Sample Design of Trough

18
MANNING FORMULA = Q = V = A = B = D = Z Q=AxV = n = S = P Duty = = r = Vc 1 Acre = sq.m = Ha 1 cubic meter = 1000 Litre 1 litre = 0.001 cu.m 1 litre =0.03529 cu.ft Duty = 2 l/s/Ha = 0.07078 cu.ft/s/Ha = 0.02877 cu.ft/s/Ac Discharge for 1000 Acs = 28.772358 cu.ft/s 1 cusec = 30 Acs 1000 Acs = 33.3333 Cusecs = 0.94462 Cu.m/s 35.2876 = 944.62 l/s STATION B(m) D(m) Z n A(sq.m) P(m) r(m) Kaudulla Branch canal 1.5 0.9 0 0.015 1.350 3.300 0.409 say Free Board= 0.3 m 0.0004 Cement Mortar 0.018 Excavation ea 4048.9292 0.404892921 S Earth Excavated 0.025 Canals of cap Concrete- Average 0.015 Canals of cap Canal Bed slope Wetted Perimeter Hydraulic mean radius Critical Velocity Type of canal Coeff.(n) Rock Excavated 0.035 Bed width V = (r^0.6666 Full Supply depth Vc =0.5477 x Side slope of canal (ss) Rugosity coefficient for the material Critical Veloc Sample canal desi Discharge P =B+2xDx((1 Mean Velocity A =(B+DxZ)xD Mean area r = A/P

description

Design of Trough

Transcript of Sample Design of Trough

Page 1: Sample Design of Trough

MANNING FORMULA

= Q

= V

= A

= B

= D

= Z Q=AxV

= n

= S

= P Duty =

= r

= Vc

1 Acre = sq.m = Ha

1 cubic meter = 1000 Litre

1 litre = 0.001 cu.m

1 litre =0.03529 cu.ft

Duty = 2 l/s/Ha

= 0.07078 cu.ft/s/Ha

= 0.02877 cu.ft/s/Ac

Discharge for 1000 Acs = 28.772358 cu.ft/s

1 cusec = 30 Acs

1000 Acs = 33.3333 Cusecs

= 0.94462 Cu.m/s 35.2876

= 944.62 l/s

STATION B(m) D(m) Z n A(sq.m) P(m) r(m)

Kaudulla Branch canal 1.5 0.9 0 0.015 1.350 3.300 0.409

say Free Board= 0.3 m

0.0004

Cement Mortar 0.018 Excavation earth canals for drains

4048.9292 0.404892921

S

Earth Excavated 0.025 Canals of capacity up to and including 100 cu.ft/s

Concrete- Average 0.015 Canals of capacity over 100 cu.ft/s

Canal Bed slope

Wetted Perimeter

Hydraulic mean radius

Critical Velocity

Type of canal Coeff.(n)Size of Canal

Rock Excavated 0.035

Bed width V = (r^0.6666)x(S^0.5)/n , or V = 1.486x(r^0.6666)x(S^0.5)/n

Full Supply depth Vc =0.5477 x( D^0.64) , or Vc= 0.84x (D^0.64)

Side slope of canal (ss)

Rugosity coefficient for the material Critical Velocity Ratio, CVR =V/Vc = 0.9 to 1.25

Sample canal design

Discharge P =B+2xDx((1+Z^2)^0.5)

Mean Velocity A =(B+DxZ)xD

Mean area r = A/P

Page 2: Sample Design of Trough

Width B=1.5 m

Depth d=0.9 m

FB=0.3 m

Size of trough= 1.5 x 1.2 m

Page 3: Sample Design of Trough

0.02833 cu.ft/s/Ac

Side Slope (ss)

1 on 1.5

1 on 2.0

2.0 on 1

V(sq.m/s)Vc

(sq.m/s)CVR Qd (l/s)

Extent

(Ha)Qr (l/s)

0.735 0.512 1.435 992.000 1000.00 40.000

Excavation earth canals for drains

Canals of capacity up to and including 100 cu.ft/s

Canals of capacity over 100 cu.ft/s

Size of Canal

V = (r^0.6666)x(S^0.5)/n , or V = 1.486x(r^0.6666)x(S^0.5)/n

Vc =0.5477 x( D^0.64) , or Vc= 0.84x (D^0.64)

