Bridge Design 11635z

BRIDGE DESIGN (SHEAR FORCE AND BENDING MOMENT CALCULATION) INPUT VARIABLES: CLEAR SPAN: BEARING WIDTH: FOOTPATH WIDTH: WEARING COAT: NO. OF LANES: CONCRETE GRADE: STEEL GRADE: CLEAR COVER: CURB HEIGHT:

5.5 400 1 100 2 25 415 30 225

m mm m mm N/mm2 N/mm2 mm mm

1) DESIGN DATA AND CONSTANTS: 8.3 200 10 0.29 0.903 1.09

m= k= j= Q=

2) DEPTH DETERMINATION: O/D FOR 5.5m SPAN = EFFECTIVE DEPTH = EFFECTIVE SPAN (Maximum of): I) ln+d = II) C/C dist. b/w s = =

5.9

440 mm 397.5 mm

…(Assuming 80mm depth required for 1m span) …(Assuming 25mm dia bars to be used) 5.8975 m 5.9 m

m

3) dead load bending moment and dead load shear force: D.L OF SLAB = D.L OF WEARING COAT = TOTAL= BENDING MOMENT = SHEAR FORCE =

10.56 KN 2.2 KN 12.76 KN

…(Assuming unit wt. of concrete 24 KN/m3) …(Assuming unit wt. of wearing coat 22 KN/m3)

12.76 X 5.9^2/8 = 12.76 X 5.5 /2 =

55.52 KNm/m 35.09 KNm/m

IRC CLASS AA TRACKED VEHICLE: 4) LIVE load bending moment and shear force CALCULATION: EFFECTIVE LENGTH OF DISPERSION =

3.6+2 x (440+100)/1000 =

4.68 m

EFFECTIVE BREADTH OF DISPERSION FOR SINGLE WHEEL (MAX BENDING MOMENT CRITERIA):

TOTAL BREADTH OF CARRIAGEWAY = ASPECT RATIO = 1.61 VALUE OF α FOR ASPECT RATIO 1.61 FROM CHART IS = x= 2.95 m b1= 1.05 m 2.88x2.95x(1-2.95/5.9)+1.05=

7.5 m 2.88

5.3 m

TOTAL EFFECTIVE BREADTH OF DISPERSION : Bf = 2.625+2.05+2.65= 7.325 m 2.625

2.05

2.65

2.625

2.05

2.56

EFFECTIVE BREADTH OF DISPERSION FOR SINGLE WHEEL (MAX SHEAR FORCE CRITERIA):

ASPECT RATIO = 1.61 VALUE OF α FOR ASPECT RATIO 1.61 FROM CHART IS = x= 2.34 m b1= 1.05 m 2.88x2.34x(1-2.34/5.9)+1.05= TOTAL EFFECTIVE BREADTH OF DISPERSION : Bf = 2.56+2.05+2.56=

5) impact factor:

7.17 m

2.88

5.12 m

IMPACT FACTOR = TOTAL LIVE LOAD =

0.21625 851.375 KN 0.61

4.68

6) CALCULATION OF MAXIMUM BENDING MOMENT : 24.84 TOTAL DISPERSION AREA = 7.325 X 4.68 = UNIT LIVE LOAD = 851.375 / 34.281 =

34.281 m2 24.84 KN/m2 5.9

MAX BENDING MOMENT(L.L)=

(24.84x4.68/2X5.9/2)-(24.84x4.68/2x4.68/4) = 103.46 KNm/m TOTAL BENDING MOMENT AT MID-SPAN (DL+LL)= 103.46 + 55.52 =

158.99 KNm/m

6) CALCULATION OF MAXIMUM SHEAR FORCE : TOTAL DISPERSION AREA = UNIT LIVE LOAD = 851.375 / 33.56 = MAX SHEAR FORCE (L.L) = TOTAL S.F AT (DL+LL)

33.56 m2 25.37 KN/m2

24.84x4.68x(5.9-4.68/2)/5.9 70.145 KN/m = 70.14 +35.09 =

4.68 105.24 KN/m

24.84

5.9

IRC CLASS AA TRACKED VEHICLE: WEIGHT PER WHEEL = 350 NO. OF WHEELS = 2 BREADTH OF WHEEL = 0.85 LENGTH OF WHEEL = 3.6 C= BREADTH OF CARRIAGEWAY= VALUE OF "C" = 1.2 DIST. B/W C/L OF WHEELS = 2.05 IMPACT FACTOR : FOR SPAN UPTO 9m: SPAN = 5.9 m If =

