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Debris Control Structures Evaluation and Countermeasures

Value of B  Value of F_{r}  C_{D} 

B < 0.36  F_{r} < 0.4  1.8 
B < 0.36  0.4 < F_{r} < 0.8  2.6  2.0F_{r} 
0.36 < B < 0.77  F_{r} < 1  3.1  3.6B 
B > 0.77  F_{r} < 1  1.4 1.4B 
Value of B  Value of F_{r}  C_{D} 

B < 0.33  F_{r} < 0.4  1.9 
B < 0.33  0.4 < F_{r} < 0.8  2.8  2.25F_{r} 
0.33 < B < 0.77  F_{r} < 1  3.1  3.6B 
B > 0.77  F_{r} < 1  1.4 1.4B 
The drag coefficient as provided in these tables is related to the blockage ratio and Froude number as defined below.
(4.2)
where:
B = Blockage ratio
A_{d} = Crosssectional flow area blocked by debris in the contracted bridge section, m^{2} (ft^{2})
A_{c} = Unobstructed crosssectional flow in the contracted section, m^{2} (ft^{2})
(4.3)
where:
Fr = Froude number
V_{r} = Reference velocity, see discussion in Subsection 4.3.3.1, m/s (ft/s)
g = Acceleration of gravity, 9.81 m/s^{2} (32.2 ft/s^{2})
y_{r} = Average flow depth corresponding with the reference velocity, m (ft)
The total force on the structure that is caused by the hydrostatic pressure difference can be approximated as:
(4.4)
where:
F_{h} = Horizontal hydrostatic force on area A_{h}, N (lbs)
γ = Specific weight of water, N/m^{3} (lbs/ft^{3})
h_{cu} = Vertical distance from the upstream water surface to the centroid of area A_{hu}, m (ft)
A_{hu} = Area of the vertically projected, submerged portion of the debris accumulation below the upstream water surface, m^{2} (ft^{2})
h_{cd} = Vertical distance from the downstream water surface to the centroid of area A_{hd}, m (ft)
A_{hd} = Area of the vertically projected, submerged portion of the debris accumulation below the downstream water surface, m^{2} (ft^{2})
The total resultant force is computed as the summation of the drag force (Equation 4.1) and the differential hydrostatic force (Equation 4.4). The loads computed using these equations corresponds to the pressure forces of the water on the debris accumulation. The transfer of the load from the debris to the structure depends on many factors, including the characteristics of the debris accumulation and the degree to which streambed and banks support the debris accumulation. Thus, a conservative approach of applying the resultant force as a point load is recommended in evaluating the forces on the structure. The vertical and horizontal location of the resultant hydrostatic and drag forces and that of the total force can be determined by adding the moments about convenient axes. A less conservative distribution of the load to the structure may be warranted where there is more information available on the debris configuration and structural susceptibility.
Three scenarios should be evaluated when debris accumulation exists on two piers as a result of the opening between the piers being less than the length of the design log:
Debris accumulations typically align themselves with the direction of the flow. There is a lot of uncertainty associated with debris accumulation geometry and the direction of the flood flows. Therefore, the resultant force should be applied using both consideration of the anticipated range of possible flow directions and the structure's susceptibility to the resultant forces over the range of flow direction. For example, if the possible direction of flow is 20 degrees to the axis of the pier and the pier is most susceptible to a force applied at 15 degrees, then the force should be applied at 15 degrees to the axis of the pier. For superstructures and debris accumulations that span adjacent piers, the forces should be applied in at least two directions: (1) perpendicular to the face of the bridge and (2) in the direction of the flow with consideration to the structures susceptibility.
Given:
Design flow rate = 195 m^{3}/s
Minimum upstream main channel width = 13.7 m; design log length of 13.7 m
Depth of debris is fullflow depth
Main channel width at the bridge = 60 m
Debris accumulation only on Pile Bent 2 (see Figure 4.8)
Superstructure is not submerged
Ineffective flow areas from the debris defined by 1:1 contraction and 2:1 expansion
Bottom elevation of Pile Bent 2 = 61.78 m
Left station of debris = 160.33 m; Right station of debris = 174.03
Hydraulic computation results are provided in Table 4.3 and shown in Figure 4.9
Upstream water surface elevation, WS_{US} = 65.43 m (Table 4.3)
Downstream water surface elevation, WS_{DS} = 65.06 m (Table 4.3, see discussion Downstream Water Surface Elevation in Subsection 4.3.3.1)
Figure 4.8. Upstream face of the bridge for Example 1.
