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Wize University Statics Textbook (Master) > Force System Resultants (Moment, Couple, Dist. Load)

Distributed Loads

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Distributed loads refer to loading that is not concentrated at one location on the body, but rather acts over a certain portion of the body. Distributed loads are usually expressed by intensity ω, with units of kN/m or N/m.

You will often be required to simplify distributed loading in questions. The most general method works as follows:

1. Find the area under the intensity curve, and that is the magnitude of the equivalent force:




Some common geometric loading you should be familiar with:

1. Rectangular, or uniform loads will have a resultant equal to the shaded area (length x width) and acting at the centroid the mid-point.

2. Triangular, or linearly increasing loads will have a resultant equal to the shaded area (1/2 x length x width) and acting at the centroid one-third of the width away from the big side.



2. For curved and parabolic shapes, the magnitude of FR is equivalent to the sum of all the forces in the system. In this case, integration must be used since there is an infinite number of parallel forces dF acting on the beam as highlighted in the figure below. Since dF is acting on an element of length dx, and w(x) is a force per unit length, then dF = w(x) dx = dA. This means that the magnitude of dF is determined from the colored differential area dA under the loading curve. For the entire length L.









Resultant Force:

FR=−∫Lw(x)dx=−∫AdAF_R=-\int_L^{ }w(x)dx=-\int_A^{ }dA

FR=−∫03m15x3dxF_R=-\int_0^{3m}\frac{1}{5}x^3dx

Resultant Moment:

(MR)O=∫(3−x)dFR=∫03m(3−x)(15x3dx)\left(M_R\right)_O=\int_{ }^{ }(3-x)dF_R=\int_0^{3m}(3-x)\left(\frac{1}{5}x^3dx\right)

Centroid: This coordinate xˉ\bar{x}, locates the geometric center or centroid of the area under the distributed loading.


xˉ=∫Lxw(x)dx∫Lw(x)dx=∫AxdA∫AdA\bar{x}=\frac{\int_{L} x w(x) d x}{\int_{L} w(x) d x}=\frac{\int_{A} x d A}{\int_{A} d A}




Wize Tip
Most distributed loading questions provide you with common geometric shapes. If you’re not provided with a geometric shape you may need to do one of two things:
  • Break down the loading into common geometric shapes; or
  • Integrate the loading to find the equivalent force and its location


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Distributed loads refer to loading that is not concentrated at one location on the body, but rather acts over a certain portion of the body. Distributed loads are usually expressed by intensity ω, with units of kN/m or N/m.

You will often be required to simplify distributed loading in a questions. The most
general method works as follows:
  1. Find the area under the intensity curve, and that is the magnitude of the equivalent force
  2. Find the moment caused by the loading around a certain point, and locate the single equivalent force in such a way that is creates an equivalent moment

FR=∫Lw(x)dx=∫AdA=AF_R=\int_L^{ }w(x)dx=\int_A^{ }dA=A
x‾=∫Lxw(x)dx∫Lw(x)dx=∫AdA∫AdA\overline{x}=\frac{\int_L^{ }xw(x)dx}{\int_L^{ }w(x)dx}=\frac{\int_A^{ }dA}{\int_A^{ }dA}

You may recognize that these are the two steps involved in equivalent systems, which in essence what distributed loading is.

Most distributed loading questions provide you with common geometric shapes. If you’re not provided with a geometric shape you may need to do one of two things:
  • Break down the loading into common geometric shapes; or
  • Integrate the loading to find the equivalent force and its location

Some common geometric loading you should be familiar with:

  1. Rectangular, or uniform loads will have a resultant equal to the shaded area (length x width) and acting at the mid-point.
  2. Triangular, or linearly increasing loads will have a resultant equal to the shaded area (0.5 x length x width) and acting one third of the width away from the big side
Simplify the following distributed loads into a single force.

Set origin at A, positive x to the right.
On the left is a triangular force, F1=0.5∗(6kN/m)∗(3m)=9kNF_1 = 0.5*(6 kN/m)*(3m) = 9 kN acting at x1=−1mx_1 = -1 m
On the right side is another triangular force, F2=0.5∗(6kN/m)∗(6m)=18kNF_2 = 0.5 * (6kN/m)*(6m) = 18 kN acting at x2=2mx_2 = 2m

FR=F1+F2=9+18=27kNF_R = F_1 + F_2 = 9 + 18 = 27 kN

ΣMA=(9)(1)−(18)(2)=−(27)(xR)   →xR=1m\Sigma M_A = (9)(1) - (18)(2) = -(27)(x_R) \ \ \ \rightarrow x_R = 1m to the right of A
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Simplify the following distributed loads into a single force.

