Please copy and paste this embed script to where you want to embed

Example 9.2 Design an RCC abutment for supporting a 55 m single span bridge. Geotechnical investigations indicated that sound rock is available at 13 m depth from the top of the backfill to be retained by the abutment. The allowable rock pressure is 1500 kN/m2. Consider IRC class 40 R loading. Assume suitable data not given. Maximum total dead load of the bridge is 1400 kN. Consider horizontal seismic force in the direction of the traffic, and vertical in the downward direction. The bridge is located in seismic zone IV. Solution I.

It is proposed to rest the foundation of the abutment on the rock. Therefore ,height of abutment = 13m Keeping the height of abutment in view, the design is carried out providing front and back counterforts as shown in Figs 9.8 and 909. The base of the foundation is rested on rock. Tentative proportioning of the abutment, foundation slab and counterforts are shown in Fig 9.8. The design has been carried out assuming, Unit weight of concrete

= 25 kN/ m3

Unit weight of boulder fill = 20 kN/ m3 Unit weight of backfill

= 18 kN/ m3

Angle of internal friction of backfill soil, φ = 360 Coefficient of base friction, μ = 0.65 Angle of wall friction δ = 2/3 φ = 240 Horizontal seismic coefficient Ah from Eq. (1.3) =

z I S a 0.24 1.2 × × 2.5 = 0.144 = 2R g 2 2.5

Vertical seismic coefficient, Av = 2/3 × 0.144 (IS 1893 (part 1) -2002) II.

Dead load of the bridge = 1400 kN Reaction on each abutment = 1400/2 = 700 kN Loading due to one lane IRC class 40 R wheeled load train has been shown in Fig 9.10. Influence line for reaction on right abutment having fixed bearing will be as shown in Fig 9.11.

Fig. 9.8 Front elevation of the abutment with front and back counterfort

Fig. 9.9 Abutment with counterforts

Fig 9.10 A class 40 R Load train

1.0

A

5.5 m

B

Fig. 9.11 Influence line diagram for the reaction of fixed bearing Due to one lane of IRC class 40 R wheeled load train, with a clear gap of 20 m between two load trains, the reaction will be, maximum when the first load 120 kN coincides with point A of influence line diagram (Fig. 9.11). 120 × 55 + 120 × 53.93 + 120 × 49.66 + 70 × 46.66 + 50 × 41.73 + 120 × 21.73 RRT = + 120 × 20.66 + 120 × 16.39 + 70 × 13.37 + 70 × 12.12 + 50 × 8.46 = 668.5 kN 55 Reaction on the left abutment, RLT = 1100 – 668.5 = 431.50 kN III.

As per IRC: 6 – 200, for bridges having span more than 45 m , impact factor is 0.154 Therefore impact load on rocker end = 0.154 × 668.5 = 102.95 kN Impact load on roller end = 0.154 × 431.50 = 66.45 kN

IV. Longitudinal forces a)

Braking effect

The braking effect will be 20 percent of the first class 40 R load train (plus 10 % of the other train),

= (0.2 + 0.1) × 550 =165 kN This force acts along line parallel to roadway and 1.2 m above it. Therefore moment due to this

braking force is = 165 × (1.2 + 1.75) = 486.75 kN-m

Where, 1.75 m is the height of the road above the bearing pin. Increase/ decrease in reaction in the right abutment (fixed bearing knuckle)/ left abutment (free knuckle) respectively = 486.75/55 = moment/ span =8.85 kN b)

Seismic force horizontal

As the seismic force is taken in the direction of traffic , seismic forces will not be considered on the live load. Horizontal seismic force on the dead load transferred from bridge deck = 0.144 × 1400 = 201.6 kN Considering that it acts at the centre of bearing, Increase in reaction due to their force on rocker bearing = decrease in reaction on roller bearing = 201.6 × (0.25 + 0.75)/ 55 = 3.66 kN Total reaction on roller bearing = 700 + 431.50 + 66.45 – 8.85 – 3.66 + 700 × 0.096 = 1252.64 kN Coefficient of friction at roller bearing, μ = 0.03 Frictional force at roller bearing = 0.03 × 1252.64= 37.58 kN Therefore longitudinal force at rocker bearing = (165 + 201.6)/2 + 37.58 = 220.88 kN V.

