Retaining walls without surcharged
In above fig.11.a shows
section of retaining wall. Fig.11.b shows earth pressure distribution diagram.
Fig.11.c shows plan for unit width.
Let,
a = top width of retaining
wall;
b = bottom width of retaining
wall;
H = height of retaining
wall;
ρs = density
of earth or soil;
ρm = density
of retaining wall masonry;
W = self-weight of retaining
wall;
P = force due to earth;
R = resultant force of earth
force and self-weight;
e = eccentricity due to
resultant force;
Now,
we know that resultant stress

Where, AD = AC + CD

centroidal distance form vertical face of retaining wall.
To find CD take moment at D

-\left(%20W%20\times%20CD%20\right)=0)

Where,
P = earth force;
P = Area of water pressure diagram (as shown in fig.11.b) X force applied on area X Angle of repose
%20\times%20\left(%20h%20\times%201%20\right)\times \left( \frac{1-sin \theta}{1+sin \theta} \right))
)


Here ‘d’ is unit length = 1

By putting all these values we can find min and max stresses.
First discuss about direct stress σ0 which is developed by total downward force i.e. self-weight of retaining wall.
We know that
(Note=Total downward force = W)
Where,
A = Base area of retaining wall;
W = Self-weight;
W = Volume X density of retaining wall masonry
Here volume is found for unit length of retaining wall.
Volume = Cross sectional area X unit length
A = Base area A = bottom width of retaining wall X unit length; A = b X 1 = b
Now, discuss about bending stress σb which is developed by earth pressure.
We know that
Here,
W = self weight;
e = eccentricity due to resultant force;
e = eccentricity due to resultant force;
Where, AD = AC + CD
centroidal distance form vertical face of retaining wall.
To find CD take moment at D
Where,
P = earth force;
P = Area of water pressure diagram (as shown in fig.11.b) X force applied on area X Angle of repose
h = height of earth level;
W = self-weight of retaining wall;
Here ‘d’ is unit length = 1
By putting all these values we can find min and max stresses.
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