Question

A 4-ft tall, 8-in wide concrete (150lb/ft3)

retainingwall is built. During a heavy rain, water fills the

spacebetween the wall and the earth behind it to a depth, h.

Determine the maximum depth of water possible without the

walltipping over. The wall simply rests on the ground withoutbeing

anchored to it.

Answer 1

Answer:

maximum possible height of water is 2.34 ft without wall tipping

Explanation:

given data

wall height = 4 ft

wall width = 8 in = 0.667 ft

weight density = 150 lb/ft³

to find out

the maximum depth of water possible without the  wall tipping

solution

we find here first weight of concrete wall that is

weight = volume × weight density

weight = length × width × height × weight density

weight = 150 × 4 × 0.667 × L

weight = 400 L lb

here L is length

now find the resulting force acting due to hydrostatic force per unit length

Fr = density of water × hc × ( h×L)

heer hc is distance between fluid surface and centroid area = $\frac{h}{2}$ and L is length and h is height

and density of water is 62.4 lb/ft³

so Fr = 62.4 × $\frac{h}{2}$ × ( h×L)

Fr = 31.2 h²L

now

point of position of apply resultant force that is Yr

Yr = $\frac{moment interia}{hc* area} + hc$

Yr = $\frac{L*\frac{h^3}{12}}{\frac{h}{2} * h*L} + \frac{h}{2}$

Yr = $\frac{2h}{3}$

so

moment about about point A will be  zero to avoid tapping

so

∑Ma = 0

$W*\frac{0.667}{2} - Fr*( h -Yr ) = 0$

$400*\frac{0.667}{2} - 31.2 h^2 * ( h -\frac{2h}{2}) = 0$

h = 2.34

so maximum possible height of water is 2.34 ft without wall tipping