Inclined face dam

In fig.10.a shows section of dam. Fig.10.b shows water pressure distribution diagram. Fig.10.c shows plan for unit width.

Let,

a = top width of dam;

b = bottom width of dam;

H = height of dam;

h = height of water level;

H = height of dam;

ρ_m = density of dam masonry;

ρ_w = density of water;

W = self-weight of dam;

P = force due to water applied on inclined face (AE);

Pcosθ = horizontal component of water force;

Psinθ = vertical component of water force;

R = resultant force of water force and self-weight;

e = eccentricity due to resultant force;

$\bar{x}=centroidal\: distance\: form\: outer\: edge\: of\: base$

Now, we know that resultant stress

$\sigma_{r}=\sigma_{0} \pm \sigma_{b}$

First discuss about direct stress σ₀ which is developed by total downward force i.e. self-weight of dam and water on inclined face ‘AE’ of dam.

We know that

$\sigma_{0}=\frac {P}{A}=\frac {W+ Psin \theta }{A}$

Where,

A = Base area of dam;

W = Self-weight;

W = Volume X density of dam masonry

Here volume is found for unit length of dam.

Volume = Cross sectional area X unit length

$Volume=\left(\frac{a+b}{2} \times H \right) \times 1$

$\therefore W =\left(\frac{a+b}{2} \times H \right) \times 1 \times \rho_{m}$

$\therefore W =\frac{a+b}{2} \times H \times \rho_{m}$

A = Base area A = bottom width of dam X unit length; A = b X 1 = b

P = water force;

P = Area of water pressure diagram (as shown in fig.10.b) X force applied on area

$P = \left(\frac {1}{2} \times \rho_{w} \times h \right) \times \left( h \times 1 \right)$

$From\: \Delta AFE$

$cos \theta = \frac{AF}{AE}=\frac{h}{AE}$

$\therefore AE = \frac{h}{cos \theta}$

$\therefore P = \frac{\rho_{w}h^{2}}{2 cos \theta}$
h = height of water level;

$P = \frac {\rho_{w}h^{2}}{2}$

h = height of water level;

Now, discuss about bending stress σ_b which is developed due to total downward force i.e. self-weight and water on inclined face ‘AE’ of dam.

We know that
$\sigma_{b} =\frac{M}{Z}=\frac{\left( W+P sin \theta \right) \cdot e}{Z}$

Here,

W = self weight;
e = eccentricity due to resultant force;

$e =AD-\frac {b}{2}=\frac {W \cdot e}{Z}$

Where, AD = AC + CD

$AC = \bar{x}$

centroidal distance form vertical face contact with water

To find CD take moment at D

$\sum M_{D} = 0$

$\left( Pcos \theta \times \frac{h}{3} \right)-\left( W \times CD \right)=0$

$CD = \frac {Pcos \theta \cdot h}{3W}$

$Z = \frac {I_{yy}}{y_{max}}=\frac {\frac {db^{3}}{12}}{b/2}=\frac {db^{3}}{12} \times \frac {2}{b}$

$Z =\frac {db^{2}}{6}$

Here ‘d’ is unit length = 1

$\therefore Z =\frac {b^{2}}{6}$

By putting all these values we can find min and max stresses.

Vinayak Hulwane

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Inclined face dam

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Inclined face dam

Vinayak Hulwane

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MSBTE

University

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