Set origin at A, positive x to the right.
On the left is a triangular force, F1=0.5∗(6kN/m)∗(3m)=9kNF_1 = 0.5*(6 kN/m)*(3m) = 9 kN acting at x1=−1mx_1 = -1 m
On the right side is another triangular force, F2=0.5∗(6kN/m)∗(6m)=18kNF_2 = 0.5 * (6kN/m)*(6m) = 18 kN acting at x2=2mx_2 = 2m

FR=F1+F2=9+18=27kNF_R = F_1 + F_2 = 9 + 18 = 27 kN

ΣMA=(9)(1)−(18)(2)=−(27)(xR)   →xR=1m\Sigma M_A = (9)(1) - (18)(2) = -(27)(x_R) \ \ \ \rightarrow x_R = 1m to the right of A
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Simplify each of the loadings into a single equivalent force:


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Quiz: Distributed Loads - Quiz Practice Questions (Geometric Shapes)
Simplify the following loading into a single force.



Simplify the following loading into a single force.


Simplify the following loading into a single force. Note: the couple moment at O is equal to 50 kNm.


Simplify the following distributed loading into a single force acting on segment AB.


Quiz: Distributed Loads Practice - Geometric Shapes Quiz
Replace the loading with a single equivalent force.


Replace the loading with a single equivalent force.


Quiz: Distributed Loads - Quiz Practice Questions (Geometric Shapes)
Simplify the following loading into a single force.



Simplify the following loading into a single force.


Simplify the following loading into a single force. Note: the couple moment at O is equal to 50 kNm.


Simplify the following distributed loading into a single force acting on segment AB.


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Example


Simplify the following distributed loads into a single force and specify the location from point A.

Set origin at A, positive x to the right.
On the left is a triangular force, F1=12∗(6kN/m)∗(3m)=9kNF_1=\frac{1}{2}*(6kN/m)*(3m)=9kN acting at x1=−1mx_1 = -1 m
On the right side is another triangular force, F2=12∗(6kN/m)∗(6m)=18kNF_2=\frac{1}{2}*(6kN/m)*(6m)=18kN acting at x2=2mx_2 = 2m





FR=F1+F2=9+18=27kNF_R = F_1 + F_2 = 9 + 18 = 27 kN




ΣMA=(9)(1)−(18)(2)=−(27)(xR)   →xR=1m\Sigma M_A = (9)(1) - (18)(2) = -(27)(x_R) \ \ \ \rightarrow x_R = 1m to the right of A
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Example


Simplify the following distributed loads into a single force and specify the location from point A.


Resultant Force,


Set the origin at A, positive to the right.



FR=∫04wdx=∫04(2.5x3)dx=2.54 x4∣04=160NF_R = \int_{0}^{4} wdx = \int_{0}^{4} (2.5x^3)dx = \frac{2.5}{4} \ x^4 \Big|_0^4 = 160 N

Resultant Moment at A,




MR=∫04xwdx=∫04x(2.5x3)dx=2.55 x5∣04=512NmM_R = \int_{0}^{4} xwdx = \int_{0}^{4} x(2.5x^3)dx = \frac{2.5}{5} \ x^5 \Big|_0^4 = 512 Nm



Location of Resultant Force,





xR=MRFR=512Nm160N=3.2mx_R = \frac{M_R}{F_R} = \frac{512 Nm}{160N} = 3.2 m


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Practice

Simplify each of the loadings into a single equivalent force and determine the location from point A:




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Practice:


Simplify each of the loadings into a single equivalent force and determine the location from point A:




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checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.
Simplify each of the loadings into a single equivalent force:




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Simplify the following distributed loads into a single force.


Set the origin at A, positive to the right.

FR=∫04wdx=∫04(2.5x3)dx=2.54 x4∣04=160NF_R = \int_{0}^{4} wdx = \int_{0}^{4} (2.5x^3)dx = \frac{2.5}{4} \ x^4 \Big|_0^4 = 160 N

MR=∫04xwdx=∫04x(2.5x3)dx=2.55 x5∣04=512NmM_R = \int_{0}^{4} xwdx = \int_{0}^{4} x(2.5x^3)dx = \frac{2.5}{5} \ x^5 \Big|_0^4 = 512 Nm

xR=MRFR=512Nm160N=3.2mx_R = \frac{M_R}{F_R} = \frac{512 Nm}{160N} = 3.2 m


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Simplify the following distributed loads into a single force.


Set the origin at A, positive to the right.

FR=∫04wdx=∫04(2.5x3)dx=2.54 x4∣04=160NF_R = \int_{0}^{4} wdx = \int_{0}^{4} (2.5x^3)dx = \frac{2.5}{4} \ x^4 \Big|_0^4 = 160 N

MR=∫04xwdx=∫04x(2.5x3)dx=2.55 x5∣04=512NmM_R = \int_{0}^{4} xwdx = \int_{0}^{4} x(2.5x^3)dx = \frac{2.5}{5} \ x^5 \Big|_0^4 = 512 Nm

xR=MRFR=512Nm160N=3.2mx_R = \frac{M_R}{F_R} = \frac{512 Nm}{160N} = 3.2 m


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checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
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Simplify the loading into a single equivalent force:




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  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
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Simplify the loading into a single equivalent force:




checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.
Simplify the loading into a single equivalent force:

checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.
Simplify the loading into a single equivalent force:

Quiz: Distributed Loads Practice - Integration Quiz
Replace the loading with a single equivalent force.


Quiz: Distributed Loads Practice - Integration Quiz
Replace the loading with a single equivalent force.