Formulas are simply short methods of indicating operations otherwise expressed by rules, by using letters and signs in place of words. The letters are usually those of the English alphabet, and the signs are those previously given. Besides showing at a glance the various steps, formulas are much more convenient than rules to memorize. Many people are unnecessarily deterred from using a formula, because a few letters are used instead of many words; but a formula can really be more readily followed than a rule. We shall confine our attention to explanations of some simple formulas, and to showing how a formula may be transposed so as to obtain any required term.

To show the similarity between a rule and a formula, let it be required to find the area of footing necessary to sustain a certain load, having given the safe load per square foot. The rule is: Divide the total load in pounds by the safe load in pounds per square foot; the quotient is the required area square feet. Not much space can be saved by expressing such a simple rule in a formula, but the operation is more quickly comprehended. Let the total load in pounds be represented by the letter P, the safe load per square foot by the letter and the required area by the letter A. Then the rule is expressed by the formula A =P/S.

A formula is used by substituting for the letters the quantities known in the problem, and finding the unknown by performing the indicated operations. Sometimes the formula as written does not give the desired quantity directly, but must be rearranged so as to enable it to be found. This is done by simple operations, the aim being to obtain the desired quantity by itself on one side of the equality sign. Thus, in the above formula, suppose it is desired to find the value of P, A and S being given. By multiplying A and

P/S= by S (such operation being called clearing of fractions) there results AX S = P. in which the desired quantity P is by itself and is found to be equal to the product of A and S

The formula is further shortened by omitting the X sign, and writing AS = P. When two or more letters are thus written, it means that they are to be multiplied together. In a similar manner, S may be obtained, by dividing the last found formula through by A; thus, S =P/A

### Example

The safe bearing power of a soil is 1,500 lb. per sq. ft. What area of footing is required to sustain a load of 30,000 lb.?

Here A is wanted, and Pand S are given. Hence, A =P/S

30 000 = = 20 sq. ft. Again, suppose it was desired to ascer-l,500 tain how much per sq. ft. a footing 20 ft. in area carried, the total load, 30,000 lb., being known. Then, A and P are given to findS; or S = P/A=30000/20= 1,5001b.

The following formula shows how to determine the safe load that a stone beam or lintel will carry, if uniformly loaded:

2bd2A /fL = W, in which 6 = breadth in inches; d = depth in inches; A = number from a table; f = factor of safety; L = length in feet; W = safe distributed load, in pounds. Suppose any other quantity than W is wanted; it may be found by transposing the formula; thus, if the depth to sustain a certain load is required, the formula is thus arranged: Multiplying through by fL,

2bd2A = WfL. Arranging so as to have d2 alone on one side of the = sign, d2 = Wfl/2bA

Extracting the square root, d = ### Example

It is desired to carry a distributed load of 3.840 lb. on a piece of bluestone flagging 4 ft. wide and 6 ft. span, using a factor of safety of 10. How thick must the stone beam be? Here, W = 3,840 lb.; 6 = 4 ft. X12 = 48 in.; L = 6 ft.; f = 10; A is found from a table to be 150 for blue-stone; then d is the required term. Substituting the values, there results d = = √16 = 4 in.

A rule for finding the strength of a wooden column is: From the ultimate compressive strength of the material in pounds per square inch, subtract the fraction: the ultimate strength multiplied by the length of column in inches, and divided by 100 times the least side in inches. The difference is the ultimate strength of the column in pounds per square inch. This rule is rendered much more intelligible by using a formula; thus: Let S = ultimate strength of column in lb. per sq. in.; U= ultimate compressive strength of material in lb. per sq. in.; l = length of column, in inches; d = least side of column, in inches.

Then, S=U- (Ul / 100d)

The values of U for different woods are found in tables.

### Example

The ultimate compressive stress of white pine, parallel to the grain, is 3,500 lb. per sq. in. What is the ultimate load a 10" X 10" column 20 ft. long will carry?

Looking through the problem, it is found that all the quantities in the formula are given except S; the length expressed in inches = 240 in.; and the least side is 10 in., as the column is square. Substituting these values in the formula

S = U - (Ul / 100d')

S = 3,500-[(3500x240)/(100x10)]=3500-840=2660 lb.

Suppose it is desired to find the side of a square wooden column to carry a known load. The preceding formula contains the required term d. but as it also involves 8, which depends on d, that term must be eliminated. Let P represent the total load on the column; then, since the sectional area is d2, P = d2S; or, substituting for S its value.

P = Multiplying by 100 and dividing by U,

100P/U = 100 d2 - d l.

Such a quantity must be added to both sides of this equation (so as not to change its value) as will make the right side a perfect square. This quantity is found by trial to be (lxl) /400

Then, Extracting the square root of both members, placing l/20 on the left of the equality sign, and dividing by 10, = 10d-l/20;andl = d.

1/10 = ; multiplying the expression under the radical by it, = d*

In using any formula, care must be taken to have all dimensions, weights, etc. expressed in the units required by the formula.