abutment jem ba an

7/28/2019 Abutment Jem Ba An

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ABUTMENT : Bendung ….

1. Input

Dimension

+ 80.45 HT

B1

B2

B3

+ 79.00 B4

B5

B6

BT

H1

H1 max

+ 75.00 H2

H3

H4

H5

+ 74.00 H6

H10

ho

Abutment wi

+ 69.00 Support from P

Rh1

Rb1

Hw1

+ 67.40 Hw2

Slope

(in case no soi

Design Load f

Unit Weight Wheel load

Soil 1.80 t/m3 T

Soil (saturated) 2.00 t/m3 Contact widConcrete 2.40 t/m3 a

Effective width

Reaction of superstructure Thickness of pav

Normal Vn=Rd+Rl 137.96 ton

Seismic Ve=Rd 71.27 ton With Impa

He= 12.83 ton (He=2 kh Rd, for fixed bearing)

Type of bearing Movable (He=kh Rd, for movable bearing) Width of Co

(Input Fixed or Movable) Thickness of Impac

Surcharge Load 0.70 t/m2 Length of Imp

Soil depth ab

Parameters

q : surcharge load (t/m2) 0.70

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

g : unit weight of earth (t/m3) 1.80

w : ground surface angle (degree) 0.00f : internal friction angle (degree) 30.00

d 1: friction angle between earth and wall (degree) normal 20.00 (=2/3f)

d 2: friction angle between earth and earth (degree) normal 0.00

d E1: friction angle between earth and wall (degree) seismic 15.00 (=1/2f)

d E2: friction angle between earth and earth (degree) seismic 24.20

b : wall angle (degree) 0.00

c : cohesion of soil (t/m2) (do not consider) 0.00

kh : 0.18

Uc: Uplift coefficient 1.00

f : Friction Coefficient =Tan f b = 0.60

N-SPT : 50.00

Qa : Allowable bearing capacity normal t/m2 20.83 ( max. 20.83 t/m

Qae: Allowable bearing capacity seismic t/m2 31.25 ( max. 31.25 t/m

Normal condition Seismic condition

Concrete Design Strength sc kgf/m2 175 175

Creep strain coefficient (concrete) 0.0035 0.0035

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

Overturning e= -0.32 m e= -0.62 m

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

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

Sliding Fs =Hu / H= 2.65 Fs =Hu / H= 1.26Fs > 2.00 OK Fs > 1.25 OK

Settlement Qmax = 13.19 t/m2 Qmax = 10.56 t/m2

(bearing capacity) Qmax < Qa OK Qmax < Qa OK

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

2.2 Structural Analysis

(1) Body Section A-A S

Normal Seismic Normal

Bar arrangement

Back face (tensile bar) f (mm) 25 25 25

(vertical) spacing (mm) 90 90 180

As (cm2) 273 273 136

Front face (compressive bar) f (mm) 25 25 25

(vertical) spacing (mm) 180 ok 180 ok 180

( As' > 0.5 As, cm2 ) 136 >=136.4 136 >=136.4 136 >

Hoop bar (horizontal) f (mm) 16

interval (mm) 200

Max interval (mm) 300

Design dimensions

Effective width (whole width) (cm) 500 500 500

Concrete cover : d1(cm) 7.5 7.5 7.5

d2(cm) 7.5 7.5 7.5

Effective height (cm) : d-d1(cm) 172.5 172.5 109.7

Design load Mf (t m) 704.25 1210.07 144.44

Nd (t) 321.86 255.17 219.11

S (t) 118.55 240.87 42.67

Checking of minimum reinforcement bar

Required bar (cm2) 258 356 53Checking of allowable stress as rectangular beam as column as rectangular bea

Compressive stress sc kgf/cm2 33 ok 58 ok 22 o

Bending stress ss kgf/cm2 1676 ok 2212 ok 1050 o

ss' kgf/cm2 - 787 ok -

Mean shearing stress tm kgf/cm2 1.54 ok 3.03 ok 3.04 o

(2) Footing Toe side Heel side

(Normal / Seismic) Normal

Bar arrangement

Upper (tensile bar) f (mm) 25

(bridge axis) spacing (mm) 150

As1 (cm2)

(compressive bar) f (mm) 25

(bridge axis) spacing (mm) 300 ok

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

spacing (mm) 300 ok 100 ok

Aso (cm2, > As /3) 9.45 >=5.45 28.35 >=4.50

Lower (tensile bar) f (mm) 25

(bridge axis) spacing (mm) 300

As2 (cm2) 16.36


(bridge axis) spacing (mm) 300 ok

As1' (cm2, >0.5 As1) 16.36 >=4.50

(distribution bar) f (mm) 19 19

spacing (mm) 150 ok 200 ok

Aso (cm2, >As /3) 18.90 >=5.45 14.18 >=5.45

Design dimensions

Effective width (unit width) (cm) 100 100

Concrete cover : d1(cm) 7.5 7.5

d2(cm) 7.5 7.5

Effective height (cm) : d-d1(cm) 92.5 92.5




(3) Parapet With Impact Plate Without Impact Plate With Impact Plate

Normal Normal Seismic

Bar arrangement

Back face (tensile bar) f (mm) 12 16

(vertical) spacing (mm) 250 250

As1 (cm2) 8.04


(vertical) spacing (mm) 250 ok

As1 (cm2, >As3/2) 8.04 >=8.04(distribution bar) f (mm) 12 12 12

(horizontal) spacing (mm) 250 250 ok 250

As2 (cm2, >As1/3) 4.52 >=4.50 4.52 >

Front face (tensile bar) f (mm) 16

(vertical) spacing (mm) 125

As3 (cm2, >As1/2) 16.08

(compressive bar) f (mm) 12 16

(vertical) spacing (mm) 125

As3 (cm2, >As1/2) 16.08 >

(distribution bar) f (mm) 12 16 16

(horizontal) spacing (mm) 250 ok 250

As6 (cm2, >As3/3) 8.04 >=5.36 8.04 >

Design dimensions

Effective width (unit width) (cm) 100 100

Concrete cover of fronf face (cm) 7

Concrete cover of back face (cm) 10 10

Effective height (cm) 33 30

Design load Mf (t m) 5.189 1.570

Nd (t) 0.000 0.000

S (t) 0.000 1.599

Checking of minimum reinforcement bar

Required bar (cm2) 10.78 3.14

Checking of allowable stress

Compressive stress 19.04 ok 12.92 o

Bending stress 1239.88 ok 722.95 o

Mean shearing stress 0 ok 0.59 o

(4) Impact Plate and Corbel

Impact PlateUpper Lower Upper

Bar arrangement

(main bar) f (mm) 16 19 19

spacing (mm) 300 ok 150 300

As1 (cm2) 6.70 >=4.50 18.90 9.45

(distribution bar) f (mm) 16 12 12

spacing (mm) 300 ok 250 ok 250

As2 (cm2, >As1/6) 6.70 >=4.50 4.52 >=4.50 4.52 >

Design dimensions

Effective width (unit width) (cm) 100 100 100

Concrete cover (cm) 5 5 7

Effective height (cm) 25 25 23

Design load Mf (t m) 5.818 3.146

Nd (t) - -

S (t) - -Checking of minimum reinforcement bar

Required bar (cm2) 14.73 8.66

Checking of allowable stress

Compressive stress 48.88 ok 37.88 o

Bending stress 1447.22 ok 1642.35 o

Mean shearing stress - -



DIMENSIONS OF ABUTMENT

Super structure Type T-Beam

Dimensions

HT B1 B2 B3 B4 B5 B6 BT H1 H2 H3

13.05 0.90 0.40 0.40 2.25 1.80 4.95 9.00 1.45 10.00 0.60

H1 max 1.45

Hw1 Hw2 Rh1 Rb1 Width of impact plate (Corbel)

