Retaining walls

Contents - Previous - Next

Wall Failure

Walls are commonly used to retain soil on sloping sites, water in a pond or bulk products within a storage area. There are several limiting conditions which, if exceeded, can lead to the failure of a retaining wall. Each must be addressed in designing a wall.

1 Overturning - This occurs when the turning moment due to lateral forces exceeds that due to the self-weight of the wall. The factor of safety against overturning should be at least two.

2 Sliding - The wall will slide if the lateral thrust exceeds the frictional resistance developed between the base of the wall and the soil. The factor of safety against sliding should be about two.

3 Bearing on Ground - The normal pressure between the base of the wall and the soil beneath can cause a bearing failure of the soil, if the ultimate bearing capacity is exceeded. Usually the allowable bearing pressure will be one-third of the ultimate value. Note that the pressure distribution across the base is not constant.

Bearing pressure

4 Rotational Slip - The wall and a large amount of the retained material rotate about some point O. if the shear resistance developed along a circular arc is exceeded. The analysis is too complex to include here.

Rotation

5 Wall Material Failure - The structure itself must be capable of withstanding the internal stresses set up, that is, the stresses must not exceed allowable values. Factors of safety used here depend on the material and the level of the designer's knowledge in respect to the loads actually applied. Naturally, both shear and bending must be considered, but the most critical condition is likely to be tension failure of the 'front' facet

Joint failure in block work

Gravity walls and dams are dependent on the effect of gravity, largely from self-weight of the wall itself, for stability. Other types of walls rely on a rigid base, combined with a wall designed against bending to provide an adequate structure.

Tension bending failure

Pressure Exerted by Retained Material

Liquid Pressure

The pressure in a liquid is directly proportional to both the depth and the specific weight of the liquid (w) which is the weight per unit volume. w = pg (N/m³) where: p = density of liquid ( kg/m³) g = gravitational acceleration (9.81m/s2)

Liquid Pressure I

The pressure at a given depth acts equally in all directions, and the resultant force on a dam or wall face is normal to the face. The pressure due to the liquid can be treated as a distributed load with linear variation in a triangular load form, having a centroid twothirds of the way down the wet face.

p = r gH = wH (N/m²) and:

P = WH² / 2 acting at a depth of 2/3 H

Liquid Pressure II

It should be noted that a wall retaining a material that is saturated (water-logged) must resist this liquid pressure in addition to the lateral pressure from the retained material.

Example 25

Gravity Wall Retaining Water

Consider a mass concrete dam with the cross section shown which retains water to 3m depth. Assume: Ground safe bearing capacity 300kN/m². Coefficient of sliding friction at base 0.7. Specific weight of concrete 23kN/m³.

1 Find water force P:

All calculations per metre length of wall:

p = wH² / 2 = (9.8 x 10³ x 3²) = 44.1kN (acting 1m up face)

2 Find mass of 1m Length of Wall:

W = A x 1 x specific weight

= 3 x (0.6 + 1.8)/2 x 23 = 82.8kN

3 Find line of action of w: Taking moments of area about vertical face:

Hence self-weight of wall acts 0.25m to left of base centre line.

4 Find vertical compressive stress on base:

P_c = W / A = 82.8 / (I x 8) = 46kN/m²

5 Find moment about centre line of base

M = (1 x 44.1) - (0.25 x 82.8); (clockwise) (anticlockwise) M = 23.4 kNm

6 Find bending stresses/pressures

s _b = P_b = MI / Y_max where;

I = bd³ / 12 = (I x 1.8³) / 12 = 0.486m⁴

Y_max = ± 1.8 / 2 = ± 0.9m

s _b = P_b = ± (23.4 x 0.486) / 0.9 = 12.6kN/m²

7 Find actual stresses/pressures

s = p = W/A + M_y/I

s _E = P_E = 46 + 12.6 = 58.6kN/m² (comp)

s _D = P_D = 46 - 12.6 = 33.4kN/m² (comp)

(Note: Compression only indicates the resultant P and W would intersect the base line within its middle third).

8 Compare maximum pressure with allowable bearing capacity:

P_max = 58.6kN/m²

This is less than the allowable safe bearing capacity of the soil. Hence wall-soil interface is safe in bearing.

