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pressure on a inclined surface

pressure on a inclined surface

pressure on a inclined surface

(OP)
Hi all!

I have a ''wall'' which is 58 degrees from the horizontal plane and it is said that a force of 200kN shall be applied horizontally on that wall surface according to a standard, i.e. the force is in 0 degree direction and the wall is in 58 degree direction.

My question is then if I want to apply a pressure on that wall which shall correspond to the force of 200kN, how should I do? I cannot just take force / area because the force is not perpendicular to the area, i.e. it is not normal to the wall.

Thank you

RE: pressure on a inclined surface

If the sloping height of the wall is 'h' the vertical height is h*sin58 = 0.85h.  

A force of 200 kN applied horizontally is equivalent to a horizontal pressure of 200/(0.85*h*w) = 286/(h.w) where w is the width of wall.

If you want to apply a normal pressure to the wall such that the horizontal component is 200 kN, the normal pressure is 286/(h.w) and the vertical component is 200*tan58 = 125 kN.

Don't forget that any soil lying above the sloping wall produces a variable downward pressure on the wall.  It varies from γz to 0 where z is the maximum depth of fill and γ is the unit weight of soil.

BA

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