Figure 4.9. Water surface profile for Example 1.
River Station (m)  Water Surface Elevation (m)  Flow Area (m^{2})  Main Channel Velocity (m/s)  Cross Section Average Velocity (m/s)  Average Flow Depth^{1} (m) 

0  64.27  230.98  2.48  0.84  2.34 
127  64.70  300.10  1.59  0.65  2.08 
254  64.89  307.65  1.03  0.63  1.55 
406.4  65.05  123.84  1.66  1.57  1.95 
421.7 BR D  65.06  110.19  1.77  1.77  1.94 
421.7 BR U  64.99  67.96  2.87  2.87  1.59 
436.9  65.43  149.06  1.39  1.31  2.32 
488.7  65.56  415.13  1.23  0.47  2.77 
628.9  65.63  439.48  1.29  0.44  2.73 
769.1  65.71  464.02  1.19  0.42  2.55 
Notes:
1. For this example, the entire bridge opening was defined as the main channel. So, the average depth of the main channel is the same as the average depth of the entire cross section.
Determine:
Compute the hydrostatic and drag forces for a debris accumulation on a single pier.
Solution:
Hydrostatic Force on Single Pier Accumulation
A_{hu} = Area of the debris accumulation below the upstream water surface
A_{hu} = (WS_{US}  Debris bottom, DB_{EL})(Width of debris accumulation, W_{D})
A_{hu} = (65.43  61.78)(13.7) = 50.01 m^{2}A_{hd} = Area of the debris accumulation below the downstream water surface
A_{hd} = (WS_{DS}  DB_{EL})( W_{D})
A_{hd} = (65.06  61.78)(13.7) = 44.93 m^{2}
h_{cu} = Vertical distance to centroid of A_{hu} = 0.5(WS_{US}  DB_{EL})
h_{cu} = 0.5(65.43  61.78) = 1.83 mh_{cd} = Vertical distance to centroid of A_{hd} = 0.5(WS_{DS}  DB_{EL})
h_{cd} = 0.5(65.06  61.78) = 1.64 mF_{hu} = Hydrostatic force upstream = γh_{cu}A_{hu}
F_{hu} = (9.81)(1.83)(50.01) = 898 kNF_{hd} = Hydrostatic force downstream = γh_{cd}A_{hd}
F_{hd} = (9.81)(1.64)(44.93) = 723 kenF_{h} = Total hydrostatic force on Pile Bent 2 = F_{hu}  F_{hd}
F_{h} = 898  723 = 175 kNDrag Force on Single Pier Accumulation
B is greater than 0.3, therefore V_{r} should be based on the average velocity in the contracted section.
V_{r} = 2.87 m/s (Table 4.3)
C_{D} = Drag coefficient = 3.1  3.6B (Table 4.1)
C_{D} = 3.1  (3.6)(0.42) = 1.59
Total Force on Single Pier Accumulation
F = Total segment force = F_{h} + F_{D}
F = 175 + 327 = 502 kNLocation of Forces on Single Pier Accumulation
F_{DEL} = Elevation of drag force = 0.5(WS_{US} + DB_{EL})
F_{DEL} = 0.5(65.43 + 61.78) = 63.61 m
F_{hST} = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
F_{DST} = Station of drag force = F_{hST}
F_{ST} = Station of total force = F_{DST} = F_{hST}
F_{hST} = F_{DST} = F_{ST} = 0.5(160.33 + 174.03) = 167.18 m
Given:
Design flow rate = 85 m^{3}/s
Minimum upstream main channel width = 13.7 m; design log length = 13.7 m
Depth of debris is fullflow depth
Main channel width at the bridge = 60 m
Superstructure is not submerged
Ineffective flow areas from debris defined by 1:1 contraction and 2:1 expansion
Bottom elevation of Pile Bent 2 = 61.98 m; Pile Bent 3 = 61.92 m
Total accumulation width = 25.4 m (defined by assuming that the accumulation extends laterally half the design log length beyond each pier).