6.60 7.60 0.73 0.45

WEIGHT OF ABUTMENT

Bo

1.45

10.00

1: 0

2.60

13.05 11.450.00 0.00

10.00

6 00 8 60 So

H5

H6B3

B6B5B4

HT

H10

H2

H1

H4

H3

B1 B2

BT

Rh1

Rb1

B1 B2

ho

H5

H6

B3HT H10

H2

H1

Bimp

1

2

1

3

4

5

10

11

12

13

15

16

1718

19

2021

22'

22



EARTH PRESSUREqs Parameters

qs : surcharge loa

q' : surcharge loa

g : unit weight o

1.45 g sat: unit weight o

w : ground surfa

f : internal fricti

d 1: friction angle

d 2: friction angle

d E1: friction angle

11.45 d E2: friction angle

13.05 b : wall angle (d

10.00 c : cohesion of s

kh :

w : ground surfa

f : internal fricti

6.60 7.60 d 1: friction angle

0.60 d 2: friction angle

1.00 d E1: friction angle

d E2: friction angle

b : wall angle (r

a : tan-1 kh (ra

9.00 d E: internal fricti

Uc: uplift coeffic

Earth Pressure

Normal Condition Seismic Condition

Description Description

Pa : active earth pressure (t/m2/m) 56.061 Pea : active earth p

Ka 1: coefficient of active earth pressure, earth and earth 0.333 Kea1 : coefficient ofKa 2: coefficient of active earth pressure, wall and earth 0.297

y : Vertical acting point 4.406 y : Vertical actin

X: Horizontal acting point = BT 9.0 X: Horizontal a

PV : Vertical Pressure = Pa sin d 0 PV : Vertical Pres

PH : Horizontal Pressure = Pa cos d 56.061 PH : Horizontal P

qa1 = 0.333 x 0.70 = 0.233 qa1 = 0.438

qa2 = 0.333 x 5.45 x 1.80 = 3.270 qa2 = 0.438

qa3 = 0.333 x 7.60 x 2.00 = 5.067 qa3 = 0.438

qw2 = -6.600 x 1.00 = -6.600 qw2 = -6.600

qu1 = -6.60 x 1.0 = -6.60 qu1 = -6.60

qu2 = -7.60 x 1.0 = -7.60 qu2 = -7.60


No. H Y HY No.

Pa1 = 0.233 x 13.050 = 3.045 6.525 19.869 Pa1= 0.000

Pa2 = 3.270 x 5.450 x 0.500 = 8.911 9.417 83.910 Pa2 = 4.301

Pa2' = 3.270 x 7.600 = 24.852 3.800 94.438 Pa2' = 4.301

Pa3 = 5.067 x 7.600 x 0.500 = 19.253 2.533 48.775 Pa3 = 6.664

Total (Pa) 56.061 4.406 246.991 Total

Pw2 = -6.600 x 6.600 x 0.500 = -21.780 2.200 -47.916 Pw2 = -6.600

Total -21.780 2.200 -47.916 Total

No. V X VX No.

Pu1 = -6.600 x 9.000 x 0.500 = -29.700 3.000 -89.100 Pu1 = -6.60

Pu2 = -7.600 x 9.000 x 0.500 = -34.200 6.000 -205.200 Pu2 = -7.60Total -63.900 4.606 -294.300 Total

Description

Description

HTH10

H2

H1

H4

H3

BT

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

Soil Toe 6.75 0.75 1.40 5.06 0.00

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

Earth Pressure 0.00 280.31 9.00 4.41 0.00 1234.95

Hydrostatic pressure 0.00 (108.90) 2.20 (239.58)

Uplift pressure (319.50) 4.61 (1471.50)

Surcharge Load 20.48 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? OK

BT/6= 1.500 m

2.2 Sliding

Friction Coefficient =Tan f b = 0.60

Shearing registance force at base

Hu=V . Tan f b 454.06 tf

Safety Rate against sliding

Fs=Hu/H 2.649 Fs is larger than 2.0 ? OK

2.3 Settlement

Allowale bearing capacity Qa 20.83 tf/m2

Reaction from the Foundation

Maximiun Fondation Reaction Q max

Q max = 13.193 tf/m2 Qmax is smaller than Qa? OK

Minimum Fondation Reaction Q max

Q 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

Soil Toe 6.75 0.75 1.40 5.06 0.00

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

Earth Pressure 142.91 317.99 9.00 4.28 1286.2 1362.3

Hydrostatic pressure (108.90) 2.20 (47.92)

Uplift pressue (319.50) 4.61 (294.30)

Surcharge 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? OK

2max(min)BTBL

M6

BTBL

VQ

×

±×

=



Bearing Capacity of soil

(1) Design Data

fB = 30.00o

cB = 0.00 t/m2

gs' =

B = 9.00 m z = 1.60 m L =

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

Calculation of ultimate bearing capacity will be obtained by appliying the following

Terzaghi's formula :

