Counterfort Design (2018.10.30)

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE

W E IR

CODES

BS5950 & BS2573

Reference

ELEMENT DESIGNED BY

KUSHAN

COUNTERFORT DESIGN CHEC HECKED KED BY

Calculations

1.0 COUNTERFORT DESIGN .

MAHES HESH

REF 2018‐10‐16 Output

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE

WEIR

CODES

ELEMENT

BS5950 & BS2573

DESIGNED BY

Reference

KUSHAN

COUNTERFORT DESIGN CHECKED BY

MAHESH

Calculations Design Information General Soil Data Density of retained soil Angle of internal friction Angle of soil face Ground cover over base at front of wall Bearing capacity of soil Level difference of Ground

s   tc q h'

= = = = = =

18.0 kN/m³ 30.0 ° 0.0 ° 0 mm 200.0 kN/m² 7600 mm

ws = 10.0 kN/m²

Wall details Density of wall base Grade of concrete Characteristic strength of steel,

b = 24.0 kN/m³ f cu = 35.0 N/mm² f y = 460.0 N/mm²

= 1.5

Proportioning of wall componets Height of wall above the base Height of stem Assume Base width (0.6H to 0.7H) Assume Toe projection (b/4) Therefore heel length Assume thickness of vertical wall Assume thickness of base slab Assume c/c spacing between counterforts Assume width of counterforts Therefore clear spacing provided

H = 7600 mm h = 7150 mm b = 4950 mm = 1200 mm lheel = 3400 mm = 350 mm tb = 450 mm = 3027 mm = 400 mm L = 2627 mm

350 mm

    m     m      0      0      6      7    =      '      h

    m     m      0

0.0 °

3400 mm     m     m      0      5      1      7    =      h

2018‐10‐16 Output

Loads on wall and ground Surcharge load (horizontal plan area)

Load factors Ultimate load factor

REF

    m     m      0      0      6      7    =      H

1200 mm

b = 4950mm Height of active soil (at back of stem and heel) hsoil = H + l heel × tan( = 7600 mm

    m     m      7      2      0      3

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE

WEIR

CODES

BS5950 & BS2573

ELEMENT DESIGNED BY

Reference

KUSHAN

COUNTERFORT DESIGN CHECKED BY

MAHESH

Calculations Coefficients using Rankine Active coefficient (Sloping surface)

_  

 

   

 

   

 

1   1  

= 0.333 For a 1m width (Horizontal components ‐ Active Pressure) At virtual back of the wall (back of heel) Surcharge Ps = ka_s × ws Pressure = 3.3 kN/m²

force ‐ horizontal

overturning moment about toe,

Backfill Pressure

force ‐ horizontal

Fs = (ps × hsoil) × cos² = 25.3 kN/m Ms = Fs × hsoil/2 = 96.3 kNm/m

Pa = ka_s × s × hsoil = 45.6 kN/m² Fa = (pa × hsoil/2) × cos ² = 173.28 kN/m

overturning moment about toe,

Ma = Fa × hsoil/3 = 439.0 kNm/m

Total horizontal force (unfactored)

Ft = Fs + Fa = 198.6 kN/m Mt = Ms + Ma =  535.2 kNm/m

Total overturning moment (unfactored)

No 1 2 3 4

Description of loads

2018‐10‐16 Output

= 0.333 Active coefficient (level surface)

REF

Location of line of Loads (kN/m width) action from T 24×0.35×1×7.15 1.2+0.35/2 Weight of stem = 60.06 = 1.375 m 24×0.45×1×4.95 4.95/2 Weight of base = 53.46 = 2.475 m slab 18×3.4×1×7.15 1.2+0.35+3.4/2 Weight of earth = 437.58 = 3.250 m over heel slab 24×0.4×3.4×7.15/(2×3.02 1.2+0.35+3.4/3 Weight of = 38.55 = 2.683 m counterfort Total ΣW = 589.65

Moment about T (kNm/ m width) 82.58 132.31 1422.14 103.44 ΣM = 1740.47

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE CODES

WEIR BS5950 & BS2573

ELEMENT DESIGNED BY

Reference

KUSHAN

COUNTERFORT DESIGN CHECKED BY

MAHESH

Calculations

REF 2018‐10‐16 Output

Stability Check (Unfactored Loads ‐ Active Soil Pressures) Sliding Coefficient of friction

