(a) Draw the Mohr’s circle for the centroidal moments and products of inertia for the L-shaped area in Fig. 1 of Example C-2, given that:

I_y = \frac{5894}{147}t^4 = 40.10t^4, I_z = \frac{33,103}{294}t^4 = 112.60t^4I_{yz} = \frac{-270}{7}t^4 = -38.57t^4

(b) Use the Mohr’s circle constructed in Part (a) to compute the principal moments of inertia I_{p_1} and I_{p_2} and to locate the principal axes. Show the orientation of the principal axes on a sketch.

Step-by-Step

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(a) Sketch Mohr’s circle and calculate the principal moments of inertia. Points Y and Z are plotted and Mohr’s circle is then drawn (Fig. 1). From the circle,

I_{avg.} = \frac{40.10t^4 + 112.60t^4}{2} = 76.35t^4R = \sqrt{\left(\frac{112.60t^4 – 40.10t^4}{2}\right)^2+ (38.57t^4)^2} = 52.93t^4

I_{p_1} = I_{avg.} + R = 129.3t^4

I_{p_2} = I_{avg.} – R = 23.4t^4 (a)

(b) Determine the orientation of the principal axes and show them on a sketch.

\tan |2θ_{zp_1}| = \frac{38.57t^4}{\left(\frac{112.60t^4 – 40.10t^4}{2}\right)} = 1.064Therefore, 2θ_{zp} = 46.78° (clockwise), so

θ_{zp_1} = θ_{yp_2} = 23.4° clockwise (b)

Note that the orientations of the principal axes in Fig. 2 are such that the contributions to I_{p_1p_2} of the areas in the four quadrants cancel out, giving I_{p_1p_2} = 0.

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