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

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

1 strip 1.002 square 1.30

3 rectangular, B x L 1.21

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

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

4 circular, diameter = B 1.30

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

Nc = 36.0 Nq = 23.0

f Nc Nq

0 5.7 0.0

5 7.0 1.4

10 9.0 2.7

15 12.0 4.520 17.0 7.5

25 24.0 13.0

30 36.0 23.0

35 57.0 44.0

37 70.0 50.0

39 > 82.0 50.0

a c Nc = 0.000

gs' z Nq = 36.800

b gs' B Ng = 72.000

qu = 108.80 t/m2

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

qa = qu / 3 = 36.27 t/m2

(safety factor = 3 , no

qae = qu / 2 = 54.40 t/m2

(safety factor = 2 , se



STRUCTURAL CALCULATION

A ABUTMENT BODY

1 Load and Bending Moment of Abutment Body

Dimensions

B1 B2 B3 B5 H1

0.90 0.40 0.40 1.80 1.45

Hw1 Hw2

5.00 6.00

Abutment Width

Bw= 5.00 m Plat Form of Impact P

1.45 Seismic Coefficient kh= 0.18

Support from Parapet hs= 0.45

A-A section

Normal Seismic

Vertical Distance Moment Horizontal

No. Load X Mx Load

t/m m tf.m/m t/m

1 1.392 0.200 0.278 0.251

10.00 2 6.048 0.450 2.722 1.089

8.60 3 0.960 0.200 0.192 0.173

5.73 4 0.192 0.133 0.026 0.035

or 5 18.576 0.450 8.359 3.344

5.00 6.00 5' 9.288 -0.300 -2.786 1.672

10 0.216 -0.550 -0.119 0.039

11 0.108 -0.500 -0.054 0.019

S 36.780 8.618 6.620

Total 183.900 43.088 33.102

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 Seismic

Vertical Distance Moment Horizontal

No. Load X Mx Load

t/m m tf.m/m t/m

1 1.392 -0.514 -0.715 0.251

2 6.048 0.136 0.823 1.089

3 0.960 -0.514 -0.493 0.173

4 0.192 -0.447 -0.086 0.035

5 5.616 0.136 0.764 1.011

5' 1.698 -0.405 -0.687 0.306

10 0.216 -0.864 -0.187 0.039

11 0.108 -0.814 -0.088 0.019

S 16.230 -0.669 2.921

Total 81.149 -3.347 14.607

2 Load and Bending Moment due to Super Structure Normal 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/m

Section A-A

Normal Rd + Rl 137.962 0.45 62.083 0.00 0.00 0.00 62.083

Seismic Rd 71.272 0.45 32.072 12.83 10.97 140.69 172.763

Section B-B

Normal Rd + Rl 137.962 0.45 62.083 0.00 0.00 0.00 62.083

Seismic 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 Condition

ho

H5

H6

B3

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 Condition

qe1= q'*Kea qe1= 0.000 tf/m

qe2= (q'+g*H)*Kea qe2= 4.301 tf/m

EeH= (qe1+qe2)*(1/2)*H*cosd EaH= 11.320 tf/m

y= [H*(2qe1+qe2)]/[3*(qe1+qe2)] y= 1.817 m

S= EeH*Bw S= 56.599 tf

My= EeH*Bw*y My= 102.822 tf.m/m

where, q' Surcharge Load (seismic) 0 tf/m2

Kea coefficient of active earth pressure 0.438

g Unit weight of soil 1.8 tf/m3

d friction angle between earth and wall 0.2618 radian

H 5.45 m

y Acting Point (m)

S Shearing Force(tf)

Bw Abutment Width 5.00 m

2) Section A-A

a) Normal Condition

Description

EaH : active earth pressure (t/m2/m) = Pa . cos d 36.210

Ka2 : coefficient of active earth pressure 0.297y : Vertical acting point y= 3.884

S=EaH*Bw S= 181.051

My=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. H Y HY

Pa1 = 0.208 x 11.450 = 2.383 5.725 13.64

Pa2 = 2.917 x 5.450 x 0.500 = 7.948 7.817 62.12

Pa2' = 2.917 x 6.000 = 17.500 3.000 52.50

Pa3 = 3.568 x 6.000 x 0.500 = 10.703 2.000 21.40Total (Pa),per m 38.534 3.884 149.67

Pw2 = -5.000 x 5.000 x 0.500 = -12.500 1.667 -20.83

Total, per m -12.500 -20.83

Total (per width) width = 5.00 m -62.500 1.667 -104.16

b) Seismic Condition

Description

Pea : active earth pressure (t/m2/m) = Pa . cos d 51.489

Kea : coefficient of active earth pressure 0.438

y : Vertical acting point y= 3.763

S=EaH*Bw S= 257.443

My=S*y My= 968.728

qa1 = 0.438 x 0.00 = 0.000

qa2 = 0.438 x 5.45 x 1.80 = 4.301qa3 = 0.438 x 6.00 x 2.00 = 5.261

qw2 = -5.000 x 1.00 = -5.000

No. H Y HY

Pa1= 0.000 x 11.450 = 0.000 5.725 0.00

Pa2 = 4.301 x 5.450 x 0.500 = 11.719 7.817 91.60

Pa2' = 4.301 x 6.000 = 25.804 3.000 77.41

Pa3 = 5.261 x 6.000 x 0.500 = 15.782 2.000 31.56

Total (per m) 53.305 3.763 200.58

Pw2 = -5.000 x 5.000 x 0.500 = -12.500 1.667 -20.83

Total (per m) -12.500 -20.83

Total (per width) width= 5.00 m -62.500 1.667 -104.16

4 Summary of Intersectional Force

Normal Condition Seismic Condition

Description

Description



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

1) Cracking Moment Section A-A

Mc= Zc*(s'ck + N/Ac) Mc= 90516242 kgf.cm/m

= 905 tf.m/m

where, Mc Cracking Moment kgf.cm

Zc Section Modulus

Zc=1/6*b*h12

b=100 cm 2,700,000 cm3

s'ck Tensile strength of Concrete (bending)

s'ck = 0.5*sck 2/3

15.6 kgf/cm2

s ck= 175 kgf/cm2

N Axial force 321,862 kg

Ac Area of Concrete = b*h1 18000 cm2

h1 thickness of section, B5 180 cm

b 500 cm

Section A-A

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

2) Required Bar Area As_req= Mf / (s sa*j*d)

As_req= 258 cm2

s sa= Allowable Stress R-bar 1850 kgf/cm2

j= 1 - k/3 0.854

k= n / (n + ssa / sca) 0.438

n= Young's modulus ratio 24

sca= Allowable Stress Concrete 60 kgf/cm2

d= Effective height = h1-d1 173 cm

d1= 7.5 cm

h1= 180 cm

3) Ultimate Bending Moment Mu= As*ssy { d - 0.5*[As*ssy]/[0.85*sck*b]}

Mu= 1,297 tf.m

Mu= Ultimate Bending Moment

As= Area of Tensile Bar

ssy= Yielding point of Tensile Bar (Spec >295 N/mm2) 3000 kgf/cm2

s'ck= Design Compressive Strength of Concrete 175 kgf/cm2

b= Effective Width 500 cm

4) Checking : Single or Double Bar Arrangement M1=

M1= 1,669 tf.m

M1= Resistance moment

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

12.844

j= 1 - k/3 0.854

k= n / (n + ssa / sca) 0.438

n= ssa/sca 31

ssa= 1850 kgf/cm2

sca= 60 kgf/cm2

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

Mf= 704 tf.m

Mf < M l : Tensile Bar Only M f < M l : Tens

(a) Tensile Bar

Max Bar Area : 2%*b*d As max = 1,725 cm2

Min Bar Area : 4.5%*b As min = 23 cm2

Estimation of Required Bar Area : As_req= 258 cm2

Apply f :

Required Bar Nos : b/pitch= 56 nos

Bar Area : As= 273 cm2

ok

Resistance Moment by Tensile bar As2 As1= Mf / {ssa*j*d}

As1= 258 cm2

As2= 14 cm2

25 @ 90

(d/Cs)2*ssa*b



Bar Area : As' = 0 cm2 ok



5) Checking of Allowable Stress

(a) Tensile Bar Only

Mf= 704 tf.m

S= 119 tf

ss = Mf/(As*j*d)= 1,676 kgf/cm2 ok

sc = 2*Mf/(k*j*b*d

2

)= 33 kgf/cm2 ok tm = S/(b*j*d)= 2 kgf/cm2 ok

p= As/(b*d)= 0.003

k= {(n*p)2+2*n*p}

1/2- n*p= 0.321

j= 1-k/3= 0.893

b= 500 cm

d= 172.5 cm

n= 24

(b) Tensile Bar & Compressive Bar

Mf= 0 tf.m

S= 0 tf

sc = Mf/(b*d2*Lc)= 0 kgf/cm2 ok

ss = n*sc*(1-k)/k= 0 kgf/cm2 ok

ss' = n*sc*(k-d2/d)/k= 0 kgf/cm2 ok

tm = S/(b*j*d)= 0 kgf/cm2 ok

p= As/(b*d)= 0.003

p'= p'=As'/(b*d)= 0.000

k= {n2(p+p')

