8.5 Connect a distributed load resultant to an area centroid | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Centroids and Centres of Gravity

Replacing a load-intensity graph with one equivalent force

A distributed load acts along a length instead of at one point. In planar statics, we can replace it with one force if that force preserves the original load’s total force and moment on the body. Its magnitude is the area under the load-intensity graph, and its location is the horizontal coordinate of that area’s centroid. This lesson develops the connection for downward loads on straight beams. Treat each beam as a rigid body, and measure load intensity in force per unit length.

What you will learn

Find a distributed load’s resultant magnitude from the area under its load-intensity graph.
Locate the resultant using the centroid of that area.
Replace simple distributed loads with equivalent forces and verify their total force and moment.

1. Load intensity and resultant force

Define the beam segment being considered and choose a coordinate

x

along it. Let

w(x)

be the positive magnitude of the downward load intensity. For example,

w

may have units of

\mathrm{kN/m}

. A short segment of length

dx

carries a small downward force with magnitude

dF=w(x)\,dx

. Adding these small forces over the loaded interval gives the resultant magnitude.

On a graph with distance along the beam on the horizontal axis and load intensity on the vertical axis, the area under the graph has units of force. A uniform load gives a rectangle. A load that changes linearly gives a triangle or trapezoid. The resultant points in the same direction as the distributed load.

The area gives the force magnitude, but not its location. That location matters because a force farther from a reference point creates a larger moment. The resultant must act at the centroid of the load-intensity area.

R=\int_a^b w(x)\,dx

The area under the load-intensity graph gives the resultant’s magnitude.
The centroid’s horizontal coordinate gives the line of action along the beam.
Keep intensity, force, and distance units distinct.

2. Why the centroid locates the resultant

Take moments about the left end of the loaded interval. Each small downward force at position

x

contributes a clockwise moment with magnitude x dF. Adding those contributions gives the total moment magnitude of the distributed load. A single downward force

R

at

x_R

has moment magnitude

R x_R

. For the replacement to preserve the moment, these magnitudes must be equal.

Combining the force and moment relations gives the centroid coordinate of the load-intensity area. The numerator is the area’s first moment about the origin: each small area is multiplied by its distance from the origin. Dividing by the total area gives a distance.

For a rectangular intensity graph, the centroid is at the midpoint. For a triangular graph that rises from zero at the left to a maximum at the right, its centroid is two-thirds of the length from the zero-intensity end. If the triangle falls from a maximum at the left to zero at the right, the centroid is one-third of the length from the left.

For a combination of simple shapes, calculate each area and its centroid. Add the force magnitudes and add their moments about the same origin. Dividing the total moment magnitude by the total force gives the combined resultant’s location.

x_R=\frac{\int_a^b xw(x)\,dx}{R}

The resultant’s moment must equal the distributed load’s moment.
A triangle’s centroid is closer to its larger-intensity end.
For combined areas, add moments before finding the combined location.

3. A consistent statics procedure

Define the beam segment and draw its load-intensity shape, marking the interval and endpoint intensities. Choose the left end as the origin unless another reference is more convenient. For a downward load, use its positive intensity magnitude

w(x)

in area calculations.

Use

+x

to the right and

+y

upward. Take counterclockwise moments as positive. The distributed load and its equivalent force both point downward, so their signed vertical forces are negative. Their moments about the left end are also negative because they turn the beam clockwise.

Find the resultant magnitude from the area or integral, then find its location from the area centroid or moment relation. Finally compare the original and equivalent total vertical forces and moments about the same point. Matching both confirms that the replacement is valid for planar statics.

The replacement is a statics model: it preserves the total force and moment of the distributed load on the rigid body. It does not claim that the load is physically applied at one point.

\sum F_y=-R,\quad \sum M_O=-Rx_R

State the origin and sign convention before calculating moments.
A downward resultant has negative vertical force and, to the right of the origin, a negative moment.
Verify both force and moment, not just the area.

