Explanation Given information: The maximum and minimum moment of inertia are I min = 0.23 × 10 − 6 mm 4 and I max = 2.28 × 10 − 6 mm 4 . The angle made by the principal axis are 25.7 ° and 64.3 ° . Calculation: Calculate the moment of inertia ( I u ) and ( I v ) as follows: I v = I min = 0.23 × 10 − 6 mm 4 I u = I max = 2.28 × 10 − 6 mm 4 Show the cross-section of the cross-section as shown in Figure 1. Refer Figure 1. Consider the moment about the principal axes are M u and M v . Calculate the moment M u about the principal axes as using the relation: M u = M 0 sin θ Substitute 64.3 ° for θ . M u = M 0 sin 64.3 ° Calculate the moment M v about the principal axes as using the relation: M v = M 0 cos θ Substitute 64.3 ° for θ . M v = M 0 cos 64.3 ° Consider the angle made by the neutral axis with the principal axis is denoted by ϕ . Calculate the value of the angle ϕ using the relation: tan ϕ = I v I u tan θ (1) Substitute 0.23 × 10 − 6 mm 4 for I u , 2.28 × 10 − 6 mm 4 for I v , and 64.3 ° for θ in Equation (1). tan ϕ = 0.23 × 10 − 6 2.28 × 10 − 6 tan ( 64.3 ° ) = 0.100877 × 2.077 tan ϕ = 0.20952 ϕ = tan − 1 ( 0.20952 ) = 11.84 ° Consider the points D and E are the farthest points from neutral axis. Calculate the distance y D and z D as follows: y D = − 5 mm z D = 45 mm Calculate the perpendicular distance ( u D ) from principal axis to point D using the relation: u D = y D cos θ − z D sin θ (2) Substitute − 5 mm for y D , 45 mm for z D , and 25.7 ° for θ in Equation (2). u D = − 5 cos ( 25.7 ° ) − 45 sin ( 25.7 ° ) = − 5 × 0.9010 − 45 × 0.4336 = − 4.505 − 19.512 = − 24.02 mm Calculate the perpendicular distance ( v D ) from principal axis to point D using the relation: v D = z D cos θ + y D sin θ (3) Substitute 45 mm for z D , − 5 mm for y D , and 25.7 ° for θ in Equation (3). v D = 45 cos ( 25

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Chapter 4.9, Problem 153P

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The largest permissible value of moment M0 such that the maximum stress in the beam is 80 MPa.

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Chapter 4.9, Problem 152P

Chapter 4.9, Problem 154P

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The largest permissible value of moment M 0 such that the maximum stress in the beam is 80 MPa .

Solution Summary: The author calculates the moment of inertia (I_u), and the angle made by the principal axes are 25.7°.

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The largest permissible value of moment M0 such that the maximum stress in the beam is 80 MPa.

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Chapter 4 Solutions