 R 8 l = 0.79 e = 1*0.09 + 21*0.22 (m)

3.7 Material survey:

Concrete volume = 0.36 (m3)

Formwork = 4.74 (m2)

Steel T Total weight = 36.69 (kG)

Density = 100.79 (kG/m3) Average diameter = 12.0 (mm)

Survey according to diameters:

Diameter Length Weight NumberTotal weight (mm) (m) (kG) (No.) (kG) 12 1.23 1.09 3 3.28 12 1.73 1.54 3 4.61 12 1.88 1.67 1 1.67 12 4.51 4.00 3 12.01 12 5.68 5.04 3 15.12

Steel R

Total weight = 13.77 (kG) Density = 37.84 (kG/m3)

Average diameter = 8.0 (mm) Survey according to diameters:

Diameter Length Weight NumberTotal weight

13

4. BEAM 200X200 (Ring beam)

Fire rating : 1.5 (h) Maximum cracking: 0.30 (mm) Environment class : moderate Concrete creep coefficient : p = 2.00

4.1 Material properties:

Concrete: CONCR fcu = 25 (KN)



4.2 Geometry:

2.2.1 Span Position L.supp. L R.supp. (m) (m) (m) P1 Span 0.20 4.80 0.20

Span length: Lo = 5.00 (m) Section from 0.00 to 4.80 (m)

20 x 20 (cm) without left slab without right slab

FAUSTIN PROJECT

14

Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)

P1 11.68 -0.00 -17.72 -5.00 21.35 -15.96

[kN*m]


Regulation of combinations : BS5950 Calculations according to : BS 8110

Precast beam : no Cover : bottom c = 3 (cm)

: side c1 = 3 (cm) : top c2 = 3 (cm)


4.4.1 Internal forces in ULS

-25

-20

-15

-10

-5

0

5

10 [m]

15 0 1 2 3 4 5

Bending Moment ULS: M Mr Mc

80

[kN] 60

40

20

0

-20

-40

-60

-80

[m]

0 1 2 3 4 5

Shear Force ULS: V Vr Vc(stirrups) Vc(total)

4.4.2 Internal forces in SLS

Span Mtmax. Mtmin. Ml Mr Ql Qr (kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN) P1 0.00 0.00 0.00 0.00 0.00 0.00

4.4.3 Required reinforcement area

Span Span (mm2) Left support (mm2) Right support (mm2) bottom top bottom top bottom top P1 205 0 8 349 4 80

15

500

400

300

[mm2]

200

100

0

100

200

300

[m]

0 1 2 3 4 5

Reinforcement Area for Bending: Abt Abr Abmin

1500

1000

[mm2/m]

500

0

500

1000

1500

[m]

0 1 2 3 4 5

Reinforcement Area for Shear: Ast Asr AsHang

4.4.4 Deflection and cracking

at(s-t) - initial deflection due to total load ap(s-t) - initial deflection due to long-term load ap(l-t) - long-term deflection due to long-term load a - total deflection aall - allowable deflection

Cw - width of perpendicular cracks

-25

-20

-15

-10

-5

0

5

10

15

20

25

[mm]

Span at(s-t) ap(s-t) ap(l-t) a aall Cw (mm) (mm) (mm) (mm) (mm) (mm)

P1 0 0 0 0=(L0/--) -20 0.00

[m]

0 1 2 3 4 5

Deflections: ap(s-t) ap(l-t) at(s-t) a a adm

0.3

0.2

[mm]

0.1

0

0.1

0.2

0.3

[m]

0 1 2 3 4 5

Cracking: Cw Cw adm

16

4.5 Theoretical results - detailed results:

4.5.1 P1 : Span from 0.20 to 5.00 (m) Abscissa

ULS M max.

M min.

SLS M max.

M min.

A bottom

A top

A compressive

(m) (kN*m) (kN*m) (kN*m) (kN*m) (mm2) (mm2) (mm2) 0.20 0.00 -17.72 0.00 0.00 8 349 8 0.60 0.00 -9.64 0.00 0.00 0 164 0 1.10 0.00 -1.49 0.00 0.00 0 24 0 1.60 4.72 -0.00 0.00 0.00 76 0 0 2.10 8.98 -0.00 0.00 0.00 152 0 0 2.60 11.30 -0.00 0.00 0.00 197 0 0 3.10 11.68 -0.00 0.00 0.00 205 0 0 3.60 10.11 -0.00 0.00 0.00 174 0 0 4.10 6.60 -0.00 0.00 0.00 108 0 0 4.60 1.15 -0.00 0.00 0.00 18 0 0 5.00 0.23 -5.00 0.00 0.00 4 80 0

Abscissa

ULS Q max.