2+2n(p+p'*d2/d)}

1/2-n(p+p')= 0.321

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

j= (1-d2/d)+k 2/{2*n*p*(1-k)}*(d2/d-k/3)= 0.893

b= 500 cm

d2= 0 cmd= 172.5 cm

n= 24

h d

x=kd

b

As

d1

h d

x=kd

b

As

As'

d2

d1



6 Calculation of Required Reinforcement Bar as Column, Seismic Condition

1) Minimum Area as Column Section A-A

Acmin = N / (0.008*ssa+sca) 2,274 cm2

ssa= Allowable stress of Reinforcement bar 2775 kgf/m2

sca= Allowable stress of Concrete 90 kgf/m2

N= Axial force 255 tf

Acdes= b*h 90,000 > Acmin ok

Minimum Reinforcement Bar

(a) As a beam 4.5% * b As min= 23 cm2

(b) As a column 0.8% * Acmin As min= 18 cm2

Maximum Reinforcement Bar

(a) As a beam 2%*b*d As max = 1,725 cm2

(b) As a column 6% * Ac As max = 5,400 cm2

2) Required Reinforcement Bar

Design Bending Moment Mf= 1,210 tf.m

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

As_req= 237 cm2

sc= Stress of concrete sc= 72.09 kg.cm2

Eq1= sc3

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

2- 6*Ms/(n*b*d

2)*ssa*sc - 3*Ms/(n

2*b*d

2)*ssa

2= 0

ssa= Allowable stress of Reinforcement Bar 2775 kg.cm2

Ms= Eccentric Moment, Ms=N(e+c) 1.42E+08 kgf.cm

e= Essentric Distance e=M/N 474.22 cm

M= Design Bending Moment 1,210 tf.m

N= Axial Force 255 tf

n= Young's Modulus Ratio 16

c= c=h/2 - d1 82.5 cm

h= Height of Section 180 cm

b= Width of section 500 cm

d1= Concrete Cover 8 cm

d= Effective Width of section d=h-d1 173 cms= n*sc/(n*sc+ssa) 0.294

[3*ssa/(2*n)-3*Ms/(b*d2)]= 232

6*Ms/(n*b*d2)*ssa= 9,936

3*Ms/(n2*b*d^2)*ssa

2= 861,643

sc (trial)= 72 sc (trial)

Eq1 (trial)= 0 ok Eq1 (trial

cross check 0 ok cross che

3) Ultimate Bending Moment

Mu= c*(h/2-0.4X)+Ts'(h/2-d2)+Ts(h/2-d1)

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

Mu= Min (Mu1,Mu2) in case X>0 in case X<0

Mu1= 1.46E+08 1.86E+08

c= 0.68*sck*b*X 672859 -1155.34

sck= design strength of Concrete 175 kg/cm2

b= Width of section 500 cm

X= solve the equation Eq2 below 11.309 -9.709

Ts'= As'*Es*ecu*(X-d2)/X 293344 1543872

As'= Compressive Bar, As'=0.5 As 119

As= Required Reinforcement Bar (Tensile) 237

Es= Young's modulus (reinforcement bar) 2,100,000

ecu= Creep strain coefficient (concrete) 0

h= Height of Section 180 cm

d1= Concrete Cover (tensile side) 7.5 cm

d2= Concrete Cover (compressive side) 7.5 cm

Eq2 = a*X2

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

a= 0.68*sck*b 50 59500

b= As'*Es*ecu 871013

T A * 711031



4) Bar Arrangement

Max Bar Area : As max = 1,725 cm2

Min Bar Area : As min = 23 cm2

(a) Tensile Bar

Required Bar Area As req= 237 cm2

Apply f =

Column width b= 500 cm2

Bar Area As = 273 cm2 ok

(b) Compressive Bar

Required Bar Area As' req= 119 cm2

Apply f =

Column width b= 500 cm2

Bar Area As' = 136 cm2 ok

(c ) Hoop Bar

Minimum Diameter f' min = 12 mm

Apply f' = 16 mm

Bar Interval shall satisfy the following conditions:

d = 1725 mm

t < 12*f = 300 mm

t < 48*f '= 768 mm

then t = 200 mm ok

5) Checking of Allowable Stress, Seismic Condition


Mf= 1,210 tf.m

S= 241 tf

ss = Mf/(As*j*d)= 2,828 kgf/cm2 check

sc = 2*Mf/(k*j*b*d2)= 66 kgf/cm2 ok

tm = S/(b*j*d)= 3 kgf/cm2 ok p= As/(b*d)= 0.00316

k= {(n*p)^2+2*n*p}^0.5 - n*p= 0.27149

j= 1-k/3= 0.90950

b= 500 cm

d= 172.5 cm

n= 16


Mf= 1,210 tf.m

S= 241 tf

sc = Mf/(b*d2*Lc)= 60 kgf/cm2 ok

ss = n*sc*(1-k)/k= 2,793 kgf/cm2 check

ss' = n*sc*(k-d2/d)/k= 790 kgf/cm2 ok

tm = S/(b*j*d)= 3 kgf/cm2 ok

p= As/(b*d)= 0.003

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

k= {n2(p+p')

2+2n(p+p'*d2/d)}

1/2-n(p+p')= 0.254

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

j= (1-d2/d)+k 2/{2*n*p*(1-k)}*(d2/d-k/3)= 0.921

b= 500 cm

d2= 7.5 cm

d= 172.5 cm

n= 16

25 @ 90

25 @ 180

d1

d d2

h d

x=kd

b

As

d1

h d

x=kd

b

As'

