# 2- How to find the area and Cg for a circular shaft?

## Area and Cg for a circular shaft.

### Reference handbook 10.00 value for the area and Cg for a circular shaft.

There is a list of the common round shapes area and CG value. Our second case is the case of a circular shaft.

You can click on any picture to enlarge, then press the small arrow at the right to review all the other images as a slide show.

The external axes X&Y are tangents to the circle which has a radius value of a.

### Area and Cg for a circular shaft- select an area dA.

We have a circular shaft, or two circles inside each other the inner circle is hollow. The Largest circle is a circle with a radius a, while the second circle is empty with a radius b. For the hollow shaft, it is required to get the area and CG. We can find that there are two axes X’ and Y’ dividing the shaft into four similar parts. Due to that symmetry, we expect that the Cg or the center of gravity will be at the point of intersections of these two axes X’ and Y’.

We have Two axes X&Y are tangents to the circle and apart from the X’s axis by a value of a, which is the shaft outer radius. We will select a small area dA that has a radius of ρ from the intersection of the two axes X’ and Y’.

The small area dA has an angle from the X’ axis that is equal to θ and is enclosed by an angle dθ and a width equal to . The area of the strip dA=( ρ*dρ* dθ).

### Area and Cg for a circular shaft- first moment of area about the x’-axis.

Area and Cg for a circular shaft can be obtained by estimating the first moment of the area about the x’-axis by using a small area dA about the x’-axis as the product of that rea by the vertical distance to the X’- axis. The vertical distance is y which is equal to ρ*sin θ. The moment dMx=dA*(y)=(ρ*dρ*d θ)*(ρ*sin θ). it will be simplified to (ρ^2*dρ*sin θ*dθ).

The expression of the first moment of area about X’ axis is shown in the next slide image.

For the first moment of area for the circular shaft, we will use double integration since we have to integrate for ρ from ρ=b to ρ=a. The second integration os for the angle dθ from θ equals to zero to dθ equal to 2*π or 360 degrees. dA*y=∬(ρ^2*dρ*sin θ*dθ).

ρ^2*dρ from b to a will give 1/3*ρ^3 after substitution will lead to 1/3*(a^3-b^3)=(a^3-b^3)/3, while the ∫(sin θ*dθ)=-cos(θ), after substitute from zero to 2*π. The value will be (- (cos(2*π)- cos(0))=zero.

But the circular shaft area can be found from the integration of dA= ∬(ρ*dρ*dθ) from ρ=b to ρ=a and for dθ from θ equals zero to dθ=2*π, the final expression for the area is dA=ρ^2*0.5*(θ), substitute to get A=0.50*(a^2-0)*(2*π-0)=π*(a^2-b^2).

the details of the calculation for the area estimation are shown in the next slide image.

### Area and Cg for a circular shaft- first moment of area about the Y’-axis.

Area and Cg for a circular shaft can be obtained by estimating the first moment of the area about the Y’-axis by using a small area dA, the first moment of area for that small area is dMy=dA*(x)=(ρ*dρ*d θ)*(ρ*cos θ). It will be simplified to (ρ^2*dρ*cos θ*dθ).

For the first moment of area for the whole circle, we will use double integration since we have to integrate for ρ from ρ=b to ρ=a. The second integration os for the angle dθ from θ equals to zero to dθ equal to 2*π or 360 degrees. dA*y=∬(ρ^2*dρ*cos θ*dθ).

ρ^2*dρ from b to a=1/3*ρ^3 after substitution will lead to 1/3*(a^3-b^3), while the ∫(cos θ*dθ)=-cos(θ), after substitute from zero to 2*π. The value will be (+ (sin(2*π)-sin(0))=zero.

We have estimated the area of the circle as equal to π*(a^2–b^2), The x bar is equal to A*X/A= zero/π*a^2 =0. This includes that the Cg lies on the Y’ axis.

Divide A*y/A to get Y bar will lead to ybar=0/π*(a^2-b^2)=0. This includes that the Cg lies on the Y’ axis.

The Cg distances are x bar=a and y bar=a and are shown in the next slide image.

We have completed the subject of the area and Cg for a semi-circle.

This is a link to a very useful site: Engineering statics open and interactive.

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