Worked example

1. A triangular load rising from zero

A beam carries a downward triangular load over

4\,\mathrm{m}

. The intensity is zero at the left end and rises linearly to

6\,\mathrm{kN/m}

at the right end. Replace the load with one equivalent force and verify its force and moment about the left end.

Define the load and origin
Isolate the beam segment and take its left end as $x=0$ . The intensity graph is a triangle with base $4\,\mathrm{m}$ and height $6\,\mathrm{kN/m}$ . The equivalent force acts downward.
Find the area and centroid
The triangular area gives the force magnitude. Because the triangle rises from zero at the left, its centroid is two-thirds of the length from that end.
$R=\frac{1}{2}(4\,\mathrm{m})(6\,\mathrm{kN/m})=12\,\mathrm{kN},\quad x_R=\frac{2}{3}(4\,\mathrm{m})=2.67\,\mathrm{m}$
Verify force and moment
With upward force and counterclockwise moment positive, the resultant has negative vertical force and negative moment about the left end. The triangular load’s moment magnitude is its area times its centroid distance, giving the same value.
$\sum F_y=-12\,\mathrm{kN},\quad \sum M_O=-(12\,\mathrm{kN})(2.67\,\mathrm{m})\approx-32.0\,\mathrm{kN\cdot m}$

Answer: The equivalent force is

12\,\mathrm{kN}

downward at

2.67\,\mathrm{m}

from the left end.

Check: The original triangular load has moment magnitude

12\,\mathrm{kN}\times 2.67\,\mathrm{m}\approx32.0\,\mathrm{kN\cdot m}

. Its signed force and moment match those of the equivalent force.

Worked example

2. A uniform load plus a triangular load

A

3\,\mathrm{m}

beam segment carries a uniform downward load of

2\,\mathrm{kN/m}

over its full length. It also carries a separate triangular downward load that rises from zero at the left to

4\,\mathrm{kN/m}

at the right. Find one equivalent resultant and its location from the left end.

Split the load-intensity graph
Treat the uniform part as a rectangle and the additional varying part as a triangle. The rectangle’s centroid is at the midpoint. The rising triangle’s centroid is two-thirds of the length from the left.
$R_1=(2\,\mathrm{kN/m})(3\,\mathrm{m})=6\,\mathrm{kN},\quad R_2=\frac{1}{2}(3\,\mathrm{m})(4\,\mathrm{kN/m})=6\,\mathrm{kN}$
Add force and moment contributions
Both component forces are downward. Their magnitudes add, and their clockwise moment magnitudes about the left end add. Dividing total moment magnitude by total force gives the resultant location.
$R=12\,\mathrm{kN},\quad x_R=\frac{(6\,\mathrm{kN})(1.5\,\mathrm{m})+(6\,\mathrm{kN})(2.0\,\mathrm{m})}{12\,\mathrm{kN}}=1.75\,\mathrm{m}$
Verify the equivalent
The signed vertical force is downward. The resultant’s clockwise moment equals the sum of the rectangle’s and triangle’s clockwise moments.
$\sum F_y=-12\,\mathrm{kN},\quad \sum M_O=-(12\,\mathrm{kN})(1.75\,\mathrm{m})=-21\,\mathrm{kN\cdot m}$

Answer: The combined resultant is

12\,\mathrm{kN}

downward at

1.75\,\mathrm{m}

from the left end.

Check: The separate moment magnitudes sum to

(6\,\mathrm{kN})(1.5\,\mathrm{m})+(6\,\mathrm{kN})(2.0\,\mathrm{m})=21\,\mathrm{kN\cdot m}

, matching the resultant.

Worked example

3. A trapezoidal load using integration

A

6\,\mathrm{m}

beam segment carries a downward load whose intensity increases linearly from

2\,\mathrm{N/m}

at the left end to

5\,\mathrm{N/m}

at the right end. Find the resultant and location, then verify the moment using integration.

