abutment jem bat an

1/39 document.xls,Input

ABUTMENT : Bendung ….

1. Input

Dimension

+ 80.45 HT = 13.05 mB1 = 0.90 mB2 = 0.40 mB3 = 0.40 m

+ 79.00 B4 = 2.25 mB5 = 1.80 mB6 = 4.95 mBT = 9.00 mH1 = 1.45 m

H1 max = 1.45 m+ 75.00 H2 = 10.00 m

H3 = 0.60 mH4 = 1.00 mH5 = 1.00 m

+ 74.00 H6 = 0.40 mH10 = 11.45 m

ho = 8.60 mAbutment width BL = 5.00 m

+ 69.00 Support from Parapet = 0.45 mRh1 = 0.73 mRb1 = 0.45 m

Hw1 = 6.60 m+ 67.40 Hw2 = 7.60 m

Slope = 0.0 slope 1:n

Design Load for ParapetUnit Weight Wheel load of T-Load

Soil 1.80 t/m3 T = 10.0 tonSoil (saturated) 2.00 t/m3 Contact width of T-loadConcrete 2.40 t/m3 a = 0.20 m

Effective width of road = 3.00 mReaction of superstructure Thickness of pavement = 0.05 m

Normal Vn=Rd+Rl 137.96 tonSeismic Ve=Rd 71.27 ton With Impact Plate? Yes

He= 12.83 ton (He=2 kh Rd, for fixed bearing)Type of bearing Movable (He=kh Rd, for movable bearing) Width of Corbel Lp= 0.30 m

Thickness of Impact Plate = 0.30 mSurcharge Load 0.70 t/m2 Length of Impact plate 3.00 m

Soil depth above plate 0.70 mParameters

q : surcharge load (t/m2) 0.70q' : surcharge load (t/m2) (=0) 0.00

unit weight of earth (t/m3) 1.80ground surface angle (degree) 0.00internal friction angle (degree) 30.00friction angle between earth and wall (degree) normal 20.00friction angle between earth and earth (degree) normal 0.00friction angle between earth and wall (degree) seismic 15.00friction angle between earth and earth (degree) seismic 24.20wall angle (degree) 0.00

c : cohesion of soil (t/m2) (do not consider) 0.00kh : 0.18Uc: Uplift coefficient 1.00

f : 0.60N-SPT : 50.00

Qa : Allowable bearing capacity normal t/m2 20.83 ( max. 20.83Qae: Allowable bearing capacity seismic t/m2 31.25 ( max. 31.25 t/m2 for soil foundation)

Normal condition Seismic conditionConcrete Design Strength kgf/m2 175 175Creep strain coefficient (concrete) 0.0035 0.0035Reinforcement concrete

Allowable stressConcrete kgf/m2 60 90Re-Bar kgf/m2 1850 2775Shearing kgf/m2 (concrete) 5.5 8.25

kgf/m2 (stirrup) 14 21Yielding Point of Reinforcement Bar

kgf/cm2 3000 3000Young's modulus (reinforcement bar) 2100000 2100000Young's Modulus Ratio n 24 16

(in case no soil on toe side, input "0")

(Input Yes or No)

(Input Fixed or Movable)

g :w :f :d 1: (=2/3f)d 2:

d E1: (=1/2f)d E2:

b :

Friction Coefficient =Tan f b =

t/m2 for soil foundation)

sc

scassatatma

ssy

ho

H5

H6B3

B6B5B4

HT

H10

H2

H1

H4

H3

B1 B2

BT

Hw1

Hw2

Slope 1:n

Rh1

Rb1


2. Check

2.1 Stability Analysis

Normal condition Seismic conditionOverturning e= -0.32 m e= -0.62 m

BT / 6 = 1.50 m BT / 3 = 3.00 mOK OK

Sliding Fs =Hu / H= 2.65 Fs =Hu / H= 1.262.00 OK 1.25 OK

Settlement Qmax = 13.19 t/m2 Qmax = 10.56 t/m2(bearing capacity) OK OK

Qa = 20.8 t/m2 Qae= 31.3 t/m2

2.2 Structural Analysis

(1) Body Section A-A Section B-BNormal Seismic Normal Seismic

Bar arrangementBack face (tensile bar) 25 25 25 25

(vertical) spacing (mm) 90 90 180 180 As (cm2) 273 273 136 136

Front face (compressive bar) 25 25 25 25 (vertical) spacing (mm) 180 ok 180 ok 180 ok 180 ok

( As' > 0.5 As, cm2 ) 136 >=136.4 136 >=136.4 136 >=68.2 136 >=68.2Hoop bar (horizontal) 16 16

interval (mm) 200 200 Max interval (mm) 300 300

Design dimensions500 500 500 500

Concrete cover : d1(cm) 7.5 7.5 7.5 7.5 d2(cm) 7.5 7.5 7.5 7.5

Effective height (cm) : d-d1(cm) 172.5 172.5 109.7 109.7 Design load Mf (t m) 704.25 1210.07 144.44 235.25

Nd (t) 321.86 255.17 219.11 152.42 S (t) 118.55 240.87 42.67 84.04

Checking of minimum reinforcement barRequired bar (cm2) 258 356 53 84

Checking of allowable stress as rectangular beam as column as rectangular beam as columnCompressive stress kgf/cm2 33 ok 58 ok 22 ok 32 ok

Bending stress kgf/cm2 1676 ok 2212 ok 1050 ok 1194 okkgf/cm2 - 787 ok - 393 ok

Mean shearing stress kgf/cm2 1.54 ok 3.03 ok 3.04 ok 1.66 ok

(2) Footing Toe side Heel side(Normal / Seismic) Normal Seismic

Bar arrangementUpper (tensile bar) 25 25

(bridge axis) spacing (mm) 150 150As1 (cm2) 32.72

(compressive bar) 25(bridge axis) spacing (mm) 300 ok

As2' (cm2, >0.5 As2) 16.36 >=8.18(distribution bar) 19 19 19

spacing (mm) 300 ok 100 ok 100 ok9.45 >=5.45 28.35 >=4.50 28.35 >=10.91

Lower (tensile bar) 25 (bridge axis) spacing (mm) 300

As2 (cm2) 16.36(compressive bar) 25 25(bridge axis) spacing (mm) 300 ok 300 ok

As1' (cm2, >0.5 As1) 16.36 >=4.50 16.36 >=16.36(distribution bar) 19 19 19

spacing (mm) 150 ok 200 ok 200 okAso (cm2, >As /3) 18.90 >=5.45 14.18 >=5.45 14.18 >=5.45

Design dimensions100 100 100

Concrete cover : d1(cm) 7.5 7.5 7.5d2(cm) 7.5 7.5 7.5

Effective height (cm) : d-d1(cm) 92.5 92.5 92.5Design load Mf 27.63 128.15 75.76

Nd 0.00 0.00 0.00 S 24.70 56.47 38.67

Checking of minimum reinforcement barRequired bar (cm2) 11.47 53.18 20.20

Checking of allowable stressCompressive stress 12.58 ok 44.33 ok 30.68 ok

Bending stress 1,187.67 ok 2,825.55 Check 1,643.98 okMean shearing stress 1.74 ok 4.07 ok 2.75 ok

(e < BT/6) (e < BT/3)

Fs > Fs >

Qmax < Qa Qmax < Qa

f (mm)

f (mm)

f (mm)

Effective width (whole width) (cm)

scssss'tm

f (mm)

f (mm)

f (mm)

Aso (cm2, >As /3)f (mm)

f (mm)

f (mm)

Effective width (unit width) (cm)


(3) Parapet With Impact Plate Without Impact Plate With Impact PlateNormal Normal Seismic

Bar arrangementBack face (tensile bar) 12 16

(vertical) spacing (mm) 250 250As1 (cm2) 8.04

(compressive bar) 16(vertical) spacing (mm) 250 ok

8.04 >=8.04(distribution bar) 12 12 12(horizontal) spacing (mm) 250 250 ok 250 ok

4.52 >=4.50 4.52 >=4.50Front face (tensile bar) 16

(vertical) spacing (mm) 12516.08

(compressive bar) 12 16(vertical) spacing (mm) 125 ok

16.08 >=4.50(distribution bar) 12 16 16 (horizontal) spacing (mm) 250 ok 250 ok

8.04 >=5.36 8.04 >=5.36Design dimensions

100 100Concrete cover of fronf face (cm) 7Concrete cover of back face (cm) 10 10

Effective height (cm) 33 30Design load Mf (t m) 5.189 1.570

Nd (t) 0.000 0.000S (t) 0.000 1.599

Checking of minimum reinforcement barRequired bar (cm2) 10.78 3.14

Checking of allowable stressCompressive stress 19.04 ok 12.92 ok

Bending stress 1239.88 ok 722.95 okMean shearing stress 0 ok 0.59 ok

(4) Impact Plate and CorbelImpact Plate Corbel

Upper Lower Upper LowerBar arrangement

(main bar) 16 19 19 16spacing (mm) 300 ok 150 300 300 ok

As1 (cm2) 6.70 >=4.50 18.90 9.45 6.70 >=4.50(distribution bar) 16 12 12 12

spacing (mm) 300 ok 250 ok 250 ok 250 ok6.70 >=4.50 4.52 >=4.50 4.52 >=4.50 4.52 >=4.50

Design dimensions100 100 100 100

Concrete cover (cm) 5 5 7 7Effective height (cm) 25 25 23 23

Design load Mf (t m) 5.818 3.146Nd (t) - -S (t) - -

Checking of minimum reinforcement barRequired bar (cm2) 14.73 8.66

Checking of allowable stressCompressive stress 48.88 ok 37.88 ok

Bending stress 1447.22 ok 1642.35 okMean shearing stress - -

f (mm)

f (mm)

As1 (cm2, >As3/2)f (mm)

As2 (cm2, >As1/3)f (mm)

As3 (cm2, >As1/2)f (mm)

As3 (cm2, >As1/2)f (mm)

As6 (cm2, >As3/3)


f (mm)

f (mm)

As2 (cm2, >As1/6)


4/39 document.xls,Stability

DIMENSIONS OF ABUTMENT

Abutment TypeSuper Structure Type

Super structure Type T-Beam

Dimensions AbutmentHT B1 B2 B3 B4 B5 B6 BT H1 H2 H3 H4 H5 H6 H10 ho Width BL

13.05 0.90 0.40 0.40 2.25 1.80 4.95 9.00 1.45 10.00 0.60 1.00 1.00 0.40 11.45 8.60 5.00 H1 max 1.45

Hw1 Hw2 Rh1 Rb1 Width of impact plate (Corbel), Lp= 0.30 m Thickness of impact plate Ft=