FOS against sliding Factor of safety required

 =   0.9tan = 0.520

 

FOS_reqdsliding = 1.5

Conservatiely, it is assumed that the surcharge load does not prvide downward pressure on the heel in sliding calculation Total horizontal force tending to slide wall Resisting force

= 198.6 kN/m = ΣW = 306.4 kN/m

Therefore FOS against sliding

= ΣW/Ft = 1.543

Overturning FOS against overturning Factor of safety required

OK

  FOS_reqdoverturning = 2

Conservatiely, it is assumed that the surcharge load does not prvide downward pressure on the heel in overturning calculation Overturning moment Restoring moment Therefore FOS against overturning

Bearing pressure under base Net moment

Mt = 535.2 kNm/m ΣM = 1740.5 kNm/m =   ΣM/Mt = 3.252

OK

M =  ΣM ‐ Mt = 1205.2 kNm/m

Let x be the distance from toe where the resultant R acts. Therefore x = M/ΣW = 2043.98 mm b/6 = 825.00 mm e = b/2 ‐ x = 431.02 mm ≤ b/6 Therefore whole base is under compression PA   = (ΣW/b)×(1+6e/b) Maximum pressure at toe = 181.4 kN/m² < 200.0 kN/m² PD   = (ΣW/b)×(1‐6e/b) Minimum pressure at heel = 56.9 kN/m² > 0.0 kN/m² Intensity of pressure at junction of stem with PB   = 56.89+(181.36‐56.89)×3.75/4.95 toe i.e. under B = 151.2 kN/m² Intensity of pressure at junction of stem with PC   = 56.89+(181.36‐56.89)×3.4/4.95 toe i.e. under C = 142.4 kN/m² Middle third rule Eccentricity

OK

OK

OK

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE

WEIR

CODES

BS5950 & BS2573

ELEMENT DESIGNED BY

Reference

KUSHAN

COUNTERFORT DESIGN CHECKED BY

MAHESH

Calculations

REF 2018‐10‐16 Output

350 mm

    m     m      0      0      6      7

1200 mm     m     m      0

3400 mm

      ²     m      /      N      k      4  .      1      8      1

      ²       ²     m     m      /      /      N      N      k      k      0      2  .  .      1      1      6      5      1      1

      ²     m      /      N      k      9  .      6      5

      ²     m      /      N      k      4  .      2      4      1

4950 mm Design of Toe Slab Since the projection of the toe is small, it is designed as a cantilever fixed at the stem. Intensity of base pressure at B = 151.2 kN/m² Neglecting the weight of soi above the toe slab, the forces acting on the toe slab are 1. downward force due to weight of toe slab TB 2. upward soil pressure on length AB

Therefore moment at B

Cover of r/f Dia of base r/f Effective depth

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

MB = 1.5[151.18 × 1.2²/2 + (181.36 ‐ 151.18) × 1.2² × 1/3] = 185.0 kN/m² = 50 mm = T 16 d = tb ‐ cover ‐ bar dia/2 = 392 mm K =   M/(f cud2) = 0.0344 ≤ K' (=0.156) Hence no compresion reinforcement required

OK

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE CODES Reference BS8110:1985 Part ‐ 1 Clause 3.4.4.4

WEIR BS5950 & BS2573

ELEMENT DESIGNED BY

KUSHAN

COUNTERFORT DESIGN CHECKED BY

MAHESH

2018‐10‐16

Calculations

Z

Output 1/2

=   [0.5+(0.25‐K/0.9) ]d = 0.96d = 372 mm

Limiting Z=0.95d,

Z

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

Required steel area to carry bending moment,

As, reqd =   (M/0.87f yZ) = 1241.3 mm²/m

BS8110:1985 Part ‐ 1 Table 3.27

Minimum r/f required in rectangular beams As, min =  0.13 × A c/100 for crack control = 585.0 mm²/m = 125

Therefore suitable spacing

bars

As, prov = π(Ø/2)^2 × No of bars = 1608.0 mm²/m

Provided r/f area

Clear Spacing between adjacent r/f

Clause 3.12.11.1

Minimum spacing of r/f

equation 8

Design service stress in tension r/f 

Clause 3.12.11.2.4

REF

= 125 ‐ 16 = 109 mm

OK

=  Maximum of (h agg + 5 mm) = 25 mm f s

5f yAs, req 1 × 8As, prov βb = 221.9 N/mm² =

=  Min of (47000/f s or 300) = 212 mm Provided Reinfocement and spacing are adequate

