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Hydraulics Engineering

 

Debris Control Structures Evaluation and Countermeasures
Hydraulic Engineering Circular No. 9

Chapter 4 - Analyzing and Modeling Debris IMpacts to Structures

4.1 Introduction

After the location and extent of the debris accumulation on the bridge structure has been determined using the procedures discussed in the previous Chapter, a hydraulic analysis should be conducted to evaluate the affects the accumulation would have on the hydraulic characteristics through and upstream of the bridge structure, local scour at the piers, and hydraulic loading on the structure. General information for performing such an analysis is presented in this Chapter. This Chapter also includes information for estimating local pier scour and hydraulic loading on the bridge structure associated with debris accumulation.

4.2 Hydraulic Analyses of Debris

Hydraulic analyses of affects of debris upon a drainage structure are often conducted using a one-dimensional (1-D) water surface model. However, such analyses can also be performed using:

  • Hand calculations;
  • Two-dimensional (2-D) numerical (computer) models;
  • Three-dimensional (3-D) computer models; or
  • Physical (laboratory) modeling.

The selection of any such analytical technique is based on the complexity of the hydraulics and debris, risk and importance of the drainage structure, and other project site characteristics.

4.3 One-Dimensional Debris Analysis Modeling

One-dimensional programs available for performing debris analyses include the U.S. Army Corps of Engineers (USACE) River Analysis System (HEC-RAS) and the FHWA Water Surface Profile (WSPRO) (among others). For culverts, the FHWA HY-8 program allows evaluation of complex hydraulic conditions.

4.3.1 Data Requirements

Data required for the hydraulic analyses include geometric and flow data. The geometric data consists of cross section data, reach length, energy loss coefficients, and hydraulic structure data. Flow data includes the discharges used in the analyses and the associated boundary conditions and regimes. A thorough discussion of this information, as well as modeling approaches is beyond the scope of this document. However, such information can be found in appropriate user manuals and model documentation. A brief description of each of these modeling data follows:

Cross Section Geometryis the representation of the ground surface perpendicular to the direction of flow. Cross sections are located along a watercourse to define the conveyance capacity of the main channel and the adjacent floodplain. Cross sections are required at representative locations throughout the watercourse and at distinct locations where changes occur in discharge slope, shape, or roughness, or at location where hydraulic structures are located. In modeling debris, the cross section data can be modified to include or simulate ineffective flow areas, levees, and/or blocked obstructions.

Reach Length is the measured distance between cross sections. Reach lengths are provided for the main channel, measured along the thalweg, and for the left and right overbanks, measured along the anticipated path of the center of mass of the overbank flow. In debris analyses, reach lengths serve to allow refined characterization of the extent of the debris field and the associated effects.

Energy Loss Coefficients estimate losses caused by the resistance to flow from bed-surface and vegetative roughness (i.e., Manning's n coefficient) (8,15,28,3,15), channel irregularities, channel alignment, obstructions, and by the contraction and expansion of the flow. Adjusting these coefficients allows simulation of the presence and extent of debris. These also provide a means to simulate the impacts and effects of debris.

Hydraulic Structure Data is the geometric representation of structures that influence the water surface profile within a watercourse. The hydraulic structures can include bridges, culverts, spillways, diversion structures, weirs, etc. The information required to define the bridge structure are the dimensions of the bridge deck, piers, and bridge abutments. The geometry of the debris accumulation should also be accounted for when defining these features. Typically, the dimensions of these features have to be manually adjusted to account for the debris accumulation.

Flow Data required for the model is the discharge in the watercourse and the flow conditions at the boundaries of the model. The discharge is based on the peak discharge for the design flood event of the bridge or for a specific flood event that is being used to estimate the hydraulic loads on the bridge structure. For this discharge, the flow depth is required at the downstream boundary for subcritical flow (Froude number less than 1) and at the upstream boundary for supercritical flow (Froude number greater than 1) to initiate the water surface profile computations.

4.3.2 Background of Modeling Methods and Approaches

There are several methods available for evaluating the hydraulics through a bridge structure for a one-dimensional flow analysis. The type of methods available depends on the flow conditions through the bridge. The flow through the bridge could be classified as either low or high flow conditions.

4.3.2.1 Low Flow Conditions

Low flow conditions exist when the flow through the bridge opening is open channel flow, i.e., the water surface is below the highest point on the low chord of the bridge opening. Three types of flow classes can exist for this condition:

  1. Class A exists when the water surface through the bridge is completely subcritical, i.e., above critical depth;
  2. Class B exists when the water surface profile passes through critical depth within the bridge structure, which can occur for either supercritical or subcritical flow; and
  3. Class C exists when the water surface profile through the bridge structure is completely supercritical, i.e., below critical depth.

There are three methods commonly available for computing the hydraulics through the bridge for low flow conditions. These methods are:

  1. Energy Equation - This method uses the conservation of energy to determine water surface elevations, velocities, and losses in a waterway. This method is best used when the bridge piers are a small obstruction to the flow and the friction losses are the predominate consideration. This method can be used for both supercritical and subcritical flow (Class A, B, and C).
  2. Momentum Equation -The momentum equation uses the second law of thermodynamics to describe how change of momentum per unit of time in the body of water in a flowing channel is equal to the resultant of all the external forces that are acting on the body. Unlike the energy equation, this method does not account for non-uniform velocity distributions. The momentum equation method is best used when the bridge piers are the dominant contributor to energy losses or when the pier losses and friction losses are both predominant. As in the energy method, this method can be used for both supercritical and subcritical flow (Class A, B, and C). FHWA does not recommend this method for 1-D bridge hydraulics.
  3. Yarnell Equation - The Yarnell equation empirically predicts the change in water surface from just downstream of the bridge to just upstream of the bridge. The equation is based on about 2,600 lab experiments in which the researchers varied the shape of the piers, the width, the length, the angle, and the flow rate.(66) This method is most applicable when the piers are the dominant contributor to energy losses and the flow through the bridge remains subcritical (Class A). FHWA does not recommend this method for 1-D bridge hydraulics.

The energy and momentum equation methods can be used for all of the classes for the low flow condition (Class A, B, and C), while the Yarnell equation method is intended only for Class A low flow.

4.3.2.2 High Flow Conditions

High flow conditions exist when the flow through the bridge opening comes in contact with the maximum low chord of the bridge deck. The type of flow conditions that can occur for high flow include pressure flow, a combination of pressure and weir flow, and a combination of weir flow and open channel flow through the bridge. Generally, three computational approaches exist to evaluate high flow conditions.

