chapter 15 answers

chapter 15 answers - Problem 15.1 The beam consists of...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 15.1 The beam consists of material with modulus of elasticity E = 70 GPa and is subjected to couples M = 250 kN − m at its ends. (a) What is the resulting radius of curvature of the neutral axis? (b) Determine the maximum tensile stress due to bending. Solution: The moment of inertia for the cross-section is: I = bh 3 12 = (0 . 16 m)(0 . 32 m) 3 12 = 4 . 369 × 10 − 4 m 4 (a) Using Equation (15.10) to determine the magnitude of the radius of curvature: 1 ρ = M EI = 250 , 000 N − m (70 × 10 9 N / m 2 )(4 . 369 × 10 − 4 m 4 ) = 8 . 174 × 10 − 3 m − 1 ANS: ρ = 122 . 34 m (b) The maximum normal stress due to the bending moment is: σ MAX = My MAX I = (250 , 000 N − m)(0 . 16 m) 4 . 369 × 10 − 4 m 4 ANS: σ MAX = 91 . 6 MPa Problem 15.2 The material of the beam in Prob- lem 15.1 will safely support a tensile stress of 180 MPa and a compressive stress of 200 MPa . Based on these criteria, what is the largest couple M to which the beam can be subjected? Solution: The symmetry of the cross-section tells us that: ( σ MAX ) TENSILE = ( σ MAX ) COMPRESSIVE The moment of inertia for the cross-section is: I = bh 3 12 = (0 . 16 m)(0 . 32 m) 3 12 = 4 . 369 × 10 − 4 m 4 The maximum normal stress due to the bending moment is: σ MAX = My MAX I → M = σ MAX I y MAX = (180 × 10 6 N / m 2 )(4 . 369 × 10 − 4 m 4 ) . 16 m ANS: M = 492 kN − m Problem 15.3 The material of the beam in Prob- lem 15.1 will safely support a tensile stress of 180 MPa and a compressive stress of 200 MPa . Suppose that the beam is rotated 90 ◦ about its axis, so that the width of its cross section is . 32 m and its height is . 16 m . What is the largest couple M to which the beam can be subjected? Compare your answer to the answer to Problem 15.2. Solution: The moment of inertia for the cross-section is: I = bh 3 12 = (0 . 32 m)(0 . 16 m) 3 12 = 1 . 092 × 10 − 4 m 4 Using Equation (15-12) to determine the applied moment: σ MAX = My MAX I → M = σ MAX I y MAX = (180 × 10 6 N / m 2 )(1 . 092 × 10 − 4 m 4 ) . 08 m ANS: M = 246 kN − m This moment is 1 / 2 the result in Prob- lem 15.2. Problem 15.4 The beam consists of material with modulus of elasticity E = 14 x 10 6 psi and is subjected to couples M = 150 , 000 in − lb at its ends. (a) What is the resulting radius of curvature of the neutral axis? (b) Determine the maximum tensile stress due to bending? Solution: The moment of inertia for the cross-section is: I = πr 4 4 = π (2 in ) 4 4 = 12 . 57 in 4 (a) Using Equation (15-10) to determine the magnitude of the radius of curvature: 1 ρ = M EI = 150 , 000 in − lb (14 × 10 6 lb / in 2 )(12 . 57 in 4 ) = 8 . 524 × 10 − 4 in − 1 ANS: ρ = 1173 . 2 in = 97 . 7 ft Using Equation (15-12) to determine the applied moment: σ MAX = My MAX I = (150 , 000 in − lb)(2 in ) 12 . 57 in 4 ANS: σ MAX = 23 . 9ksi Problem 15.5 The material of the beam in Prob- lem 15.4 will safely support a tensile or compressive stress of 30 , 000 psi . Based on this criterion, what is the....
View Full Document

This homework help was uploaded on 04/03/2008 for the course MAE 101 taught by Professor Orient during the Winter '08 term at UCLA.

Page1 / 32

chapter 15 answers - Problem 15.1 The beam consists of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online