Accumulation width on Pile Bent 2 = 13.7 m for Case 1 and 11.7 for Case 2
Accumulation width on Pile Bent 3 = 11.7 m for Case 1 and 13.7 for Case 2
Pile Bent 2, left station of debris = 154.69 m; Right station of debris = 168.39 m
Pile Bent 3, left station of debris = 168.39 m; Right station of debris = 180.09 m
Hydraulic computation results are provided in Table 4.4 and shown in Figure 4.11
Upstream water surface elevation, WS_{US} = 65.28 m (Table 4.4)
Downstream water surface elevation, WS_{DS} = 64.59 m (Table 4.4, see discussion of Downstream Water Surface Elevation in Subsection 4.3.3.1)
Figure 4.10. Upstream face of the bridge for Example 2.
Figure 4.11. Water surface profile for Example 2.
River Station (m)  Water Surface Elevation (m)  Flow Area (m^{2})  Main Channel Velocity (m/s)  Cross Section Average Velocity (m/s)  Average Flow Depth^{1}(m) 

0  63.22  28.81  3.52  2.95  1.38 
127  64.25  176.00  1.22  0.48  1.63 
254  64.42  176.31  0.82  0.48  1.08 
402.33  64.55  90.49  0.96  0.94  1.63 
421.7 BR D  64.56  83.47  1.02  1.02  1.63 
421.7 BR U  64.54  27.41  3.10  3.10  0.99 
440.46  65.30  142.17  0.64  0.60  2.20 
488.7  65.32  308.93  0.69  0.28  2.53 
628.9  65.35  308.60  0.79  0.28  2.44 
769.1  65.38  315.16  0.78  0.27  2.22 
Notes:
1. For this example, the entire bridge opening was defined as the main channel. So, the average depth of the main channel is the same as the average depth of the entire cross section.
Determine:
Compute the hydrostatic and drag forces for a debris accumulation on two piers. The calculations are for only Case 1, which is based on the width of the debris accumulation on Pile Bent 2 being equal to the design log length and Pile Bent 3 having a smaller accumulation width. Case 2 is the reverse of Case 1, i.e., the width of the accumulation on Pile Bent 3 would be equal to the design log length and Pile Bent 2 would have a smaller accumulation width. Case 3 is based on the assumption that the design log length spans the opening approximately in the middle of the two piers and the resulting load on the accumulation is transferred equally to each pier.
Solution:
Hydrostatic Forces on Two Adjacent Piers (Case 1)
A_{hu} = Area of the debris accumulation below the upstream water surface
A_{hu} = (WS_{US}  Debris bottom, DB_{EL})(Width of debris accumulation, W_{D})
A_{hu} = (65.30  61.98)(13.70) = 45.48 m^{2} for Pier Bent 2
A_{hu} = (65.30  61.92)(11.70) = 39.55 m^{2} for Pier Bent 3A_{hd} = Area of the debris accumulation below the downstream water surface
A_{hd} = (WS_{DS}  DB_{EL})( W_{D})
A_{hd} = (64.55 61.98)(13.70) = 35.21 m^{2} for Pier Bent 2
A_{hd} = (64.55 61.92)(11.70) = 30.77 m^{2} for Pier Bent 3h_{cu} = Vertical distance to centroid of A_{hu} = 0.5(WS_{US}  DB_{EL})
h_{cu} = 0.5(65.30  61.98) = 1.66 m for Pier Bent 2
h_{cu} = 0.5(65.30  61.92) = 1.69 m for Pier Bent 3h_{cd} = Vertical distance to centroid of A_{hd} = 0.5(WS_{DS}  DB_{EL})
h_{cd} = 0.5(64.55  61.98) = 1.29 m for Pier Bent 2
h_{cd} = 0.5(64.55  61.92) = 1.32 m for Pier Bent 3F_{hu} = Hydrostatic force upstream = γh_{cu}A_{hu}
F_{hu} = (9.81)(1.66)(45.48) = 741 kN for Pier Bent 2
F_{hu} = (9.81)(1.69)(39.55) = 656 kN for Pier Bent 3F_{hd} = Hydrostatic force downstream = γh_{cd}A_{hd}
F_{hd} = (9.81)(1.29)(35.21) = 446 kN for Pier Bent 2
F_{hd} = (9.81)(1.32)(30.77) = 399 kN for Pier Bent 3F_{h} = Total hydrostatic force on Pile Bent = F_{hu}  F_{hd}
F_{h} = 741  446 = 295 kN for Pier Bent 2
F_{h} = 656  399 = 257 kN for Pier Bent 3Drag Force on Two Adjacent Piers (Case 1)
B is greater than 0.3, therefore V_{r} should be based on the average velocity in the contracted section.
