243110432 Abutment Jembatan

BAB III. PERHITUNGAN ABUTMENT :

1. Input

Dimension

+ 80.30 HT = 12.20 mB1 = 0.70 mB2 = 0.35 mB3 = 0.40 m

+ 79.00 B4 = 1.20 mB5 = 1.00 mB6 = 2.80 mBT = 5.00 mH1 = 1.30 m

H1 max = 1.30 m+ 75.00 H2 = 10.00 m

H3 = 0.40 mH4 = 0.50 mH5 = 0.40 m

+ 74.00 H6 = 0.30 mH10 = 11.30 m

ho = 9.30 mAbutment width BL = 5.00 m

+ 69.00 Support from Parapet = 0.35 mRh1 = 0.50 mRb1 = 0.35 m

Hw1 = 2.00 m+ 68.10 Hw2 = 4.00 m

Slope = 0.0 slope 1:n

Design Load for ParapetUnit Weight Wheel load of T-Load

Soil 1.30 t/m3 T = 7.0 tonSoil (saturated) 1.60 t/m3 Contact width of T-loadConcrete 2.40 t/m3 a = 0.30 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

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

(Input Yes or No)

ho

H5

H6B3

B6B5B4

HT

H10

H2

H1

H4

H3

B1 B2

BT

Hw1

Hw2

Slope 1:n

Rh1

Rb1

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

Soil depth above plate 0.40 mParameters

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

unit weight of earth (t/m3) 1.30ground surface angle (degree) 0.00internal friction angle (degree) 40.00friction angle between earth and wall (degree) normal 26.70friction angle between earth and earth (degree) normal 0.00friction angle between earth and wall (degree) seismic 20.00friction angle between earth and earth (degree) seismic 33.80wall 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 : 60.00

Qa : Allowable bearing capacity normal t/m2 25.00 ( max. 25.00Qae: Allowable bearing capacity seismic t/m2 37.50 ( max. 37.50 t/m2 for soil foundation)

Normal condition Seismic conditionConcrete Design Strength kgf/m2 200 200Creep strain coefficient (concrete) 0.0035 0.0035Reinforcement concrete

Allowable stressConcrete kgf/m2 70 105Re-Bar kgf/m2 2500 3750Shearing kgf/m2 (concrete) 7 10.5

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

(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

2. Check

2.1 Stability Analysis

Normal condition Seismic conditionOverturning e= 0.30 m e= 0.84 m

BT / 6 = 0.83 m BT / 3 = 1.67 mOK OK

Sliding Fs =Hu / H= 2.62 Fs =Hu / H= 1.392.00 OK 1.25 OK

Settlement Qmax = 25.38 t/m2 Qmax = 38.34 t/m2(bearing capacity) Check Check

Qa = 25.0 t/m2 Qae= 37.5 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) 92.5 92.5 81.7 81.7 Design load Mf (t m) 351.37 739.57 148.35 352.57

Nd (t) 254.41 187.72 225.62 158.93 S (t) 79.42 158.95 44.14 94.96

Checking of minimum reinforcement barRequired bar (cm2) 175 82 74 9

Checking of allowable stress as rectangular beam as column as rectangular beam as columnCompressive stress kgf/cm2 47 ok 101 ok 32 ok 77 ok

Bending stress kgf/cm2 1613 ok 1331 ok 1485 ok 1829 okkgf/cm2 - 1381 ok - 947 ok

Mean shearing stress kgf/cm2 1.99 ok 3.83 ok 3.84 ok 2.59 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

(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) 42.5 42.5 42.5Design load Mf 16.41 30.10 84.98

Nd 0.00 0.00 0.00 S 26.51 16.86 59.40

Checking of minimum reinforcement barRequired bar (cm2) 9.19 16.85 30.63

Checking of allowable stressCompressive stress 19.98 ok 28.51 ok 92.92 ok

Bending stress 1,332.96 ok 1,262.70 ok 3,495.12 okMean shearing stress 3.52 ok 2.31 ok 8.00 ok

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) 28 25Design load Mf (t m) 3.348 0.920

Nd (t) 0.000 0.000S (t) 0.000 1.073

Checking of minimum reinforcement barRequired bar (cm2) 6.07 1.65

Checking of allowable stressCompressive stress 17.07 ok 10.21 ok

Bending stress 942.86 ok 513.06 okMean shearing stress 0 ok 0.48 ok

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)


(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) 4.097 2.147Nd (t) - -S (t) - -

Checking of minimum reinforcement barRequired bar (cm2) 7.57 4.31

Checking of allowable stressCompressive stress 34.42 ok 25.85 ok

Bending stress 1018.96 ok 1120.83 okMean shearing stress - -

f (mm)

f (mm)

As2 (cm2, >As1/6)


7/32 document.xls,Stability

DIMENSIONS OF ABUTMENT

Abutment Type BM-100Super Structure Type T-Beam

Super structure Type T-BeamUnit :m

Dimensions Abutment SupportHT B1 B2 B3 B4 B5 B6 BT H1 H2 H3 H4 H5 H6 H10 ho Width BL from Parapet

12.20 0.70 0.35 0.35 1.20 1.00 2.80 5.00 1.30 10.00 0.40 0.50 1.00 0.35 11.30 8.65 5.00 0.35 H1 max 1.30

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

2.00 4.00 0.50 0.35

WEIGHT OF ABUTMENT LOAD AND MOMENT

Area unit Vertical Distance MomentNo. weight Load X Y X Y

m2 t/m3 t/m m m t.m/m t.m/mBody 1 0.455 2.40 1.092 2.075 11.550 2.266 12.613

2 1.350 2.40 3.240 1.550 10.225 5.022 33.129

1.30 3 0.350 2.40 0.840 2.075 10.400 1.743 8.736

4 0.061 2.40 0.147 2.017 9.450 0.296 1.389

5 6.055 2.40 14.532 1.550 5.225 22.525 75.930

5' 1.298 2.40 3.114 2.000 3.783 6.228 11.781

6 0.240 2.40 0.576 0.800 0.633 0.461 0.365

10.00 7 0.400 2.40 0.960 1.700 0.700 1.632 0.672

1: 0 8 0.560 2.40 1.344 3.133 0.633 4.211 0.851

9 2.500 2.40 6.000 2.500 0.250 15.000 1.500

5.55 10 0.090 2.40 0.216 2.400 11.750 0.518 2.538

12.20 11.30 11 0.045 2.40 0.108 2.350 11.500 0.254 1.242

0.00 0.00 Sub-total 13.404 32.169 60.156 150.746

10.00 Total 67.02 160.845 300.78 753.73

3.10 8.65 SoilToe 12 0.000 1.30 0.000 0.800 0.000 0.000 0.000

0.00 13 0.000 1.30 0.000 0.600 0.900 0.000 0.000

14 0.000 1.30 0.000 0.400 0.767 0.000 0.000

13' 0.000 1.60 0.000 0.600 0.900 0.000 0.000

14' 0.240 1.60 0.384 0.400 0.767 0.154 0.294

2.00 0.40 Sub-total 0.240 0.384 0.154 0.294

Total 1.200 1.920 0.768 1.472

0.50 4.00 Heel 10 0.090 0.00 0.000 2.400 11.500 0.000 0.000

5.00 11 0.045 0.00 0.000 2.350 11.500 0.000 0.000 15 1.925 1.30 2.503 3.625 12.050 9.072 30.155

