Explanation Given information: The beam is made of elastic perfectly plastic material. The allowable stress in the section is σ Y = 345 MPa . Explanation: Determination of Elastic moment: Moment of inertia: Show the free-body diagram of the beam as in Figure 1. The section is symmetrical about both the x and y directions. Determine the moment of inertia of the sectionusing the formula. I = b 1 ( d 1 ) 3 12 + b 2 ( d 2 ) 3 12 + b 3 ( d 3 ) 3 12 (1) Here, b 1 is width of section 1, d 1 is depth of section 1, b 2 is width of section 2, d 2 is depth of section 2, b 3 is width of section 3, and d 3 is the depth of section 3. Substitute 100 mm b 1 , 300 mm for d 1 , 100 mm for b 2 , 100 mm for d 2 , 100 mm for b 3 , and 100 mm for d 3 in Equation (1). I = 100 × 300 3 12 + 100 × 100 3 12 + 100 × 100 3 12 = 2.417 × 10 8 mm 4 × ( 1 m ) 4 ( 1,000 mm ) 4 = 0.2417 × 10 − 3 m 4 Elastic moment: Determine the elastic moment using the equation. M Y = σ Y I c (2) Here, σ Y is allowable stress in the beam and c is the distance between the centroidal distance. Substitute 345 MPa for σ Y , 0.2417 × 10 − 3 m 4 for I and 150 mmfor c in Equation (2). M Y = 345 MPa × 10 6 N/m 2 1 MPa × 0.2417 × 10 − 3 150 mm × 1 m 1,000 mm = 555 , 8333

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Author: Russell C. Hibbeler

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Chapter 6.10, Problem 6.168P

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The maximum elastic moment and the plastic moment that can be applied to the cross section.

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Chapter 6.10, Problem 6.167P

Chapter 6.10, Problem 6.169P

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