## LESSON 31. Stability Analysis of Gravity Dams: Stresses

31.1 Introduction:  In this lesson we will derive expressions for the base pressure and stresses developed in a gravity dam.

Fig. 31.1.

In the above figure let R be the resultant force cutting the base at a distance $\bar x$ from the toe of the dam. The components of R in x and y direction are obtained as,

${R_x}={F_H}$                                (31.1)

${R_y}=W + {F_V} - U$        (31.2)

$\bar x$ is obtained as,

$\bar x={{{M_{toe}}} \over {R{}_y}}$                        (31.3)

The eccentricity of from the centre of the base is given by, $e = {b {\left/2} - \bar x$   .The nominal stress at any point on the base is the sum of direct stress and bending stress.

The direct stress is always compressive and given by,

${\sigma _{cc}} = {{{R_y}} \over b}$      [per unit length of the dam]     (31.4)

Bending moment about the centre of the base is, $M={R_y} \times e$  . Corresponding bending stress at a distance x from the centre of the base is given by,

${\sigma _{bc}}=\pm {{Mx} \over I}$            (31.5)

Where, I is the second moment of area of the base per unit length of the dam. I is given by,

$I={{1 \times {b^3}} \over {12}}={{{b^3}} \over {12}}$          (31.6)

Therefore total normal stress at a distance x from the centre of the base is,

${p_n}={\sigma _{cc}} + {\sigma _{bc}}={{{R_y}} \over b} \pm {{Mx} \over I} = {{{R_y}} \over b} \pm {{12Mx} \over {{b^3}}}$         (31.7)

The resulting moment produces tension at heel and compression at toe.

Therefore,

${p_{nheel}}={{{R_y}} \over b} - {{12\left( {{R_y} \times e} \right)\left( {{b {\left/2}})} \over {{b^3}}}={{{R_y}} \over b}\left({1-{{6e} \over b}} \right)$             (31.8)

${p_{toe}}={{{R_y}} \over b} + {{12\left( {{R_y} \times e} \right)\left( {{b{\left/2}})} \over {{b^3}}}={{{R_y}} \over b}\left( {1 + {{6e} \over b}} \right)$            (31.9)

The distributions of normal stress at the base of the dam for three different situations are shown in Figure 31.2.

Fig. 31.2.