Mechanics of Materials Homework Set 12 Spring 2020 Due, Schemes and Mind Maps of Mechanics

Download Mechanics of Materials Homework Set 12 Spring 2020 Due and more Schemes and Mind Maps Mechanics in PDF only on Docsity! ME 323: Mechanics of Materials Homework Set 12 Spring 2020 Due: Wednesday, Apr. 29 Problem 12.1 (10 points) A section is subjected to an axial load 𝑃, a torque 𝑇 and a moment 𝑀. It has a diameter 2D. The material is ductile and has a yield stress 𝜎% = 150 𝑀𝑃𝑎. Determine the factors of safety using the maximum shear stress theory and the maximum distortion energy theory. Use: 𝐷 = 10 𝑚𝑚, 𝑃 = 100 𝑁, 𝑇 = 80 𝑁 ∙ 𝑚,𝑀 = 50 𝑁 ∙ 𝑚 Solution: Point A has the absolute maximum normal stress. 𝜎2 = 𝑃 𝐴 + 𝑀𝑦 𝐼7 = 𝑃 𝜋𝐷9 − 𝑀 ∗ 𝐷 𝜋 4 𝐷 = = − 100 𝜋 ∗ 0.019 − 50 ∗ 0.01 𝜋 4 0.01 = = −63.98 𝑀𝑃𝑎 𝜏27 = 𝑇𝐷 𝐼C = 𝑇 ∗ 𝐷 𝜋 2 𝐷 = = − 80 ∗ 0.01 𝜋 2 0.01 = = −50.93 𝑀𝑃𝑎 𝜎EFG = 𝜎2 2 = −31.99 𝑀𝑃𝑎 𝑅 = 𝜎2 2 9 + 𝜏2I9 = − 63.98 2 9 + 50.939 = 60.14 𝑀𝑃𝑎 Principal stresses: 𝜎J = 𝜎EFG + 𝑅 = −31.99 + 60.14 = 28.15 𝑀𝑃𝑎 𝜎9 = 0 𝜎K = 𝜎EFG − 𝑅 = −31.99 − 60.14 = −92.13 𝑀𝑃𝑎 Absolute maximum shear stress: 𝜏LE2,EMN = 𝜎J − 𝜎K 2 = 60.14 𝑀𝑃𝑎 Solution: 𝜎2 = −10 𝑀𝑃𝑎, 𝜎I = 30 𝑀𝑃𝑎, 𝜏2I = 15 𝑀𝑃𝑎 𝜎EFG = 𝜎2 + 𝜎I 2 = 10 𝑀𝑃𝑎 𝑅 = 𝜎2 − 𝜎I 2 9 + 𝜏2I9 = 209 + 159 = 25 𝑀𝑃𝑎 (1) Principal stresses: 𝜎J = 𝜎EFG + 𝑅 = 35 𝑀𝑃𝑎 𝜎9 = 0 𝜎K = 𝜎EFG − 𝑅 = −15 𝑀𝑃𝑎 Absolute maximum shear stress: 𝜏LE2,EMN = 𝜎J − 𝜎K 2 = 25 𝑀𝑃𝑎 Von mises stress: 𝜎O = 1 2 𝜎J − 𝜎9 9 + 𝜎9 − 𝜎K 9 + 𝜎K − 𝜎J 9 = 44.44 𝑀𝑃𝑎 (2) Ductile material: For maximum-shear-stress theory: Safety factor: 𝑆𝐹 = 𝜎% 𝜏LE2,EMN ∗ 2 = 75 50 = 1.5 For maximum distortion energy theory: Safety factor: 𝑆𝐹 = 𝜎% 𝜎O = 75 44.44 = 1.69 (3) Brittle material: For maximum normal stress theory: In tension: 𝜎ST = 𝜎J = 35 𝑀𝑃𝑎 Safety factor: 𝑆𝐹T = 100 35 = 2.86 In compression: 𝜎SU = 𝜎K = −15 𝑀𝑃𝑎 Safety factor: 𝑆𝐹U = 120 15 = 8 So, 𝑆𝐹 = 𝑆𝐹T = 2.86 For Mohr’s failure criterion: 𝜎J 100 − 𝜎K 120 = 35 100 + 15 120 = 0.475 Safety factor: 𝑆𝐹 = 1 0.475 = 2.11 Problem 12.3 (10 points) A pipe has outer diameter 4𝑑 and inner diameter 2𝑑 . The material has a yield stress 𝜎% = 200 𝑀𝑃𝑎. If the factor of safety at point A is 2, determine the minimum value of 𝑑 using the maximum shear stress theory and the maximum distortion energy theory. 𝐹𝑆 = 𝜎% 𝜎O = 2 𝜎O = 200 ∗ 10_ 2 = 100 𝑀𝑃𝑎 𝑑 = 8.20 𝑚𝑚 Problem 12.4 (10 points) A horizontal rigid bar DCB is supported by a pin-fixed column AC and is subjected to a uniformly distributed load 𝑞. The column AC has a Young’s modulus 𝐸 and a rectangular cross section as is shown. Consider supports A and C to act as pinned-pinned when buckling in x-y plane and fixed- fixed when buckling in y-z plane. Determine the maximum distributed load 𝑞 can be applied without buckling. Use: 𝐿 = 2 𝑚, 𝑏 = 40 𝑚𝑚, ℎ = 60 𝑚𝑚, 𝐸 = 150 𝐺𝑃𝑎 Solution: FBD: 𝑀 = 0f : 𝑞 ∗ 2𝐿 ∗ 2𝐿 − 𝐹U ∗ 2𝐿 = 0 (1) solving for equation (1): 𝐹U = 2𝑞𝐿 For buckling in x-y plane, 𝐼7 = 1 12ℎ𝑏 K 𝑃gh = 𝜋9𝐸𝐼i 𝐾𝐿 9 = 𝜋9150 ∗ 10k ∗ 112 0.06 ∗ 0.04 K 1 ∗ 2𝐿 9 = 29.61 𝑘𝑁 𝐹U = 2𝑞𝐿 = 𝑃gh 𝑞gh = 𝑃gh 2𝐿 = 7.4 𝑘𝑁/𝑚 For buckling in y-z plane, 𝐼2 = 1 12 𝑏ℎ K 𝑃gh = 𝜋9𝐸𝐼i 𝐾𝐿 9 = 𝜋9150 ∗ 10k ∗ 112 0.04 ∗ 0.06 K 0.5 ∗ 2𝐿 9 = 266.48 𝑘𝑁 𝐹U = 2𝑞𝐿 = 𝑃gh 𝑞gh = 𝑃gh 2𝐿 = 66.62 𝑘𝑁/𝑚 So, the maximum distributed load 𝑞 = 7.4 𝑘𝑁/𝑚