6.60 7.60 0.73 0.45

WEIGHT OF ABUTMENT LOAD AND MOMENT

Area unit Vertical DistanceNo. weight Load X Y

m2 t/m3 t/m m mBody 1 0.580 2.40 1.392 3.350 12.325

2 2.520 2.40 6.048 2.700 10.900

1.45 3 0.400 2.40 0.960 3.350 11.100

4 0.080 2.40 0.192 3.283 10.133

5 7.740 2.40 18.576 2.700 5.900

5' 3.870 2.40 9.288 3.450 4.467

6 0.675 2.40 1.620 1.500 1.200

10.00 7 1.080 2.40 2.592 3.150 1.300

1: 0 8 1.485 2.40 3.564 5.700 1.200

9 9.000 2.40 21.600 4.500 0.500

2.60 10 0.090 2.40 0.216 3.700 12.600

13.05 11.45 11 0.045 2.40 0.108 3.650 12.350

0.00 0.00 Sub-total 27.565 66.156

10.00 Total 137.83 330.780

6.00 8.60 SoilToe 12 0.000 1.80 0.000 1.500 0.000

0.00 13 0.000 1.80 0.000 1.125 1.600

14 0.000 1.80 0.000 0.750 1.400

13' 0.000 2.00 0.000 1.125 1.600

14' 0.675 2.00 1.350 0.750 1.400 6.60 0.60 Sub-total 0.675 1.350

Total 3.375 6.750

1.00 7.60 Heel 10 0.090 0.00 0.000 3.700 12.350

9.00 11 0.045 0.00 0.000 3.650 12.350 15 4.632 1.80 8.339 6.275 12.900

16 1.425 1.80 2.565 6.425 12.600

17 1.425 1.80 2.565 6.425 12.300

2.25 1.80 4.95 18 0.045 1.80 0.081 3.750 12.250

19 5.450 1.80 9.810 6.275 11.175

20 2.180 1.80 3.924 6.275 10.400

Unit Weight 21 0.080 1.80 0.144 3.417 10.333

Soil 1800 kg/m3 22 12.870 1.80 23.166 6.525 8.900

Soil (saturated) 2000 kg/m3 Total Surcharge Load (t) 20.475 22' 1.720 1.80 3.096 4.317 7.333

Concrete 2400 kg/m3 Acting Point (m) 6.725 23 29.700 2.00 59.400 6.525 4.600

23' 1.485 2.00 2.970 7.350 1.200

Sub-total 61.15 116.06

Total 305.74 580.30

Canal Slope 0 Slope 1:nReaction Body 3.572 m 4.1 mNormal Vn= Rd+Rl= 137.96 ton Hn=0 Soil Toe 0.75 m 1.4 mSeismic Ve= =Rd= 71.27 ton Heel 7.617 m 8.733 m

He= 12.83 tonSurcharge Load : qs= 0.70 t/m2

Xo=SMx/SV Yo=SMy/SV

ho

H5

H6B3

B6B5B4

HT

H10

H2

H1

H4

H3

B1 B2

BT

Rh1

Rb1

B1 B2

ho

H5

H6

B3

B6B5B4

HT H10

H2

H1

BT

Bimp

Hw1

Hw2

1

2

1

3

4

5

5'

6

9

8

10

11

12

13

14

15

16

17 18

19

20 21

22'

22

23

7


EARTH PRESSUREqs Parameters

qs : surcharge load (t/m2)q' : surcharge load (t/m2) (=0)

unit weight of earth (t/m3)1.45 unit weight of earth (t/m3)

ground surface angle (degree)internal friction angle (degree)friction angle between earth and wall (degree) normalfriction angle between earth and earth (degree) normalfriction angle between earth and wall (degree) seismic

11.45 friction angle between earth and earth (degree) seismic13.05 wall angle (degree)

10.00 c : cohesion of soil (t/m2) (do not consider)kh :

ground surface angle (radian)

internal friction angle (radian)6.60 7.60 friction angle between earth and wall (radian) normal

0.60 friction angle between earth and earth (radian) normal

1.00 friction angle between earth and wall (radian) seismicfriction angle between earth and earth (radian) seismicwall angle (radian)tan-1 kh (radian)

9.00 Uc: uplift coefficient

Earth Pressure

Normal Condition Seismic Condition

Description DescriptionPa : active earth pressure (t/m2/m) 56.061 Pea : active earth pressure (t/m2/m)Ka 1: coefficient of active earth pressure, earth and earth 0.333 Kea1 : coefficient of active earth pressureKa 2: coefficient of active earth pressure, wall and earth 0.297y : Vertical acting point 4.406 y : Vertical acting point X: Horizontal acting point = BT 9.0 X: Horizontal acting point = BT

0

56.061

qa1 = 0.333 x 0.70 = 0.233 qa1 = 0.438 x 0.00 = 0.000 qa2 = 0.333 x 5.45 x 1.80 = 3.270 qa2 = 0.438 x 5.45 x 1.80 = 4.301 qa3 = 0.333 x 7.60 x 2.00 = 5.067 qa3 = 0.438 x 7.60 x 2.00 = 6.664

qw2 = -6.600 x 1.00 = -6.600 qw2 = -6.600 x 1.00 = -6.600 qu1 = -6.60 x 1.0 = -6.60 qu1 = -6.60 x 1.0 -6.60 qu2 = -7.60 x 1.0 = -7.60 qu2 = -7.60 x 1.0 -7.60

Normal condition Seismic condition

No. Description H Y HY No. Description HPa1 = 0.233 x 13.050 = 3.045 6.525 19.869 Pa1= 0.000 x 13.050 = 0.000 Pa2 = 3.270 x 5.450 x 0.500 = 8.911 9.417 83.910 Pa2 = 4.301 x 5.450 x 0.500 = 11.719 Pa2' = 3.270 x 7.600 = 24.85 3.800 94.438 Pa2' = 4.301 x 7.600 = 32.685 Pa3 = 5.067 x 7.600 x 0.500 = 19.253 2.533 48.775 Pa3 = 6.664 x 7.600 x 0.500 = 25.321

Total (Pa) 56.061 4.406 246.991 Total 69.725 Pw2 = -6.600 x 6.600 x 0.500 = -21.780 2.200 -47.916 Pw2 = -6.600 x 6.600 x 0.500 = -21.780 Total -21.780 2.200 -47.916 Total -21.780

No. Description V X VX No. Description VPu1 = -6.600 x 9.000 x 0.500 = -29.700 3.000 -89.100 Pu1 = -6.60 x 9.000 x 0.500 = -29.700 Pu2 = -7.600 x 9.000 x 0.500 = -34.200 6.000 -205.200 Pu2 = -7.60 x 9.000 x 0.500 = -34.200 Total -63.900 4.606 -294.300 Total -63.900

g :g sat:w :f :d 1:d 2:d E1:d E2:b :

w :

f :d 1:

d 2:

d E1:d E2:b :a :

d E: internal friction angle seismic condition f/2(radian)

PV : Vertical Pressure = Pa sin d PV : Vertical Pressure = Pea sin dE

PH : Horizontal Pressure = Pa cos d PH : Horizontal Pressure = Pea cos dE

HTH10

H2

H1

H4

H3

BT

q

qa1

qa2 qa3qw2qu1

qu2


STABILITY ANALYSIS

Case 1 Normal Condition

1 Moment and Acting Point

Description V Load HLoad Distance (m) Moment (t.m)V (t) H (t) X Y Mx My

Body 330.78 3.57 4.10 1181.55 0.00 Combined Acting PointSoil Toe 6.75 0.75 1.40 5.06 0.00 4.823 m

Heel 580.30 7.62 8.73 4420.13 0.00

Reaction (bridge) 137.96 0.00 2.70 11.60 372.50 0.00 Eccentric DistanceEarth Pressure 0.00 280.31 9.00 4.41 0.00 1234.95 e=(BT/2-Xo) -0.323 mHydrostatic pressure 0.00 (108.90) 2.20 (239.58)

Uplift pressure (319.50) 4.61 (1471.50) -244.61 t.mSurcharge Load 20.47 0.00 6.73 0.00 137.69 0.00

S 756.76 171.41 4645.43 995.37

2 Stability Analysis

2.1 Over Turnng

e<=(BT/6) e= -0.323 m BT/6 is larger than e? OKBT/6= 1.500 m

2.2 Sliding

0.60Shearing registance force at base

454.06 tfSafety Rate against slidingFs=Hu/H 2.649 Fs is larger than 2.0 ? OK

2.3 Settlement

Allowale bearing capacity Qa 20.83 tf/m2Reaction from the Foundation

Maximiun Fondation Reaction Q maxQ max = 13.193 tf/m2 Qmax is smaller than Qa? OK

Minimum Fondation Reaction Q maxQ min = 20.441 tf/m2

Case 2 Seismic Condition

1 Moment and Acting Point

Description V Load HLoad Distance (m) Moment (t.m)V (t) H (t) X Y Mx My

Body 330.78 59.54 3.57 4.10 1181.5 244.12 Combined Acting PointSoil Toe 6.75 0.75 1.40 5.06 0.00 5.123 m

Heel 580.3 104.45 7.62 8.73 4420.1 912.2

Reaction (bridge) 71.27 12.83 2.70 12.33 192.43 158.12 Eccentric DistanceEarth Pressure 142.91 317.99 9.00 4.28 1286.2 1362.3 e=(BT/2-Xo) -0.623 mHydrostatic pressure (108.90) 2.20 (47.92)

Uplift pressue (319.50) 4.61 (294.30) -505.98 t.mSurcharge Load 0.00 0.00 9.00 4.41 0.00 0.00

S 812.5 385.91 6791 2628.8

2 Stability Analysis

2.1 Over Turnng

e<=(BT/3) e= -0.623 m BT/3 is larger than e? OKBT/3= 3.000 m

2.2 Sliding

0.6Shearing registance force at base

487.51 tfSafety Rate anaginst slidingFs=Hu/H 1.263 Fs is larger than 1.25 ? OK

2.3 Settlement

Allowale bearing capacity Qa 31.25 tf/m2Reaction from the Foundation e > BT/6 ?

in case e<BT/6 in case e>BT/6

X=3(BT/2-e) = 15.368 m

Maximiun Fondation Reaction Q max Foundation Reaction QeQ max = 10.560 tf/m2 Q=2V/(BL.X)= 21.148 tf/m2

Minimum Fondation Reaction Q maxQ min = 25.552 tf/m2

Qmax is smaller than Qa? OK Qe is smaller than Qa? OKQmax = 10.56 tf/m2Qmin = 25.55 tf/m2X = 9.00 m

Xo=(SMx-SMy)/SV

Bending Moment M =SV x e


Hu=V . Tan f b

Xo=(SMx-SMy)/SV

Bending Moment M =SV x e


Hu=V . Tan f b

BT

e

Q max

VH

BT

e

Q

VH

X

Qmax (min )=V

BL⋅BT±

6 M

BL⋅BT 2

Qmax (min )=V

BL⋅BT±

6 M

BL⋅BT 2


Bearing Capacity of soil

(1) Design Data

= 30.00 = 0.00 = 1.00B = 9.00 m z = 1.60 m L = 5.00 m

(2) Ultimate Bearing Capacity of soil, (qu)

Calculation of ultimate bearing capacity will be obtained by appliying the following Terzaghi's formula :

qu =

Shape factor (Table 2.5 of KP-06)

a = 1.21 b = 0.40

Shape of footing : rectangular, B x L

Shape of footing a b1 strip 1.00 0.502 square 1.30 0.403 rectangular, B x L 1.21 0.40

(= 1.09 + 0.21 B/L)(B > L) (= 1.09 + 0.21 L/B)

4 circular, diameter = B 1.30 0.30

Bearing capacity factor (Figure 2.3 of KP-06, by Capper)

Nc = 36.0 Nq = 23.0 = 20.0

f Nc Nq0 5.7 0.0 0.05 7.0 1.4 0.0

10 9.0 2.7 0.215 12.0 4.5 2.320 17.0 7.5 4.725 24.0 13.0 9.530 36.0 23.0 20.035 57.0 44.0 41.037 70.0 50.0 55.039 > 82.0 50.0 73.0