  1. Pressure Flow Condition - Pressure flow occurs when the flow comes into contact with the low chord of the bridge deck. The backwater upstream of the bridge associated with this type of flow condition causes the flow through the structure to behave as orifice flow. In general, there are two types of pressure flow that can exist (details of which are described in other FHWA documents).(10) Depending on the conditions, pressure flow can describe a bridge with full submergence of the low chord at the upstream side and open channel flow at the downstream side of the bridge (acting as a sluice gate - Figure 4.1). The second type of pressure flow exists when both the upstream and downstream sides of the bridge are submerged and uses the standard full flowing orifice equation (Figure 4.2).

    For a bridge profile, the sluice gate type of pressure flow will have water build up on the upstream bridge face. The flow passes under the bridge behaving as a sluice or orifice opening. The water elevation downstream of the bridge is less than the low chord.
    Figure 4.1. Sketch of the sluice gate type of pressure flow.

    A bridge may experience pressure flow with partial bridge superstructure submergence both upstream and downstream bridge. Both the upstream and downstream face (as well as low chord) has been submerged, but no overtopping has yet occurred.
    Figure 4.2. Sketch of fully submerged pressure flow.

  2. Weir Flow Condition - Weir flow exists when water flows over the bridge structure and/or the roadway approaches for the bridge. Typically this condition is modeled using the standard weir equation (Figure 4.3). This illustration depicts pressure flow occurring through the bridge structure, which might not always be the case. Note that pressure flow may also occur through the bridge opening. Typically, some balancing of flow through the opening and over the "roadway" weir occurs before the model converges to a solution.

    A bridge may exhibit submerged (and pressure flow) through the normal opening and weir flow from overtopping.
    Figure 4.3. Sketch of pressure and weir flow.

  3. Energy Equation Approach - as with low-flow conditions, the energy method balances the total energy from the downstream side to the upstream side of the bridge structure. All of the computations are performed as though the flow is open channel flow, and the area obstructed by the bridge structure is subtracted from the flow area and the wetted perimeter is increased for the portions of the structure in contact with the water. This method should be used when the bridge is highly submerged and the flow over the road is not controlled by weir flow (low water bridges) or when the bridge deck is a small obstruction to the flow and the bridge opening is not acting like a pressurized orifice. This method is also best used for bridges that are perched above the floodplain.
4.3.3 Scenarios for Hydraulic Modeling of Debris Accumulation

These 1-D model data, methods, and approaches permit computation of water surface profiles at a hydraulic structure with debris accumulation. As previously mentioned, the general concepts discussed would apply to most 1-D models commonly used in the highway hydraulics community.

4.3.3.1 Bridge Debris Scenarios

Scenarios for analyzing debris accumulation at a bridge structure involve relocating cross sections, redefining the ineffective flow boundaries, modifying the cross section and bridge geometry, and changing the contraction and expansion coefficients.

Scenario 1: Relocation of Downstream Wake. As depicted in Figure 4.4, relocating an "expansion" cross section further downstream would attempt to simulate an ineffective flow zone (downstream wake) created by the debris accumulation. The ineffective flow created by the debris accumulation should extend downstream from the upstream face. Assuming a 2:1 to 4:1 expansion ratio for the reach downstream of the bridge, this distance would range from the width (2:1) to twice the width (4:1) of the debris accumulation. This scenario might be required where most of the flow downstream of the bridge is conveyed within the main channel, the overbank areas are not extremely wide, or the bridge structure and/or roadway do not significantly constrict the flow.

Relocating an "expansion" cross section further downstream simulates an ineffective flow zone created by the debris accumulation. The ineffective flow created by the debris accumulation should extend downstream from the upstream face. The figure uses a 2:1 expansion ratio for the reach downstream of the bridge, so this distance would equal the width of the debris accumulation.
Figure 4.4. Changing downstream expansion cross section location.

Scenario 2: Creating additional downstream ineffective flow boundaries. As seen in Figure 4.5, adding ineffective flow boundaries to downstream bridge face cross section would simulate the downstream wake created by the debris accumulation. Once again, assume a 2:1 to 4:1 expansion ratio to create the locations of the ineffective flow. Adding additional downstream cross sections would assist in the transition. This scenario could be used alone or in conjunction with other scenarios.

Adding ineffective flow boundaries to downstream bridge face cross section simulates the downstream wake created by the debris accumulation. The figure uses a 2:1 expansion ratio to create the locations of the ineffective flow.
Figure 4.5. Adding downstream ineffective flow locations.

Scenario 3: Creating upstream ineffective flow boundaries. An additional upstream cross section would simulate an ineffective flow zone created by the debris accumulation (Figure 4.6). Assuming a 1:1 to 2:1 contraction ratio the reach upstream of the bridge to the debris accumulation, the point where flow is not affected by the accumulation would be located upstream about one-half (1:1) to the entire (2:1) of the debris accumulation width. Depending on the accumulation width, adding additional upstream cross sections would assist in the transition. This scenario could be used alone or in conjunction with other scenarios.

Adding upstream ineffective flow locations allows evaluation of where the upstream water surface is no longer influenced by the debris accumulation.
Figure 4.6. Adding upstream ineffective flow locations.

Scenario 4: Modifying Bridge Geometry. Modification of the bridge geometry could be used to reflect debris accumulation. The modification could range from changing local structural elements, such as piers and abutments, to actual changes in the bridge opening area or low chord elevations. For piers, debris accumulation would change how pier width and height would be described in model input. The debris may collect either symmetrically or asymmetrically about the pier centerline. For an asymmetrical debris accumulation, the centerline of the pier would have to be moved to the centerline of the debris accumulation or an additional "dummy" bridge pier with an extremely narrow width would have to be defined at the centerline of the debris accumulation. This is illustrated in Figure 4.7.

Scenario 5: Modifying Contraction and Expansion Losses. In some cases, increasing the contraction and expansion loss coefficients may be appropriate if the debris accumulation causes an abrupt contraction and expansion, respectively, of the flow.

Adding a "Dummy" bridge pier to simulate an asymmetrical debris accumulation involves either adds or relocates a pier to the center of the debris accumulation width.
Figure 4.7. "Dummy" bridge pier used to simulate an asymmetrical debris accumulation.

The results of the hydraulic analysis can be used to estimate the hydraulic loading on the bridge structure associated with the debris accumulation. Information required from a 1-D model to estimate the loads is briefly discussed in the following paragraphs.

Upstream Water Surface Elevation. The upstream water surface elevation is used to compute the blockage ratio, B, the hydrostatic and hydrodynamic forces at the upstream side of bridge, and the location of the forces. This elevation should be selected at a location upstream of where the flow accelerates in response to the debris accumulation, causing flow-separation and a zone of ineffective flow. The upstream boundary of this zone is depicted in Figure 4.6. As shown in this figure, the boundary is based on the flow contraction ratio typically assumed near a bridge structure. Based on this assumption, the upstream water surface elevation should be obtained from a cross section located upstream of the debris accumulation at a minimum distance of at least one half of the total width of the debris accumulation.

