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Weight of the dam acting vertically. This is equal to (Hb/2) x w x p (where w is the weight of water per unit) acting at a distance of b/3 from the upstream vertical face.

**THE FORCES THAT ARE ACTING ON THE DAM ARE**

Weight
of the dam acting vertically. This is
equal to (Hb/2) x w x p (where w is the weight of water per unit) acting at a
distance of b/3 from the upstream vertical face.

Horizontal
water pressure:

This
is equal to wH^{2}/2 acting at a height of H/3 above the base of the
dam.

Net
Moment = Hb/2 x wp x 2b/3 x wH2/2 x H/3

The
vertical component of the resultant is HB/2 x wp

If
the resultant cuts the base at the outer mid third point, the lever arm of the
resultant force about the Toe has to be ' b'/3

Thus
the equation giving the relationship is

[ Hb/2 wp x 2b/3 - wH2/H x
H/3 ] / [ Hb/2 x wp ]

b/2

This
is reduced to

2Hb^{2} wp - Hp^{2} wp - wH^{3}
= 0

i.e., Hb^{2} wp = wH^{3}

b^{2} p = H^{2}

Value

The base
width of the elementary profile of a dam storage to a height H, is a right
angled triangle of base width

B= H/Rt(p)

it is on the assumption, that there
are no uplift pressures base of the dam from the foundation. If uplift is taken
count, the value of p will be (p â€'1), thus giving a base

B= H/Rt(p-1)

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Civil - Water Resources and Irrigation Engineering - Diversion and Impounding Structures : The Forces that are acting on the Dam Are |

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