This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.

Similarly G F would become tension if the wind were from the left.

We will consider the tendency of wind to overturn roofs, and this can best be done by calculating one or two practical examples of steeples. Before doing this, however, the student should be warned to always arrange for wind braces, that is, diagonal ties between the truss and in a plane at right angles to the trusses. The object of these, is to make all the trusses (that is the entile roof), act as one mass and thus keep the wind from blowing over each truss individually, and thus collapsing the roof. The arrangement of these ties varies with circumstances. They are usually placed immediately under the roof surface, that is from foot of one rafter to apex of next rafter on each side. If the rafters are long they arc placed diagonally in the individual panels. Sometimes a longitudinal truss is made the entire length of ridge, by wind bracing from both sides of each apex to centre of each neighboring main tie.

Wind tendency to overturn roofs.

Wind braces.

Example III.

A wooden, slate-covered steeple, 27' 6" high, covers a 24' square brick tower. The walls are 16" thick at the top and the steeple is anchored down four feet into the walls. Is this sufficient to keep it from blowing overt

It makes no difference just how the anchoring is done, it can be either by means of iron anchors bolting the plate down to the masonry below, or by means of wooden trestle work built inside and against the walls, which will also force the steeple to lift the walls from their bearings, before the steeple can topple over.

The wind pressure on one side of the roof will be the area of this side multiplied by 40 pounds per square foot (see Table XLIV) and it will act or can be considered as centralized at the centre of gravity of the side, which, being a triangle would be at one-third its height, or at D in Figure 254. The length of rafter will be 30', therefore area of one side = 30.12 = 360 square feet, and wind pressure FD = 360.40 = 14400 or say 15000 pounds total wind pressure which we consider as centralized at D and normal to rafter. Resisting this, we have the dead load, that is the weight of steeple and of the masonry as far as steeple will have to lift it.

Taking the weight of steeple at 20 pounds per square foot (making, of course, no allowance for snow) and weight of brickwork at 112 pounds we have weight of (four sides of) steeple, = 4.360.20 = 28800 pounds.

Weight of masonry = 4. (24 - l 1/3).4.112 = 54208 pounds, or total dead (vertical) load = 83000 pounds.

We now draw c a parallel and = 15000 pounds, the wind pressure; and a b vertically and = 83000 pounds, the dead load. Draw c b and from intersection G of FD with the main vertical neutral axis of the dead load draw G K parallel c b till it intersects the horizontal joint line A B, which we draw four feet below the tops of walls or at the anchor level. We now use Formulæ (44) and (45) to obtain the extreme edge strains at A and B.

Wooden steeple.

FIC.254.

Fig.254a.

We have these values : p = the pressure = c b = 91000 pounds. x = M K = 18" (by measurcment) d = AB = 24.12 = 288" a = area of wall at A B = 4.16". (288 - 16) = 17408 square inches.

We have then for pressure at nearer edge B = 91000/17408 + 6.18.91000/17408.288 = +7,21 pounds compressive pressure, per square inch ; and at A

_ = 91000/17408 - 6.18.91000/17408.288 = + .3,25 pounds compressive pressure, per square inch.

If the latter value had been a negative one, we should have had to rely on the quality of the mortar not to tear apart at A and thus allow the steeple to fall. It would be better, however, in such a case to carry the anchoring process further down, and thus gain more dead vertical load to resist the wind pressure.

Example IV.

A square stone steeple has 12" stone sides at the top ; 19 feet from the top vertically, the length of side is 12 feet. Is the steeple safe against wind pressure at this point ?

This example is intended to show that we can examine any point of steeple similarly to the manner of examining the base.

In Figure 255, A D measures 12 feet = 144" ; MC= 19 feet and CD scales 20 feet, hence area of each side= 6.20= 120 square feet, and weight of each side approximately = 120.150 = 18000 pounds and weight of four sides or vertical load = 4.18000 = 72000 pounds.

The wind pressure will be

= 120.40 = 4800 pounds and will be centralized at B or one-third the height of A C.

We draw B G normal to A C till it intersects C M at G.

Make a b = 72000 pounds and vertical, draw c a = 4800 pounds and parallel B G and draw a b which scales 74000 pounds. Draw G K parallel c b and we find K is 4" distant from centre of joint M. The area of joint is = 4.11.144 = 6336 square inches. We have, therefore, pressure at nearer edge of joint D,

Stone steeple.

__= 74000/6336 +6. 4.74000/6336.144 = + 13,65 pounds, per square inch, and at edge A

_ 74000/6336 - 6. 4.74000/6336.7144 = +9,75 pounds per square inch.

In a similar manner we might examine any other joint of the steeple.

It will be found that at the very top or near it the greatest danger exists. The finial frequently exposes a large surface to the wind and almost at right angles to the vertical dead load, deflecting this line much more in proportion. Then, too, the mortar is so much exposed that it cannot he relied on. For this reason there is placed usually a vertical iron tie-rod from the finial to some point below, frequently even below the springing line of the steeple. This arrangement is all right, provided the rod is properly put in. The writer has seen ponderous rods 2" diameter or more and perhaps, fifty or sixty feet long, intended as tie-rods, that had become so loosened by contraction and expansion that they could be easily swayed back and forth by the hand. Hence their only service was their own dead weight.

Continue to: