VERTICAL PRESSURE CAUSED BY
RECTANGULAR LOADING
A rectangular concrete footing 6 x 8 ft (182.9 x 243.8 cm) carries a total load of 180 kips
(800.6 kN), which may be considered to be uniformly distributed. Determine the vertical
pressure caused by this load at a point 7 ft (213.4 cm) below the center of the footing.
Calculation Procedure:
- State the equation for az
Referring to Fig. 5, let/? denote the uniform pressure
on the rectangle abed and (T 2 the resulting vertical
pressure at a point A directly below a vertex of the
rectangle. Then
(T 2 O ABH /1 1 \
7
= 0
25
-W
+
^(c
+
DJ
(8)
- Substitute given values and solve
for az
Resolve the given rectangle into four rectangles hav-
ing a vertex above the given point. Then p =
180,000/[6(8)] = 3750 lb/ft^2 (179.6 kPa). With A = 3
ft (91.4 cm); B = 4 ft (121.9 cm); H = 7 ft (213.4
FIGURE 5 cm); C = 58; D = 65; E = 74; O = snr^1 0.9807 =
78.7°; (T 2 Jp = 4(0.25 - 0.218 + 0.051) = 0.332; (T 2 =
3750(0.332) = 1245 lb/ft^2 (59.6 kPa).
APPRAISAL OF SHEARING CAPACITY OF
SOIL BY UNCONFINED COMPRESSION TEST
In an unconfmed compression test on a soil sample, it was found that when the axial
stress reached 2040 lb/ft^2 (97.7 kPa), the soil ruptured along a plane making an angle of
56° with the horizontal. Find the cohesion and angle of internal friction of this soil by
constructing Mohr's circle.
Calculation Procedure:
- Construct Mohr's circle in Fig. 6b
Failure of a soil mass is characterized by the sliding of one part past the other; the failure
is therefore one of shear. Resistance to sliding occurs from two sources: cohesion of the
soil and friction.
Consider that the shearing stress at a given point exceeds the cohesive strength. It is
usually assumed that the soil has mobilized its maximum potential cohesive resistance
plus whatever frictional resistance is needed to prevent failure. The mass therefore re-
mains in equilibrium if the ratio of the computed frictional stress to the normal stress is
below the coefficient of internal friction of the soil.
Loaded
area
