Thrust Block Calculations

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DESIGN CONSIDERATION :

STA 0 + 67.904 (N 41301.17 E 62710.408)

BEND: X q = FGL

Hydraulic gradeline ; = m

Surge Pressure

r

s = Bar W1

Pipe outside Diameter; D = = =

Pipe thickness; t = H1 H3

Inverted Level : EL = m =

Finish Ground Level = m H2 P

Weight of the pipe Wr = =

Unit weight of soil; gs = kN/m3 H E

Soil friction coeffiecient m =

Angle of internal friction ; f = B =

2.18 45.88degrees 2.00 1.8 DN1200 45.88 0.5 25 HGL 2.12 26.128 2.284 1224.9 12.45 18 23.54 3.8 W3 W2 kN/m3 mm mm 2.0 124.46kg/m 2.11 FGL 5.50

Unit weight of concrete; gc =

Compressive strenght of concrete; f'c = 2.12

23.54 27.5 24647 MPa kN/m3 l =

Concrete Elastic Modulus Ec =

413

24647

2.0

l =

Yield strenght of steel; fy = MPa L2 =

Elastic Modulus of steel Es = GPa

2 200

413 2.0

2.0 L2 =

Width of the block ; B = m

Height of the block ; H = m q

L1 = m 2 L2 = m 1.2 1.2 2.00 = 22.9 2 2 1.15 1.79 L2 =

Min. Length block of the per direction ; Max. Length block of the per direction ;

Depth of the earth cover; H1 = m L1 =

Height from top of the block to NGL ; H2 = m

1.2 1.2 L1 = 3.79 2.18 1.79

Height from top of pipe to NGL ; H3 = m L1 =

2.1 2.18

Total length of the block ; l = 2.1 m

1.) Determine design presure ; d

d = design (total) pressure = [HGL x Kn/m3 + surge pressure]

= m x kN + x

m3

d =

2.) Determine Thrust force; R (Horizontal bend) AS PER AWWA M 45 7.1

where d= Design (total) pressure A = Area of the Nominal area of the pipe

P = 2dA sin ( q ) = 2 x kN x[( m)2x( p )] x sin

2 m2 ₄

P = kN

m ***Assume block size to be;

B = m H = m L = m l = m

3.) Weight of soil on the block; W1

W1 = gs X h1 X 2L X B

= kN X m X 2( m) X m

W1 = kN

4.) Weight of water and pipe in the block; W2

W2 = gw 2AL + Wr = kN X 2( m2) x( m) + kg X m X m x 2 m m3 _m W2 = kN 2 346.225 kPa 1.79 1.58 26.13 2.28 BarG 100 kPa 1BarG 312.263 203.213 38.838 9.81 9.81 1.4 346.225 45 2 1.22 1.13 2 2 1.58 2.125 1.4 1.58

(

)

1000 1.58 124.46 9.81 sec2 18.00 9.81

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HORIZONTAL BENDS / 5.) Weight of block; W3 = kN [( m) - ( p )( m) 2] 2 x m m3 ₄ W3 = kN 6.) Total weight ; W W = W1 + W2 + W3 = kN + kN + kN m m m W = kN 7.)Friction force ; F F = mW = x kN F = kN m

8.) Passive Pressure earth pressure at the back of the block ; E

E = 1 Cw gs l

2

= 1 x tan2₍ _{45 + 25 )x} _{kN x[(} _m

2 2 m3

E = kN

9.) Total resistance force ; R

R = F + E = kN + kN R = kN 10.) Safety Factor ;F.S. F.S. = R = kN = P kN 2.12 1.22 m)2- ( m)2]x

The calculated factor is larger than 1.5, therefore this block is satisfactory

312.263 203.213 38.838 209.459 451.510 615.96 615.964 1.97 1.58 390.2094 390.209 225.755 (h2 2 -h1 2 ) 18 3.79 1.79 0.5 451.510 225.755 23.54 2 2 209.459 1.58 m)x(

11.) Check base stability

SOIL BEARING CAPACITY 98.100 2.12 l = Pb= kN/m2 2.0 98.100 l = 2.0 L2 = REQUIRED AREA Areq 2.0 = 451.510 kN/m 2.0 L2 = .= W total Soil Bearing kN/m2 q = B = m2 Areq Areq 1.2 45.88 = 451.510 kN/m 98.100 4.603 L2 = 1.2 ACTUAL AREA L1 = Aact. 1.2 L1 = 1.2 = 1 B ( L1 + L2) x 2 2 = 1 x m( m + m ) x 2 2

= m2 since Aact.> Areq. THEREFORE Use Aact.

