Cellar wall deeperthan adjoining building

Cellar wall deeperthan adjoining building

Fig. 67.

The weight of the wall C G, per running foot of length, including floors and roofs, we find to be 13000 pounds, but to this we must add the possible loads coming on floors, which we find to be 6000 pounds additional, or 19000 pounds total, possible maximum load. This load will be distributed over the area of C D E. In calculating the weight of ABCD resisting the pressure, we must lake, of course, only the minimum weight; that is, the actual weight of construction and omit all loads on floors, as these may not always be present. The weight of walls and unloaded floors coming on A B C D, and including the weight of A B C D itself, we find to be 21500 pounds per running foot. Now to find the pressure p, proceed as follows: Make angle E1 D M= 17°, the angle of friction of wet loam (See Table X), and prolong D E1 till it intersects C E1 at E. Now C E, we find, measures 52 feet; C D or L is 15 feet; then, instead of using w in formula (51), we must use w1, as found from formula (55), viz.: , 2. 19000 w1 = w + 2.1900/CE.L w for wet loam (Table X) is 130 pounds; therefore, w1= 130 + 2.19000/52.15=130+48,7=179

Inserting this value for to in formula (51) we have: p = 179. 152. 0,138 = 5558. The height X from D at which P O is applied is found from formula 5G, and is:

X = 15. (179 - 2/3.150) = 15. (179 - 100) = 2.179 - 150 358 - 150

5',697 = 5'8 1/2"

Make D 0 = X = 5'8 1/2"; draw P O parallel to E D till it intersects the vertical neutral axis of wall (centre line) at F; make F II (vertically) at any scale equal to 21500, draw H I parallel to P O and make H I=p = 5558 pounds at same scale, draw I F, then is its point of intersection with the prolongation of A D at K a point of the curve of pressure, and F I measured at same scale as F H is the amount of pressure on joint. The distance K from centre of joint N we find is 14 1/2", F I measures 23800 units or pounds; the stress (v) at A, therefore, will be, formula (44): v=23800/28.12+6. 14 1/2.23800/28.12.28=+202

While the stress at D would be formula (45) v=23800/28.12 - 6. 14 1/2.23800/28.12.28 = -150

Or the edge at A would be subjected to a compression of 292 pounds, while the edge at D would be submitted to a tension of 150 pounds per square inch, both strains much beyond the safe limit of even the best masonry. The wall will, therefore, have to be thickened and a new calculation made.

Example III.

A stage pit 30 feet deep is to be enclosed by a stonewall, 3 feel thick at (he top and increasing 4 inches in thickness for every 5 feet of depth. The wall, etc., coming over this wall weighs 25000 pounds per running foot, but cannot be included in the calculation, as peculiar circumstances will not allow braces to be kept against the cellar wall, until the superimposed weight is on it. The surrounding ground to be taken as the average, that is, 120 pounds weight per cubic foot, and with an angle of friction of 33°.

Wall to stage pit.

Wall to stage pit

Fig. 68.

Find B M (= Z) and Q T (= Y) by making angles T A E = 33° and B A U = 66° and then proceeding as explained for Figure 60. We scale B M and Q T at same scale as height of wall A B is drawn, and find:

BM = Z = 25ft. 6"==25 1/2

Q T = Y = 9 ft. 8" = 9 2/3; assuming each cubic foot of wall to weigh 150 pounds we find Q from Formula (57)

Q= 9 2/3.25 1/2.120/30.150=6,573=6'7".

Make A E = Q = 6' 7" and draw B E.

At equal heights, that is, every 5 feet, in this case, draw the joint lines D E, D1 E1 D11, E11, etc. Find the centres of gravity F, F1, F11, etc., of the six parts of A E B, (see foot-note, p. 101) and also the centres of gravity G, G1, G11, etc., of the six parts of the wall itself, which, in the latter case, will be at the centre of each part.

Horizontally, opposite the centres F, F1 F11, etc., apply the pressures P O, P1, O1, etc., against wall, and parallel to M A. Through centres G, G1, G11 ,etc., draw vertical axes.

Draw the lines S N, S1, N1, S11, N11, etc., at half the vertical height of each section. Now in Figure 69 make a b1 = Rv Nv; b1,d1, = RlV NlV; d1 f1 = R111 N111; f1, h1, = R11 N11,; h1 j1 = R1 N1; and j1 I1 = R N. Draw the vertical lines through these points. Now begin at a, make a b parallel to P O, make b c = Rv Sv; draw c d parallel P O; make d e = RlV SlV; draw e f parallel P O, make fg = R111 S111, and similarly g h, i j and h I parallel P O, and h i = R11 S11; j k = R1 S1 and l m = R S. Draw from a lines to all the points c, d, e, f, g, etc. Now in Figure G8 begin at Pv Ov, prolong it till it intersects vertical axis Gv at Iv, draw Iv IIV parallel a c till it intersects PlV OlV at IIV; draw Hv IlV parallel to a d till it intersects vertical axis GlV at IlV; draw IIv IIlV parallel to a e till it intersects P111 O111 at HlV and similarly HlV I111, parallel af; I111 II111 parallel a g; H111, I11 parallel a h; I11 H11 parallel to a i; H11 I, parallel to a j; I, II, parallel to a k; II, I parallel to a I, and IK parallel to a m. The points of intersection K, K1 K11, etc., are points of the curve of pressure. To find the amount of the pressure at each point, find weight per running foot of length of any part of wall, say, the bottom part (A A1 D1 D) the contents are 5' high, 4' 8" wide, 1' thick = 5.4 2/3. 1 = 23 1/3 cubic feet a 150 pounds = 3500 pounds.

Graphical Method 100142

Fig. 69.

Divide the weight by the length of m I in inches, and we have the number of pounds per inch, by which to measure the pressures. As m I measures 56 inches, each inch will represent 3500/56 = 62 1/2 lbs.

Now let us examine any joint, say, A111 D111; I111 H111, which intersects A111, D111 at K111, is parallel to a g. Now a g scales 1G6 inches, therefore, pressure at K111 = 166. 62 1/2 = 10375 pounds. In measuring the distance of K111 from centre of joint in the following, remember that the width of A111 D111 is 44 inches, the width of masonry above joint, and not 48" (the width of masonry below). A111 K111 scales 38", therefore, distance x of K111 from centre of joint is x= 38 - 22 = 16".