= 0.000

= 36.800

= 72.000

qu = 108.80

(3) Allowable Bearing Capacity of soil, (qa)

qa = qu / 3 = 36.27 (safety factor = 3 , normal condition)

qae = qu / 2 = 54.40 (safety factor = 2 , seismic condition)

fBo cB t/m2 gs' t/m3 (=gsat-gw)

a c Nc + gs' z Nq + b gs' B Ng

(B < L)

Ng

Ng

a c Nc

gs' z Nq

b gs' B Ng

t/m2

t/m2

t/m2


BM-100T-Beam

Unit :mSupport

from Parapet

0.45

0.30 m

MomentX Y

t.m/m t.m/m4.663 17.156

16.330 65.923

3.216 10.656

0.630 1.946

50.155 109.598

32.044 41.486

2.430 1.944

8.165 3.370

20.315 4.277

97.200 10.800

0.799 2.722

0.394 1.334

236.341 271.212

1181.71 1356.06

0.000 0.000

0.000 0.000

0.000 0.000

0.000 0.000

1.013 1.890

1.013 1.890

5.063 9.450

0.000 0.000 0.000 0.000 52.324 107.567

16.480 32.319

16.480 31.549

0.304 0.992

61.558 109.627

24.623 40.810

0.492 1.488

151.158 206.177

151.158 206.177

387.585 273.240

21.830 3.564

883.99 1013.51

4419.96 5067.55


0.7 0.0 1.8

20.0

30.0 20.0

015

24.20.0 0.0

0.18

0.000

0.524 0.349

0.000

0.262 0.422 0.000 0.178

0.262 1.0

69.725 0.4380.4334.284

9.0

28.582

63.598

Y HY6.525 0.000 9.417 110.356 3.800 124.202 2.533 64.148 4.284 298.705 2.200 -47.916 2.200 -47.916

X VX3.000 -89.100 6.000 -205.200 4.606 -294.300

(=2/3f)

(=1/2f)


Q min

11/39 document.xls, Body

STRUCTURAL CALCULATION

A ABUTMENT BODY

1 Load and Bending Moment of Abutment BodyKlik untuk menghitung

DimensionsB1 B2 B3 B5 H1 H2 H5 H6

0.90 0.40 0.40 1.80 1.45 10.00 1.00 0.40 Hw1 Hw25.00 6.00

Abutment WidthBw= 5.00 m Plat Form of Impact Plate (Lp) = 0.30 m

1.45 Seismic Coefficient kh= 0.18Support from Parapet hs= 0.45 m

A-A sectionNormal Seismic

Vertical Distance Moment Horizontal Distance Moment Mi=No. Load X Mx Load Y My Mx+My

t/m m tf.m/m t/m m tf.m/m tf.m/m1 1.392 0.200 0.278 0.251 10.725 2.687 2.966

10.00 2 6.048 0.450 2.722 1.089 9.300 10.124 12.846 8.60 3 0.960 0.200 0.192 0.173 9.500 1.642 1.834

5.73 4 0.192 0.133 0.026 0.035 8.867 0.306 0.332 or 5 18.576 0.450 8.359 3.344 4.300 14.378 22.737

5.00 6.00 5' 9.288 -0.300 -2.786 1.672 2.867 4.793 2.006 10 0.216 -0.550 -0.119 0.039 10.700 0.416 0.297 11 0.108 -0.500 -0.054 0.019 10.450 0.203 0.149 S 36.780 8.618 6.620 34.549 43.167

Total 183.900 43.088 33.102 172.746 215.834 Section B-B

Where Hw1' is considered, this section is same as groundwater level.Wehre Hw1' is not considered, this section is just half of (H1 + H2).Distance AB: 6.00 m B-B section

Normal SeismicVertical Distance Moment Horizontal Distance Moment Mi=

No. Load X Mx Load Y My Mx+Myt/m m tf.m/m t/m m tf.m/m tf.m/m

1 1.392 -0.514 -0.715 0.251 4.725 1.184 0.468 2 6.048 0.136 0.823 1.089 3.300 3.593 4.415 3 0.960 -0.514 -0.493 0.173 3.500 0.605 0.111 4 0.192 -0.447 -0.086 0.035 2.867 0.099 0.013 5 5.616 0.136 0.764 1.011 1.300 1.314 2.078 5' 1.698 -0.405 -0.687 0.306 0.867 0.265 -0.422 10 0.216 -0.864 -0.187 0.039 4.700 0.183 -0.004 11 0.108 -0.814 -0.088 0.019 4.450 0.087 -0.001 S 16.230 -0.669 2.921 7.329 6.659

Total 81.149 -3.347 14.607 36.643 33.296

2 Load and Bending Moment due to Super StructureNormal SeismicVertical Distance Moment Horizontal Distance Moment Mi=

No. Load X Mx Load Y My Mx+Myt/m m tf.m/m t/m m tf.m/m tf.m/m

Section A-ANormal Rd + Rl 137.962 0.45 62.083 0.00 0.00 0.00 62.083Seismic Rd 71.272 0.45 32.072 12.83 10.97 140.69 172.763Section B-BNormal Rd + Rl 137.962 0.45 62.083 0.00 0.00 0.00 62.083Seismic Rd 71.272 0.45 32.072 12.83 4.97 63.72 95.790

3 Load and Bending Moment due to Earth and Water Pressure

1) Section B-B

a) Normal Conditionq1= qs*Ka q1= 0.208 tf/mq2= q2= 3.125 tf/mEaH= Eah= 8.534 tf/my= [H*(2q1+q2)]/[3*(q1+q2)] y= 1.930 m H= 5.45 S= EaH*Bw S= 42.672 tfMy= EaH*Bw*y My= 82.362 tf.m

where, qs Surcharge Load 0.70 tf/m2 6.00 Ka2 coefficient of active earth pressure 0.297 5.00 g Unit weight of soil 1.8 tf/m3

Unit weight of saturated soil 2.0 tf/m3d friction angle between earth and wall 0.3491 radianH depth11 5.45 my Acting Point (m)S Shearing Force(tf)Bw Abutment Width 5.00 m

(qs+g*H)*Ka(q1+q2)*(1/2)*H*cosd

qa2

H1+H2

qa1

qw2 qa3A

B

ho

H5

H6B3

B5

H2

H1

B1 B2

X

0.60Lp

A A

B B

(H1+H2)/2

Hw1'

(0,0)X

Y

5'

5

4

23

1

10

11


b) Seismic Conditionqe1= q'*Kea qe1= 0.000 tf/mqe2= qe2= 4.301 tf/mEeH= EaH= 11.320 tf/my= [H*(2qe1+qe2)]/[3*(qe1+qe2)] y= 1.817 mS= EeH*Bw S= 56.599 tfMy= EeH*Bw*y My= 102.822 tf.m/m

where, q' Surcharge Load (seismic) 0 tf/m2Kea coefficient of active earth pressure 0.438g Unit weight of soil 1.8 tf/m3d friction angle between earth and wall 0.2618 radianH 5.45 my Acting Point (m)S Shearing Force(tf)Bw Abutment Width 5.00 m

2) Section A-A

a) Normal ConditionDescription

EaH : active earth pressure (t/m2/m) 36.210 Ka2 : coefficient of active earth pressure 0.297 y : Vertical acting point y= 3.884 S=EaH*Bw S= 181.051My=S*y My= 703.241

qa1 = 0.297 x 0.70 = 0.208 qa2 = 0.297 x 5.45 x 1.80 = 2.917 qa3 = 0.297 x 6.00 x 2.00 = 3.568

qw2 = -5.000 x 1.00 = -5.000

No. Description H Y HYPa1 = 0.208 x 11.450 = 2.383 5.725 13.643 Pa2 = 2.917 x 5.450 x 0.500 = 7.948 7.817 62.126 Pa2' = 2.917 x 6.000 = 17.500 3.000 52.500 Pa3 = 3.568 x 6.000 x 0.500 = 10.703 2.000 21.407

Total (Pa),per m 38.534 3.884 149.675 Pw2 = -5.000 x 5.000 x 0.500 = -12.500 1.667 -20.833

Total, per m -12.500 -20.833 Total (per width) width = 5.00 m -62.500 1.667 -104.167

b) Seismic ConditionDescription

Pea : active earth pressure (t/m2/m) 51.489Kea : coefficient of active earth pressure 0.438y : Vertical acting point y= 3.763S=EaH*Bw S= 257.443My=S*y My= 968.728

qa1 = 0.438 x 0.00 = 0.000 qa2 = 0.438 x 5.45 x 1.80 = 4.301 qa3 = 0.438 x 6.00 x 2.00 = 5.261

qw2 = -5.000 x 1.00 = -5.000

No. Description H Y HYPa1= 0.000 x 11.450 = 0.000 5.725 0.000 Pa2 = 4.301 x 5.450 x 0.500 = 11.719 7.817 91.605 Pa2' = 4.301 x 6.000 = 25.804 3.000 77.411 Pa3 = 5.261 x 6.000 x 0.500 = 15.782 2.000 31.564

Total (per m) 53.305 3.763 200.580 Pw2 = -5.000 x 5.000 x 0.500 = -12.500 1.667 -20.833

Total (per m) -12.500 -20.833 Total (per width) width= 5.00 m -62.500 1.667 -104.167

4 Summary of Intersectional Force

Normal Condition Seismic ConditionDescription Moment Load Shearing Moment Load Shearing

M (tfm) N (tf) S (tf) M (tfm) N (tf) S (tf)Section A-AAbutment Body 43.09 183.90 0.00 172.75 183.90 33.10 Reaction at Abutment 62.08 137.96 0.00 172.76 71.27 12.83 Hydrostatic pressure -104.17 0.00 -62.50 -104.17 0.00 -62.50 Earth Pressure 703.24 0.00 181.05 968.73 0.00 257.44 1.3775192

Total 704.25 321.86 118.55 1,210 255.17 240.87 Section B-BAbutment Body 0.00 81.15 0.00 36.64 81.15 14.61 Reaction at Abutment 62.08 137.96 0.00 95.79 71.27 12.83 Hydrostatic pressure 0.00 0.00 0.00 0.00 0.00 0.00 Earth Pressure 82.36 0.00 42.67 102.82 0.00 56.60 1.2484236

Total 144 219.11 42.67 235 152.42 84.04

(q'+g*H)*Kea(qe1+qe2)*(1/2)*H*cosd

= Pa . cos d

= Pa . cos d


5 Calculation of Required Reinforcement Bar as Rectangular Beam, Normal Condition

1) Cracking Moment Section A-A Section B-B

Mc= Mc= 90516242.033 kgf.cm/m 42832788.401 kgf.cm/m

= 905 tf.m/m 428 tf.m/mwhere, Mc Cracking Moment kgf.cm kgf.cm

Zc Section Modulus

b=100 cm 2,700,000 cm3 1,144,835 cm3Tensile strength of Concrete (bending)

15.6 kgf/cm2 15.6 kgf/cm2175 kgf/cm2

N Axial force 321,862 kg 255,172 kgAc Area of Concrete = b*h1 18000 cm2 11721 cm2h1 thickness of section, B5 180 cm 117 cmb 500 cm 500 cm