d2



7 Check for Stress of Concrete and Reinforcement bar as Column

Section A-A

Normal Seismic

Height of section h cm 180 180

Effective width b cm 500 500

Concrete cover d1 cm 7.5 7.5

d2 cm 7.5 7.5

Reinforcement bar As cm2 273 273

As' cm2 0 136

Morment Mf kgf/cm 7.04.E+07 1.21.E+08

Axcis force N kgf 321,862 255,172

Shearing force S kgf 118,551 240,874

Young's modulus ratio n 24 16

c 83 83

e 219 474

a1 386 1153

b1 23665 39405

c1 -4082130 -5105274

Neutral line x 72 51

0 0

ok ok

a2 18005 12697

b2 66 73

b3 7497 7088

b4 65 43

b5 7497 7088

b6 100 122

Stressing sc 29.04 ok 57.70 ok

ss 972.30 ok 2212.39 ok

ss' 624.33 ok 786.83 ok

tc 1.49 ok 3.03 ok

SUMMARY OF DESIGN CALCULATIONSection A-A

Description Abbr. unit

Calculation Condition Column

Princi ple Dimensions


Effective width of section b cm 500 500

Height of Section h cm 180 180

Concrete Cover (tensile) d1 cm 7.5 7.5

Concrete Cover (compressive) d2 cm 7.5 7.5

Effective height of Section d cm 172.5 172.5

Allowable Stress Concrete sca kgf/m2 60 90

Re-Bar ssa kgf/m2 1850 2775

Shearing ta kgf/m2 5.5 8.25

Yielding Point of Reinforcement Bar ssy kgf/cm2 3000 3000

Reinfor cement Bar

Tensile Bar Required As_req. cm2 258 237Designed As cm2 273 273

Compressive Bar Required As'_req. cm2 0 119

Designed As' cm2 0 136

Hoop Bar Designed Aso

Design L oad

Design Bending Moment Mf tf.m 704 1,210

Design Axis Force Nd tf 322 255

Shearing Force S tf 119 241

Checking of M inimum Re-Bar

Ultimate Bending Moment Mu tf.m 1,297 1,460Max Re-bar As max cm2 1,725 1,725

Min Re-bar As min cm2 23 23

Required Bar As req cm2 258 356

D25@180

D16@200

D25@90 D25@90

Normal Condition

Rectangular Beam

Seismic Condition



B FOOTING

1 Load and Bending Moment of Footing

1) Computation Conditions

Normal Condition Toe Side

Heel Side

Seismic Condition

2) Footing

Dimensions

B4 B5 B6 B

2.25 1.80 4.95 9.Abutment Width

Bw= 5.00 m

Unit Weight of Material

Soil 1800 kgf/m

Concrete 2400 kgf/m

Surcharge Load q 700 kgf/m

q'= q seismic 0 kgf/m

Toe Side normal condition seismic condition

Area unit Vertical Distance Moment Horizontal Distance Moment

No. weight Load X Mx Load Y My

m2 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 0.350

F1 2.250 2.4 5.400 1.125 6.075 0.972 0.500 0.486

sub-total 7.020 7.290 1.264 0.836

Distance l1 l1 (X) 1.038 m l1 (Y) 0.662 m

Heel Side normal condition seismic condition

Area unit Vertical Distance Moment Horizontal Distance Moment

No. weight Load N X Mx Load Y My

m2 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 0.834

F3 4.950 2.400 11.880 2.475 29.403 2.138 0.500 1.069

sub-total 15.444 35.284 2.780 1.903

Distance l2 l2 (X) 2.285 m l2 (Y) 0.685 m

3) Earth Pressure, Heel Side Only

Normal Condition Seismic Conditio

Dead Load (soil) W1= 116.060 tf/m We1= 1

Distance l3 =Xo-(B4+B5) l3 = 2.379 m l3 =

Bending Moment Md Md= 276 157 tf m/m Mde= 2

BT

Q

X

BT

Q maxQ min

B4

Ra=Qmax

B6

Ra'

l 2

M

(0,0)

B6B5B4

H4

H3

BT

F1 F2 F3

7 86

9



4) Reaction from Foundation

Reaction from the Foundation

Normal Q max 13.193 tf/m2

Qmin 20.441 tf/m2

Seismic Qmax 10.560 tf/m2

Qmin 25.552

X 9.000 m

Reaction from Foundation, Toe Side

Normal Condition

Rat= Q max Rat= 13.193 tf/m2 Cantilever length

Ra't= Q max-(B4/BT)*(Q max - Qmin) Ra't= 15.005 tf/m2

Seismic Condition

Reat= Q Reat= 10.560 tf/m2 Cantilever length

Rea't= Q*(X-B4)/X Reat'= 7.920 tf/m2

Reaction from Foundation, Heel Side

Normal Condition

Rah= Q min Rah= 20.441 tf/m2 Cantilever length

Ra'h= Q max-[(B4+B5)/BT]*(Q max - Qmin) Ra'h= 16.455 tf/m2

Seismic ConditionReah'= Q*(X-B4-B5)/X Rea'h= 17.306 tf/m2 Cantilever length

Reah= Qmin Reah= 25.552 tf/m2

Reah'= Q*(X-B4-B5)/X Rea'h= 5.808 tf/m2 Cantilever length

Reah= 0 Reah= 0.000 tf/m2

Reah'= Q*(X-B4-B5)/X Rea'h= 17.306 tf/m2 Cantilever length

Reah= Qmin Reah= 25.552 tf/m2

5) Intersectional Force due Reaction

Axial Force = 0 tf/m

Toe Section Heel

Description Moment Shearing Moment Shearing Moment She

M S Me Se M

(tf.m/m) (tf/m) (tf.m/m) (tf/m) (tf.m/m) (tf

Footing -7.290 -7.020 -7.290 -7.020 35.284 1Dead Load, Earth 276.157 11

Surcharge Load 8.576

Earth Pressure 42.284 1

Reaction form Foundation 34.924 31.723 24.502 20.790 -234.147 -9

Total - per m 27.634 24.703 17.212 13.770 128.153 5

Total -5 m 138.169 123.514 86.060 68.848 640.765 28

2 Calculation of Required Reinforcement Bar :

1) Design Bending Moment ( Mf )

Normal Condition M= 27.634 tf.m/m

Seismic Condition Me= 17.212 tf.m/mMax. Design Bending Moment Mf= 27.63 tf.m/m

2) Required Bar Area

As= Mf / (s sa*j*d) 11.47 cm2


j= 1 - k/3 0.854

k= n / (n + ssa / sca) 0.438




Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]} Mu= 5.21E+06 kgf.cm

52.07 tf.mwhere, Mu Ultimate Bending Moment

As Area of Tensile Bar

s sy Yielding point of Tensile Bar 3000 kgf/cm2 (Spec >295 N/mm2)

Normal CondiSeismic Condition Normal Condition

Toe Section



(a) Tensile Bar

Max Bar Area : 2%*b*d = 305.00 cm2

Min Bar Area : 4.5%*b = 4.50 cm2

Required Bar Area As req= 11.47 cm2

Apply f =

Required Bar Nos Nos=b/pitch = 3.333333 nos

Bar Area As = 16.36 cm2 ok

(b) Compressive Bar, in case M1<Mf

M' = Mf - M1

M'= 0.00 tf.m

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

Required Bar Area As'= 0.00 cm2

d= 152.5 cm

d2= 0.00 cm

ssa= 1850 kgf/cm2

Apply f =

Bar Area As' = 0.00 cm2 ok



Mf= 27.63 tf.m/m

S= 24.70 tf/m

ss = Mf/(As*j*d)= 1,187.67 kgf/cm2 ok

sc = 2*Mf/(k*j*b*d2)= 12.58 kgf/cm2 ok

tm = S/(b*j*d)= 1.74 kgf/cm2 ok

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

k= {(n*p)2+2*n*p}

1/2- n*p= 0.20 d= 152.5 cm

j= 1-k/3= 0.93 n= 24


Mf= 27.63 tf.m

S= 24.70 tf

sc = Mf/(b*d2*Lc)= 12.58 kgf/cm2 ok

ss = n*sc*(1-k)/k= 1,187.67 kgf/cm2 ok

ss' = n*sc*(k-d2/d)/k= 301.84 kgf/cm2 ok


p= As/(b*d)= 0.00

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

k= {n2(p+p')

