Retaining Wall design

Retaining Walls

UNIT 9 RETAINING WALLS Structure 9.1

Introduction Objectives

9.2

Pressure on Retaining Walls Due to Liquids

9.3

Analysis of a Masonry Dam

9.4

Stability of Retaining Walls

9.5

Pressure on Retaining Walls Due to Granular Materials : Rankine’s Theory

9.6

Calculation of Pressures at the Base of Retaining Wall

9.7

Retaining Walls with Inclined Earth Slopes

9.8

Minimum Depth of Foundation

9.9

Summary

9.10 Answers to SAQs

9.1 INTRODUCTION When a mass of material, whether liquid or granular solid, is required to be maintained at a higher level on only one side, the dividing structure between the two is known as a ‘retaining wall’ when a solid is retained, or a ‘dam’ when a liquid is retained. Free Board

u/s Side

Retaining Wall

Dam H

Earth

h

Water

hf

d/s Side

(a)

(b) Figure 9.1

In Figure 9.1(a), water at a higher level, as in a reservoir, etc., is maintained at a height ‘h’ on the right hand (up-stream) side of the dam. In Figure 9.1(b) earth on the right hand side is retained at a level higher than on the left hand (down-stream) side, and it is maintained so by the wall. In either case the dam or wall acts as a retaining wall which prevents the material (liquid or granular solid) from slipping or flowing towards the left; and, therefore, it is under a horizontal pressure which tends to overturn it or move it towards the left. The retaining wall should be good enough to prevent any such event or fail under such a pressure. The weight of the wall and the horizontal pressure exerts pressure on the soil below foundation which should also be within safe limits.

Objectives After studying this unit, you should be able to • calculate the pressures acting on a retaining wall due to the retained fluid or granular solid, 213

Theory of Structures-I

•

design a wall which is strong enough to withstand these pressures,

•

proportion a wall so that no overturning or sliding takes place, and

•

calculate the soil pressures below the wall base and see that pressures are within safe limits.

9.2 PRESSURE ON RETAINING WALLS DUE TO LIQUIDS In Figure 9.2(a), the face of the wall in contact with the liquid is vertical, hence by the well known law of hydrostatics, the fluid pressure at any point A on the wall at a depth h is wh, where w is the density (or unit weight) of the liquid, and acts normal to the wall, (i.e. in the horizontal direction). The pressure diagram will be a triangle with zero pressure at top and a pressure wH at the bottom of the wall. The resultant force per unit length of wall will be the area of the pressure wH 2 triangle, i.e. P = and it will act at the CG of the triangle, i.e. at a height 2 H from the base of the wall. 3

h H

p = wh

Water

A

p = wH2/2

H/3 p = wH

(a)

h 90o H

α

p = wh

α

α

α

P = wH2/2 H/3

p = wH

(b) Figure 9.2

In Figure 9.2(b), the face of the wall is inclined at an angle α with the vertical, hence the pressure at a depth h is wh and acts normal to the wall, i.e. at an angle α with the horizontal. The pressure diagram is again triangular, and the resultant wH 2 inclined at α with the horizontal. The horizontal component of P will P= 2 wH 2 H be cos α acting at depth from the bottom. 2 3

9.3 ANALYSIS OF A MASONRY DAM 214

Figure 9.3 shows a masonry dam of trapezoidal cross section ABCD with water face BC vertical. The top width of the dam is ‘a’ and its base width ‘b’ (b > a). The height of the dam is H, and the density of the material of the dam (masonry) is ρ. Let the height of water retained be ‘h’ (which is less than H, due to the free-board kept over reservoir water level). a

A

Retaining Walls

B Free Board

H

h 2

P = wh /2 h/3

w e N b/2

M

D b/2

Toe

C Heel

b

w/b (Comp)

+ (i) (Tensile)

−

+

6We/b2 (Comp)

−

2ph/b2 (Tensile)

(ii) (Comp)

+ (iii)

(i)

Pressure distribution due to vertical load

(ii)

Pressure distribution due to moment W . e

(iii)

Pressure distribution due to water pressure P

Figure 9.3 : Pressure Distribution below Dam Base

The two forces acting on the dam is its self weight W which is W =ρ⋅

a+b H 2

and the horizontal pressure of water P as shown. The self weight W does not pass through the middle point M of the base BC, rather it passes through N. It can be easily seen that NC = x =

a 2 + ab + b 2 3(a + b)