V_{r} = 3.10 m/s (Table 4.4)
C_{D} = Drag coefficient = 3.1  3.6B (Table 4.1)
C_{D} = 3.1  (3.6)(0.76) = 0.36
Total Force on Two Adjacent Piers (Case 1)
F = Total segment force = F_{h} + F_{D}
F = 295 + 79 = 374 kN for Pile Bent 2
F = 257 + 68 = 325 kN for Pile Bent 3
F_{DEL} = Elevation of drag force = 0.5(WS_{US} + DB_{EL})
F_{DEL} = 0.5(65.30 + 61.98) = 63.64 m for Pile Bent 2
F_{DEL} = 0.5(65.30 + 61.72) = 63.51 m for Pile Bent 3
F_{hST} = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
F_{DST} = Station of drag force = F_{hST}
F_{ST} = Station of total force = F_{DST} = F_{hST}
F_{hST} = F_{DST} = F_{ST} = 0.5(154.69 + 168.39) = 161.54 m for Pile Bent 2
F_{hST} = F_{DST} = F_{ST} = 0.5(138.39 + 180.09) = 174.24 m for Pile Bent 3
Given:
Design flow rate = 220 m^{3}/s
Low chord elevation of bridge = 65.5 m
Depth of debris is 1.2 meters below the bridge low chord = 64.3 m
Debris accumulation extends along the entire length of the structure (see Figure 4.12)
Main channel width at the bridge = 60.0 m
Left station of debris = 137.16 m; Right station of debris = 197.21 m
Hydraulic computation results are provided in Table 4.5 and shown in Figure 4.13
Upstream water surface elevation, WS_{US} = 65.71 m (Table 4.5)
Downstream water surface elevation, WS_{DS} = 65.13 m (Table 4.5, see discussion of Downstream Water Surface Elevation in Subsection 4.3.3.1)
Figure 4.12. Upstream face of the bridge for Example 3.
Figure 4.13. Water surface profile for Example 3.
River Station (m)  Water Surface Elevation (m)  Flow Area (m^{2})  Main Channel Velocity (m/s)  Cross Section Average Velocity (m/s)  Average Flow Depth^{1} (m) 

0  64.36  255.76  2.54  0.86  2.43 
127  64.79  325.88  1.64  0.68  2.17 
254  64.98  333.72  1.07  0.66  1.64 
406.4  65.13  130.59  1.80  1.68  2.03 
421.7 BR D  64.30  75.34  2.92  2.92  1.45 
421.7 BR U  64.30  75.34  2.92  2.92  1.45 
436.9  65.71  171.53  1.39  1.28  2.61 
488.7  65.85  563.03  1.03  0.39  3.06 
628.9  65.89  573.58  1.10  0.38  2.99 
769.1  65.94  579.47  1.05  0.38  2.78 
Notes:
1. For this example, the entire bridge opening was defined as the main channel. So, the average depth of the main channel is the same as the average depth of the entire cross section.
Determine:
Compute the hydrostatic and drag forces for a debris accumulation on a superstructure.
Solution:
Hydrostatic Force on Superstructure Accumulation
A_{hu} = Area of the debris accumulation below the upstream water surface
A_{hu} = (WS_{US}  Debris bottom, DB_{EL})(Width of debris accumulation, W_{D})
A_{hu} = (65.71  64.30)(60.0) = 84.60 m^{2}A_{hd} = Area of the debris accumulation below the downstream water surface
A_{hd} = (WS_{DS}  DB_{EL})(W_{D})
A_{hd} = (65.13  64.30)(60.0) = 49.80 m^{2}h_{cu} = Vertical distance to centroid of A_{hu} = 0.5(WS_{US}  DB_{EL})
h_{cu} = 0.5(65.71  64.30) = 0.71 mh_{cd} = Vertical distance to centroid of A_{hd} = 0.5(WS_{DS}  DB_{EL})
h_{cd} = 0.5(65.13  64.30) = 0.42 mF_{hu} = Hydrostatic force upstream = γh_{cu}A_{hu}
F_{hu} = (9.81)(0.71)(84.60) = 589 kNF_{hd} = Hydrostatic force downstream = γh_{cd}A_{hd}
F_{hd} = (9.81)(0.42)(49.80) = 205 kNF_{h} = Total hydrostatic force on Pile Bent 2 = F_{hu}  F_{hd}
F_{h} = 589  205 = 384 kNDrag Force on Superstructure Accumulation
B is greater than 0.3, therefore V_{r} should be based on the average velocity in the contracted section.
