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Need help with this.

Part of a sea defense consists of a section of a concrete wall 4 meters high, and 6 meters wide. enter image description here

a) Calculate the resultant thrust on the section when the sea reaches a height of 3 metres relative to the base of the section.

b) Calculate the overturning moment experienced by the section about point A.

This is my answer. Have i done this correctly? I am not sure if i am on the right track with this or not. Especially the overturning moment.

a) Thrust

$Density of water = 1000kg/m^3$

Width = 6m

Height of fluid = 3m

$f = pg \frac{h} {2} A$

Ft = 1000 x 9.81 x 1.5 x (6x3)

Ft = 264,870 N

b) Overturning Moment

$M = Ft \frac{h} {3}$

$M = 264.870 x \frac{3} {3} = 264.870 Nm$

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  • $\begingroup$ This may be a help : engineering.stackexchange.com/questions/2516/… $\endgroup$ – Solar Mike Jan 25 '18 at 10:57
  • $\begingroup$ Is point A at the base of the sea face or the beach face or somewhere else? It is not clear from your sketch. Also if this is for a real project, the density of sea water is greater than 1000 kg/m3. Is the width along the length or cross-sectional? Are you looking for a unit loading or total load? $\endgroup$ – AsymLabs Jan 25 '18 at 13:38
  • $\begingroup$ Also when doing these types of calculations it is always good practice to show your units to ensure that they cancel out - leaving you with a means of checking the result. $\endgroup$ – AsymLabs Jan 25 '18 at 13:56
  • $\begingroup$ I think point A is sea face. This is the sketch i got to work off. The Width is across the length. I am looking for: a) Calculate the resultant thrust on the section when the sea reaches a height of 3 metres relative to the base of the section. b) Calculate the overturning moment experienced by the section about point A. Thanks for your replys $\endgroup$ – Kyle Anderson Jan 25 '18 at 14:51
  • $\begingroup$ You don't define many of the variables used. In real life the water would flow around the wall. $\endgroup$ – paparazzo Jan 25 '18 at 15:32
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Usually with this kind of problems what I would do is to analyse what effect is caused by what source. In your specific problem, there is water under gravitational load. The gravitational field imposes a vertical pressure distribution inside the water. This pressure acts on the wall.

Thus the first answer I would give is the distribution of pressure $p$ as a function of the gravity of earth $g$, the density of water $\rho$ and the vertical coordinate $z$ (I would choose $z=0$ at sea level and positive values of $z$ going downward, but you can choose a different coordinate system). This gives you a function $p(z)$. This is not directly asked for, but is required to answer the two questions.

In the second step, since the load is distributed, I would integrate this distributed load (force per unit length) along the height: $$ F=\int_0^h w \cdot p(z) \, \mathrm{d}z $$ with width $w=6\,\mathrm{m}$ and height $h=3\,\mathrm{m}$. The distributed load is the product of pressure and wall width, because the pressure does not vary with width. If it did, I would have to integrate not only along the height axis $z$, but also along the width axis.

Finally, to calculate the momentum, I would integrate the local torque, with is the product of the distributed load $w \cdot p(z)$ and the lever arm length $(h-z)$: $$ M=\int_0^h w \cdot p(z) \cdot (h-z) \, \mathrm{d}z \quad\mbox{.} $$ Now you can calculate everything and check if my results are the same as yours. :-)

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