. . . (9.1)

Thus, the eccentricity ‘e’ of the weight W is e = MN = MC − NC =

b a 2 + ab + b 2 − 2 3( a + b)

. . . (9.2)

Considering 1.0 m length of the dam, the pressures acting at the base of the dam can be divided into the following three types 215

Theory of Structures-I

(a)

Due to the vertical load W, it is ∴

(b)

p1 =

W (uniformly compressive) A

W W W = = A b ×1 b

[Figure 9.3 (i)]

Due to the moment W. e caused by eccentricity of W; ⎛b⎞ W .e.⎜ ⎟ M .y ⎝ 2 ⎠ = ± 6W . e p2 = ± =± I ⎛ 1 × b3 ⎞ b2 ⎜ ⎟ ⎜ 12 ⎟ ⎝ ⎠

. . . (9.3)

There will be compressive force at heel C and tensile at toe D, (Figure 9.3(ii)) as the point N is nearer to C. (c)

h Due to the moment P ⋅ caused by the horizontal water pressure 3 h acting at a height from the base 3 ∴

h⎞ b ⎛ P. ⎟ ⎜ 2 Ph My 3⎠ 2 p3 = ± =± ⎝ =± I ⎛ 1 × b3 ⎞ b2 ⎜ ⎟ ⎜ 12 ⎟ ⎝ ⎠

. . . (9.4)

It will cause compression at toe D and tension at heel C as shown in Figure 9.3(iii). The resultant pressure at any point is the algebraic sum of the pressures p1, p2 and p3 as calculated above. For the safety of the dam the following conditions have to be ensured : (a)

There should not be any tensile stress in the dam, as masonry is not expected to resist any tensile stress.

(b)

The total compressive stresses at any section (at any point) of the dam should not exceed the allowable compressive stress in masonry.

(c)

The maximum pressure below the foundation should not exceed the allowable bearing capacity of soil.

(d)

The dam should not fail due to lack of stability, i.e. by sliding or by overturning due to the horizontal water pressure.

This last point is discussed in the next section.

9.4 STABILITY OF RETAINING WALLS Sliding

216

As retaining walls are subjected to horizontal (as well as vertical loads) there will be a tendency of the wall to move or slide in the direction of the force. This is resisted by the frictional force acting at the base of the wall. If the coefficient of friction between the base and the soil is μ and W is the total vertical forces, then as the vertical reaction R = W, the frictional force

F = μ R = μ W.

Retaining Walls

Hence, for a condition of no sliding, total horizontal forces Ph < μ W FS =

μW Ph

. . . (9.5)

W

Ph

μR R

The ratio

μW is known as the factor of safety against sliding, and it should Ph

never be less than 1.0. Normally the factor of safety against sliding is taken as 2.0. Overturning

Another effect of the horizontal forces is its tendency to overturn the retaining wall about its toe D. The overturning moment is Ph . y , where y is the distance of resultant of the horizontal forces above the base. The stabilizing moment which is caused by the weight of the wall (or any material resting on the slopes of the wall) is given by W ⋅ z , where z = (b − x ) is the distance of the resultant from the toe D. For the stabilizing moment to be greater than overturning moment, we must W ⋅z have W ⋅ z > Ph ⋅ y and the factor of safety against overturning is Ph ⋅ y which should not be less than 1.0. Normally the safety factor is taken as 2.0. FS =

Wz Ph y

. . . (9.6) a

A

B

h

P

y

Toe W D

C x

z

Heel b

217

Theory of Structures-I

Figure 9.4

Example 9.1

A masonry dam 8 m high is 1.5 m wide at the top and 5 m wide at the base. It retains water to a depth of 7.5m. The water face of the dam is vertical. Find the maximum and minimum stresses at the base. The weight of the masonry is 22.4 kN/m3. Solution

Consider 1 m width of the dam (Figure 9.5). 1.5

A

B

8m

7.5 m p = 281.25

G W= 582.4 kN M

D 2.5 m

e

2.5 m N

C

2.5 m

1.782 m

5m

2

+ Compression

116.480 kN/m

(i) Pressure Diagram Due to W − 100.359 kN/mm2 + 100.359 kN/mm2 (ii) Pressure Diagram Due to Eccentricity ‘e’ − 168.75 kN/m2 + 168.75 kN/m2 (iii) Pressure Diagram Due to P D

