DoAssignment.ca

2.5 · Replace a simple distributed load with an equivalent resultant

Learn to replace a simple distributed load with an equivalent resultant through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Force-System Resultants

ENGG 130 Engineering Mechanics: Statics — Study topic 2.5

A distributed load acts along a length rather than at one point. To simplify a statics model, we can replace a simple distributed load with one concentrated force that has the same total force and the same moment about any point. This replacement preserves the load’s external effect on the body; it does not mean the original load was physically applied at one point. In this lesson, the body is a planar beam segment, and loads act vertically downward. We use a horizontal coordinate measured from the left end and take counterclockwise moments as positive.

What you will learn

  • Explain what it means for a concentrated resultant to be equivalent to a distributed load.
  • Find the magnitude and location of a resultant for uniform, triangular, and trapezoidal load patterns.
  • Use force and moment balance to check an equivalent-load replacement.

1. What makes a resultant equivalent?

A load intensity describes force per unit length. For example, w=4 kN/mw=4\,\mathrm{kN/m} means that each metre of the loaded length contributes a downward force of 4 kN4\,\mathrm{kN}. A distributed load can vary with position, so its intensity is written as w(x)w(x).
The equivalent resultant must have the same total force as the distributed load. Its line of action must also produce the same moment. For a downward load over 0≤x≤L0\le x\le L, the magnitude is the area under the load-intensity graph. Its location is the centroid of that area, measured from the chosen reference end.
For a small length dxdx, the force magnitude is dF=w(x) dxdF=w(x)\,dx. Adding these contributions gives the total force. Adding their moments about the left end gives the resultant’s moment. Dividing that moment by the total force gives the resultant location. This is the integral form of the area-and-centroid method.
R=∫0Lw(x) dx,xR=∫0Lxw(x) dxRR=\int_0^L w(x)\,dx,\qquad x_R=\frac{\int_0^L xw(x)\,dx}{R}
  • The area under a load-intensity graph gives force, not force per unit length.
  • The centroid location is measured from a stated reference point.
  • Use one concentrated force at the centroid of the load diagram.

2. Basic load shapes and their locations

For a uniform load of intensity w0w_0 over length LL, the load diagram is a rectangle. Its area is w0Lw_0L, and its centroid is halfway along the length. Thus the resultant has magnitude w0Lw_0L and acts at L/2L/2 from either end.
For a triangular load that rises linearly from zero at the left end to a maximum w0w_0 at the right end, the diagram is a triangle. Its area is w0L/2w_0L/2. The centroid lies one-third of the base length from the larger-intensity end, or two-thirds of the length from the zero-intensity end.
A trapezoidal load can be treated as a rectangle plus a triangle. Find each area and its centroid, then use moment balance to locate the combined resultant. This avoids memorizing a separate centroid location for every trapezoid.
The equations below assume the load points downward. With upward chosen as positive, the force component of the resultant is negative. Many diagrams instead label only the positive magnitude and show a downward arrow; either notation is acceptable if used consistently.
Runiform=w0L,Rtriangle=w0L2R_{\mathrm{uniform}}=w_0L,\qquad R_{\mathrm{triangle}}=\frac{w_0L}{2}
  • Uniform load: resultant magnitude is intensity times length; location is the midpoint.
  • Triangular load: resultant magnitude is half the maximum intensity times length.
  • For a changing load, use the area-weighted centroid or sum component-load moments.

3. A reliable replacement procedure

First identify the loaded body and the interval over which the load acts. Sketch the load-intensity shape and mark the reference coordinate. This is the load diagram; when the beam itself is being simplified for equilibrium, also show the concentrated replacement force at its computed location.
Next choose axes and signs. Here, +x+x is to the right, +y+y is upward, and counterclockwise moment is positive. A downward resultant therefore has Ry=−RR_y=-R when RR denotes its positive magnitude.
Find the total force from the diagram’s area or by integration. Then find the position by matching moments about the reference end. The governing equivalence conditions are equal force and equal moment. For a downward force located a distance xRx_R from the left end, its moment about that end is clockwise.
Finally, check units. Intensity multiplied by length gives force, while force multiplied by distance gives moment. A value reported as a moment location must have units of length, not force or moment. If a component decomposition is used, verify the total force and moment independently.
∑Fy=0,∑MO=0\sum F_y=0,\qquad \sum M_O=0
  • State the reference end before reporting the centroid location.
  • A downward force to the right of the reference point creates a clockwise moment.
  • Equivalent means both the total force and the total moment are preserved.