Critical Velocity Ratio, CVR =V/Vc = 0.9 to 1.25

Sample canal design

P =B+2xDx((1+Z^2)^0.5)

A =(B+DxZ)xD

r = A/P

Page 4: Sample Design of Trough

Refer:BS 8110- Clauses :3.3.1.6, 3.4.12, 3.4.1.6, 3.4.4.4, 3.5.5.2, 3.12.8.4,and 3.4.5.4

tw

D

L

tb

Bedwidth of rough = L

Length of base = Lb

Depth of rough = D

Full supply depth = Df

Weight of wall = Ww

Thickness of wall = tw

Weight of base = Wb

Thickness of base = tb

Weight of concrete = 24 KN/m3

Weight of soil = 2000 Kg/m3

Angle of friction of soil,Ø = 30 degrees

Weight of vehicle = 6.3052 KN/m2 (Not necessary)

Assume the Trough is full of water at critical condition

Consider 1 m -length of Trough

A D

B C

L (m) D (m) Df (m) tw (m)Ww

(KN/m)

Wb

(KN/m)

1.5 1.2 1.28 0.2 5.76 9.12

Water pressure at the bottom of the slab, P1 = 1.9 x 1000 x9.81/1000 = KN/m2

Say the Basic wind speed, V= m/s

And the multiplying factors, S1 = 1.1 S2= 0.7 S3= 1.0

Wind load , wk =0.613 Vs2 N/m

2 and Vs =V.S1S2S3

wk= 0.408 KN/m per m-length

Case:A Flume empty with wind load

Sample Design of Trough

18.64

1.9

33.5

tb (m)

0.2

Lb (m)

Dimension should

be designed

Page 5: Sample Design of Trough

Depth of wind effect Dw= 2.1 m

(+) =

(Wind load) (self weight) (Wind+self weight)

For unit wind load BM at bottom of wall = 1 X Dw2 /2 = 2.2 KNm

For unit dead load BM at mid of slab = 1 XL2/8 = 0.281 KNm

At ulitimate condition, BM at bottom of wall = 1.4 X wk x 2 = KNm

At ultimate condition BM at mid of slab =1.4 X(Wb/L)X1.022

= 2.39 KNm

Resultant BM at mid of slab= 3.65 KNm

A D

B C

1.26 1.26

Case:B

Flume with water and wind

Take partial safety factors as 1.2 for each loading cases

(+) (+) (=)

BM at A, due to water=-[1.2 x(1/2)x1.92x9.81x(1.9/3+0.1)] = KNm

BM at A, due to wind=1.2 x(1/2)x(1.9+0.2)x2.0x0.408 = KNm

Resultant BM at A, due to wind & water = KNm

Design load on slab = 1.2x(0.2x1x1x24)+1.2x1.9x9.81 = KNm

BM at mid of slab =28.127x3.262/8 = KNm

Resultant BM at mid span of slab = KNm

A D

1.02816

3.65

-15.582

1.26

-14.554

28.1268

28.7582

14.2042

Page 6: Sample Design of Trough

-14.554 -14.55404

B C

Case:C Flume only with water

A C

B D

At ultimate condition

Water pressure at the bottom of wall=1.4 X1.9 x9.81= 26.09 KN/m per m- length

BM at bottom of wall =-[(1/2) X 26.09 X1.9 X(0.1+1.9/3) ] =-18.18 KNm

Ultimate load on slab=1.4 X 0.2 X1 X24 +26.09 = 32.81 KN/m

BM at mid of slab= 32.81X 2.862/8 = KNm

Resultant BM at mid of slab = 15.37 KNm

A D

-18.18 -18.18

B C

Case

A

B

C

Take ultimate BM at mid span of slab Mu = KNm

Depth of slab =200 mm

Clear cover to r/f = 40 mm

Dia. Of Tor r/f = 10 mm

Yield strenght of r/f =460 N/mm2

Strength of concrete fcu = 25 N/mm2

Effecive depth d= 155 mm

K= Mu /(b d2).fcu = 0.0256 < 0.156 =K

1

No compression r/f is required

15.37

14.204

33.5466

BM at end of

slab(KNm)

BM at mid of slab

(KNm)