Total vertical seismic forces on items [2-10] of the Table 9.2 = 0.096 × (181.12 + 17.875 + 7 + 42.70 + 40.95 + 672.75 + 51 + 87.75 + 135 + 39 + 85.40 + 81.90 + 1345.50) = 0.096 × 3138.94 = 301.34 kN Total counter-clockwise moment of vertical seismic forces on items [2-10] of Table 9.2 about point C = 0.096 × (878.46 + 86.69 + 37.80 + 369.35 + 342.38 + 5684.74 + 43.86 + 1039.84 + 405 + 2340 + 738.71 + 700.24 + 11369.4) =0.096 × 24036.54 = 2307.51 kN

VERTICAL FORCES Table 9.2 The computations of vertical forces Vertical Loads 1)Total dead load, Live load, impact load , increase in reaction due to braking effect, seismic force and vertical seismic force on dead load 2)Main abutment portion 3)Abutment cap 4)Parapet wall 5)Back counterfort (Thickness = 0.4 m) 6)Approach slab 7)Back wall 8)Front counterfort (Thickness = 0.4 m) 9)Base slab

Loads in kN 700 + 668.5 +102.95 + 8.85 + 3.66 +0.096 × 700 = 1551.16 10.35 × 0.7 × 1 × 25 = 181.12 1 .5 + 0 .7 0.65 × 1.0 × × 25 = 17.875 2 0.4 × 0.7 × 1 × 25 =7 (i) 6.1× 3 × 0.4 × 0.7 × 25 = 42.7 6.1 + 6.5 × 0.65 × 0.4 × 25 = 40.95 (ii) 2 (iii) 10.35 × 6.5 × 0.4 × 25 = 672.75 25 × 6.8 × 0.3 × 1.0 = 51.0 25 × 11.7 × 0.3 × 1.0 = 87.75 25 × ½ × 6.0 × 4.5 × 0.4 = 135 25 × 12 × 1.3 × 1.0 = 390.0 (i)

10)Boulder fill

20 × 6.1 × 0.7 × 1.0 = 85.40 6.1 + 6.5 × 0.65 × 1.0 × 20 = 81.90 (ii) 2 (iii) 20 × 10.35 × 6.5 × 1.0 = 1345.50

Distance from toe C (m)

Moment ( kN-m)

4.5 + 0.35 = 4.85

7523 .13

4.85

878.46

4.85

86.69

(4.5 + 0.7 + 0.2) = 5.4 (i) 4.5 + 0.7 + 0.4 + 6.1 2 =8.65

37.80 369.35

(ii)

4.5+ 0.7 + 6.5×0.65×3.25− 0.5×0.4×0.65×0.4 / 3 6.5×0.65− 0.5×0.4×0.65

= 8.55 (iii) 4.5 + 0.7 + 3.25 = 8.45 4.5 + 0.7 + 6.8/2 = 8.6 (12 – 0.15 ) = 11.85 2/3 × 4.5 = 3.0 6.0 8.65 8.55

342.38 5684.74 43.86 1039.84 405.0 2430 738.71 700.24 11369.47

8.45

Horizontal seismic forces on items [2- 10] of table = 0.144 × 3138.94 = 452.00 kN Moment of horizontal seismic forces on items (2- 10) of table about point C 181.12 × (1.3 + 10.35 / 2 ) + 17.875 × 2 × 1.5 + 0.70 × 0.65 + 10.35 + 1.3 3 1.5 + 0.7 + 7.0 × (1.3 + 10.35 + 0.65 + 0.35 ) + 42.7 × (1.3 + 10.35 + 0.65 + 0.2 ) 10.35 2 × 6.1 + 6.5 0.65 = 0.144 × 40.95 × 1.3 + 10.35 + × + 672.75 × 1.3 + 3 2 6.1 + 6.5 + 51 × (1.3 + 11.7 + 0.15) + 87.75 × 1.3 + 11.7 + 135 × 1.33 + 6 + 139 × 1.3 2 3 2 + 85.40 × 1.3 + 10.35 + 2 × 6.1 + 6.5 × 0.65 + 1345.50 × 1.3 + 10.35 3 3 6.1 + 6.5 = 0.144 × [181.12 × 6.475 + 17.875 × 12.014 + 7 × 12.65 + 42.7 × 12.5 + 40.95 × 11.97 + 672.75 × 6.475 + 51.0 × 13.15 + 87.75 × 7.15 + 135 × 3.3 + 390 × 0.65 + 85.4 × 12.5 + 81.9 × 11.97 + 1345.50 × 6.475] = 0.144 × [1172.75 + 214.014 + 88.55 + 533.75 + 490.17 + 4356.06 + 670.65 + 627.41 + 445.5 + 253.5 + 1067.5 + 980.34 + 8712.11] = 0.144 × 19613.31 = 2824.71 kN m VI. Substituting the values of φ= 36o , δ= 24 o , α = 0 o , i = 00 , Ah = 0.144 and Av = 0.096 in Eq 3.3, we get Ca = 0.3478. Equation (3.3) gives the corresponding static earth coefficient ka as 0.2348. Total seismic active earth pressure = ½ × 18 × 0.3478 × 132 = 529.0 kN Total static active earth pressure = ½ × 18 × 0.2349 × 132 = 357.28 kN Moment of earth pressure forces about point C = 357.28 × 0.42 × 13 + (529.0 – 357.28) × 0.5 × 13 = 1950.76 + 1116.18 = 3066.94 kN-m

