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Total Pressure on Plane Areas

The total hydrostatic pressure on a plane surface is equal to the pressure at the centroid times the area of the plane surface.

$$P = \gamma \bar{h} A$$ $$\bar{h} = \text{vertical depth of centroid below liquid surface}$$

For inclined surfaces, pressure still depends on the vertical depth of the centroid, not the sloping distance alone.

$$\bar{h} = \bar{y}\sin\theta$$ $$P = \gamma A\bar{y}\sin\theta$$

Center of Pressure

The resultant hydrostatic force acts below the centroid for vertical and inclined plane areas. The distance from centroid to center of pressure is commonly denoted by $e$.

$$e = \frac{I_G}{A\bar{y}}$$ $$y_{cp} = \bar{y} + e$$ $$h_{cp} = y_{cp}\sin\theta$$
For vertical plates, $\bar{y}$ and $h$ are the same vertical distance. For inclined plates, use distance measured along the plane for $y$ and vertical depth for $h$.

Problem: Rectangular Gate

A rectangular gate is 1.5 m wide and 3 m high, vertically submerged in water with its top edge 2 m below the water surface. Find the total force and the center of pressure depth.

$$\bar{h} = 2 + \frac{3}{2} = 3.5 \text{ m}$$ $$A = 1.5(3) = 4.5 \text{ m}^2$$ $$P = 9.81(3.5)(4.5) = 154.51 \text{ kN}$$ $$I_G = \frac{1.5(3)^3}{12} = 3.375 \text{ m}^4$$ $$e = \frac{I_G}{A\bar{h}} = \frac{3.375}{4.5(3.5)} = 0.214 \text{ m}$$ $$h_{cp} = 3.5 + 0.214 = 3.714 \text{ m}$$

Answer: $P = 154.51 \text{ kN}$ and the center of pressure is 3.714 m below the water surface.

Problem: Square Plate Location

A square plate 3 m on each side is vertical in water. If the center of pressure is 8 cm from the centroid, find how far the upper side is below the water surface and the hydrostatic force.

$$I_G = \frac{3(3)^3}{12} = 6.75 \text{ m}^4$$ $$e = \frac{I_G}{A\bar{h}}$$ $$0.08 = \frac{6.75}{(3)(3)\bar{h}}$$ $$\bar{h} = 9.375 \text{ m}$$ $$y_{top} = 9.375 - 1.5 = 7.875 \text{ m}$$ $$P = 9.81(9.375)(3)(3) = 827.72 \text{ kN}$$

Answer: The upper side is 7.875 m below the surface, and $P = 827.72 \text{ kN}$.

Problem: Circular Plate — Total Force and Center of Pressure

A circular plate with diameter 1.20 m is submerged vertically in fresh water with its center at a depth of 2.50 m below the free surface. Find the total hydrostatic force on one face and the depth of the center of pressure.

$$A = \frac{\pi(1.20)^2}{4} = 1.131 \text{ m}^2, \quad \bar{h} = 2.50 \text{ m}$$ $$P = \gamma \bar{h} A = 9.81(2.50)(1.131) = 27.73 \text{ kN}$$ $$I_G = \frac{\pi d^4}{64} = \frac{\pi(1.20)^4}{64} = 0.1018 \text{ m}^4$$ $$e = \frac{I_G}{A\bar{h}} = \frac{0.1018}{1.131(2.50)} = 0.036 \text{ m}$$ $$h_{cp} = 2.50 + 0.036 = 2.536 \text{ m}$$

Answer: Total force is 27.73 kN. The center of pressure is 2.536 m below the surface — only 36 mm below the centroid because the plate is deep relative to its size.