4. Reading the replacement in a statics model

When a distributed load on a beam is replaced by its resultant, remove the original distributed-load arrows from that simplified model and draw one force at the centroid location. Do not keep both the original load and the resultant in the same equilibrium equation; that would count the same loading twice.
If other forces or supports are present, the resultant can be used in the beam’s usual planar equilibrium equations. The load replacement itself does not determine support reactions; it only makes the distributed load easier to represent while preserving its external force and moment.
For a load that is not a basic shape, the same integral definitions apply when the intensity is given as a function of position. In this topic, use them only to find the total resultant and its location.
MO=RxRM_O=Rx_R
  • The resultant replaces, rather than supplements, the distributed load.
  • Keep the force direction and moment sign consistent with the chosen axes.
  • Check the replacement about more than one point when useful; matching force and one moment fixes the moment about any other point.

Worked example

Uniform load on a beam segment

A 3.0 m beam segment carries a uniform downward load of 2.4 kN/m. Replace the distributed load by an equivalent concentrated force, and verify the force and moment about the left end.
  1. Define the system and load
    Take the beam segment as the system. The load-intensity diagram is a rectangle with height 2.4 kN/m2.4\,\mathrm{kN/m} and width 3.0 m3.0\,\mathrm{m}. The replacement resultant is downward.
  2. Find the total force
    The rectangular area gives the resultant magnitude. Its units reduce to force.
    R=(2.4 kN/m)(3.0 m)=7.2 kNR=(2.4\,\mathrm{kN/m})(3.0\,\mathrm{m})=7.2\,\mathrm{kN}
  3. Locate the force
    A rectangle’s centroid is at its midpoint, so the resultant acts 1.5 m from the left end. With upward positive, its vertical component is negative.
    xR=3.02=1.5 m,Ry=−7.2 kNx_R=\frac{3.0}{2}=1.5\,\mathrm{m},\qquad R_y=-7.2\,\mathrm{kN}
  4. Verify force and moment
    The concentrated force equals the total distributed force. Its moment about the left end is clockwise; the distributed load gives the same moment because its centroid is at the same location.
    ∑Fy=−7.2 kN,MO=−(7.2)(1.5)=−10.8 kN⋅m\sum F_y=-7.2\,\mathrm{kN},\qquad M_O=-(7.2)(1.5)=-10.8\,\mathrm{kN\cdot m}
Answer: Replace the load by a 7.2 kN downward force at 1.5 m from the left end.
Check: The original load’s total force is 7.2 kN, and its moment about the left end is 10.8 kN·m clockwise. The replacement has the same force and moment.

Worked example

Triangular load increasing toward the right

A 4.0 m beam segment carries a downward triangular load that is zero at the left end and rises linearly to 6.0 kN/m at the right end. Find the equivalent resultant and verify its moment about the left end.
  1. Define the load function
    Let x=0x=0 at the zero-load end. Because the intensity rises linearly, it is proportional to xx and reaches 6.0 kN/m6.0\,\mathrm{kN/m} at x=4.0 mx=4.0\,\mathrm{m}.
    w(x)=6.04.0x kN/mw(x)=\frac{6.0}{4.0}x\,\mathrm{kN/m}
  2. Find the resultant magnitude
    The load diagram is a triangle, so its area gives the total downward force.
    R=12(4.0 m)(6.0 kN/m)=12 kNR=\frac{1}{2}(4.0\,\mathrm{m})(6.0\,\mathrm{kN/m})=12\,\mathrm{kN}
  3. Locate the resultant
    A triangular area’s centroid is one-third of its base from the high-intensity end. Here that is one-third of 4.0 m from the right, or two-thirds from the left.
    xR=23(4.0 m)=2.67 mx_R=\frac{2}{3}(4.0\,\mathrm{m})=2.67\,\mathrm{m}
  4. Check the moment
    The resultant produces a clockwise moment about the left end. Integration of the linearly varying intensity gives the same moment magnitude.
    MO=−(12)(2.67)=−32.0 kN⋅mM_O=-(12)(2.67)=-32.0\,\mathrm{kN\cdot m}
Answer: Replace the triangular load by a 12 kN downward force 2.67 m from the left end, equivalently 1.33 m from the right end.
Check: The triangular load’s area is 12 kN. Its moment magnitude about the left end is 32.0 kN·m, matching the resultant’s moment.