Section A-A Section B-B

1) Design Bending Moment Mf= 704 tf.m/m 144 tf.m/m

2) Required Bar Area As_req=As_req= 258 cm2 53 cm2

Allowable Stress R-bar 1850 kgf/cm2 1850 kgf/cm2j= 1 - k/3 0.854 0.854k= 0.438 0.438n= Young's modulus ratio 24 24

Allowable Stress Concrete 60 kgf/cm2 60 kgf/cm2d= Effective height = h1-d1 173 cm 173 cm

d1= 7.5 cmh1= 180 cm

3) Ultimate Bending Moment Mu=

Mu= 1,297 tf.m 273 tf.m

Mu= Ultimate Bending MomentAs= Area of Tensile Bar

Yielding point of Tensile Bar (Spec >295 N/mm2) 3000 kgf/cm2 3000 kgf/cm2Design Compressive Strength of Concrete 175 kgf/cm2 175 kgf/cm2

b= Effective Width 500 cm 500 cm

4) Checking : Single or Double Bar Arrangement M1=

M1= 1,669 tf.m 1,669 tf.mM1= Resistance moment

Cs= 12.844 12.844j= 1 - k/3 0.854 0.854k= n / (n + ssa / sca) 0.438 0.438n= 31 31

1850 kgf/cm2 1850 kgf/cm260 kgf/cm2 60 kgf/cm2

Check : M1 > Mf ? M1= 1,669 tf.m 1,669 tf.mMf= 704 tf.m 144 tf.m

Mf < Ml : Tensile Bar Only Mf < Ml : Tensile Bar Only

(a) Tensile Bar

Max Bar Area : 2%*b*d As max = 1,725 cm2 1,725 cm2Min Bar Area : 4.5%*b As min = 23 cm2 23 cm2Estimation of Required Bar Area : As_req= 258 cm2 53 cm2

25 @ 90 25 @ 180Required Bar Nos : b/pitch= 56 nos 28 nosBar Area : As= 273 cm2 136 cm2

ok ok

As1=As1= 258 cm2 83 cm2As2= 14 cm2 53 cm2

1850 kgf/cm2 1850 kgf/cm2d= 172.5 cm 109.70930233 cm

d2= 7.5 cm 7.5 cm

Mrs=Mrs= 44 tf.m 100 tf.m

As=As1+As2

M' = Mf - M1

M'= 0 tf.m 0 tf.m

As'_req=As'_req= 0 cm2 0 cm2

As' M1= 1,669 tf.m 1,669 tf.mMf= 704 tf.m 144 tf.m

d= 172.5 cm 172.5 cmd2= 0 cm 0 cm

As 1850 kgf/cm2 1850 kgf/cm2Required Bar Area : As'_req= 0 cm2 0 cm2

Zc*(s'ck + N/Ac)

Zc=1/6*b*h12

s'ck

s'ck = 0.5*sck2/3

s ck=

Mf / (s sa*j*d)

s sa=

n / (n + ssa / sca)

sca=

As*ssy { d - 0.5*[As*ssy]/[0.85*sck*b]}

ssy=s'ck=

(d/Cs)2*ssa*b

{ 2n / ( k*j ) }1/2

ssa/scassa=sca=

Apply f :

Resistance Moment by Tensile bar As2 Mf / {ssa*j*d}

ssa=

ssa*As2*(d-d2)

(b) Compressive Bar ( in case Mf > Ml )

M' / [ssa*(d - d2)]

ssa=

Apply f =

d1h d

hd1

d d2


Bar Area : As' = 0 cm2 ok 0 cm2 ok


5) Checking of Allowable Stress

(a) Tensile Bar OnlyMf= 704 tf.m 144 tf.m

S= 119 tf 241 tfMf/(As*j*d)= 1,676 kgf/cm2 ok 1,050 kgf/cm2 ok

33 kgf/cm2 ok 22 kgf/cm2 okS/(b*j*d)= 2 kgf/cm2 ok 3 kgf/cm2 ok

p= As/(b*d)= 0.003 0.002

k= 0.321 0.240j= 1-k/3= 0.893 0.920

b= 500 cm 500 cmd= 172.5 cm 172.5 cmn= 24 24

(b) Tensile Bar & Compressive Bar

Mf= 0 tf.m 0 tf.mS= 0 tf 0 tf

0 kgf/cm2 ok 0 kgf/cm2 ok0 kgf/cm2 ok 0 kgf/cm2 ok0 kgf/cm2 ok 0 kgf/cm2 ok

S/(b*j*d)= 0 kgf/cm2 ok 0 kgf/cm2 okp= As/(b*d)= 0.003 0.002p'= p'=As'/(b*d)= 0.000 0.000

k= 0.321 0.240Lc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.143 0.110

j= 0.893 0.920b= 500 cm 500 cm

d2= 0 cm 0 cmd= 172.5 cm 172.5 cmn= 24 24

ss =

sc = 2*Mf/(k*j*b*d2)=tm =

{(n*p)2+2*n*p}1/2 - n*p=

sc = Mf/(b*d2*Lc)=ss = n*sc*(1-k)/k=ss' = n*sc*(k-d2/d)/k=tm =

{n2(p+p')2+2n(p+p'*d2/d)}1/2-n(p+p')=

(1-d2/d)+k2/{2*n*p*(1-k)}*(d2/d-k/3)=

h dx=kd

b

As

d1

h dx=kd

b

As

As'

d2

d1


6 Calculation of Required Reinforcement Bar as Column, Seismic Condition

1) Minimum Area as Column Section A-A Section B-B

Acmin = 2,274 cm2 1,358 cm2Allowable stress of Reinforcement bar 2775 kgf/m2 2775 kgf/m2Allowable stress of Concrete 90 kgf/m2 90 kgf/m2

N= Axial force 255 tf 152 tfAcdes= b*h 90,000 > Acmin ok 58,605 > Acmin

Minimum Reinforcement Bar(a) As a beam 4.5% * b As min= 23 cm2 23 cm2(b) As a column 0.8% * Acmin As min= 18 cm2 11 cm2

Maximum Reinforcement Bar(a) As a beam 2%*b*d As max = 1,725 cm2 1,097 cm2(b) As a column 6% * Ac As max = 5,400 cm2 3,516 cm2

2) Required Reinforcement Bar

Design Bending Moment Mf= 1,210 tf.m 235 tf.m

As_req=As_req= 237 cm2 56 cm2

Stress of concrete 72.09 kg.cm2 50.15 kg.cm2

Eq1=

Allowable stress of Reinforcement Bar 2775 kg.cm2 2775 kg.cm2Ms= Eccentric Moment, Ms=N(e+c) 142058774.13 kgf.cm 31314870.727 kgf.cm

e= Essentric Distance e=M/N 474.22 cm 154.34 cmM= Design Bending Moment 1,210 tf.m 235 tf.mN= Axial Force 255 tf 152 tfn= Young's Modulus Ratio 16 16 c= c=h/2 - d1 82.5 cm 51.104651163 cmh= Height of Section 180 cm 117 cmb= Width of section 500 cm 500 cm

d1= Concrete Cover 8 cm 8 cmd= Effective Width of section d=h-d1 173 cm 110 cms= 0.294 0.224

232 245

9,936 5,415

861,643 469,571 72 50

Eq1 (trial)= 0 ok Eq1 (trial)= 0 re-checkcross check 0 ok cross check 0 ok

3) Ultimate Bending MomentMu= c*(h/2-0.4X)+Ts'(h/2-d2)+Ts(h/2-d1)

Mu= Ultimate Bending Moment (tf.m) Mu= 1,460 tf.m 284 tf.m

Mu= Min (Mu1,Mu2) in case X>0 in case X<0 in case X>0 in case X<0Mu1= 146031328.36 185930779.09 28402758.532 37876729.0759641

c= 672859 -1155.34 366281.82813 -503.65design strength of Concrete 175 kg/cm2 175 kg/cm2

b= Width of section 500 cm 500 cmX= solve the equation Eq2 below 11.309 -9.709 6.156 -4.232

Ts'= 293344 1543872 -45127 572980As'= 119 28As= Required Reinforcement Bar (Tensile) 237 56Es= Young's modulus (reinforcement bar) 2,100,000 2,100,000

Creep strain coefficient (concrete) 0 0h= Height of Section 180 cm 117 cm

d1= Concrete Cover (tensile side) 7.5 cm 7.5 cmd2= Concrete Cover (compressive side) 7.5 cm 7.5 cm

Eq2 =

a= 50 59500 59500b= 871013 206698Ts= 711031 168733

Yielding point of Tensile Bar (Spec >295 N/mm2) 3000 kgf/cm2 3000 kgf/cm2N= Axial Force 255172 kgf 152421 kgf

put a,b,c as follows : a= a = 59500 59500b= -95190 -114456c= -6532600 -1550237

then, X= 11.309 6.156

X= -9.709 -4.232

N / (0.008*ssa+sca)ssa=sca=

{[sc*(s/2)-N/(b*d)]/ssa}*b*d

sc= sc=

sc3 + [3*ssa/(2*n)-3*Ms/(b*d2)]*sc2 - 6*Ms/(n*b*d2)*ssa*sc - 3*Ms/(n2*b*d2)*ssa2 = 0

ssa=

n*sc/(n*sc+ssa)

[3*ssa/(2*n)-3*Ms/(b*d2)]=

6*Ms/(n*b*d2)*ssa=

3*Ms/(n2*b*d^2)*ssa2=sc (trial)= sc (trial)=

0.68*sck*b*Xsck=

As'*Es*ecu*(X-d2)/XCompressive Bar, As'=0.5 As

ecu=

a*X2 + (b-Ts-N)*X-b*d2 = 0

0.68*sck*bAs'*Es*ecuAs*ssy

ssy=

b-Ts-N =- b*d2 =

(-b + (b2-4ac)0.2 ) / (2*a)=

(-b - (b2-4ac)0.2 ) / (2*a)=


4) Bar Arrangement

Max Bar Area : As max = 1,725 cm2 1,097 cm2Min Bar Area : As min = 23 cm2 23 cm2

(a) Tensile BarRequired Bar Area As req= 237 cm2 56 cm2

25 @ 90 25 @ 180Column width b= 500 cm2 500 cm2Bar Area As = 273 cm2 ok 136 cm2 ok

(b) Compressive BarRequired Bar Area As' req= 119 cm2 28 cm2

25 @ 180 25 @ 180Column width b= 500 cm2 500 cm2Bar Area As' = 136 cm2 ok 136 cm2 ok