Problem: Inclined Rectangular Gate — Force and Location

A rectangular gate is 1.0 m wide and 2.0 m long, set at 60° from the horizontal. The upper edge of the gate is 1.5 m (vertical) below the water surface. Find: (a) the total hydrostatic force on the gate, and (b) the center of pressure measured along the inclined gate from the water-surface intersection.

$$\bar{h} = 1.5 + \frac{2.0\sin 60°}{2} = 1.5 + 0.866 = 2.366 \text{ m (vertical depth to centroid)}$$ $$A = 1.0(2.0) = 2.0 \text{ m}^2$$ $$P = 9.81(2.366)(2.0) = 46.43 \text{ kN}$$ $$\bar{y} = \frac{\bar{h}}{\sin 60°} = \frac{2.366}{0.866} = 2.732 \text{ m (along incline)}$$ $$I_G = \frac{1.0(2.0)^3}{12} = 0.667 \text{ m}^4$$ $$e = \frac{I_G}{A\bar{y}} = \frac{0.667}{2.0(2.732)} = 0.122 \text{ m}$$ $$y_{cp} = 2.732 + 0.122 = 2.854 \text{ m (along incline)}$$

Answer: Total force is 46.43 kN. The center of pressure is 2.854 m from the water-surface line measured along the inclined face of the gate.

Problem: Gate Hinged at the Bottom — Reaction at the Top Stop

A rectangular gate is 1.20 m wide and 2.40 m tall, hinged at its bottom edge and held by a horizontal stop (latch) at its top edge. The gate is vertical, and the water surface is 1.00 m above the top of the gate. Determine: (a) the total hydrostatic force on the gate, (b) the depth of the center of pressure, and (c) the horizontal reaction force at the top stop.

$$\bar{h} = 1.00 + \frac{2.40}{2} = 2.20 \text{ m}$$ $$A = 1.20(2.40) = 2.88 \text{ m}^2$$ $$P = 9.81(2.20)(2.88) = 62.11 \text{ kN}$$ $$I_G = \frac{1.20(2.40)^3}{12} = 1.382 \text{ m}^4$$ $$e = \frac{1.382}{2.88(2.20)} = 0.218 \text{ m}$$ $$h_{cp} = 2.20 + 0.218 = 2.418 \text{ m below surface}$$

Distance of resultant from hinge: gate bottom is $(1.00 + 2.40) = 3.40$ m below surface, so $d = 3.40 - 2.418 = 0.982$ m from hinge.

$$\sum M_{hinge} = 0: \quad R(2.40) = P(0.982)$$ $$R = \frac{62.11(0.982)}{2.40} = 25.40 \text{ kN}$$

Answer: Total force is 62.11 kN; center of pressure at 2.418 m depth. The reaction at the top stop is 25.40 kN outward.

Problem: Triangular Plate in Sea Water

An isosceles triangular plate has a base of 1.80 m and height of 2.40 m. It is submerged vertically in sea water (SG = 1.025) with its base (top edge) 0.80 m below the free surface and the apex pointing downward. Find the total hydrostatic force on the plate and the depth of the center of pressure.

The centroid of a triangle is 1/3 of height from the base, so the centroid is 0.80 + (2.40/3) = 0.80 + 0.80 = 1.60 m below the surface.

$$A = \frac{1}{2}(1.80)(2.40) = 2.16 \text{ m}^2$$ $$\gamma_{sw} = 9.81(1.025) = 10.055 \text{ kN/m}^3$$ $$P = 10.055(1.60)(2.16) = 34.75 \text{ kN}$$ $$I_G = \frac{bh^3}{36} = \frac{1.80(2.40)^3}{36} = 1.382 \text{ m}^4$$ $$e = \frac{I_G}{A\bar{h}} = \frac{1.382}{2.16(1.60)} = 0.400 \text{ m}$$ $$h_{cp} = 1.60 + 0.400 = 2.00 \text{ m below surface}$$

Answer: Total force is 34.75 kN. The center of pressure is 2.00 m below the surface (0.40 m below the centroid).

Problem: Plate with Fluid on Both Faces — Net Resultant

A vertical rectangular gate 2.0 m wide and 3.0 m tall is set in a wall between two water reservoirs. On the left (high) side, the water surface is 1.0 m above the top of the gate. On the right (low) side, the water surface is at the same elevation as the top of the gate (i.e., the right water depth equals the gate height). Find the net hydrostatic force on the gate and the location of the net resultant from the bottom of the gate.

