A synopsis of a paper in the Proceedings of the 1996 EME Conference in August 15--18, Baltimore, MD


There is a systematic approach for sizing the concrete piers sufficient to prevent the overturning of towers in high winds. The force of the wind F on the antenna produces an overturning moment Fh using the (effective) leverarm of the tower height h. This moment induces an upward pullout force Pon the upwind piers(s).

For the tower to remain upright, the overturning moment Fh must be exceeded by a resisting moment Rw at the upwind leg(s):

Fh < Rw (1)

where w is the width of the tower and where R is the force of resistance of the soil to shearing at the concrete-soil interface plus the weight of the pier(s).


Soil shear strength tau (the stress at which the soil will fail by shearing) is given for each infinitesimal horizontal slice by the Mohr-Coulomb model

tau = c + sigma tan(phi) (2)

where c is soil cohesion, where sigma is the normal stress exerted by the surrounding soil, and where phi is the angle of shearing resistance. Each of the three quantities tau, c, and sigma are given in units of pressure (lb/ft^2). See Tables 1 and 2.

Let A be the area of the side(s) of the concrete pier(s) that will besubjected to shear --- not the bottom or the top. Then the shear resistance of the soil will be

R = A {tau} = Ac + A{sigma} tan(phi), (3)

where {.} represents average value.

Our design task is to select the effective area A that will yield a resistingmoment Rw to exceed the overturning moment Fh.

The stress {sigma} is the average pressure on the sides of the pier exerted by the weight of the surrounding soil.

A table of values for c and phi are given in Tables 1 and 2 for clay and sand.


Suppose a four-legged tower of base width w = 4 ft supports a tightly meshed 12 ft diameter parabolic dish whose center is h = 10 feet above ground. Assume worst case: the dish is looking at the horizon and at a 70 MPH wind. The standard assumed wind pressure for 70.7 MPH is 20 lb/ft^2. Thus

Fh = 20 \pi x 6^2 x 10 = 22 620 ft-lb. (4)

To counterbalance this overturning moment the soil shear resistance R must satisfy before failure

R > Fh/w = 22620/4 = 5 655 lb. (5)

Assuming public safety is not at risk, we take a small safety factor ofsay f = 1.5, giving a design resistance of

R = f Fh/w = 1.5 x 5655 = 8 483 lb. (6)

The piers are to be cylindrical of diameter d with L = 3 ft of concrete below the organic topsoil. This diameter d is determined by the soil type.

Case 1. Clay

Suppose the upwind piers are in medium clay. From Table 1 we see c = 1500lb/ft^2 and phi = 0. Therefore the required area at failure is

A = R/{tau} = R/(c + 0) = 8483/750 = 11.3 ft^2. (7)

This resistance is shared by the two legs and thus each pier must be ofdiameter only

d = (A/2)/(L pi) = (11.3/2)/(3 pi) = 0.6 ft (8)

illustrating the enormous shear strength of clay. (For a triangular towerthe single upwind pier must supply all the resistance, and so in contrast to (8), d = A/(L pi).)

The two concrete piers (of density delta0 = 150 lb/ft^3) weigh

Rpiers = 2 pi (d/2)^2 L delta0 = 254 lb (9)

which is an additional resistance of 3% of the required (6). The tower and dish weigh (say) 1000 lbs, half of which contributes an additional 6% to the required resistance (6).

On the other hand the tower itself provides windload proportional to half of the square of the height h times the cross sectional area per unit height. Let us say the tower is built from 2x 3 x 1/4 inch angle iron giving an additional overturning moment of

20 x 0.1667 x 10^2/2 = 167 ft-lb (10)

per leg, together amounting to only 3% of the moment (4) due to the dish.

Case 2. Sand

Suppose the upwind piers are in dense dry sand. From Table 2 we take phi= 40 deg. One assumes that sand has negligible cohesion, i.e., that c =0. The density delta of this sand from Table 3 is 109 lb/ft^3. Therefore the average soil induced stress on the sides of the piers is

{sigma} = delta L/2 = 109 x 3/2 = 163 lb/ft^2. (11) Thus the required area is

A = R{sigma} tan(phi) = 8483/137 = 62 ft^2. (12)

This resistance is shared by the two legs and thus each pier must be ofdiameter at least

d = (A/2)/(L pi) = (62/2)/(3 pi) = 3.3 ft . (13)

The two concrete piers (of density delta0 = 150 lb/ft^3) weigh

R_piers = 2 pi (d/2)^2 L delta0 = 7 698 lb, (14)

yielding an additional resistance of 91% of the required (6). We have oversized.

Perform a second iteration by returning to equation (12). Choose a smaller A, (say A = 45), recompute R from R = A{sigma}tan(phi), then d, then Rpiers, repeating until

R + Rpiers \approx 8483. (15)

Case 3. Mixed soil

Suppose the upwind legs are in mixed soil. Read the horizontal lines ofFigure 4 to find the percentage of clay versus sand.

Rule 1. If the percentage of clay exceeds 40%, design as if the soil wereall clay as in Case 1.

Rule 2. If the percentage of clay is less than 12%, design as if the soilwere all sand as in Case 2.

Rule 3. If the percentage of clay lies between 12% and 40%, linearly interpolatec and phi.

For example, suppose the soil is 70% dense dry sand, 30% medium clay. From Table 1, c = 1500, and so interpolating between 12% and 40% gives

c' = 750x(30 - 12)/(40-12)= 482 lb/ft^2. (16)

Consulting Table 2, we interpolate phi = 42 deg to

phi' = 42x(40 - 30)/(40 - 12) = 15 deg. (17)

From Table 3 we choose the interpolated sand density

delta = 0.7x109 + 0.30x106 = 108 lb/ft^3 (18)

giving the average soil induced stress of

sigma' = delta l/2= 108 x 3/2 = 162 lb/ft^2. (19)

Therefore the required area is

A = R/(c' + sigma' tan(phi'))= 8483/(483 + 43) = 16.2 ft^2. (20)

Proceed as before for sizing.

Case 4. Layered soil

Sum up the resistance from each layer. For the average soil induced stress {sigma} of a layer of sand, use the average weight of the soil above this layer.


Any realistic design must incorporate a safety factor f of from 1.5 to 4 in the value of R in (6) depending on your soil knowledge and risk to public safety. Careful design would necessitate an on-site shear vane test in clay soil or a standard penetration test in sand.

For amateur application, we may assume that the dish is placed in theupright `birdbath' position during heavy wind. This would significantly reducethe effective wind load, yielding an additional safety factor of perhaps 6.


J.E. Bowles, Foundation Analysis and Design, 4th ed., McGraw-Hill, New York, 1988.

R.B. Peck, W.E. Hanson, and T.H. Thornburn, Foundation Engineering, 2nd. ed., J. Wiley and sons, New York, 1973.

TABLE 1. (from Peck et al.) Cohesive soils (clay)

Use \tau = c (assumes \phi = 0.) Consistency Field Identification

Note: adhesion bonding of clay to pipe is often stronger than the cohesivebond but strength decreases with time under constant stress.

TABLE 2. (from Bowles) Angle phi of internal friction (in degrees)

* 14--20 if consolidated and undrained.

TABLE 3. (from Peck et al.) Soil density