Represent the linear intensity
Take the left end as $x=0$ . The intensity increases by $3\,\mathrm{N/m}$ over $6\,\mathrm{m}$ , so its slope is $0.5\,\mathrm{N/m^2}$ . The resulting function describes the stated endpoint values.
$w(x)=2+0.5x\;\mathrm{N/m},\quad 0\le x\le6\,\mathrm{m}$
Integrate to find the resultant
Integrate intensity over the length. The product of force per length and length has units of force.
$R=\int_0^{6\,\mathrm{m}}(2+0.5x)\,dx=21\,\mathrm{N}$
Find its location
The first moment of the load-intensity area about the left end is the integral of position times intensity. Divide it by the resultant to obtain a distance.
$x_R=\frac{\int_0^{6\,\mathrm{m}}x(2+0.5x)\,dx}{21\,\mathrm{N}}=\frac{72\,\mathrm{N\cdot m}}{21\,\mathrm{N}}=3.43\,\mathrm{m}$
Verify force and moment
The integrated force is downward, and its moment about the left end is clockwise. The equivalent force has the same signed force and moment.
$\sum F_y=-21\,\mathrm{N},\quad \sum M_O=-(21\,\mathrm{N})(3.43\,\mathrm{m})\approx-72\,\mathrm{N\cdot m}$

Answer: The resultant is

21\,\mathrm{N}

downward at approximately

3.43\,\mathrm{m}

from the left end.

Check: The resultant lies to the right of the midpoint, consistent with the higher intensity toward the right. Its moment magnitude is approximately

72\,\mathrm{N\cdot m}

, equal to the integrated first moment.

Common mistakes and how to avoid them

Using maximum intensity times the full length for a triangular load.

Correction: Use the triangular area: one-half times base times height.

Placing every resultant at the midpoint of its interval.

Correction: The midpoint is the centroid of a uniform rectangular intensity graph. Other shapes generally have different centroids.

Measuring a rising triangle’s centroid from its high-intensity end when using the two-thirds rule.

Correction: Measure two-thirds of the length from the zero-intensity end, or one-third from the high-intensity end.

Checking the resultant force but not its moment.

Correction: Compare both total force and moment about the same reference point. Matching the area alone does not establish the correct location.

Lesson summary

The area under a load-intensity graph gives the resultant force magnitude.
The graph area’s centroid gives the resultant’s location along the beam.
For combined shapes, add the force contributions and their moments before locating the combined resultant.
Use consistent units and verify both force and moment.

Check your understanding

Question 1

A triangular downward load rises from zero to

8\,\mathrm{kN/m}

over

3\,\mathrm{m}

. What is its resultant magnitude?

$12\,\mathrm{kN}$
$24\,\mathrm{kN}$
$8\,\mathrm{kN}$
$4\,\mathrm{kN}$

Show answer and explanation

12\,\mathrm{kN}

Its intensity graph is a triangle, so its area and resultant magnitude are

\frac{1}{2}(3\,\mathrm{m})(8\,\mathrm{kN/m})=12\,\mathrm{kN}

.

Question 2

For that same triangle, where does the resultant act measured from the zero-intensity end?

$1.0\,\mathrm{m}$
$1.5\,\mathrm{m}$
$2.0\,\mathrm{m}$
$2.5\,\mathrm{m}$

Show answer and explanation

2.0\,\mathrm{m}

The centroid of a triangle that rises from zero is two-thirds of the length from the zero-intensity end:

\frac{2}{3}(3\,\mathrm{m})=2.0\,\mathrm{m}

.

Key terms

Distributed load: A load spread along a length and described by force per unit length.
Load intensity: The force applied per unit length at a position along the beam.
Resultant: A single force that preserves the total force and moment of the distributed load.
Centroid: The geometric centre of an area; for a load-intensity graph, its horizontal coordinate locates the resultant.

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8.5 · Connect a distributed load resultant to an area centroid

University of Alberta ENGG 130: Engineering Mechanics: Statics

Replacing a load-intensity graph with one equivalent force

What you will learn

1. Load intensity and resultant force

2. Why the centroid locates the resultant

3. A consistent statics procedure

1. A triangular load rising from zero

2. A uniform load plus a triangular load

3. A trapezoidal load using integration

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through ENGG 130

About this lesson and its review