Explanation Given information: The moment of inertia of the beam are I y = I z = 11.3 in . 4 . The area of the beam is A = 4.75 in . 2 . The maximum stress in the beam is σ = 12 ksi . The value of radius of gyration is k min = 0.983 in . Show the expression for minimum moment of inertia ( I min ) as follows: I min = A k min 2 (1) Show the relation between I min and I max as follows: I min + I max = I y + I x (2) The angle between the principal axes and the coordinate axes is θ = 45 ° . Calculation: 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 45 ° for θ . M u = M 0 sin 45 ° = 0.70711 M 0 Calculate the moment M v about the principal axes as using the relation: M v = − M 0 cos θ Substitute 45 ° for θ . M v = − M 0 cos 45 ° = − 0.70711 M 0 Calculate the minimum moment of inertia ( I min ) as follows: Substitute 4.75 in . 2 for A and 0.983 in . for k min in Equation (1). I min = 4.75 × 0.983 2 = 4.75 × 0.966289 = 4.59 in . 4 Calculate the maximum moment of inertia ( I max ) as follows: Substitute 4.59 in . 4 for I min , 11.3 in . 4 for I y , and 11.3 in . 4 for I z in Equation (2). 4.59 + I max = 11.3 + 11.3 I max = 22.6 − 4.59 = 18.01 in . 4 Calculate the distance y B and z B as follows: z D = − ( 5 in . − 1.43 in . ) = − 3.57 in . y D = 1.43 in . − 0.5 in . = 0.93 in . 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 (3) Substitute − 3

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Chapter 4.9, Problem 151P

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The largest permissible value of moment M0 acting on the horizontal plane such that the maximum stress in the beam is 12 ksi.

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Chapter 4.9, Problem 150P

Chapter 4.9, Problem 152P

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The largest permissible value of moment M 0 acting on the horizontal plane such that the maximum stress in the beam is 12 ksi .

Solution Summary: The author shows the expression for minimum moment of inertia and radius of gyration.

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The largest permissible value of moment M0 acting on the horizontal plane such that the maximum stress in the beam is 12 ksi.

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Chapter 4 Solutions