AENG260 - Problem set 4, chapter 3.pdf - PERIDO JOHN PAULO...

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Unformatted text preview: PERIDO, JOHN PAULO M. MSAE – FPM 2018-68796 AENG 260 -EXPERIMENTAL STRESS ANALYSIS PROBLEM SET IV DR. DELFIN C. SUMINISTRADO 3.31. A steel (E = 207 GPa and v = 0.30) ring is shrunk onto another steel ring, as shown in the figure below. Determine the maximum interference possible without yielding one of the rings if the yield strength of both rings is 900 MPa and the dimensions of the rings are as follows: θ , , , Case 1 100 125 150 Case 2 100 115 200 Case 3 100 150 175 Case 4 0 100 150 Interference between two different materials = [ + + ( + ) + ( − )] − − Where: - interference, mm - pressure generated with the designed connection or yield strength, MPa - Modulus of elasticity of the outer materials, GPa - Modulus of elasticity of the inner materials, GPa - outer radius of the larger ring, mm - inner radius of the smaller ring, mm - inner radius of the larger ring or outer radius of the smaller ring, mm - Poisson’s ratio of the inner ring - Poisson’s ratio of the outer ring Interference of same the material − = [ ( − )( − ) Where: - interference, mm - pressure generated with the designed connection or yield strength, MPa - Modulus of elasticity of the inner materials, GPa - outer radius of the larger ring, mm - inner radius of the smaller ring, mm - inner radius of the larger ring or outer radius of the smaller ring, mm CASE 1. ( )( ) ( ) − ( ) = [ (() − ( ) )(( ) − ( ) ) = . CASE 2. ( )( ) ( ) − ( ) = [ (() − ( ) )(( ) − ( ) ) = . CASE 3. = ( )( ) ( ) − ( ) [ (( ) − ( ) )(( ) − ( ) ) = . CASE 4. ( )( ) ( ) − ( ) = [ (( ) − ( ) )(( ) − ( ) ) = . Reference Press Fit Pressure Calculator – Optimize Your Interference/Transition Fit Design (2012, October 0000021). Retrieved from - calculator00000optimize.html. 3.16. Determine the stresses in the uniformly loaded cantilever beam shown in the figure below by using Airy’s stress function approach. Assume a unit thickness. Boundary conditions At x = 0; = 0 = 0 At x = L; ℎ ∫ = −ℎ ℎ ∫ = 0 −ℎ ℎ ∫ = −ℎ At y = h; = − = 0 1 2 2 At y = -h; = 0 = 0 Stress function, superimposing polynomials = 2 + 3 + 4 + 5. Also, since the cross section is symmetric about the y-axis and = − at y = h and = 0 at y = -h, should be an odd function of y. Thus the stress function should contain odd functions of y, hence, should also contain odd functions of y since the net axial force is zero. Second, since is constant as a function of x on the top and bottom faces, the stress function should not contain powers of x greater than 2 . = 2 + + 2 + 3 + 2 + 2 + 3 + 4 + 3 + 2 2 + 3 + 4 + 5 + 4 + 3 2 + 2 3 + 4 + 5 Removing all terms with a power of x greater than two and y with even power. Rearranging all terms will be. = + 2 + 2 + 3 + 3 + 2 3 + 5 Since ∇4 = 0, ∇4 = 24 + 120 = 0 Thus = −5 Equations of Stresses 2 = = 6 + 6 − 30 2 + 20 3 2 = 2 = 2 + 2 − 10 3 2 2 = = −( + 2 + 3 2 − 30 2 ) For at = 0, + 3 2 = 0 this can be true for all values of y when, ==0 For at = ±ℎ and equating to zero, 0 = −(2 − 30ℎ2 ) = 15ℎ2 Substituting to equation of , = 2 + 30ℎ2 − 10 3 For = − and = 0 at = ℎ and = −ℎ, − = 2 + 30ℎ3 − 10ℎ3 = 2 + 20ℎ3 And 0 = 2 − 30ℎ3 + 10ℎ3 = 2 − 20ℎ3 Solving simultaneously, =− =− 4 40ℎ3 Substituting to the equation G, = 15 (− 3 2 ) ℎ = − 40ℎ3 8ℎ Substituting E and G to = 6 + 6(0) − 30(− = 6 + ℎ ) 2 + 20(− ) 3 3 40ℎ 40ℎ3 3 2 − 3 3 4ℎ 2ℎ3 1 For ∫−ℎ = 2 2 at = , ℎ ∫ (6 + −ℎ 3 2 1 − 3 ) = 2 3 3 4ℎ 2ℎ 2 ℎ ∫ (6 2 + −ℎ (2 3 + 3 2 2 1 2 4 ) − = 4ℎ3 2ℎ3 2 2 3 1 2 5 − ) = 4ℎ3 10ℎ3 2 {2[ℎ3 − (−ℎ3 )] + 2 3 1 [ℎ − (−ℎ3 )] − [ℎ5 − (−ℎ5 )]} = 2 3 3 4ℎ 10ℎ 2 {2[ℎ3 − (−ℎ3 )] + 2 3 1 [ℎ − (−ℎ3 )] − [ℎ5 − (−ℎ5 )]} = 2 3 3 4ℎ 10ℎ 2 2 ℎ2 1 2 4ℎ + − = 2 5 2 3 4ℎ3 = = ℎ2 5 20ℎ Substituting the constants back to the equations for stresses, 2 ) + 6(0) − 30 (− ) + 20 (− ) 3 20ℎ 40ℎ3 40ℎ3 = 6 ( = 3 3 2 + − 3 3 10ℎ 4ℎ 2ℎ3 = ( + ) = 2 (− − 3 ) − 10 (− ) + 2 (− ) 3 4 8ℎ 40ℎ3 = − = − = −(0 + 2 (− 3 3 − + 2 4ℎ 4ℎ3 ( + − ) 3 ) + 3(0) 2 − 30 (− ) 2 ) 8ℎ 40ℎ3 = = 3 3 2 − 4ℎ 4ℎ3 ( − ) ...
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