a considerable weight is to be carried, buttresses can be constructed at the columns, and the parapet between those buttresses reinforced as a COUNTERFORT

WALL.

8. Basement Walls. When the basement of a building is located below grade it is necessary to design the exterior walls to resist EARTH-PRESSURE. Under ordinary conditions, the minimum thickness of such walls is limited to 10 in, and many building codes require a minimum of 12 in. When the height of the wall is not appreciably over one-half the clear span between columns, it is desirable to carry the thrust on the vertical reinforcement spanning from the basement floor, or the wall-footing acting as a horizontal beam, to the first structural floor. If there is no footing beneath the wall, the latter must also be designed to carry its own weight, as a deep beam, spanning between columns. In Fig. 27 is shown the load diagram, and details of design, for a basement wall subjected to earth-pressure.

Let

P

total pressure per linear foot against the back of the wall, in pounds; h the height of the wall, that is, span length, in feet; equivalent fluid-pressure of the soil, in pounds per sq ft.

Ρ

=3

=

the bottom of the wall, Fig. 27, the reactions per linear foot of wall are then,

Considering any section, such as A-A', it is apparent

that the total pressure P', on the part of the wall above this section equals

This is the expression for the bending moment at any section of the wall.

The bending moment is a maximum when the section A-A' is that of zero shear or, since the shear at any section is equal to the sum of the reactions minus the sum of the loads to the right or left of the section, when R2 — P' = 0.

and the depth of the section of the maximum bending moment is

For purposes of computing the SECTIONAL AREA OF THE CONCRETE AND REINFORCEMENT required for walls subjected to earth-pressure, the WEIght of THE SOIL may be assumed to be 100 lb per cu ft, and to exert a pressure equal to that of a fluid weighing 30 lb per cu ft, p is then equal to 30.

Example 1. If it is required to design a basement wall 10 ft high, without surcharge, the procedure is as follows:

By Formula (1), the section of maximum bending moment is at

The concrete-section and steel-area are then designed as for ordinary slabs, placing a few horizontal bars, up to 0.3% of the concrete sectional area, to provide for shrinkage and to support the vertical reinforcement which is placed 2 in from the inside face of the wall. As such walls are ordinarily built monolithic with the floor-slabs, a small amount of negative reinforcement should be

placed near the outside of the wall at top and bottom. The bending moment due to the weight of the wall, acting as a simple beam 10 ft deep, is then computed and the necessary steel placed 3 or 4 in from the bottom of the wall to carry the load between columns.

The above design contemplates a basement floor at the base of the wall capable of resisting the horizontal thrust which is equal to the reaction R1. If there is not a floor at this level, the base of the wall, or wall-footing, should be designed as a HORIZONTAL BEAM, spanning between columns. For continuous bays, both the positive and negative moments may be taken as equal to

and is the clear span between columns. At the mid-span the reinforcement is placed near the inside face of the footing and near the outside face at the columns.

If the height of the basement wall is considerably more than ONE HALF OF THE SPAN BETWEEN COLUMNS, or openings in the wall interfere with the vertical reinforcement, it is usually more economical to design the wall as a SLAB SPANNING HORIZONTALLY BETWEEN COLUMNS. As in the former case, the wall must be designed not only to resist the soil-pressure, but also to carry its own weight between columns, unless other means of support are provided. Under this condition, the wall is designed as a number of horizontal strips, each one foot in height, and spanning between columns. The PRESSURE AT THE BASE OF THE WALL is equal to ph per sq ft of wall surface and diminishes uniformly to zero at the top of the wall, except as augmented by surcharge. For CONTINUOUS BAYS, both the positive and negative bending moments are then equal to wl2/12, when w equals the pressure per sq ft of wall surface at the height of the particular horizontal strip under design. The value of is usually taken as the clear distance between columns. The positive reinforcement is placed near the inside face of the wall, the negative reinforcement,, at columns, near the outside face. Vertical bars of small diameter, in size and number equivalent to the distributing bars in floor-slab design, see page 69, are used to support the horizontal reinforcement.

When the relation between column-spacing and wall-height results in WALL-PANELS APPROXIMATELY SQUARE, it is sometimes desirable to employ a two-way reinforcement under which condition the same stress distribution is followed as for floor slabs, see page 68. In all types of wall design the momentfactors are increased for conditions of semi-continuity, or where simple support is assumed, as in the case of floor-slab design.

In building construction basement walls are often subjected to SURCHARGE, as shown in Fig. 28. Under this condition, the load-diagram is not a triangle and it is necessary to compute the position of the RESULTANT OF THE EXTERNAL HORIZONTAL FORCES, or the CENTER OF GRAVITY of the applied loads, which may be found by Formula (3),

y =

h h+3h'
X

(3)

in which

y = the distance from bottom of span to center of gravity of loads in feet;

h

= the height of wall, that is, span-length, in feet;

h' the height of surcharge, in feet.

=

Example 2. It is required to design a basement wall 10 ft high and subjected to a surcharge of 300 lb per sq ft of ground surface, or the equivalent of 3 ft of earth, as shown in Fig. 29.

By Formula (3), the distance y from the bottom of the span to the center of gravity of the load is

Fig. 28. Vertical Section of Basement Wall, Showing Trapezoid of Earth-Pressure Fig. 29. Typical Design of a Basement Wall Subjected to Earth-Pressure

=

=

90 lb

390 lb per

The pressure at the top of the wall, due to the surcharge, is 30 X 3 per linear ft. At the bottom of the wall the pressure is 30 X 13 linear ft. As the pressure is assumed to vary uniformly, the average is (390 + 90)/2 240 lb per linear ft. The total load on a vertical strip of wall 1 ft wide is then 240 X 10 = 2400 lb. Taking moments about the top and bottom of the wall,

=

As the maximum moment occurs where the shear equals zero, the SECTION OF MAXIMUM MOMENT is found by equating to zero an expression for the shear

at any section. In this case, the shear at a section x distance from the top of the wall, is

In this equation the first term to the right of the equality sign represents the uniform portion of the load due to the surcharge, 90 lb per ft, multiplied by the distance x or 90x lb; the second term is the gradually increasing load represented by the area of the triangle abc, shown in Fig. 29. This load is zero at the top of the wall and increases to 30x at the section under consideration, since for every foot of depth there is an increment of 30 lb pressure; the average for the height x is, therefore, 30x/2 and the load 30x2/2; the third term is the reaction per linear ft at the top of the wall.

Completing the quadratic and solving for the value of x,

Knowing then that the section of maximum moment is 5.5 ft below the top of the wall, moments are taken about this section,

In this equation the first term to the right of the equality sign is the moment

(5.5)27

of the reaction; 90 X is the moment of the uniform portion of the 2 load due to the surcharge; the last term is the moment of the gradually increasing load represented by the area of the triangle abc. The load, as above, is equal to 30x2/2 or 30 X (5.5)2/2. As the center of gravity of a triangle is at 1/3 the height, the load is multiplied by 5.5/3 to obtain the moment.

Solving this equation, the maximum bending moment is

Knowing the maximum bending moment and maximum shear, the thickness of the wall and its reinforcement are determined by the ordinary formulas for rectangular beams and slabs.

9. Pits and Areas. Elevator-pits must provide sufficient depth to accommodate the buffer-installation, door travel, and the requirements of local codes. In plan they conform to the size of the elevator-shaft, all details of which are obtained from the manufacturers. As they generally occur below the basement floor-level it is not usually practicable to install drains, consequently provision must often be made for water-proofing.