16 0.630 1.30 0.819 3.775 11.750 3.092 9.623

17 0.630 1.30 0.819 3.775 11.450 3.092 9.378

1.20 1.00 2.80 18 0.045 1.30 0.058 2.450 11.400 0.143 0.667

19 2.750 1.30 3.575 3.625 10.425 12.959 37.269

20 0.962 1.30 1.251 3.625 9.725 4.536 12.168

Unit Weight 21 0.061 1.30 0.080 2.133 9.667 0.170 0.770

Soil 1300 kg/m3 22 15.540 1.30 20.202 3.600 6.775 72.727 136.869

Soil (saturated) 1600 kg/m3 Total Surcharge Load (t) 10.85 22' 1.514 1.30 1.968 2.433 6.667 72.727 136.869

Concrete 2400 kg/m3 Acting Point (m) 3.775 23 8.680 1.60 13.888 3.600 2.450 49.997 34.026

23' 0.560 1.60 0.896 4.067 0.633 3.644 0.567

Sub-total 33.43 46.06 232.16 408.36

Total 167.16 230.29 1160.79 2041.80

Canal Slope 0 Slope 1:nReaction Body 1.87 m 4.686 mNormal Vn= Rd+Rl= 137.96 ton Hn=0 Soil Toe 0.4 m 0.767 mSeismic Ve= =Rd= 71.27 ton Heel 5.04 m 8.866 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) 0.7 q' : surcharge load (t/m2) (=0) 0.0

unit weight of earth (t/m3) 1.3 1.30 unit weight of earth (t/m3) 1.6

ground surface angle (degree) 0.0 internal friction angle (degree) 40.0 friction angle between earth and wall (degree) normal 26.7 friction angle between earth and earth (degree) normal 0friction angle between earth and wall (degree) seismic 20

11.30 friction angle between earth and earth (degree) seismic 33.812.20 wall angle (degree) 0.0

10.00 c : cohesion of soil (t/m2) (do not consider) 0.0 kh : 0.18

ground surface angle (radian) 0.000

internal friction angle (radian) 0.698 2.00 4.00 friction angle between earth and wall (radian) normal 0.466

0.40 friction angle between earth and earth (radian) normal 0.000

0.50 friction angle between earth and wall (radian) seismic 0.349 friction angle between earth and earth (radian) seismic 0.590 wall angle (radian) 0.000 tan-1 kh (radian) 0.178

5.00 0.349 Uc: uplift coefficient 1.0

Earth Pressure

Normal Condition Seismic Condition

Description DescriptionPa : active earth pressure (t/m2/m) 23.416 Pea : active earth pressure (t/m2/m) 32.035 Ka 1: coefficient of active earth pressure, earth and earth 0.217 Kea1 : coefficient of active earth pressure 0.323Ka 2: coefficient of active earth pressure, wall and earth 0.200 0.304y : Vertical acting point 4.167 y : Vertical acting point 4.001 X: Horizontal acting point = BT 5.0 X: Horizontal acting point = BT 5.0

0 17.821

23.416 26.62

qa1 = 0.217 x 0.70 = 0.152 qa1 = 0.323 x 0.00 = 0.000 qa2 = 0.217 x 8.20 x 1.30 = 2.318 qa2 = 0.323 x 8.20 x 1.30 = 3.444 qa3 = 0.217 x 4.00 x 1.60 = 1.392 qa3 = 0.323 x 4.00 x 1.60 = 2.068

qw2 = -2.000 x 1.00 = -2.000 qw2 = -2.000 x 1.00 = -2.000 qu1 = -2.00 x 1.0 = -2.00 qu1 = -2.00 x 1.0 -2.00 qu2 = -4.00 x 1.0 = -4.00 qu2 = -4.00 x 1.0 -4.00

Normal condition Seismic condition

No. Description H Y HY No. Description H Y HYPa1 = 0.152 x 12.200 = 1.857 6.100 11.327 Pa1= 0.000 x 12.200 = 0.000 6.100 0.000 Pa2 = 2.318 x 8.200 x 0.500 = 9.504 6.733 63.991 Pa2 = 3.444 x 8.200 x 0.500 = 14.122 6.733 95.086 Pa2' = 2.318 x 4.000 = 9.272 2.000 18.544 Pa2' = 3.444 x 4.000 = 13.777 2.000 27.554 Pa3 = 1.392 x 4.000 x 0.500 = 2.783 1.333 3.711 Pa3 = 2.068 x 4.000 x 0.500 = 4.136 1.333 5.514

Total (Pa) 23.416 4.167 97.573 Total 32.035 4.001 128.155 Pw2 = -2.000 x 2.000 x 0.500 = -2.000 0.667 -1.333 Pw2 = -2.000 x 2.000 x 0.500 = -2.000 0.667 -1.333 Total -2.000 0.667 -1.333 Total -2.000 0.667 -1.333

No. Description V X VX No. Description V X VXPu1 = -2.000 x 5.000 x 0.500 = -5.000 1.667 -8.333 Pu1 = -2.00 x 5.000 x 0.500 = -5.000 1.667 -8.333 Pu2 = -4.000 x 5.000 x 0.500 = -10.000 3.333 -33.333 Pu2 = -4.00 x 5.000 x 0.500 = -10.000 3.333 -33.333 Total -15.000 2.778 -41.667 Total -15.000 2.778 -41.667

g :g sat:w :f :d 1: (=2/3f)d 2:d E1: (=1/2f)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 160.84 1.87 4.69 300.78 0.00 Combined Acting PointSoil Toe 1.92 0.40 0.77 0.77 0.00 2.201 m

Heel 230.29 5.04 8.87 1160.68 0.00

Reaction (bridge) 137.96 0.00 1.55 10.90 213.84 0.00 Eccentric DistanceEarth Pressure 0.00 117.08 5.00 4.17 0.00 487.87 e=(BT/2-Xo) 0.299 mHydrostatic pressure 0.00 (10.00) 0.67 (6.67)

Uplift pressure (75.00) 2.78 (208.33) 139.69 t.mSurcharge Load 10.85 0.00 3.78 0.00 40.96 0.00