KN m m 7.5 m m

0.21625 ...(FOR ALL BRIDGE TYPES)

FOR SPAN OVER 9m: If =

If=

0.1 …(FOR SPAN UPTO 40M FOR RC BRIDGE) 0.1 …(FOR ALL SPANS FOR STEEL BRIDGE) 0.21625

350

C

350

0.85

3.6

IRC 70R TRACKED VEHICLE: WEIGHT PER WHEEL = NO. OF WHEELS = BREADTH OF WHEEL = LENGTH OF WHEEL = C= BREADTH OF CARRIAGEWAY= NO. OF LANES = VALUE OF "C" = DIST. B/W C/L OF WHEELS = IMPACT FACTOR : FOR SPAN UPTO 9m: SPAN = 5.9 m If =

350 KN 2 0.84 m 4.57 m 7.5 m 2 1.2 m 2.06 m

0.21625 ...(FOR ALL BRIDGE TYPES)

FOR SPAN OVER 9m: If =

If=

0.1 …(FOR SPAN UPTO 40M FOR RC BRIDGE) 0.1 …(FOR ALL SPANS FOR STEEL BRIDGE) 0.21625

350

C

350

0.84

4.57

IRC CLASS AA WHEELED VEHICLE: TOTAL WEIGHT = 400 NO. OF WHEELS = 8 BREADTH OF WHEEL = 0.3 LENGTH OF WHEEL = 0.15 C= BREADTH OF CARRIAGEWAY= VALUE OF "C" = 1.2

KN m m 7.5 m m

IMPACT FACTOR : SPAN= 5.9 m If= 0.25

1200 1050

300

2.2 1

37.5

62.5

62.5

37.5 150

300

300

700

300

300

300

IRC CLASS A WHEELED VEHICLE: TOTAL WEIGHT = NO. OF WHEELS =

554 KN 16

VALUE OF LENGTH & BREADTH: LOAD B W 27 0.15 0.2 68 0.2 0.38 114 0.25 0.5

1.1 27

B VALUE OF "C" = WIDTH OF CARRIAGEWAY =

0.15 m 7.5 m

GAP BETWEEN TWO TRAINS OF VEHICLE =

1.2 m W

IMPACT FACTOR : SPAN = 5.9 m FOR R.C BRIDGE, If= 0.378

3.2 27

1.2 114

4.3 114

3 68

3 68

3 68

68

1.8

IRC 70R WHEELED VEHICLE:

0.91

3.96 80

COLUMN "l": TOTAL WEIGHT = NO. OF WHEELS = C=

1000 KN 28

BREADTH OF CARRIAGEWAY= NO. OF LANES = VALUE OF "C" = 1.2 m

7.5 m 2

IMPACT FACTOR : SPAN= 5.9 m If= 0.25

COLUMN "m": TOTAL WEIGHT = NO. OF WHEELS = C=

1000 KN 28

BREADTH OF CARRIAGEWAY= NO. OF LANES = VALUE OF "C" = 1.2 m

IMPACT FACTOR : SPAN= 5.9 m If= 0.25

COLUMN "n":

7.5 m 2

1.52 120

120

TOTAL WEIGHT = NO. OF WHEELS = C=

1000 KN 56

BREADTH OF CARRIAGEWAY= NO. OF LANES = VALUE OF "C" = 1.2 m

IMPACT FACTOR : SPAN= 5.9 m If= 0.25

7.5 m 2

2.13

1.37 170

170

3.05 170

1.37

0.91

170

2.79

0.61

0.41

0.04

2.79

0.61

0.41

0.385

0.38

2.79

0.51

0.05 0.23

0.25

0.76

DETAILING: 1

100mm thick wearing coat 12mm dia @200mmC/C

20mm dia @140mm C/C 12mm dia @185mm C/C 9.5 CROSS SECTION ALONG BREADTH

20mm dia @280mm C/C

12mm dia @185mm C/C 0.85l

20mm dia @140mm C/C

5.9 CROSS SECTION ALONG LENGTH

2 - 12mm dia bars

12mm dia @200mmC/C

20mm dia @140mm C/C

16mm dia @170mm C/C

12mm dia @200mmC/C 0.1 0.44

12mm dia @185mm C/C 20mm dia @140mm C/C

NOTE: ALL DIMENSIONS ARE IN metres