SLS Q max.

Cw

(m) 0.20

(kN) 21.35

(kN) 0.00

(mm) 0.00

0.60 18.25 0.00 0.00 1.10 14.36 0.00 0.00 1.60 10.47 0.00 0.00 2.10 6.58 0.00 0.00 2.60 2.70 0.00 0.00 3.10 -1.19 0.00 0.00 3.60 -5.08 0.00 0.00 4.10 -8.96 0.00 0.00 4.60 -12.85 0.00 0.00 5.00 -15.96 0.00 0.00

4.6 Reinforcement:

4.6.1 P1 : Span from 0.20 to 5.00 (m) Longitudinal reinforcement: bottom

2 T 12 l = 5.80 from 0.03 to 5.17

assembling (top) 2 T 12 l = 4.01 from 1.10 to 5.10

support (T) 2 T 12 l = 2.10 from 0.03 to 1.78 2 T 12 l = 1.25 from 0.08 to 0.88 2 T 12 l = 1.10 from 4.42 to 5.17

Transversal reinforcement: main

stirrups 48 R 8 l = 0.61 e = 1*0.05 + 47*0.10 (m)

pins 48 R 8 l = 0.61 e = 1*0.05 + 47*0.10 (m)


Concrete volume = 0.21 (m3) Formwork = 3.12 (m2)

Steel T



Diameter Length Weight NumberTotal weight

17

(mm) (m) (kG) (No.) (kG) 12 1.10 0.97 2 1.95 12 1.25 1.11 2 2.21 12 2.10 1.86 2 3.72 12 4.01 3.56 2 7.12 12 5.80 5.15 2 10.29

Steel R



Diameter Length Weight NumberTotal weight (mm) (m) (kG) (No.) (kG)

8 0.61 0.24 48 11.46

FAUSTIN PROJECT

18

5. ANALYSIS AND DESIGN OF COLUMN

5.1 Column: C1 (200x200mm)

Fire rating : 1.5 (h)

Environment class : mild

5.1.1 Material properties:

Concrete : CONCR fcu = 25 (N/mm2) Concrete Unit weight : 2501.36 (Kg/m3)



5.2 Geometry:

2.2.1 Rectangular 20 x 20 (cm) 2.2.2 Height: L = 3.38 (m) 2.2.3 Slab thickness = 0.12 (m) 2.2.4 Beam height = 0.35 (m) 2.2.5 Cover = 2 (cm)


Calculations according to : BS 8110 Precast column : no Pre-design : no Slenderness taken into account : yes Ties :to slab Non-sway structure

5.4 Loads:

Case Nature Group f N Myu Myl Myi Mzu Mzl Mzi (kN) (kN*m) (kN*m) (kN*m) (kN*m) (kN*m) (kN*m) COMB1 design 45 1.00 248.41 4.14 -2.15 1.66 3.55 -1.67 1.46 f - load factor


5.5.1 ULS Analysis

Design combination: COMB1 (A) Internal forces:

NSd = 248.41 (kN) MSdy = 4.14 (kN*m) MSdz = 3.55 (kN*m) Design forces: Upper node

NSd = 248.41 (kN) NSd*etotz = 6.77 (kN*m) NSd*etoty= 3.55 (kN*m)

5.5.1.1 Eccentricity:

Eccentricity: static

ee:

ez (My/N) 2 (cm)

ey (Mz/N) 1 (cm)

II order eadd: 1 (cm) 1 (cm)

19

Minimal emin: 1 (cm) 1 (cm) total etot: 3 (cm) 1 (cm)

5.5.1.2 Detailed analysis-Direction Y:

5.5.1.2.1 Slenderness analysis

Non-sway structure lo (m) le (m)