S 466.87 107.08 1508.70 481.21

2 Stability Analysis

2.1 Over Turnng

e<=(BT/6) e= 0.299 m BT/6 is larger than e? OKBT/6= 0.833 m

2.2 Sliding

0.60Shearing registance force at base

280.12 tfSafety Rate against slidingFs=Hu/H 2.616 Fs is larger than 2.0 ? OK

2.3 Settlement

Allowale bearing capacity Qa 25.00 tf/m2Reaction from the Foundation

Maximiun Fondation Reaction Q maxQ max = 25.380 tf/m2 Qmax is smaller than Qa? Check

Minimum Fondation Reaction Q maxQ min = 11.970 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 160.84 28.95 1.87 4.69 300.8 135.67 Combined Acting PointSoil Toe 1.92 0.40 0.77 0.77 0.00 1.664 m

Heel 230.3 41.45 5.04 8.87 1160.7 367.5

Reaction (bridge) 71.27 12.83 1.55 11.40 110.47 146.25 Eccentric DistanceEarth Pressure 89.11 133.10 5.00 4.00 445.5 532.5 e=(BT/2-Xo) 0.836 mHydrostatic pressure (10.00) 0.67 (1.33)

Uplift pressue (75.00) 2.78 (41.67) 400.11 t.mSurcharge Load 0.00 0.00 5.00 4.17 0.00 0.00

S 478.4 206.33 1977 1180.6

2 Stability Analysis

2.1 Over Turnng

e<=(BT/3) e= 0.836 m BT/3 is larger than e? OKBT/3= 1.667 m

2.2 Sliding

0.6Shearing registance force at base

287.06 tfSafety Rate anaginst slidingFs=Hu/H 1.391 Fs is larger than 1.25 ? OK

2.3 Settlement

Allowale bearing capacity Qa 37.50 tf/m2Reaction from the Foundation e > BT/6 ?

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

X=3(BT/2-e) = 4.991 m

Maximiun Fondation Reaction Q max Foundation Reaction QeQ max = 38.342 tf/m2 Q=2V/(BL.X)= 38.343 tf/m2

Minimum Fondation Reaction Q maxQ min = -0.068 tf/m2

Qmax is smaller than Qa? Check Qe is smaller than Qa? CheckQmax = 38.34 tf/m2Qmin = 0.00 tf/m2X = 4.99 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

Q min

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

= 40.00 = 0.00 = 0.60B = 5.00 m z = 0.90 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 = 1.30 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 1.30 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 = 82.0 Nq = 50.0 = 73.0

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

= 27.000

= 87.600

qu = 114.60

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

qa = qu / 3 = 38.20 (safety factor = 3 , normal condition)

qae = qu / 2 = 57.30 (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

11/32 document.xls, Body

STRUCTURAL CALCULATION

A ABUTMENT BODY

1 Load and Bending Moment of Abutment BodyKlik untuk menghitung

Dimensions unit:mB1 B2 B3 B5 H1 H2 H5 H6 ho

0.70 0.40 0.35 1.00 1.30 10.00 1.00 0.35 8.65 Hw1 Hw25.00 6.00

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

1.30 Seismic Coefficient kh= 0.18Support from Parapet hs= 0.35 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 1.092 0.400 0.437 0.197 10.650 2.093 2.530

10.00 2 3.240 0.150 0.486 0.583 9.325 5.438 5.924 8.65 3 0.840 0.375 0.315 0.151 9.500 1.436 1.751

5.65 4 0.147 0.317 0.047 0.026 8.883 0.235 0.282 or 5 14.532 0.150 2.180 2.616 4.325 11.313 13.493

1.10 3.10 5' 3.114 -0.300 -0.934 0.561 2.883 1.616 0.682 10 0.216 -0.750 -0.162 0.039 10.550 0.410 0.248 11 0.108 -0.700 -0.076 0.019 10.300 0.200 0.125 S 23.289 2.292 4.192 22.743 25.035

Total 116.445 11.462 20.960 113.715 125.176 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: 3.10 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 1.092 -0.454 -0.496 0.197 7.550 1.484 0.989 2 3.240 0.096 0.312 0.583 6.225 3.630 3.942 3 0.840 -0.429 -0.360 0.151 6.400 0.968 0.608 4 0.147 -0.370 -0.054 0.026 5.783 0.153 0.099 5 9.324 0.096 0.897 1.678 2.775 4.657 5.555 5' 2.564 -0.318 -0.815 0.462 1.850 0.854 0.039 10 0.216 -0.804 -0.174 0.039 7.450 0.290 0.116 11 0.108 -0.754 -0.081 0.019 7.200 0.140 0.059 S 17.531 -0.771 3.156 12.176 11.405

Total 87.655 -3.855 15.778 60.879 57.024

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 0.15 20.694 0.00 0.00 0.00 20.694Seismic Rd 71.272 0.15 10.691 12.83 10.87 139.41 150.099Section B-BNormal Rd + Rl 137.962 0.15 20.694 0.00 0.00 0.00 20.694Seismic Rd 71.272 0.15 10.691 12.83 7.77 99.64 110.329

3 Load and Bending Moment due to Earth and Water Pressure

1) Section B-B

a) Normal Conditionq1= qs*Ka q1= 0.140 tf/mq2= q2= 2.270 tf/mEaH= Eah= 8.828 tf/my= [H*(2q1+q2)]/[3*(q1+q2)] y= 2.892 m H= 8.20 S= EaH*Bw S= 44.142 tf 11.30 My= EaH*Bw*y My= 127.659 tf.m

where, qs Surcharge Load 0.70 tf/m2 3.10 Ka2 coefficient of active earth pressure 0.200 1.10 g Unit weight of soil 1.3 tf/m3

Unit weight of saturated soil 1.6 tf/m3d friction angle between earth and wall 0.4660 radianH depth11 8.20 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= 3.444 tf/mEeH= EaH= 13.270 tf/my= [H*(2qe1+qe2)]/[3*(qe1+qe2)] y= 2.733 mS= EeH*Bw S= 66.350 tfMy= EeH*Bw*y My= 181.357 tf.m/m

where, q' Surcharge Load (seismic) 0 tf/m2Kea coefficient of active earth pressure 0.323g Unit weight of soil 1.3 tf/m3d friction angle between earth and wall 0.3491 radianH 8.20 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) 16.489 Ka2 : coefficient of active earth pressure 0.200 y : Vertical acting point y= 3.885 S=EaH*Bw S= 82.444My=S*y My= 320.322

qa1 = 0.200 x 0.70 = 0.140 qa2 = 0.200 x 8.20 x 1.30 = 2.130 qa3 = 0.200 x 3.10 x 1.60 = 0.991

qw2 = -1.100 x 1.00 = -1.100

No. Description H Y HYPa1 = 0.140 x 11.300 = 1.581 5.650 8.932 Pa2 = 2.130 x 8.200 x 0.500 = 8.735 5.833 50.954 Pa2' = 2.130 x 3.100 = 6.605 1.550 10.237 Pa3 = 0.991 x 3.100 x 0.500 = 1.537 1.033 1.588

