Explanation Given information: The moment of inertia of the cross-section as follows: I y = 1.894 × 10 6 mm 4 I z = 0.614 × 10 6 mm 4 I y z = 0.800 × 10 6 mm 4 Consider the coordinates of the points Y , Z , and E are ( I y , I y z ) , ( I z , I y z ) , and ( I y + I z 2 , I y z ) . Show the points Y, Z, and E as follows: Y ( 1.894 × 10 6 mm 4 , 0.800 × 10 6 mm 4 ) Z ( 1.894 × 10 6 mm 4 , 0.800 × 10 6 mm 4 ) E ( 1.254 × 10 6 mm 4 , 0 ) Plot the points Y, Z, and E and construct the Mohr circle as shown in Figure 1. Refer to Figure 1. Calculate the radius R of the Mohr circle using the relation: R = E F ¯ 2 + F Z ¯ 2 (1) Substitute 0.640 for E F ¯ , and 0.800 for F Z ¯ in Equation (1). R = ( 0.640 × 10 6 ) 2 + ( 0.800 × 10 6 ) 2 = 1.0496 × 10 12 = 1.0245 × 10 6 mm 4 Calculate the moment of inertia about the principal axis u and v as follows: I v = E − R = 1.254 × 10 6 − 1.0245 × 10 6 mm 4 = 0.2295 × 10 6 mm 4 I u = E + R = 1.254 × 10 6 + 1.0245 × 10 6 mm 4 = 2.2785 × 10 6 mm 4 Find the orientation of the principal axis as follows: tan 2 θ m = F Z F E = 0.800 × 10 6 0.640 × 10 6 2 θ m = tan − 1 ( 1.25 ) θ m = tan − 1 ( 1.25 ) 2 θ m = 25.67 ° Calculate the moments about the principal axis v as follows: M v = M cos θ m Substitute 1.2 kN ⋅ m for M and 25.67 ° for θ m . M v = 1.2 kN ⋅ m × ( 10 6 N ⋅ mm 1 kN ⋅ m ) cos ( 25.67 ° ) = 1.2 × 10 6 ( 0.9013 ) = 1.0816 × 10 6 N ⋅ mm Calculate the moments about the principal axis u as follows: M u = − M sin θ m Substitute 1

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Chapter 4.9, Problem 141P

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Solution Summary: The author analyzes the moment of inertia of the cross-section as follows: lY(I_y

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