Maximum spacing of r/f

Therefore use T16 @ 125 C/C

Check for Shear Since the soil pressure induces compression in the wall the critical section for shear is taken at a distance d from the face of stem. Intensity of pressure at distance d

PE   = 56.89+(181.36‐56.89)(3.75+392)/4.95 = 161.0 kN/m²

Intensity of pressure at E

Shear due to downward force of slab in length of = (181.36+161.04)×0.808/2‐(24×0.45×0.808) = 129.6 kN/m 0.808m (= 1.2 ‐ 0.392) Therefore net ultimate shear V = 1.5×129.6 = 194.40 kN/m Shear stress in section

BS8110:1985 Part ‐ 1 Table 3.9 Conditions to Satisfy BS8110:1985 Part ‐ 1 Table 3.9

= V/d = 0.50 N/mm²

² Minimum of (0.8√ f cu or 5 N/mm ) = → 100A s/bd = = → 400/d = = →

Design shear stress,

vc

2 4.73  N/mm > v Ok (100x1608)/(1000x392) 0.4102 < 3 Ok 400/392 1.020 > 1 Ok

=   0.79[100As/bd]1/3[400/d]1/4[f cu/25]1/3/γm = {0.79[0.4]^(1/3)x[1]^(1/4)x[35/25]^(1/3)}/1.25 = 0.53 N/mm²

Ok

Ok

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE CODES

WEIR BS5950 & BS2573

ELEMENT DESIGNED BY

Reference

KUSHAN

COUNTERFORT DESIGN CHECKED BY

MAHESH

Calculations

REF 2018‐10‐16 Output

Design of Heel Slab

The heel slab is designed as a continuous slab supported on counterforts. The downward force will be maximum at the edge of the slab where intensity of soil pressure is minimum. Therefore conside 1m wide strip near outer edge D The forces acting near edge are 1. Downward weight of soil of height 7.15 m = = 2. Downward weight of heel slab = = 3. Upward soil pressure = Therefore net downward force at D Also net downward force at C

Let the width of counterfort Clear spacing between counterforts Maximum ultimate moment in heel slab at counterfort

p = = = =

128.7 + 15.75 ‐ 56.89 87.56 kN/m 128.7 + 15.75 ‐ 142.38 2.07 kN/m

= L = Mu   = =

400 mm 2627 mm 1.5×87.56×2.627²/12 75.5 kNm/m

= 50 mm = T 12

Cover of r/f Dia of base r/f

d = tb ‐ cover ‐ bar dia/2 = 394 mm

Effective depth

K =   M/(f cud2) = 0.0139 ≤ K' (=0.156) Hence no compresion reinforcement required

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

18×7.15 128.7 kN/m 35×0.45 15.75 kN/m 56.89 kN/m

Z

=   [0.5+(0.25‐K/0.9)1/2]d = 0.98d = 374 mm

Limiting Z=0.95d,

Z

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

Required steel area to carry bending moment,

As, reqd =   (M/0.87f yZ) = 504.3 mm²/m

BS8110:1985 Part ‐ 1 Table 3.27

Minimum r/f required in rectangular beams As, min =  0.13 × A c/100 for crack control = 585.0 mm²/m Therefore suitable spacing Provided r/f area

Clear Spacing between adjacent r/f

Clause 3.12.11.1

Minimum spacing of r/f

OK

= 125

bars

As, prov = π(Ø/2)^2 × No of bars = 905.0 mm²/m = 125 ‐ 12 = 113 mm =  Maximum of (h agg + 5 mm) = 25 mm

OK

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE CODES

WEIR BS5950 & BS2573

ELEMENT DESIGNED BY

Reference equation 8

Clause 3.12.11.2.4

CHECKED BY

MAHESH

Calculations

Design service stress in tension r/f 

f s

Output

Therefore use T12 @ 125 C/C

= V/d = 0.44 N/mm²

² Minimum of (0.8√ f cu or 5 N/mm ) = → 100A s/bd = = → 400/d = = →

vc

Design shear stress,

2 4.73  N/mm > v Ok (100x905)/(1000x394) 0.2297 < 3 Ok 400/394 1.015 > 1 Ok

=   0.79[100As/bd]1/3[400/d]1/4[f cu/25]1/3/γm = {0.79[0.2]^(1/3)x[1]^(1/4)x[35/25]^(1/3)}/1.25 = 0.43 N/mm²