Downstream Water Surface Elevation. The downstream water surface elevation is used to compute the downstream hydrostatic forces on the bridge structure that is used to determine the net hydrostatic forces on the structure. As in the upstream water surface elevation, this elevation should be obtained sufficiently far downstream from the structure that the flow is not affected by the wake (ineffective flow zone) created by the debris accumulation (see Figure 4.5). Based on the assumptions discussed for the upstream water surface elevation, the downstream water surface elevation should be obtained from a cross section located downstream of the debris accumulation at a distance equal to the total width of the debris accumulation. The typical response within this reach is for the water surface elevation to recover from the drop in the water surface elevation (increased velocities) through the contracted section by increasing downstream of the bridge. However, there are some conditions (high rates of energy dissipation, large channel slopes, or large changes in channel geometry) where the water surface elevation downstream does not recover (rise in elevation) from the flow contraction. For these conditions, the water surface elevation downstream from the effects of the debris may be lower than the elevations in the contraction of the bridge section. For such cases, the water surface elevation in the contracted section should be used to compute the downstream hydrostatic forces. To summarize, the water surface elevation used to compute the downstream hydrostatic forces should be based on the higher water surface elevation at either the contracted section or the cross section immediately downstream of the wake created by the debris accumulation.

Area within the Contracted Section. The flow area within the contracted section is used to compute the blockage ratio, B. The area should be based on the smallest area within the bridge section.

Flow Velocity within the Contracted Section. The flow velocity within the contracted section is used to compute the dynamic forces on the structure. For accumulations on piers, the reference location for the velocity depends on the percentage of blockage of the debris and bridge structure. If the reduction in area is anticipated to be greater than 30 percent of the entire wetted cross-sectional flow area in the bridge opening, then the reference velocity is based on the maximum average velocity in the contracted section of the entire bridge opening. If the reduction is anticipated to be less than 30 percent, then the reference velocity is based on the maximum local average velocity near the pier and debris accumulation, i.e., maximum average flow velocity in the main channel for piers located in the main channel and maximum average flow velocity in the left overbank for piers located in the left overbank. For accumulations on superstructures, the reference velocity is the maximum contracted flow velocity in main channel under the superstructure for any degree of blockage.

Average Flow Depth within the Contracted Section. The average flow depth within the contracted section is used to compute the Froude number that is utilized in selecting the drag coefficient. This depth should be based on the same area used to define the reference velocity, i.e., the average depth in the main channel should be used if the reference velocity is based on the average flow velocity in the main channel.

4.3.3.2 Culvert Debris Scenarios

Most of the common 1-D models have some culvert hydraulic capabilities incorporated within their algorithms. In several cases, these capabilities would allow application of the bridge-based scenarios to these culvert structures. However, these 1-D models do not simulate every barrel type or configuration. Additionally, for certain hydraulic and discharge conditions, these 1-D models may not replicate underlying assumptions of culvert hydraulics.

Additionally, there is not a great deal of research available into the effects of debris upon hydraulic performance of culverts. Therefore, FHWA recommends use of specific culvert models, specifically HY-8, for both culvert hydraulic and debris analyses. In such use, debris analysis scenarios would modify barrel parameters to reflect changes in inlet efficiency (i.e., entrance loss coefficients), additional roughness, or reduced equivalent barrel area.

4.4 Advanced Modeling

Flow, geometric, boundary, topographic, or other conditions near the bridge structure may necessitate use of 2-D numeric, 3-D numeric, or physical models. Such models allow a more detailed prediction of the flow-separation regions and hydraulic pressure variations near the bridge structure and debris accumulation than what is assumed in the one-dimensional analysis.

These models can also be very useful in defining the locations of the stream channel where high-debris transport would most likely occur. The numerical (i.e., 2-D and 3-D) models need approximately the same type of information as the 1-D model, however they use different means to represent data including:

  1. Using a finite element mesh of nodes and links to represent the topography, bathometry, and drainage structures;
  2. Requiring the resistance coefficient and turbulence parameter be defined at each node; and
  3. Defining the boundary conditions differently.

The physical models require scaled design of the site to allow predictions of flow and debris characteristics and tendencies.

Use of such models would be predicated on the relative importance of simulating the debris accumulation for a project. Several research projects are investigating the appropriate use of such advanced models in situations such as these. Additionally, increased power of computers and hydraulic software packages are making such analyses more cost effective for the transportation community.

4.5 Local Pier Scour Associated with Debris Accumulation

Debris accumulations on a bridge pier can increase local scour at the pier as a result of increased pier width and downward flow component upstream of the pier. When debris accumulates on a pier, the scour depth can be estimated by assuming that the pier width is larger than the actual width. A width equal to the design log length as defined by Diehl(17) and presented in the previous chapter of this manual can be assumed for estimating the scour at the pier. This assumption could be on the conservative side at large depths because the effect of the debris on scour depth diminishes.

Only limited research exists on local scour at piers with debris accumulation. Melville and Dongol have conducted a limited quantitative study of the effect of debris accumulation on local pier scour and have made some recommendations which support the approach suggested above(39). An interim procedure for estimating the effect of debris accumulation on local scour at piers is presented in Appendix D of HEC-18(55).

4.6 Hydraulic loading Associated with Debris Accumulation

There are three steps for computing the hydraulic loading on a bridge structure with debris accumulation. The first step is to define the geometry of the debris accumulation using the procedures and recommendations presented in Chapter 3 of this manual. The second step is to compute the flow hydraulics through the bridge structure using the procedures and recommendations presented in the previous sections of this chapter. The last step is to compute the hydrodynamic loads using the hydraulic characteristics associated with the presence of the debris accumulation and the following equations and general procedure developed by Parola(49).

The hydrodynamic drag force is based on the general form of the drag equation and the drag coefficient relationship developed from a model study investigation by Parola at the University of Louisville(49).

F sub D equals C sub D times gamma times A sub D times V sub r squared divided by 2g (4.1)

where:

FD = Drag force, N (lbs)

CD = Drag coefficient, see Tables 4.1 and 4.2

γ = Specific weight of water, N/m3 (lbs/ft3)

AD = Area of wetted debris based on the upstream water surface elevation projected normal to the flow direction, m2 (ft2)

Vr = Reference velocity, see discussion in Subsection 4.3.3.1, m/s (ft/s)

g = Acceleration of gravity, 9.81 m/s2 (32.2 ft/s2)

Drag coefficient for debris on piers is provided in Table 4.1 and for debris on superstructures in Table 4.2.

