Structural Design of a Commercial

Building

Structural
January 2024

Client: MUTABAZI FRANCIS XAVIER

Design Code: BS 8110-1997, BS 8007-1987

& EN1993-1997
Protastructures Software

CONSULTANTS:
beacon ltd 1
NCSE LTD

Done at Kigali, January 2024

 0. INTRODUCTION

The aim of design is the achievement of an acceptable probability that

structures being designed will perform satisfactorily during their intended life.
With an appropriate degree of safety, they should sustain all the loads and
deformations of normal construction and use and have adequate durability and
resistance to the effects of misuse and fire. Once the building form and
structural arrangement have been finalized the design problem consists of the
following:
1. Idealization of the structure into loadbearing frames and elements for
analysis and design
2. Estimation of loads
3. Analysis to determine the maximum moments, thrusts and shears for
design
4. Design of sections and reinforcement arrangements for slabs, beams,
columns and walls using the results from 3.
5. Production of arrangement and detail drawings and bar schedules

This structural design process has been carried out under use of BS8110
design code of practice. Especially, computations have been made by use of
Protostructures 2021 Software.

beacon ltd 2
 I. NOTATIONS

The symbolic notation used in this project is in accordance with the BS code of
practice. Other symbols not defined here, have been defined alongside the
particular place where they have been applied.
A: cross section area Qk: imposed load
Asmin: minimum required S: spacing of shear reinforcement
reinforcement section V: shear force in concrete section
B: width of foundation footing, Øt: shear reinforcement diameter
Beam Ø: reinforcement diameter
b: width reinforced concrete section B.S: British standard
bf: width of flange in a beam C.P: Code of Practice
bw: width of web of a flanged a beam RC: Reinforced concrete
C: cover m.f: modification facto
d: effective depth of tensile
reinforcement
H: depth of foundation
fcu: characteristic yield strength of
concrete at 28 days
fy: characteristic yield strength of
steel
GK: dead load
h: overall depth of a concrete section
hf: thickness of flange in a T-beam
L: span length
lx: short-span length
ly: long-span length
M: bending moment
P: perimeter
qadm: bearing pressure

beacon ltd 3
 II. ASSUMPTIONS

Design standards used

Design standard used to determine section of steel bars of different structural
elements of concerned building are BS 8110.
Units
Volumetric load: kN/m3
Surface load: kN/m2
Linear load: kN/m
Point load: kN
Loads
Floor imposed load (for all floor): 2 kN/m2
Accessible Roof Imposed load: 1.0 kN/ m2
Floor finishes (all floor): 0.25 kN/ m2
Masonry Wall: 2.0 kN/ m2

Cover conditions
Slabs, beams and columns [mild condition]: 20mm
Foundation pads [moderate condition]: 40mm
Soil characteristics
Sandy-gravel subsoil of unit weight: 18kN/m3
Allowable bearing pressure: 300kpa=300KN/m2

Mix proportions [BS 5328-2]

Mix ratio: 350 kg/ m3
Elasticity limit for construction materials
Strength of reinforcement:
 Hot rolled mild steel: 250 N/mm2
 High yield steel (hot rolled or cold worked): 500 N/mm2
 Concrete ƒcu: 16/20 N/mm2
Partial safety magnification factors

beacon ltd 4
  For dead load: 1.4
 For live load: 1.6
Basic span-effective depth ration: 20.8

NCSE LTD 5
 III. SLAB DESIGN

RC SLAB DESIGN
In accordance with EN1992-1-1:2004 incorporating corrigendum January 2008 and the recommended
values
Tedds calculation version 1.0.17

Slab definition
Slab reference name; CRITICAL SLAB
Type of slab; Two way spanning with restrained edges
Overall slab depth; h = 150 mm
Shorter effective span of panel; lx = 4800 mm
Longer effective span of panel; ly = 4900 mm
Support conditions; Four edges continuous (interior panel)
Top outer layer of reinforcement; Long span direction
Bottom outer layer of reinforcement; Long span direction
Loading
Characteristic permanent action; Gk = 3.6 kN/m2
Characteristic variable action; Qk = 1.5 kN/m2
Partial factor for permanent action; G = 1.40
Partial factor for variable action; Q = 1.60
Quasi-permanent value of variable action; 2 = 0.30
Design ultimate load; q = G  Gk + Q  Qk = 7.4 kN/m2
Quasi-permanent load; qSLS = 1.0  Gk + 2  Qk = 4.1 kN/m2

Concrete properties
Concrete strength class; C20/25
Characteristic cylinder strength; fck = 20 N/mm2
Partial factor (Table 2.1N); C = 1.50
Compressive strength factor (cl. 3.1.6); cc = 1.00
Design compressive strength (cl. 3.1.6); fcd = 13.3 N/mm2

NCSE LTD 6
Mean axial tensile strength (Table 3.1); fctm = 0.30 N/mm2  (fck / 1 N/mm2)2/3 = 2.2 N/mm2
Maximum aggregate size; dg = 20 mm
Effective strength factor – exp.3.21;  = 1.00
Effect. compr. zone height factor – exp.3.19;  = 0.80
Ultimate strain - Table 3.1; cu2 = 0.0035
Shortening strain - Table 3.1; cu3 = 0.0035
K1 = 0.44
K2 = 1.25  (0.6 + 0.0014/cu2) = 1.25
Design value modulus of elasticity reinf – 3.2.7(4) Es = 200000 N/mm2
Reinforcement properties
Characteristic yield strength; fyk = 500 N/mm2
Partial factor (Table 2.1N); S = 1.15
Design yield strength (fig. 3.8); fyd = fyk / S = 434.8 N/mm2

Concrete cover to reinforcement

Nominal cover to outer top reinforcement; cnom_t = 30 mm
Nominal cover to outer bottom reinforcement; cnom_b = 30 mm
Fire resistance period to top of slab; Rtop = 60 min
Fire resistance period to bottom of slab; Rbtm = 60 min
Axia distance to top reinft (Table 5.8); afi_t = 10 mm
Axia distance to bottom reinft (Table 5.8); afi_b = 10 mm
Min. top cover requirement with regard to bond; cmin,b_t = 10 mm
Min. btm cover requirement with regard to bond; cmin,b_b = 10 mm
Reinforcement fabrication; Not subject to QA system
Cover allowance for deviation; cdev = 10 mm
Min. required nominal cover to top reinft; cnom_t_min = 20.0 mm
Min. required nominal cover to bottom reinft; cnom_b_min = 20.0 mm
PASS - There is sufficient cover to the top reinforcement
PASS - There is sufficient cover to the bottom reinforcement
Reinforcement design at midspan in short span direction (cl.6.1)
Bending moment coefficient; sx_p = 0.0248
Design bending moment; Mx_p = sx_p  q  lx2 = 4.3 kNm/m
Reinforcement provided; 10 mm dia. bars at 200 mm centres
Area provided; Asx_p = 393 mm2/m
Effective depth to tension reinforcement; dx_p = h - cnom_b - y_p - x_p / 2 = 105.0 mm
K factor; K = Mx_p / (b  dx_p2  fck) = 0.019
Redistribution ratio;  = 1.0
K’ factor; K’ = (2    cc / C)  (1 -   ( - K1) / (2  K2))  (  ( -
K1) / (2  K2)) = 0.196
K < K' - Compression reinforcement is not required
Lever arm; z = min(0.95  dx_p, dx_p/2  [1 + (1 - 2  K / (  cc / C))0.5])
= 99.7 mm
Area of reinforcement required for bending; Asx_p_m = Mx_p / (fyd  z) = 98 mm2/m
Minimum area of reinforcement required; Asx_p_min = max(0.26  (fctm/fyk)  b  dx_p, 0.0013bdx_p) =
137 mm2/m
Area of reinforcement required; Asx_p_req = max(Asx_p_m, Asx_p_min) = 137 mm2/m