Dia of shear links for base = T 10 Since SF varies linearly along span of 3.027 m, the zone of design shear r/f can be determined Let x1 be the distance from counterfort where SF = 0.43 N/mm² x1   = 1.3135×(0.44‐0.43)/0.44 = 25.00 mm BS8110:1985 Part ‐ 1 Table 3.17

2018‐10‐16

Vumax   = 1.5×87.56×2.627/2 = 172.52 kN/m

Shear stress in section

BS8110:1985 Part ‐ 1 Table 3.9 Conditions to Satisfy

REF

5f yAs, req 1 = × 8As, prov βb = 185.8 N/mm²

=  Min of (47000/f s or 300) = 253 mm Provided Reinfocement and spacing are adequate

Maximum spacing of r/f

Check for shear Maximum shear

BS8110:1985 Part ‐ 1 Table 3.9

KUSHAN

COUNTERFORT DESIGN

No of shear legs rq per 250×250 mm²  

Distribution steel For Base

Dia of distribution bars for base Suitable spacing

≥ ≥ ≥ =

bsv(v‐vc)/[0.87f y(πD²/4)] 250×250×(0.44‐0.43)/(0.87×460)π×10²/4 0.01 1.00

= = = =

0.13/100 × 1000 × 450 585.0 mm²/m T 12 125.0 mm

Ok

Shear links required

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE CODES

WEIR BS5950 & BS2573

ELEMENT DESIGNED BY

Reference

KUSHAN

COUNTERFORT DESIGN CHECKED BY

MAHESH

Calculations

REF 2018‐10‐16 Output

Design of Stem (Vertical Slab) The stem acts as a continuous slab spanning between the counterforts. It is subjected to linearly varing earth pressure having maximum intensity at bottom.

Consider 1m wide strip at bottom of stem at C The intensity of earth pressure. Maximum ultimate moment

ka_sh 42.9 kN/m² 1.5×42.9×2.627²/12 37.0 kN/m²

= 54 mm = T 12

Cover of r/f Dia of stem r/f

d = tb ‐ cover ‐ bar dia/2 = 290 mm

Effective depth

K =   M/(f cud2) = 0.0126 ≤ K' (=0.156) Hence no compresion reinforcement required

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

= = = =

Z

=   [0.5+(0.25‐K/0.9)1/2]d = 0.99d = 276 mm

Limiting Z=0.95d,

Z

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

Required steel area to carry bending moment,

As, reqd =   (M/0.87f yZ) = 335.7 mm²/m

BS8110:1985 Part ‐ 1 Table 3.27

Minimum r/f required in rectangular beams As, min =  0.13 × A c/100 for crack control = 455.0 mm²/m = 125

Therefore suitable spacing Provided r/f area

Minimum spacing of r/f

equation 8

Design service stress in tension r/f 

Clause 3.12.11.2.4

Maximum spacing of r/f

mm

As, prov = π(Ø/2)^2 × No of bars = 905.0 mm²/m

Clear Spacing between adjacent r/f

Clause 3.12.11.1

OK

= 125 ‐ 12 = 113 mm

OK

=  Maximum of (h agg + 5 mm) = 25 mm f s

5f yAs, req 1 × 8As, prov βb = 144.5 N/mm² =

=  Min of (47000/f s or 300) = 300 mm Provided Reinfocement and spacing are adequate

Therefore use T12 @ 125 C/C

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE CODES

WEIR BS5950 & BS2573

ELEMENT DESIGNED BY

Reference

KUSHAN

CHECKED BY

MAHESH

Calculations Check for shear Maximum lateral pressure at stem‐base interface

Output

= V/d = 0.59 N/mm² ²

Minimum of (0.8√ f cu or 5 N/mm ) = → 100A s/bd = = → 400/d = = → vc

Design shear stress,

2 4.73  N/mm > v Ok (100x905)/(1000x290) 0.31207 < 3 Ok 400/290 1.379 > 1 Ok

=   0.79[100As/bd]1/3[400/d]1/4[f cu/25]1/3/γm = {0.79[0.3]^(1/3)x[1.4]^(1/4)x[35/25]^(1/3)}/1.25 = 0.52 N/mm²

