How are point loads, UDL, UVL and couples on a beam converted into equivalent forces?

In one line

Each distributed load is replaced by a single force equal to its area acting through the centroid of the load diagram; couples are kept as moments.

Explained simply

Sand spread evenly on a plank presses all along it, but for finding the support reactions, it acts like one big sack placed at the middle of the sand. If the sand is piled more on one side (triangular pile), the 'sack' sits closer to the heavy side.

Technically, the resultant of a distributed load equals the area under the load-intensity diagram and acts through its centroid.

How it works

1. Point load: acts at a point (kN).
2. UDL of intensity w (kN/m) over length L: equivalent load \( wL \) at L/2.
3. UVL (triangular) from 0 to w over L: equivalent load \( wL/2 \) at L/3 from the larger end (2L/3 from the zero end).
4. Trapezoidal: split into a UDL plus a triangle.
5. Applied couple / moment (kN·m): no net force; include it only in the moment equation with its sense.
6. Inclined point load: resolve into components along and perpendicular to the beam.

Key formula

UDL: \( P = wL \) at \( L/2 \). UVL: \( P = \tfrac12 wL \) at \( L/3 \) from the maximum-intensity end. w in kN/m, L in m, P in kN.

Quick example

A triangular load rising from 0 at A to 6 kN/m at 3 m: \( P = \tfrac12 \times 6 \times 3 = 9 \) kN acting 2 m from A (1 m from the 6 kN/m end).

Common mistakes

• Placing the UVL resultant at L/3 from the zero end instead of from the maximum end.
• Writing the UDL resultant as w instead of wL.
• Adding an applied couple to \( \Sigma F_y \).

How MU asks it

Not usually asked alone, but every beam-reaction numerical (6 to 8 marks) has mixed loads. Show the equivalent loads on a separate line or sketch, e.g. 'UDL: 2 × 3 = 6 kN at 1.5 m from C', before writing the equations.