NCSE LTD 7
 PASS - Area of reinforcement provided exceeds area required
Check reinforcement spacing
Reinforcement service stress; sx_p = (fyk / S)  min((Asx_p_m/Asx_p), 1.0)  qSLS / q = 59.2
N/mm2
Maximum allowable spacing (Table 7.3N); smax_x_p = 300 mm
Actual bar spacing; sx_p = 200 mm
PASS - The reinforcement spacing is acceptable
Reinforcement design at midspan in long span direction (cl.6.1)
Bending moment coefficient; sy_p = 0.0240
Design bending moment; My_p = sy_p  q  lx2 = 4.1 kNm/m
Reinforcement provided; 10 mm dia. bars at 200 mm centres
Area provided; Asy_p = 393 mm2/m
Effective depth to tension reinforcement; dy_p = h - cnom_b - y_p / 2 = 115.0 mm
K factor; K = My_p / (b  dy_p2  fck) = 0.016
Redistribution ratio;  = 1.0
K’ factor; K’ = (2    cc / C)  (1 -   ( - K1) / (2  K2))  (  ( -
K1) / (2  K2)) = 0.196
K < K' - Compression reinforcement is not required
Lever arm; z = min(0.95  dy_p, dy_p/2  [1 + (1 - 2  K / (  cc / C))0.5])
= 109.2 mm
Area of reinforcement required for bending; Asy_p_m = My_p / (fyd  z) = 87 mm2/m
Minimum area of reinforcement required; Asy_p_min = max(0.26  (fctm/fyk)  b  dy_p, 0.0013bdy_p) =
149 mm2/m
Area of reinforcement required; Asy_p_req = max(Asy_p_m, Asy_p_min) = 149 mm2/m
PASS - Area of reinforcement provided exceeds area required
Check reinforcement spacing
Reinforcement service stress; sy_p = (fyk / S)  min((Asy_p_m/Asy_p), 1.0)  qSLS / q = 52.2
N/mm2
Maximum allowable spacing (Table 7.3N); smax_y_p = 300 mm
Actual bar spacing; sy_p = 200 mm
PASS - The reinforcement spacing is acceptable
Reinforcement design at continuous support in short span direction (cl.6.1)
Bending moment coefficient; sx_n = 0.0323
Design bending moment; Mx_n = sx_n  q  lx2 = 5.5 kNm/m
Reinforcement provided; 10 mm dia. bars at 200 mm centres
Area provided; Asx_n = 393 mm2/m
Effective depth to tension reinforcement; dx_n = h - cnom_t - y_n - x_n / 2 = 105.0 mm
K factor; K = Mx_n / (b  dx_n2  fck) = 0.025
Redistribution ratio;  = 1.0
K’ factor; K’ = (2    cc / C)  (1 -   ( - K1) / (2  K2))  (  ( -
K1) / (2  K2)) = 0.196
K < K' - Compression reinforcement is not required
Lever arm; z = min(0.95  dx_n, dx_n/2  [1 + (1 - 2  K / (  cc / C))0.5])
= 99.7 mm

NCSE LTD 8
 Area of reinforcement required for bending; Asx_n_m = Mx_n / (fyd  z) = 127 mm2/m
Minimum area of reinforcement required; Asx_n_min = max(0.26  (fctm/fyk)  b  dx_n, 0.0013bdx_n) =
137 mm2/m
Area of reinforcement required; Asx_n_req = max(Asx_n_m, Asx_n_min) = 137 mm2/m
PASS - Area of reinforcement provided exceeds area required
Check reinforcement spacing
Reinforcement service stress; sx_n = (fyk / S)  min((Asx_n_m/Asx_n), 1.0)  qSLS / q = 76.8
N/mm2
Maximum allowable spacing (Table 7.3N); smax_x_n = 300 mm
Actual bar spacing; sx_n = 200 mm
PASS - The reinforcement spacing is acceptable
Reinforcement design at continuous support in long span direction (cl.6.1)
Bending moment coefficient; sy_n = 0.0320
Design bending moment; My_n = sy_n  q  lx2 = 5.5 kNm/m
Reinforcement provided; 10 mm dia. bars at 200 mm centres
Area provided; Asy_n = 393 mm2/m
Effective depth to tension reinforcement; dy_n = h - cnom_t - y_n / 2 = 115.0 mm
K factor; K = My_n / (b  dy_n2  fck) = 0.021
Redistribution ratio;  = 1.0
K’ factor; K’ = (2    cc / C)  (1 -   ( - K1) / (2  K2))  (  ( -
K1) / (2  K2)) = 0.196
K < K' - Compression reinforcement is not required
Lever arm; z = min(0.95  dy_n, dy_n/2  [1 + (1 - 2  K / (  cc / C))0.5])
= 109.2 mm
Area of reinforcement required for bending; Asy_n_m = My_n / (fyd  z) = 115 mm2/m
Minimum area of reinforcement required; Asy_n_min = max(0.26  (fctm/fyk)  b  dy_n, 0.0013bdy_n) =
149 mm2/m
Area of reinforcement required; Asy_n_req = max(Asy_n_m, Asy_n_min) = 149 mm2/m
PASS - Area of reinforcement provided exceeds area required
Check reinforcement spacing
Reinforcement service stress; sy_n = (fyk / S)  min((Asy_n_m/Asy_n), 1.0)  qSLS / q = 69.6
N/mm2
Maximum allowable spacing (Table 7.3N); smax_y_n = 300 mm
Actual bar spacing; sy_n = 200 mm
PASS - The reinforcement spacing is acceptable
Shear capacity check at short span continuous support
Shear force; Vx_n = q  lx / 2 = 17.9 kN/m
Effective depth factor (cl. 6.2.2); k = min(2.0, 1 + (200 mm / dx_n)0.5) = 2.000
Reinforcement ratio; l = min(0.02, Asx_n / (b  dx_n)) = 0.0037
Minimum shear resistance (Exp. 6.3N); VRd,c_min = 0.035 N/mm2  k1.5  (fck / 1 N/mm2)0.5  b  dx_n
VRd,c_min = 46.5 kN/m
Shear resistance constant (cl. 6.2.2); CRd,c = 0.18 N/mm2 / C = 0.12 N/mm2
Shear resistance (Exp. 6.2a);
VRd,c_x_n = max(VRd,c_min, CRd,c  k  (100  l  (fck / 1 N/mm2))0.333  b  dx_n) = 49.3 kN/m