V_{r} = 2.92 m/s (Table 4.5)
C_{D} = Drag coefficient = 3.1  3.6B (Table 4.1)
C_{D} = 3.1  (3.6)(0.53) = 1.19
Total Force on Superstructure Accumulation
F = Total segment force = F_{h} + F_{D}
F = 384 + 429 = 813 kNLocation of Forces on Superstructure Accumulation
F_{DEL} = Elevation of drag force = 0.5(WS_{US} + DB_{EL})
F_{DEL} = 0.5(65.71 + 64.30) = 65.00 m
F_{hST} = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
F_{DST} = Station of drag force = F_{hST}
F_{ST} = Station of total force = F_{DST} = F_{hST}
F_{hST} = F_{DST} = F_{ST} = 0.5(137.16 + 197.21) = 167.19 m
Given:
Design flow rate = 6,890 ft^{3}/s
Minimum upstream main channel width = 45 ft; design log length of 45 ft
Depth of debris is fullflow depth
Main channel width at the bridge = 197 ft
Debris accumulation only on Pile Bent 2 (see Figure 4.14)
Superstructure is not submerged
Ineffective flow areas from the debris defined by 1:1 contraction and 2:1 expansion
Bottom elevation of Pile Bent 2 = 202.69 ft
Left station of debris = 526.02 ft; Right station of debris = 570.97 ft
Hydraulic computation results are provided in Table 4.6 and shown in Figure 4.15
Upstream water surface elevation, WS_{US} = 214.66 ft (Table 4.6)
Downstream water surface elevation, WS_{DS} = 213.44 ft (Table 4.6, see discussion of Downstream Water Surface Elevation in Subsection 4.3.3.1)
Figure 4.14. Upstream face of the bridge for Example 4.
Figure 4.15. Water surface profile for Example 4.
River Station (miles)  Water Surface Elevation (ft)  Flow Area (ft^{2})  Main Channel Velocity (ft/s)  Cross Section Average Velocity (ft/s)  Average Flow Depth^{1} (ft) 

5.13  210.86  2487.36  8.12  2.77  7.69 
5.21  212.27  3231.47  5.23  2.13  6.81 
5.29  212.90  3312.69  3.39  2.08  5.08 
5.38  213.41  1333.33  5.46  5.17  6.40 
5.39 BR D  213.44  1186.32  5.81  5.81  6.36 
5.39 BR U  213.21  731.66  9.42  9.42  5.21 
5.40  214.66  1605.03  4.56  4.29  7.62 
5.43  215.09  4472.12  4.04  1.54  9.08 
5.52  215.34  4733.89  4.22  1.46  8.95 
5.60  215.58  4997.62  3.91  1.38  8.36 
Notes:
1. For this example, the entire bridge opening was defined as the main channel. So, the average depth of the main channel is the same as the average depth of the entire cross section.
Determine:
Compute the hydrostatic and drag forces for a debris accumulation on a single pier.
Solution:
Hydrostatic Force on Single Pier Accumulation
A_{hu} = Area of the debris accumulation below the upstream water surface
A_{hu} = (WS_{US}  Debris bottom, DB_{EL})(Width of debris accumulation, W_{D})
A_{hu} = (214.66  202.69)(45) = 538.65 ft^{2}A_{hd} = Area of the debris accumulation below the downstream water surface
A_{hd} = (WS_{DS}  DB_{EL})(W_{D})
A_{hd} = (213.44  202.69)(45) = 483.75 ft^{2}h_{cu} = Vertical distance to centroid of A_{hu} = 0.5(WS_{US}  DB_{EL})
h_{cu} = 0.5(214.66  202.69) = 5.98 fth_{cd} = Vertical distance to centroid of A_{hd} = 0.5(WS_{DS}  DB_{EL})
h_{cd} = 0.5(213.44  202.69) = 5.38 ftF_{hu} = Hydrostatic force upstream = γh_{cu}A_{hu}
F_{hu} = (62.4)(5.98)(538.65) = 200,998 lbs (100.5 tons)F_{hd} = Hydrostatic force upstream = γh_{cd}A_{hd}
F_{hd} = (62.4)(5.38)(483.75) = 162,401 lbs (81.2 tons)F_{h} = Total hydrostatic force on Pile Bent 2 = F_{hu}  F_{hd}
F_{h} = 200,998  162,401 = 38,597 lbs (19.3 tons)Drag Force on Single Pier Accumulation
B is greater than 0.3, therefore V_{r} should be based on the average velocity in the contracted section.
