Calculation of forces on immersed surfaces

There are many surfaces that have to be designed and built by engineers to achieve purposes that require the surface to be submerged. These surfaces are frequently submerged in liquid that is either at rest or effectively at rest and, almost without exception, they are made to simple geometrical shapes. As the forces exerted on such surfaces may have to be calculated for design purposes this makes calculation relatively easy.


Surfaces that work when submerged, such as dams, sluices, weirs, flow control gates, lock gates and entraining walls can be seen in rivers and canals. Examples of submerged surfaces that are not so obvious are tanks holding water or chemicals, ships, (which are inside out tanks), shuttering for shaping concrete and moulds for casting metal.


These surfaces do not share a mode of construction. The plates of a ship or a water tank have to be supported at intervals both horizontally and vertically and deeply submerged plates need stronger support than similar plates at a smaller depth. Ships and water tanks need internal support structures. In order to design such structures we need to be able to calculate the force exerted on a part of the surface and we may need to know both the magnitude and direction of the resultant force on this part of the surface. By contrast, in civil engineering, water is ducted and stored using surfaces that have to be anchored into the ground. Indeed a dam that may be made by dumping earth and rock across a river and then sealing the water face has no formal structure as such. Clearly we need methods to calculate forces that are exerted on submerged surfaces but which forces we need to calculate depends on the application.


We need some basic method for calculating these forces and then find ways to apply it sensibly. The method is simple but the application is part of the skill that an engineer brings to design and is best learnt by example. For the purpose of examination the question setter will effectively define the application and expect the examinee to use whatever methods appear to be appropriate. All sorts of methods have evolved, mainly it must be said, to avoid integration but now software like Mathcad make this unnecessary in practice. Examiners still have to devise examination questions for students who may not in examination have access to computers. In this text I will use Mathcad and concentrate on the basic methods.


The method can be illustrated in stages as follows. Usually this involves dividing the surface into elemental strips, finding elemental forces and then integrating. Some thought is needed to find a suitable way of subdividing but the process is usually straightforward.


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Figure 2-11
Figure 2-11 shows a vertical surface that is rectangular with one edge parallel to the free surface. At a depth h below the free surface the pressure p is given by , where r is the density of the liquid and, if we take an elemental strip of the surface of area , all at this depth h it will have a force on it . This elemental force acts at the mid-point of the length  and at right angles to the surface. The surface can be regarded as being made up of many such strips and we can sum the elemental forces to give the total force exerted on the surface or sum the moments of such forces about some point to locate the resultant force or make any other summation that suits our purpose


For the surface in figure 2-11 the force on the element  so the total force exerted on the surface is given by:-               


For this surface this can be integrated[1] to give .


Putting some values to the dimensions and to  for water Mathcad can be used to integrate and give the force in Newtons.[2]

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Any force that acts over an area in this way might usefully be called a distributed force. We have found its value and there could be some advantage in knowing where a single concentrated force equal to the distributed force[3] would act. The depth at which this concentrated force acts can be found by taking moments of the elemental forces about the free surface and dividing by . The moment of the elemental force =  and the total moment M about the free surface is given by:-  which equals  Dividing the moment by the force gives the depth of the point of action of the total force =2/3.H.


This could be evaluated for the dimensions above by using Mathcad again. It gives:-

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Note that the numerical answer agrees with the outcome of the mathematics.[4]


These simple calculations are valuable in the design of tanks for holding liquid and the inverse for tanks used as floating vessels, both of which must be stiffened. It is also useful in dealing with other surfaces which are immersed in a liquid at rest. We can also note that the force F can be written as = b.H.(r.g.H/2) which can be interpreted as the area multiplied by the pressure at the depth of the centre of area. Whilst this is of no special value here, it turns out to be a general statement for all plane surfaces. As areas and the position of centres of area are known for many geometrical plane shapes the expression can be useful.


We have looked at the most simple surface that we are likely to use in engineering. As we have seen it might form part of a tank. Now we must look at sloping or curved surfaces because the weight of the water that is supported by the surface is used as an aid to holding the surface in place. By way of exercise one might choose to calculate any one of the forces that might be regarded as acting on the face but the important ones to know are the horizontal and vertical forces that act on the surface. Let us look at the surface shown in figure 2-13 that could be a length b of one wall of a trapezoidal channel. The wall is inclined at q  to the horizontal.


An elemental area, all at depth h, is shown in figure 2-13. For convenience the vertical height of the element is dh although the slant height is greater than this. The horizontal force on the element is given by :-F = å r.g.h.b.dh as before  this can be integrated to give the same answer.


In figure 2-14 The triangle of forces for the element shows that the vertical force on the elemental area is equal to the horizontal force/tanq  which can also be evaluated using say Mathcad.


This force will be equal to the gravitational force on the water that is vertically above it. This can be found quite easily by calculating the volume and multiplying by .








Figure 2-15 Shows a curved surface replacing the plane wall.


This surface could be produced by moving a part of a curve such as y = a.xn along a horizontal line. We have to describe the curve in some way. Let the upper point  have co-ordinates X and E relative to the vertex of the curve. Given this we can find a value for the constant a for any value of n. It will be . Now we have a surface where the distributed force is made up of lots of elemental forces all acting in different directions. This complicates the finding of the forces on the surface. The magnitude and position of the resultant force on the whole face could be found as in figure 2-15 provided that the volume and the position of the centre of mass of the water above the curved face is known. However this is unlikely and will not help if the force on a part of the surface is required. It is more likely that we would need to find the vertical and horizontal forces on the surface. We have to go back to the elemental areas.


An elemental area all at depth h and therefore subject to the same pressure is shown in figure 2-15 and in end elevation in figure 2-16 The plan area of this element is b.dx and the pressure on it is  Then the horizontal force on the elemental strip s given by  We have to decide how to calculate. If we choose to use a computer package we can easily make it general.


In frame 2-17,  which is a copy of a Mathcad calculation, I have put values to the various lengths and constants so that they can be changed easily. Then the first calculation gives the horizontal force in Newtons.


In order to find the vertical force on the curved face we must find the projected area of the elemental strip. It is. . Then the vertical force on the elemental strip is:-


This is evaluated in the second calculation also in Newtons. It is a very simple extension of this computation to give a graph of the curved face and to show the water level. This has been added.


The system can then be explored for any depth, any value of n and for any values of the other dimensions. If this were to be a really significant application the ability to do this would be very powerful.


When looking at examples of submerged surfaces it is sometimes important to realise that they have been designed to withstand collision rather than water forces. When looking at control works in rivers it is useful to have observed that the resultant force on any part of a submerged cylinder passes through the axis of the cylinder. This simple fact permits a rotating drum gate to be mounted on a shaft so that the force exerted by the water has no moment about the shaft.


[1]  [1] This brings us to a serious problem. Most students of engineering will be expected to be able to integrate and to do so in examination. Practising engineers will be able to use computer programs to differentiate or integrate without any fuss. Such programs make the manipulation of differential equations so simple as to be no longer a problem. Add to this the facility to draw graphs and explore mathematical models almost instantly and it becomes difficult to justify spending time on learning how to differentiate or integrate. These programs liberate the engineer from the burden of mathematics.

[2] Mathcad can work out units but it thinks that a force is in units of kg.m/sec2 and not Newtons.

[3] The position at which this concentrated force acts is called the centre of pressure or centre of effort.

[4] This idea of putting a distributed force equal to a single force acting at a point is used extensively in aerodynamics where it becomes the centre of pressure and in sailing where it is the centre of effort.