Total (Pa),per m 18.457 3.885 71.711 Pw2 = -1.100 x 1.100 x 0.500 = -0.605 0.367 -0.222

Total, per m -0.605 -0.222 Total (per width) width = 5.00 m -3.025 0.367 -1.109

b) Seismic ConditionDescription

Pea : active earth pressure (t/m2/m) 25.638Kea : coefficient of active earth pressure 0.323y : Vertical acting point y= 3.720S=EaH*Bw S= 128.189My=S*y My= 476.862

qa1 = 0.323 x 0.00 = 0.000 qa2 = 0.323 x 8.20 x 1.30 = 3.444 qa3 = 0.323 x 3.10 x 1.60 = 1.603

qw2 = -1.100 x 1.00 = -1.100

No. Description H Y HYPa1= 0.000 x 11.300 = 0.000 5.650 0.000 Pa2 = 3.444 x 8.200 x 0.500 = 14.122 5.833 82.376 Pa2' = 3.444 x 3.100 = 10.677 1.550 16.550 Pa3 = 1.603 x 3.100 x 0.500 = 2.484 1.033 2.567

Total (per m) 27.283 3.720 101.493 Pw2 = -1.100 x 1.100 x 0.500 = -0.605 0.367 -0.222

Total (per m) -0.605 -0.222 Total (per width) width= 5.00 m -3.025 0.367 -1.109

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 11.46 116.45 0.00 113.71 116.45 20.96 Reaction at Abutment 20.69 137.96 0.00 150.10 71.27 12.83 Hydrostatic pressure -1.11 0.00 -3.03 -1.11 0.00 -3.03 Earth Pressure 320.32 0.00 82.44 476.86 0.00 128.19 1.4886957

Total 351.37 254.41 79.42 740 187.72 158.95 Section B-BAbutment Body 0.00 87.65 0.00 60.88 87.65 15.78 Reaction at Abutment 20.69 137.96 0.00 110.33 71.27 12.83 Hydrostatic pressure 0.00 0.00 0.00 0.00 0.00 0.00 Earth Pressure 127.66 0.00 44.14 181.36 0.00 66.35 1.42064

Total 148 225.62 44.14 353 158.93 94.96

(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= 35450383 kgf.cm/m 25311625 kgf.cm/m

= 355 tf.m/m 253 tf.m/mwhere, Mc Cracking Moment kgf.cm kgf.cm

Zc Section Modulus

b=100 cm 833,333 cm3 663,775 cm3Tensile strength of Concrete (bending)

17.1 kgf/cm2 17.1 kgf/cm2200 kgf/cm2

N Axial force 254,407 kg 187,717 kgAc Area of Concrete = b*h1 10000 cm2 8925 cm2h1 thickness of section, B5 100 cm 89 cmb 500 cm 500 cm

Section A-A Section B-B

1) Design Bending Moment Mf= 351 tf.m/m 148 tf.m/m

2) Required Bar Area As_req=As_req= 175 cm2 74 cm2

Allowable Stress R-bar 2500 kgf/cm2 2500 kgf/cm2j= 1 - k/3 0.866 0.866k= 0.402 0.402n= Young's modulus ratio 24 24

Allowable Stress Concrete 70 kgf/cm2 70 kgf/cm2d= Effective height = h1-d1 93 cm 93 cm

d1= 7.5 cmh1= 100 cm

3) Ultimate Bending Moment Mu=

Mu= 471 tf.m 203 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 200 kgf/cm2 200 kgf/cm2

b= Effective Width 500 cm 500 cm

4) Checking : Single or Double Bar Arrangement M1=

M1= 521 tf.m 521 tf.mM1= Resistance moment

Cs= 14.325 14.325j= 1 - k/3 0.866 0.866k= n / (n + ssa / sca) 0.402 0.402n= 36 36

2500 kgf/cm2 2500 kgf/cm270 kgf/cm2 70 kgf/cm2

Check : M1 > Mf ? M1= 521 tf.m 521 tf.mMf= 351 tf.m 148 tf.m

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

(a) Tensile Bar

Max Bar Area : 2%*b*d As max = 925 cm2 925 cm2Min Bar Area : 4.5%*b As min = 23 cm2 23 cm2Estimation of Required Bar Area : As_req= 175 cm2 74 cm2

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

ok ok

As1=As1= 175 cm2 84 cm2As2= 97 cm2 53 cm2

2500 kgf/cm2 2500 kgf/cm2d= 92.5 cm 81.748555 cm

d2= 7.5 cm 7.5 cm

Mrs=Mrs= 207 tf.m 98 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= 521 tf.m 521 tf.mMf= 351 tf.m 148 tf.m

d= 92.5 cm 92.5 cmd2= 0 cm 0 cm

As 2500 kgf/cm2 2500 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= 351 tf.m 148 tf.m

S= 79 tf 159 tfMf/(As*j*d)= 1,613 kgf/cm2 ok 1,485 kgf/cm2 ok

47 kgf/cm2 ok 32 kgf/cm2 okS/(b*j*d)= 2 kgf/cm2 ok 4 kgf/cm2 ok

p= As/(b*d)= 0.006 0.003

k= 0.409 0.312j= 1-k/3= 0.864 0.896

b= 500 cm 500 cmd= 92.5 cm 92.5 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)= 0.006 0.003p'= p'=As'/(b*d)= 0.000 0.000

k= 0.409 0.312Lc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.177 0.140

j= 0.864 0.896b= 500 cm 500 cm

d2= 0 cm 0 cmd= 92.5 cm 92.5 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 = 1,390 cm2 1,177 cm2Allowable stress of Reinforcement bar 3750 kgf/m2 3750 kgf/m2Allowable stress of Concrete 105 kgf/m2 105 kgf/m2

N= Axial force 188 tf 159 tfAcdes= b*h 50,000 > Acmin ok 44,624 > Acmin ok

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= 11 cm2 9 cm2

Maximum Reinforcement Bar(a) As a beam 2%*b*d As max = 925 cm2 817 cm2(b) As a column 6% * Ac As max = 3,000 cm2 2,677 cm2