V_{r} = 9.42 ft/s (Table 4.6)
C_{D} = Drag coefficient = 3.1  3.6B (Table 4.1)
C_{D} = 3.1  (3.6)(0.42) = 1.59
Total Force on Single Pier Accumulation
F = Total segment force = F_{h} + F_{D}
F = 38,597 + 73,638 = 112,235 lbs (56.1 tons)Location of Forces on Single Pier Accumulation
F_{DEL} = Elevation of drag force = 0.5(WS_{US} + DB_{EL})
F_{DEL} = 0.5(214.66 + 202.69) = 208.68 ft
F_{hST} = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
F_{DST} = Station of drag force = F_{hST}
F_{ST} = Station of total force = F_{DST} = F_{hST}
F_{hST} = F_{DST} = F_{ST} = 0.5(526.02 + 570.97) = 548.50 ft
Given:
Design flow rate = 3,000 ft^{3}/s
Minimum upstream main channel width = 45 ft; design log length = 45 ft
Depth of debris is fullflow depth
Main channel width at the bridge = 197 ft
Superstructure is not submerged
Ineffective flow areas from debris defined by 1:1 contraction and 2:1 expansion
Bottom elevation of Pile Bent 2 = 203.35 ft; Pile Bent 3 = 203.15 ft
Total accumulation width = 83.4 ft (defined by assuming that the accumulation extends laterally half the design log length beyond each pier).
Accumulation width on Pile Bent 2 = 45 ft for Case 1 and 38.4 ft for Case 2
Accumulation width on Pile Bent 3 = 38.4 ft for Case 1 and 45 ft for Case 2
Pile Bent 2, left station of debris = 507.53 ft; Right station of debris = 552.48 ft
Pile Bent 3, left station of debris = 552.48 ft; Right station of debris = 590.87 ft
Hydraulic computation results are provided in Table 4.7 and shown in Figure 4.17
Upstream water surface elevation, WS_{US} = 214.17 ft (Table 4.7)
Downstream water surface elevation, WS_{DS} = 211.89 ft (Table 4.7, see discussion of Downstream Water Surface Elevation in Subsection 4.3.3.1)
Figure 4.16. Upstream face of the bridge for Example 5.
Figure 4.17. Water surface profile for Example 5.
River Station (miles)  Water Surface Elevation (ft)  Flow Area (ft^{2})  Main Channel Velocity (ft/s)  Cross Section Average Velocity (ft/s)  Average Flow Depth^{1} (ft) 

5.13  207.42  310.14  11.55  9.68  4.53 
5.21  210.80  1894.63  4.00  1.57  5.35 
5.29  211.36  1897.97  2.69  1.57  3.54 
5.38  211.79  974.12  3.15  3.08  5.35 
5.39 BR D  211.82  898.55  3.35  3.35  5.35 
5.39 BR U  211.76  295.07  10.17  10.17  3.25 
5.40  214.25  1530.45  2.10  1.97  7.22 
5.43  214.31  3325.62  2.26  0.92  8.30 
5.52  214.41  3322.07  2.59  0.92  8.01 
5.60  214.51  3392.69  2.56  0.89  7.28 
Notes:
1. For this example, the entire bridge opening was defined as the main channel. So, the average depth of the main channel is the same as the average depth of the entire cross section.
Determine:
Compute the hydrostatic and drag forces for a debris accumulation on two piers. The calculations are for only Case 1, which is based on the width of the debris accumulation on Pile Bent 2 being equal to the design log length and Pile Bent 3 having a smaller accumulation width. Case 2 is the reverse of Case 1, i.e., the width of the accumulation on Pile Bent 3 would be equal to the design log length and Pile Bent 2 would have a smaller accumulation width. Case 3 is based on the assumption that the design log length spans the opening approximately in the middle of the two piers and the resulting load on the accumulation is transferred equally to each pier.
