It can be carried out by following methods:

1. Gravity method – Graphical method. Analytical method.

2. Trail load twist method.

3. Slab analogy method.

4. Lattice analogy method.

5. Experimental methods – Direct method. Indirect method.

1. Gravity method:

a] Graphical method.

b] Analytical method.

a] Graphical method.

• In this method, the dam is divided into sections – horizontal sections 1-1, 2-2, 3-3 etc. at suitable interval or at places where slope changes.

• For each section, sum of all horizontal and vertical forces. Acting above the section are calculated 4 and their line of action is located graphically.

• This is done for each section and finally its line of action is drawn joining all points at which resultant cuts each sections.

• The line of resultant should lie in the middle third portion, for no tension to develop.

• This method is used for both reservoir full and reservoir empty.

b] Analytical method: The procedure is

(1) Consider unit length of the dam, (2) Calculate all the vertical forces acting on the dam, viz, self weight of the dam, weight of water acting on the inclined face, uplift pressure and inertia forces due to vertical acceleration. Compute their algebraic sum $(\sum V)$, (3) Calculate all the horizontal forces acting on the dam and their algebraic sum $(\sum H)$ and the horizontal pressure due to hydrodynamic pressure, (4) Calculate the lever arm of all these forces form the toe, (5) Calculate the sum of overturning moments $(\sum M_o$ and the sum of righting moments $(\sum M_R)$, at the toe, (6) Calculate the algebraic sum of all those moments $(\sum M = \sum M_R – M_o)$, (7) Calculate the location of resultant force by determining its

Distance from the toe, $\bar x = \frac{\sum M}{ \sum V}$, (8) Calculate the eccentricity e of the resultant R from the centre of the base width b, i.e. $(b/2 - \bar x)$. It must be < b/6 for no tension to develop. (9) Calculate the normal stress at the toe $p_n = \frac{\sum V}{b} ( 1 + \frac{6e}{b})$, (10) Calculate the normal stress at the heel $p_n = \frac{\sum V}{b} (1 - \frac{6e}{b})$. Normal compressive stresses are sometimes calculated ignoring uplift. (11) Calculate the principal stress $(\sigma)$ ans shear stresses $(\pi)$ at the toe and heel.

$\sigma_{toe} = P_n \ sec^2 \ \theta – (P – P_e) \ tan^2 \ \theta$ considering hydrodynamic pressure $(P_e)$ exerted by tail water during earthquake.

$\sigma_{heel} = P_n \ sec^2 \ \theta – (P – P_e) \ tan^2 \ \theta$

$\sigma_{toe} = (P_n – P) \ tan \ \theta$

$\sigma_{heel} = - (P_n – P) tan \ \theta$

$\sigma_{toe} = P_n \ tan \ \theta$. If tail water is neglected.

$\sigma_{toe} = [P_n – (P – P_e)] \ tan \ \theta$, considering hydrodynamic pressure due to earthquake.

$\sigma_{heel} = [P – (P + P_e)] \ tan \ \theta$, considering hydrodynamic pressure due to earthquake.

(12) Calculate factor of safety against sliding = $\mu \ \frac{\sum V}{\sum H}$ it must be greater than unity, (13) Calculate factor of safety against overturning = $\frac{\sum M_B}{\sum M_b}$, (14) Calculate shear friction factor = $\frac{(\mu \sum V + b4)}{\sum H}$ Where, $\sum V$ is the net vertical force, inclusive of uplift. It must be greater than 3 but less than 5. The procedure is followed for reservoir full and reservoir empty conditions.

2. Trail load twist method.

• In gravity method, the assumption is made that dam behaves as vertical cantilevers independent of each other.

• The assumption is true if dam is constructed in a V – Shaped valley and joints are not keyed and grouted.

• If the joints are keyed and grouted, the dam will behave as a monolithic structure and gravity method will not be applicable.

• Trail load twist method is based on the assumption that horizontal elements of dam will behave as a beam and will transfer the water pressure acting upon it to the abutments by beam action.

3. Slab Analogy Method.

• According to this method, dam is considered as a slab.

• Analysis is carried out is similar to the calculation of stresses in bridge slab.

4. Lattice Analogy Method.

• According to this method, the dam is considered to be composed of diagonally braced square frames.

• The analysis is simpler than slab analogy.

5. Experimental methods.

a] Direct method.

b] Indirect method.

a] Direct method.

• Three dimensional models are made of materials of the prototype.

• Models are located in similar manner as the prototype.

• Models are acted upon various pressures proportional to those of prototype.

• Deflections are measured in the model to pressure and then it is correlated with the prototype.

b] Indirect method.

Photo elastic method and methods of magnetic and electrical analogy are usually considered.

ELEMENTARY PROFILE OF A GRAVITY DAM.

• Elementary profile of a dam subjected only to external water pressure on upstream side, will be right angled triangle, having zero width at water level and a base width ($\beta$) at bottom i.e. the point at which maximum hydrostatic pressure acts.

When reservoir is empty.

• When reservoir is empty only self wt of dam acts at distance B/3 from heel.

• This is the maximum possible innermost position of the resultant for no tension to develop.

• Hence such a line of action of W is most ideal as it gives maximum possible stabilizing moment about the toe without causing tension at toe, when reservoir is empty.

• The vertical stress distribution at the base, when reservoir is empty is given as

$P_{max/min} = \frac{\sum V}{B} [ 1 \pm \frac{6e}{B}]$

Here, $\sum V = w$

e = B/6

$\therefore$ $P_{max/min} = \frac{W}{B} \ [ 1 \pm \frac{6.B}{B.6}]$

$\therefore$ $P_{max} = \frac{2W}{B}$

$\therefore$ $P_{max} = 0$

Hence, maximum vertical stress equal to $\frac{2W}{B}$ at eh heel. ($\because$ resultant is near the heel) as vertical stress at toe will be zero.

When reservoir is full,

The base width is governed by

1] The resultant of all forces, i.e. P, Wan U passed through the outermost middle third point.

2] The dam is safe in sliding.

First condition.

Taking moments of all forces about the middle third point, we get

$W (\frac{B}{3}) – U (\frac{B}{3}) – P \frac{H}{3} = R \times 0$

Or $(w – u) \frac{B}{3} - \frac{PH}{3} = 0$ $\rightarrow$ [1]

But $w = ½ \times B \times H \times S_c \times r_w$

Sc = SP.gravity

Rw = unit weight of water

= 9.8 $KN/m^3$

$U = (\frac{1}{2} . c . rw . H) B$

If c is 1,

$B = \frac{H}{ \mu (Sc – 1)}$

If C is 0, that is if there is no uplift pressure,

$B \geq \frac{H}{ \mu . Sc}$

The value B chosen should be greater of the two values from above two equations.