2) Required Reinforcement Bar

Design Bending Moment Mf= 740 tf.m 353 tf.m

As_req=As_req= 55 cm2 6 cm2

Stress of concrete 72.09 kg.cm2 50.15 kg.cm2

Eq1=

Allowable stress of Reinforcement Bar 3750 kg.cm2 3750 kg.cm2Ms= Eccentric Moment, Ms=N(e+c) 81934590 kgf.cm 41156602 kgf.cm

e= Essentric Distance e=M/N 393.98 cm 221.84 cmM= Design Bending Moment 740 tf.m 353 tf.mN= Axial Force 188 tf 159 tfn= Young's Modulus Ratio 16 16 c= c=h/2 - d1 42.5 cm 37.124277 cmh= Height of Section 100 cm 89 cmb= Width of section 500 cm 500 cm

d1= Concrete Cover 8 cm 8 cmd= Effective Width of section d=h-d1 93 cm 82 cms= 0.235 0.176

294 315

26,932 17,321

3,156,150 2,029,800 72 50

Eq1 (trial)= -3194502 re-check Eq1 (trial)= -1981096 re-checkcross check -3194502 ok cross check -1981096 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= 255 tf.m 88 tf.m

Mu= Min (Mu1,Mu2) in case X>0 in case X<0 in case X>0 in case X<0Mu1= 25511888 32609824 8808908 9128431

c= 403914 -506.00 207358.08 -104.68design strength of Concrete 200 kg/cm2 200 kg/cm2

b= Width of section 500 cm 500 cmX= solve the equation Eq2 below 5.940 -3.721 3.049 -0.770

Ts'= -52627 604287 -31060 228639As'= 27 3As= Required Reinforcement Bar (Tensile) 55 6Es= Young's modulus (reinforcement bar) 2,100,000 2,100,000

Creep strain coefficient (concrete) 0 0h= Height of Section 100 cm 89 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 68000 68000b= 200373 21281Ts= 163570 17372

Yielding point of Tensile Bar (Spec >295 N/mm2) 3000 kgf/cm2 3000 kgf/cm2N= Axial Force 187717 kgf 158927 kgf

put a,b,c as follows : a= a = 68000 68000b= -150914 -155018c= -1502799 -159605

then, X= 5.940 3.049

X= -3.721 -0.770

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 = 925 cm2 817 cm2Min Bar Area : As min = 23 cm2 23 cm2

(a) Tensile BarRequired Bar Area As req= 55 cm2 6 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= 27 cm2 3 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 = 925 817.48555300 300768 768

t = 200 mm ok 200 mm ok

5) Checking of Allowable Stress, Seismic Condition

(a) Tensile Bar Only

Mf= 740 tf.m 353 tf.mS= 159 tf 95 tf

Mf/(As*j*d)= 3,319 kgf/cm2 ok 3,486 kgf/cm2 ok

112 kgf/cm2 check 84 kgf/cm2 okS/(b*j*d)= 4 kgf/cm2 ok 3 kgf/cm2 ok

p= As/(b*d)= 0.00590 0.00334k= {(n*p)^2+2*n*p}^0.5 - n*p= 0.35016 0.27768j= 1-k/3= 0.88328 0.90744

b= 500 cm 500 cmd= 92.5 cm 81.748555 cmn= 16 16


Mf= 740 tf.m 353 tf.mS= 159 tf 95 tf

98 kgf/cm2 ok 72 kgf/cm2 ok3,269 kgf/cm2 ok 3,458 kgf/cm2 ok1,172 kgf/cm2 ok 735 kgf/cm2 ok

S/(b*j*d)= 4 kgf/cm2 ok 3 kgf/cm2 okp= As/(b*d)= 0.006 0.003p'= p'=As'/(b*d)= 0.003 0.003

k= 0.324 0.251Lc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.177 0.146

j= 0.897 0.915b= 500 cm 500 cm

d2= 7.5 cm 7.5 cmd= 92.5 cm 81.748555 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 100 100 89 89Effective 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 3.51.E+07 7.40.E+07 1.48.E+07 3.53.E+07Axcis force N kgf 254,407 187,717 225,617 158,927 Shearing force S kgf 79,419 158,953 44,142 94,957 Young's modulus ratio n 24 16 24 16

c 43 43 37 37e 138 394 66 222

a1 264 1032 63 532b1 14185 32056 4040 11615c1 -1312152 -2183026 -330258 -590486

Neutral line x 72 51 56 331454126 2237649 273329 399385

Check Check Check Checka2 18005 12697 14036 8208b2 26 33 26 34b3 3862 3651 2164 2467b4 65 43 49 25b5 3862 3651 2164 2467b6 20 42 26 49

Stressing 44.12 ok 101.27 ok 28.29 ok 76.71 ok301.14 ok 1330.89 ok 309.70 ok 1828.60 ok948.71 ok 1381.09 ok 588.34 ok 947.00 ok

1.91 ok 3.83 ok 1.20 ok 2.59 ok

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 200 200 200 200Effective width of section b cm 500 500 500 500Height of Section h cm 100 100 89 89 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 92.5 92.5 81.748555 81.748555Allowable Stress Concrete kgf/m2 70 105 70 105

Re-Bar kgf/m2 2500 3750 2500 3750Shearing kgf/m2 7 10.5 7 10.5

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

Reinforcement BarTensile Bar Required As_req. cm2 175 55 74 6

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

Compressive Bar Required As'_req. cm2 0 27 0 3 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 351 740 148 353 Design Axis Force Nd tf 254 188 226 159 Shearing Force S tf 79 159 44 95

Checking of Minimum Re-Bar

Ultimate Bending Moment Mu tf.m 471 255 203 88 Max Re-bar As max cm2 925 925 925 817 Min Re-bar As min cm2 23 23 23 23 Required Bar As req. cm2 175 82 74 9 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 92.5 92.5 81.748555 81.748555Compressive Stress kgf/cm2 47 ok 32 okBending Tensile Stress kgf/cm2 1,613 ok 1,485 ok

kgf/cm2 - - Mean Shearing Stress kgf/cm2 2 ok 4 ok

ColumnCompressive Stress kgf/cm2 44 ok 101 ok 28 ok 77 okBending Stress kgf/cm2 301 ok 1,331 ok 310 ok 1,829 ok

kgf/cm2 949 ok 1,381 ok 588 ok 947 okShearing Stress kgf/cm2 2 ok 4 ok 1 ok 3 ok

scssss'tc

sc

scassatassy

scssss'tm

scssss'tm

19/32 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 HT1.20 1.00 2.80 5.00 0.40 0.50 12.20

Abutment Width HT-H3-H4Bw= 5.00 m 11.30

Unit Weight of MaterialSoil 1300 kgf/m3 1.30 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 0.240 2.4 0.576 0.400 0.230 0.104 0.633 0.066F1 0.600 2.4 1.440 0.600 0.864 0.259 0.250 0.065