NCSE LTD 9
 PASS - Shear capacity is adequate
Shear capacity check at long span continuous support
Shear force; Vy_n = q  lx / 2 = 17.9 kN/m
Effective depth factor (cl. 6.2.2); k = min(2.0, 1 + (200 mm / dy_n)0.5) = 2.000
Reinforcement ratio; l = min(0.02, Asy_n / (b  dy_n)) = 0.0034
Minimum shear resistance (Exp. 6.3N); VRd,c_min = 0.035 N/mm2  k1.5  (fck / 1 N/mm2)0.5  b  dy_n
VRd,c_min = 50.9 kN/m
Shear resistance constant (cl. 6.2.2); CRd,c = 0.18 N/mm2 / C = 0.12 N/mm2
Shear resistance (Exp. 6.2a);
VRd,c_y_n = max(VRd,c_min, CRd,c  k  (100  l  (fck / 1 N/mm2))0.333  b  dy_n) = 52.3 kN/m
PASS - Shear capacity is adequate
Basic span-to-depth deflection ratio check (cl. 7.4.2)
Reference reinforcement ratio; 0 = (fck / 1 N/mm2)0.5 / 1000 = 0.0045
Required tension reinforcement ratio;  = max(0.0035, Asx_p_req / (b  dx_p)) = 0.0035
Required compression reinforcement ratio; ’ = Ascx_p_req / (b  dx_p) = 0.0000
Stuctural system factor (Table 7.4N); K = 1.5
Basic limit span-to-depth ratio (Exp. 7.16);
ratiolim_x_bas = K  [11 +1.5(fck/1 N/mm2)0.50/ + 3.2(fck/1 N/mm2)0.5(0/ -1)1.5] = 32.50
Mod span-to-depth ratio limit;
ratiolim_x = min(1.5, (500 N/mm2/ fyk)  (Asx_p / Asx_p_m))  ratiolim_x_bas = 48.75
Actual span-to-eff. depth ratio; ratioact_x = lx / dx_p = 45.71
PASS - Actual span-to-effective depth ratio is acceptable

Reinforcement sketch
The following sketch is indicative only. Note that additional reinforcement may be required in accordance with
clauses 9.2.1.2, 9.2.1.4 and 9.2.1.5 of EN 1992-1-1:2004 to meet detailing rules.

NCSE LTD 10
 IV. BEAMS DESIGN

Beam Reinforcement Design

Axis: 5 Storey: 1 (/ 2)
Materials: C20/25 / Grade 500 (Type 2) (Links: Grade 500 (Type 2)) Concrete Cover: 25.0 mm

Diagrams

Bending
1B41 L= 700mm 1B40 L= 3700mm 1B39 L= 2200mm
Bw x H (mm) 200 x 400 200 x 400 200 x 400
Flange Bf x Hf --- --- ---
(Left)
(Right)
Top Edge
M (kN.m) 9.9 10.2 13.0 21.1 0.0 21.7 15.8 20.9 39.0
d (mm) 360.0 360.0 360.0 360.0 360.0 360.0 360.0 360.0 360.0
K/K' 0.10 0.10 0.13 0.21 0.00 0.21 0.16 0.21 0.39
x (mm) 40.0 40.0 40.0 40.0 40.0 40.0 40.0 40.0 57.6
As (mm2) 66.82 68.29 87.27 141.89 0.00 145.70 106.07 140.84 268.37
As' (mm2) 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
As,min (mm2) 104.00 104.00 104.00 104.00 104.00 104.00 104.00 104.00 104.00
Bottom Edge
M (kN.m) 0.0 0.0 0.0 6.1 26.1 4.5 0.0 0.0 0.0
d (mm) 360.0 360.0 342.0 342.0 360.0 342.0 342.0 360.0 342.0
K/K' 0.00 0.00 0.00 0.07 0.26 0.05 0.00 0.00 0.00
x (mm) 40.0 40.0 38.0 38.0 40.0 38.0 38.0 40.0 38.0
As (mm2) 0.00 0.00 0.00 42.97 175.60 31.79 0.00 0.00 0.00
As' (mm2) 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
As,min (mm2) 104.00 104.00 104.00 104.00 104.00 104.00 104.00 104.00 104.00

Shear And Torsion Design

Vd (kN) 15.7 20.4 46.0 43.4 11.5 35.5
v (MPa) 0.22 0.28 0.64 0.60 0.16 0.49
vc (MPa) 0.56 0.56 0.56 0.56 0.56 0.56
vMax (MPa) 4.00 4.00 4.00 4.00 4.00 4.00
Vnom (kN) 107.5 107.5 107.5
Td (kN.m) 0.2 0.0 0.0
vt (MPa) 0.00 0.00 0.00
vt,min (MPa) 0.00 0.00 0.00
bsupport (mm) 0.0 0.0 0.0 0.0 0.0 0.0
Links 1H8-200 1H8-200 1H8-200 1H8-200 1H8-200 1H8-200 1H8-200 1H8-200 1H8-200

NCSE LTD 11
Deflection Check
L/d --- 10.28 ≤ ---
61.16 √

Steel Areas (mm2)

Required
Top Edge 66.82 68.29 87.27 141.89 0.00 145.70 106.07 140.84 268.37
Bottom Edge 0.00 0.00 0.00 42.97 175.60 31.79 0.00 0.00 0.00
Supplied
Top Edge 461.81 461.81 461.81 461.81 461.81 461.81 461.81 461.81 461.81
Bottom Edge 461.81 461.81 461.81 461.81 461.81 461.81 461.81 461.81 461.81
Steel Bars
Top Bars 3H14 3H14 3H14
Top.Sup.Bars 3H14 3H14 3H14 3H14 3H14
Top.Sup.Bars
Bottom Bars 3H14 3H14 3H14
Bottom Bars
Bot.Sup.Bars
Side Bars

NCSE LTD 12
Axis: 5 Storey: 1 (/ 2) ...
Materials: C20/25 / Grade 500 (Type 2) (Links: Grade 500 (Type 2)) Concrete Cover: 25.0 mm