Left side — water 1.0 m above gate top, so centroid depth = 1.0 + 1.5 = 2.5 m:

$$F_L = 9.81(2.5)(2.0)(3.0) = 147.15 \text{ kN}$$ $$I_{G,L} = \frac{2.0(3.0)^3}{12} = 4.5 \text{ m}^4$$ $$e_L = \frac{4.5}{6.0(2.5)} = 0.30 \text{ m} \Rightarrow h_{cp,L} = 2.80 \text{ m from surface}$$ $$d_L = (1.0+3.0) - 2.80 = 1.20 \text{ m from gate bottom}$$

Right side — water surface at top of gate, centroid depth = 1.5 m:

$$F_R = 9.81(1.5)(2.0)(3.0) = 88.29 \text{ kN}$$ $$e_R = \frac{4.5}{6.0(1.5)} = 0.50 \text{ m} \Rightarrow h_{cp,R} = 2.00 \text{ m from right surface}$$ $$d_R = 3.0 - 2.0 = 1.00 \text{ m from gate bottom}$$

Net force and resultant location:

$$F_{net} = 147.15 - 88.29 = 58.86 \text{ kN (leftward)}$$ $$\bar{d} = \frac{147.15(1.20) - 88.29(1.00)}{58.86} = \frac{176.58 - 88.29}{58.86} = 1.50 \text{ m from bottom}$$

Answer: Net hydrostatic force is 58.86 kN acting toward the lower reservoir. The resultant acts at 1.50 m from the bottom of the gate.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q354

HGE - Hydraulics / Hydrostatic Forces on Plane Surfaces / Engr. Janclyde Espinosa (Clidez)

A corner at the bottom of the cross-section of a tank is shown in figure. The tank is 2m wide into the plane of this paper. Neglecting atmospheric pressure, evaluate:

q354

The total horizontal force on the corner plate AB, in kN.

  1. 323.73
  2. 346.82
  3. 341.87
  4. 320.64

The total vertical force on the corner plate AB, in kN.

  1. 323.73
  2. 346.82
  3. 341.87
  4. 320.64

The resultant force acting on the plate, in kN.

  1. 457.8
  2. 490.48
  3. 483.48
  4. 453.45

Solution pending in psadquestions/q354.json.

Question Bank: q359

HGE - Hydraulics / Hydrostatic Forces on Plane Surfaces / Engr. Janclyde Espinosa (Clidez)

A triangular gate or height 1.2m and base 0.9m is installed in a position that its plane is inclined 60 degrees with the horizontal with its vertex at the top and the base parallel to the water surface. The vertex is at a depth of 2m vertically below the water surface. Fresh water is on one side of the gate.

Evaluate the total hydrostatic force on the gate in kN.

  1. 14.3
  2. 10.6
  3. 12.7
  4. 12.4

Locate the point of action of the total hydrostatic force from the vertex on the plane of the gate.

  1. 0.8257
  2. 0.8357
  3. 0.8457
  4. 0.8557

If the gate is hinged at the bottom, evaluate the force normal to the gate at its vertex that will be required to open it in kN.

  1. 4.45
  2. 4.56
  3. 4.67
  4. 4.78

Part 1.

The centroid of the triangular gate is two-thirds of the height from the vertex along the inclined plane:
$\bar h=2+\frac{2}{3}(1.2)\sin60^\circ=2.693$ m
$A=\frac{1}{2}(0.9)(1.2)=0.54$ m2
Total hydrostatic force:
$F=\gamma A\bar h=9.81(0.54)(2.693)=14.26$ kN
$\boxed{F\approx14.3\text{ kN}}$

Part 2.