Table 4.1. Drag Coefficient for Debris on Piers.
Value of B Value of Fr CD
B < 0.36 Fr < 0.4 1.8
B < 0.36 0.4 < Fr < 0.8 2.6 - 2.0Fr
0.36 < B < 0.77 Fr < 1 3.1 - 3.6B
B > 0.77 Fr < 1 1.4 -1.4B

Table 4.2. Drag Coefficient for Debris on Superstructure.
Value of B Value of Fr CD
B < 0.33 Fr < 0.4 1.9
B < 0.33 0.4 < Fr < 0.8 2.8 - 2.25Fr
0.33 < B < 0.77 Fr < 1 3.1 - 3.6B
B > 0.77 Fr < 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.

B equals A sub d divided by (A sub d plus A sub c) (4.2)

where:

B = Blockage ratio

Ad = Cross-sectional flow area blocked by debris in the contracted bridge section, m2 (ft2)

Ac = Unobstructed cross-sectional flow in the contracted section, m2 (ft2)


F sub r equals V sub r divided by square root quantity g times y sub r (4.3)

where:

Fr = Froude number

Vr = Reference velocity, see discussion in Subsection 4.3.3.1, m/s (ft/s)

g = Acceleration of gravity, 9.81 m/s2 (32.2 ft/s2)

yr = 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:

F sub r equals gamma times quantity h sub cu times A sub hu minus h sub cd times A sub hg (4.4)

where:

Fh = Horizontal hydrostatic force on area Ah, N (lbs)

γ = Specific weight of water, N/m3 (lbs/ft3)

hcu = Vertical distance from the upstream water surface to the centroid of area Ahu, m (ft)

Ahu = Area of the vertically projected, submerged portion of the debris accumulation below the upstream water surface, m2 (ft2)

hcd = Vertical distance from the downstream water surface to the centroid of area Ahd, m (ft)

Ahd = Area of the vertically projected, submerged portion of the debris accumulation below the downstream water surface, m2 (ft2)


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:

  1. Debris accumulation of maximum effective width (design log length) forms on Pile Bent 1, with a smaller effective accumulation on Pile Bent 2;
  2. Debris accumulation of maximum effective width forms on Pile Bent 2, with a smaller effective accumulation on Pile Bent 1; and
  3. A large log spans the opening and transfers or divides the load on the accumulation between the piers almost equally to each pier. Although the pressures on the debris accumulation are almost identical for each scenario, the distribution of the total force to each of the piers may be substantially different for each of the scenarios.

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.

4.7 Hydraulic Loading Example Problems

4.7.1 Example 1 - Hydraulic Loading on a Single Pier (SI)

Given:

Design flow rate = 195 m3/s
Minimum upstream main channel width = 13.7 m; design log length of 13.7 m
Depth of debris is full-flow 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, WSUS = 65.43 m (Table 4.3)
Downstream water surface elevation, WSDS = 65.06 m (Table 4.3, see discussion Downstream Water Surface Elevation in Subsection 4.3.3.1)


In this HEC-RAS model (Example 1), the center pier is substituted with the single pier debris accumulation that is wider than the original pier. The units are in metric (SI). The bridge shape is the same for SI Examples 1, 2, and 3. Examples 4, 5, and 6 are the same items, but use customary units (CU) as opposed to SI.
Figure 4.8. Upstream face of the bridge for Example 1.

The water surface profile for SI Example 1 shows a rapid backwater effect. The water surface elevation is still below the low chord, so no pressure flow exists.
Figure 4.9. Water surface profile for Example 1.

River Station (m) Water Surface Elevation (m) Flow Area (m2) Main Channel Velocity (m/s) Cross Section Average Velocity (m/s) Average Flow Depth1 (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

Ahu = Area of the debris accumulation below the upstream water surface
Ahu = (WSUS - Debris bottom, DBEL)(Width of debris accumulation, WD)
Ahu = (65.43 - 61.78)(13.7) = 50.01 m2

Ahd = Area of the debris accumulation below the downstream water surface

Ahd = (WSDS - DBEL)( WD)

Ahd = (65.06 - 61.78)(13.7) = 44.93 m2

hcu = Vertical distance to centroid of Ahu = 0.5(WSUS - DBEL)
hcu = 0.5(65.43 - 61.78) = 1.83 m

hcd = Vertical distance to centroid of Ahd = 0.5(WSDS - DBEL)
hcd = 0.5(65.06 - 61.78) = 1.64 m

Fhu = Hydrostatic force upstream = γhcuAhu
Fhu = (9.81)(1.83)(50.01) = 898 kN

Fhd = Hydrostatic force downstream = γhcdAhd
Fhd = (9.81)(1.64)(44.93) = 723 ken

Fh = Total hydrostatic force on Pile Bent 2 = Fhu - Fhd
Fh = 898 - 723 = 175 kN

Drag Force on Single Pier Accumulation

B equals Blockage ratio equals A sub d divided by quantity A sub d plus A sub c
B equals 50.01 divided by quantity 50.01 plus 67.96 equals 0.42

B is greater than 0.3, therefore Vr should be based on the average velocity in the contracted section.

Vr = 2.87 m/s (Table 4.3)

F sub r equals Froude number equals V sub r divided by square root quantity g times y sub r
F sub r equals 2.87 divided by square root quantity 9.81 times 1.59 equals 0.73

CD = Drag coefficient = 3.1 - 3.6B (Table 4.1)
CD = 3.1 - (3.6)(0.42) = 1.59

F sub D equals Drag force on Pile Bent 2 equals C sub D times gamma times A sub hu times V sub r squared divided by 2g
F sub D equals 1.59 times 9810 times 50.01 times 2.87 squared divided by 2 times 9.81 equals 327 kN

Total Force on Single Pier Accumulation

F = Total segment force = Fh + FD
F = 175 + 327 = 502 kN

Location of Forces on Single Pier Accumulation

F sub hEL equals Elevation of hydrostatic force equals DB sub EL plus (F sub hu times ((WS sub US minus DB sub EL) divided by 3) minus F sub hd times ((WS sub DS minus DB sub EL) divided by 3) divided by F sub h)
F sub hEL equals 61.78 plus (898 times ((65.43 minus 61.78) divided by 3) minus 723 times ((65.06 minus 61.78) divided by 3) divided by 175) equals 63.51 m

FDEL = Elevation of drag force = 0.5(WSUS + DBEL)
FDEL = 0.5(65.43 + 61.78) = 63.61 m

F sub EL equals Elevation of total force equals (F sub D times F sub DEL) plus (F sub h times F sub hEL) divided by F
F sub EL equals quantity (327 times 63.61) plus (175 times 63.51) divided by 502 equals 63.58