Summary of computations: Total vertical force = 1551.16 + 3138.94 + 301.34 = 4991.44 kN Total moment of the vertical forces = 7523.13 + 878.46 + 86.69 + 37.80 + 369.35 + 342.38 + 5684.74 + 43.86 + 1039.84 + 405 + 738.71 + 700.24 + 11369.47 + 2307.51 = 21529.18 kN m (counter clockwise) Total horizontal force = 452 + 529 + 220.88 = 1201.88 kN Total moment of horizontal forces = 220.88 × (0.25 + 0.65 + 10.35 + 1.3) + 2824.17 + 3066.94 = 8863.15 kN m (clockwise) Factor of safety against sliding = μ V / H = 0.65 × 4991.44 / 1201.88 = 2.7 [> 1.5 , safe ] Overturnin g moment 21529.18 = = 2.43 [ > 1.5 , safe] Stabilizin g moment 8863.15 Net moment = 21529.18 – 8863.15 = 12666.03 kN-m Factor of safety against overturning =

Eccentricity with respect to centre of base = B/2 – 12666.03/ 4991.44 = 12/2 – 2.53 = 3.46 m Maximum base pressure =

4991 .44 6 × 3.46 1 + = 1135 kN/m2 12 12

Minimum base pressure =

4991 .44 6 × 3.46 1 − = 303 kN m2 12 12

View more...
It is proposed to rest the foundation of the abutment on the rock. Therefore ,height of abutment = 13m Keeping the height of abutment in view, the design is carried out providing front and back counterforts as shown in Figs 9.8 and 909. The base of the foundation is rested on rock. Tentative proportioning of the abutment, foundation slab and counterforts are shown in Fig 9.8. The design has been carried out assuming, Unit weight of concrete

= 25 kN/ m3

Unit weight of boulder fill = 20 kN/ m3 Unit weight of backfill

= 18 kN/ m3

Angle of internal friction of backfill soil, φ = 360 Coefficient of base friction, μ = 0.65 Angle of wall friction δ = 2/3 φ = 240 Horizontal seismic coefficient Ah from Eq. (1.3) =

z I S a 0.24 1.2 × × 2.5 = 0.144 = 2R g 2 2.5

Vertical seismic coefficient, Av = 2/3 × 0.144 (IS 1893 (part 1) -2002) II.

Dead load of the bridge = 1400 kN Reaction on each abutment = 1400/2 = 700 kN Loading due to one lane IRC class 40 R wheeled load train has been shown in Fig 9.10. Influence line for reaction on right abutment having fixed bearing will be as shown in Fig 9.11.

Fig. 9.8 Front elevation of the abutment with front and back counterfort

Fig. 9.9 Abutment with counterforts

Fig 9.10 A class 40 R Load train

1.0

A

5.5 m

B

Fig. 9.11 Influence line diagram for the reaction of fixed bearing Due to one lane of IRC class 40 R wheeled load train, with a clear gap of 20 m between two load trains, the reaction will be, maximum when the first load 120 kN coincides with point A of influence line diagram (Fig. 9.11). 120 × 55 + 120 × 53.93 + 120 × 49.66 + 70 × 46.66 + 50 × 41.73 + 120 × 21.73 RRT = + 120 × 20.66 + 120 × 16.39 + 70 × 13.37 + 70 × 12.12 + 50 × 8.46 = 668.5 kN 55 Reaction on the left abutment, RLT = 1100 – 668.5 = 431.50 kN III.

