Rigid-body equilibrium

A beam resting on two walls, a crane boom held by a cable, a ladder leaning against a building, a truck bed being tipped by a hydraulic cylinder: these bodies have size, and the forces on them act at different points. Balancing the forces is no longer enough. The moments must balance too, or the body would start to rotate.

This chapter puts together everything from the first three lessons. You will learn what force each type of support exerts, draw free-body diagrams of rigid bodies, handle distributed loads, and solve for the reactions that keep a body in place. You will also meet two shortcuts (two-force and three-force members), learn to tell whether a structure can be solved by statics at all, and extend the method to three dimensions. Support reactions are the starting point for nearly every structural and machine design calculation you will do.

The equations of equilibrium

A rigid body is in equilibrium when the resultant force and the resultant moment about any point are both zero:

In two dimensions this gives three independent scalar equations:

So a planar rigid body allows you to solve for three unknowns. The moment equation can be taken about any point, on the body or off it, and that freedom is your most powerful tool. Taking moments about a point where two unknown forces act removes both from the equation, often leaving a single unknown that you can find directly.

You may replace one or both force equations with extra moment equations (for example, \(\sum M_A = 0\), \(\sum M_B = 0\) and \(\sum F_x = 0\)), provided the moment points are not on a line perpendicular to the remaining force direction. In practice, two moment equations and one force equation is often the cleanest set.

Supports and their reactions

Supports prevent motion, and each kind of support prevents particular motions. The rule is simple: a support exerts a force in any direction in which it prevents translation, and a couple about any axis about which it prevents rotation.

• Roller, rocker or smooth surface: one unknown, a force perpendicular to the surface. Bridges sit on rollers or sliding bearings at one end so they can expand and contract with temperature without building up large horizontal forces.
• Cable or short link: one unknown, a force along the cable or link. A cable can only pull.
• Pin or hinge: two unknowns, usually written as horizontal and vertical components. A pin prevents translation in any direction but lets the body rotate.
• Fixed support (built-in end): three unknowns, two force components and a couple. It prevents both translation and rotation, like the wall end of a balcony slab or a flagpole set in a concrete base.

If you do not know the direction of a reaction, assume one. If the answer comes out negative, the reaction acts the other way. Leave the sign in your equations; do not redraw the diagram halfway through.

Free-body diagrams of rigid bodies

The procedure is the same as for a particle, with two additions: forces are drawn where they actually act, and couples are included.

• Sketch the body's outline, cut free from its supports and surroundings.
• Replace each support with its reactions, from the table above.
• Add the applied loads at their points of application, including the body's own weight at its centre of gravity if it is significant.
• Replace distributed loads with their resultants (but keep the original on the sketch for later use in beam analysis).
• Mark dimensions, angles and a coordinate system.

Distributed loads

Many loads are spread out: the weight of a floor slab on a beam, snow on a roof, wind on a wall, water pressure on a dam. They are described by a load intensity \(w\), in kN/m. For equilibrium, a distributed load is equivalent to a single force equal to the area under the load diagram, acting through the centroid of that area.

• Uniform load \(w\) over length \(L\): resultant \(wL\) at the middle of the loaded length.
• Triangular load rising from zero to \(w\) over length \(L\): resultant \(\tfrac{1}{2}wL\) at \(L/3\) from the tall end (\(2L/3\) from the zero end).
• Trapezoidal load: split it into a rectangle and a triangle and treat each separately.

Worked solutions

Notice what the overhang does. If the end load grew to 36 kN, \(A_y\) would drop to zero, and beyond that A would have to hold the beam down. A pin can do that if it is anchored; a beam simply resting on a wall at A would lift off. Checking for uplift at supports is a routine part of design.

Bodies with cables and inclined forces

This shows a theme from Lesson 2 in a new setting. The cable rises at only 36.9°, so most of its tension goes into horizontal force that does nothing to hold up the load. A steeper cable (attached higher on the wall) would cut both the tension and the compression in the boom.

Contact and friction

Rigid-body problems often involve contact with surfaces. A smooth surface provides only a normal force. A rough surface also provides friction, which can be as large as \(\mu_s N\) before slipping starts (from PHYS-130). In equilibrium problems, find the friction force needed from the equations, then compare it with \(\mu_s N\).

Safety guidance on ladder angle follows directly from this kind of analysis. The common "4 to 1" rule (foot one unit out for every four units up) gives a steeper ladder than the example, which reduces the friction needed at the foot.

Two-force and three-force members

Two observations save a lot of algebra.

A two-force member has forces applied at only two points and no couples. For it to be in equilibrium, the two forces must be equal, opposite and collinear: both act along the line joining the two points. Links, struts, hydraulic cylinders (ignoring their weight) and the members of a pin-jointed truss are two-force members. You know the direction of the force immediately, so each one adds only one unknown, its magnitude.

A three-force member has forces at three points. For moment equilibrium, the three lines of action must meet at a single point (or all be parallel). If you know two of the lines, the third must pass through their intersection, which fixes its direction.

The large cylinder force comes from its short lever arm: it acts close to the hinge to lift a load much farther out. That is a typical trade-off in machine design: a short stroke near the hinge gives a large rotation, at the price of a large force.

Can statics solve it? Determinacy and constraints

Count the unknown reactions and compare them with the three equilibrium equations.

• Statically determinate: exactly three unknowns, properly arranged. Statics alone gives every reaction. The examples above are all determinate.
• Statically indeterminate: more unknowns than equations, such as a beam on three supports or a cantilever with an extra prop at its tip. The extra supports are called redundants. Equilibrium alone cannot find the reactions; you need the deformation of the members, which is covered in mechanics of materials and structural analysis. Indeterminate structures are common in practice because the extra supports add safety and stiffness.
• Improperly constrained: enough unknowns, but arranged so they cannot prevent some motion. If all reactions are parallel (a beam on three rollers with an inclined load) the body can slide. If all reaction lines pass through one point, the body can rotate about it. Such a body is not stable, whatever the numbers say.

Rigid bodies in three dimensions

In 3D, equilibrium requires six scalar equations:

so up to six unknowns can be found. Supports add more possible reactions: a ball-and-socket joint gives three force components, a journal bearing gives forces perpendicular to its shaft, and a fixed support gives three forces and three couples. The strategy stays the same: choose moment axes that pass through as many unknowns as possible. A force parallel to an axis, or crossing it, has no moment about it, so a well-chosen axis can isolate one unknown.

That result is a general rule worth remembering. A body supported at three points is stable only if the resultant of its loads falls inside the triangle formed by the supports. The same rule explains why a tripod is steady on uneven ground, why a loaded shelf bracket can tip forward, and why mobile cranes put their outriggers out as wide as possible.