2.016 1.094 0.363 0.1300.543 m 0.360 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 0.560 2.400 1.344 0.933 1.254 0.242 0.700 0.169F3 1.400 2.400 3.360 1.400 4.704 0.605 0.250 0.151

4.704 5.958 0.847 0.3211.267 m 0.379 m

3) Earth Pressure, Heel Side OnlyNormal Condition Seismic Condition

Dead Load (soil) W1= 46.059 tf/m We1= 46.059 tf/m1.365 m 1.365 m

Bending Moment Md Md= 62.890 tf.m/m 62.890 tf.m/m

Surcharge Load W2=q*B6 W2= 1.960 tf/m We2= 0.000 tf/m1.400 m 1.400 m

Bending Moment Ms 2.744 tf.m/m 0.000 tf.m/m

Vertical Earth Pressure Pv= Pv= 8.164 tf/m Pve= 8.639 tf/m

Coefficient of active earth pressure Ka= 0.200 Kea= 0.304Angle between earth and wall 0.4660 radian 0.3491 radian

1.867 1.867 mBending Moment Me 15.239 tf.m/m 16.126 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 25.380 tf/m2

Qmin 11.970 tf/m2Seismic Qmax 0.000 tf/m2

Qmin 0.000 X 4.991 m

Reaction from Foundation, Toe SideNormal ConditionRat= Q max Rat= 25.380 tf/m2 Cantilever length =B4= 1.200 mRa't= Q max-(B4/BT)*(Q max - Qmin) Ra't= 22.161 tf/m2Seismic ConditionReat= Q Reat= 0.000 tf/m2 Cantilever length =B4= 1.200 mRea't= Q*(X-B4)/X Reat'= 0.000 tf/m2

Reaction from Foundation, Heel SideNormal ConditionRah= Q min Rah= 11.970 tf/m2 Cantilever length =B6= 2.800 mRa'h= Q max-[(B4+B5)/BT]*(Q max - Qmin) Ra'h= 19.479 tf/m2Seismic ConditionReah'= Q*(X-B4-B5)/X Rea'h= 0.000 tf/m2 Cantilever length =X-B4-B5= 2.800 mReah= Qmin Reah= 0.000 tf/m2Reah'= Q*(X-B4-B5)/X Rea'h= 0.000 tf/m2 Cantilever length =X-B4-B5= 2.791 mReah= 0 Reah= 0.000 tf/m2Reah'= Q*(X-B4-B5)/X Rea'h= 0.000 tf/m2 Cantilever length =X-B4-B5= 2.791 mReah= Qmin Reah= 0.000 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 -1.094 -2.016 -1.094 -2.016 5.958 4.704 5.958 4.704 Dead Load, Earth 62.890 46.059 62.890 46.059 Surcharge Load 2.744 1.960 0.000 0.000 Earth Pressure 15.239 8.164 16.126 8.639 Reaction form Foundation 17.501 28.525 0.000 0.000 -56.734 -44.029 0.000 0.000

Total - per m 16.407 26.509 -1.094 -2.016 30.097 16.857 84.975 59.402 Total -5 m 82.033 132.544 -5.472 -10.080 150.486 84.287 424.876 297.010

2 Calculation of Required Reinforcement Bar : Toe Section

1) Design Bending Moment ( Mf )

Normal Condition M= 16.407 tf.m/mSeismic Condition Me= -1.094 tf.m/m

Max. Design Bending Moment Mf= 16.41 tf.m/m

2) Required Bar Area

As= 9.19 cm2Allowable Stress R-bar 2500 kgf/cm2

j= 1 - k/3 0.866k= 0.402n= Young's modulus ratio 24

Allowable Stress Concrete 70 kgf/cm2

3) Ultimate Bending Moment

Mu= 2.25E+06 kgf.cm22.51 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 82.5 cm

cover d1= 7.5 cmh1= 90 cm

Design Compressive Strength of Concrete 200 kgf/cm2b Effective Width 100 cm

4) Checking of Single or Double bar arrangementM1=M1= 8.29E+06 kgf.cm/m

= 82.92 tf.m/mM1= Resistance moment

Cs= 14.325k= 0.402n= Young's modulus ratio 35.714

2500 kgf/cm270 kgf/cm2

n= 24Check M1 > Mf? M1= 82.917 tf.m

Mf= 16.407 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 = 165.00 cm2Min Bar Area : 4.5%*b = 4.50 cm2Required Bar Area As req= 9.19 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= 82.5 cmd2= 0.00 cm

2500 kgf/cm2

Bar Area As' = 0.00 cm2 ok


(a) Tensile Bar OnlyMf= 16.41 tf.m/m

S= 26.51 tf/mMf/(As*j*d)= 1,332.96 kgf/cm2 ok

19.98 kgf/cm2 okS/(b*j*d)= 3.52 kgf/cm2 ok

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

k= 0.26 d= 82.5 cmj= 1-k/3= 0.91 n= 24

(b) Tensile Bar & Compressive BarMf= 16.41 tf.m

S= 26.51 tf

19.98 kgf/cm2 ok1,332.96 kgf/cm2 ok


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

k= 0.26 d= 82.5 cmLc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.12 d2= 0.00 cm

j= 0.91 n= 24

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

1) Design Bending Moment Me>0 Me<0

Normal Condition M= 30.097 tf.m/mSeismic Condition Me= 84.975 tf.m/m

Mf= 30.10 tf.m/m






Mu= 4095213 kgf.cm= 40.95 tf.m





4) Checking of Single or Double bar arrangement

M1= M1= 12338772.5 kgf.cm/m= 123.39 tf.m/m

M1= Resistance moment Mf= 30.097 tf.m M1>Mf : Design Tensile Bar Only

Cs= 11.74 k= 0.402 n= Young's modulus ratio 24

(a) Tensile Bar

Max Bar Area : 2%*b*d = 180.00 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= 16.85 cm2

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

(b) Compressive Bar, in case M1<Mf

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

d= 82.5 cmd2= 0.00 cm

Required Bar Area As' req= 0.00 cm2




Mf= 30.10 tf.m/mS= 16.86 tf/m

Mf/(As*j*d)= 1,262.70 kgf/cm2 ok ok

28.51 kgf/cm2 ok okS/(b*j*d)= 2.31 kgf/cm2 ok ok

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

k= 0.35 d= 82.5 cmj= 1-k/3= 0.88 n= 24


Mf= 30.10 tf.mS= 16.86 tf

28.51 kgf/cm2 ok1,262.70 kgf/cm2 ok


p= As/(b*d)= 0.00

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

k= 0.35 b= 100 cmLc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.16 d= 82.5 cm

j= 0.88 d2= 0.00 cmn= 24

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

1) Design Bending Moment Me>0 Me<0

Normal Condition M= 30.097 tf.m/mSeismic Condition Me= 84.975 tf.m/m

Mf= 84.98 tf.m/m





3) Ultimate Bending Moment Mu= 7.33E+06 kgf.cm= 73.31 tf.m





4) Checking of Single or Double bar arrangement

M1= M1= 2.21E+07 kgf.cm/m= 221.32 tf.m/m

M1= Resistance moment Mf= 84.975 tf.m M1>Mf : Design Tensile Bar Only

Cs= 10.739 k= 0.309 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 = 3.30 cm2Required Bar Area As req= 30.63 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= 82.5 cm