C + 48.089 kN/m2

+ 184.871 kN/m2

(iv) Final Pressure Diagram

Figure 9.5

Weight of masonry, W =

1.5 + 5.0 × 8.0 × 22.4 = 582.4 kN 2

Its line of action meet the base DC at N here NC is NC = x =

218

a 2 + ab + b 2 1.52 + 1.5 × 5 + 52 = = 1.782 m 3(a + b) 3(1.5 + 5)

∴ Eccentricity, e = MN = MC – x = 2.5 – 1.782 = 0.718 m

Retaining Walls

The horizontal water pressure P is P=

Wh 2 10 × (7.5)2 = = 281.25 kN 2 2 7.5 = 2.5 m above the base DC. 3

and it acts at a height of ∴

(a)

Pressure due to weight W is p1 =

W 582.4 = = 116.48 kN/m2 5 b

and is uniform at the base DC. (b)

Pressure due to the eccentricity of weight W is p2 = ±

6 We b

2

=±

6 × 582.4 × 0.718 5

2

= ± 100.359 kN/mm 2

It varies from a tensile stress – 100.359 kN/m2 at D to a compressive Stress + 100.359 kN/m2 at C at base DC. (c)

Pressure due to the moment of the horizontal water pressure p is p3 = ±

2 ph b

2

=±

2 × 281.25 × 7.5 5

2

= ± 168.750 kN/m 2

which varies from a compressive stress at D to tensile at C. The final pressure diagram is shown in Figure 9.5(iv) as an algebraic sum of p1, p2, p3 and is 184.871 kN/m2 (comp.) at D to 48.089 kN/m2 (comp.) at C. Example 9.2

Find the factor of safety against (a) sliding and (b) overturning for the dam section in Example 9.1 (coefficient of friction between soil and dam base is 0.6). Solution

(a)

The vertical load is the weight (W) of the masonry dam which has been calculated as 582.4 kN per metre of the wall The frictional force at base is F = μ W or

F = 0.6 × 582.4 kN = 349.44 kN

The sliding force is the horizontal pressure P = 281.25 kN. Hence, the safety factory against sliding = (b)

349.44 = 1.243 281.25

The moment of the vertical load W about the toe D is W (b − x ) = 582.4 (5 – 1.782) = 1874.16 kNm

which is the stabilizing moment. The overturning moment due to water pressure is

h P ⋅ = 281.25 × 2.5 = 703.125 kNm 3 219

Theory of Structures-I

∴ The factor of safety against overturning =

1874.16 = 2.67 . 703.125

Example 9.3

Design the section of a trapezoidal masonry dam (with water face vertical) to impound water up to 29 m depth on the upstream side, with a free-board of 1 m. The maximum allowable pressure on base is 900 kN/m2. Assume no tension in masonry (weight of masonry = 23 kN/m3). Solution

(Refer Figure 9.6) Total height of dam = 29 + 1 = 30 m Let the top width

= a metre

Base width

= b metre

Considering one metre length of dam. a

A

B

1 m (Free Board)

30 m

29 m O

p

w= (a+b/2)H D

M

e

b/2 Toe

29/3 m N C

b/2

b

Heel

Figure 9.6

Weight of masonry, W = =

a+b ρ⋅H 2

a+b × 23 × 30 2

= 345 (a + b) kN Maximum water pressure P=

wh 2 10 × 292 = = 4205 kN 2 2

Maximum pressure at toe D pD =

6 We 2 Ph W − + = 900 kN/m 2 (maximum allowable pressure) 2 2 b b b

Minimum pressure at heel C pC =

W 6 We 2 Ph + − = 0 (for no tension) b b2 b2

Substituting the values of W, P and h in the above equation 220

345 (a + b) 6 × 345 (a + b) ⋅ e 2 × 4205 × 29 − + = 900 b b2 b2

. . . (A)

345 ( a + b) 6 × 345 (a + b) ⋅ e 2 × 4205 × 29 + − =0 b b2 b2

...

