This preview has intentionally blurred sections. Sign up to view the full version.
View Full 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 crosssection 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 crosssection tells us that: ( σ MAX ) TENSILE = ( σ MAX ) COMPRESSIVE The moment of inertia for the crosssection 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 crosssection is: I = bh 3 12 = (0 . 32 m)(0 . 16 m) 3 12 = 1 . 092 × 10 − 4 m 4 Using Equation (1512) 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 crosssection is: I = πr 4 4 = π (2 in ) 4 4 = 12 . 57 in 4 (a) Using Equation (1510) 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 (1512) 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.
 Winter '08
 ORIENT
 Statics, Stress

Click to edit the document details