Solution:
Hydrostatic Forces on Two Adjacent Piers (Case 1)
A_{hu} = Area of the debris accumulation below the upstream water surface
A_{hu} = (WS_{US}  Debris bottom, DB_{EL})(Width of debris accumulation, W_{D})
A_{hu} = (214.25  203.35)(45) = 490.50 ft^{2} for Pile Bent 2
A_{hu} = (214.25  203.15)(38.4) = 426.24 ft^{2} for Pile Bent 3A_{hd} = Area of the debris accumulation below the downstream water surface
A_{hd} = (WS_{DS}  DB_{EL})(W_{D})
A_{hd} = (211.82  203.35)(45) = 381.15 ft^{2} for Pile Bent 2
A_{hd} = (211.82  203.15)(38.4) = 332.93 ft^{2} for Pile Bent 3h_{cu} = Vertical distance to centroid of A_{hu} = 0.5(WS_{US}  DB_{EL})
h_{cu} = 0.5(214.25  203.35) = 5.45 ft for Pile Bent 2
h_{cu} = 0.5(214.25  203.15) = 5.55 ft for Pile Bent 3h_{cd} = Vertical distance to centroid of A_{hd} = 0.5(WS_{DS}  DB_{EL})
h_{cd} = 0.5(211.82  203.35) = 4.24 ft for Pile Bent 2
h_{cd} = 0.5(211.82  203.15) = 4.34 ft for Pile Bent 3F_{hu} = Hydrostatic force upstream = γh_{cu}A_{hu}
F_{hu} = (62.4)(5.45)(490.50) = 166,809 lbs (83.4 tons) for Pile Bent 2
F_{hu} = (62.4)(5.55)(426.24) = 147,615 lbs (73.8 tons) for Pile Bent 3F_{hd} = Hydrostatic force upstream = γh_{cd}A_{hd}
F_{hd} = (62.4)(4.24)(381.15) = 100,843 lbs (50.4 tons) for Pile Bent 2
F_{hd} = (62.4)(4.34)(332.93) = 90,163 lbs (45.1 tons) for Pile Bent 3F_{h} = Total hydrostatic force on Pile Bent 2 = F_{hu}  F_{hd}
F_{h} = 166,809  100,843 = 65,966 lbs (33.0 tons) for Pile Bent 2
F_{h} = 147,615  90,163 = 57,452 lbs (28.7 ton) for Pile Bent 3Drag Force on Two Adjacent Piers (Case 1)
B is greater than 0.3, therefore V_{r} should be based on the average velocity in the contracted section.
V_{r} = 10.17 ft/s (Table 4.7)
C_{D} = Drag coefficient = 3.1  3.6B (Table 4.1)
C_{D} = 3.1  (3.6)(0.76) = 0.36
Total Force on Two Adjacent Piers (Case 1)
F = Total segment force = F_{h} + F_{D}
F = 65,966 + 19,589 = 85,555 lbs (42.8 tons) for Pile Bent 2
F = 57,452 + 17,023 = 74,475 lbs (37.2 tons) for Pile Bent 3Location of Forces on Single Pier Accumulation
F_{DEL} = Elevation of drag force = 0.5(WSEL_{US} + DBEL)
F_{DEL} = 0.5(214.25 + 203.35) = 208.80 ft for Pile Bent 2
F_{DEL} = 0.5(214.25 + 203.15) = 208.70 ft for Pile Bent 3
F_{hST} = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
F_{DST} = Station of drag force = F_{hST}
F_{ST} = Station of total force = F_{DST} = F_{hST}
F_{hST} = F_{DST} = F_{ST} = 0.5(507.53 + 552.48) = 530.00 ft for Pile Bent 2
F_{hST} = F_{DST} = F_{ST} = 0.5(552.48 + 590.87) = 571.67 ft for Pile Bent 3
Given:
Design flow rate = 7,770 ft^{3}/s
Low Chord Elevation = 214.91 ft
Depth of debris is 3.94 feet below the bridge low chord = 210.97 ft
Debris accumulation extends along the entire length of the structure (see Figure 4.18)
Main channel width at the bridge = 197 ft
Left station of debris = 450.00 ft; Right station of debris = 647.01 ft
Hydraulic computation results are provided in Table 4.8 and shown in Figure 4.19
Upstream water surface elevation, WS_{US} = 215.59 ft (Table 4.8)
Downstream water surface elevation, WS_{DS} = 213.69 ft (Table 4.8, see discussion of Downstream Water Surface Elevation in Subsection 4.3.3.1)
Figure 4.18. Upstream face of the bridge for Example 6.