3.20 1.00 3.20

ley/h = 16.00 > 15.00 (10-7) lez/b = 16.00 > 15.00 (10-7) Slender column

5.5.1.2.2 Buckling analysis

M2 = 4.14 (kN*m) M1 = -2.15 (kN*m) Case: Cross-section at the column end (Upper node), Slenderness taken into account M = 4.14 (kN*m) emin = max (20mm ; 0.05 *hy) = 1 (cm) (3.8.2.4)

hy = 20 (cm) Mmin = N*emin = 2.48 (kN*m) au/2 = a*K*h/2 = 1 (cm) (32)

a = 0.13 (34) K = min((Nuz-N)/(Nuz-Nbal) ; 1) = 0.83 (33)

Nuz = 2/3*fcu/c*Ac+fy/s*Asc = 642.63 (kN) Ac = 0.04 (m2) Asc = 452 (mm2)

Nbal = 165.82 (kN) h = 20 (cm)

Madd/2 = au/2 * N = 2.63 (kN*m) (35) Md = max (Mmin;M+Madd/2) = 6.77 (kN*m)

5.5.1.3 Detailed analysis-Direction Z:

M2 = 3.55 (kN*m) M1 = -1.67 (kN*m) Case: Cross-section at the column end (Upper node), Slenderness not taken into account M = 3.55 (kN*m) emin = max (20mm ; 0.05 *hz) = 1 (cm) (3.8.2.4)

hz = 20 (cm) Mmin = N*emin = 2.48 (kN*m) Md = max (Mmin;M) = 3.55 (kN*m)

5.5.2 Reinforcement:

Real (provided) area Asr = 452 (mm2)

Ratio: = 1.13 %

5.6 Reinforcement:

Main bars:

4 T 12 l = 3.25 (m)

Transversal reinforcement: stirrups: 21 R 8 l = 0.61 (m)

pins 21 R 8 l = 0.61 (m)

20




Steel T Total weight = 11.53 (kG)

Density = 98.53 (kG/m3)

Average diameter = 12.0 (mm) Reinforcement survey:

Diameter Length Weight Number Total weight (m) (kG) (No.) (kG) 12 3.25 2.88 4 11.53

Steel R


Average diameter = 8.0 (mm) Reinforcement survey:

Diameter Length Weight Number Total weight (m) (kG) (No.) (kG) 8 0.61 0.24 21 5.02

FAUSTIN PROJECT

24

A = 0.90 (m) a = 0.20 (m) B = 0.90 (m) b = 0.20 (m) h1 = 0.30 (m) ex = 0.00 (m) h2 = 0.60 (m) ey = 0.00 (m) h4 = 0.05 (m)

7. ANALYSIS AND DESIGN OF FOUNDATION

7.1 Basic data

7.1.1 Assumptions

Geotechnic calculations according to : BS 8004 Concrete calculations according to : BS 8110 Shape selection : without limits

7.1.2 Geometry:

a' = 20 (cm) b' = 20 (cm) c1 = 5 (cm) c2 = 5 (cm)

7.1.3 Materials

Concrete : C25; Characteristic strength = 25000.00 kPa

Unit weight = 2501.36 (kG/m3) Longitudinal reinforcement

460000.00 kPa : type T Characteristic strength =

Transversal reinforcement 250000.00 kPa

: type R Characteristic strength =

7.1.4 Loads:

Foundation loads: Case Nature Group N Fx Fy Mx My (kN) (kN) (kN) (kN*m) (kN*m) COMB1 design ---- 257.38 -1.96 -1.63 1.05 1.39

Backfill loads: Case Nature Q1

25

(kN/m2)

7.1.5 Combination list

1/ ULS : COMB1 N=257.38 Mx=1.05 My=1.39 Fx=-1.96 Fy=-1.63 2/* ULS : COMB1 N=257.38 Mx=1.05 My=1.39 Fx=-1.96 Fy=-1.63

7.2 Geotechnical design

7.2.1 Assumptions

Foundation design for: • Capacity • Rotation

7.2.2 Soil:

Soil level: N1 = 0.00 (m) Column pier level: Na = 0.00 (m) Minimum reference level: Nf = -0.50 (m)

well graded gravels • Soil level: 0.00 (m) Unit weight: 2000 (kG/m

3)

• Unit weight of solid: 2502.25 (kG/m3)

• Internal friction angle: 42.0 (Deg) • Cohesion: 0.00 (kPa)

