Determine the second moment of area for theshaded region shown in Fig.with respect toThe x-axis.y= 2bx2b

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International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter9: Moments And Products Of Inertia Of Areas

Section: Chapter Questions

Problem 9.49P: Determine the dimension b of the square cutout so that Ixy=0 for the region shown.

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Question

Determine the second moment of area for the
shaded region shown in Fig.
with respect to
The x-axis.
y= 2bx
2b

Expert Solution

Step 1

Consider the region under the parabola $y^{2} = 2 b x$ as area (1) and consider a strip of thickness dy at a distance of y from the x-axis as shown in the below figure.

The expression for the area of strip is given by;

$d A = (2 b - x) d y$

Eliminate x from the equation of the parabola $y^{2} = 2 b x$ .

$\begin{array}{rcl} y^{2} & = & 2 b x \\ x & = & \frac{y^{2}}{2 b} \end{array}$

The expression for the second moment of the area below the parabola about x-axis is given by;

${(I_{x})}_{1} = \int_{0}^{2 b} y^{2} d A$

Substitute all the values in the above equation.

$\begin{array}{rcl} {(I_{x})}_{1} & = & \int_{0}^{2 b} y^{2} (2 b - x) d y \\ = & \int_{0}^{2 b} y^{2} (2 b - \frac{y^{2}}{2 b}) d y \\ = & \int_{0}^{2 b} (2 b y^{2} - \frac{y^{4}}{2 b}) d y \\ = & {[\frac{2 b y^{3}}{3} - \frac{y^{5}}{10 b}]}_{0}^{2 b} \\ = & \frac{2 b {(2 b)}^{3}}{3} - \frac{{(2 b)}^{5}}{10 b} \\ = & \frac{16 b^{4}}{3} - \frac{16 b^{4}}{5} \\ = & \frac{32 b^{4}}{15} \end{array}$

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Determine the second moment of area for the shaded region shown in Fig. with respect to The x-axis. y= 2bx 2b