A gravity dam of 84 m height is built over a river. The upstream and downstream height of the water level is 80 m and 6 m respectively. Base width of the dam is 56 m and a drainage gallery is provided at 8 m from the heel of the dam. The total uplift force due to seepage forces only per unit length of the dam (expressed in kN/m) is ________
12975.36

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Solution

The correct option is A 12975.36

$γ_{w} H_{2} + \frac{1}{3} γ_{w} (H_{1} - H_{2})$

$u_{1} = \frac{1}{2} [γ_{w} H_{1} + γ_{w} H_{2} + \frac{1}{3} γ_{w} (H_{1} - H_{2})] \times 8$

$= \frac{1}{2} \times 9.81 [80 + 6 + \frac{1}{3} (80 - 6)] \times 8$

$= 4342.56 k N$

$u_{2} = \frac{1}{2} [γ_{w} H_{2} + \frac{1}{3} γ_{w} (H_{1} - H_{2}) + γ_{w} H_{2}] \times 48$

$= 24 \times 9.81 \times [6 + \frac{1}{3} (80 - 6) + 6]$

$= 8632.8 k N$
$∴ Total uplift force, u = u_{1} + u_{2}$
$= 4342.56 + 8632.8$
$= 12975.36 k N$

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The correct option is A 12975.36 γwH2+13γw(H1−H2) u1=12[γwH1+γwH2+13γw(H1−H2)]×8 =12×9.81[80+6+13(80−6)]×8 =4342.56kN u2=12[γwH2+13γw(H1−H2)+γwH2]×48 =24×9.81×[6+13(80−6)+6] =8632.8kN ∴Total uplift force, u=u1+u2 =4342.56+8632.8 =12975.36 kN