the rod and the bending moment in the piling are reduced; the net result is a saving in ma-
terial despite the increased length.
Investigation of this problem discloses that the most economical depth of penetration
is that for which the tangent to the elastic curve at the lower end passes through the an-
chorage point. If this point is considered as remaining stationary, this condition can be de-
scribed as one in which the elastic curve is vertical at Z), the surrounding soil acting as a
fixed support. Whereas an equation can be derived for the depth associated with this con-
dition, such an equation is too cumbersome for rapid solution.
When the elastic curve is vertical at D, the lower point of contraflexure lies close to
the point where the net pressure (the difference between active pressure to the right and
passive pressure to the left of the wall) is zero. By assuming that the point of contraflex-
ure and the point of zero pressure are in fact coincident, this problem is transformed to
one that is statically determinate. The method of analysis based on this assumption is
termed the equivalent-beam method.
When the piling is driven to a depth greater than the minimum needed for stability, it
deflects in such a manner as to mobilize passive pressure to the right of the wall at its low-
er end. However, the same simplifying assumption concerning the pressure distribution as
made in the previous calculation procedure is made here.
Let C denote the point of zero pressure. Consider a 1-ft (30.5-cm) length of wall, and
let T- reaction at anchorage point and V= shear at C.
Locate C and construct the net-pressure diagram for AC as shown in Fig. 136.
Thus, w = 110 lb/ft^3 (17.28 kN/m^3 ) and 0 = 32°. Then Ca = tan^2 (45° - 16°) - 0.307;
Cp = tan^2 (45° + 16°) = 3.26; Cp-Ca = 2.953; pA = 0.307(200) = 61 lb/ft^2 (2.9 kPa);
pB = 61 + 0.307(2O)(IlO) = 737 lb/ft
2
(35.3 kPa); a = 737/[2.953(11O)] = 2.27 ft (0.69
m).
- Calculate the resultant forces P 1 and P 2
Thus, P 1 =
1
/ 2 (20)(61 + 737) = 7980 Ib (35,495.0 N); P 2 =
1
/ 2 (2.27)(737) = 836 Ib (3718.5
N); PI + P 2 = 8816 Ib (39,213.6 N). - Equate the bending moment at C to zero to find T, V,
and the tension in the tie rod
Thus b = 2.27 + (^2 %)(737 + 2 x 61)/(737 + 61) = 9.45 ft (2.880 m); c =^2 /s(2.27) = 1.51 ft
(0.460 m); SMC = 19.27T- 9.45(7980) - 1.51(836) = O; T= 3980 Ib (17,703.0 N); V =
8816 - 3980 = 4836 Ib (21,510.5 N). The tension in the rod = 3980(8) - 31,840 Ib
(141,624.3N). - Construct the net-pressure diagram for CD
Refer to Fig. 13c and calculate the distance Jt. (For convenience, Fig. 13c is drawn to a
different scale from that of Fig. lib.) ThuspD = 2953(1 IQx) - 324.8; R 1 = ^(324.S^2 ) =
162.4^2 ; SMD = RlX/3 -Vx = 0;Rl = 3V; 162 Ax^2 = 3(4836); x = 9.45 ft (2.880 m).
5. Establish the depth of penetration
To provide a factor of safety and to compensate for the slight inaccuracies inherent in this
method of analysis, increase the computed depth by about 20 percent. Thus, penetration =
1.20(0+) =14 ft (4.3m). - Locate the point of zero shear; calculate the piling maximum
bending moment
Refer to Fig. 136. Locate the point E of zero shear. Thus pE = 61 + 0.307(11Oy) = 61 +
33.11y; y 2 y(pA + pE) = T; or '/2X122 + 33.1Iy) = 3980; y = 13.6 ft (4.1 m), and/^ = 520
lb/ft^2 (24.9 kPa); Mmax = ME = 3980(10.6 - (13.6/3)(520 + 2 x 61)/(520 + 61)] - 22,300
ft-lb/ft (99.190 N-m/m) of piling. Since the tie rods provide intermittent rather than con-
tinuous support, the piling sustains biaxial bending stresses.