In pieces under compression the load is directly applied to the material. In short pieces, therefore, which cannot give sideways, the strain will just equal the load, or we have: s = w.

Where s = the strain in pounds.

And where w = the load in pounds.

The stress will be equal to the area of cross-section of the piece being compressed, multiplied by the amount of resistance to compression its fibres are capable of.1 This amount of resistance to compression which its fibres are capable of is found by tests, and is given for each square inch cross-section of a material. A table of constants for the resistance to crushing of different materials will be given later on.

In all the formulae these constants are represented by the letter c.

We have, then, for the stress of short pieces under compression: - v = a.c

Where v is the ultimate stress in pounds.

Where a is the area of cross-section of the piece in inches.

And where c is the ultimate resistance to compression in pounds per square inch.

Inserting this value for v, and to for s in the fundamental formula (1), we have for short pieces under compression, which cannot yield sideways: a. c = w. f, or: w=a.(c/f). (2)

Where w = the safe total load in pounds. Where a = the area of cross-section in inches.

And where (c/f) = the safe resistance to crushing per square inch.

Example. What is the safe load which the granite cap of a 12" x 12" pier will carry, the cap being twelve inches thick?

The cap being only twelve inches high, and as wide and broad as high, is evidently a short piece under compression, therefore the above formula (2) applies.

1 This is not theoretically correct, as there is in every case a tendency for the material under compression to spread; but it is near enough for all practical purposes.

The area is, of course: a = 12.12 = 144 square inches.

The ultimate resistance of granite to crushing per square inch is, say, fifteen thousand pounds, and using a factor-of-safety of ten, we have for the safe resistance: c/f = 15000/10 = 1500lbs

Therefore the safe load w on the block would be: w=144. 1500 = 216000 pounds. Where long pieces (pillars) are under compression, and are not secured against yielding sideways, it is evident they would be liable to bend before breaking. To ascertain the exact strain in such pieces is probably one of the most difficult calculations in strains, and in consequence many authors have advanced different theories and formula?. The writer has always preferred to use Rankine's formula, as in his opinion it is the most reliable. According to this, the greatest strain would be at the centre of the length of the pillar, and would be equal to the load, plus an amount equal to the load multiplied by the square of the length in inches, and again multiplied by a certain constant, n, the whole divided by the "square of the radius of gyration" of the cross-section of the pillar. We have therefore for the total strain at the centre of long pillars: s = w+w.l2n /p2

Compression, Long Columns.

Where s = the strain in pounds. " w = the total load in pounds. " 1 = the length in inches.

" p2 = the square of the radius of gyration of the cross-section. " n = a constant, as follows: -