Dia of shear links for wall = T 10 Since SF varies linearly along span of 3.027 m, the zone of design shear r/f can be determined Let x1 be the distance from counterfort where SF = 0.52 N/mm² x1   = 1.3135×(0.59‐0.52)/0.59 = 175.00 mm BS8110:1985 Part ‐ 1 Table 3.17

2018‐10‐16

= Ps + Pa = 48.9 kN/m²

Shear stress in section

BS8110:1985 Part ‐ 1 Table 3.9 Conditions to Satisfy

REF

Vumax   = 1.5×48.93×2.627/2 = 172.52 kN/m

Maximum shear (per 1m width)

BS8110:1985 Part ‐ 1 Table 3.9

COUNTERFORT DESIGN

No of shear legs rq per 250×250 mm²  

For stem Dia of distribution bars for stem Suitable spacing

≥ ≥ ≥ =

bsv(v‐vc)/[0.87f y(πD²/4)] 250×250×(0.59‐0.52)/(0.87×460)π×10²/4 0.15 1.00

= = = =

0.13/100 × 1000 × 350 455.0 mm²/m T 10 125.0 mm

Ok

Shear links required

PROJECT STRUCTURE CODES Reference

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT WEIR BS5950 & BS2573

ELEMENT DESIGNED BY

KUSHAN

COUNTERFORT DESIGN CHECKED BY

MAHESH

Calculations

REF 2018‐10‐16 Output

Design of Counterforts Width of Counterfort = 400 mm C/C Spacing of counterforts = 3027 mm They are subjected to earth pressure and downward reaction from heel slab At ane section at any depth h bellow the top E the total horizontal earth pressure acting on the counterfort. =   (1/2sh²k + Ps)×C/C distcnce beteween counterforts = 9.081h² + 10.09 Therefore BM at any depth h = 9.081h² × h/3 + 10.09 × h/2 = 3.027h³ + 5.05h BM at the base at C = 3.027 × 7.15³ + 5.05 × 7.15 = 1142.5 kNm M Ultimate moment u = 1713.8 kNm

Net downward pressure on heel slab at D

=

wt due to earth pressure + wt of heel slab ‐ upward soil pressure + surcharge

= 18 × 7.15 + 24 × 0.45 ‐ 56.89 + 10 = 92.6 kN/m Net downward pressure on heel slab at C = 18 × 7.15 + 24 × 0.45 ‐ 142.38 + 10 = 7.1 kN/m Total downward force at D = 92.61 × 3.027 = 280.3 kN/m Total downward force at C = 7.12 × 3.027 = 18.7 kN/m As mentioned earlier the counterfore acts as a T beam. As can be seen that the depth available is much more than required from BM considerations

    m     m      0      5      1      7

3400 mm Effective cover = 50 mm Dia of counterfort r/f = T 16 The effective depth is taken at right angle to the reinforcement tan   = 7150/3400  = 64.6 ° Effective depth d = 3400 × sinθ ‐ eff cover ‐ bar dia/2 = 3012.5 °

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE CODES

WEIR BS5950 & BS2573

ELEMENT DESIGNED BY

Reference

KUSHAN

CHECKED BY

MAHESH

2018‐10‐16 Output

2

K =   M/(f cud ) = 0.0054 ≤ K' (=0.156) Hence no compresion reinforcement required Z Z

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

Required steel area to carry bending moment,

As, reqd =   (M/0.87f yZ) = 1496.3 mm²

BS8110:1985 Part ‐ 1 Table 3.27

Minimum r/f required in rectangular beams As, min =  0.13 × A c/100 for crack control = 1596.7 mm² = 8

Therefore req no of bars Provided r/f area

Minimum spacing of r/f

equation 8

Design service stress in tension r/f 

Maximum spacing of r/f

bars

As, prov = π(Ø/2)^2 × No of bars = 1608.0 mm²/m

Clear Spacing between adjacent r/f

Clause 3.12.11.1

OK

=   [0.5+(0.25‐K/0.9)1/2]d = 0.99d = 2862 mm

Limiting Z=0.95d,

Clause 3.12.11.2.4

REF

Calculations

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

BS8110:1985 Part ‐ 1 Clause 3.4.4.4

COUNTERFORT DESIGN

= 400 ‐ 50×2 ‐ 10×2 ‐ 10×2 ‐ 16×8 = 132 mm =  Maximum of (h agg + 5 mm) = 25 mm 5f yAs, req 1 × 8As, prov βb = 285.5 N/mm² =  Min of (47000/f s or 300) = 165 mm Provided Reinfocement and spacing are adequate f s

