Distributed loads and fluid pressure

Few real loads act at a single point. Snow lies across a whole roof; people, furniture and the floor itself spread their weight over a slab; wind presses on the face of a building or a sign; water and soil push on dams, tanks, gates and retaining walls; a beam's own weight runs along its length. Lesson 4 replaced a uniform and a triangular load by single forces to find reactions. This lesson develops that idea fully. It shows how to find the resultant of any distributed load and where it acts, how engineers turn floor and roof loads into loads on beams, and how to calculate the forces of fluids at rest on walls, gates and submerged bodies, including buoyancy. These calculations underlie the design of nearly every structure that holds back water or soil, and every beam that carries a floor.

Line loads and their resultants

A load spread along a beam is described by its intensity w(x), the force per unit length, in kN/m. The total force on a short length dx is w dx, so the resultant force is the integral of the intensity, which is the area under the load diagram. Its line of action, found by matching moments, passes through the centroid of that area (Lesson 6):

For the shapes that occur most often:

• A uniform load w over length L has resultant wL at the middle of the loaded length.
• A triangular load rising from zero to w over length L has resultant wL/2, one-third of the length from the high end (two-thirds from the zero end).
• A trapezoidal load is split into a rectangle and a triangle, each replaced by its own resultant.

From area loads to line loads

Building codes specify loads on floors and roofs as pressures, force per unit area, in kPa (kN/m²). A beam or joist collects the load from the floor area it supports, its tributary area. For a series of parallel beams, each one carries the strip of floor halfway to the beams on either side, a width called the tributary width. Multiplying the area load by the tributary width gives the line load on the beam.

ASCE 7, adopted through the International Building Code, specifies, for example, a live load of 40 psf (1.92 kPa) on residential floors, 50 psf (2.4 kPa) on office floors, and 100 psf (4.79 kPa) on assembly areas such as gyms and auditoriums, along with ground snow loads that depend on the site's climate. Dead loads come from the weights of the structure and permanent finishes. Designers combine them with load factors for strength design, such as 1.2 × dead + 1.6 × live.

Pressure in a fluid at rest

A fluid at rest cannot resist shear, so the force it exerts on any surface is perpendicular to that surface. Its pressure at a point is the same in all directions, and in a liquid of density ρ it increases linearly with depth h below the free surface:

For fresh water, ρg = 9.81 kN/m³, so pressure rises by 9.81 kPa for every meter of depth; seawater is about 3 % denser. Engineers usually work with gauge pressure, ignoring atmospheric pressure, because the atmosphere acts on both sides of most walls and gates and cancels out.

Because pressure grows linearly with depth, the load on a vertical wall is a triangle, and on a submerged surface whose top is below the water it is a trapezoid. The resultant force equals the volume of the pressure diagram, and it acts through that volume's centroid, called the center of pressure.

For any plane surface, vertical or inclined, the resultant force equals the pressure at the surface's centroid times its area, \(F = \rho g h_c A\), where \(h_c\) is the depth of the centroid. The center of pressure lies below the centroid, by an amount that depends on the area's second moment, a property introduced in Lesson 11. For rectangles, the trapezoidal pressure diagram gives the same result directly.

Retaining walls and dams

Walls that hold back water or soil must not slide or overturn. The horizontal force of the water tries to slide the wall along its base and to tip it about its downstream toe; the wall's weight, and friction at its base, resist. Concrete gravity dams are triangular or trapezoidal in section for exactly this reason: their width grows with depth to match the pressure.

Curved surfaces

On a curved surface, the pressure acts perpendicular to every element, so the direction of force varies around the curve. It is easiest to find the horizontal and vertical components separately. The horizontal component equals the force on the vertical projection of the surface, calculated exactly as for a plane wall. The vertical component equals the weight of the water directly above the surface, up to the free surface (or, if the water is below the surface, the weight of the water that would fill that space, acting upward).

Buoyancy

A body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces (Archimedes' principle), \(F_B = \rho g V\), acting through the centroid of the displaced volume. Buoyancy explains why ships float and why submerged objects seem lighter. It also creates hazards: empty underground tanks, pipelines and manholes in high groundwater can float out of the ground, and engineers must anchor them.

Wind pressure

Air in motion also exerts distributed pressure. The basic quantity is the velocity pressure \(q = \tfrac{1}{2}\rho V^2\), with air density ρ ≈ 1.25 kg/m³. The pressure on a surface is q times a pressure or drag coefficient that depends on the shape, about 1.2 to 2.0 for flat signs. At 100 km/h (27.8 m/s), q = 0.48 kPa; because pressure grows with the square of speed, a 140 km/h gust produces almost twice that. Building codes give reference wind pressures for each location and rules for coefficients, heights and gusts. A 3 m × 2 m road sign with a coefficient of 1.2 in a 100 km/h wind carries about 0.48 × 1.2 × 6 = 3.5 kN, acting at the center of the sign, which its posts and foundations must resist in bending and overturning.