(c ) Hoop BarMinimum Diameter 12 12

16 16

d = 1725 1097.0930233300 300768 768

t = 200 mm ok 200 mm ok

5) Checking of Allowable Stress, Seismic Condition

(a) Tensile Bar Only

Mf= 1,210 tf.m 235 tf.mS= 241 tf 84 tf

Mf/(As*j*d)= 2,828 kgf/cm2 check 1,713 kgf/cm2 ok


p= As/(b*d)= 0.00316 0.00249k= {(n*p)^2+2*n*p}^0.5 - n*p= 0.27149 0.24505j= 1-k/3= 0.90950 0.91832

b= 500 cm 500 cmd= 172.5 cm 109.70930233 cmn= 16 16


Mf= 1,210 tf.m 235 tf.mS= 241 tf 84 tf

60 kgf/cm2 ok 30 kgf/cm2 ok2,793 kgf/cm2 check 1,697 kgf/cm2 ok


p= As/(b*d)= 0.003 0.002p'= p'=As'/(b*d)= 0.002 0.002

k= 0.254 0.223Lc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.137 0.129

j= 0.921 0.927b= 500 cm 500 cm

d2= 7.5 cm 7.5 cmd= 172.5 cm 109.70930233 cmn= 16 16

Apply f =

Apply f =

f' min = mm mmApply f' = mm mmBar Interval shall satisfy the following conditions:

mm mmt < 12*f = mm mmt < 48*f '= mm mm

then

ss =



{n2(p+p')2+2n(p+p'*d2/d)}1/2-n(p+p')=

(1-d2/d)+k2/{2*n*p*(1-k)}*(d2/d-k/3)=

d1

d d2

h dx=kd

b

As

d1

h dx=kd

b

As

As'

d2

d1


7 Check for Stress of Concrete and Reinforcement bar as ColumnSection A-A Section B-B

Normal Seismic Normal SeismicHeight of section h cm 180 180 117 117Effective width b cm 500 500 500 500Concrete cover d1 cm 7.5 7.5 7.5 7.5

d2 cm 7.5 7.5 7.5 7.5Reinforcement bar As cm2 273 273 136 136

As' cm2 0 136 0 136 Morment Mf kgf/cm 7.04.E+07 1.21.E+08 1.44.E+07 2.35.E+07Axcis force N kgf 321,862 255,172 219,111 152,421 Shearing force S kgf 118,551 240,874 42,672 84,035 Young's modulus ratio n 24 16 24 16

c 83 83 51 51e 219 474 66 154

a1 386 1153 22 287b1 23665 39405 4596 8081c1 -4082130 -5105274 -504172 -610343

Neutral line x 72 51 56 330 0 0 0

ok ok ok oka2 18005 12697 14036 8208b2 66 73 40 48b3 7497 7088 2979 3396b4 65 43 49 25b5 7497 7088 2979 3396b6 100 122 54 77

Stressing 29.04 ok 57.70 ok 16.71 ok 31.87972.30 ok 2212.39 ok 382.67 ok 1193.79624.33 ok 786.83 ok 347.50 ok 393.38

1.49 ok 3.03 ok 0.84 ok 1.66

SUMMARY OF DESIGN CALCULATIONSection A-A Section B-B

Description Abbr. unit Normal Condition Seismic Condition Normal Condition Seismic ConditionCalculation Condition Rectangular Beam Column Rectangular Beam Column

Principle DimensionsConcrete Design Strength kgf/m2 175 175 175 175Effective width of section b cm 500 500 500 500Height of Section h cm 180 180 117 117 Concrete Cover (tensile) d1 cm 7.5 7.5 7.5 7.5Concrete Cover (compressive) d2 cm 7.5 7.5 7.5 7.5Effective height of Section d cm 172.5 172.5 109.70930233 109.709302325581Allowable Stress Concrete kgf/m2 60 90 60 90

Re-Bar kgf/m2 1850 2775 1850 2775Shearing kgf/m2 5.5 8.25 5.5 8.25

Yielding Point of Reinforcement Bar kgf/cm2 3000 3000 3000 3000

Reinforcement BarTensile Bar Required As_req. cm2 258 237 53 56

Designed As cm2 273 273 136 136 D25@90 D25@90 D25@180 D25@180

Compressive Bar Required As'_req. cm2 0 119 0 28 Designed As' cm2 0 136 0 136

D25@180 D25@180Hoop Bar Designed Aso D16@200 D16@200

Design LoadDesign Bending Moment Mf tf.m 704 1,210 144 235 Design Axis Force Nd tf 322 255 219 152 Shearing Force S tf 119 241 43 84

Checking of Minimum Re-Bar

Ultimate Bending Moment Mu tf.m 1,297 1,460 273 284 Max Re-bar As max cm2 1,725 1,725 1,725 1,097 Min Re-bar As min cm2 23 23 23 23 Required Bar As req. cm2 258 356 53 84 Area of Re-bar for Design As_tot=As+As' cm2 273 409 136 273

Checking of Allowable Stress

Rectangular BeamYoung's Modulus Ratio n 24 16 24 16Effective height d cm 172.5 172.5 109.70930233 109.709302325581Compressive Stress kgf/cm2 33 ok 22 okBending Tensile Stress kgf/cm2 1,676 ok 1,050 ok

kgf/cm2 - - Mean Shearing Stress kgf/cm2 2 ok 3 ok

ColumnCompressive Stress kgf/cm2 29 ok 58 ok 17 ok 32 Bending Stress kgf/cm2 972 ok 2,212 ok 383 ok 1,194

kgf/cm2 624 ok 787 ok 347 ok 393 Shearing Stress kgf/cm2 1 ok 3 ok 1 ok 2

scssss'tc

sc

scassatassy

scssss'tm

scssss'tm


unit:mho

8.60

11.45 H1+H2


ok


okokokok

Seismic Condition

D25@180

D25@180

okokokok

22/39 document.xls, Footing

B FOOTING

1 Load and Bending Moment of Footing

1) Computation Conditions

Normal Condition Toe Side

Heel Side

Seismic Condition

2) FootingDimensions unit:m

B4 B5 B6 BT H3 H4 HT2.25 1.80 4.95 9.00 0.60 1.00 13.05

Abutment Width HT-H3-H4Bw= 5.00 m 11.45

Unit Weight of MaterialSoil 1800 kgf/m3 1.80 tf/m3Concrete 2400 kgf/m3 2.40 tf/m3Surcharge Load q 700 kgf/m3 0.70 tf/m3q'= q seismic 0 kgf/m3 0 tf/m3

Toe Side normal condition seismic conditionArea unit Vertical Distance Moment Horizontal Distance Moment Seismic Coefficient

No. weight Load X Mx Load Y My Kh= 0.18m2 tf/m3 tf/m m tf.m/m tf/m m tf.m/m

6 0.675 2.4 1.620 0.750 1.215 0.292 1.200 0.350F1 2.250 2.4 5.400 1.125 6.075 0.972 0.500 0.486

7.020 7.290 1.264 0.8361.038 m 0.662 m

Heel Side normal condition seismic conditionArea unit Vertical Distance Moment Horizontal Distance Moment

No. weight Load N X Mx Load Y Mym2 tf/m3 tf/m m tf.m/m tf/m m tf.m/m

8 1.485 2.400 3.564 1.650 5.881 0.642 1.300 0.834F3 4.950 2.400 11.880 2.475 29.403 2.138 0.500 1.069

15.444 35.284 2.780 1.9032.285 m 0.685 m

3) Earth Pressure, Heel Side OnlyNormal Condition Seismic Condition

Dead Load (soil) W1= 116.059 tf/m We1= 116.059 tf/m2.379 m 2.379 m

Bending Moment Md Md= 276.157 tf.m/m 276.157 tf.m/m

Surcharge Load W2=q*B6 W2= 3.465 tf/m We2= 0.000 tf/m2.475 m 2.475 m

Bending Moment Ms 8.576 tf.m/m 0.000 tf.m/m

Vertical Earth Pressure Pv= Pv= 12.813 tf/m Pve= 13.237 tf/m

Coefficient of active earth pressure Ka= 0.297 Kea= 0.433Angle between earth and wall 0.3491 radian 0.2618 radian

3.300 3.300 mBending Moment Me 42.284 tf.m/m 43.683 tf.m/m

sub-totalDistance l1 l1 (X) l1 (Y)

sub-totalDistance l2 l2 (X) l2 (Y)

Distance l3=Xo-(B4+B5) l3= l3=Mde=

Distance l4=B6/2 l4= l4=Ms= Mse=

Pv=sin d*{ (1/2)*Ka*g*HT^2 + q*Ka*HT}

d = d e=Distance l5 =B6*(2/3) l5 = l5 =

Me= Mee=

BT

Q

X

BT

Q maxQ min

B4

Ra=Qmax Ra'

l1

M(0,0)

B6

Ra' Ra=Qmin

l2

M

(0,0)

B6B5B4

H4

H3

BT

F1 F2 F3

7 86

9


4) Reaction from Foundation

Reaction from the FoundationNormal Q max 13.193 tf/m2

Qmin 20.441 tf/m2Seismic Qmax 10.560 tf/m2

Qmin 25.552 X 9.000 m

Reaction from Foundation, Toe SideNormal ConditionRat= Q max Rat= 13.193 tf/m2 Cantilever length =B4= 2.250 mRa't= Q max-(B4/BT)*(Q max - Qmin) Ra't= 15.005 tf/m2Seismic ConditionReat= Q Reat= 10.560 tf/m2 Cantilever length =B4= 2.250 mRea't= Q*(X-B4)/X Reat'= 7.920 tf/m2

Reaction from Foundation, Heel SideNormal ConditionRah= Q min Rah= 20.441 tf/m2 Cantilever length =B6= 4.950 mRa'h= Q max-[(B4+B5)/BT]*(Q max - Qmin) Ra'h= 16.455 tf/m2Seismic ConditionReah'= Q*(X-B4-B5)/X Rea'h= 17.306 tf/m2 Cantilever length =X-B4-B5= 4.950 mReah= Qmin Reah= 25.552 tf/m2Reah'= Q*(X-B4-B5)/X Rea'h= 5.808 tf/m2 Cantilever length =X-B4-B5= 11.318 mReah= 0 Reah= 0.000 tf/m2Reah'= Q*(X-B4-B5)/X Rea'h= 17.306 tf/m2 Cantilever length =X-B4-B5= 4.950 mReah= Qmin Reah= 25.552 tf/m2

5) Intersectional Force due ReactionAxial Force = 0 tf/m per m

Toe Section Heel SectionNormal Condition Seismic Condition Normal Condition Seismic Condition

Description Moment Shearing Moment Shearing Moment Shearing Moment ShearingM S Me Se M S Me Se

(tf.m/m) (tf/m) (tf.m/m) (tf/m) (tf.m/m) (tf/m) (tf.m/m) (tf/m)Footing -7.290 -7.020 -7.290 -7.020 35.284 15.444 35.284 15.444 Dead Load, Earth 276.157 116.059 276.157 116.059 Surcharge Load 8.576 3.465 0.000 0.000 Earth Pressure 42.284 12.813 43.683 13.237 Reaction form Foundation 34.924 31.723 24.502 20.790 -234.147 -91.316 -279.368 -106.073

Total - per m 27.634 24.703 17.212 13.770 128.153 56.465 75.755 38.667 Total -5 m 138.169 123.514 86.060 68.848 640.765 282.327 378.775 193.336

2 Calculation of Required Reinforcement Bar : Toe Section

1) Design Bending Moment ( Mf )

Normal Condition M= 27.634 tf.m/mSeismic Condition Me= 17.212 tf.m/m

Max. Design Bending Moment Mf= 27.63 tf.m/m

2) Required Bar Area

As= 11.47 cm2Allowable Stress R-bar 1850 kgf/cm2

j= 1 - k/3 0.854k= 0.438n= Young's modulus ratio 24

Allowable Stress Concrete 60 kgf/cm2

3) Ultimate Bending Moment

Mu= 5.21E+06 kgf.cm52.07 tf.m

where, Mu Ultimate Bending MomentAs Area of Tensile Bar

Yielding point of Tensile Bar 3000 kgf/cm2 (Spec >295 N/mm2)d Effective height = h1-cover 152.5 cm

cover d1= 7.5 cmh1= 160 cm

Design Compressive Strength of Concrete 175 kgf/cm2b Effective Width 100 cm

4) Checking of Single or Double bar arrangementM1=M1= 2.61E+07 kgf.cm/m

= 260.82 tf.m/mM1= Resistance moment

Cs= 12.844k= 0.438n= Young's modulus ratio 30.833

1850 kgf/cm260 kgf/cm2

n= 24Check M1 > Mf? M1= 260.818 tf.m

Mf= 27.634 tf.m M1 > Mf : Design Tensile Bar Only

(a) Tensile Bar

Mf / (s sa*j*d)s sa=

n / (n + ssa / sca)

sca=

Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]}

s sy

s'ck

(d/Cs)2*ssa*b

{ 2n / ( k*j ) }1/2

n / (n + ssa / sca)

ssa=sca=

Tensile

Compressive


Max Bar Area : 2%*b*d = 305.00 cm2Min Bar Area : 4.5%*b = 4.50 cm2Required Bar Area As req= 11.47 cm2

