75/3848–12FE Reference Handbook 10.6 · Statics · p. 96 Centroid of area, applied to a uniform load diagram (no separate distributed-load formula is printed)
1. Moment & Equilibrium2. Centroid of a Composite Area3. Coulomb Friction4. Force Resultant5. Rectangle Centroidal Moment of Inertia6. Parallel-Axis Theorem7. Distributed Load Resultant8. Method of Joints9. Two-Force Member10. Ideal Pulley11. Moment of a Force12. Simply Supported Beam Reactions13. Load–Shear–Moment Relations14. Method of Sections (Truss)15. Sagging Cable (Small Sag)16. Overhanging-Beam Reactions17. Radius of Gyration18. Triangle Area Moment of Inertia19. Triangular Distributed Load Resultant20. Three-Force Member21. Couple Moment22. Frames: Multi-Force Members23. Zero-Force Truss Members

Handbook formula

Distributed Load Resultant

A distributed line load w(x) is equivalent to a concentrated force W equal to the area under the load diagram, acting at the centroid of that diagram. For a uniform w on length L, W = wL at midspan.

(N/m)
Load intensity
(N)
Resultant force
(m)
Location of the resultant

Step-by-step solved example

Simply supported beam, uniform 4 kN/m over 6 m. Find the equivalent load and its location.

4 kN/m3 m3 mW = 24 kNW = 4 × 6 = 24 kNResultant at midspan, 3 mReplay
Uniform w is a resultant W = wL at mid-length of the loaded patch.

  1. 1. Area

    W = 4×6 = 24 kN.

  2. 2. Centroid

    Uniform rectangle → resultant at 3 m from either end.

Answer: 24 kN at midspan

10 practice questions

0/10 correct

1.Uniform 2 kN/m on 5 m → W =

2.Triangular load 0 to w0 over L has W =

3.That triangular resultant sits

0w0L
Beam of span L with a triangular load, zero at the left end and w0 at the right end.

4.w has units of

5.A 3 kN/m load over 2 m starting 1 m from the left support of a 6 m beam acts

3 kN/m1 m2 m3 m6 m
6 m beam. Uniform 3 kN/m covers 2 m, starting 1 m from the left support. The other 3 m is unloaded.

6.Replacing w(x) by W at the wrong location

7.Hydrostatic on a vertical wall is a

8.Two uniform patches may be

9.w = 8 kN/m, L = 3 m, simply supported. Each reaction =

10.The internal shear just inside the left support equals