9 Compare actual stresses in wall with allowable values:

Max. stress = 58.6 kN/m² (Compression) and no tensile stress at any point across wall. Hence wall material is safe.

10 Check overturning:

Overturning moment about D = 44.1 x 1 = 44.1 kNm

Stabilising moment about D = 82.8 x 1.15 = 95.22kNm

Factor of safety overturning = 94.22 / 44.1 = 2.16

Wall safe in overturning.

11 Check sliding

Frictional resistance = m W

m W = 0.7 x 82.8 = 58kN

Horizontal thrust = P = 44.1kN

Since required factor against sliding is 2, there is a deficiency of (2 x 44.1) - 58 = 30.2 kN.

Additional anchorage against sliding should be provided.

Example 26

Circular Water Tank

Diameter 5m, depth of water 3m

Water weighs 9.8 x 10³N/m³

Pressure (P) at depth of am

P₃=wH=9.8 x 10³ x 3=294kN/m²

This acts vertically over the whole base, thus design base for u.d.1. of 29.4kN/m².

Pressure P₃ also acts laterally on the side wall at its bottom edge. This pressure decreases linearly to zero at the water surface.

Total force on base = P₃A_B = 29,4 x (p x 5²) = 577.3 kN

(acting at centre of base)

Total force on side per metre of perimeter wall:

P₃H / 2 = (29.24 x 3) / 2 = 44.1 kN/m run (acting 1m above base)

Pressure Due to Granular Materials

Granular materials such as sandy soils, gravelly soils and grain possess the property of internal friction (friction between adjacent grains), but are assumed not to possess the property of cohesion. If a quantity of such material in a dry condition is tipped on to a flat surface, it will form a conical heap, the shape maintained by this internal friction between grains. The angle of the sloping side is known as the angle of repose.

For a dry material the angle of repose is usually equal to the angle of shearing resistance of the material. This angle of shearing resistance is called the angle of internal friction (a). The angle of friction is the essential property of a granular material on which Rankine's theory is based. This theory enables the lateral pressure to be expressed as a proportion of the vertical pressure, which was shown (above) to depend on specific weight and depth only.

In this case at a depth h, the active lateral pressure is given by:

P=k x w x h where:

k = a constant dependent on the materials involved

There exists some friction between the retained material and the wall face, but usually this is disregarded giving a relatively simple relationship for k:

where

q = the angle of friction

where: p_a = total force per m of wallface (N)

(N/m lenght of fall)

P_a = total force per m of wall face (N)

This gives the approximate horizontal resultant force on a vertical wall face, when retaining material that is level with the top of the wall. If the surface of the retained material is sloping up from the wall at an angle equal to its angle of repose, a modification is required.

Example 27

Wall retaining soil

Consider the wall shown retaining loose sandy soil to a depth of 2 metres. Tables provide angle of friction 35°, specific weight 18.6kN/m³. Assuming smooth vertical surface and horizontal soil surface using Rankine's theory gives:

P = 10.1 kN/m length of wall

If steel posts are placed at 2.5m centres, each post can be approximated to a vertical cantilever beam 2.5m long carrying a total distributed load of 10.1 x 2.5 = 25.25kN of linear variation from zero at the top to a maximum at the base. The steel post and foundation concrete must be capable of resisting the applied load, principally in bending but also in shear.

The timber crossbeams can be analyzed as beams simply supported over a span of 2.5m, each carrying a uniformly distributed load. This load is equal to the product of the face area of the beam and the pressure in the soil at a depth indicated by the centroid of area of the beam face.

Totalu.d.l.onbeam=9.29 x 0.3 x 2.5=6.97 kN

The maximum bending moment at the centre of the span can be determined and the beam section checked.

Example 28

Grain Storage Bin

Pressure diagrams

(The theory given does not apply to deep bins). A shallow bin can be defined as one having a sidewall height of less than

S/2 tan (45° + q / 2 ) for a square bin of side length S.

Consider a square bin of side length 4m and retaining shelled maize corn to a depth of 2m. Assume q = 27°; specific weight is 7.7kN/m³.

Critical height is:

Design as shallow bin since depth of grain is only 2m.