Diagrams

Bending
1B38 L= 800mm
Bw x H (mm) 200 x 400
Flange Bf x Hf ---
(Left)
(Right)
Top Edge
M (kN.m) 48.8 26.3 4.5
d (mm) 360.0 361.0 361.0
K/K' 0.48 0.26 0.04
x (mm) 73.8 40.1 40.1
As (mm2) 343.59 0.00 0.00
As' (mm2) 0.00 0.00 0.00
As,min (mm2) 104.00 0.00 0.00
Bottom Edge
M (kN.m) 0.0 7.8 20.3
d (mm) 342.0 360.0 342.0
K/K' 0.00 0.08 0.22
x (mm) 38.0 40.0 38.0
As (mm2) 0.00 0.00 0.00
As' (mm2) 0.00 0.00 0.00
As,min (mm2) 104.00 0.00 0.00

Shear And Torsion Design

Vd (kN) 91.5 83.7
v (MPa) 1.27 1.16
vc (MPa) 0.56 0.56
vMax (MPa) 4.00 4.00
Vnom (kN) 91.4
Td (kN.m) 0.3
vt (MPa) 0.00
vt,min (MPa) 0.00
bsupport (mm) 0.0 0.0
Links 1H8-200

Deflection Check

NCSE LTD 13
 L/d 2.22 ≤
10.17 √

Steel Areas (mm2)

Required
Top Edge 343.59
Bottom Edge 0.00
Supplied
Top Edge 461.81
Bottom Edge 461.81
Steel Bars
Top Bars
Top.Sup.Bars 3H14
Top.Sup.Bars
Bottom Bars 3H14
Bottom Bars
Bot.Sup.Bars
Side Bars

NCSE LTD 14
Axis: 5 Storey: 3 (/ 2)
Materials: C20/25 / Grade 500 (Type 2) (Links: Grade 500 (Type 2)) Concrete Cover: 25.0 mm

Diagrams

Bending
3B41 L= 700mm 3B40 L= 3700mm 3B39 L= 2200mm
Bw x H (mm) 200 x 300 200 x 300 200 x 300
Flange Bf x Hf --- --- ---
(Left)
(Right)
Top Edge
M (kN.m) 3.9 0.7 0.0 0.0 0.0 2.9 0.8 0.6 1.5
d (mm) 260.0 260.0 260.0 260.0 260.0 260.0 260.0 260.0 260.0
K/K' 0.07 0.01 0.00 0.00 0.00 0.05 0.01 0.01 0.03
x (mm) 28.9 28.9 28.9 28.9 28.9 28.9 28.9 28.9 28.9
As (mm2) 36.68 6.46 0.00 0.00 0.00 26.78 7.02 5.23 13.83
As' (mm2) 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
As,min (mm2) 78.00 78.00 78.00 78.00 78.00 78.00 78.00 78.00 78.00
Bottom Edge
M (kN.m) 0.0 3.7 6.2 3.2 3.4 0.2 0.0 0.2 0.0
d (mm) 260.0 260.0 242.0 242.0 260.0 242.0 242.0 260.0 242.0
K/K' 0.00 0.07 0.14 0.07 0.07 0.00 0.00 0.00 0.00
x (mm) 28.9 28.9 26.9 26.9 28.9 26.9 26.9 28.9 26.9
As (mm2) 0.00 34.59 61.82 31.91 32.00 1.78 0.00 1.53 0.00
As' (mm2) 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
As,min (mm2) 78.00 78.00 78.00 78.00 78.00 78.00 78.00 78.00 78.00

Shear And Torsion Design

Vd (kN) 15.0 14.3 2.2 4.9 1.7 2.4
v (MPa) 0.29 0.28 0.04 0.09 0.03 0.05
vc (MPa) 0.53 0.53 0.53 0.53 0.53 0.53
vMax (MPa) 4.00 4.00 4.00 4.00 4.00 4.00
Vnom (kN) 101.2 101.2 101.2
Td (kN.m) 0.0 0.0 0.0
vt (MPa) 0.00 0.00 0.00
vt,min (MPa) 0.00 0.00 0.00
bsupport (mm) 0.0 0.0 0.0 0.0 0.0 0.0
Links 1H8-175 1H8-175 1H8-175 1H8-175 1H8-175 1H8-175 1H8-175 1H8-175 1H8-175

Deflection Check

NCSE LTD 15
 L/d 2.69 ≤ 14.23 ≤ 8.46 ≤
60.57 √ 60.57 √ 60.57 √

Steel Areas (mm2)

Required
Top Edge 36.68 6.46 0.00 0.00 0.00 26.78 7.02 5.23 13.83
Bottom Edge 0.00 34.59 61.82 31.91 32.00 1.78 0.00 1.53 0.00
Supplied
Top Edge 307.88 307.88 307.88 307.88 307.88 307.88 307.88 307.88 307.88
Bottom Edge 307.88 307.88 307.88 307.88 307.88 307.88 307.88 307.88 307.88
Steel Bars
Top Bars 2H14 2H14 2H14
Top.Sup.Bars 2H14 2H14 2H14 2H14 2H14
Top.Sup.Bars
Bottom Bars 2H14 2H14 2H14
Bottom Bars
Bot.Sup.Bars
Side Bars

NCSE LTD 16
Axis: 5 Storey: 3 (/ 2) ...
Materials: C20/25 / Grade 500 (Type 2) (Links: Grade 500 (Type 2)) Concrete Cover: 25.0 mm

Diagrams

Bending
3B38 L= 800mm
Bw x H (mm) 200 x 300
Flange Bf x Hf ---
(Left)
(Right)
Top Edge
M (kN.m) 3.0 1.4 0.1
d (mm) 260.0 260.0 260.0
K/K' 0.06 0.03 0.00
x (mm) 28.9 28.9 28.9
As (mm2) 28.25 13.11 0.57
As' (mm2) 0.00 0.00 0.00
As,min (mm2) 78.00 78.00 78.00
Bottom Edge
M (kN.m) 0.0 0.9 1.8
d (mm) 242.0 260.0 260.0
K/K' 0.00 0.02 0.03
x (mm) 26.9 28.9 28.9
As (mm2) 0.00 7.99 16.88
As' (mm2) 0.00 0.00 0.00
As,min (mm2) 78.00 78.00 78.00

Shear And Torsion Design

Vd (kN) 6.7 5.9
v (MPa) 0.13 0.11
vc (MPa) 0.53 0.53
vMax (MPa) 4.00 4.00
Vnom (kN) 101.2
Td (kN.m) 0.1
vt (MPa) 0.00
vt,min (MPa) 0.00
bsupport (mm) 0.0 0.0
Links 1H8-175 1H8-175 1H8-175

Deflection Check

NCSE LTD 17
 L/d 3.08 ≤
60.57 √

Steel Areas (mm2)

Required
Top Edge 28.25 13.11 0.57
Bottom Edge 0.00 7.99 16.88
Supplied
Top Edge 307.88 307.88 307.88
Bottom Edge 307.88 307.88 307.88
Steel Bars
Top Bars 2H14
Top.Sup.Bars 2H14
Top.Sup.Bars
Bottom Bars 2H14
Bottom Bars
Bot.Sup.Bars
Side Bars

