Strength of Materials and Structure

In this section we will combine several subjects covered in previous modules with an emphasis on their applications to the cistern. The cistern must serve at least one function, to hold water. In many cases it also serves as a foundation for at least a part of the house.

Pressure and the Cistern

As you know from the plumbing module, water pressure increases linearly with depth. The pressure is greatest at the bottom of a full cistern. The picture to the left is of a water tower in Modesto, CA. It is a wooden tower held together by steel rings. Note how the rings are spaced closer together near the bottom. Since the pressure increases linearly with depth, the number of steel rings per unit height also increases linearly. The next time you pass along the road from upper campus to the Physical Plant, look at the large cisterns next to the road. The support buttresses are wider at the bottom than at the top for exactly the same reason.

For a typical cistern for a house, it is not worth the increase in time and labor to pour a non-uniform wall. Instead, additional steel rebar is used. We can estimate what is needed. For a modest cistern with 2 meters of water, the gauge pressure at the bottom is

The ultimate compressive strength for concrete is about 20 MPa. The standard reinforced 4" concrete wall can easily withstand the compressive pressures, provided it is supported. If the cistern is buried, then the surrounding soil may be able to provide the support to the outside of the walls. But if, as is often the case, one or more of the walls is free standing, then the outward pressure on the walls will cause them to bulge and pull away from each other. This is flexural stress, just like that you studied in an earlier lab. Recall that flexural stress is a combination of compression on the inner concave surface and tension on the outer convex surface. Since concrete is very weak in ultimate tensile strength (only about 2 MPa), a concrete wall cannot bulge or flex very much before it cracks. While that may not be catastrophic for house walls, it is very undesireable for cistern walls.

A properly constructed cistern will have angled steel rebar around the corners protruding into each wall. Additional rebar is needed at the bottom where the pressure is greatest. (In pracrtice, many contractors simply install additional rebar throughout the cistern walls.) Just how important is it to have rebar in the walls? Let's estimate how far a standard 4" concrete wall without rebar would flex when full of water. Since this is simply an estimate, let's use the same equation we used in the lab for the simple beam. It was

This equation is appropriate for a point load at the center of a the beam supported only on two ends. The flexure Dy is at the center of the beam. Recall that L is length between the support points. A cistern wall is supported on all four sides and the total force is spread out. So, it is not clear which dimension is L and which is w for this case. Since we are calculating only an estimate, we'll try both and look at the range of values.

Let's assume a typical house in which one wall of the cistern runs the length of the house, about 10 m and is 2 meters high. The thickness (t) of a 4" wall is about 10 cm. Taking L= 10 m and w = 2 m will give the largest flexure. To calculate the load F, we use the average gauge pressure on the wall times the area, F = P_aveA. Since the gauge pressure is zero at the top and 19.6 kPa at the bottom, we'll use P_ave = ¹/₂ x 19.6 kPa = 9.8 kPa. The total force will be 1.96 x 10⁵ N. Substituting into the deflection equation above,

Of course, this is an absurd answer. The concrete wall would fail long before flexing a distance of 1.2 meters at the center. What if we assigned the 2 m length to L and 10 m to w? The equation above would yield a central flexure of 1.96 mm. The actual flexure would probably fall somewhere between these values. While a 2 mm flexure might be acceptable, it represents the lowest possible value. In addition to the force exerted by the water, the cistern walls are also subjected to forces from the house structure above it. Clearly, structural reinforcement with rebar is needed.

Compressive Forces on the Cistern

In most cases, the cistern also supports the part of the house directly over it. What kind of compressive forces can we expect? Consider a house constructed of concrete floors, walls and roof. Let's estimate the weight of the house structure using typical dimensions, layout, and the density of concrete.

A common small home consists of two bedrooms, a bathroom, and a living/dining room. Assume the house is 10 m by 10 m with a 2.4 m (8 ft) high ceiling. The floor plan shows four inner walls that form the bedrooms and bathroom. This will be the heavier portion of the house and require a stronger foundation to carry the weight. But we are interested in an estimate, so we'll calculate the total weight of the house and consider the case where the cistern walls must carry at least part of the total. This is common for houses built on a steep hillside. The cistern provides the foundation on one side and pillars on the other.

Concrete has an average density of 2,300 kg/m³, but a range of 1,700 to 2,500 is common. Let's assume all walls, the floor and the ceiling are 10 cm thick (standard 4" walls). What is the total volume of concrete in the house? Use the dimensions given in the diagram and see if you can obtain a value of about 40 m³. The total weight of the house would be w = mg = rVg = 2,300 x 40 x 9.8 ~ 9 x 10⁵N.

The ultimate compressive strength of concrete is about 20 MPa. The minimum total area required to support 9 x 10⁵N of weight would be 9 x 10⁵N / 20 x 10⁶N/m² = .0045 m². The total cross sectional area of only one of the cistern walls would be 10 m x .1 m = 1.0 m². The conclusion is clear that the cistern walls should be able to handle the compressive stress from the weight of the house.

Compressive Stress Become Flexural But there is one more consideration of which engineers are well aware. While the 4" walls can easily withstand the static compressive forces from the weight of the house structure, the situation changes drastically during earthquakes ... or even during minor tremors. The walls will flex during these vibrational events and the vertical compressive load quickly becomes a flexural load. The effect is easily understood by a simple experiment. Place a meter stick vertically with one end on the floor. Push vertically downward on the top end with your weight, being careful that the stick does not flex. You will find that it's very difficult to do. Once the stick flexes even slightly, the vertical force is no longer directly along the length and it can create a tremendous flexural stress. A friend can supply a very small horizontal force that will easily keep the stick straight. But once that stabilizing horizontal force is removed, flexural stress is very difficult to avoid. A building designed to be acceptably safe under a static load may not be able to withstand flexural stress resulting from even normal vibrational conditions.