14.2042

15.37

-18.18 15.37

1.26 3.65

-14.55

Page 7: Sample Design of Trough

Z = d [ 0.5 + [0.25-(k/0.9)]0.5

] = 149.49 mm

0.95 d = 147.25 <149.43 hence take Z = 147.25 mm

As =Mu/(0.87fy. Z) = 260.775 mm2

Cross section of 10mm dia. Bar = mm2

Hence No. of bars required =

Use T12@ 250 -B

At the bottom of wall Mu= 18.18 KNm

K= 0.0303 < K1

Z= 149.6 >0.95d mm

Take Z= 147.25 mm

As = 308.49 mm2

No. of bars required = 3.9263

Use T10@ 250 - Inner face of wall

Provide 0.13% of Secondary r/f = 195 mm2 per 1-m length

No.of r/f = 2.4818

Spacing = 402.93

Provide T10 mm dia. Distribution r/f at 300 mm spacing

Check for shear

UDL on slab = 1.4 x0.2 x 1 x 24 +1.6 x 1.9 x 9.81 x1= KN/m per 1-m width

Maximum shear force =36.54*2.86/ 2 = 27.407 KN

Shear stress , v =52.256x1000 /(1000 x 155) = N/mm2

Concrete shear stress, vc =0.79(100As/(bvd))0.33

(400/d)0.25

/rm

As (mm2) bv (mm)

Effective

depth

d(mm)

314.29 1000 155

vc > v, hence shear r/f is not required

Design of wall as beam -simply supported on piers

Wt.of slab = 0.2x2.86 x24/2 = 6.864 KN/m

Wt. of wall = 0.2 x(1.9 +0.2) x1 x 24 = 10.08

Total dead load = 16.944 KN/m

Wt.of water = 1.9 x9.81 x2.86/2 = 26.654 KN/m

Ultimate design load = 1.4 x 16.944 +1.6 x 26.654 = KN/m

Say the length of beam = 6.686 m

KN/m

1.25 0.47

0.17682

rm

66.368

vc (mm2)

36.54

78.5714

3.31896

Page 8: Sample Design of Trough

Max. BM at mid of span = 66.368 x 6.686 2 /8 = KNm

Max. shear force at the end of beam = 66.368 x 6.686/2 = 221.867 KN

Effective depth of beam = 1900+200 -40-10/2 = mm

K =M/bd2 fcu = 0.0176 <0.156

No compression bars are required

Z = 2055 x ( 0.5 +(0.25- 0.0176/9)0.5

) = 2051 mm

0.95 d = 1952.3 < 2051 mm, hence take Z= 1952.25 mm

As = M/(0.87 fy Z) = 474.664 mm2

Use T20 mm bar hence the cross sectional area = mm2

No. of r/f required = 1.5103

Use 2 T20 - B

Asmini = 0.13 x 200 x 2055/100 = 534.3 mm2

Asprovided = 628.57 >534.3 mm2

Max. shear stress at support = 221.87 x 1000/(200x 2055) = 0.540 N/ mm2

100 As /bd =(100 x 628.57/200 x 2055) = mm2

400/d = 400/2055 = 0.1946

Vc =0.79 x (0.1529)0.33

x(0.1946)0.25

/1.25 = N/mm2

0.5Vc = 0.11293466 N/mm2

Vc +0.4 = 0.6259 N/mm2

Hence 0.5Vc < V<Vc +0.4

Max.spacing = 0.75 deff = 1541.3 mm

Spacing of links, Sv < 0.87 fyv Asv /(0.4 bv)

Use T10 links , Asv = 157.143 mm2

Sv < 786.093 mm

Provide T10@400 links

370.851

2055

314.29

0.15294

0.22587

Page 9: Sample Design of Trough

Design of Pier

Refer:BS8110, cl3.8.2.4,3.8.1.6.2,cl3.9.3.7.2,cl3.8.1.3,cl3.8.3.1,cl3.8.4.5,cl3.8.4.3,cl3.12.11.2.3

cl5.2.3.2,cl5.2.3.4,cl5.2.3.3,cl3.4.4.4,cl5.2.7.2.3,cl3.4.5.2,cl3.11.3.2,cl3.12.11.2.7,cl3.4.5.10,cl3,7.6.6