25 @ 300Required Bar Nos Nos=b/pitch = 3.3333333 nosBar Area As = 16.36 cm2 ok

(b) Compressive Bar, in case M1<MfM' = Mf - M1M'= 0.00 tf.m

As' =Required Bar Area As'= 0.00 cm2

d= 152.5 cmd2= 0.00 cm

1850 kgf/cm2

Bar Area As' = 0.00 cm2 ok


(a) Tensile Bar OnlyMf= 27.63 tf.m/m

S= 24.70 tf/mMf/(As*j*d)= 1,187.67 kgf/cm2 ok

12.58 kgf/cm2 okS/(b*j*d)= 1.74 kgf/cm2 ok

p= As/(b*d)= 0.00 b= 100 cm

k= 0.20 d= 152.5 cmj= 1-k/3= 0.93 n= 24

(b) Tensile Bar & Compressive BarMf= 27.63 tf.m

S= 24.70 tf

12.58 kgf/cm2 ok1,187.67 kgf/cm2 ok


p= As/(b*d)= 0.00 p'= p'=As'/(b*d)= 0.00 b= 100 cm

k= 0.20 d= 152.5 cmLc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.09 d2= 0.00 cm

j= 0.93 n= 24

3 Calculation of Required Reinforcement Bar : Heel Section ( Normal Condition )

1) Design Bending Moment Me>0 Me<0


Mf= 128.15 tf.m/m






Mu= 23475776 kgf.cm= 234.76 tf.m





4) Checking of Single or Double bar arrangement

M1= M1= 33507898.8 kgf.cm/m= 335.08 tf.m/m

M1= Resistance moment Mf= 128.153 tf.m M1>Mf : Design Tensile Bar Only

Cs= 11.33 k= 0.438 n= Young's modulus ratio 24

(a) Tensile Bar

Max Bar Area : 2%*b*d = 320.00 cm2

Apply f =

M' / [ssa*(d - d2)]

ssa=Apply f =

ss =


{(n*p)2+2*n*p}1/2 - n*p=


{n2(p+p')2+2n(p+p'*d2/d)}1/2-n(p+p')=

(1-d2/d)+k2/{2*n*p*(1-k)}*(d2/d-k/3)=


n / (n + ssa / sca)

sca=


s sy

s'ck

(d/Cs)2*ssa*b

{ 2n / ( k*j ) }1/2

n / (n + ssa / sca)

d

d1h d

hd1

d d2

h dx=kd

b

As

d1

h d

x=kd

b

As

As'

d2

d1

Tensile

Compressivee Tensile

Compressive


Min Bar Area : 4.5%*b = 4.50 cm2Required Bar Area As req= 53.18 cm2

25 @ 150Required Bar Nos Nos=b/pitch = 6.6666667 nos Pitch shall be same as that of toeBar Area As = 32.72 cm2 check or arrange compressive bar

(b) Compressive Bar, in case M1<Mf

M' = Mf - M1 M'= 0.00 tf.mAs'= 0.00 cm2

d= 152.5 cmd2= 0.00 cm

Required Bar Area As' req= 0.00 cm2




Mf= 128.15 tf.m/mS= 56.47 tf/m

Mf/(As*j*d)= 2,825.55 kgf/cm2 check check

44.33 kgf/cm2 ok okS/(b*j*d)= 4.07 kgf/cm2 ok ok

p= As/(b*d)= 0.00 b= 100 cm

k= 0.27 d= 152.5 cmj= 1-k/3= 0.91 n= 24


Mf= 128.15 tf.mS= 56.47 tf

44.33 kgf/cm2 ok2,825.55 kgf/cm2 check1,063.96 kgf/cm2 ok

S/(b*j*d)= 4.07 kgf/cm2 ok

p= As/(b*d)= 0.00

p'= p'=As'/(b*d)= 0.00

k= 0.27 b= 100 cmLc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.12 d= 152.5 cm

j= 0.91 d2= 0.00 cmn= 24

4 Calculation of Required Reinforcement Bar : Heel Section ( Seismic Condition )

1) Design Bending Moment Me>0 Me<0


Mf= 75.76 tf.m/m





3) Ultimate Bending Moment Mu= 9.12E+06 kgf.cm= 91.19 tf.m





4) Checking of Single or Double bar arrangement

M1= M1= 6.11E+07 kgf.cm/m= 610.54 tf.m/m

M1= Resistance moment Mf= 75.755 tf.m M1>Mf : Design Tensile Bar Only

Cs= 10.281 k= 0.342 n= Young's modulus ratio 16

Apply f =

As' = M'/[ssa*(d - d2)]

Apply f =

ss =


{(n*p)2+2*n*p}1/2 - n*p=


{n2(p+p')2+2n(p+p'*d2/d)}1/2-n(p+p')=

(1-d2/d)+k2/{2*n*p*(1-k)}*(d2/d-k/3)=


n / (n + ssa / sca)

sca=

s sy

s'ck

(d/Cs)2*ssa*b

{ 2n / ( k*j ) }1/2

n / (n + ssa / sca)

d1h d

hd1

d d2

h dx=kd

b

As

d1

h dx=kd

b

As

As'

d2

d1

Tensile

Compressivee Tensile

Compressive


(a) Tensile BarMax Bar Area : 2%*b*d = 0.00 cm2Min Bar Area : 4.5%*b = 4.10 cm2Required Bar Area As req= 20.20 cm2

25 @ 150Required Bar Nos Nos=b/pitch = 6.6666667 nosBar Area As = 32.72 cm2 check or arrange compressive bar

(b) Compressive Bar, in case M1<MfM' = Mf - M1 M'= 0.00 tf.m

As'= 0.00 cm2d= 152.5 cm

d2= 0.00 cmRequired Bar Area As' req= 0.00 cm2



(a) Tensile Bar OnlyMf= 75.76 tf.m/m

S= 38.67 tf/mMf/(As*j*d)= 1,643.98 kgf/cm2 ok


p= As/(b*d)= 0.00 b= 100 cm

k= 0.23 d= 152.5 cmj= 1-k/3= 0.92 n= 16

(b) Tensile Bar & Compressive BarMf= 75.76 tf.m

S= 38.67 tf

30.68 kgf/cm2 ok1,643.98 kgf/cm2 ok


p= As/(b*d)= 0.00 p'= p'=As'/(b*d)= 0.00

k= 0.23 b= 100 cmLc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.11 d= 152.5 cm

j= 0.92 d2= 0.00 cmn= 16

SUMMARY OF DESIGN CALCULATION, FOOTING

Description Abbr. unit Toe Side Heel Side

Calculation Condition Normal Normal Seismic

Principle DimensionsConcrete Design Strength kgf/m2 175 175 175Effective width of section b cm 100 100 100Height of Section, H4 h cm 100 100 100Concrete Cover (tensile) d1 cm 7.5 7.5 7.5Concrete Cover (compressive) d2 cm 0 0 0 Effective height of Section d cm 92.5 92.5 92.5Allowable Stress Concrete kgf/m2 60 60 90

Re-Bar kgf/m2 1850 1850 2775Shearing kgf/m2 5.5 5.5 8.25

kgf/m2 14 14 21Yielding Point of Reinforcement Bar kgf/cm2 3000 3000 3000

Reinforcement BarTensile Bar Required As req. cm2 11.47 53.18 20.20

Designed As cm2 16.36 32.72 32.72 D25@300 D25@150 25 @ 150

Compressive Bar Required As' req. cm2 0.00 0.00 0.00 Designed As' cm2 0.00 0.00 0.00

Design LoadDesign Bending Moment Mf tf.m/m 27.63 128.15 91.19 Design Axis Force Nd tf/m 0.00 0.00 0.00 Shearing Force S tf/m 24.70 56.47 38.67 Ultimate Bending Moment Mu tf.m 52.07 234.76 91.19 Resistance moment (single bar) Ml tf.m 260.82 335.08 610.54 Max Re-bar As max cm2 305.00 320.00 2.75 Min Re-bar As min cm2 4.50 4.50 4.50 Required Bar As req. cm2 11.47 53.18 20.20 Area of Re-bar for Design As cm2 16.36 32.72 32.72

Checking of Allowable StressYoung's Modulus Ratio n 24 24 16 Effective height d-d1 cm 92.5 92.5 92.5Compressive Stress kgf/cm2 12.58 ok 44.33 ok 30.68 okBending Tensile Stress kgf/cm2 1,187.67 ok 2,825.55 check 1,643.98 ok

kgf/cm2 Mean Shearing Stress kgf/cm2 1.74 ok 4.07 ok 2.75 ok

Apply f =

As' = M'/[ssa*(d - d2)]

Apply f =

ss =


{(n*p)2+2*n*p}1/2 - n*p=


{n2(p+p')2+2n(p+p'*d2/d)}1/2-n(p+p')=

(1-d2/d)+k2/{2*n*p*(1-k)}*(d2/d-k/3)=

sc

scassatatmassy

scssss'tm

d1h d

hd1

d d2

h dx=kd

b

As

d1

h dx=kd

b

As

As'

d2

d1

27/39 document.xls, Parapet

C PARAPET WALL

1 Sectional Force without Impact Plate

(1) Rear Face Reinforcement Bar Mo= Mp + MeSo= Sp + Ph

where,Mo Bending Moment at Parapet (tfm/m)So Shearing Force at Platform (tf/m)Mp Bending Moment due T-Load (tfm/m)Me Bending Moment due Earth Pressure (tfm/m)Sp Shearing Force dueT-Load (tf/m)Ph Earth Pressure (tf/m)

1) Sectional Force due to T-LoadMp= Ka T/1.375*{-H1+(H1+a)*ln[(a+H1)/a] 4.393 tfmSp= Ka T/1.375*ln[(a+H1)/a] 4.563 tf

where, Ka Coefficient of active earth pressure 0.297T Wheel load of T-Load 10.000 tfa contact width of T-load 0.200 mH1 Height of Parapet 1.450 m

2) Sectional Force due to Earth PressurePh= Ph= 0.554 tfMe= (1/3)*Ph*H1 Me= 0.268 tfm

where, Kah 0.293d 0.175 (radian)g Unit Weight of Soil 1.800 tf/m3H1 Height of Parapet 1.450 m

3) Summary of Sectional ForceMo= Mp + Me 4.661 tfmSo= Sp + Ph 5.117 tf

2. Sectional Force with Impact Plate

(1) Front Face Reinforcement Bar (compute under normal condition)