Let $y$ be distance along the gate from the vertex. The pressure varies with vertical depth $2+y\sin60^\circ$, and the triangular width is proportional to $y$. Thus:
$y_R=\frac{\int_0^{1.2}y[y(2+y\sin60^\circ)]dy}{\int_0^{1.2}y(2+y\sin60^\circ)dy}$
$y_R=0.8257$ m from the vertex along the gate
$\boxed{0.8257}$

Part 3.

Take moments about the bottom hinge. The resultant force is $F=14.3$ kN and acts $0.8257$ m from the vertex, so its distance from the hinge is $1.2-0.8257$ m:
$P(1.2)=14.3(1.2-0.8257)$
$P=4.45$ kN
$\boxed{P=4.45\text{ kN}}$

Question Bank: q360

HGE - Hydraulics / Hydrostatic Forces on Plane Surfaces / Engr. Janclyde Espinosa (Clidez)

A circular gate 1.5m in diameter is inclined at an angle of 45º. Fresh water stands on one side of the gate to a height of 10m. above the center of the gate.

Evaluate the total force on the gate.

  1. 173.4
  2. 173.8
  3. 174.2
  4. 174.6

Locate the point of action of the total force from the bottom on the plane of the gate.

  1. 0.74
  2. 0.73
  3. 0.72
  4. 0.71

If the gate is hinged at the top, evaluate the force normal to the gate at the bottom required to open it in kN.

  1. 87.8
  2. 88.9
  3. 86.5
  4. 84.7

Part 1.

For a plane surface, total force is $F=\gamma A\bar h$. The center of the circular gate is 10 m below the water surface:
$A=\frac{\pi(1.5)^2}{4}=1.767$ m2
$F=9.81(1.767)(10)=173.36$ kN
$\boxed{F\approx173.4}$

Part 2.

For an inclined circular gate:
$s=\frac{I_G\sin\theta}{A\bar h}$
where $s$ is the shift of the center of pressure below the centroid along the plane.
$I_G=\frac{\pi(0.75)^4}{4},\quad A=\pi(0.75)^2,\quad \bar h=10$ m
$s=0.00994$ m
Distance from bottom to point of action:
$0.75-s=0.740$ m
$\boxed{0.74}$

Part 3.

Take moments about the top hinge. The force acts $0.75+0.00994=0.75994$ m from the top along the gate, and the bottom force has lever arm 1.5 m:
$P(1.5)=173.4(0.75994)$
$P=87.8$ kN
$\boxed{87.8}$

Question Bank: q365

HGE - Hydraulics / Hydrostatic Forces on Plane Surfaces / Engr. Janclyde Espinosa (Clidez)

A triangular plate of height h = 0.90 m and base b = 2 m is submerged vertically in a water with its top edge (base) at the liquid surface and parallel to it.

Evaluate the total force acting on one side of the plate in Newtons.

  1. 2648.7
  2. 2578.3
  3. 2785.3
  4. 2674.8

Obtain the location of the force from the center of gravity of the plate, in mm.

  1. 150
  2. 140
  3. 160
  4. 170

Obtain the location of the force from the liquid surface, in mm.

  1. 450
  2. 440
  3. 460
  4. 470

Part 1.

The triangular plate has its base at the water surface, so its centroid is $h/3$ below the surface:
$A=\frac{1}{2}(2)(0.90)=0.90$ m2
$\bar h=0.90/3=0.30$ m
$F=\gamma A\bar h=9810(0.90)(0.30)=2648.7$ N
$\boxed{F=2648.7\text{ N}}$

Part 2.

For the vertical triangular plate, the centroid is at $\bar h=0.90/3=0.30$ m below the surface. The center of pressure shift is:
$y_{cp}-\bar y=\frac{I_G}{A\bar y}$
$I_G=\frac{bh^3}{36}=\frac{2(0.9)^3}{36}=0.0405$ m4, $A=0.90$ m2
$y_{cp}-\bar y=\frac{0.0405}{0.90(0.30)}=0.150$ m
$\boxed{150\text{ mm}}$

Part 3.

