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An engineer is designing a structural beam that contains a lengthwise conduit th

ID: 1844771 • Letter: A

Question

An engineer is designing a structural beam that contains a lengthwise conduit that will be used for routing cable. As shown, the cross section of the beam has the dimensions a = 0.740 ft , b = 1.18 ft , c = 0.290 ft , and d = 0.410 ft . (Figure 1) The beam's centroid is located at point C (4.65×10?2 ft , 2.92×10?2 ft ). The engineer wishes to determine the beam's principal moments of inertia and the orientation of the principal axes using Mohr's circle. First, find the beam's moment of inertia and product of inertia. What are the beam's moments of inertia, Ixand Iy, about the centroidal x' and y' axes, respectively? What is the beam's product of inertia about its centroid?

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Explanation / Answer

beam's moment of inertia = Ia = (1/12)bh3

=> Ix = (1/12)1.18 x (0.740)3

=> 0.08 x 1.18 x 0.40

=> 0.08 x 0.47 = 0.037

moment of inertia of the beam = 0.037 kg/m2

moment of inertia of beam, Ib = (1/12) b3h

=> (1/12)(0.410)3 x 0.290

=> 0.08 x 0.06 x 0.290

=> 0.08 x 0.01 = 8x10-04

moment of inertia of beam = 8x10-04 kg/m2

product of inertia = (Ix)(Iy) = I xy

=> Ixy = 0.037 x (8x10-04)

product of both moment of inertia = 2.96x10-05

Since it is a rectangular structure

length of line x is half of the line b = 1.18ft

we have , length of line from centre = (1/4)b = (1/4)(1.18) = 0.29

0.29 x 2 = 0.59ft (length of half the rectangle according to centroid is 0.59ft)

so the length of line on x axis is 0.59ft so the lx = 0.59/2 = 0.29ft

lx = 0.29ft

similarly , ly =

we have length from centre = (1/4)a = 0.185ft

=> 0.185 x 2 ( it is a 1/4 of length a)

=>0.37ft (half of legth a)

length of y axis = 0.37ft

therefore

ly = 0.37- (1/2)length y = (1/2)0.37

=> 0.18ft

ly = 0.1ft

now the moment of inertia of beam Ix = (1/12)bh3

(1/12)0.59x(0.37)3

=> 0.08x0.02 = 1.6x10-03

moment of inertia ob beam Ix = 1.6x10-03

same for Iy = (1/12)b3h

= (1/12) 0.293x 0.1

= 0.08x 0.02x0.01

moment of inertia Iy=1.6x10-05

product of inertia = (Ix)(Iy) = Ixy

=>beams product of inertia about its centroid is = 2.56x10-08