Retaining Walls

(B) Adding Eqs. (A) and (B), we get ⎛a + b⎞ 2 × 345 ⎜ ⎟ = 900, ⎝ b ⎠

giving

a+b = 1.304 b

or

a = 0.304 b

Subtracting Eq. (B) from Eq. (A) we get 2 × 6 × 345 ( a + b)

e b

2

−

4 × 4205 × 29 b2

= 900

– 4140 (a + b) e + 487780 = 900 b2

b2 + 4.6 (a + b) e – 542 = 0

or (C)

...

Also we know that e+ x= x =

where

b 2

[∵ MN + NC = MC =

DC ] 2

a 2 + ab + b 2 3(a + b)

. . . (D)

Substituting the value of a

∴

x =

(0.304 b) 2 + 0.304 b 2 + b 2 = 0.357 b 3(0.304 b + b)

e=

b − x = 0.5 b − 0.357 b = 0.143 b 2

Substituting the value of a and e in Eq. (C), we get b 2 + 4.6 (0.304 b + b) (0.143 b) − 542 = 0

giving

b = 17.17 m ,

say 17.2 m

and

a = 0.304 b

5.2 m (say)

,

Check for Stability

Assuming coefficient of friction μ = 0.6

W = 345 (a + b) = 345 (17 + 5.2) = 7659 kN ∴

μ W = 0.6 × 7659 = 4595.4 kN

P = 4205 kN/m2 ∴ Factor of safety against sliding =

4595.4 = 1.09 > 1 4205

and factor of safety against overturning

221

b⎞ ⎛ W ⎜e + ⎟ 2 ⎠ 7659 (0.143 + 0.5) × 17 = ⎝ = = 2.06 > 1 29 h 4205 × P⋅ 3 3

Theory of Structures-I

∴

OK.

9.5 PRESSURES ON RETAINING WALLS DUE TO GRANULAR MATERIALS – RANKINE’S THEORY Earth material (or any other granular material like sand, cement grains, etc.) when collected in a heap tends to adopt a slope, which is characteristics of the material and is known as its “angle of repose”. This is due to the internal friction between the grains of the material. A French engineer, Rankine, developed the theory of earth pressures in the 18th century. Rankine’s earth pressure equation gives the horizontal earth pressure on the vertical face of a retaining wall due to a level earth fill behind if the earth pressure (p) can be expressed as p = wh

where k =

1 − sin φ = kwh 1 + sin φ

. . . (9.7)

1 − sin φ is the Rankine’s earth pressure coefficient, 1 + sin φ

w = Unit weight of the material, h = Depth of the point at which pressure is sought, and φ = Angle of internal friction (or angle of repose) of the material.

This is shown in Figure 9.7 where the maximum pressure on the vertical face of the retaining wall AB at depth H is p = wH

1 − sin φ 1 + sin φ

. . . (9.8)

and the pressure diagram is a triangle. The total pressure is the area of the triangle and is given as P=

and acts at a depth

1 − sin φ 1 wH2 2 1 + sin φ

. . . (9.9)

H from the bottom 3 a A

h

p = wh . H

P=

G H/3

222

B b

wH .

1 − sin φ 1 + sin φ

1 − sin φ 1 + sin φ

1 − sin φ 1 wH 2 2 1 + sin φ

Retaining Walls

Figure 9.7

9.6 CALCULATION OF PRESSURES AT THE BASE OF RETAINING WALL In all these calculations we consider 1 m width of the wall. Here P is the total horizontal pressure acting on a retaining wall and W is the total weight of all vertical loads acting on it including its own weight. If the resultant of these forces meet the base at point E and O is the midpoint of the base then OE = e is the eccentricity of the resultant R. A

B

P

P

H W

h R D

E e

C

O

b

(a) Section of Wall E

D

C

O

1m D′

O′ Plan

C′

(b) Pressure Diagram Figure 9.8

Therefore, the base will be subjected to an eccentric load caused by a direct vertical load W and a moment M (= Ph ± W.e). The stress at any point distant x from the centre O will be given by p=

M .x W ± A I0

. . . (9.10)

where b is the width of base DC, and I0 is the moment of inertia of the base ⎛

rectangle CC′DD′ about the centre line OO′ ⎜⎜ Here I 0 = ⎝

1 × b3 b3 ⎞ = ⎟ 12 12 ⎟⎠

and A (= 1 × b = b) is area of the rectangle CC′DD′. The extreme pressure (maximum and minimum) acts at the point D and C where, x =

b . 2

Substituting these values in Eq. (9.10), we have 223

Theory of Structures-I

b W W 6M = ± 32 = ± 2 b b b b 12 M

pext

We have

pext =

. . . (9.11)

