Last Updated on September 10, 2026 by Maged kamel
Area and CG for a circular shaft.
Reference handbook 10.00 value for the Area and CG for a circular shaft.
A list of common round shapes, their areas, and CG values. Our second case is a circular shaft. The external axes X and Y are tangent to the circle, which has a radius value of a.
Area and CCG 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, we need the Area and CG. We can find two axes, X’ and Y’, that divide 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 intersection of these two axes X’ and Y’.
We have two axes, X and Y, that are tangent to the circle and apart from the X-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 dρ. The Area of the strip is 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 Area 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).
This simplifies to (ρ^2*dρ*sin θ*dθ). The Expression of the first Moment of Area about the X’ axis is shown in the next slide image.
For the first Moment of Area of the circular shaft, we will use double integration, since we have to integrate with respect to ρ from ρ=b to ρ=a. The second integration is over the angle dθ, from θ = to θ = 2π (30 degrees). A*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=0 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 Area estimate calculation are shown on the next slide.
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 Area about the Y’-axis using a small Area dA; the first Moment of Area for that small Area is dMy = dA*(x) = (ρ*dρ*dθ)*(ρ*cos θ). It will be simplified to (ρ^2*dρ*cos θ*dθ).
For the first Moment of Area of the whole circle, we will use double integration, since we have to integrate with respect to ρ from ρ=b to ρ=a. The second integration is over the angle dθ, from θ = 0 to θ = 2π (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 π*(a^2 – b^2). The x bar is equal to A*X/A= zero/π*a^2 =0. This implies that the Cg lies on the Y-axis.
DDividingA*y/A to get Y bar will lead to ybar=0/π*(a^2-b^2)=0. This implies that the Cg lies on the Y’- axis.
The Cg distances are x bar = a and y bar = a, as shown in the image on the next slide.
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We have completed the subject of the Area and CG for a semicircle.
This is a link to a very useful site: Engineering statics,, openand inractive.