FhST = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)

FDST = Station of drag force = FhST

FST = Station of total force = FDST = FhST

FhST = FDST = FST = 0.5(160.33 + 174.03) = 167.18 m

4.7.2 Example 2 - Hydraulic Loading on Two Adjacent Piers, Case 1 (SI)

Given:

Design flow rate = 85 m3/s
Minimum upstream main channel width = 13.7 m; design log length = 13.7 m
Depth of debris is full-flow 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, WSUS = 65.28 m (Table 4.4)
Downstream water surface elevation, WSDS = 64.59 m (Table 4.4, see discussion of Downstream Water Surface Elevation in Subsection 4.3.3.1)

At the upstream face of the bridge in SI Example 2, HEC-RAS shows that debris accumulation has covered the two piers middle piers of a four pier bridge opening. The bridge shape is the same for SI Examples 1, 2, and 3.
Figure 4.10. Upstream face of the bridge for Example 2.

The water surface profile for SI Example 2 shows a rapid and larger backwater effect than SI Example 1. This is reasonable given the larger debris accumulation. The water surface elevation is still below the low chord.
Figure 4.11. Water surface profile for Example 2.

Table 4.4. Results of Hydraulic Calculations for Example 2.
River Station (m) Water Surface Elevation (m) Flow Area (m2) Main Channel Velocity (m/s) Cross Section Average Velocity (m/s) Average Flow Depth1(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)

Ahu = Area of the debris accumulation below the upstream water surface
Ahu = (WSUS - Debris bottom, DBEL)(Width of debris accumulation, WD)
Ahu = (65.30 - 61.98)(13.70) = 45.48 m2 for Pier Bent 2
Ahu = (65.30 - 61.92)(11.70) = 39.55 m2 for Pier Bent 3

Ahd = Area of the debris accumulation below the downstream water surface
Ahd = (WSDS - DBEL)( WD)
Ahd = (64.55- 61.98)(13.70) = 35.21 m2 for Pier Bent 2
Ahd = (64.55- 61.92)(11.70) = 30.77 m2 for Pier Bent 3

hcu = Vertical distance to centroid of Ahu = 0.5(WSUS - DBEL)
hcu = 0.5(65.30 - 61.98) = 1.66 m for Pier Bent 2
hcu = 0.5(65.30 - 61.92) = 1.69 m for Pier Bent 3

hcd = Vertical distance to centroid of Ahd = 0.5(WSDS - DBEL)
hcd = 0.5(64.55 - 61.98) = 1.29 m for Pier Bent 2
hcd = 0.5(64.55 - 61.92) = 1.32 m for Pier Bent 3

Fhu = Hydrostatic force upstream = γhcuAhu
Fhu = (9.81)(1.66)(45.48) = 741 kN for Pier Bent 2
Fhu = (9.81)(1.69)(39.55) = 656 kN for Pier Bent 3

Fhd = Hydrostatic force downstream = γhcdAhd
Fhd = (9.81)(1.29)(35.21) = 446 kN for Pier Bent 2
Fhd = (9.81)(1.32)(30.77) = 399 kN for Pier Bent 3

Fh = Total hydrostatic force on Pile Bent = Fhu - Fhd
Fh = 741 - 446 = 295 kN for Pier Bent 2
Fh = 656 - 399 = 257 kN for Pier Bent 3

Drag Force on Two Adjacent Piers (Case 1)

B equals Blockage ratio equals A sub d divided by quantity A sub d plus A sub c equals (A sub hu Pier2 plus A sub hu Pier 3) divided by  (A sub hu pier 2 plus A sub pier 3 plus A sub c)
B equals (45.48 plus 39.55) divided by (45.48 plus 39.55 plus 27.41) equals 0.76

B is greater than 0.3, therefore Vr should be based on the average velocity in the contracted section.

Vr = 3.10 m/s (Table 4.4)

F sub r equals Froude number equals V sub r divided by square root quantity g times y sub r
F sub r equals 3.10 divided by square root quantity 9.81 times 0.99 equals 0.99

CD = Drag coefficient = 3.1 - 3.6B (Table 4.1)
CD = 3.1 - (3.6)(0.76) = 0.36

F sub D equals Drag force on pile bent equals C sub D times gamma times A sub hu times V sub r squared divided by 2g
F sub D equals 0.36 times 9810 times 45.48 times 3.10 squared divided by 2 times 9.81 equals 79kN for Pile Bent 2
F sub D equals 0.36 times 9810 times 39.55 times 3.10 squared divided by 2 times 9.81 equals 68kN for Pile Bent 3

Total Force on Two Adjacent Piers (Case 1)

F = Total segment force = Fh + FD
F = 295 + 79 = 374 kN for Pile Bent 2
F = 257 + 68 = 325 kN for Pile Bent 3

F sub hEL equals Elevation of hydrostatic force equals DB sub EL plus (F sub hu times ((WS sub us minus DB sub EL) divided by 3)) minus (F sub hd times (WS sub DS minus DB sub EL divided by 3)) divided by F sub h)
F sub HEL equals 61.98 plus (741 times ((65.30 minus 61.98) divided by 3) minus 446 times ((64.55 minus 61.98) divided by 3) divided by 295) equals 63.46 m for Pile Bent 2
F sub hEL equals 61.92 plus (656 times ((65.30 minus 61.92) divided by 3) minus 399 times ((64.55 minus 61.92) divided by 3) divided by 257) equals 63.43 m for Pile Bent 3

FDEL = Elevation of drag force = 0.5(WSUS + DBEL)
FDEL = 0.5(65.30 + 61.98) = 63.64 m for Pile Bent 2
FDEL = 0.5(65.30 + 61.72) = 63.51 m for Pile Bent 3

F sub EL Elevation of total force equals ((F sub D times F sub DEL) plus (F sub h times F sub hEL)) divided by F
F sub EL equals ((79 times 63.64) + (295 times 63.46)) divided by 374 equals 63.50 m for Pile Bent 2
F sub EL equals ((68 times 63.51) plus (257 times 63.43)) divided by 325 equals 63.45 m for Pile Bent 3

FhST = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
FDST = Station of drag force = FhST
FST = Station of total force = FDST = FhST
FhST = FDST = FST = 0.5(154.69 + 168.39) = 161.54 m for Pile Bent 2
FhST = FDST = FST = 0.5(138.39 + 180.09) = 174.24 m for Pile Bent 3

4.7.3 Example 3 - Hydraulic Loading on a Superstructure (SI)

Given:

Design flow rate = 220 m3/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, WSUS = 65.71 m (Table 4.5)
Downstream water surface elevation, WSDS = 65.13 m (Table 4.5, see discussion of Downstream Water Surface Elevation in Subsection 4.3.3.1)

At the upstream face of the bridge in SI Example 3, debris accumulation extends over the entire bridge length. The channel below the bridge is still open below elevation 64.3 meters, indicating the debris does not extend below that depth. Otherwise, the bridge shape is the same for SI Examples 1, 2, and 3.
Figure 4.12. Upstream face of the bridge for Example 3.

