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Concrete Gravity Examples
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Example

A concrete gravity dam had maximum reservoir level 150.0m, base level of dam 100.0m, tail water elevation 110.0m, base width of dam 40m, location of drainage gallery 10m from u/s face which may be assumed as vertical. Compute hydrostatic thrust and uplift force per metre length of dam at its base level. Assume 50% reduction in net seepage head at the location of the drainage gallery.

Solution Consider 1m length of dam enter image description here

Free board =5% of dam height = 0.05 x 50 = 2.5 say 3m

Dam Height = 50+3 = 53m

Top width = 0.14H or 0.55 $H^1{}^/{}^2$, where H is height of the dam

=0.14 x 53 = 7.5 or 0.55 $\sqrt 53 1/ 2$ = 4m, adopt 7.5m

enter image description here

Uplift pressure at drainage with 50% reduction =10.0 +1/2(50.0-10.0) = 30t

Uplift pressure $U_1$ = 20 x 10/2 = 100t = (30 + 10 X 2/3) = 36.67 3667t.m.

Uplift pressure $U_2$ = 10 x 30 x 1 = 300t (30+10/2) = 35 10500t.m.

Uplift pressure $U_3$= 10 x 30 x 1= 300t (30/2) = 15 4500t.m.

Uplift pressure $U_4$= 20 x 30/2 x 1=300t (30 x 2/3) = 20 6000t.m.

Uplift pressure $U= 1000t$

=$M_2$ = -24,667t.m.

=$V = W-U = 2671 - 1000 =1671t, M= M-M_2 = 71,031 - 24,667 = 46364t.m.$

Water pressure, u/s face $P'=\frac{wh^2}{2} = \frac{(1 * 50 * 50)}{2} = 1250 \frac{50}{3} =(-)20,833$

d/s face, $P' =\frac{wh^2}{2} = 1 * 10 * \frac{10}{2} = 50$

=$\frac{10}{3} + 167$

Pressure acting downstream = 1250 - 50 = 1200, $M_3 = (-)20833 + 167$

=(-)20,666

Net moment = 46,364 - 20,666 = 25,698t.m.

Position of resultant from toe, $\bar x =\frac{M}{V} = \frac{25698}{1671} = 15.38m$

Eccentricity, $e=\frac{B}{2}-\bar x = \frac{40}{2} - 15.38m = 4.62m$

Normal compressive stress at toe, $P_n =\frac{V}{B}(1+\frac{6e}{B})$

=$\frac{1671}{40}(1+ 6*\frac{4.62}{40}) = 70.73t/m^2$

Normal compressive stress at heel $=\frac{V}{B}(1-\frac{6e}{B})$

=$\frac{1671}{40}(1-\frac{6*4.62}{40}) = 12.82t/m^2$

Maximum principal stress at toe, $\sigma = P_n\sec^2\alpha - P'\tan^2\alpha$, here $P' = 10$

$\tan\alpha = 40/50 and \sec^2\alpha =1.64$

Therefore, $\sigma =70.73 * 1.64 - 10(40/50)^2 = 110t/m^2$

Intensity of shear stress on a horizontal plane near toe,

$\tau_0 = (P_n - P')\tan\alpha$

$\tau_0 = (70.73-10)40/50 =49t/m^2$

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