=

Design of Horizontal Ties Due to horizontal earth pressure, the vertical stem has a tendancy of separating out from the counterforts, Hence it should be tied to it by horizontal ties. The direct pull by wall on counterfort for 1m = (Pa + Ps) × C/C distance height at the base. = 148.1 kN Area of steel req to resist direct pull = 1.5 × 148.12 × 10³/(0.87 × 460) = 555.2 mm²/m R/f dia for vertical tie (2 legged) = T 10 Suitable spacing = 100.0 mm Design of Vertical Ties Due to net vertical downward force acting on the base slab, it has a tendancy to separate out from the counterfort. This is prevented by providing vertical ties. The maximum pull will be exerted at the end of = 92.6 kN/m heel slab where net downward force Consider 1m strip Total downward force at D = 92.61 × C/C distance between counterforts = 280.34 kN Required r/f = 1.5 × 280.34 × 10³/0.87 × 460 = 1050.8 mm² R/f dia for vertical tie (2 legged) = T 10 Suitable spacing = 100.0 mm

OK

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE CODES

WEIR BS5950 & BS2573

ELEMENT DESIGNED BY

Reference

KUSHAN

COUNTERFORT DESIGN CHECKED BY

Calculations

8T16

T10 @ 125 C/C

T10 @ 100 C/C

T12 @ 125 C/C

2018‐10‐16 Output

350 mm

T12 @ 125 C/C

MAHESH

REF

8T16 T10 @ 100 C/C

T16 @ 125 C/C T12 @ 125 C/C T12 @ 125 C/C Cross sectional details of wall through the counterfort

350 mm

T12 @ 125 C/C

T10 @ 125 C/C

T16 @ 125 C/C T12 @ 125 C/C T12 @ 125 C/C Cross section between counterfort

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE CODES Reference

BS8007:1987 Table 5.1 BS8007:1987 Table A.1 BS8007:1987 Table A.1 BS8007:1987 Clause A.5 BS8007:1987 Clause 2.2.3.3 BS8007:1987 BS8007:1987 Clause A.3

BS8007:1987 Figure A.1

WEIR BS5950 & BS2573

ELEMENT DESIGNED BY

KUSHAN

MAHESH

2018‐10‐16 Output

Assume continuous constructions without movement joints [Option ‐1] Critical steel ratio, Ratio of tensile strength to average bond strength between concrete and steel, Restraint factor, Maximum allowable crack width,

ρcrit

  = 0.0035

f ct/f b

= 0.67

R

= 0.5

wmax

= 0.3

mm

Thermal Requirement for toe slab Fall in temperature between the hydration peak and ambient, For toe slab (Thickness = 450 mm ) T1 = 40 °C smax Maximum crack spacing, = [f ct/f b][Ф/2ρ] wmax = smaxRα[T1+T2] Maximum crack width, [f ct/f b]Rα[T1+T2]Ф Required steel ratio based on the areas ρ = 2wmax of surface zone, = 0.00030 Ф ρ For Main r/f = 0.0003x16 1 = 0.0048 > ρcrit → Ok ρ2 For distribution steel = 0.0003x12 = 0.0036 > ρcrit → Ok

Thickness of surface zone,

tzone

= 225

Therefore Required minimum steel area of surface zone in Main r/f 

As,req

= 0.0048x225x1000

Provided main r/f 

Therefore Required minimum steel area of surface zone in distribution steel Provided r/f for distribution

BS8007:1987 Clause A.3

CHECKED BY

REF

Calculations Reinforcement to Resist Shrinkage and Thermal Movement in Immature Concrete T2 Fall in temperature due to seasonal variations, = 11 °C α Coefficient of thermal expansion of mature concrete, = 1.0E‐05 oC‐1