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



(a) Tensile Bar OnlyMf= 84.98 tf.m/m

S= 59.40 tf/mMf/(As*j*d)= 3,495.12 kgf/cm2 ok

92.92 kgf/cm2 checkS/(b*j*d)= 8.00 kgf/cm2 ok

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

k= 0.30 d= 82.5 cmj= 1-k/3= 0.90 n= 16

(b) Tensile Bar & Compressive BarMf= 84.98 tf.m

S= 59.40 tf

92.92 kgf/cm2 check3,495.12 kgf/cm2 ok1,486.66 kgf/cm2 ok

S/(b*j*d)= 8.00 kgf/cm2 okp= As/(b*d)= 0.00 p'= p'=As'/(b*d)= 0.00

k= 0.30 b= 100 cmLc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.13 d= 82.5 cm

j= 0.90 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 200 200 200Effective width of section b cm 100 100 100Height of Section, H4 h cm 50 50 50Concrete Cover (tensile) d1 cm 7.5 7.5 7.5Concrete Cover (compressive) d2 cm 0 0 0 Effective height of Section d cm 42.5 42.5 42.5Allowable Stress Concrete kgf/m2 70 70 105

Re-Bar kgf/m2 2500 2500 3750Shearing kgf/m2 7 7 10.5

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

Reinforcement BarTensile Bar Required As req. cm2 9.19 16.85 30.63

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 16.41 30.10 73.31 Design Axis Force Nd tf/m 0.00 0.00 0.00 Shearing Force S tf/m 26.51 16.86 59.40 Ultimate Bending Moment Mu tf.m 22.51 40.95 73.31 Resistance moment (single bar) Ml tf.m 82.92 123.39 221.32 Max Re-bar As max cm2 165.00 180.00 8.00 Min Re-bar As min cm2 4.50 4.50 4.50 Required Bar As req. cm2 9.19 16.85 30.63 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 42.5 42.5 42.5Compressive Stress kgf/cm2 19.98 ok 28.51 ok 92.92 okBending Tensile Stress kgf/cm2 1,332.96 ok 1,262.70 ok 3,495.12 ok

kgf/cm2 Mean Shearing Stress kgf/cm2 3.52 ok 2.31 ok 8.00 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

24/32 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] 1.402 tfmSp= Ka T/1.375*ln[(a+H1)/a] 1.703 tf

where, Ka Coefficient of active earth pressure 0.200T Wheel load of T-Load 7.000 tfa contact width of T-load 0.300 mH1 Height of Parapet 1.300 m

2) Sectional Force due to Earth PressurePh= Ph= 0.214 tfMe= (1/3)*Ph*H1 Me= 0.093 tfm

where, Kah 0.194d 0.233 (radian)g Unit Weight of Soil 1.300 tf/m3H1 Height of Parapet 1.300 m

3) Summary of Sectional ForceMo= Mp + Me 1.495 tfmSo= Sp + Ph 1.917 tf

2. Sectional Force with Impact Plate

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

Rf= (1/2)*Wd*Lo Rf= 1.958 tf/mRp= T/1.375 Rp= 5.091 tf/mR= Rf + Rp R= 7.048 tf/m B2= 0.35 mLx= (1/2)*B2+Bu Lx= 0.475 m Bu= 0.30 m (=Lp)Mf= R*Lx Mf= 3.348 tfm/m

where, Rf Reaction due self weight Wd tfWd w1 + w2 1.31 tf/m2w1 weight of soil on impact plate 0.59 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= 0.40 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 0.705 tf/m Lp= 0.300 m (= Bu)Mr = Rh*Yr 0.705 tfm/m

where, Kh seismic coefficient 0.18Yr = H1-Lp = 1.000 m

2) Sectional Force due to Earth Pressure

H1 > Ds ? H1= 1.300 m0.900 m

Peh 1= 0.011 tf/mMe 1= Peh*[(Lp)*(1/3)+(H1-Lp)] 0.013 tfm/mPeh 2= 0.102 tf/mMe 2= (1/3)*Peh*he 0.031 tfm/m

then Peh =Peh 1+ Peh2 0.114 tf/mMe =Me 1+Me2 0.043 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 0.455 2.4 1.092 0.18 0.197 0.650 0.128

10 0.090 2.4 0.216 0.18 0.039 0.750 0.02911 0.045 2.4 0.108 0.18 0.019 0.800 0.016S 0.255 0.172

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

4) Summary of Intersectional Force

Mo= Mr+Me+Mg 0.920 tfm/mSo= Rh+Reh+Gh 1.073 tf/m

3. Calculation of Required Reinforcement Bar

1) Cracking Moment

Mc= Mc= 349120.1 kgf.cm = 3.491 tf.m

where, Mc Cracking Moment kgf.cmZc Sectional Coefficient

Zc=b*B2^2/6 b=100 cm 20417 cm3Tensile strength of Concrete (bending)

17.10 kgf/cm2200 kgf/cm2

N Axis force (=0) 0Ac Area of Concrete = b*B2 3500 cm2b Effective Width 100 cmB2 Thickness of parapet 35 cm

2) Checking of Cracking Moment and Design Bending Moment

Design Bending Moment Mf 3.348 tf.m

Check Mf & Mc 1.7*Mf>Mc?, if yes check ultimate bending moment1.7*Mf = 5.692 tf.mMc= 3.491 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= 455836 kgf.cm = 4.558 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 28 cm

cover d1= 7 cmB2 = 35 cm


5.523 cm2Allowable Stress Rbar 2500 kgf/cm2

j= 1 -k/3 (=8/9 ) 0.866or

n= Young's modulus ratio 24Allowable Stress Concrete 70 kgf/cm2

Check Mu & Mc Mu = 4.558 tf.mMc = 3.491 tf.m Mu>Mc? ok

4) Bar Arrangement

(a) Front Face, with Impact Plate

Max Bar Area As max = 0.02*b*d = 56.0 cm2 Concrete Cover d1= 7 cmMin Bar Area As min = b*4.5%= 4.5 cm2 Then d= 28 cmRequired Bar Area As req= 6.066 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 = 50.0 cm2 Concrete Cover d1= 10 cmMin Bar Area As min = b*4.5%= 4.5 cm2 Then d= 25 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 = 50.0 cm2 Concrete Cover d1= 10 cmMin Bar Area As min = b*4.5%= 4.5 cm2 Then d= 25 cmRequired Bar Area As req= 1.650 cm2