2+2n(p+p'*d2/d)}

1/2-n(p+p')= 0.20 d= 152.5 cm

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

j= (1-d2/d)+k 2/{2*n*p*(1-k)}*(d2/d-k/3)= 0.93 n= 24

3 Calculation of Required Reinforcement Bar : ( Normal Condition )

1) Design Bending Moment Me>0


Seismic Condition Me= 75.755 tf.m/m

Mf= 128.15 tf.m/m


As= Mf / (s sa*j*d) 53.18 cm2


j= 1 - k/3 0.854

k= n / (n + ssa / sca) 0.438




Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]} Mu= 23475776 kgf.cm

= 234 76 tf m

Heel Section

25 @ 300 d1

h d

h

d1

d

h

h

Tensile

Com ressivee






Apply f =

Required Bar Nos Nos=b/pitch = 6.666667 nos Pitch shall be same as that of toe

Bar Area As = 32.72 cm2 check or arrange compressive bar


M' = Mf - M1 M'= 0.00 tf.m

As' = M'/[ssa*(d - d2)] As'= 0.00 cm2

d= 152.5 cm

d2= 0.00 cm

Required Bar Area As' req= 0.00 cm2

Apply f =




Mf= 128.15 tf.m/m

S= 56.47 tf/mss = Mf/(As*j*d)= 2,825.55 kgf/cm2 check check

sc = 2*Mf/(k*j*b*d2)= 44.33 kgf/cm2 ok ok

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

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

k= {(n*p)2+2*n*p}

1/2- n*p= 0.27 d= 152.5 cm

j= 1-k/3= 0.91 n= 24


Mf= 128.15 tf.m

S= 56.47 tf


ss = n*sc*(1-k)/k= 2,825.55 kgf/cm2 check

ss' = n*sc*(k-d2/d)/k= 1,063.96 kgf/cm2 ok

tm = S/(b*j*d)= 4.07 kgf/cm2 ok p= As/(b*d)= 0.00

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

k= {n2(p+p')

2+2n(p+p'*d2/d)}

1/2-n(p+p')= 0.27 b= 100 cm

Lc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.12 d= 152.5 cm

j= (1-d2/d)+k 2/{2*n*p*(1-k)}*(d2/d-k/3)= 0.91 d2= 0.00 cm

n= 24

4 Calculation of Required Reinforcement Bar : ( Seismic Condition )

1) Design Bending Moment Me>0


Seismic Condition Me= 75.755 tf.m/m

Mf= 75.76 tf.m/m


As= Mf / (s sa*j*d) 20.20 cm2


j= 1 - k/3 0.886

k= n / (n + ssa / sca) 0.438



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

= 91.19 tf.m

where, Mu Ultimate Bending Moment

As Area of Tensile Bar

s sy Yielding point of Tensile Bar 3000 kgf/cm2 (Spec >295 N/mm2)

Heel Section

25 @ 150

d1

h d

h

d1

d

h

h

Tensile

Com ressivee



(a) Tensile Bar




Apply f =Required Bar Nos Nos=b/pitch = 6.666667 nos

Bar Area As = 32.72 cm2 check or arrange compressive bar


M' = Mf - M1 M'= 0.00 tf.m

As' = M'/[ssa*(d - d2)] As'= 0.00 cm2

d= 152.5 cm

d2= 0.00 cm

Required Bar Area As' req= 0.00 cm2

Apply f =




Mf= 75.76 tf.m/m

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

sc = 2*Mf/(k*j*b*d2)= 30.68 kgf/cm2 ok


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

k= {(n*p)2+2*n*p}

1/2- n*p= 0.23 d= 152.5 cm

j= 1-k/3= 0.92 n= 16


Mf= 75.76 tf.m

S= 38.67 tf


ss = n*sc*(1-k)/k= 1,643.98 kgf/cm2 ok

ss' = n*sc*(k-d2/d)/k= 490.93 kgf/cm2 ok

tm = S/(b*j*d)= 2.75 kgf/cm2 ok p= As/(b*d)= 0.00

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

k= {n2(p+p')

2+2n(p+p'*d2/d)}

1/2-n(p+p')= 0.23 b= 100 cm

Lc= 0.5 k (1-k/3)+(np'/k) (k-d2/d)(1-d2/d)= 0.11 d= 152.5 cm

j= (1-d2/d)+k 2/{2*n*p*(1-k)}*(d2/d-k/3)= 0.92 d2= 0.00 cm

n= 16

SUMMARY OF DESIGN CALCULATION, FOOTING

Description Abbr. unit

Calculation Condition Normal Normal S

Principle Dimensions

Concrete Design Strength sc kgf/m2 175 175Effective width of section b cm 100 100

Height of Section, H4 h cm 100 100

Concrete Cover (tensile) d1 cm 7.5 7.5

Concrete Cover (compressive) d2 cm 0 0

Effective height of Section d cm 92.5 92.5




tma kgf/m2 14 14


Reinforcement Bar

Tensile Bar Required As req. cm2 11.47 53.18

Designed As cm2 16.36 32.72

Compressive Bar Required As' req. cm2 0.00 0.00

Designed As' cm2 0.00 0.00

Design Load

D25@300 D25@150

Toe Side Heel Side

25 @ 150

d1

h d

h

d1

d

h

h



C PARAPET WALL

1 Sectional Force without Impact Plate

(1) Rear Face Reinforcement Bar Mo= Mp + Me

So= Sp + Ph

where,

Mo Bending Moment at Parapet (

So Shearing Force at Platform (tf

Mp Bending Moment due T-Loa

Me Bending Moment due Earth P

Sp Shearing Force dueT-Load (tf

Ph Earth Pressure (tf/m)

1) Sectional Force due to T-Load

Mp= Ka T/1.375*{-H1+(H1+a)*ln[(a+H1)/a] 4.393 tfm

Sp= Ka T/1.375*ln[(a+H1)/a] 4.563 tf

where, Ka Coefficient of active earth pressure 0.297

T Wheel load of T-Load 10.000 tf

a contact width of T-load 0.200 m

H1 Height of Parapet 1.450 m

2) Sectional Force due to Earth Pressure

Ph= 0.5 g *Kah*H1^2 Ph= 0.554 tf

Me= (1/3)*Ph*H1 Me= 0.268 tfm

where, Kah Ka*cos d 0.293

d friction angle between the parapet wall and soil = (1/3)f 0.175 (radian)g Unit Weight of Soil 1.800 tf/m3