The water surface profile for SI Example 3 shows pressure flow and larger backwater effect than SI Examples 1 or 2. This is reasonable given the larger debris accumulation.
Figure 4.13. Water surface profile for Example 3.

Table 4.5. Results of Hydraulic Calculations for Example 3.
River Station (m) Water Surface Elevation (m) Flow Area (m2) Main Channel Velocity (m/s) Cross Section Average Velocity (m/s) Average Flow Depth1 (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

Ahu = Area of the debris accumulation below the upstream water surface
Ahu = (WSUS - Debris bottom, DBEL)(Width of debris accumulation, WD)
Ahu = (65.71 - 64.30)(60.0) = 84.60 m2

Ahd = Area of the debris accumulation below the downstream water surface
Ahd = (WSDS - DBEL)(WD)
Ahd = (65.13 - 64.30)(60.0) = 49.80 m2

hcu = Vertical distance to centroid of Ahu = 0.5(WSUS - DBEL)
hcu = 0.5(65.71 - 64.30) = 0.71 m

hcd = Vertical distance to centroid of Ahd = 0.5(WSDS - DBEL)
hcd = 0.5(65.13 - 64.30) = 0.42 m

Fhu = Hydrostatic force upstream = γhcuAhu
Fhu = (9.81)(0.71)(84.60) = 589 kN

Fhd = Hydrostatic force downstream = γhcdAhd
Fhd = (9.81)(0.42)(49.80) = 205 kN

Fh = Total hydrostatic force on Pile Bent 2 = Fhu - Fhd
Fh = 589 - 205 = 384 kN

Drag Force on Superstructure Accumulation

B equals Blockage ratio equals A sub d divided by quantity A sub d plus A sub c
B equals 84.60 divided by quantity 84.60 plus 75.34 equals 0.53

B is greater than 0.3, therefore Vr should be based on the average velocity in the contracted section.

Vr = 2.92 m/s (Table 4.5)

F sub r equals Froude number equals V sub r divided by square root quantity g times y sub r
F sub r equals 2.92 divided by square root quantity 9.81 times 1.45 equals 0.77

CD = Drag coefficient = 3.1 - 3.6B (Table 4.1)
CD = 3.1 - (3.6)(0.53) = 1.19

F sub D equals Drag force on superstructure equals C sub D gamma times A sub hu times V sub r squared divided by 2g
F sub D equals 1.19 times 9810 times 84.60 times 2.92 squared divided by 2 times 9.81 equals 429 kN

Total Force on Superstructure Accumulation

F = Total segment force = Fh + FD
F = 384 + 429 = 813 kN

Location of Forces on Superstructure Accumulation

F sub hEL equals Elevation of hydrostatic force equals DB sub EL plus ((F sub hu times ((WS sub US minus DB sub EL) divided by 3) minus F sub hd times ((WS sub DS minus DB sub EL) divided by 3)) divided by F sub h)

F sub hEL equals 64.30 plus (589 times ((65.71 minus 64.30) divided by 3) minus 205 times ((65.13 minus 64.30) divided by 3) divided by 384) equals 64.87m

FDEL = Elevation of drag force = 0.5(WSUS + DBEL)
FDEL = 0.5(65.71 + 64.30) = 65.00 m

F sub EL Elevation of Total force equals ((F sub D times F sub DEL) plus (F sub h times F sub hEL)) divided by F
F sub EL equals ((429 times 65.00) plus (384 times 64.87)) divided by 813 equals 64.94 m

FhST = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
FDST = Station of drag force = FhST
FST = Station of total force = FDST = FhST
FhST = FDST = FST = 0.5(137.16 + 197.21) = 167.19 m

4.7.4 Example 4 - Hydraulic Loading on a Single Pier (CU)

Given:

Design flow rate = 6,890 ft3/s
Minimum upstream main channel width = 45 ft; design log length of 45 ft
Depth of debris is full-flow 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, WSUS = 214.66 ft (Table 4.6)
Downstream water surface elevation, WSDS = 213.44 ft (Table 4.6, see discussion of Downstream Water Surface Elevation in Subsection 4.3.3.1)

In this HEC-RAS model (Example 4), the center pier is substituted with the single pier debris accumulation that is wider than the original pier. The units are in English or customary units (CU). The bridge shape is the same for CU Examples 4, 5, and 6. Examples 1, 2, and 3 are the same items, but use metric units (SI) as opposed to CU.
Figure 4.14. Upstream face of the bridge for Example 4.

The water surface profile for CU Example 4 shows a rapid backwater effect. The water surface elevation is still below the low chord, so no pressure flow exists.
Figure 4.15. Water surface profile for Example 4.

Table 4.6. Results of Hydraulic Calculations for Example 4.
River Station (miles) Water Surface Elevation (ft) Flow Area (ft2) Main Channel Velocity (ft/s) Cross Section Average Velocity (ft/s) Average Flow Depth1 (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

Ahu = Area of the debris accumulation below the upstream water surface
Ahu = (WSUS - Debris bottom, DBEL)(Width of debris accumulation, WD)
Ahu = (214.66 - 202.69)(45) = 538.65 ft2

Ahd = Area of the debris accumulation below the downstream water surface
Ahd = (WSDS - DBEL)(WD)
Ahd = (213.44 - 202.69)(45) = 483.75 ft2

hcu = Vertical distance to centroid of Ahu = 0.5(WSUS - DBEL)
hcu = 0.5(214.66 - 202.69) = 5.98 ft

hcd = Vertical distance to centroid of Ahd = 0.5(WSDS - DBEL)
hcd = 0.5(213.44 - 202.69) = 5.38 ft

Fhu = Hydrostatic force upstream = γhcuAhu
Fhu = (62.4)(5.98)(538.65) = 200,998 lbs (100.5 tons)

Fhd = Hydrostatic force upstream = γhcdAhd
Fhd = (62.4)(5.38)(483.75) = 162,401 lbs (81.2 tons)

Fh = Total hydrostatic force on Pile Bent 2 = Fhu - Fhd
Fh = 200,998 - 162,401 = 38,597 lbs (19.3 tons)

Drag Force on Single Pier Accumulation

B equals Blockage ratio equals A sub d  divided by quantity A sub d plus A sub c
B equals 538.65 divided by quantity 538.65 plus 731.66 equals 0.42

B is greater than 0.3, therefore Vr should be based on the average velocity in the contracted section.

