Problem_09_20 - Problem 9.20 Given: A load P = 12.0 kN is...

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Problem 9.20 Given: A load P = 12.0 kN is applied to the clamp shown. Determine the circumferential stresses at points B and C , assuming that the curved beam formula is valid at that section. Solution: The cross section is divided into the 3 sub-sections shown. λ b p 2 rt = λ i 12 2 29 14 0.355 = := λ o 7 2 72 8 0.085 = := 1 2 3 From Table 9.3 Interpolating gives: α i 0.950 0.917 .950 () λ i 0.3 0.1 + 0.932 = := β i 0.836 1.056 .836 λ i 0.3 0.1 + 0.956 = := b pi α i 12 mm 11.183 mm = := The outer flange is not wide so we assign: α o 1.0 := β o 0. := A 1 12 mm 2b pi + 14 mm 481.137 mm 2 = := R 1 22 mm 7mm + 29 mm = := Am 1 12 mm pi + ln 36 22 16.925 mm = := A 2 32 mm 12 mm 384 mm 2 = := A 3 26 mm 8 mm 208 mm 2 = := R 2 36 mm 16 mm + 52 mm = := R 3 68 mm 4mm + 72 mm = := Am 2 12 mm ln 68 36 7.632 mm = := Am 3 26 mm ln 76 68 2.892 mm = := AA 1 A 2 + A 3 + 1.073 10 3 × mm 2 = := Am Am 1 Am 2 + Am 3 + 27.449 mm = :=
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R R 1 A 1 R 2 A 2 + R 3 A 3 + A 45.565 mm = := N1 2 k N := M x 12 kN 73 mm 22 mm R + () 0.097 N m = := σ θθ r N A M x Ar A m RAm A + := For point B r2 2 m m := σ θθ r ( ) 140.908 MPa = For point C r7 6 m m := σ θθ r ( ) 69.878 MPa = What is the stress in the x direction. σ bar σ θθ R 1 69.3 MPa = := σ xx β i σ
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This note was uploaded on 01/11/2011 for the course MAE 3040 taught by Professor Mechanicsofsolids during the Spring '10 term at Utah State University.

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Problem_09_20 - Problem 9.20 Given: A load P = 12.0 kN is...

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