7.2.3 Limit states

7.3 RC design

7.3.1 Assumptions

Exposure : mild

7.3.2 Analysis of punching and shear

Shear

Design combination ULS : COMB1 N=257.38 Mx=1.05 My=1.39 Fx=-1.96 Fy=-1.63 Load factors: 1.00 * Foundation weight

1.00 * Soil weight Design load:

Nr = 274.09 (kN) Mx = 2.52 (kN*m) My = -0.38 (kN*m) Length of critical circumference: 0.90 (m) Shear force: 33.19 (kN) Section effective height heff = 0.24 (m) Shear area: A = 0.22 (m2) Reinforcement ratio: = 0.16 % Shear stress: 153.67 (kPa) Admissible shear stress: 391.86 (kPa) Safety factor: 2.55 > 1

26

7.3.3 Required reinforcement

Spread footing:

bottom:

ULS : COMB1 N=257.38 Mx=1.05 My=1.39 Fx=-1.96 Fy=-1.63 My = 17.65 (kN*m) Asx = 390 (mm2/m)

ULS : COMB1 N=257.38 Mx=1.05 My=1.39 Fx=-1.96 Fy=-1.63 Mx = 18.37 (kN*m) Asy = 390 (mm2/m)

As min = 390 (mm2/m)

top: A'sx = 0 (mm2/m)

A'sy = 0 (mm2/m)

As min = 0 (mm2/m)

Column pier: Longitudinal reinforcement A = 160 (mm2) A min. = 160 (mm2)

A = 2 * (Asx + Asy) Asx = 30 (mm2) Asy = 50 (mm2)

7.3.4 Provided reinforcement

2.3.1 Spread footing: Bottom: Along X axis:

4 T 12 l = 0.80 (m) e = 1*-0.40

Along Y axis: 4 T 12 l = 0.80 (m) e = 0.24

Top:

2.3.2 Pier Longitudinal reinforcement Along X axis:

2 T 12 l = 1.76 (m) e = 1*-0.02 + 1*0.03

Along Y axis: 2 T 12 l = 1.80 (m) e = 1*-0.05

Transversal reinforcement 5 R 8 l = 0.45 (m) e = 1*0.05

7.4Material survey:



Steel T

27

Total weight = 5.68 (kG) Density = 21.29 (kG/m3) Average diameter = 12.0 (mm) Survey according to diameters:

Diameter Length Number: (m) 12 0.80 8

Steel R



Diameter Length Number: (m) 8 0.45 5 12 1.76 2 12 1.80 2

FAUSTIN PROJECT

28

A = 1.20 (m)

a1 = 0.20 (m) a2 = 0.20 (m)

B = 1.20 (m) b1 = 0.20 (m) b2 = 0.20 (m) h1 = 0.30 (m) e1 = 0.40 (m) h2 = 0.60 (m) e2 = 0.20 (m) ey = 0.20 (m) h4 = 0.05 (m)

Case Nature Group Pier N Fx Fy Mx My (kN) (kN) (kN) (kN*m) (kN*m) COMB1 design ---- 1 -44.57 -5.05 0.24 -0.25 -5.35 2 185.26 -5.62 0.02 -0.01 -5.95

8. FOUNDATION ANALYSIS & DESIGN DOUBLE COLUMN (F2)

8.1 Basic data

8.1.1 Assumptions

Geotechnic calculations according to : BS 8004 Concrete calculations according to : BS 8110 Shape selection : without limits

8.1.2 Geometry:

a1' b1'

= 20 (cm) = 20 (cm)

a2' b2'

= 20 (cm) = 20 (cm)

c1 c2

= 5 (cm) = 5 (cm)

8.1.3

Materials

Concrete : C25; Characteristic strength = 25000.00 kPa Unit weight = 2501.36 (kG/m3)

Longitudinal reinforcement 460000.00 kPa

: type T Characteristic strength =

Transversal reinforcement 250000.00 kPa

: type R Characteristic strength =

8.1.4 Loads:

Foundation loads:

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Backfill loads: Case Nature Q1

(kN/m2)

8.1.5 Combination list

1/ ULS : COMB1 N=140.69 Mx=-0.26 My=11.68 Fx=-10.67 Fy=0.25 2/* ULS : COMB1 N=140.69 Mx=-0.26 My=11.68 Fx=-10.67 Fy=0.25