H1 Height of Parapet 1.450 m

3) Summary of Sectional Force

Mo= Mp + Me 4.661 tfm

So= 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/m

Rp= T/1.375 Rp= 7.273 tf/m

R= Rf + Rp R= 10.378 tf/m B2= 0.40 m

Lx= (1/2)*B2+Bu Lx= 0.500 m Bu= 0.30 mMf= R*Lx Mf= 5.189 tfm/m

where Rf Reaction due self weight Wd tf

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 m

H1-Ds= 0.750 m

Peh 1= (1/2)*g*Kah*Lp^2 0.024 tf/m

Me 1= Peh*[(Lp)*(1/3)+(H1-Lp)] 0.030 tfm/m

Peh 2= (1/2)*g*Kah*he^2 0.148 tf/m

Me 2= (1/3)*Peh*he 0.037 tfm/m

then Peh =Peh 1+ Peh2 0.172 tf/m

Me =Me 1+Me2 0.067 tfm/m

3) Intersectional Force of Parapet due to Earth Pressure

Section Area Unit Weight Kh Se

(m2) Weight (tf) W

G

1 0.580 2.4 1.392 0.18

10 0.090 2.4 0.216 0.18

11 0.045 2.4 0.108 0.18

S

Note: if H1 < Dc, areas 10 & 11 w

4) Summary of Intersectional Force

Mo= Mr+Me+Mg 1.570 tfm/m

So= Rh+Reh+Gh 1.599 tf/m

3. Calculation of Required Reinforcement Bar

1) Cracking Moment

Mc= Zc*(s'ck + N/Ac) Mc= 417155 kgf.cm = 4.172 tf.m

where, Mc Cracking Moment kgf.cm

Zc Sectional Coefficient

Zc=b*B2^2/6 b=100 cm 26667 cm3

s'ck Tensile strength of Concrete (bending)s'ck = 0.5*sck^(2/3) 15.64 kgf/cm2

s ck= 175 kgf/cm2

N Axis force (=0) 0

B1 B2 Bu

Ds

FtR H

Lo

heYr

H1 H1

g K ah he

B5

B2

Lp

Lp

Lp

H1 110

11

B1




Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]} Mu= 955213 kgf.cm = 9.552

where, Mu Ultimate Bending Moment tf.m

As Area of Tensile Bar cm2

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

cover d1= 7 cm

B2 = 40 cm

s'ck Design Compressive Strength of Concrete 175 kgf/cm2

b Effective Width 100 cm

As=Mf/(s sa*j*d) 9.951 cm2

s sa= Allowable Stress Rbar 1850 kgf/cm2

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

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


s ca Allowable Stress Concrete 60 kgf/cm2

Check Mu & Mc Mu = 9.552 tf.m

Mc = 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

Min Bar Area As min = b*4.5%= 4.5 cm2 Then d= 33


Apply f = 16 @ 125 mm


(b) Rear Face, without Impact Plate



Required Bar Area As req= cm2

Apply f = 12 @ 250 mm spacing of body, tensile f 25

Bar Area As = cm2

(c) Rear Face, with Impact Plate




Apply f = 16 @ 250 mm spacing of body, tensile f 25



(a) Front Face, with Impact Plate

Mf (front) 5.189 tf.m Concrete Cover d1= 7

S 0.000 tf Then d= 33

ss = Mf/(As*j*d) 1239.88 kgf/cm2 check ss < ssa ? ok

sc = 2*Mf/(k*j*b*d^2) 19.04 kgf/cm2 check sc < sca ? ok

p=As/(b*d) 0.0230

k={(n*p)^2+2*n*p}^0.5 - n*p 0.6347

j= 1-k/3 0.7884

(b) Rear Face, without Impact Plate

Mf (rear) 4.661 tf.m Concrete Cover d1= 10

So 5.117 tf Then d= 30

ss = Mf/(As*j*d) kgf/cm2 check ss < ssa ?

sc = 2*Mf/(k*j*b*d^2) kgf/cm2 check sc < sca ?

tm = S/(b*j*d) kgf/cm2 check tm < ta ?

p=As/(b*d)

k={(n*p)^2+2*n*p}^0.5 - n*p

j= 1-k/3



4. Summary of Design Calculation

Description Abbr. unit Front Face Back Face

Provision on Impact Plate Yes No Yes

Calculation Condition Normal Normal Seismic


Concrete Design Strength sc kgf/m2 175 175 175

Effective width of section b cm 100 100 100

Height of Parapet B2 cm 40 40 40

concrete cover (tensile) d1 cm 7 10 10

concrete cover (compressive) d2 cm 0 0 0

Effective width of Parapet d cm 33 30 30

Allowable Stress Concrete sca kgf/m2 60 60 90

Re-Bar ssa kgf/m2 1850 1850 2775

Shearing ta kgf/m2 5.5 5.5 8.25

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

Reinforcement Bar


Tensile Bar Designed As cm2 16.08 8.04D16@125 D16@250

Design Load

Design Bending Moment Mf tf.m 5.189 4.661 1.570

Design Axis Force Nd tf 0.000 0.000 0.000

Shearing Force S tf 0.000 5.117 1.599

Checking of Minimum Re-Bar

Cracking Moment Mc tf.m 4.172 4.172 4.172

1.7*Mf 8.821 7.924 2.669

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

Ultimate Bending Moment Mu tf.m 9.552

Mu > Mc ? ok

Max Re-bar As max cm2 66.0 60.0 60.0

Min Re-bar As min cm2 4.5 4.5 4.5

Required Bar As req. cm2 10.780 3.143

Area of Re-bar for Design As cm2 16.085 8.042

Checking of Allowable Stress

Young's Modulus Ratio n 24 24 16

Effective height d cm 33 30 30

Compressive Stress sc kgf/cm2 19.04 12.92

Bending Tensile Stress ss kgf/cm2 1239.88 722.95

Mean Shearing Stress tm kgf/cm2 0 0.59



ABUTMENT : Bendung ….

0.90 0.40

D16@200 D25@180

+ 80.45

D16@200

1.45

+ 79.00

0.40

0.40

D16@200

10.00

11.45

13.05

8.60

5.00

D19@100

+

0.60

2.50

1.00+

D25@300 D25@300

9.00

1.80 4.952.25

D19@150

1.00

D16@200

D16@200

D25@150

D16@200

D25@150

D16@200

D19@100

D25@90

D25@180

D25@180

D25@180

D16@200



D IMPACT PLATE AND CORBEL

1 Design Parameters

Active load T-Load

Impact plate Length L= 3.00 m Span length Ls = 0.7*L= 2.10 mThickness h1= 0.30 m Width of corbel Lp= 0.30 m

Effective width of road B= 3.00 m Height of corbel h2= 0.30 m

Unit weight of plate gc = 2.40 tf/m3

Unit weight of soil gs = 1.80 tf/m3 Cover of R-bar

Soil depth above plate Ds= 0.05 m Impact plate d1= 5 cm

Thickness of pavement Dp= 0.05 m Corbel d3= 7 cm

2 Computation of Intersectional Force, Corbel

1) Dead Load

Impact plate 0.72 tf/m2

Soil 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 0.446513 tf.m

3) Intersectional Force due Live Load

wL=2*T*(1+ i)/{2.75*(a+2*d)} wL = 33.39051 tf/m2

ML={(1/4)*wL*Ls*(a+2*d)-(1/8)*wL*(a+2*d)^2}*a ML= 5.371698 tf.m

where, ML bending moment due live load

T: wheel load of T-load 10.00 tf

a contact width of T-load 0.20 m

i: impact coefficient

i=20/(50+L)= 0.377

Dp: thickness of pavement 0.05 m

a: 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/m

bu width of corbel =Lp 0.300 m

2) Intersectional Force due Corbel

M2=(1/6)*(2w1+w2)*gc*bu^2 M2= 0.032 tf.m

where, w1 0.15 m

w2 0.60 m

gc = 2.40 tf/m3

bu 0.30 m

3) Total Intersectional Force

M=M1+M2 3.146 tf.m

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


h M ki t k f



2) Checking of Cracking Moment and Design Bending Moment

Design bending moment Mf 5.818 tf.m

Check Mf & Mc 1.7*Mf>Mc?, if yes check ultimate bending moment

1.7*Mf = 9.891 tf.mMc= 2.346 tf.m 1.7*Mf>Mc? Yes, check ultimate bending mome



where, Mu ultimate bending moment tf.m

As area of tensile bar cm2

s sy yielding point of tensile bar 3000 kgf/cm2 (Spec >295 N/mm2)

d effective height = h1-cover 25 cm

cover d1= 5 cm

h1= 30 cm

s'ck design compressive strength of concrete 175 kgf/cm2

b effective width 100 cm

As=Mf/(s sa*j*d) 14.729 cm2s sa= allowable stress of reinforcement bar 1850 kgf/cm2

