written 5.3 years ago by |

- 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 (B) at bottom, i.e., the point at which maximum hydrostatic pressure acts.

**When reservoir is empty**

- When reservoir is empty, only self weight of the dam acts at a distance of B/3 from the 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 stabilising 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

$\cfrac{P_{max}}{P_{min}}=\cfrac{\sum Y}{B} \left[ 1 \pm \cfrac{6e}{B}\right ]$ Here, $\sum Y=W$ and $e=B/6$

$\cfrac{P_{max}}{P_{min}}=\cfrac{W}{B} \left[ 1 \pm \cfrac{6 \times B/6}{B}\right ]$

$P_{max}=\cfrac{2W}{B}$

$P_{min}=0$

Hence, the maximum vertical stress is 2W/B at the heel, since the resultanat is near the heel and the vertical stress at the toe will be zero.

**When the reservoir is full**

The base width is governed by

(i) The resultant of all forces, i.e., P wan U passed through the outermost middle third point.

(ii) The dam is safe in sliding.

First condition,

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

$W \left( \cfrac{B}{3} \right)- U \left( \cfrac{B}{3} \right)-\cfrac{pH}{3}=R \times 0$

$(W-U) \left( \cfrac{B}{3} \right)-\cfrac{pH}{3}= 0$ ----------(1)

But $W=\cfrac{1}{2} \times B \times H \times S_{c} \times r_{w}$

$S_{c}=$specific gravity

$r_{w}=$unit weight of water=9.81kN/m^{3}

$U=\cfrac{1}{2} \cdot C \cdot r_{w} \cdot H \cdot B$

If C is 1, $B=\cfrac{H}{\mu (S_{c}-1)}$

If C is 0, that is if there is no uplift pressure, $B \ge \cfrac{H}{\mu \cdot S_{c}}$

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