W 6We ± 2 b b

. . . (9.11a)

Masonry is assumed to be capable of taking compressive stresses but no tensile stresses. As tensile stresses will occur first at the point C (the point of smallest stress) if pc < 0, stress at C is tensile. 6W . e W ± < 0 for tensile stress. b b2

or

1<

or or

6e b

. . . (9.12)

writing it in the other way e >

b . 6

. . . (9.12a)

b to the left of O there will be 6 b tensile stresses at O. (Similarly, if the eccentricity is more than to the right of 6 O there will be tensile stresses at D.) Thus, we see that as long as the resultant lies within the middle third of the base there will not be any tensile stress in the section. This is the famous middle third rule. In other words if the eccentricity is more than

Example 9.4

A masonry retaining wall of trapezoidal section retains level earth 6 metres high. The retaining wall is 1 m wide at the top, determine the bottom width so that no tension is induced in the base. The unit weight of masonry is 23 kN/m3 and of soil 15 kN/m3. The angle of repose of the soil is 30o and the back face of the wall is vertical. 1m

B

A

G

6m

P W 2m

O D

E e

b/2

D′

x

C

ph (max)

b/2

O′

C′

Figure 9.9

Find the value of the maximum pressure at the base.

224

Solution The horizontal pressure ph at any depth h is given by

1 − sin φ 1 − sin 30o ph = w . h ⋅ = 15 ⋅ h 1 + sin φ 1 + sin 30o ph = 15 ×

Retaining Walls

1 h = 5 h (kN/m 2 ) 3

ph(max) = 5 × 6 = 30 kN/m2 (at h = 6 m) at base Considering 1 m width of the wall. Total horizontal pressure P =

30 × 6 6 = 90 kN acts at a height of = 2 m 3 2

from the bottom edge C. If the width of the base of the wall for no tension is ‘b’ then weight of 1 m ⎛1 + b ⎞ × 6 ⎟ kN = 69 (1 + b) kN. ⎝ 2 ⎠

width of wall 23 × 1 × ⎜

It acts through x where x is the distance of CG of this trapezium ABCD from line BC. 6 ⎛b − 1 ⎞ 1 × 6 × 0.5 + (b − 1) × × ⎜ + 1⎟ 2 ⎝ 3 ⎠ x = 6 1 × 6 + (b − 1) × 2

B

A1 A2 CG G1 x1 G2 x2 D x

b2 + b + 1 = 3(b + 1)

6m C

b

The weight line GE meets the wall base CD; at E such that EC =

1m

A

A x + A2 x2 x = 1 1 A1 + A2

2b 2 + 2b − 1 b2 + b + 1 , DE = b − EC = 3(b + 1) 3(b + 1)

Hence, eccentricity (e) of the weight, OE = DE − DO =

2b 2 + 2b − 1 b b 2 + b − 2 − = 3(b + 1) 2 6 (b + 1) h 3

The net bending moment at base, M = P ⋅ − W ⋅ OE = 90 × 2 − 69 (1 + b)

b2 + b − 2 = 180 − 11.5 (b 2 + b − 2) kNm 6 (1 + b)

b M ⋅ W 2 = W − 6M ∴ Minimum bending stress at C = − b × 1 1 × b3 b b2 12 =

69 (1 + b) ⎛ 1080 69 (b 2 + b − 2) ⎞ −⎜ − ⎟ ⎜ b2 ⎟ b b2 ⎝ ⎠

For no tension at base DC the above stress at C must be zero ∴

69 (1 + b) 1080 69 (b 2 + b − 2) = − b b2 b2

giving, or,

b (1 + b) = 15.652 – (b2 + b – 2) b2 + b – 8.826 = 0 225

Theory of Structures-I

Giving

b=

− 1 ± 6.025 = 2.513 m (taking the positive value) 2

Hence take the base width as 2.6 m. Maximum pressure at base at point D is =

69 (1 + 2.513) 6 × [180 − (2.5133 + 2.513 − 2) × 11.5 + 2.513 2.5132 2

= 96.46 + 96.46 = 192.92 kN/m

SAQ 1 (a)

In Example 9.4, find the factor of safety against sliding and overturning if the coefficient of friction between wall base and soil is 0.5.