(Dimension should be designed) 0.2

0.95

wk 1.9

0.2

0.95 0.15 2.86

62.698 m MSL 0.35

0.55 0.2

3.66 0.2

0.3

T T

5.0

57.0 m MSL

0.75

0.3 0.3

0.6

1.155

0.955

Page 10: Sample Design of Trough

1.22

1.9

0.2

0.2

0.3

0.75

S S

Actual height of pier, lo =62.698-57.0-0.2 = m

Effective height of pier, le=βlo , where β = 1.3 , hence le = m

For unbraced column , le < 30 x Pt , where Pt = Thickness of pier

Pt > 7.147/30 = m

Say, Pt=600 mm and le/ Pt < > 10 hence the pier is slender

Take the distance between piers = 7.5 m

Weight of flume= (0.2x1.9x2 +(2.86+0.2x2))x7.5x24 = KN

Assume the flume is full of water

Hence the weight of water =7.5x1.9x2.86x9.81 = KN

Say the width of head =1.22m

Hence the weight of pier head =1.22 x0.35x2x0.2X24+1.22x0.2x3.66X24+(3.66X1.22+0.75X0.6)X24/2 KN

= 84.515 KN

Total dead load = 254.16+84.515 = KN

Design load at ultimate condition,N =1.4Gk+1.6Qk =1.4x338.675+1.6x399.81 = KN

Nmax=(0.4xfcu)bd/2=(0.4x25)x750x600/2 = KN

Nmax/2= 2250/2 =1125 KN> 1113.8 KN

Computation of forces due to water current

Force due to water current Pw = KVD2

0.235

5.498

7.147

0.23823

11.9123

338.675

0.15

1113.84

2250

254.16

399.81

Page 11: Sample Design of Trough

Velocity of stream VD=3.232 m/s

Constant K=0.35 , hence Pw= KN/m2

Height of water at stream =HFL- Bed level =61.5 -58.6= 2.9 m

3.656

Pw

2x2.9/3

Force due to water current, fw =3.656x(2.9x0.6 )/2= KN

20% of fw is acting on lateral direction

Hence force on lateral direction= (20/100)x(3.181)= 0.64 KN

Force due to debris= Pd =0.56Vd2 = 0.56x3.232

2= KN/m

2

Assume the average depth of debris=1.2 m

Force due to debris, fD =1.2x0.6x5.85 = KN

Force due to debris on lateral direction =(20/100)x4.212 = KN

1.22

1.9

5.850

4.212

0.8424

3.656

3.181

Page 12: Sample Design of Trough

0.2

0.2

0.3

0.75

61.5 mMSL HFL

fD 0.6

fw 2.3

58.6 mMSL stream Bed 1.93

1.15

P

5.7 mMSL

Y

0.3

b=0.6 m 1.35 X 0.75 X

h=0.75 m

0.3

Y

0.6

Designed ultimate capacity of a section when subjected axial load only,

Nuz=0.45fcuAc +0.87fy Asc

and Madd=Nau=N βaKh =NKh (le/b')2/2000

where, K= (Nuz - N)/(Nuz-Nbal) < or = 1

Take K=1,

and the depth of cross section measured in the plane under consideration, h=1.35 m

Hence, Madd= 1113.8 x 1x1.35x (7.147/0.6)2/2000 = KNm

Self weight of pier =5.0x24x(0.75x0.6+2x0.3x0.3/2) = KN

Self weight of flume = KN

Weight of pier head = KN

Total dead load Gk= KN

254.16

106.67

0.15

84.515

414.28

75.6

Page 13: Sample Design of Trough

Weight of water Qk= KN

Ultimate axial load, N=1.4Gk+1.6Qk =1.4x414.28 +1.6x399.81 =

Application of this load at the edge of the pier is considered to be the critical condition

BM at point P, along Y-Y axis, Minitial= (1.4x(254.16/2)x1.22/2)+(1.6x0.64x(2x2.9/3)+1.15)

+(1.6x0.8424x(1.15+2.9-0.6))

Minitial=

My =Minitial +Madd = 116.306 +106.67 = KNm

MB at point P,along X-X axis due to wind, water current and debris

Mx =1.4x0.408x1.9x7.5x(5+0.3+0.2+0.2+0.95)+1.6x3.181x((2/3)x2.9+1.15)

+1.6x4.212x(1.15+2.3)

Mx = KNm

Say the cover to r/f = 50 mm, hence b'=600-50 =550 mm and h'=750-50=700 mm

My/b' =222.98x106/550 =0.4054 x10

6

Mx/h' =69.82x106/700 =0.0997x10

6< My/b'

