Material of pillar. Both ends of pillar smooth(turned or planed.) One end smooth(turned or planed) other end a pin end. Both ends pin ends. Cast-iron 0.0003 0.0001 0.00057 Wrought-iron 0.000025 0.000033 0.00005 Steel 0.00002 0.000025 0.000033 Wood 0.00033 0.00044 0.00067 Stone 0.002 Brick 0.0033

The stress of course will be as before: - v = a. c.

Where v = the ultimate stress in pounds. " a = the area of cross-section in inches. " c = the ultimate resistance to crushing per square inch.

Inserting the values for strain, s, and stress, v, in the fundamental formula (1) we have: c.a= (w+w.l2.n) p2 or: a.(c/f) = w (1+l2n) p2 or: w = a.(c/f)

1+l2n/p2 (3)

Long Columns.

 Where w = the safe total load on the pillar. ,, a = the area of cross-section in inches. ,, P2 = the square of the radius of gyration of the cross-section. ,, l = the length in inches. ,, = : the safe resistance to crushing per square inch.

Example.

What safe load will a 12" x 12" brick pier carry, if the pier is ten feet long, and of good masonry?

The area of cross-section will be: a = 12.12 = 144 square inches.

The square of the radius of gyration according to Section No. 1 in Table I would be: d2/12 and as d=V2, we have Q2= 12.12 = 12

12

For the safe resistance to crushing per square inch, we have, using a factor-of-safety of ten, and considering the ultimate resistance to be 2,000 pounds per square inch,

(c/f) = 2000 = 200lbs.

The length will be ten feet, or one hundred and twenty inches; therefore: l2= 14400 For n we must use (according to Table II), for brickwork: n = 0.0033;

Therefore the safe total load on the pier would be: w = 144.200 = 28800 = 5806 lbs.

1+14400.0,0033 1+3,96

In all formulae where constants and factors of safety are used, it will be found simpler and avoiding confusion to immediately reduce the constant by dividing it by the factor-of-safety, and then using only the reduced or safe constant.

Thus if c = 48,000 pounds, and if f = 4, do not write into your formula for (c/f) = 48000 but use at once for (c/f) = 12000.

Materials in compression that have an even bearing on all parts of the bed will stand very much more compression to the square inch than materials with rough, uneven or rounded beds, or where the bearing is on part of the cross-section only, as in the case of pins (in trusses) bearing on eye-bars. It is usual in calculating to make allowance for this. Columns with perfectly even bearing on all parts of the bed (planed or turned iron or dressed stone) will stand the largest amount of compression. Columns with rough, rounded or uneven ends are calculated the same as for pin-ends of eye-bars. In the table (II) giving the values for n of Rankine's formula for compression, the different values for smooth and also for pin ends are given.