(b)

A masonry retaining wall of trapezoidal section is 10 m high and has 1.5 m top width and 6.5 m bottom width. The earth face of the retaining wall has a better of 1 in 10, and the soil surface retained is level at the top of the wall. Find the maximum and minimum normal stresses at the base. Masonry weights 23 kN/m3 and soil 18 kN/m3. The angle of repose of the soil is 30o. 1.5 m

A

B

E

10 m

1m

4m D

G 6.5mm 6.5

Figure 9.10

226

F

C

9.7 RETAINING WALLS WITH INCLINED EARTH SLOPES

Retaining Walls

Sometimes the earth retained behind a retaining wall is not horizontal but sloping at an angle α with it (Figure 9.11). It has been proved that in such cases the pressure on the retaining wall at a depth x is given by p x = w . x cos α

where

cos α − cos α +

cos 2 α − cos 2 φ 2

. . . (9.13)

2

cos α − cos φ

w = Density of the soil, and φ = Angle of internal friction (or angle of repose) of soil.

A

α

B

Pv α

D

h

P Ph

C

Figure 9 .11

It acts at an angle α with the horizontal (i.e. parallel to the earth slope). The total pressure P acting on the surface BC of the wall will be P=

⎡ cos α − Wh 2 ⎢ cos α 2 ⎢ cos α + ⎣

P acts at a height

cos 2 α − cos 2 φ ⎤ ⎥ 2 2 ⎥ cos α − cos φ ⎦

. . . (9.14)

h from the base DC of the retaining wall. 3

[Note : It can be seen that if we put α = 0 in Eqs. (9.13) and (9.14), it reduces to the formulae which are Eqs. (9.8) and (9.9) for the case of soil with horizontal surface.] Example 9.5

A masonry retaining wall of trapezoidal section and earth side face vertical has a top width of 1.5 m and base width of 3.5 m. The wall is 6 m high and retains earth at a slope of 1 vertical to 2 horizontal. The weight of 3 1 repose is 30o. soil retained is 18 kN/m and the angle of 1.5 m 2 Find the maximum and minimum pressure α A intensities at the base of the wall. Weight of B masonry is 23 kN/m3. x

Pv

P

6m

α Ph

W C

D 3.5 m

227

Theory of Structures-I

C′

D′

108.291

133.33

(a) Retaining Wall

(b) Pressure Diagram Figure 9.12

Solution

Considering 1 m width of wall (Figure 9.12). tan α =

1 2

α = 26.57o and cos α = 0.8944

∴

φ = 30o ∴ cos φ = 0.866 ∴ Coefficient of active earth pressure K a = cos α

cos α −

cos 2 α − cos 2 φ

cos α +

cos 2 α − cos 2 φ

= 0.5366 ∴

P = Ka

Wh 2 62 = 0.5366 × 18 × = 173.87 kN 2 2

which is inclined at an angle α = 26.57o with horizontal. Horizontal component of P, Ph = P cos α = 173.87 cos 26.57o = 155.5 kN Vertical component of P; Pv = P sin α = 173.87 sin 26.57o =77.7 kN and P acts at a distance

h 6 = = 2 m above the base. 3 3

Weight of masonry wall (per m width), W =

a+b 1.5 + 3.5 ⋅h⋅ρ= × 6 × 23 = 345 kN 2 2

This acts at a distance of x from C, where ∴ 228

x =

a 2 + ab + b 2 = 1.317 m 3(a + b)

Total pressure acting at point C of the base will be

⎛b ⎞ ⎛b⎞ 6 W ⎜ − x ⎟ 6 Pv ⎜ ⎟ P W ⎝2 ⎠ + ⎝ 2 ⎠ − 2 Ph × h pc = + v + 2 2 b b b2 b b =

Retaining Walls

345 77.7 6 × 345(1.75 − 1.317) 6 × 77.7 × 1.75 2 × 155.5 × 6 + + + − 3.5 3.5 (3.5) 2 (3.5) 2 (3.5)2