N/bhfcu=1219.69x103/(600x750x25) =

β =0.88 +(0.77-0.88)x(0.10842-0.1)/(0.2-0.1) = 0.87

My'=My+β(b'/h')Mx =222.98+0.87x(550/700)x93.07 = KNm

M/bh2 =286.6x10

6/(600x750

2) = N/mm

2

N/bh =1219.69x103/(600x750)= N/mm

2

Since these values are very small nominal r/f can be provided

Hence, 100Asc/bh =0.4, where Asc is the area of steel in compression

Asc=0.4*600*750/100 = mm2

Minimum area required by concrete Ac=N/0.4fcu=1219.96x103/(0.4x25) = mm

2

<750x600= mm2

Ac-required < Ac -provided

Provide r/f for 1% of crushing area, hence Ac = 1x121996/100 = mm2

Diameter of bar=20 mm and cross sectional area = mm2

No of bar required =1219.96/314.286 =

use 4T20@165

Design of Corbal

Self weight of flume = 254.16 KN

2.71042

1800

1219.69

286.60

0.84919

450000

314.28571

121996

KN

116.306 KNm

1219.96

3.88164

399.81

222.98

93.07

0.108417

Page 14: Sample Design of Trough

Weight of water = 399.81 KN

Ultimate load =1.4Gk+1.6Qk =1.4x254.16+1.6x399.81= KN

Self weight of corbal beam =24x[3.66x1.22x0.2 +{3.66x1.22+0.75x0.6}/2] = KN

Ultimate bearing stress=0.4fcu = 0.4x25 = 10 N/mm2

610 av

200

300 300

310

l=1220/2=610 mm

Ultimate design load of beam = KN/m

Reaction at pier =66.368x7.5/2 =

Effective bearing length =(610/2)+100 = mm< 600 mm

Net bearing width = 248.88x103/(405x10) = mm <200 mm (actual bearing width)

Ultimate weight of corbal/area =1.4 x 39.128/(3.66x1.22) = KN/m2

Let the load transmitted at the one third of bearing area from load face

av = (2/3)x(1220/2)-300 = mm

BM at S=(995.5/2)x0.106 +(3.66x0.2x0.31x24)x(0.31/2) +(0.31x0.3/2)x3.66x24x0.31/3 KN

