Introduction: In this lesson we will derive expressions for the base pressure and stresses developed in a gravity dam.
Module 5 Lesson 31 Fig.31.1
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.
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