Aact.

2.00 1.15 2.00

Aact. 6.307

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NET ULTIMATE UPWARD SOIL PRESSURE

= x kN/m x

m2

= kN/m3

ALLOWABLE ULTIMATE SOIL PRESSURE

qult = x X

qult = q ult > q net SAFE !!!!

I.] DESIGN SPECIFICATIONS :

1.1 ACI 318-02 & PCA NOTES ON ACI 318-02

II.] MATERIALS : (Input Data)

2.1 Loads : Unit Weight of concrete, gc = kN/m3

: Unit Weight of soil, gs = kN/m3

2.2 Concrete : Compressive Strength, f'c = MPa

Cover to Main Bars, c = mm

2.3 Deformed Bars : Yield Strength, fy = MPa

Main Bar Diameter, DM = mm

T & S Bar Diameter, DT = mm

2.4 Slab : Length, L = m Width, W = m Slab Thickness, t = m Height of soil, H1 = m III.] CALCULATIONS : FIGURE : DESIGN FORCES : Min. t = ( L / 20 ) [ 0.40 + ( fy / 700 ) ] = ( / [ + ( / ) ] = mm use mm Consider 1m strip : Wt. of slab = = kN/m Surcharge = x x = kN/m Overburden Soil = x x = kN/m2 1.7 6.31 PB 1.7 Wtotal Wtotal Wtotal 1.7 qnet qnet Aact. = 451.51 121.70 166.77 23.540 18.000 27.500 75.000 413.000 16 16 1.612 3.154 0.400 1.790 1612 20 ) 0.40 413 700 79.82 400 ( 0.40 ) ( 3.15 ) ( 23.54 ) 29.69 1.00 m 3.154 18.000 56.763 1.79 3.154 18.000 101.606 1.79 m 1.61 m 2.00 m

L

W

L

Wu

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HORIZONTAL BENDS /

Total design load :

Wu = + = x + x = + = kN/m Mu = Wu * L2 / 12 = x )2 / = kN-m = where : d = t - c - DM/2 øbd2fy = mm 2 x = m = fy / 0.85f'c = rreq'd = [1 - (1-2mx)] / m = ( + ) ( + ) = use r = As req = rbd = mm2 Ab = p (DM) 2 / 4 S = (Ab / As req'd) = mm2 ₌ _mm Min S = mm Max S = ( , ) \ USE : @ TEMPERATURE BARS As = 0.0020bh = x x = mm2 Ab = p (DT) 2 / 4 S = (Ab / Min As) (b) = mm2 = mm Min S = mm Max S = ( , ) \ USE : @

CHECK FOR SHEAR :

Vu = + = kN 1.4 WDL 1.7 WS 1.40 29.69 1.70 158.37 41.57 269.23 310.80 310.80 ( 1.61 101.01 x 317.000 = 101.010 x 10-3 (0.9) ( 1.00 ) ( 0.317 ) 413 2.7043E-03 8 17.668 0.002772 rmax = (0.75) (0.85) ( b ) Mu ( fc' ) (600) = 0.75 (0.85) (0.85) ( 28 ) ( 600 ) fy 600 fy 413 600 413 ( 0.50 ) 878.78 201.062 228.80 100 600 mm 450 mm ø16mm 220mm O.C. 0.0028 ( 1.70 ) 219.49 ø16mm 250mm O.C. ( 29.69 ) ( 1.41 ) ( 0.50 ) ( 1.40 ) ( 158.4 ) ( 1.41 ) ( ø16mm ) 0.002 1000 400.000 800 201.062 251.327 100 2000 mm 450 mm 0.021

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Using 0.50 m thickness of concrete at face of support : Vc = fc bw d = fVc = x = kN therefore : VU < VC ; Safe !!! @ @ = X H B = ø16mm 220mm O.C. ø16mm 250mm O.C. 27.50 ( 1000.00 ) 317.00 ₌ _277.06 _kN 6 6 2.00 m 2 .0 0 m DN1200 45.9 0.85 277.06 235.50