NCSE LTD 18
 V. COLUMNS ANALYSIS AND DESIGN

Column Reinforcement Design

1C10 (F-5) (200/200)
Materials: C20/25 / Grade 500 (Type 2) (Links: Grade 500 (Type 2))

Section

Combinations

No NTop M1 Top M2 Top NBot M1 Bot M2 Bot

(kN) (kN.m) (kN.m) (kN) (kN.m) (kN.m)
1 224.3 -6.3 -0.2 228.3 3.1 0.4
2 197.2 -5.5 0.2 201.3 2.7 0.2
3 160.9 -4.5 -0.6 165.0 2.2 0.5
4 189.0 -5.3 -3.8 192.4 2.6 4.1
5 187.0 -5.2 3.5 190.4 2.6 -3.4
6 190.0 -8.3 0.1 193.4 6.0 0.1
7 186.0 -2.2 -0.4 189.4 -0.8 0.6

Interaction Diagram

Critical Loading: 4 -
(G+Q+Nx)
Min Design

NCSE LTD 19
N 192.4 - 192.4 kN
M11 -5.3 -1.9 -16.5 kN.m
M22 4.1 1.9 0.0 kN.m
NMax 569.6

Concrete Cover = 25.0 mm

(BS8110 - Cl. 3.8.4.5)

N/bhFcu =0.192

Beta =0.78

Shear Rebars
Vd(1/2) = 0.0 / 0.4 kN Slender Column... As (Req): %0.54 214.81 mm2
vc'(1/2) = 0.71 / 0.80 N/mm2 Le1/b1 = 15.2 ≥ 10.0 As (Sup): %1.54 615.75 mm2
v(1/2) = 0.00 / 0.01 N/mm2 Le2/b2 = 15.5 ≥ 10.0
MAdd(1)/2) = 4.6 / 4.4 kN.m 4H14
Links = H8-125

2C10 (F-5) (200/200)

Materials: C20/25 / Grade 500 (Type 2) (Links: Grade 500 (Type 2))

Section

Combinations

No NTop M1 Top M2 Top NBot M1 Bot M2 Bot

(kN) (kN.m) (kN.m) (kN) (kN.m) (kN.m)
1 121.7 -9.5 -0.5 125.7 9.4 0.6
2 107.0 -8.2 0.2 111.0 8.2 0.0
3 87.7 -6.8 -0.9 91.7 6.8 1.0
4 102.3 -7.9 -2.5 105.7 7.9 2.6
5 101.8 -7.9 1.7 105.3 7.9 -1.6
6 102.4 -9.7 -0.2 105.8 9.5 0.3
7 101.7 -6.2 -0.6 105.1 6.4 0.7

Interaction Diagram

NCSE LTD 20
Critical Loading: 6 -
(G+Q+Ny)
Min Design
N 105.8 - 105.8 kN
M11 -9.7 -1.1 -10.6 kN.m
M22 0.3 1.1 0.0 kN.m
NMax 569.6

Concrete Cover = 25.0 mm

(BS8110 - Cl. 3.8.4.5)

N/bhFcu =0.106

Beta =0.87

Shear Rebars
Vd(1/2) = 0.1 / 6.4 kN Short Column... As (Req): %0.40 (min) 160.00 mm2
vc'(1/2) = 0.81 / 0.87 N/mm2 Le1/b1 = 9.2 < 15.0 √ As (Sup): %1.54 615.75 mm2
v(1/2) = 0.00 / 0.17 N/mm2 Le2/b2 = 9.5 < 15.0 √
MAdd(1)/2) = 0.0 / 0.0 kN.m 4H14
Links = H8-125

3C10 (F-5) (200/200)

Materials: C20/25 / Grade 500 (Type 2) (Links: Grade 500 (Type 2))

Section

Combinations

NCSE LTD 21
 No NTop M1 Top M2 Top NBot M1 Bot M2 Bot
(kN) (kN.m) (kN.m) (kN) (kN.m) (kN.m)
1 14.2 -3.9 0.3 18.3 7.0 -0.2
2 12.6 -3.3 0.5 16.7 5.9 -0.6
3 10.8 -3.1 0.0 14.8 5.2 0.3
4 12.1 -3.3 0.0 15.5 5.8 0.1
5 12.1 -3.3 0.6 15.5 5.9 -0.5
6 12.0 -3.6 0.3 15.4 6.0 -0.2
7 12.1 -3.0 0.2 15.6 5.7 -0.1

Interaction Diagram

Critical Loading: 1 - (G+Q)

Min Design
N 18.3 - 18.3 kN
M11 7.0 0.2 7.3 kN.m
M22 0.3 0.2 0.0 kN.m
NMax 569.6

Concrete Cover = 25.0 mm

(BS8110 - Cl. 3.8.4.5)

N/bhFcu =0.018

Beta =0.98

Shear Rebars
Vd(1/2) = 0.4 / 3.6 kN Short Column... As (Req): %0.43 172.63 mm2
vc'(1/2) = 0.72 / 0.73 N/mm2 Le1/b1 = 9.6 < 15.0 √ As (Sup): %1.13 452.39 mm2
v(1/2) = 0.01 / 0.10 N/mm2 Le2/b2 = 9.9 < 15.0 √
MAdd(1)/2) = 0.0 / 0.0 kN.m 4H12
Links = H8-125

NCSE LTD 22
 VI. FOOTINGS ANALYSIS AND DESIGN

FOOTING 1 DESIGN
Tedds calculation version 2.0.07

Pad footing details

Length of pad footing; L = 1700 mm
Width of pad footing; B = 1700 mm
Area of pad footing; A = L  B = 2.890 m2
Depth of pad footing; h = 400 mm
Depth of soil over pad footing; hsoil = 1200 mm
Density of concrete; conc = 23.6 kN/m3

Column details Column A Column B

Column base length; lA = 200 mm; lB = 200 mm
Column base width; bA = 200 mm; bB = 200 mm
Column eccentricity in x; ePxA = 0 mm; ePxB = 0 mm
Column eccentricity in y; ePyA = -400 mm; ePyB = 400 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 = 300 kN/m2
Axial loading on columns Column A Column B
Dead axial load on column; PGA = 22.7 kN; PGB = 35.8 kN
Imposed axial load on column; PQA = 50.0 kN; PQB = 93.8 kN
Wind axial load on column; PWA = 0.0 kN; PWB = 0.0 kN

NCSE LTD 23
Total axial load on column; PA = 72.7 kN; PB = 129.6 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 = 9.440 kN/m2
Soil self weight; Fsoil = hsoil  soil = 24.000 kN/m2
Total foundation load; F = A  (FGsur + FQsur + Fswt + Fsoil) = 96.6 kN