Figure 4.19. Water surface profile for Example 6.
River Station (miles)  Water Surface Elevation (ft)  Flow Area (ft^{2})  Main Channel Velocity (ft/s)  Cross Section Average Velocity (ft/s)  Average Flow Depth Depth^{1} (ft) 

5.13  211.17  2753.25  8.33  2.82  7.97 
5.21  212.58  3508.09  5.38  2.23  7.12 
5.29  213.20  3592.48  3.51  2.17  5.38 
5.38  213.69  1405.80  5.91  5.51  6.66 
5.39 BR D  210.97  811.03  9.58  9.58  4.76 
5.39 BR U  210.97  811.03  9.58  9.58  4.76 
5.40  215.59  1846.51  4.56  4.20  8.56 
5.43  216.05  6061.00  3.38  1.28  10.04 
5.52  216.19  6174.57  3.61  1.25  9.81 
5.60  216.35  6237.97  3.45  1.25  9.12 
Notes:
1. For this example, the entire bridge opening was defined as the main channel. So, the average depth of the main channel is the same as the average depth of the entire cross section.
Determine:
Compute the hydrostatic and drag forces for a debris accumulation on a superstructure.
Solution:
Hydrostatic Force on Superstructure Accumulation
A_{hu} = Area of the debris accumulation below the upstream water surface
A_{hu} = (WS_{US}  Debris bottom, DB_{EL})(Width of debris accumulation, W_{D})
A_{hu} = (215.59  210.97)(197.0) = 910.14 ft^{2}A_{hd} = Area of the debris accumulation below the downstream water surface
A_{hd} = (WS_{DS}  DB_{EL})(W_{D})
A_{hd} = (213.69  210.97)(197.0) = 535.84 ft^{2}h_{cu} = Vertical distance to centroid of A_{hu} = 0.5(WS_{US}  DB_{EL})
h_{cu} = 0.5(215.59  210.97) = 2.31 fth_{cd} = Vertical distance to centroid of A_{hd} = 0.5(WS_{DS}  DB_{EL})
h_{cd} = 0.5(213.69  210.97) = 1.36 ftF_{hu} = Hydrostatic force upstream = γh_{cu}A_{hu}
F_{hu} = (62.4)(2.31)(910.14) = 131,191 lbs (65.6 tons)F_{hd} = Hydrostatic force upstream = γh_{cd}A_{hd}
F_{hd} = (62.4)(1.36)(535.84) = 45,474 lbs (22.7 tons)F_{h} = Total hydrostatic force on Pile Bent 2 = F_{hu}  F_{hd}
F_{h} = 131,191  45,474 = 85,717 lbs (42.9 tons)Drag Force on Superstructure Accumulation
B is greater than 0.3, therefore V_{r} should be based on the average velocity in the contracted section.
V_{r} = 9.58 ft/s (Table 4.8)
C_{D} = Drag coefficient = 3.1  3.6B (Table 4.1)
C_{D} = 3.1  (3.6)(0.53) = 1.19
Total Force on Superstructure Accumulation
F = Total segment force = F_{h} + F_{D}
F = 85,717 + 96,313 = 182,030 lbs (91.0 tons)Location of Forces on Superstructure Accumulation
F_{DEL} = Elevation of drag force = 0.5(WS_{US} + DB_{EL})
F_{DEL} = 0.5(215.59 + 210.97) = 213.28 ft
F_{hST} = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
F_{DST} = Station of drag force = F_{hST}
F_{ST} = Station of total force = F_{DST} = F_{hST}
F_{hST} = F_{DST} = F_{ST} = 0.5(450.00 + 647.01) = 548.51 ft
Brian Beucler
Office of Bridges and Structures
2023664598
brian.beucler@dot.gov
Joe Krolak
Office of Bridges and Structures
2023664611
joe.krolak@dot.gov