Rf= (1/2)*Wd*Lo Rf= 3.105 tf/mRp= T/1.375 Rp= 7.273 tf/mR= Rf + Rp R= 10.378 tf/m B2= 0.40 mLx= (1/2)*B2+Bu Lx= 0.500 m Bu= 0.30 m (=Lp)Mf= R*Lx Mf= 5.189 tfm/m

where, Rf Reaction due self weight Wd tfWd w1 + w2 2.07 tf/m2w1 weight of soil on impact plate 1.35 tf/m2w2 self-weight of impact plate 0.72 tf/m2Lo Length of Impact Plate 3.00 m Thickness of impact plate Ft= 0.30 mRp Reaction due to T-Load tf Soil depth above plate Ds= 0.70 mR Total Reaction tf Thickness of pavement Dp= 0.05 mLx Distance from center of parapet to reaction point m Effective width of road B= 3.00 mBu width of corbel 0.30 mMf Bending Moment at Parapet tfm/m

0.5 g *Kah*H1^2

Ka*cos dfriction angle between the parapet wall and soil = (1/3)f

a

b

H1

gKaH1

T-Load Earth Pressure

B1 B2 Bu

Ds

Ft

R

Lx

Lo

L=0.7 Lo

w1

w2

Lo-Bu


(2) Rear Face Reinforcement Bar (compute under seismic condition)

1) Sectional Force due to Impact Plate

Rh= 2*Rf*Kh 1.118 tf/m Lp= 0.300 m (= Bu)Mr = Rh*Yr 1.285 tfm/m

where, Kh seismic coefficient 0.18Yr = H1-Lp = 1.150 m

2) Sectional Force due to Earth Pressure

H1 > Ds ? H1= 1.450 m0.750 m

Peh 1= 0.024 tf/mMe 1= Peh*[(Lp)*(1/3)+(H1-Lp)] 0.030 tfm/mPeh 2= 0.148 tf/mMe 2= (1/3)*Peh*he 0.037 tfm/m

then Peh =Peh 1+ Peh2 0.172 tf/mMe =Me 1+Me2 0.067 tfm/m

3) Intersectional Force of Parapet due to Earth Pressure

Section Area Unit Weight Kh Seismic Distance Moment(m2) Weight (tf) Weight

Gh (tf) y(m) Mg (tfm)1 0.580 2.4 1.392 0.18 0.251 0.725 0.182

10 0.090 2.4 0.216 0.18 0.039 0.600 0.02311 0.045 2.4 0.108 0.18 0.019 0.650 0.013S 0.309 0.218

Note: if H1 < Dc, areas 10 & 11 would be counted as "0"

4) Summary of Intersectional Force

Mo= Mr+Me+Mg 1.570 tfm/mSo= Rh+Reh+Gh 1.599 tf/m

3. Calculation of Required Reinforcement Bar

1) Cracking Moment

Mc= Mc= 417155 kgf.cm = 4.172 tf.m

where, Mc Cracking Moment kgf.cmZc Sectional Coefficient

Zc=b*B2^2/6 b=100 cm 26667 cm3Tensile strength of Concrete (bending)

15.64 kgf/cm2175 kgf/cm2

N Axis force (=0) 0Ac Area of Concrete = b*B2 4000 cm2b Effective Width 100 cmB2 Thickness of parapet 40 cm

2) Checking of Cracking Moment and Design Bending Moment

Design Bending Moment Mf 5.189 tf.m

Check Mf & Mc 1.7*Mf>Mc?, if yes check ultimate bending moment1.7*Mf = 8.821 tf.mMc= 4.172 tf.m 1.7*Mf>Mc? Yes, check ultimate bending moment

H1-Ds=

(1/2)*g*Kah*Lp^2

(1/2)*g*Kah*he^2

Zc*(s'ck + N/Ac)

s'cks'ck = 0.5*sck^(2/3)

s ck=

B1 B2 Bu

Ds

FtRH

Lo

heYr

H1 H1

g Kah he

B5

B2

Lp

Lp

Lp

H1 1 10

11

B1



Mu= 955213 kgf.cm = 9.552 tf.m

where, Mu Ultimate Bending Moment tf.mAs Area of Tensile Bar cm2

Yielding point of Tensile Bar 3000 kgf/cm2 (Spec >295 N/mm2)d Effective height = B2-cover 33 cm

cover d1= 7 cmB2 = 40 cm


9.951 cm2Allowable Stress Rbar 1850 kgf/cm2

j= 1 -k/3 (=8/9 ) 0.854or

n= Young's modulus ratio 24Allowable Stress Concrete 60 kgf/cm2

Check Mu & Mc Mu = 9.552 tf.mMc = 4.172 tf.m Mu>Mc? ok

4) Bar Arrangement

(a) Front Face, with Impact Plate

Max Bar Area As max = 0.02*b*d = 66.0 cm2 Concrete Cover d1= 7 cmMin Bar Area As min = b*4.5%= 4.5 cm2 Then d= 33 cmRequired Bar Area As req= 10.780 cm2

16 @ 125 mmBar Area As = 16.085 cm2 ok

(b) Rear Face, without Impact Plate

Max Bar Area As max = 0.02*b*d = 60.0 cm2 Concrete Cover d1= 10 cmMin Bar Area As min = b*4.5%= 4.5 cm2 Then d= 30 cmRequired Bar Area As req= cm2

12 @ 250 mm 25 @ 180Bar Area As = cm2

(c) Rear Face, with Impact Plate

Max Bar Area As max = 0.02*b*d = 60.0 cm2 Concrete Cover d1= 10 cmMin Bar Area As min = b*4.5%= 4.5 cm2 Then d= 30 cmRequired Bar Area As req= 3.143 cm2

16 @ 250 mm 25 @ 180Bar Area As = 8.042 cm2 ok


(a) Front Face, with Impact Plate

Mf (front) 5.189 tf.m Concrete Cover d1= 7 cmS 0.000 tf Then d= 33 cm

Mf/(As*j*d) 1239.88 kgf/cm2 ok2*Mf/(k*j*b*d^2) 19.04 kgf/cm2 ok

p=As/(b*d) 0.0230k={(n*p)^2+2*n*p}^0.5 - n*p 0.6347j= 1-k/3 0.7884

(b) Rear Face, without Impact Plate

Mf (rear) 4.661 tf.m Concrete Cover d1= 10 cmSo 5.117 tf Then d= 30 cm

Mf/(As*j*d) kgf/cm22*Mf/(k*j*b*d^2) kgf/cm2S/(b*j*d) kgf/cm2

p=As/(b*d)k={(n*p)^2+2*n*p}^0.5 - n*pj= 1-k/3

(c) Rear Face, with Impact Plate

Mf (rear) 1.570 tf.m Concrete Cover d1= 10 cmSo 1.599 tf Then d= 30 cm

Mf/(As*j*d) 722.95 kgf/cm2 ok2*Mf/(k*j*b*d^2) 12.92 kgf/cm2 okS/(b*j*d) 0.59 kgf/cm2 ok

p=As/(b*d) 0.0027k={(n*p)^2+2*n*p}^0.5 - n*p 0.3001j= 1-k/3 0.9000


s sy

s'ck

As=Mf/(s sa*j*d)s sa=

k = n/{n+s sa/s ca)

s ca

Apply f =

Apply f = spacing of body, tensile f

Apply f = spacing of body, tensile f

ss = check ss < ssa ?sc = check sc < sca ?

ss = check ss < ssa ?sc = check sc < sca ?tm = check tm < ta ?

ss = check ss < ssa ?sc = check sc < sca ?tm = check tm < ta ?


4. Summary of Design Calculation

Description Abbr. unit Front Face Back FaceProvision on Impact Plate Yes No Yes RemarksCalculation Condition Normal Normal Seismic

Principle DimensionsConcrete Design Strength kgf/m2 175 175 175Effective width of section b cm 100 100 100Height of Parapet B2 cm 40 40 40concrete cover (tensile) d1 cm 7 10 10concrete cover (compressive) d2 cm 0 0 0Effective width of Parapet d cm 33 30 30Allowable Stress Concrete kgf/m2 60 60 90 (K175)

Re-Bar kgf/m2 1850 1850 2775Shearing kgf/m2 5.5 5.5 8.25

Yielding Point of Reinforcement Bar kgf/cm2 3000 3000 3000 (Spec >295 N/mm2)

Reinforcement BarTensile Bar Required As req. cm2 10.78 3.14Tensile Bar Designed As cm2 16.08 8.04

D16@125 D16@250Design Load

Design Bending Moment Mf tf.m 5.189 4.661 1.570Design Axis Force Nd tf 0.000 0.000 0.000Shearing Force S tf 0.000 5.117 1.599

Checking of Minimum Re-BarCracking Moment Mc tf.m 4.172 4.172 4.172

1.7*Mf 8.821 7.924 2.6691.7*Mf < Mc ? If no, check Mu check Mu ok

Ultimate Bending Moment Mu tf.m 9.552Mu > Mc ? ok

Max Re-bar As max cm2 66.0 60.0 60.0Min Re-bar As min cm2 4.5 4.5 4.5Required Bar As req. cm2 10.780 3.143Area of Re-bar for Design As cm2 16.085 8.042

Checking of Allowable StressYoung's Modulus Ratio n 24 24 16Effective height d cm 33 30 30Compressive Stress kgf/cm2 19.04 12.92Bending Tensile Stress kgf/cm2 1239.88 722.95Mean Shearing Stress kgf/cm2 0 0.59

sc

scassatassy

scsstm

ABUTMENT : Bendung ….

0.90 0.40

D16@200 D25@180+ 80.45

D16@200D16@200

1.45+ 79.00

[email protected]

D16@200 0.40D16@200

10.00 [email protected] D25@180

13.05

D16@200 [email protected]

5.00D25@180

D25@90

D25@150D19@100 D19@100 D25@150

+ 69.000.60

2.501.00

+ 67.40

D25@300 D19@150 D25@300

9.00

2.25 1.80 4.95

32/39 document.xls, Imp-Plate

D IMPACT PLATE AND CORBEL

1 Design Parameters

Active load T-LoadImpact plate Length L= 3.00 m Span length Ls = 0.7*L= 2.10 m

Thickness h1= 0.30 m Width of corbel Lp= 0.30 mEffective width of road B= 3.00 m Height of corbel h2= 0.30 mUnit weight of plate 2.40 tf/m3Unit weight of soil 1.80 tf/m3 Cover of R-barSoil depth above plate Ds= 0.05 m Impact plate d1= 5 cmThickness of pavement Dp= 0.05 m Corbel d3= 7 cm

2 Computation of Intersectional Force, Corbel

1) Dead Load

Impact plate 0.72 tf/m2Soil above plate 0.09 tf/m2

Total dead load Wd= 0.81 tf/m2

2) Intersectional Force due Dead Load

Md= (1/8)*Wd*Ls^2 0.446512 tf.m

3) Intersectional Force due Live Load

33.39051 tf/m2ML= 5.371698 tf.m

where, ML bending moment due live loadT: wheel load of T-load 10.00 tfa contact width of T-load 0.20 mi: impact coefficient

i=20/(50+L)= 0.377Dp: thickness of pavement 0.05 m

coefficient 1.10

4) Total Intersectional Force

M=Md+ML 5.818211 tf.m

3 Corbel1) Intersectional Force due Impact Plate

M1=R*bu M1= 3.113 tf.m

where, R total reaction form corbel 10.378 tf.m/mbu width of corbel =Lp 0.300 m

2) Intersectional Force due CorbelM2= 0.032 tf.m

where, 0.15 m0.60 m2.40 tf/m3

bu 0.30 m

3) Total Intersectional ForceM=M1+M2 3.146 tf.m

3 Calculation of Required Reinforcement Bar for Impact Plate1) Cracking Moment

Mc= Mc= 234650 kgf.cm = 2.346 tf.m

where, Mc cracking moment kgf.cmZc section modulus

Zc=b*h1^2/6 15000 cm3b= 100 cm

tensile strength of concrete (bending)15.643 kgf/cm2

175 kgf/cm2N axial force 0.000 tfAc area of concrete = b*h1 3000 cm2h1 thickness of section, impact plate 30 cm

gc =gs =

wL=2*T*(1+ i)/{2.75*(a+2*d)} wL =ML={(1/4)*wL*Ls*(a+2*d)-(1/8)*wL*(a+2*d)^2}*a

a:

