Question

Chapter 01 Homework 12 6 Help Save & Exh Submit You received no credit for this question in the previous attempt. Check my work View previous attempt Required information The Frydraulic cylinder CF, which partially controls the position of rod DE has been locked in the position shown. Member BD is 15 mm thick and is connected at Cto the vertical rod by a 9-mm-diameter bolt. Know that P-2.4 KN and 8 75" Determine the average sharing stress in the boit The average shearing stress in the bolt is MPa < Prey 5 6 of 8 Next >

This is my 3rd time posting the questions. Please give me the right answer, I am trying to figure out how to solve the questions.

Answer 1

Common calculation for both (5) and (6):

The length of the link AB is: L = V200² +452 = 205 mm Draw the free body diagram of link BCD: BD is 15 mm thick and is connected at to the vertical X10 Plose E14 AL05 Sing

Consider moment about C. Mc=0 (2015 (100cos 2009 ( 1 (100 in 2009 LO 1-(2.4 cos 75°)(175sin 20°)–(2.4 sin 75°)(175cos 20°)] 84.17 FAB - 418.4 = 0 FAB = 4.971 kN Consider force equilibrium in horizontal direction: Fx = 0 C4 - 45 FAB+2.4 cos 75° = 0 C4 - 4-(4.971)+2.4 cos 75º = 0 Ct = 0.47 KN 205

Consider force equilibrium in vertical direction: EF, = 0 C, -200 F4B - 2.4 sin 75º = 0 ) - 2.4 sin 75º = 0 C, = 7.168 kN The net reaction force at C: C= VC+C; C = V0.472 +7.1682 C = 7.18339 KN

(5) The shear stress in bolt is: T Anelt 7.18339x103 =112.916 N/mm² t=112.916 MPa

The bearing stress at C is: 7.18339x103 9x15 = 53.21 N/mm2 100 = 53.21 MPa