Calculate pad base reaction

Total base reaction; T = F + PA + PB = 298.9 kN
Eccentricity of base reaction in x; eTx = (PA  ePxA + PB  ePxB + MxA + MxB + (HxA + HxB)  h) / T
= 0 mm
Eccentricity of base reaction in y; eTy = (PA  ePyA + PB  ePyB + MyA + MyB + (HyA + HyB)  h) / T
= 76 mm
Check pad base reaction eccentricity
abs(eTx) / L + abs(eTy) / B = 0.045
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) =
75.651 kN/m2
q2 = T / A - 6  T  eTx / (L  A) + 6  T  eTy / (B  A) =
131.194 kN/m2
q3 = T / A + 6  T  eTx / (L  A) - 6  T  eTy / (B  A) =
75.651 kN/m2
q4 = T / A + 6  T  eTx / (L  A) + 6  T  eTy / (B  A) =
131.194 kN/m2
Minimum base pressure; qmin = min(q1, q2, q3, q4) = 75.651 kN/m2
Maximum base pressure; qmax = max(q1, q2, q3, q4) = 131.194 kN/m2
PASS - Maximum base pressure is less than allowable bearing pressure

NCSE LTD 24
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 columns

Ultimate axial load on column A; PuA = PGA  fG + PQA  fQ + PWA  fW = 111.8 kN
Ultimate axial load on column B; PuB = PGB  fG + PQB  fQ + PWB  fW = 200.1 kN

Ultimate foundation loads

Ultimate foundation load; Fu = A  [(FGsur + Fswt + Fsoil)  fG + FQsur  fQ] = 135.3 kN

Ultimate horizontal loading on column A

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 horizontal loading on column B

Ultimate horizontal load in x direction; HxuB = HGxB  fG + HQxB  fQ + HWxB  fW = 0.0 kN
Ultimate horizontal load in y direction; HyuB = HGyB  fG + HQyB  fQ + HWyB  fW = 0.0 kN

Ultimate moment on column A

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

Ultimate moment on column B

Ultimate moment on column in x direction; MxuB = MGxB  fG + MQxB  fQ + MWxB  fW = 0.000 kNm
Ultimate moment on column in y direction; MyuB = MGyB  fG + MQyB  fQ + MWyB  fW = 0.000 kNm

Calculate ultimate pad base reaction

Ultimate base reaction; Tu = Fu + PuA + PuB = 447.2 kN

NCSE LTD 25
Eccentricity of ultimate base reaction in x; eTxu = (PuAePxA+PuBePxB+MxuA+MxuB+(HxuA+HxuB)h)/Tu = 0
mm
Eccentricity of ultimate base reaction in y; eTyu = (PuAePyA+PuBePyB+MyuA+MyuB+(HyuA+HyuB)h)/Tu =
79 mm
Calculate ultimate pad base pressures
q1u = Tu/A - 6TueTxu/(LA) - 6TueTyu/(BA) = 111.586
kN/m2
q2u = Tu/A - 6TueTxu/(LA) + 6Tu eTyu/(BA) = 197.894
kN/m2
q3u = Tu/A + 6TueTxu/(LA) - 6TueTyu/(BA) = 111.586
kN/m2
q4u = Tu/A + 6TueTxu/(LA) + 6TueTyu/(BA) = 197.894
kN/m2
Minimum ultimate base pressure; qminu = min(q1u, q2u, q3u, q4u) = 111.586 kN/m2
Maximum ultimate base pressure; qmaxu = max(q1u, q2u, q3u, q4u) = 197.894 kN/m2
Calculate rate of change of base pressure in x direction
Left hand base reaction; fuL = (q1u + q2u)  B / 2 = 263.058 kN/m
Right hand base reaction; fuR = (q3u + q4u)  B / 2 = 263.058 kN/m
Length of base reaction; Lx = L = 1700 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 = 850 mm
Right hand length; LR = L / 2 - ePxA = 850 mm
Calculate ultimate moments in x direction
Ultimate moment in x direction; Mx = fuL  LL2 / 2 + Cx  LL3 / 6 - Fu  LL2 / (2  L) = 66.279
kNm
Calculate rate of change of base pressure in y direction
Top edge base reaction; fuT = (q2u + q4u)  L / 2 = 336.420 kN/m
Bottom edge base reaction; fuB = (q1u + q3u)  L / 2 = 189.696 kN/m
Length of base reaction; Ly = B = 1700 mm
Rate of change of base pressure; Cy = (fuB - fuT) / Ly = -86.308 kN/m/m
Calculate pad lengths in y direction
Top length; LT = B / 2 - max(ePyA, ePyB) = 450 mm
Middle length; LC = max(ePyA, ePyB) - min(ePyA, ePyB) = 800 mm
Bottom length; LB = B / 2 + min(ePyA, ePyB) = 450 mm
Calculate shear forces in y direction
Shear at top column; ST = fuT  LT + Cy  LT2 / 2 - Fu  LT / B = 106.836 kN
Shear at bottom column; SB = fuT  (LT + LC) + Cy  (LT + LC)2 / 2 - PuB - Fu  (LT + LC) /
B = 53.492 kN
Calculate ultimate moments in y direction
Ultimate positive moment in y direction; My = fuT  LT2 / 2 + Cy  LT3 / 6 - Fu  LT2 / (2  B) = 24.693
kNm
Position of maximum negative moment; Lz = 922 mm

NCSE LTD 26
Ultimate negative moment in y direction; Myneg = fuB  Lz2 / 2 - Cy  Lz3 / 6 - PuB  (Lz - LB) - Fu  Lz2 /
(2  B) - HyuB  h - MyuB
Myneg = -36.383 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 = 500 N/mm2
Nominal cover to reinforcement; cnom = 40 mm
Moment design in x direction
Diameter of tension reinforcement; xB = 14 mm
Depth of tension reinforcement; dx = h - cnom - xB / 2 = 353 mm

Design formula for rectangular beams (cl 3.4.4.4)

Kx = Mx / (B  dx2  fcu) = 0.013
Kx’ = 0.156
Kx < Kx' compression reinforcement is not required
Lever arm; zx = dx  min([0.5 + (0.25 - Kx / 0.9)], 0.95) = 335 mm
Area of tension reinforcement required; As_x_req = Mx / (0.87  fy  zx) = 454 mm2
Minimum area of tension reinforcement; As_x_min = 0.0013  B  h = 884 mm2
Tension reinforcement provided; 10 No. 14 dia. bars bottom (175 centres)
Area of tension reinforcement provided; As_xB_prov = NxB    xB2 / 4 = 1539 mm2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Moment design in y direction
Diameter of tension reinforcement; yB = 14 mm
Depth of tension reinforcement; dy = h - cnom - xB - yB / 2 = 339 mm

Design formula for rectangular beams (cl 3.4.4.4)

Ky = My / (L  dy2  fcu) = 0.005
Ky’ = 0.156
Ky < Ky' compression reinforcement is not required
Lever arm; zy = dy  min([0.5 + (0.25 - Ky / 0.9)], 0.95) = 322 mm
Area of tension reinforcement required; As_y_req = My / (0.87  fy  zy) = 176 mm2
Minimum area of tension reinforcement; As_y_min = 0.0013  L  h = 884 mm2
Tension reinforcement provided; 10 No. 14 dia. bars bottom (175 centres)
Area of tension reinforcement provided; As_yB_prov = NyB    yB2 / 4 = 1539 mm2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Negative moment design in y direction
Diameter of tension reinforcement; yT = 14 mm
Depth of tension reinforcement; dy = h - cnom - yT / 2 = 353 mm