M2=(1/6)*(2w1+w2)*gc*bu^2

w1w2gc =

Zc*(s'ck + N/Ac)

s'cks'ck = 0.5*sck^(2/3)

s ck=



Design bending moment Mf 5.818 tf.m



Mu= 1039032 kgf.cm = 10.390 tf.m

where, Mu ultimate bending moment tf.mAs area of tensile bar cm2

yielding point of tensile bar 3000 kgf/cm2 (Spec >295 N/mm2)d effective height = h1-cover 25 cm

cover d1= 5 cmh1= 30 cm

design compressive strength of concrete 175 kgf/cm2b effective width 100 cm

14.729 cm2allowable stress of reinforcement bar 1850 kgf/cm2

j= 1 -k/3 (=8/9 ) 0.854or

n= Young's modulus ratio 24allowable stress of concrete 60 kgf/cm2


4) Bar Arrangement

Checking of single or double bar arrangement

M1= 700936 kgf.cm = 7.009 tf.m

where, M1 resistance momentCs ={2m/[s*(1-s/3)]}^(1/2) 12.844s 0.438m 30.833


n 24

Check M1 > Mf? M1= 7.009 tf.m M1>Mf: Design Tensile Bar OnlyMf= 5.818 tf.m

(a) Tensile Bar

Max bar area As max = 0.02*b*d = 50.0 cm2Min bar area As min = b*4.5%= 4.5 cm2Required bar area As req= 14.729 cm2

19 @ 150 mmBar area As = 18.902 cm2 ok

(b) Compressive Bar, in case M1<Mf

As'= 0.000 cm2

d= 25 cm M1= 7.009 tf.md2= 0 cm Mf= 5.818211 tf.m

1850 kgf/cm2

Required bar area As' req= 0.000 cm216 @ 300 mm

Bar area As' = 6.702 cm2 ok


s sy

s'ck


k = n/{n+s sa/s ca)

s ca

M1= (d/Cs)^2*ssa*b >Mf?

(n*sca)/(n*sca+ssa)ssa/sca

ssasca

Apply f =

M' = Mf - M1=ssa*As'*(d - d2)As' = M'/[ssa*(d - d2)]

ssa=

Apply f =

d1h d

h

d1

d d2




Mf 5.818211 tf.mS 0 tf


p=As/(b*d) 0.0076 b= 100 cmk={(n*p)^2+2*n*p}^0.5 - n*p 0.4477 d= 25 cmj= 1-k/3 0.8508 n= 24


Mf 5.818211 tf.mS 0.000 tf

Mf/(b*d^2*Lc) 38.88 kgf/cm2 ok1371.70 kgf/cm2 ok933.07 kgf/cm2 ok

p=As/(bd) 0.0076p'=As'/(bd) 0.0027k={n^2(p+p')^2+2n(p+p'*d2/d)}^05-n(p+p') 0.4048Lc=(1/2)k(1-k/3)+(np'/k)(k-d2/d)(1-d2/d) 0.2394

b= 100 cmd= 25 cm

d2= 0 cmn= 24

4 Calculation of Required Reinforcement Bar for Corbel1) Cracking Moment

Mc= Mc= 234650 kgf.cm = 2.346 tf.m

where, Mc cracking moment kgf.cmZc section modulus

Zc=b*h2^2/6 15000 cm3b= 100 cm

tensile strength of concrete (bending)15.643 kgf/cm2

175 kgf/cm2N axial force 0 tfAc area of concrete = b*h1 3000 cm2h2 thickness of section, impact plate 30 cm


Design bending moment Mf 3.146 tf.m



Mu= 574637 kgf.cm = 5.746 tf.m

where, Mu ultimate bending moment tf.mAs area of tensile bar cm2

yielding point of tensile bar 3000 kgf/cm2 (Spec >295 N/mm2)d effective height = h1-cover 23 cm

cover d1= 7 cmh2= 30 cm

design compressive strength of concrete 175 kgf/cm2b effective width 100 cm

8.657 cm2allowable stress of reinforcement bar 1850 kgf/cm2

j= 1 -k/3 (=8/9 ) 0.854or

n= Young's modulus ratio = 12 24allowable stress of concrete 60 kgf/cm2



sc = check sc < sca ?ss = n*sc*(1-k)/k check ss < ssa ?ss' = n*sc*(k-d2/d)/k check ss' < ssa ?

Zc*(s'ck + N/Ac)

s'cks'ck = 0.5*sck^(2/3)

s ck=


s sy

s'ck


k = n/{n+s sa/s ca)

s ca

h dx=kd

b

As

d1

h dx=kd

b

As

As'

d2

d1


4) Bar Arrangement

Checking of single or double bar arrangement

M1= 593272 kgf.cm = 5.933 tf.m

where, M1 resistance momentCs ={2m/[s*(1-s/3)]}^(1/2) 12.844s 0.438m 30.833


n 24

Check M1 > Mf? M1= 5.933 tf.m M1>Mf: Design Tensile Bar OnlyMf= 3.146 tf.m

(a) Tensile Bar

Max bar area As max = 0.02*b*d = 46.0 cm2Min bar area As min = b*4.5%= 4.5 cm2Required bar area As req= 8.657 cm2

19 @ 300 mmBar area As = 9.451 cm2 ok

5) Checking of Allowable StressMf 3.146 tf.mSo 0.000 tf


p=As/(b*d) 0.0041k={(n*p)^2+2*n*p}^0.5 - n*p 0.3563j= 1-k/3 0.8812

SUMMARY OF DESIGN CALCULATION

Description Abbr. unit Impact Plate Corbel Remarks

Principle DimensionsConcrete Design Strength kgf/m2 175 175Effective width of section b cm 100 100Height of Section h cm 30 30concrete cover (tensile) d1 cm 5 7concrete cover (compressive) d2 cm - -Effective height of Section d cm 25 23Allowable Stress Concrete kgf/m2 60 60 (K175)

Re-Bar kgf/m2 1850 1850Shearing kgf/m2 5.5 5.5

Yielding Point of Reinforcement Bar kgf/cm2 3000 3000 (Spec >295 N/mm2)

Reinforcement BarTensile Bar Required As req. cm2 14.73 8.66

Designed As cm2 18.90 D19@150 9.45 D19@300

Compressive Bar Required As' req. cm2 0.00 Designed As' cm2 6.70 D16@300

Design LoadDesign Bending Moment Mf tf.m 5.818 3.146 Design Axis Force Nd tf - -Shearing Force S tf - -

Checking of Minimum Re-BarCracking Moment Mc tf.m 2.346 5.746

1.7*Mf 9.891 5.348 1.7*Mf<Mc? If "no" check Mu check Mu ok

Ultimate Bending Moment Mu tf.m 10.390 Mu>1.7*Mf? ok

Max Re-bar As max cm2 50.00 46.00 Min Re-bar As min cm2 4.50 4.50 Required Bar As req. cm2 14.73 8.66 Area of Re-bar for Design As cm2 18.90 9.45

Checking of Allowable StressYoung's Modulus Ratio n 24 24Effective height d cm 25 23Compressive Stress kgf/cm2 48.88 ok 37.88 okBending Tensile Stress kgf/cm2 1,447.22 ok 1,642.35 ok

kgf/cm2 - - Mean Shearing Stress kgf/cm2 - -

M1= (d/Cs)^2*ssa*b >Mf?

(n*sca)/(n*sca+ssa)ssa/sca

ssasca

Apply f =


sc

scassatassy

scssss'tm

d1h d

36/39 document.xls,Read me

Remarks

1 Input DataFigure in red are input data.

2 There are two try and error calculation except input data such as dimension and condition.To calculate them, click on the macro button at their right sides after inputting data.

Sheet name "Body", H530 to N531.

3 Allowable stress for "Body"

4 Minimum reinforcement bar

Step 1

Step 2

Step 3

Step 4

5 Wing Wall

Revised

4-Nov-02 1 Bering capacity calculation was added.

1.5 to 2.01.2 to 1.25

9-Nov-02 2

15-Nov-02 3

Water pressure calculated per m, there revised as per width.

29-Nov-02 4 All sheets

previous figure1.50 2.001.20 1.25

Reinforcement bar arrangements for additional bars and distribution bars have been added.

3-Dec-02 5

"Input":

(1) Sheet name "Body", G391 to K392 Button Calculate sc(2) Button Calculate x

If allowable stress is "Check" for body, see Sheet "Body", calculation of stress

Calculation of Md', find 1.7Md

Comparison between Md' and Mc

if Md' > Mc go to step 3

Calculation of Mu

Comparison between Mu and Mc

Mu > Mc OK

Wing Wall are not considered.

Safety factor was revised as follows.Normal conditionSeismic condition

Sheet " Footing"Cell "H548" & "L548" is changed.Figure "Mu" is not referred correctly. Therefore revised.

Sheet "Body"Cell "I123" and "I150"

Some of inputting cells and calculation result cells have been rearranged.Safety factors against sliding have been set at the same figures as retaining wall.

revised figurenormal conditionseismic condition

Modification of drawings


4-Dec-02 6

12-Dec-02 7 Minor change: sheets "Stability", "Body" and "Footing"

28-Dec-02 8 Minor correction: sheet "Body" (calculation results are same as revision 7)

20-Jan-03 9

2-Feb-03 10

26-Feb-03 11

Modification of additional bar arrangement

Sheet "Parapet"Cells "K65", "L65" and "M65" were added.Cells "C63" and "H66" were corrected.Cells "D71", "D72", "H71", "I70", "I71" and "I72" were corrected.Figure 600 on Cell "C86" was deleted.Cells "K127" and "D150" were corrected.Cells "C151", "D151", "H151" and "I151" were inserted.Cell "L287" was corrected.

Sheet "Input"Cell "L40"was modified.Calculation formulae in Cells "J226", "J227" and "J228" were deleted.Calculation formulae in Cells "L230", "L231" and "L232" were corrected.

Sheet "Input"Input Cell "E41"was modified to calculation cell.Input Cell "E42" was added.Cells "B42", "E43", "G41" and "G42" were added.

Sheet "Stability"Calculation formula in Cell "J267" was changed.

Sheet "Body"Calculation formula in Cell "H66" was corrected.

Sheet "Footing"Calculation formula in Cells "K120", "L120", "M120" and "N120" were corrected.

Sheet "Input"Cell "G128" was corrected. Top side ------>> Toe side


Reinforcement bar arrangements for additional bars and distribution bars have been added.


Calculation formula in Cells "K120", "L120", "M120" and "N120" were corrected.

abutment jem bat an

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