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


(a) Front Face, with Impact Plate

Mf (front) 3.348 tf.m Concrete Cover d1= 7 cmS 0.000 tf Then d= 28 cm

Mf/(As*j*d) 942.86 kgf/cm2 ok2*Mf/(k*j*b*d^2) 17.07 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) 1.495 tf.m Concrete Cover d1= 10 cmSo 1.917 tf Then d= 25 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) 0.920 tf.m Concrete Cover d1= 10 cmSo 1.073 tf Then d= 25 cm

Mf/(As*j*d) 513.06 kgf/cm2 ok2*Mf/(k*j*b*d^2) 10.21 kgf/cm2 okS/(b*j*d) 0.48 kgf/cm2 ok

p=As/(b*d) 0.0032k={(n*p)^2+2*n*p}^0.5 - n*p 0.3233j= 1-k/3 0.8922


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 200 200 200Effective width of section b cm 100 100 100Height of Parapet B2 cm 35 35 35concrete cover (tensile) d1 cm 7 10 10concrete cover (compressive) d2 cm 0 0 0Effective width of Parapet d cm 28 25 25Allowable Stress Concrete kgf/m2 70 70 105 (K200)

Re-Bar kgf/m2 2500 2500 3750Shearing kgf/m2 7 7 10.5

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

Reinforcement BarTensile Bar Required As req. cm2 6.07 1.65Tensile Bar Designed As cm2 16.08 8.04

D16@125 D16@250Design Load

Design Bending Moment Mf tf.m 3.348 1.495 0.920Design Axis Force Nd tf 0.000 0.000 0.000Shearing Force S tf 0.000 1.917 1.073

Checking of Minimum Re-BarCracking Moment Mc tf.m 3.491 3.491 3.491

1.7*Mf 5.692 2.541 1.5651.7*Mf < Mc ? If no, check Mu check Mu ok

Ultimate Bending Moment Mu tf.m 4.558Mu > Mc ? ok

Max Re-bar As max cm2 56.0 50.0 50.0Min Re-bar As min cm2 4.5 4.5 4.5Required Bar As req. cm2 6.066 1.650Area 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 28 25 25Compressive Stress kgf/cm2 17.07 10.21Bending Tensile Stress kgf/cm2 942.86 513.06Mean Shearing Stress kgf/cm2 0 0.48

sc

scassatassy

scsstm

BAB III. PERHITUNGA 0

0.70 0.35

D16@200 D25@180+ 80.30

D16@200D16@200

1.30+ 79.00

[email protected]

D16@200 0.40D16@200

10.00 [email protected] D25@180

12.20

D16@200 [email protected]

5.00D25@180

D25@90

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

+ 69.000.40

1.800.50

+ 68.10

D25@300 D19@150 D25@300

5.00

1.20 1.00 2.80

29/32 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 1.30 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 0.07 tf/m2

Total dead load Wd= 0.78 tf/m2

2) Intersectional Force due Dead Load

Md= (1/8)*Wd*Ls^2 0.432731 tf.m

3) Intersectional Force due Live Load

17.53002 tf/m2ML= 3.663774 tf.m

where, ML bending moment due live loadT: wheel load of T-load 7.00 tfa contact width of T-load 0.30 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 4.096505 tf.m

3 Corbel1) Intersectional Force due Impact Plate

M1=R*bu M1= 2.115 tf.m

where, R total reaction form corbel 7.048 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 2.147 tf.m

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

Mc= Mc= 256496 kgf.cm = 2.565 tf.m

where, Mc cracking moment kgf.cmZc section modulus

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

tensile strength of concrete (bending)17.100 kgf/cm2

200 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 4.097 tf.m



Mu= 552464 kgf.cm = 5.525 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 200 kgf/cm2b effective width 100 cm

7.568 cm2allowable stress of reinforcement bar 2500 kgf/cm2

j= 1 -k/3 (=8/9 ) 0.866or

n= Young's modulus ratio 24allowable stress of concrete 70 kgf/cm2


4) Bar Arrangement

Checking of single or double bar arrangement

M1= 761401 kgf.cm = 7.614 tf.m

where, M1 resistance momentCs ={2m/[s*(1-s/3)]}^(1/2) 14.325s 0.402m 35.714


n 24

Check M1 > Mf? M1= 7.614 tf.m M1>Mf: Design Tensile Bar OnlyMf= 4.097 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= 7.568 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= 7.614 tf.md2= 0 cm Mf= 4.096505 tf.m

2500 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 4.096505 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 4.096505 tf.mS 0.000 tf

Mf/(b*d^2*Lc) 27.37 kgf/cm2 ok965.79 kgf/cm2 ok656.96 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= 256496 kgf.cm = 2.565 tf.m

where, Mc cracking moment kgf.cmZc section modulus

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

tensile strength of concrete (bending)17.100 kgf/cm2

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


Design bending moment Mf 2.147 tf.m



Mu= 292575 kgf.cm = 2.926 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 200 kgf/cm2b effective width 100 cm

4.312 cm2allowable stress of reinforcement bar 2500 kgf/cm2

j= 1 -k/3 (=8/9 ) 0.866or

n= Young's modulus ratio = 12 24allowable stress of concrete 70 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= 644450 kgf.cm = 6.444 tf.m

where, M1 resistance momentCs ={2m/[s*(1-s/3)]}^(1/2) 14.325s 0.402m 35.714


n 24

Check M1 > Mf? M1= 6.444 tf.m M1>Mf: Design Tensile Bar OnlyMf= 2.147 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= 4.312 cm2

19 @ 300 mmBar area As = 9.451 cm2 ok

5) Checking of Allowable StressMf 2.147 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 200 200Effective 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 70 70 (K200)

Re-Bar kgf/m2 2500 2500Shearing kgf/m2 7 7

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

Reinforcement BarTensile Bar Required As req. cm2 7.57 4.31

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 4.097 2.147 Design Axis Force Nd tf - -Shearing Force S tf - -

Checking of Minimum Re-BarCracking Moment Mc tf.m 2.565 2.926

1.7*Mf 6.964 3.650 1.7*Mf<Mc? If "no" check Mu check Mu check Mu

Ultimate Bending Moment Mu tf.m 5.525 2.926 Mu>1.7*Mf? ok 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 7.57 4.31 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 34.42 ok 25.85 okBending Tensile Stress kgf/cm2 1,018.96 ok 1,120.83 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

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