Vr = 9.42 ft/s (Table 4.6)

F sub r equals Froude number equals V sub r divided by square root quantity g times y sub r
F sub r equals 9.42 divided by square root quantity 32.2 times 5.21

CD = Drag coefficient = 3.1 - 3.6B (Table 4.1)
CD = 3.1 - (3.6)(0.42) = 1.59

F sub D equals Drag force on pile bent 2 equals C sub D times gamma times A sub hu times V sub r squared divided by 2g
F sub D equals 1.59 times 62.4 times 538.65 times 9.42 squared divided quantity 2 times 32.2 equals 73,638 lbs (36.8 tons)

Total Force on Single Pier Accumulation

F = Total segment force = Fh + FD
F = 38,597 + 73,638 = 112,235 lbs (56.1 tons)

Location of Forces on Single Pier Accumulation

F sub hEL equals Elevation of hydrostatic force equals DB sub EL plus (F sub hu times ((WS sub US minus DB sub EL) divided by 3) minus F sub hd times ((WS sub DS minus DB sub EL) divided by 3) divided by F sub h)
F sub hEL equals 202.69 plus (100.5 times (214.66 minus 202.69) divided by 3) minus 81.2 times (213.44 minus 202.69) divided by 3 divided by 19.3) equals 208.39 ft

FDEL = Elevation of drag force = 0.5(WSUS + DBEL)
FDEL = 0.5(214.66 + 202.69) = 208.68 ft

F sub EL equals Elevation of total force equals ((F sub D times F sub DEL) plus (F sub h times F sub hEL)) divided by F
F sub EL equals ((36.8 times 208.68) plus (19.3 times 208.39)) divided by 56.1 equals 208.58 ft

FhST = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
FDST = Station of drag force = FhST
FST = Station of total force = FDST = FhST
FhST = FDST = FST = 0.5(526.02 + 570.97) = 548.50 ft

4.7.5 Example 5 - Hydraulic Loading on Two Adjacent Piers, Case 1 (CU)

Given:

Design flow rate = 3,000 ft3/s
Minimum upstream main channel width = 45 ft; design log length = 45 ft
Depth of debris is full-flow 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, WSUS = 214.17 ft (Table 4.7)
Downstream water surface elevation, WSDS = 211.89 ft (Table 4.7, see discussion of Downstream Water Surface Elevation in Subsection 4.3.3.1)

At the upstream face of the bridge in CU Example 5, HEC-RAS shows that debris accumulation has covered the two piers middle piers of a four pier bridge opening. The bridge shape is the same for CU Examples 4, 5, and 6.
Figure 4.16. Upstream face of the bridge for Example 5.

The water surface profile for CU Example 5 shows a rapid and larger backwater effect than CU Example 4. This is reasonable given the larger debris accumulation. The water surface elevation is still below the low chord.
Figure 4.17. Water surface profile for Example 5.

Table 4.7. Results of Hydraulic Calculations for Example 5.
River Station (miles) Water Surface Elevation (ft) Flow Area (ft2) Main Channel Velocity (ft/s) Cross Section Average Velocity (ft/s) Average Flow Depth1 (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)

Ahu = Area of the debris accumulation below the upstream water surface
Ahu = (WSUS - Debris bottom, DBEL)(Width of debris accumulation, WD)
Ahu = (214.25 - 203.35)(45) = 490.50 ft2 for Pile Bent 2
Ahu = (214.25 - 203.15)(38.4) = 426.24 ft2 for Pile Bent 3

Ahd = Area of the debris accumulation below the downstream water surface
Ahd = (WSDS - DBEL)(WD)
Ahd = (211.82 - 203.35)(45) = 381.15 ft2 for Pile Bent 2
Ahd = (211.82 - 203.15)(38.4) = 332.93 ft2 for Pile Bent 3

hcu = Vertical distance to centroid of Ahu = 0.5(WSUS - DBEL)
hcu = 0.5(214.25 - 203.35) = 5.45 ft for Pile Bent 2
hcu = 0.5(214.25 - 203.15) = 5.55 ft for Pile Bent 3

hcd = Vertical distance to centroid of Ahd = 0.5(WSDS - DBEL)
hcd = 0.5(211.82 - 203.35) = 4.24 ft for Pile Bent 2
hcd = 0.5(211.82 - 203.15) = 4.34 ft for Pile Bent 3

Fhu = Hydrostatic force upstream = γhcuAhu
Fhu = (62.4)(5.45)(490.50) = 166,809 lbs (83.4 tons) for Pile Bent 2
Fhu = (62.4)(5.55)(426.24) = 147,615 lbs (73.8 tons) for Pile Bent 3

Fhd = Hydrostatic force upstream = γhcdAhd
Fhd = (62.4)(4.24)(381.15) = 100,843 lbs (50.4 tons) for Pile Bent 2
Fhd = (62.4)(4.34)(332.93) = 90,163 lbs (45.1 tons) for Pile Bent 3

Fh = Total hydrostatic force on Pile Bent 2 = Fhu - Fhd
Fh = 166,809 - 100,843 = 65,966 lbs (33.0 tons) for Pile Bent 2
Fh = 147,615 - 90,163 = 57,452 lbs (28.7 ton) for Pile Bent 3

Drag Force on Two Adjacent Piers (Case 1)

B equals Blockage Ratio equals A sub d divided by quantity A sub d plus A sub c plus A sub c equals (A sub hu pier 2 plus A sub hu pier 3) divided by (A sub hu pier 2 plus A hu pier 3 plus A sub c)
B equals (490.50 plus 426.24) divided by (490.50 plus 426.24 plus 295.07) equals 0.76

B is greater than 0.3, therefore Vr should be based on the average velocity in the contracted section.

