Question

Locate the centroid of the composite cross-sectional area shown in the figure below. Also, determine the moments of inertia for the area about its x’and y' centroidal axes. y=y' Note: all dimensions in (mm). 

Answer 1

Solution:

In the next figure I have drawn the diagram of the given problem in which I have divided the given composite area into 3 areas namely 1,2&3.

In the same figure I have calculated the value of centroids of the different areas from bottom x-axis.

From figure,

y₁=700 mm ; y₂=300 mm ; y₃=375 mm.

{y₁,y₂ & y₃ are the centroids of the areas 1,2 &3 respectively which are measured from the bottom x-axis.}

a₁ =800*200 =160000 mm² ,

a₁ =600*400 =240000 mm² ,

a₃ =250*200 =50000 mm² , {a₁ ,a₂ & a₃ are the calculated areas of 1,2 & 3 respectively}

Now centroid of any composite area is given as:

{ is the centroid of the given problem measured from the bottom x-axis.}

In the given problem,area 3 is cut from the composite area.So the contribution of area 3 is subtracted from the above centroid equation.

So formula for the given problem becomes:

Putting the values in the above eq.

In the next figure I have written the moment of inertia(I_xx & I_yy) for a rectangle.

Also in the same figure I have used parallel axis theorem to calculate moment of inertia about any X-axis parallel to the centroidal axis.

So moment of inertia(I_x'x') about x'-axis is given as:

where d₁,d₂ & d₃ are the distances between centroids of a₁ ,a₂ & a₃ respectively from the centroidal axis of the composite area.

d₁=y₁-=700-472.143 => d₁=227.857 mm

d₂=-y₂ =472.143-300 => d₂=172.143 mm

d₃=-y₃ =472.143-375 => d₃=97.143 mm

Putting the values in the moment of inertia equation.

Similarly, for moment of inertia about y' axis

Note:Here parallel axis theorem is not used because centroids of all the areas lie on y'-axis.

Putting the values in the above eq.

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