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

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


s ca allowable stress of concrete 60 kgf/cm2

Check Mu & Mc Mu = 10.390 tf.m

Mc = 2.346 tf.m Mu>Mc? ok

4) Bar Arrangement

Checking of single or double bar arrangement

M1= (d/Cs)^2*ssa*b >Mf? M1= 700936 kgf.cm = 7.009 tf.m

where, M1 resistance moment

Cs ={2m/[s*(1-s/3)]}^(1/2) 12.844

s (n*sca)/(n*sca+ssa) 0.438

m ssa/sca 30.833

ssa 1850 kgf/cm2

sca 60 kgf/cm2

n 24

Check M1 > Mf? M1= 7.009 tf.m M1>Mf : Design Tensil e Bar Onl y

Mf= 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 cm2

Required bar area As req= 14.729 cm2

Apply f = 19 @ 150 mm

Bar area As = 18.902 cm2 ok


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

As' = M'/[ssa*(d - d2)] As'= 0.000 cm2

d= 25 cm M1= 7.009 tf.m

d2= 0 cm Mf= 5.818211 tf.m

ssa= 1850 kgf/cm2

Required bar area As' req= 0.000 cm2Apply f = 16 @ 300 mm

Bar area As' = 6.702 cm2 ok

d1

h d

h

d1

d





Mf 5.818211 tf.m

S 0 tf



p=As/(b*d) 0.0076 b= 100 cm

k={(n*p)^2+2*n*p}^0.5 - n*p 0.4477 d= 25 cm

j= 1-k/3 0.8508 n= 24


Mf 5.818211 tf.m

S 0.000 tf

sc = Mf/(b*d^2*Lc) 38.88 kgf/cm2 check sc < sca ? ok

ss = n*sc*(1-k)/k 1371.70 kgf/cm2 check ss < ssa ? ok

ss' = n*sc*(k-d2/d)/k 933.07 kgf/cm2 check ss' < ssa ? ok

p=As/(bd) 0.0076

p'=As'/(bd) 0.0027

k={n^2(p+p')^2+2n(p+p'*d2/d)}^05-n(p+p') 0.4048

Lc=(1/2)k(1-k/3)+(np'/k)(k-d2/d)(1-d2/d) 0.2394

b= 100 cm

d= 25 cm

d2= 0 cm

n= 24

4 Calculation of Required Reinforcement Bar for Corbel1) Cracking Moment


where, Mc cracking moment kgf.cm

Zc section modulus

Zc=b*h2^2/6 15000 cm3

b= 100 cm

s'ck tensile strength of concrete (bending)

s'ck = 0.5*sck^(2/3) 15.643 kgf/cm2

s ck= 175 kgf/cm2

N axial force 0 tf

Ac area of concrete = b*h1 3000 cm2

h2 thickness of section, impact plate 30 cm

2) Checking of Cracking Moment and Design Bending Moment

Design bending moment Mf 3.146 tf.m

Check Mf & Mc 1.7*Mf>Mc?, if yes check ultimate bending moment

1.7*Mf = 5.348 tf.m

Mc= 2.346 tf.m 1.7*Mf>Mc? Yes, check ultimate bending momen



where, Mu ultimate bending moment tf.m

As area of tensile bar cm2

s sy yielding point of tensile bar 3000 kgf/cm2 (Spec >295 N/mm2)

d effective height = h1-cover 23 cm

cover d1= 7 cm

h2= 30 cm

s'ck design compressive strength of concrete 175 kgf/cm2 b effective width 100 cm

A Mf/( *j*d) 8 657 2

h d

b

As

d1

h d

b

As

As'

d1



4) Bar Arrangement

Checking of single or double bar arrangement

M1= (d/Cs)^2*ssa*b >Mf? M1= 593272 kgf.cm = 5.933 tf.m

where, M1 resistance moment

Cs ={2m/[s*(1-s/3)]}^(1/2) 12.844

s (n*sca)/(n*sca+ssa) 0.438

m ssa/sca 30.833

ssa 1850 kgf/cm2

sca 60 kgf/cm2

n 24

Check M1 > Mf? M1= 5.933 tf.m M1>Mf : Design Tensil e Bar Only

Mf= 3.146 tf.m

(a) Tensile Bar

Max bar area As max = 0.02*b*d = 46.0 cm2

Min bar area As min = b*4.5%= 4.5 cm2

Required bar area As req= 8.657 cm2

Apply f = 19 @ 300 mmBar area As = 9.451 cm2 ok


Mf 3.146 tf.m

So 0.000 tf



p=As/(b*d) 0.0041

k={(n*p)^2+2*n*p}^0.5 - n*p 0.3563

j= 1-k/3 0.8812

SUMMARY OF DESIGN CALCULATION

Description Abbr. unit Impact Plate Corbel



Effective width of section b cm 100 100

Height of Section h cm 30 30

concrete cover (tensile) d1 cm 5 7

concrete cover (compressive) d2 cm - -

Effective height of Section d cm 25 23





Reinforcement Bar


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 Load

Design Bending Moment Mf tf.m 5.818 3.146

Design Axis Force Nd tf - -

Shearing Force S tf - -

Checking of Minimum Re-Bar

Cracking 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

d1

h d



Remarks

1 Input Data

"Input": Figure 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.

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

(2) Sheet name "Body", H530 to N531. Button Calculate x

3 Allowable stress for "Body"

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

4 Minimum reinforcement bar

Step 1 Calculation of Md', find 1.7Md

Step 2 Comparison between Md' and Mc

if Md' > Mc go to step 3

Step 3 Calculation of Mu

Step 4 Comparison between Mu and Mc

Mu > Mc OK

5 Wing Wall

Wing Wall are not considered.

Revised

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

Safety factor was revised as follows.

Normal condition 1.5 to 2.0

Seismic condition 1.2 to 1.25

9-Nov-02 2 Sheet " Footing"

Cell "H548" & "L548" is changed.

Figure "Mu" is not referred correctly. Therefore revised.

15-Nov-02 3 Sheet "Body"

Cell "I123" and "I150"

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

29-Nov-02 4 All sheets

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.

previous figure revised figure

normal condition 1.50 2.00

seismic condition 1.20 1.25

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

3-Dec-02 5 Modification of drawings

31/34 159572688.xlsx.ms_office,Read me



4-Dec-02 6 Modification of additional bar arrangement

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 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.

2-Feb-03 10 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 corre

26-Feb-03 11 Sheet "Input"

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




ted.

abutment jem ba an

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