Design formula for rectangular beams (cl 3.4.4.4)

Ky = -Myneg / (L  dy2  fcu) = 0.007
Ky’ = 0.156
Ky < Ky' compression reinforcement is not required
Lever arm; zy = dy  min([0.5 + (0.25 - Ky / 0.9)], 0.95) = 335 mm
Area of tension reinforcement required; As_y_req = -Myneg / (0.87  fy  zy) = 249 mm2

NCSE LTD 27
Minimum area of tension reinforcement; As_y_min = 0.0013  L  h = 884 mm2
Tension reinforcement provided; 6 No. 14 dia. bars top (325 centres)
Area of tension reinforcement provided; As_yT_prov = NyT    yT2 / 4 = 924 mm2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Calculate ultimate shear force at d from right face of column A
Ultimate pressure for shear; qsu = (q1u + Cx  (L / 2 + ePxA + lA / 2 + dx) / B + q4u) / 2
qsu = 154.740 kN/m2
Area loaded for shear; As = B  min(3  (L / 2 - eTx), L / 2 - ePxA - lA / 2 - dx) = 0.675
m2
Ultimate shear force; Vsu = As  (qsu - Fu / A) = 72.838 kN

Shear stresses at d from right face of column A (cl 3.5.5.2)

Design shear stress; vsu = Vsu / (B  dx) = 0.121 N/mm2

From BS 8110:Part 1:1997 - Table 3.8

Design concrete shear stress; vc = 0.79 N/mm2  min(3, [100  As_xB_prov / (B  dx)]1/3) 
max((400 mm / dx)1/4, 0.67)  (min(fcu / 1 N/mm2, 40) / 25)1/3
/ 1.25 = 0.414 N/mm2
Allowable design shear stress; vmax = min(0.8 N/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 A
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 = 134.432 kN/m2
Average effective depth of reinforcement; d = (dx + dy) / 2 = 353 mm
Area loaded for punching shear at column; ApA = (lA)(bA) = 0.040 m2
Length of punching shear perimeter; upA = 2(lA)+2(bA) = 800 mm
Ultimate shear force at shear perimeter; VpuA = PuA + (Fu / A - qpuA)  ApA = 108.275 kN
Effective shear force at shear perimeter; VpuAeff = VpuA = 108.275 kN
Punching shear stresses at face of column A (cl 3.7.7.2)
Design shear stress; vpuA = VpuAeff / (upA  d) = 0.383 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 face of column B
Ultimate pressure for punching shear; qpuB = q1u+[(L/2+ePxB-lB/2)+(lB)/2]Cx/B-[(B/2+ePyB-
bB/2)+(bB)/2]Cy/L = 175.048 kN/m2
Average effective depth of reinforcement; d = (dx + dy) / 2 = 353 mm
Area loaded for punching shear at column; ApB = (lB)(bB) = 0.040 m2
Length of punching shear perimeter; upB = 2(lB)+2(bB) = 800 mm
Ultimate shear force at shear perimeter; VpuB = PuB + (Fu / A - qpuB)  ApB = 194.991 kN
Effective shear force at shear perimeter; VpuBeff = VpuB = 194.991 kN
Punching shear stresses at face of column B (cl 3.7.7.2)
Design shear stress; vpuB = VpuBeff / (upB  d) = 0.690 N/mm2

NCSE LTD 28
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 A and B
Ultimate pressure for punching shear; qpuC1.5d = q1u+[(L/2+(LL+LM/2-L/2)-(LM+(lA+lB)/2)/2-
1.5d)+((LM+(lA+lB)/2)+21.5d)/2]Cx/B-[B/2]Cy/L =
154.740 kN/m2
Average effective depth of reinforcement; d = (dx + dy) / 2 = 353 mm
Area loaded for punching shear at column; ApC1.5d = ((LM+(lA+lB)/2)+21.5d)B = 2.140 m2
Length of punching shear perimeter; upC1.5d = 2B = 3400 mm
Ultimate shear force at shear perimeter; VpuC1.5d = PuA + PuB + (Fu / A - qpuC1.5d)  ApC1.5d = 80.911 kN
Effective shear force at shear perimeter; VpuC1.5deff = VpuC1.5d  1.25 = 101.138 kN

Punching shear stresses at perimeter of 1.5 d from face of column A and B (cl 3.7.7.2)
Design shear stress; vpuC1.5d = VpuC1.5deff / (upC1.5d  d) = 0.084 N/mm2
From BS 8110:Part 1:1997 - Table 3.8
Design concrete shear stress; vc = 0.79 N/mm2  min(3, [100  (As_xB_prov / (B  dx) +
As_yB_prov / (L  dy)) / 2]1/3)  max((800 mm / (dx + dy))1/4,
0.67)  (min(fcu / 1 N/mm2, 40) / 25)1/3 / 1.25 = 0.414 N/mm2
Allowable design shear stress; vmax = min(0.8N/mm2  (fcu / 1 N/mm2), 5 N/mm2) = 4.000
N/mm2
PASS - vpuC1.5d < vc - No shear reinforcement required

NCSE LTD 29
FOOTING 2 DESIGN
Tedds calculation version 2.0.07

Pad footing details

Length of pad footing; L = 1700 mm
Width of pad footing; B = 1700 mm
Area of pad footing; A = L  B = 2.890 m2
Depth of pad footing; h = 400 mm
Depth of soil over pad footing; hsoil = 1200 mm
Density of concrete; conc = 23.6 kN/m3

Column details
Column base length; lA = 200 mm
Column base width; bA = 200 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 = 300 kN/m2
Axial loading on column
Dead axial load on column; PGA = 39.7 kN
Imposed axial load on column; PQA = 90.0 kN

NCSE LTD 30
Wind axial load on column; PWA = 0.0 kN
Total axial load on column; PA = 129.7 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 = 9.440 kN/m2
Soil self weight; Fsoil = hsoil  soil = 24.000 kN/m2
Total foundation load; F = A  (FGsur + FQsur + Fswt + Fsoil) = 96.6 kN

Calculate pad base reaction

Total base reaction; T = F + PA = 226.3 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) =
78.319 kN/m2
q2 = T / A - 6  T  eTx / (L  A) + 6  T  eTy / (B  A) =
78.319 kN/m2
q3 = T / A + 6  T  eTx / (L  A) - 6  T  eTy / (B  A) =
78.319 kN/m2
q4 = T / A + 6  T  eTx / (L  A) + 6  T  eTy / (B  A) =
78.319 kN/m2
Minimum base pressure; qmin = min(q1, q2, q3, q4) = 78.319 kN/m2
Maximum base pressure; qmax = max(q1, q2, q3, q4) = 78.319 kN/m2
PASS - Maximum base pressure is less than allowable bearing pressure

