Distributed Loads
Chapters 4.9 Statics
Distributed Loads
Often structures or bodies are subjected to loads that are
spread out over its surface, or part of it.
Think of things like books packed on a shelf, wind pressure
on a billboard, bricks on the back of a truck, or water pressure
on a damn wall.
These are all distributed loadings. The intensity of the
pressure at each point is given by pascals 𝑃𝑃𝑃𝑃 (or 𝑁𝑁/𝑚𝑚2 )
Distributed Loads
Think about a shelf (beam) holding up a number of books, or a beam
under a concrete floor.
If we assume the beam has a constant width,
it may have a pressure distribution as shown
of 𝑝𝑝 = 𝑝𝑝(𝑥𝑥) [𝑁𝑁/𝑚𝑚2 ]
As the beam has constant width the only
variable is 𝑥𝑥, and as such the pressure varies
only along the beam's length
We can therefore represent it as a coplanar
distribution
Most distributed loadings that we encounter in
engineering can be represented in 2D, or in
terms of a single axis.
Distributed Loads
The loading now only varies with relation to the
x axis.
To represent it in 2D we can say 𝑤𝑤 𝑥𝑥 = 𝑝𝑝 𝑥𝑥 𝑏𝑏,
where b is the beam width
The resultant of the distributed loading
can be found as follows:
𝐹𝐹𝑅𝑅 = ∑𝐹𝐹;
Simply, the resultant force is equal to the
area under the distributed load graph!
Distributed loads
Now that we have the magnitude of the resultant force, we need to know
where it acts.
We can do this by calculating moments
about point O,
̅ 𝑅𝑅 = − � 𝑥𝑥𝑥𝑥 𝑥𝑥 𝑑𝑑𝑑𝑑
(𝑀𝑀𝑅𝑅 )𝑂𝑂 = ∑𝑀𝑀𝑂𝑂 ; −𝑥𝑥𝐹𝐹
𝐿𝐿
This coordinate 𝑥𝑥,̅ locates the geometric center or centroid of the
area under the distributed loading. In other words, the resultant
force has a line of action which passes through the centroid C
(geometric center) of the area under the loading diagram
Distributed Loads
In most cases, the distributed loadings diagram will take the form of
rectangles, triangles, or trapezoids.
They are represented by the letter 𝒘𝒘 or 𝑤𝑤0 (or 𝑤𝑤(𝑥𝑥) if it’s a
function) and has the units N/m
To find the equivalent force we simply find the area under the load
by multiplying 𝒘𝒘 by the length of the load (in meters, m)
This equivalent force acts through the centroid of the area under
the curve
Let’s do a quick example!
Class Problem 1
In this case it may be easier to split the
distributed load into 2 components:
1. Rectangle
𝐹𝐹𝑅𝑅𝑅 = 2 6 = 12 𝑘𝑘𝑘𝑘
And it acts at 3m to the left of A
1. Triangle
1
3
𝐹𝐹𝑅𝑅𝑅 =
2
And it acts at
2
3
6 = 9 𝑘𝑘𝑘𝑘
6 = 4m to the left of A
𝐹𝐹𝑅𝑅𝑅
𝐹𝐹𝑅𝑅𝑅
Class Problem 1
Once the components are resolved we can move onto solving the reactions.
The reactions are shown on the diagram
To solve we can first solve for 𝐵𝐵𝑦𝑦 , by taking
moments about point 𝐴𝐴.
3m
⤽ ∑𝑀𝑀𝐴𝐴 = 0; 𝐵𝐵𝑦𝑦 (4) − 𝐹𝐹𝑅𝑅𝑅 (3) − 𝐹𝐹𝑅𝑅𝑅 (4) = 0
𝐴𝐴𝑥𝑥
⟹ 𝐵𝐵𝑦𝑦 = 18 𝑘𝑘𝑘𝑘
The we can solve for equilibrium in the x
and y directions
→ ∑𝐹𝐹𝑥𝑥 = 0;
⟹ 𝐴𝐴𝑥𝑥 = 0 𝑘𝑘𝑘𝑘
↑ ∑𝐹𝐹𝑦𝑦 = 0; 𝐴𝐴𝑦𝑦 +𝐵𝐵𝑦𝑦 −𝐹𝐹𝑅𝑅𝑅 − 𝐹𝐹𝑅𝑅𝑅 = 0
⟹ 𝐴𝐴𝑦𝑦 = 3 𝑘𝑘𝑘𝑘
𝐴𝐴𝑦𝑦
12𝑘𝑘𝑘𝑘
9𝑘𝑘𝑘𝑘
𝐵𝐵𝑦𝑦
Class Problem 2
To find the magnitude and location of the
resultant force we simply apply the
formulas
↑ (𝐹𝐹𝑅𝑅 )𝑦𝑦 = ∑𝐹𝐹𝑦𝑦 ;
𝐹𝐹𝑅𝑅 = − �
3𝑚𝑚
0
1 3
1 4 3𝑚𝑚
𝑥𝑥 = − 𝑥𝑥 �
3
12
0
𝐹𝐹𝑅𝑅 = −6.75 𝑘𝑘𝑘𝑘
31 3
1 5 3
𝑥𝑥 ∫0
𝑥𝑥
𝑑𝑑𝑑𝑑
𝑥𝑥
∫𝐿𝐿 𝑥𝑥 𝑤𝑤(𝑥𝑥) 𝑑𝑑𝑑𝑑 ∫0 3
15
=
=
𝑥𝑥̅ =
31 3
1 4 3
∫𝐿𝐿 𝑤𝑤(𝑥𝑥) 𝑑𝑑𝑑𝑑
𝑥𝑥
∫0 3 𝑥𝑥 𝑑𝑑𝑑𝑑
12 ∫0
= 2,4 𝑚𝑚
Common Load Types
https://learnaboutstructures.com/Common-Load-Types-for-Beams-and-Frames