= 108.21 kN/m2 And total pressure acting at point D of the base will be

pD

P W = + v − b b

⎛b ⎞ ⎛b⎞ 6 W ⎜ − x ⎟ 6 ( Pv ) ⎜ ⎟ 2 ⎝ ⎠ − ⎝ 2 ⎠ + 2 Ph . h b2 b2 b2

= 133.33 kN/m2

9.8 MINIMUM DEPTH OF FOUNDATION Consider a column carrying a load P to be transmitted to the soil through a footing of area (A) as in Figure 9.13. P

h

p1 M

P3 p2

N

Figure 9.13

Hence, the vertical pressure on soil just below the footing is p1 =

P A

. . . (9.15)

So on a soil element ‘M’ just below the footing the horizontal pressure p2 caused by the footing is p 2 = p1

1 − sin φ 1 + sin φ

. . . (9.16)

On a wedge ‘N’ just outside the footing this horizontal pressure p2 will cause a vertical pressure p3 such that p 3 = p2

1 − sin φ 1 + sin φ

. . . (9.17)

Substituting Eq. (9.16) in Eq. (9.17), we get 2

⎛ 1 − sin φ ⎞ P ⎛ 1 − sin φ ⎞ p 3 = p1 ⎜ ⎟ = ⎜ ⎟ A ⎝ 1 + sin φ ⎠ ⎝ 1 + sin φ ⎠

2

We know that the vertical pressure p3 = wh, where w is the unit weight of soil and h is the depth of the footing.

229

Theory of Structures-I

∴

wh =

P ⎛ 1 − sin φ ⎞ ⎜ ⎟ A ⎝ 1 + sin φ ⎠

2

or the minimum depth of footing for equilibrium will be P ⎛ 1 − sin φ ⎞ h= ⎜ ⎟ wA ⎝ 1 + sin φ ⎠

2

SAQ 2 A column carries a load of 1200 kN on a masonry footing which is 2 × 2 m in area. Find the minimum foundation depth necessary if the unit weight of soil is 18 kN/m3 and has an angle of repose of 35o (assume self weight of footing as 10% of the column load).

9.9 SUMMARY A wall used to retain some material on one or both sides of it can be termed as retaining wall. Generally, it is used to retain soil at two different levels on either side of the wall. Moreover, the materials to be retained on either side may be different, e.g. the wall of a swimming pool retains soil on one side and water on the other side. When the material retained by it is a fluid say water, it is called a dam. Gravity wall, cantilever wall, and buttress wall are some of the common types of retaining walls. In case of Gravity Wall, the self-weight of the structure provides stability against the pressure of retained earth. Cantilever wall consists of the vertical arm, which retains the earth and is held in position by the base slab. In this case, the weight of fill on top of the heel, in addition to the weight of the wall, contributes to the stability of the structure. The process of design of a retaining wall commences with preliminary proportioning of the wall, and then the design is checked against the stability requirements of wall and is revised if required.

9.10 ANSWERS TO SAQs SAQ 1

(a)

(i)

(ii)

Maximum vertical reaction = 242.4 kN ∴

Frictional force preventing sliding = 121.2 kN

∴

Factor of safety against sliding

242.4 = 1.35 . 121.2

Maximum overturning moment about toe = 180 kN. Maximum stabilizing moment about the same point = 191.55 kNm. ∴

(b) 230

Factor of safety against overturning

191.55 = 1.06 . 180

From Figure 9.14 we have (considering 1 m length of wall)

W1 =

1 × 4 × 10 × 23 = 460 kN 2

Retaining Walls

W2 = 1.5 × 10 × 23 = 345 kN W3 =

1 × 1 × 10 × 23 = 115 kN 2

W4 =

1 × 10 × 1 × 18 = 90 kN 2

x1 = 2.5 +

4 23 = m 3 6

x2 = 1.0 +

1.5 7 = m 2 4

x3 =

2 m 3

x4 =

1 m 3

∑ W = 1010 kN 1m A

1.5 m B W2

W4 x4 Earth Pressure

X2

10 m W1

X1

W3

PH =

X3 10/3 m

4m D 3.25 m

wH 2 ⎛ 1 − sin φ ⎞ ⎜ ⎟ 2 ⎝ 1 + sin φ ⎠

O

e = 0.8 6.5mm 6.5

F

C

x3 = 2.45 m

(a) Location of Retaining Wall

182 kN/m

2

128 kN/m

2

(b) Foundation Pressure Diagram Figure 9.14

∴

∑ Wi xi = x = ∑ Wi

e=

460 ×

2 7 2 1 + 345 × + 115 × + 90 × 3 4 3 3 = 2.45 m 460 + 345 + 115 + 90

6.5 − 2.45 = 3.25 − 2.45 = 0.8 m 2

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Theory of Structures-I

PH = h=

∴ ∴

W H 2 ⎛ 1 − sin φ ⎞ 18 × 102 1 − sin 30o × = 300 kN ⎜ ⎟= 2 ⎝ 1 + sin φ ⎠ 2 1 + sin 30o

