Question

The hydraulic cylinder AB partiaily controls rod EF's position. it is locKed in the position shown. DE is 15mm thick & bolted at A to the rod with a gmm diameter bolt. DererMine a) Averoge shearing b) Bearing stre ss at A in DE Stress n bolt 75 mm 100mm 75 20 1 2KN 200mm C 45mm

Answer 1

The length on cp a v 45² 200² tcd 205 mm FBD of member DAE → 2X COS15 At zsings 175x sin20 175 COS20 Aley loosin 120 100 COS20 205 200 FALD

thema=0 - 2x co SC759*1754 sin (20) - 2 x sin (951% 175 x cos(20) - 45 FCD X 100 sin20 + - 205 2002 FCD X 100x cos(20) 20 205 .: 99.184 FCD - 348.668 FCD = 3.515 KN * EFX70 Axe -45__ x 3.515 + 2X COS 9570 205 Ax = 0.254kn +1 EFY: 0 Ay - 200x3.515 - 2x sin(95)=0 205 AY = 5.362 KN As ✓ Ax²4 Ay 2

A: 10.25435.3022 S. 369 A: 5.368 KN as sheasing stress in bolt p= A= 5.368 KN I: Re: Ac= fx193%= 63.6172 mm2 T: 5.368. 5.368 x 10 63.6172 2: 84.37 mpa by Beasing stress at a Abu dt: 9x15:135 mm ob f. 5.368X10 ob 135 ob= 39.762 mpa