8.2 Geotechnical design

8.2.1 Assumptions

Foundation design for: • Capacity • Rotation

8.2.2 Soil:

Soil level: N1 = 0.00 (m) Column pier level: Na = 0.00 (m) Minimum reference level: Nf = -0.50 (m)

Silty sand • Soil level: 0.00 (m) • Unit weight:2000 (kG/m

3)

• Unit weight of solid: 2502.25 (kG/m3)

• Internal friction angle: 42.0 (Deg) • Cohesion: 0.00 (kPa)

8.2.3 Limit states

8.3 RC design

8.3.1 Assumptions

Exposure : mild

8.3.2 Analysis of punching and shear

Shear

Design combination ULS : COMB1 N=140.69 Mx=-0.26 My=11.68 Fx=-10.67 Fy=0.25 Load factors: 1.00 * Foundation weight

1.00 * Soil weight Design load:

Nr = 170.41 (kN) Mx = -28.65 (kN*m) My = 2.08 (kN*m) Length of critical circumference: 1.20 (m) Shear force: 44.07 (kN) Section effective height heff = 0.24 (m) Shear area: A = 0.29 (m2) Reinforcement ratio: = 0.16 % Shear stress: 153.02 (kPa)

30

Along X axis: 5 T 12

l = 1.10 (m)

e = 1*-0.55

Along Y axis: 5 T 12

l = 1.10 (m)

e = 0.25

Top: Along X axis:

5 T 12

l = 1.10 (m)

e = 1*-0.55 Along Y axis:

3 T 12

l = 1.10 (m)

e = 0.45

Admissible shear stress: 391.86 (kPa) Safety factor: 2.561 > 1

8.3.3 Required reinforcement

Spread footing:

bottom:

ULS : COMB1 N=140.69 Mx=-0.26 My=11.68 Fx=-10.67 Fy=0.25 My = 37.10 (kN*m) Asx = 390 (mm2/m)

ULS : COMB1 N=140.69 Mx=-0.26 My=11.68 Fx=-10.67 Fy=0.25 Mx = 10.88 (kN*m) Asy = 390 (mm2/m)

As min = 390 (mm2/m)

top:

My = 0.00 (kN*m) A'sx = 390 (mm2/m)

A'sy = 0 (mm2/m)

As min = 0 (mm2/m)

Column pier: 1 Longitudinal reinforcement A = 672 (mm2) A min. = 320 (mm2)

A = 2 * (Asx1 + Asy1) Asx1 = 223 (mm2) Asy1 = 113 (mm2)

Column pier: 2 Longitudinal reinforcement A = 160 (mm2) A min. = 160 (mm2)

A = 2 * (Asx2 + Asy2) Asx2 = 30 (mm2) Asy2 = 50 (mm2)

8.3.4 Provided reinforcement

8.3.1 Spread footing: Bottom:

8.3.2 Pier

Column pier: 1 Longitudinal reinforcement Along X axis:

2 T 12 l = 1.76 (m) e = 1*-0.12 + 1*0.03

Along Y axis: 2 T 12 l = 1.80 (m) e = 1*-0.15

Transversal reinforcement

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5 R 8 l = 0.45 (m) e = 1*-0.05

Column pier: 2 Longitudinal reinforcement Along X axis:

2 T 12 l = 1.76 (m) e = 1*0.08 + 1*0.03

Along Y axis: 2 T 12 l = 1.80 (m) e = 1*0.05

Transversal reinforcement 5 R 8 l = 0.45 (m) e = 1*0.15


Concrete volume Formwork =

= 0.48 (m3) 2.40 (m2)

Steel T Total weight = 17.58 (kG) Density = 36.63 (kG/m3) Average diameter = 12.0 (mm)

Survey according to diameters:

Diameter

12

Length (m) 1.10

Number:

18

Steel R Total weight = 14.40 (kG) Density = 30.00 (kG/m3)


Diameter Length Number: (m) 8 0.45 10 12 1.76 4 12 1.80 4 12 1.80 4

FAUSTIN PROJECT

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4

00

600

800

00

20

40

60

80

00

20

40

60

80

00

20

40

60

80

00

20

40

60

80

00

.400

.600

.800

1.00

1.20

1.40

1.60

1.80

2.00

2.20

2.40

2.60

2.80

3.00

3.20

3.40

3.60

3.80

4.00

4.20

4.40

4.60

4.80

5.00

.200

.400

.600

.800

1.00

1.20

1.40

1.60

1.80

2.00

2.20

2.40

2.60

2.80

3.00

3.20

3.40

3.60

3.80

4.00

4.20

4.40

4.60

4.80

5.00

.200

.200

9. ANALYSIS & DESIGN OF STAIR

6. DESIGN OF STAIR CASE

The waist is dl = 15cm and riser is H1 = 14.6cm.