= 1080

BS8007:1987

COUNTERFORT DESIGN

As,prov = 1608.5 → Ok

mm

2

mm /m width 2

mm /m width Ok

As,req

= 0.0036x225x1000

As,prov

= 810 = 904.8 → Ok

2

mm /m width 2

mm /m width

Thermal Requirement for heel slab Fall in temperature between the hydration peak and ambient, For heel slab (Thickness = 450 mm ) T1 = 40 °C smax Maximum crack spacing, = [f ct/f b][Ф/2ρ] w Maximum crack width, = smaxRα[T1+T2] max [f ct/f b]Rα[T1+T2]Ф Required steel ratio based on the areas ρ = 2wmax of surface zone, = 0.00030 Ф ρ For Main r/f = 0.0003x12 1 = 0.0036 < ρcrit → Not Ok use 0.0035 ρ2 For distribution steel = 0.0003x12 = 0.0036 < ρcrit → Not Ok use 0.0035

Ok

PROJECT

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT

STRUCTURE CODES Reference BS8007:1987 Figure A.1

WEIR BS5950 & BS2573

Thickness of surface zone,

ELEMENT DESIGNED BY

KUSHAN

Calculations tzone = 225

Therefore Required minimum steel area of surface zone in Main r/f  Provided main r/f 

Therefore Required minimum steel area of surface zone in distribution steel

BS8007:1987 BS8007:1987 Clause A.3

BS8007:1987 Figure A.1

2018‐10‐16

mm

As,prov

= 810 = 904.8 → Ok

2

mm /m width 2

mm /m width Ok

= 0.0036x225x1000

As,prov = 904.8 → Ok

2

mm /m width 2

mm /m width Ok

Thickness of surface zone,

tzone

= 175

Therefore Required minimum steel area of surface zone in Main r/f 

As,req

= 0.0035x175x1000

As,prov

= 612.5 = 904.8 → Ok

Therefore Required minimum steel area of surface zone in distribution steel Provided r/f for distribution

BS8007:1987 Clause A.3

MAHESH

Output

= 0.0036x225x1000

As,req

REF

Thermal Requirement for Vertical stem Fall in temperature between the hydration peak and ambient, For Vertical stem (Thickness = 350 mm ) T1 = 31 °C s [f  /f  ][Ф/2ρ] Maximum crack spacing, = ct b max wmax = smaxRα[T1+T2] Maximum crack width, [f ct/f b]Rα[T1+T2]Ф Required steel ratio based on the areas ρ = 2wmax of surface zone, = 0.00020 Ф ρ1 For Main r/f = 0.0002x12 = 0.0024 < ρcrit → Not Ok use 0.0035 ρ2 For distribution steel = 0.0002x10 = 0.002 < ρcrit → Not Ok use 0.0035

Provided main r/f 

BS8007:1987

CHECKED BY

As,req

= 810 Provided r/f for distribution

COUNTERFORT DESIGN

As,req

mm

2

mm /m width 2

mm /m width Ok

= 0.0035x175x1000 2

= 612.5

mm /m width

As,prov = 628.3 → Ok

mm /m width

2

Thermal Requirement for Counterfort Fall in temperature between the hydration peak and ambient, For Counterfort (Thickness = 400 mm ) T1 = 31 °C smax Maximum crack spacing, = [f ct/f b][Ф/2ρ] wmax = smaxRα[T1+T2] Maximum crack width, [f ct/f b]Rα[T1+T2]Ф Required steel ratio based on the areas ρ = 2wmax of surface zone, 0.00020 = Ф ρ1 For Horizontal r/f = 0.0002x10 = 0.002 < ρcrit → Not Ok use 0.0035 ρ2 For Vertical steel = 0.0002x10 = 0.002 < ρcrit → Not Ok use 0.0035

Ok

PROJECT STRUCTURE CODES Reference BS8007:1987 Figure A.1

POLGAHAWELA, POTHUHERA AND ALAWWA INTEGRATED WATER SUPPLY PROJECT WEIR BS5950 & BS2573

Thickness of surface zone,

ELEMENT DESIGNED BY

KUSHAN

Calculations tzone = 200

Therefore Required minimum steel area of surface zone in Horizontal r/f  Provided Horizontal r/f 

Therefore Required minimum steel area of surface zone in Vertical steel

CHECKED BY

2018‐10‐16

mm

= 0.0035x200x1000

As,prov

= 700 = 785.4 → Ok

As,req

MAHESH

REF Output

As,req

2

mm /m width 2

mm /m width Ok

= 0.0035x200x1000 = 700

Provided vertical r/f 

COUNTERFORT DESIGN

As,prov = 785.4 → Ok

2

mm /m width 2

mm /m width Ok