Worked example

Trapezoidal load using two simple shapes

A 5.0 m beam segment carries a downward load that increases linearly from 2.0 kN/m at the left end to 8.0 kN/m at the right end. Replace it with one resultant and verify its force and moment about the left end.
  1. Split the load diagram
    Represent the trapezoid as a uniform 2.0 kN/m load over 5.0 m plus a triangular load that rises from zero to 6.0 kN/m. The triangle accounts for the increase above the uniform part.
    w(x)=2.0+6.05.0x kN/mw(x)=2.0+\frac{6.0}{5.0}x\,\mathrm{kN/m}
  2. Add the component forces
    The rectangle contributes 10 kN at the midpoint. The triangle contributes 15 kN, located two-thirds of the length from the left. Adding these forces gives the total resultant.
    R1=(2.0)(5.0)=10 kN,R2=12(6.0)(5.0)=15 kN,R=25 kNR_1=(2.0)(5.0)=10\,\mathrm{kN},\quad R_2=\frac{1}{2}(6.0)(5.0)=15\,\mathrm{kN},\quad R=25\,\mathrm{kN}
  3. Match moments to locate it
    Take moments about the left end. The rectangle acts at 2.5 m and the triangle at 3.33 m. Their clockwise moments add; divide the total moment magnitude by the total force to get the location.
    xR=(10)(2.5)+(15)(3.33)25=3.00 mx_R=\frac{(10)(2.5)+(15)(3.33)}{25}=3.00\,\mathrm{m}
  4. Verify the replacement
    The total force is 25 kN downward. The original component moments sum to 75 kN·m clockwise, which is also the moment of the 25 kN resultant at 3.00 m.
    MO=−(10)(2.5)−(15)(3.33)=−75.0 kN⋅m=−(25)(3.00)M_O=-(10)(2.5)-(15)(3.33)=-75.0\,\mathrm{kN\cdot m}=-(25)(3.00)
Answer: Replace the trapezoidal load by a 25 kN downward force 3.00 m from the left end.
Check: The resultant preserves the 25 kN total downward force and the 75.0 kN·m clockwise moment about the left end.

Common mistakes and how to avoid them

Multiplying the load intensity by the length but reporting the answer in kN/m.
Correction: Intensity times length has units of force. For example, kN/m multiplied by m gives kN.
Putting the triangular resultant one-third of the length from the zero-intensity end.
Correction: For a triangle, the centroid is one-third of the base from the high-intensity end and two-thirds from the zero-intensity end.
Replacing the load with a force of the correct magnitude but placing it at the wrong position.
Correction: Also match the load’s moment about a reference point. The resultant must act at the load diagram’s centroid.
Keeping the distributed arrows and adding the resultant to the same equilibrium model.
Correction: The resultant replaces the distributed load. Including both counts the same loading twice.

Lesson summary

  • The equivalent resultant has the same total force and moment as the distributed load.
  • For simple load shapes, the resultant magnitude is the area of the load-intensity diagram.
  • The resultant acts through the centroid of that area.
  • For a trapezoid, combine simple shapes and match their total moment to locate the single force.

Check your understanding

Question 1

A 2.0 m beam segment carries a uniform downward load of 3.0 kN/m. What is the equivalent resultant?
  1. 6.0 kN downward at the midpoint
  2. 1.5 kN downward at the midpoint
  3. 6.0 kN downward at one-third of the length from the left
  4. 3.0 kN downward at the right end
Show answer and explanation
6.0 kN downward at the midpoint
The rectangular load area is (3.0 kN/m)(2.0 m)=6.0 kN(3.0\,\mathrm{kN/m})(2.0\,\mathrm{m})=6.0\,\mathrm{kN}. A uniform load’s centroid is at the midpoint.

Question 2

A triangular load rises from zero at the left to a maximum at the right. Where does its resultant act?
  1. One-third of the length from the left end
  2. One-half of the length from the left end
  3. Two-thirds of the length from the left end
  4. At the right end
Show answer and explanation
Two-thirds of the length from the left end
The triangular area’s centroid is two-thirds of the base from the zero-intensity end, so it lies two-thirds of the length from the left.

Question 3

A distributed load and a proposed resultant have the same total force. What else must be true for them to be equivalent?
  1. They must have the same maximum intensity
  2. They must create the same moment about a reference point
  3. The resultant must act at an end of the beam
  4. The resultant must have units of force per length
Show answer and explanation
They must create the same moment about a reference point
Equal total force alone does not preserve the load’s turning effect. Matching the moment fixes the correct line of action.

Key terms

Load intensity
Force per unit length of a distributed load, such as kN/m.
Resultant
A single force that replaces a distributed load while preserving its total force and moment.
Centroid
The balance location of an area; for a load diagram, it gives the line of action of the equivalent resultant.

Continue through ENGG 130

View the complete ENGG 130 University of Alberta ENGG 130: Engineering Mechanics: Statics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows University of Alberta ENGG 130: Engineering Mechanics: Statics, study topic 2.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question