The center of pressure is 150 mm below the centroid. Since the centroid is 300 mm below the liquid surface:
$y_{cp}=300+150=450$ mm
$\boxed{450\text{ mm}}$

Question Bank: q367

HGE - Hydraulics / Hydrostatic Forces on Plane Surfaces / Engr. Janclyde Espinosa (Clidez)

A rectangular plate is vertically submerged half in oil having a specific gravity of 0.64 and half in water such that its top edge is flushed with the top surface of the upper liquid and parallel to it. Obtain the ratio of the force exerted to the plate by the water to that by the oil.

Answer:

  1. 3.56
  2. 0.281
  3. 3.67
  4. 0.298
Let each liquid layer have depth $a$ and unit plate width. Oil force on the upper half:
$F_o=\frac{1}{2}\gamma_o a^2$
Water-side force on the lower half includes the oil pressure at the interface plus the water triangle:
$F_w=\gamma_o a^2+\frac{1}{2}\gamma_w a^2$
Thus:
$\frac{F_w}{F_o}=\frac{0.64+0.5}{0.5(0.64)}=3.5625$
$\boxed{\frac{F_w}{F_o}\approx3.56}$

Question Bank: q371

HGE - Hydraulics / Hydrostatic Forces on Plane Surfaces / Engr. Janclyde Espinosa (Clidez)

A vertical triangular surface has a horizontal base of 1.2m and an altitude of 3m, and the vertex is below the base. If the center of pressure is 0.150m below the center of gravity, how far is the base below the liquid surface?

Answer:

  1. 2.33m
  2. 2.40m
  3. 2.45m
  4. 2.66m
For a vertical plane surface:
$y_{cp}-\bar y=\frac{I_G}{A\bar y}$
For the triangle, $A=\frac{1}{2}(1.2)(3)=1.8$ m2 and $I_G=\frac{bh^3}{36}=\frac{1.2(3)^3}{36}=0.90$ m4.
Given $y_{cp}-\bar y=0.150$ m:
$0.150=\frac{0.90}{1.8\bar y}$
$\bar y=3.333$ m below the liquid surface
The centroid is 1.0 m below the base, so base depth is:
$3.333-1.0=2.333$ m
$\boxed{2.33\text{ m}}$

Question Bank: t22

HGE - Hydraulics / Hydrostatic Forces / Civil Engineering Refresher

A pyramid (W = 17.8 kN) with a 1.8 m square base and 1.2 m altitude covers a floor opening in a 1.2 m deep water tank. Find the vertical force required to lift it.

  1. 42.23 kN
  2. 25.43 kN
  3. 17.80 kN
  4. 36.15 kN

Solution pending in psadquestions/t22.json.

Question Bank: v9

HGE - Hydraulics / Hydrostatic Force / HGE May 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A circular gate of radius 0.25 m is submerged in water with its top edge flush with the water surface.

Calculate the total hydrostatic force acting on the gate, in N.

  1. 482 N
  2. 414 N
  3. 501 N
  4. 587 N

Locate the center of pressure measured from the centroid of the gate, in mm.

  1. 70 mm
  2. 50 mm
  3. 63 mm
  4. 83 mm

Locate the resultant force measured from the water surface, in mm.

  1. 209 mm
  2. 313 mm
  3. 375 mm
  4. 63 mm

Part 1 — Total force. With the top flush with the surface, the centroid lies at depth $\bar h=r$. For a plane area,

$$F=\gamma\,\bar h\,A=\gamma\,(r)\,(\pi r^{2}).$$

Part 2 — Center-of-pressure offset. Measured from the centroid,

$$e=\frac{I_g}{A\bar y}=\frac{\pi r^{4}/4}{(\pi r^{2})(r)}=\frac{r}{4}.$$

Part 3 — Depth of the resultant (from the surface):

$$y_p=\bar h+e=r+\frac{r}{4}=\frac{5r}{4}.$$

Computed answers:
1. 482 N
2. 63 mm
3. 313 mm

Question Bank: v12

HGE - Hydraulics / Hydrostatic Force / HGE May 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A rectangular gate 1.8 m wide and 2.2 m high holds water on one side and is inclined 40° with the horizontal. The water surface is 1.3 m above the top edge of the gate.