10 m 3

M = P h − W e = 300 ×

10 − 1010 × 0.8 = 192 kN/m 3

Pressure below base are

At toe,

pD =

6 M 1010 6 × 192 W + = + = 182 kN/m 2 2 2 b 6.5 b (6.5)

At heel,

pC =

6 M 1010 6 × 192 W − = − = 128 kN/m 2 2 2 b 6.5 b (6.5)

SAQ 2

Allowing 10% for self weight of footing. P Minimum depth of foundation, h = wA 1200 × 1.1 h= 18 × (2 × 2)

232

⎛ 1 − sin φ ⎞ ⎜ ⎟ ⎝ 1 + sin φ ⎠ 2

2

⎛ 1 − sin 35o ⎞ ⎜ ⎟ = 1.35 m ⎜ 1 + sin 35o ⎟ ⎝ ⎠

Retaining Walls

FURTHER READING Gupta S. P., Pandit, G. S., Gupta, R. (1999), Theory of Structures, Vol. I and II, Tata McGraw-Hill Publishing Company Limited, New Delhi. Negi, L. S. and Jangid, R. S. (1997), Structural Analysis, Tata McGraw-Hill Publishing Company Limited, New Delhi. Punmia, B. C., Jain, Ashok and Jain, Arun (1998), Design of Steel Structures, Laxmi Publications, New Delhi. Punmia, B. C., Jain, Ashok and Jain, Arun (2004), Theory of Structures, Laxmi Publications, New Delhi. Ramchandra, Dr., Ratwani, M. M. (1991), Design of Steel Structures, Standard Book House, New Delhi.

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Retaining Walls

THEORY OF STRUCTURES-I In any construction project, all components are designed properly before the execution of the project. For designing the structural components, they are being analysed first to calculate the bending moment, shear force and their variations on the basis of given load pattern. This forms the subject matter of this course. In this course, you are introduced to the concepts of analysis and design of structures. This course consists of nine units. The first three units discuss some of the standard methods of analysis of structural components. The next five units explain the method of structural design as applied to steel structures. The last unit describes the design of retaining structures, which retain the load of earthfill or water (in a reservoir). Unit 1 deals with the analysis of moving load systems, and the use of influence line diagrams. The concept of influence lines is very useful in design of all such structures which carry moving loads like bridges, flyovers, crane girder, etc. Unit 2 explains the methods of analysis of statically indeterminate structures, i.e. structures which cannot be analysed by the principles of statics only which you have learnt in earlier courses, namely Applied Mechanics and Strength of Materials. Unit 3 introduces you to the powerful tool of Moment Distribution Method, which is quite convenient for the analysis of some statically indeterminate structures. Towards the end, the unit also explains portal frames. Unit 4 deals with the design of joints, which are mainly comprise riveted or welded joints. Unit 5 deals with the design of steel members subjected to tensile load only. It explains the loads and stresses considered for the design of tension members. Unit 6 explains the design of steel members subjected to compressive forces, e.g. columns and struts. While explaining permissible stresses in compression members, the unit presents the design procedure for different types of compression member including compound compression members. Unit 7 presents the methods for design of steel members subjected to bending (flexural) forces, e.g. beams, girders, purlins, etc. The unit explains the classical theory of bending. 235

Theory of Structures-I

In Unit 8, you will find the design of steel roof trusses, which is one of the most common forms of structure, normally employed to cover large spaces. Finally, Unit 9 discusses the designs of retaining walls and dams, which are mainly massive structures used to sustain large horizontal forces due to the retained materials (earth or water). The Self-Assessment Questions (SAQs), given in each unit, are intended to help you in checking your own progress. You should study the text carefully, and then try to solve the SAQs and verify your answers with those given at the end of each unit. This will definitely develop your confidence. At the end, we wish you all the best for your all future educational endeavours.

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