"When considering the weight per cubic foot of wall, we add the proportionate share of buttress; now in Figure 63 there are 4 cubic feet of buttress to every 7 feet of wall, so that we must add to the usual weight w per cubic foot of wall 4/7 to, or w. (1 + 4/7)

To put this in a formula.

w11 = w (1+A/A1) (58)

Buttressed walls.

Buttressed walls

Fig. 63.

Graphical Method 100137

Fig. 64.

Where w11 = the weight per cubic foot, in pounds, to be used for buttressed walls, after finding the neutral axis of the whole mass.

Where w = the actual weight, in pounds, per cubic foot of the material.

Where A = the area in square feet of one buttress. Where A1= the area in square feet of wall from side of one buttress to corresponding side of next buttress.

Buttresses, however, will not be of very much value, unless they are placed quite close together. Buttresses on the back surface of a wall are of very little value, unless thoroughly bonded and anchored to walls; these latter are called counterforts. It is wiser and cheaper in most cases to use the additional masonry in thickening out the lower part of wall its entire length.

Where frost is to be resisted the back part of wall should be sloped, for the depth frost is likely to penetrate (from 3 to 4 feet in our climate), and finished smoothly with cement, and then asphalted, to allow the frozen earth to slide upwards, see Figure 65.

Walls with counterforts.

Resistance to frost.

Resistance to frost

Fig. 65.

Example I.

Cellar wall to frame dwelling

A two story and attic frame house has a 12" brick-foundation wall, the distance from cellar bottom to ground level being 6 feet. The angle of friction of ground to be assumed at 33° and the weight per cubic foot at 120 pounds. Is the wall safe?

The weight of wall and superstructure must, of course, be taken at its minimum, when calculating its resistance to the ground, we shall, therefore, examine one of the sides on which no beams or rafters rest. The weight will con-sist, therefore, only of brick wall and frame wall over. We examine only one running foot of wall, and have

8

cubic feet of brick at 112 lbs.

=

896

lbs

24

feet (high) of frame wall at 15 lbs

=

360

,,

Total weight resisting pressure

=

1256

,,

The pressure itself will be according to formula (52)

P = 16 2/3. L2=16 2/3. 36 = 600 lbs.

Now make D O = 1/3. D C. Make angle

P O C = 57°; prolong P 0 till it intersects the vertical neutral axis of wall1 at F; make F II = 1256 pounds, at any scale, draw II I parallel to P O, and make H I =p = GOO pounds at same scale. Draw I F, then is its point of intersection K (with D A) a point of the curve of pressure, and F I (measured at same scale) is the amount of pressure p to be used in formulae (14) and (45). By careful drawing we will find that K comes 1/2" beyond A (outside of A I)), or 6 1/2 from centre E of A D. F I we find measures 1GG0 units, therefore p= 1660 pounds.

Graphical Method 100139

Fig. 66.

To find the actual stress or resistance v of edge of fibres of brickwork at A use (44), viz.: v = p/a+6. x.p/a.d and as p = 1660 and x = E K = 6 1/2" and a = 12. 12 = 144 inches and d = A D = 12" we have: v=1660/144+6. 6 1/2.1660/144.12=11 1/2+37 1/2= + 49 pounds as this is a positive quantity it will be compression.

The resistance of edge-fibres at D will be according to formula (45) v= 1660/144-6. 6 1/2.1660/144.12 = 11 1/2 - 37 1/2 = -26 pounds as this is a negative quantity D will be subjected to tension; that is, there is a tendency for A B C D to tip over around the point A, the point D tending to rise. The amount of tension at D is more than ordinary brickwork will safely stand, according to Table V, still, as it would only amount to 26 pounds on the extreme edge-fibres and would diminish rapidly on the fibres nearer the centre, we can consider the wall safe, even if of but fairly good brickwork, particularly as the first-story beams and girders and the end and possible cross-walls, will all help to stiffen the wall. Had we taken a foot-slice of the wall under the side carrying the beams, we should have had an additional amount of weight resisting the pressure. If the beams were 18 ft. span, we should have three floors each 9 ft. long and with load weighing, say, 90 pounds per foot; to this must be added the roof, or about 13 ft. a 50 pounds, the additional load being:

Floors, 3. 9.

90 =

2430

Roof, 13.

50 =

650

Total

3080

Now make I M = 3080 pounds at same scale as F H, etc., draw M F and its point of intersection N with D A would be a point of

1 It should really be the vertical neutral axis of the whole weight, which would be a trifle nearer to D C than centre of wall.

the curve of pressure, and F M, at same scale the amount of pressure on D A for the bearing walls of house; E N we find measures 2 1/2", and F M measures 4600 units or pounds. The stress at A, then, would be: v= 4600/144+6. 2 1/2.4600/144.12= +72 pounds and the stress at D would be:

6=4600/144 - 6. 2 1/2.4600/144.12 = 8 pounds

There will, therefore, be absolutely no doubt about the safety of bearing walls.

Example II.

A cellar wall A B C D is to be carried 15 feet below the level of adjoining cellar; for particular reasons the neighboring wall cannot be underpinned. It is desirable not to make the wall A B C D over 2' 4" thick. Would this be safe? The soil is wet loam.

In the first place, before excavating we must sheath-pile along line C D, then as we excavate we must secure horizontal timbers along the sheath-piling and brace these from opposite side of excavation. The sheath-piling and horizontal timbers must be built in and left in wall. The braces will have to be built around and must not be removed until the whole weight is on the wall.