As per IRC: 6 – 200, for bridges having span more than 45 m , impact factor is 0.154 Therefore impact load on rocker end = 0.154 × 668.5 = 102.95 kN Impact load on roller end = 0.154 × 431.50 = 66.45 kN

IV. Longitudinal forces a)

Braking effect

The braking effect will be 20 percent of the first class 40 R load train (plus 10 % of the other train),

= (0.2 + 0.1) × 550 =165 kN This force acts along line parallel to roadway and 1.2 m above it. Therefore moment due to this

braking force is = 165 × (1.2 + 1.75) = 486.75 kN-m

Where, 1.75 m is the height of the road above the bearing pin. Increase/ decrease in reaction in the right abutment (fixed bearing knuckle)/ left abutment (free knuckle) respectively = 486.75/55 = moment/ span =8.85 kN b)

Seismic force horizontal

As the seismic force is taken in the direction of traffic , seismic forces will not be considered on the live load. Horizontal seismic force on the dead load transferred from bridge deck = 0.144 × 1400 = 201.6 kN Considering that it acts at the centre of bearing, Increase in reaction due to their force on rocker bearing = decrease in reaction on roller bearing = 201.6 × (0.25 + 0.75)/ 55 = 3.66 kN Total reaction on roller bearing = 700 + 431.50 + 66.45 – 8.85 – 3.66 + 700 × 0.096 = 1252.64 kN Coefficient of friction at roller bearing, μ = 0.03 Frictional force at roller bearing = 0.03 × 1252.64= 37.58 kN Therefore longitudinal force at rocker bearing = (165 + 201.6)/2 + 37.58 = 220.88 kN V.

Total vertical seismic forces on items [2-10] of the Table 9.2 = 0.096 × (181.12 + 17.875 + 7 + 42.70 + 40.95 + 672.75 + 51 + 87.75 + 135 + 39 + 85.40 + 81.90 + 1345.50) = 0.096 × 3138.94 = 301.34 kN Total counter-clockwise moment of vertical seismic forces on items [2-10] of Table 9.2 about point C = 0.096 × (878.46 + 86.69 + 37.80 + 369.35 + 342.38 + 5684.74 + 43.86 + 1039.84 + 405 + 2340 + 738.71 + 700.24 + 11369.4) =0.096 × 24036.54 = 2307.51 kN

VERTICAL FORCES Table 9.2 The computations of vertical forces Vertical Loads 1)Total dead load, Live load, impact load , increase in reaction due to braking effect, seismic force and vertical seismic force on dead load 2)Main abutment portion 3)Abutment cap 4)Parapet wall 5)Back counterfort (Thickness = 0.4 m) 6)Approach slab 7)Back wall 8)Front counterfort (Thickness = 0.4 m) 9)Base slab

Loads in kN 700 + 668.5 +102.95 + 8.85 + 3.66 +0.096 × 700 = 1551.16 10.35 × 0.7 × 1 × 25 = 181.12 1 .5 + 0 .7 0.65 × 1.0 × × 25 = 17.875 2 0.4 × 0.7 × 1 × 25 =7 (i) 6.1× 3 × 0.4 × 0.7 × 25 = 42.7 6.1 + 6.5 × 0.65 × 0.4 × 25 = 40.95 (ii) 2 (iii) 10.35 × 6.5 × 0.4 × 25 = 672.75 25 × 6.8 × 0.3 × 1.0 = 51.0 25 × 11.7 × 0.3 × 1.0 = 87.75 25 × ½ × 6.0 × 4.5 × 0.4 = 135 25 × 12 × 1.3 × 1.0 = 390.0 (i)

10)Boulder fill

20 × 6.1 × 0.7 × 1.0 = 85.40 6.1 + 6.5 × 0.65 × 1.0 × 20 = 81.90 (ii) 2 (iii) 20 × 10.35 × 6.5 × 1.0 = 1345.50

Distance from toe C (m)

Moment ( kN-m)

4.5 + 0.35 = 4.85

7523 .13

4.85

878.46

4.85

86.69

(4.5 + 0.7 + 0.2) = 5.4 (i) 4.5 + 0.7 + 0.4 + 6.1 2 =8.65

37.80 369.35

(ii)

