Thrust of barrels.

A D2 - D B2 = A B2 or 4. D B2 - D B2 = w12 or

Or, h = 0,578. w, (64)

Where h = the horizontal thrust, in lbs., against each running foot of wall, w1 = one-half the total load, in lbs., of barrels coming on one foot of floor in width, and half the span. In our case we should have: h = 0,578.650 = 375 lbs.

Now, to find the height at which this thrust would be applied, we see, from Figure 91, that at point 1 the thrust would be from one line of barrels; at point 2, from two lines; at point 3, from three lines, etc.; therefore, the average thrust will be at the centre of gravity of the triangle A B C, this we know would be at one-third the height A B from its base A C.

Now B C is equal to 6r or six radii of the barrels; further, A C = 3r, therefore:

A B2 = 36.r2 - 9r2= 25.r2, and A B = 5r; therefore,AB/3 = 1 2/3r.

Fie. 92

Fig.93.

To this must be added the radius A D (below A) so that the central point of thrust, O in this case, would be above the beam a distance y = 2 2/3.r.

Where y = the height, in inches, above floor at which the average thrust takes place. Where x = radius of barrels in inches. Our radius is 10", therefore: y = 26 2/3"

Now, in Figure 93 let A B C D be the 12" wall, A the floor level, G M the central axis of wall, and A O = 26 2/3; draw O G horizontally; make G H at any scale equal to the permanent load on A D, which, in this case, would be the former load less the wind and snow allowances on wall and roof, or

3S32 - (16.30 + 13.30) = 2962, or, say 3000 lbs.

Therefore, make G H = 3000 lbs., at any scale; draw HI=h = 375 lbs., at same scale, and draw and prolong G 1 till it intersects D A at K. The pressure at K will be p = G I = 3023.

*We find the distance M K measures MK = x = 3 1/2".

Therefore, from formula (44) the stress at D will be:

v = 3023/144+ 6.3 1/2.3023/12.144 = 56 lbs. (or compression)

While at A the stress would be, from formula (45): v = 3023/144 - 6.3 1/2.3023/12.144 = - 14 lbs. (or tension), so that the wall would be safe.

The writer has given this example so fully because, in a recent case, where an old building fell in New York, it was claimed that the walls had been thrust outwardly by flour barrels piled against them.

Where piers between openings are narrower than they are thick, calculate them, as for isolated piers, using for d (in place of thickness of wall) the width of pier between openings; and in place of L the height of opening. The load on the pier will consist, besides its own weight, of all walls, girders, floors, etc., coming on the wall above, from centre to centre of openings.

To calculate wind-pressure, assume it to be normal to the wall, then if A B C D, Figure

94, be the section of the whole wall above ground

(there being no beams or braces against wall).

Narrow Piers.

Wind-pressure.

Fig. 94.

Make O D = 1/2. A D== L/2; draw G H, the vertical neutral axis of the whole mass of wall, make G H, at any convenient scale, equal to the whole weight of wall; draw H I horizontally equal to the total amount of wind-pressure.

This wind-pressure on vertical surfaces is usually assumed as being equal to 30 lbs. per square foot of the surface, provided the surface is flat and normal, that is, at right angles to the wind. If the wind strikes the surface at an angle of 45° the pressure can be assumed at 15 lbs. per foot.

It will readily be seen, therefore, that the greatest danger from wind, to rectangular, or square towers, or chimneys, is when the wind strikes at right angles to the widest side, and not at right angles to the diagonal. In the latter case the exposed surface is larger, but the pressure is much smaller, and then, too, the resistance of such a structure diagonally is much greater than directly across its smaller side.

In circular structures multiply the average outside diameter by the height, to obtain the area, and assume the pressure at 15 lbs. per square foot. In the examples already given we have used 15 lbs., where the building was low, or where the allowance was made on all sides at once. Where the wall was high and supposed to be normal to wind we used 30 lbs. Referring again to Figure 94, continue by drawing G I, and prolonging it till it intersects D C, or its prolongation at K. Use formulae (44) and (45) to calculate the actual pressure on the wall at D C, remembering that, x = M K; where M the centre of D C, also that, p = G I; measured at same scale as G H.

Remember to use and measure everything uniformly, that is, all feet and tons, or else all inches and pounds. The wind-pressure on an isolated chimney or tower is calculated similarly, except that the neutral axis is central between the walls, instead of being on the wall itself; the following example will fully illustrate this.

Example.

Is the chimney, Figure 87, safe against wind-pressure?

We need examine the joints A and E only, for if these are safe the intermediate ones certainly will be safe too, where the thickening of walls is so symmetrical as it is here.

The load on A we know is 4 7 tons, while that on E is 657 tons.

Wind-pressure on Chimney.

Now the wind-pressure down to A is:

Pa = 10.30.15 = 4500 lbs., or = 2 1/4 tons.

On base joint E, the wind-pressure is:

P = 12 2/3.150.15 = 28500 lbs., or = 14 1/4 tons.

We can readily see that the wind can have no appreciable effect, but continue for the sake of illustration. Draw Pa horizontally at half the height of top part A till it intersects the central axis G1; make G, H, at any convenient scale = 47 tons, the load of the top part; draw H, I, horizontally, and (at same scale) = 2 1/4 tons = the wind-pressure on top part; draw G I,; then will this represent the total pressure (from load and wind) at K, on joint A A,.