First moment of area

First moment of area and calculations
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1. First moment of Area

The first moment of area, sometimes misnamed as the first moment of inertia, is
based in the mathematical construct moments in metric spaces, stating that the moment
of area equals the summation of area times distance to an axis [Σ(a × d)].








Considered a thick lamina or a body having area A. let x is the distance of C.G. of the area from the axis
OY and y is the distance of C.G. of the area from the axis OX.

2. Here

x = Distance of C.G. of area A from axis OY
y = Distance of C.G. of area A from axis OX

Then,




This equation (1) is known as the first moment of area about the axis OY

3. Similarly



This Equation (2) is called the first moment of area about the axis OX.
Conclusion:
First moment of area/moment of area:
The first moment of area of a lamina about an axis is defined as the product of area of the
lamina and the perpendicular distance of the C.G. of the area form the axis.
Application
It is used to determine the center of gravity of area.
Numerical Example
If we have a lamina having area 10 cm
2
and the distance of its C.G from the axis are shown
in the figure given below.

Then

The first moment of area about axis OY = 10 x 5 = 50
The first moment of area about axis OX = 10 x 7 = 70

4. How to Calculate the Statical or First Moment of
Area
The statical or first moment of area (Q) simply measures the distribution of a beam sections's
area relative to an axis. It is calculated by taking the summation of all areas, multiplied by its
distance from a particular axis (Area by Distance).
In fact, you may not have realised it, but if you've calculated the centroid of a beam
section then you would have already calculated the first moment of area. Also this property,
often denoted by Q, is most commonly used when determining the shear stress of a beam
section.
Since beam sections are usually made up many geometries we first need to split the section into
segments. After this, the area and centroid of each segment is calculated to find the overall
statical moment of area.
Consider the I-beam section shown below. In our previous tutorial we already found the
centroid to be 216.29 mm from the bottom of the section. To calculate the statical moment of
area relative to the horizontal x-axis, the section can be split into 4 segments as illustrated:
Remember that the first moment of area is the summation of the areas multiplied by the distance
from the axis. So the formula for the statical moment of area relative to the horizontal x-axis is:
QxAiyi=∑yiAi where:=The individual segment's area=The individual segment's centroid distance
from a reference line or datumQx=∑yiAi where:Ai=The individual segment's areayi=The individual
segment's centroid distance from a reference line or datum
Now for a result such as shear stress we often want the statical moment of either the TOP or
BOTTOM of the section relative to the Neutral Axis (NA) XX. Let's start with the TOP portion of
the section (i.e. Segments 1 and 2). We'll find Aiand yi for each segment of the I-beam section
above the neutral axis and then compute the statical moment of area (Qx). Remember we
measure the distances from the neutral axis!

Segment 1:A1y1=250×38=9500 mm2=159.71−382=140.71 mmSegment
1:A1=250×38=9500 mm2y1=159.71−382=140.71 mm
Segment 2:A2y2=(159.71−38)×25=3042.75 mm2=159.71−382=60.86 mmSegment
2:A2=(159.71−38)×25=3042.75 mm2y2=159.71−382=60.86 mm

5. Qx,topQx,topQx,topQx,top=∑yiAi=y1A1+y2A2=(140.71×9500)+(60.86×3042.75)≈1,521
,900 mm3Qx,top=∑yiAiQx,top=y1A1+y2A2Qx,top=(140.71×9500)+(60.86×3042.75)Qx,top≈1,52
1,900 mm3
Similarly, we can calculate the statical moment of area of the BOTTOM portion of the section. This
involes Segments 3 and 4 which are below the neutral axis.

Segment 3:A3y3=(216.29−38)×25=4457.25 mm2=216.29−382=89.15 mmSegment
3:A3=(216.29−38)×25=4457.25 mm2y3=216.29−382=89.15 mm
Segment 4:A4y4=150×38=5700 mm2=216.29−382=197.29 mmSegment
4:A4=150×38=5700 mm2y4=216.29−382=197.29 mm
Qx,bottomQx,bottomQx,bottomQx,bottom=∑yiAi=y3A3+y4A4=(89.15×4457.25)+(197.29×570
0)≈1,521,900 mm3Qx,bottom=∑yiAiQx,bottom=y3A3+y4A4Qx,bottom=(89.15×4457.25)+(197
.29×5700)Qx,bottom≈1,521,900 mm3
What you'll notice is that the statical moment of area above the neutral axis is equal to that
below the neutral axis!
Qx,top=Qx,bottomQx,top=Qx,bottom
Of course you don't need to do all these calculations manually because you can use our
fantastic Free Moment of Inertia Calculator to find the statical moment of area of beam
sections.

How is stress distributed through the depth of a beam top edge to NA
to bottom edge?

To calculate the statical moment of area relative to the horizontal x-axis, the section
can be split into 4 segments as illustrated: Remember that the first moment of area is
the summation of the areas multiplied by the distance from the axis.

How is stress distributed through the depth of a beam top edge to NA
to bottom edge?
To calculate the statical moment of area relative to the horizontal x-axis, the section can be split into 4
segments as illustrated: Remember that the first moment of area is the summation of the areas
multiplied by the distance from the axis.