4.5+ 0.7 + 6.5×0.65×3.25− 0.5×0.4×0.65×0.4 / 3 6.5×0.65− 0.5×0.4×0.65

= 8.55 (iii) 4.5 + 0.7 + 3.25 = 8.45 4.5 + 0.7 + 6.8/2 = 8.6 (12 – 0.15 ) = 11.85 2/3 × 4.5 = 3.0 6.0 8.65 8.55

342.38 5684.74 43.86 1039.84 405.0 2430 738.71 700.24 11369.47

8.45

Horizontal seismic forces on items [2- 10] of table = 0.144 × 3138.94 = 452.00 kN Moment of horizontal seismic forces on items (2- 10) of table about point C 181.12 × (1.3 + 10.35 / 2 ) + 17.875 × 2 × 1.5 + 0.70 × 0.65 + 10.35 + 1.3 3 1.5 + 0.7 + 7.0 × (1.3 + 10.35 + 0.65 + 0.35 ) + 42.7 × (1.3 + 10.35 + 0.65 + 0.2 ) 10.35 2 × 6.1 + 6.5 0.65 = 0.144 × 40.95 × 1.3 + 10.35 + × + 672.75 × 1.3 + 3 2 6.1 + 6.5 + 51 × (1.3 + 11.7 + 0.15) + 87.75 × 1.3 + 11.7 + 135 × 1.33 + 6 + 139 × 1.3 2 3 2 + 85.40 × 1.3 + 10.35 + 2 × 6.1 + 6.5 × 0.65 + 1345.50 × 1.3 + 10.35 3 3 6.1 + 6.5 = 0.144 × [181.12 × 6.475 + 17.875 × 12.014 + 7 × 12.65 + 42.7 × 12.5 + 40.95 × 11.97 + 672.75 × 6.475 + 51.0 × 13.15 + 87.75 × 7.15 + 135 × 3.3 + 390 × 0.65 + 85.4 × 12.5 + 81.9 × 11.97 + 1345.50 × 6.475] = 0.144 × [1172.75 + 214.014 + 88.55 + 533.75 + 490.17 + 4356.06 + 670.65 + 627.41 + 445.5 + 253.5 + 1067.5 + 980.34 + 8712.11] = 0.144 × 19613.31 = 2824.71 kN m VI. Substituting the values of φ= 36o , δ= 24 o , α = 0 o , i = 00 , Ah = 0.144 and Av = 0.096 in Eq 3.3, we get Ca = 0.3478. Equation (3.3) gives the corresponding static earth coefficient ka as 0.2348. Total seismic active earth pressure = ½ × 18 × 0.3478 × 132 = 529.0 kN Total static active earth pressure = ½ × 18 × 0.2349 × 132 = 357.28 kN Moment of earth pressure forces about point C = 357.28 × 0.42 × 13 + (529.0 – 357.28) × 0.5 × 13 = 1950.76 + 1116.18 = 3066.94 kN-m

Summary of computations: Total vertical force = 1551.16 + 3138.94 + 301.34 = 4991.44 kN Total moment of the vertical forces = 7523.13 + 878.46 + 86.69 + 37.80 + 369.35 + 342.38 + 5684.74 + 43.86 + 1039.84 + 405 + 738.71 + 700.24 + 11369.47 + 2307.51 = 21529.18 kN m (counter clockwise) Total horizontal force = 452 + 529 + 220.88 = 1201.88 kN Total moment of horizontal forces = 220.88 × (0.25 + 0.65 + 10.35 + 1.3) + 2824.17 + 3066.94 = 8863.15 kN m (clockwise) Factor of safety against sliding = μ V / H = 0.65 × 4991.44 / 1201.88 = 2.7 [> 1.5 , safe ] Overturnin g moment 21529.18 = = 2.43 [ > 1.5 , safe] Stabilizin g moment 8863.15 Net moment = 21529.18 – 8863.15 = 12666.03 kN-m Factor of safety against overturning =

Eccentricity with respect to centre of base = B/2 – 12666.03/ 4991.44 = 12/2 – 2.53 = 3.46 m Maximum base pressure =

4991 .44 6 × 3.46 1 + = 1135 kN/m2 12 12

Minimum base pressure =

4991 .44 6 × 3.46 1 − = 303 kN m2 12 12

Thank you for interesting in our services. We are a non-profit group that run this website to share documents. We need your help to maintenance this website.