1
1.0kviews
Concrete Gravity Examples
1 Answer
1
20views

Example A concrete dam can be assumed to be trapezoidal in section having a top width of 2 m and bottom width of 10m, Its height is 12m and the upstream face has a batter of 0.9. Give an analysis of the stability condition assuming no freeboard allowance but allowing for uplift pressures. Assume uplift intensity factor as 100%. Also determine the compressive stress at the toe and the heel and major principal and shear stress developed at the toe. Assume weight of concrete to be 2.4$t/m^3$ weight of water

enter image description here enter image description here

Vertical weight V=W-U = 178 - 60 =118 t

M=1123-688 = +435t.m

Factor of safety against overturning =+M/-M =1123/688 =1.63 >1.5 hence,safe

Factor of safety against sliding =$\mu \sum V / \sum H$ =(0.7 x 118)/72 =1.15>1, hence safe

Shear friction factor =$(\mu v +Bq)/ \sum H$

=(0.7 x 118 +10 x 140)/72 = 20.59>5, hence safe.

Stress analysis

Position of resultant $\bar x =\sum M/ \sum V =435/118 = 3.69m$

Eccentricity, $e=B/2 - \bar x =10.0/2 - 3.69 =1.31m$

which is <$\frac{b}{6}\lt\frac{10}{6} \lt1.67, hence safe Compressive stress at toe $P_n =V/B(1+6e/B)</p>

= $118/10(1 +(6 * 1.31/10) = 21.07 t/m^2$

Compressive stress at heel $P'_n =\sum V/B(1-6e/B) = 118/10(1-6 * 1.31/10) = 2.53t/m^2$

$\tan\alpha =(0.9)/12.0 =0.075$

$\tan^2\alpha = (0.075)^2 = 0.006$

$\sec^2\alpha = 1+\tan^2 = 1+0.006 = 1.006$

$\tan\beta = 7.1/12=0.59$

$\sec^2\beta =1+ \tan^2\beta = 1+(0.59)^2 =1.35$

Principal stress at toe, $\sigma =P_n\sec^2\beta - P'\tan^2\beta$,but P' is zero there being no tail water

=$21.07 * 1.35 = 28.44 t/m^2$

Shear stress at toe, $\tau_0 = P_n\tan\beta - P'\tan\beta$ ,but P' is zero

=$21.07 * 0.59 = 12.43t/m^2$

Principal stress at heel, $\sigma g =P_n' \sec^2\alpha - P\tan^2\alpha$

=2.53 x 1.006 - 1 x 12 x 0.006 = 2.47 $t/m^2$

Shear stress at heel, $\tau_0h = -(P'_n - P)\tan\alpha$

=-(2.53 - 12.0)x 0.075 = +0.71 $t/m^2$

Please log in to add an answer.