Tread is h dl

2

H1 15

2x14.6

28.9cm cos 3

a) Load on stair:

0.7826 3

- Self weight of slab : 1.4 x 0.289 x 1 x 1 x 25 = 10.115 KN/m - Finishes = 1.4x1.6=2.24KN/m - Live load : 1.6 x 5 = 8 KN/m

Total design load: P = 20.36 KN/m

tg = H /L = 175 /220 = 0.9 et = 38.5o

b) Shear forces and moment on stair

1.40 2.20 1.40

40.0

35.0

30.0

25.0

20.0

15.0

10.0 5.00

SHEAR FORCE V m ax = 38.1kN @ 0.00m

13.7

17.3

13.7

13.7

-5.00

-10.0

-15.0

-20.0

-25.0

-30.0

-35.0

I = 10.0E-3 m 4̂ E = 206E6 kPa

DEFLECTIONS D m ax = -.063m m @ 2.50m BENDING MOMENT M m ax = 50.5kNm @ 2.50m

-.005

-.010

-.015

-.020

-.025

-.030

-.035

-.040

-.045

-.050

-.055

-.060

Mmax= 50.5KNm

Vmax =38.1KN

c) Reinforcement design

5.00

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

- Effective depth of tread: h – 2.5cm = 28.9-2.5 = 26.4cm - Parameters corresponding on slab : -

k M max

50.5x1000000

0.0289 0.05 b.d 2 Fcu 1000x(264)2 x25

Z {0.5 (0.25 k / o.9) 1̂/ 2}d 0.95d

renforcementt section on top bars:

A M max

50.5x1000000

460.77mm2 / m S

0.95ZFy 0.95x0.95x264x460

Asmin=0.13xbxh/100=0.13x1000x264=375.7mm2/m

We take 5Ф12/m and As provided is 565.2mm2with spacing 150mm in principle direction and We take 5Ø10/m and As provided is 392.5mm

2 spacing 150mm in second direction.

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7. RECAPTILATION OF REINFORCEMENTS DESIGN RESULTS

7.1.1 COLUMNS SUMMARY

Floor Internal/

Edge

Column dimensions Column reinforcement Stirrups

Square circular

0 Internal 20X20 4 Ф 12 Ф8@20 cm c/c

Edge 20X20 4Ф 12 Ф8@20 cm c/c

1 Internal 20X20 4Ф12 Ф8@20 cm c/c

Edge 20X20 4Ф 12 Ф8@20 cm c/c

7.1.2 FOOTING SUMMARY

Footing type Dimensions Reinforcement in X-

direction

Reinforcement in Y-

direction

Starter column

reinforcement

Type 1 Internal 120x120x30 Ф12@200 mm c/c Ф12@200 mm c/c 4Ф14

Type 2 Edge 110x110x30 Ф12@200 mm c/c Ф12@200 mm c/c 4Ф 12

7.1.3 BEAMS SUMMARY

Beam type Dimensions Top reinforcement (at

support)

Bottom reinforcement

(mid-span)

Stirrups near

support/Mid span

Longitudinal

beams

40x20 4Φ12 4Φ14 Ф8@18cm c/c

Ф8@20cm c/c

Transversal

beams

40x20 4Φ12 4Φ14 Ф8@18cm c/c

Ф8@20cm c/c

7.1.4 SLAB SUMMARY

Element Dimensions Top

x-direction

Top

y-direction

Bottom

x-direction

Bottom

y-direction

Suspended

slab

13cm thick Ф12@18cm c/c Ф12@18cm c/c Ф10@18cm c/c Ф10@18cm c/c

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Done at Rubavu, APRIL 2018

Eng BIRASA Alexis

STUCTURAL ANALYSIS AND DESIGN REPORT

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