ENG1011 Chap.5 Distributed Loads and Equivalent Point Loads
Distributed Loads and Equivalent Point Loads
Distributed Loads and Equivalent Point Loads completes Week 3 by handling loads spread along a length. A distributed load has an intensity in kilonewtons per metre, and for finding reactions it can be replaced by a single force equal to the area under the load diagram, acting through the centroid of that area.
Uniform loads give a rectangle whose force acts at the middle; triangular loads give half the base times the height, acting one third of the base from the tall end; trapezoids are split into a rectangle and a triangle whose forces are combined with a moment balance.
The combined force always lies between the centroids of its parts, nearer the larger one, which is enough to eliminate options in a multiple choice question before calculating.
The replacement is only valid for the external reactions: the free-body diagram should still show the load as given, and the beam's internal shear and moment in Week 8 depend on the true distribution.
The 2026 Content Test contained three separate distributed-load items, so this short topic carries more marks than its size suggests.
Typical items ask for the equivalent force of a composite load, the position of that force from a named point, or a beam reaction once the load has been replaced.
The practice test solutions model a fast technique for the position question: bracket the answer between the part centroids, lean towards the larger part, and use the calculation only to confirm.
Distributed loads also prepare the ground for Week 8, where the same loads are kept as distributions to draw shear force and bending moment diagrams, and for Week 9, where a uniform load enters the standard deflection formula.
What this chapter covers
- 01
Load intensity and its units
- 02
Equivalent force equals the area under the load
- 03
Position at the centroid of the load area
- 04
Uniform, triangular and trapezoidal shapes
- 05
Combining parts with a moment balance
- 06
Estimating the resultant's position before calculating
- 07
Reactions under full-span and part-span loads
Worked example · free
Reactions under a triangular load
- 1Equivalent force: half of 6 kN/m times 4 m = 12 kN.
- 1Position: one third of the span from the tall end at B, so 8/3 = 2.67 m from A.
- 1Moments about A: 4 By = 12 times 8/3 = 32, so By = 8 kN.
- 1Vertical balance: Ay = 12 minus 8 = 4 kN.
Key terms
- Load intensity
- The force per unit length of a distributed load, usually given in kilonewtons per metre.
- Uniformly distributed load
- A load of constant intensity along a length, replaced by its total acting at the middle.
- Centroid
- The geometric centre of an area, which locates the equivalent force of a distributed load.
- Triangular load
- A load whose intensity rises linearly from zero, replaced by half its base times its height.
Distributed Loads and Equivalent Point Loads FAQ
Can I always replace a distributed load with one force?
For calculating external reactions, yes: the single force has the same total push and turning effect. For internal shear force and bending moment along the beam, no, because the true distribution changes how those actions vary with position.
How do I handle a trapezoidal load?
Split it into a rectangle and a triangle. Find each part's force from its area and its position from its centroid, then combine them: the resultant equals the sum of the forces, and its position comes from the sum of each force times its position divided by that total.
Is there a quick way to check the resultant position?
The resultant must lie between the centroids of the parts and closer to the larger part. If your calculated position falls outside that range, or nearer the smaller part, recheck the centroid distances.
Where does a part-span uniform load act?
At the middle of the loaded length, not the middle of the beam. A 3 kN/m load over the first 4 m of a 6 m beam is 12 kN acting 2 m from the loaded end, and the reactions then follow from moments about one support.
Exam move
Memorise the three standard shapes, then practise splitting composite loads until it takes seconds. Keep the free-body diagram faithful to the given load, and replace the distribution only inside the moment equation. Use the between-the-centroids estimate on every multiple choice item before calculating.
Make a three-row card of the standard shapes, with force and position for each, and test yourself until you can write both values without looking.
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