Maximum pressure at base of wall:

or resultant force P = (5.97 x 2²) / 2= 11.57kN/m

(acting 2/3m above base of wall)

Note the design of the wall is complex if it consists of a plate of uniform thickness, but if the wall is thought of as consisting of a.number of vertical members cantilevered from the floor, an approach similar to that taken in the wall retaining soil can be used.

Further reading

Morgan, W., The Elements of Structure, ed. 1. Buckle, London, Pitman, 1978.

Morgan, W. and D.T. Williams, Structural Mechanics, ed. 1. Buckle, London, Pitman, 1980.

Salvadori, M. and R. Heller, Structure in Architecture, The Building of Buildings, Englewood Cliffs, New Jersey, Pretice-Hall, 1975.

Whitlow, R., Materials and Structures, New York, Longman, 1973.

Chapter 5 Elements of construction

Elements of Construction

When designing a building, an architect plans for special, environmental and visual requirements. Once these requirements are satisfied, it is necessary to detail the fabric of the building. The choice of materials and the manner in which they are put together to form building elements such as foundation, walls, floor and roof, depend largely upon their properties relative to environmental requirements and their strength properties. The apprehension of building construction thus involve an understanding of the nature and characteristics of a number of materials, of methods to process them and form them into building units and components, of structural principles, of stability and behaviour under load, of building production operations and of building economics.

The limited number of materials available in the rural areas of east and south east Africa result in a limited number of structural forms and methods of construction. Different socioeconomic conditions and cultural beliefs are reflected in varying local building traditions. While knowledge of the indigenous building technology is widespread, a farmer and his family normally can erect a building using traditional materials and methods without the assistance of skilled or specialized craftsmen. However, population growth and external influences are gradually changing people's lives and the agricultural practices some traditional materials are getting scarce. Hence, better understanding of traditional materials and methods is needed to allow them to be used more efficiently and effectively. While complete understanding of the indigenous technology will enable the architect to design and detail good but cheap buildings, new materials with differing properties may need to be introduced to complement the older and allow for new structural form to develop.

Loads on building components

Loads are usually divided into the following categories:

Dead loads which result from the mass of all the elements of the building including footings, foundation, walls, suspended floors, frame and roof. These loads are permanent, fixed and relatively easy to calculate.

Live loads which result from the mass of animals, people, equipment and stored products. Although the mass of these loads can be readily calculated, the fact that the number or amount of components may vary considerably from time to time makes live loads more difficult to estimate than dead loads.

Also included as live loads are the forces of nature wind, earthquake and snow.

Where wind velocities have been recorded, the following equation can be used to determine the expected pressures on building walls:

q = 0,0127 V²k where:

q = basic velocity pressure, P_a

V = wind velocity, m/s

k = (h/6.1)^2/7

h = design height of building, m (eave height for low and medium roof pitches)

6.1 = height at which wind velocities were often recorded for Table 5.1.

While the use of local wind velocity data allows the most accurate calculation of wind pressures on buildings, in the absence of such data, estimates can be made from the Beaufort Scale of Winds given in Table 5.1.

Table 5.1 Beaufort Scale of Winds

From U.S. Weather Bureau

Some idea of the worst conditions to be expected can be obtained by talking to long-time residents of the area.

The effect of wind pressure on a building is influenced by the shape of the roof and by whether the building is open or completely closed. Table 5.2 gives coefficients used to determine expected pressures for low-pitch and high-pitch gable roofs and open and closed buildings. Note that there are several negative coefficients indicating that strong anchors and joint fasteners are just as critical as strong structural members.

Data on earthquake forces is very limited. The best recommendations for areas prone to earthquakes is to use building materials that have better than average tensile characteristics, to design joint fasteners with an extra factor of safety, and to include a ring beam at the top of the building wall.

Table 5.2 Wind Pressure Coefficients for Gable Roof Farm Buildings

Open on both sides < 30° 30°

H = height to eaves, W = width of building

Table 5.3 Mass of Building Materials

Table 5.4 Loads on Suspended Floors

Snow loads are a factor only in very limited areas of high elevation in East and Southeast Africa. Local information on the mass of snow loads should be used.

Table 5.3 provides information useful in determining dead loads and Tables 5.4 and 5.5 give information relevant to live loads.

Table 5.5 Mass of Farm Products

Contents - Previous - Next