Vr = 10.17 ft/s (Table 4.7)

F sub r equals Froude number equals V sub r divided by squared root quantity g times y sub r
F sub r equals 10.17 divided by square root quantity 32.2 times 3.25 equals 0.99

CD = Drag coefficient = 3.1 - 3.6B (Table 4.1)
CD = 3.1 - (3.6)(0.76) = 0.36

F sub D equals Drag force on pile bent  equals C sub D gamma A sub hu V sub r squared divided by 2g
F sub D equals 0.36 times 62.4 times 490.50 times 10.17 squared divided by quantity 2 times 32.2 equals 19,589 lbs (9.8 tons) for Pile Bent 2
F sub D equals 0.36 times 62.4 times 426.24 times 10.17 squared divided by quantity 2 times 32.2 equals 17,023 lbs (8.5 tons) for Pile Bent 3

Total Force on Two Adjacent Piers (Case 1)

F = Total segment force = Fh + FD
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 3

Location of Forces on Single Pier Accumulation

F sub hEL equals Elevation of hydrostatic force equals DB sub EL plus ((F sub hu times (WS sub US minus DB sub EL) divided by 3) minus (F sub hd times (WS sub DS minus DB sub EL) divided by 3) divided by F sub h)
F sub hEL equals 203.35 plus (((83.40 times (214.25 minus 203.35) divided by 3) minus (50.4 times (211.82 minus 203.35) divided by 3)) divided by 33.0) equals 208.22 ft for Pile bent 2
F sub hEL equals 203.15 plus (((73.8 times (214.25 minus 203.35) divided by 3) minus (45.10 times (211.82 minus 203.15) divided by 3)) divided by 28.7) equals 208.12 ft for Pile bent 3

FDEL = Elevation of drag force = 0.5(WSELUS + DBEL)

FDEL = 0.5(214.25 + 203.35) = 208.80 ft for Pile Bent 2
FDEL = 0.5(214.25 + 203.15) = 208.70 ft for Pile Bent 3

F sub EL equals Elevation of total force equals ((F sub D times F sub EDL) plus (F sub h times F sub hEL)) divided by F
F sub EL equals ((9.8 times 208.80) plus (33.0 times 208.22)) divided by 42.8 equals 208.35 ft for Pile Bent 2
F sub EL equals ((8.5 times 208.70) plus (28.7 times 208.12)) divided by 37.2 equals 208.25 ft for Pile Bent 3

FhST = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
FDST = Station of drag force = FhST
FST = Station of total force = FDST = FhST
FhST = FDST = FST = 0.5(507.53 + 552.48) = 530.00 ft for Pile Bent 2
FhST = FDST = FST = 0.5(552.48 + 590.87) = 571.67 ft for Pile Bent 3

4.7.6 Example 6 - Hydraulic Loading on a Superstructure (CU)

Given:

Design flow rate = 7,770 ft3/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, WSUS = 215.59 ft (Table 4.8)
Downstream water surface elevation, WSDS = 213.69 ft (Table 4.8, see discussion of Downstream Water Surface Elevation in Subsection 4.3.3.1)

t the upstream face of the bridge in CU Example 6, debris accumulation extends over the entire bridge length. The channel below the bridge is still open below elevation 210.97 feet, indicating the debris does not extend below that depth. Otherwise, the bridge shape is the same for CU Examples 4, 5, and 6.
Figure 4.18. Upstream face of the bridge for Example 6.

The water surface profile for CU Example 6 shows pressure flow and larger backwater effect than CU Examples 4 or 5. This is reasonable given the larger debris accumulation.
Figure 4.19. Water surface profile for Example 6.

Table 4.8. Results of Hydraulic Calculations for Example 6.
River Station (miles) Water Surface Elevation (ft) Flow Area (ft2) Main Channel Velocity (ft/s) Cross Section Average Velocity (ft/s) Average Flow Depth Depth1 (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

Ahu = Area of the debris accumulation below the upstream water surface
Ahu = (WSUS - Debris bottom, DBEL)(Width of debris accumulation, WD)
Ahu = (215.59 - 210.97)(197.0) = 910.14 ft2

Ahd = Area of the debris accumulation below the downstream water surface
Ahd = (WSDS - DBEL)(WD)
Ahd = (213.69 - 210.97)(197.0) = 535.84 ft2

hcu = Vertical distance to centroid of Ahu = 0.5(WSUS - DBEL)
hcu = 0.5(215.59 - 210.97) = 2.31 ft

hcd = Vertical distance to centroid of Ahd = 0.5(WSDS - DBEL)
hcd = 0.5(213.69 - 210.97) = 1.36 ft

Fhu = Hydrostatic force upstream = γhcuAhu
Fhu = (62.4)(2.31)(910.14) = 131,191 lbs (65.6 tons)

Fhd = Hydrostatic force upstream = γhcdAhd
Fhd = (62.4)(1.36)(535.84) = 45,474 lbs (22.7 tons)

Fh = Total hydrostatic force on Pile Bent 2 = Fhu - Fhd
Fh = 131,191 - 45,474 = 85,717 lbs (42.9 tons)

Drag Force on Superstructure Accumulation

B equals Blockage Ratio equals A sub d divided by quantity A sub d plus A sub c plus A sub c
B equals 910.40 divided by quantity 910.40 plus 811.03 equals 0.53

B is greater than 0.3, therefore Vr should be based on the average velocity in the contracted section.

Vr = 9.58 ft/s (Table 4.8)

F sub r equals Froude number equals V sub r divided by squared root quantity g times y sub r
F sub r equals 9.58 divided by square root quantity 32.2 times 4.76 equals 0.77

CD = Drag coefficient = 3.1 - 3.6B (Table 4.1)
CD = 3.1 - (3.6)(0.53) = 1.19

F sub D equals Drag force on superstructure equals C sub D times gamma times A sub hu times V sub r squared divided by 2g
F sub D equals 1.19 times 62.4 times 910.14 times 9.58 squared divided by quantity 2 times 32.2 equals 96,313 lbs (48.1 tons)

Total Force on Superstructure Accumulation

F = Total segment force = Fh + FD
F = 85,717 + 96,313 = 182,030 lbs (91.0 tons)

Location of Forces on Superstructure Accumulation

F sub hEL equals Elevation of hydrostatic force equals DB sub EL plus (((F sub hu times (WS sub US minus DB sub EL) divided by 3) minus (F sub hd times (WS sub DS minus DB sub EL) divided by 3)) divided by F sub h)
F sub hEL equals 210.97 plus (((65.6 times (215.59 minus 210.97) divided by 3) minus (22.07 times (213.69 minus 210.97) divided by 3)) divided by 42.9) equals 212.85 ft

FDEL = Elevation of drag force = 0.5(WSUS + DBEL)
FDEL = 0.5(215.59 + 210.97) = 213.28 ft

F sub EL equals Elevation of total force equals ((F sub D times F sub EDL) plus (F sub h times F sub hEL)) divided by F
F sub EL equals ((48.1 times 213.28) plus (42.9 times 212.85)) divided by 91.0 equals 213.08 ft

FhST = Station of hydrostatic force = 0.5(Left station of debris + right station of debris)
FDST = Station of drag force = FhST
FST = Station of total force = FDST = FhST
FhST = FDST = FST = 0.5(450.00 + 647.01) = 548.51 ft

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Updated: 04/07/2011
 

FHWA
United States Department of Transportation - Federal Highway Administration