= 54.0277 KNm

h=600 mm, hence d=600-50=550 mm

K=M/bd2fcu =54.028 x10

6/1000x550

2x25 =

Z=550 x[0.5+(0.25-0.0071/0.9)0.5

] = mm

0.95d =0.95x550= 522.5 mm< 545.626 mm

Hence take, Z=522.5 mm

As=M/(0.87fyZ) =54.028x106/(0.87x460x522.5) = mm

2

As-minimum=0.0015bh =0.0015 x1000x600 = mm2

Use T20 mm bars, hence No.of bars required = 900/(πd2/4) =

900

2.864

995.5

39.1277

1220

66.368

248.88 KN

405

61.45185

12.27

545.626

258.38

106.667

0.00714

Page 15: Sample Design of Trough

Use 3T20 at 300 mm spacing

BM at T due to water& flume =(995.5/2)x(0.955/2) = KNm

MB at T due to self weight of corbal =12.27 x1.155x0.235x1.155/2 = KNm

Total BM at T=237.68+2.58 = KNm

K=M/bd2fcu =240.26x10

6/(1000x550

2x25) = <0.156

Hence take Z=0.95d=522.5 mm

As =M/(0.87fyZ) =240.26x106/(0.87x460x522.5) = mm

2

Use 16 mm dia. Bars, No. of bar required =1149/(πd2/4) =

Use 6T16@160

Area of horizontal links As =(Main tension r/f)/2 = (3xπx202/4)/2 = mm

2

Depth of link provided= (2/3)x550 = mm

Use T16@175

Axial load N=1219.69 KN

Shear stress on corbal = 1219.69x103/1220x550 = N/mm

2

Allowable shear stress =0.8 fcu0.5

= >1.818 N/mm2

Design of foundation

0.6

0.3 0.3 2.6

0.3

0.3

0.6 0.6 0.3 0.375 0.3 0.6

3.3

Assume the bearing capacity of soils = KN/m2

Take the width of foundation = 1.8 m

Length of foundation = 2.55 m

Depth of foundation = 0.45 m

Self weight of foundation=2.55x1.8x0.45x24 = KN

Self weight of flume = KN

Weight of water = KN

Self weight of pier = KN

Self weight of pier head = KN

Total dead load on foundation = KN

Imposed load on foundation = KN

0.375

366.7

1.818

4

237.68

0.03177

2.58

240.26

1149.0

5.71235795

471.43

863.657

399.81

399.81

75.6

84.515

254.16

150

49.572

Page 16: Sample Design of Trough

Total service load on foundation = KN

Bearing pressure =1263.47/(1.8x2.55) = KN/m2 >150 KN/m

2 ,not allowable

Required base area =1263.47/150= m2

Say the size of foundation = 2.6x3.3 m2

Hence the bearing pressure =1263.47/(2.6x3.3) = KN/m2 < 150 KN/m

2safe

Ultimate axial load =1.4 Gk +1.6 Qk =1.4*863.657+1.6*399.81 = KN

Moment in X-direction due to water, wind & debris, Mx =93.07 KNm

Moment in Y-direction due to flume load, My=222.98 KNm

B

1.0

A A

0.3 0.3 2.6

0.3

0.3

1.0 0.975 0.3 0.375 0.3 0.975

B

3.3

Stress due to axial load =1848.82/(2.6x3.3) = KN/m2

Bending stress, σx =Mx.y/I =93.07x(3.3/2)/(2.6x3.33/12) = KN/m

2

Maximum resultant moment, σx-x =(215.48+19.72) = KNm

maximum clear cover to r/f = 40 mm

use 16 mm dia. Bar

Hence effective depth = 450-(16/2)-40 = mm

Critical BM at B-B =235.2 x0.675x2.6x0.675/2 = KNm

M/(bd2)= 139.3x10

6/(2600x402

2)=

100 As/bd2 =0.14

As =0.14x2400x402/100 = 1350.7 mm2

As -minimum=0.15 bd %=0.15x2400x402/100= mm2

No of bars required = 1447.2/(πx162/4) =

use 8T16@290

Bending stress σy =My.X/I =222.98x(2.6/2)/(3.3x2.63/12)= KN/m

2

Maximum resultant Moment σy-y =215.48+59.973 = KNm

Critical BM at A-A=275.453x0.3x3.3x0.3/2 = KNm

19.72

235.2

1447.2

139.3

0.460

0.375

402

1848.82

275.453

40.90

7.19

59.973

8.42313

147.25758

275.266

1263.47

215.4802

Page 17: Sample Design of Trough

M/bd2=40.9x10

6/(3300 x402

2) = N/mm

2very small

Hence minimum r/f required =0.15bd% =1447.2 mm2

Use 11T16@290

As-provided =201x11 = 2211 mm2

Check for distribution of r/f

X-direction, c=600 mm -width of column, d-eff.depth

(3c/4)+(9d/4) = 1354.5 > lc= 1000 mm ,distance to the edge of pad

Y-direction,3c/4+9d/4 =3x(750+600)/4+9x402/4 = 1917 > lc=975 mm

Hence the distribution is O/k

Max. spacing between bars =3d=3x402 = mm -O/K

Anchorage length =40φ =40x16 = mm

Distance between column face and of the footing =[3300-(750+600)]/2 & (2600-600)/2

=975 mm & 1000 mm -O/K

Check for vertical line shearX

d

1.5d

1.5d=1.5x402 = mm

2556

1806

603

1206

640

0.07669

Page 18: Sample Design of Trough

X

Stress at section X-X =215.48+59.973x(1.3-0.402)/1.3 = KN/m2

= 0.257 N/mm2

vc=0.79(100As/bvd)0.33

x(400/d)0.25

x1/гm =0.79x(100x2211/3300x402)0.33

x(400/402)0.25

x1/1.25

= N/mm2

safe.

Stress at section Y-Y =215.48+19.72X(1.65-0.402)/1.65 = KN/m2

= 0.23 N/mm2

vc = 0.79x(100x2211/2600x402)0.33x

(400/402)0.25

/1.25 = N/mm2

safe.

Vmax =1848.82x103/(2x(1806+2556))x402 = <0.8fcu0.5

= 4 N/mm2

safe.

Area out side the perimeter =3.3x2.6-2.556x1.806 = m2

Shear force =215.48x3.96 = 853.3 KN

Shear stress =853.3x103/(2x(1806+2556))x402 = < 4 N/mm

2safe0.24331

256.91

18.2884

230.40

19.7114555

0.52717

3.96