NCSE LTD 31
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 = 199.6 kN

Ultimate foundation loads

Ultimate foundation load; Fu = A  [(FGsur + Fswt + Fsoil)  fG + FQsur  fQ] = 135.3 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 = 334.9 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 - 6TueTxu/(LA) - 6TueTyu/(BA) = 115.875
kN/m2
q2u = Tu/A - 6TueTxu/(LA) + 6Tu eTyu/(BA) = 115.875
kN/m2

NCSE LTD 32
 q3u = Tu/A + 6TueTxu/(LA) - 6TueTyu/(BA) = 115.875
kN/m2
q4u = Tu/A + 6TueTxu/(LA) + 6TueTyu/(BA) = 115.875
kN/m2
Minimum ultimate base pressure; qminu = min(q1u, q2u, q3u, q4u) = 115.875 kN/m2
Maximum ultimate base pressure; qmaxu = max(q1u, q2u, q3u, q4u) = 115.875 kN/m2
Calculate rate of change of base pressure in x direction
Left hand base reaction; fuL = (q1u + q2u)  B / 2 = 196.987 kN/m
Right hand base reaction; fuR = (q3u + q4u)  B / 2 = 196.987 kN/m
Length of base reaction; Lx = L = 1700 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 = 850 mm
Right hand length; LR = L / 2 - ePxA = 850 mm
Calculate ultimate moments in x direction
Ultimate moment in x direction; Mx = fuL  LL2 / 2 + Cx  LL3 / 6 - Fu  LL2 / (2  L) = 42.411
kNm
Calculate rate of change of base pressure in y direction
Top edge base reaction; fuT = (q2u + q4u)  L / 2 = 196.987 kN/m
Bottom edge base reaction; fuB = (q1u + q3u)  L / 2 = 196.987 kN/m
Length of base reaction; Ly = B = 1700 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 = 850 mm
Bottom length; LB = B / 2 + ePyA = 850 mm
Calculate ultimate moments in y direction
Ultimate moment in y direction; My = fuT  LT2 / 2 + Cy  LT3 / 6 - Fu  LT2 / (2  B) = 42.411
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 = 500 N/mm2
Nominal cover to reinforcement; cnom = 40 mm
Moment design in x direction
Diameter of tension reinforcement; xB = 14 mm
Depth of tension reinforcement; dx = h - cnom - xB / 2 = 353 mm

Design formula for rectangular beams (cl 3.4.4.4)

Kx = Mx / (B  dx2  fcu) = 0.008
Kx’ = 0.156
Kx < Kx' compression reinforcement is not required
Lever arm; zx = dx  min([0.5 + (0.25 - Kx / 0.9)], 0.95) = 335 mm
Area of tension reinforcement required; As_x_req = Mx / (0.87  fy  zx) = 291 mm2
Minimum area of tension reinforcement; As_x_min = 0.0013  B  h = 884 mm2

NCSE LTD 33
Tension reinforcement provided; 10 No. 14 dia. bars bottom (175 centres)
Area of tension reinforcement provided; As_xB_prov = NxB    xB2 / 4 = 1539 mm2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Moment design in y direction
Diameter of tension reinforcement; yB = 14 mm
Depth of tension reinforcement; dy = h - cnom - xB - yB / 2 = 339 mm

Design formula for rectangular beams (cl 3.4.4.4)

Ky = My / (L  dy2  fcu) = 0.009
Ky’ = 0.156
Ky < Ky' compression reinforcement is not required
Lever arm; zy = dy  min([0.5 + (0.25 - Ky / 0.9)], 0.95) = 322 mm
Area of tension reinforcement required; As_y_req = My / (0.87  fy  zy) = 303 mm2
Minimum area of tension reinforcement; As_y_min = 0.0013  L  h = 884 mm2
Tension reinforcement provided; 10 No. 14 dia. bars bottom (175 centres)
Area of tension reinforcement provided; As_yB_prov = NyB    yB2 / 4 = 1539 mm2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Calculate ultimate shear force at d from top face of column
Ultimate pressure for shear; qsu = (q1u - Cy  (B / 2 + ePyA + bA / 2 + dy) / L + q4u) / 2
qsu = 115.875 kN/m2
Area loaded for shear; As = L  (B / 2 - ePyA - bA / 2 - dy) = 0.699 m2
Ultimate shear force; Vsu = As  (qsu - Fu / A) = 48.251 kN

Shear stresses at d from top face of column (cl 3.5.5.2)

Design shear stress; vsu = Vsu / (L  dy) = 0.084 N/mm2

From BS 8110:Part 1:1997 - Table 3.8

Design concrete shear stress; vc = 0.79 N/mm2  min(3, [100  As_yB_prov / (L  dy)]1/3) 
max((400 mm / dy)1/4, 0.67)  (min(fcu / 1 N/mm2, 40) / 25)1/3
/ 1.25 = 0.424 N/mm2
Allowable design shear stress; vmax = min(0.8 N/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 = 115.875 kN/m2
Average effective depth of reinforcement; d = (dx + dy) / 2 = 346 mm
Area loaded for punching shear at column; ApA = (lA)(bA) = 0.040 m2
Length of punching shear perimeter; upA = 2(lA)+2(bA) = 800 mm
Ultimate shear force at shear perimeter; VpuA = PuA + (Fu / A - qpuA)  ApA = 196.818 kN
Effective shear force at shear perimeter; VpuAeff = VpuA = 196.818 kN
Punching shear stresses at face of column (cl 3.7.7.2)
Design shear stress; vpuA = VpuAeff / (upA  d) = 0.711 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

NCSE LTD 34
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]Cx/B-[(B/2+ePyA-bA/2-
1.5d)+(bA+21.5d)/2]Cy/L = 115.875 kN/m2
Average effective depth of reinforcement; d = (dx + dy) / 2 = 346 mm
Area loaded for punching shear at column; ApA1.5d = L(bA+21.5d) = 2.105 m2
Length of punching shear perimeter; upA1.5d = 2L = 3400 mm
Ultimate shear force at shear perimeter; VpuA1.5d = PuA + (Fu / A - qpuA1.5d)  ApA1.5d = 54.239 kN
Effective shear force at shear perimeter; VpuA1.5deff = VpuA1.5d  1.25 = 67.798 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) = 0.058 N/mm2

From BS 8110:Part 1:1997 - Table 3.8

Design concrete shear stress; vc = 0.79 N/mm2  min(3, [100  (As_xB_prov / (B  dx) +
As_yB_prov / (L  dy)) / 2]1/3)  max((800 mm / (dx + dy))1/4,
0.67)  (min(fcu / 1 N/mm2, 40) / 25)1/3 / 1.25 = 0.419 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

NCSE LTD 35