Compute the hydrostatic force on the gate, in kN.

  1. 85.77 kN
  2. 51.46 kN
  3. 77.97 kN
  4. 75.63 kN

Locate the center of pressure measured from the top of the gate along its plane, in meters.

  1. 1.10 m
  2. 1.23 m
  3. 1.43 m
  4. 0.86 m

If the gate is hinged at the top, find the force normal to the gate at the bottom needed to open it, in kN. Neglect the gate weight.

  1. 49.66 kN
  2. 77.97 kN
  3. 43.56 kN
  4. 39.64 kN

Part 1 — Force. Depth to the centroid is $\bar h=h_{top}+\tfrac{L}{2}\sin\theta$ and $A=bL$, so

$$F=\gamma\,\bar h\,A.$$

Part 2 — Center of pressure. Along the inclined plane $\bar y=\bar h/\sin\theta$, with $I_g=\dfrac{bL^{3}}{12}$,

$$e=\frac{I_g}{A\bar y},\qquad y_F=\frac{L}{2}+e\ \text{(from the top edge).}$$

Part 3 — Opening force. Taking moments about the top hinge,

$$P\,L=F\,y_F\ \Rightarrow\ P=\frac{F\,y_F}{L}.$$

Computed answers:
1. 77.97 kN
2. 1.23 m
3. 43.56 kN

Question Bank: v15

HGE - Hydraulics / Hydrostatic Force / HGE May 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A circular gate 1.3 m in diameter is inclined at 40° to the horizontal. Fresh water stands on one side to a height of 11 m (vertical) above the center of the gate.

Evaluate the total hydrostatic force on the gate, in kN.

  1. 118.88 kN
  2. 143.23 kN
  3. 177.61 kN
  4. 95.97 kN

Locate the point of action of the total force measured from the bottom of the gate along its plane, in meters.

  1. 0.46 m
  2. 0.76 m
  3. 0.86 m
  4. 0.64 m

If the gate is hinged at the top, find the force normal to the gate at the bottom required to open it, in kN. Neglect the gate weight.

  1. 75.91 kN
  2. 62.17 kN
  3. 72.30 kN
  4. 80.97 kN

Part 1 — Force. With $A=\dfrac{\pi D^{2}}{4}$ and $\bar h$ the vertical depth of the centroid,

$$F=\gamma\,\bar h\,A.$$

Part 2 — Center of pressure. $\bar y=\bar h/\sin\theta$, $I_g=\dfrac{\pi D^{4}}{64}$, so the offset below the centroid is $e=\dfrac{I_g}{A\bar y}$ and the distance from the bottom is

$$\frac{D}{2}-e.$$

Part 3 — Opening force. Moments about the top hinge:

$$P\,D=F\left(\frac{D}{2}+e\right)\ \Rightarrow\ P=\frac{F\left(\tfrac{D}{2}+e\right)}{D}.$$

Computed answers:
1. 143.23 kN
2. 0.64 m
3. 72.30 kN

Question Bank: v58

HGE - Hydraulics / Hydrostatic Forces on Plane Surfaces / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A triangular gate has inclined height 7 m, width 3.5 m, and its upper vertex is 2.5 m below the free surface. The gate is inclined 55° and hinged along its lower base. Find the horizontal force at the upper vertex needed to open it.

  1. 507 kN
  2. 262 kN
  3. 297 kN
  4. 186 kN
The resultant normal force is $F=\gamma\bar hA$, where $A=bH/2$ and $\bar h=z+(2H/3)\sin\theta$. Its center-of-pressure shift is $e=I_G/(A\bar y)$ with $I_G=bH^3/36$ and $\bar y=\bar h/\sin\theta$. Taking moments about the hinge: $$P(H\sin\theta)=F(H/3-e),$$ giving $P=262.454195223$ kN.
Computed answer: 262 kN