Question

please hurry this is urgent, (three digits after comma)

The joystick in the figure is under the effect of loads F1 = 55 kN and F2 = 135 kN. The safety coefficient of the pins at point A and C for slip is E.K. = 1.1 and the shear damage stress is τh = 80 MPa and the connection shapes of the pins are shown next to them. According to this;

Find the pin force at point A. (Write your result in kN.)

Find the pin force at the C point. (Write your result in kN.)

Find the smallest diameter required for pin at point A. (Please write your result in mm.)

Find the smallest diameter required for the pin at point C. (Please write your result in mm.)

A B A 250 mm C С 5 4 75 100 3 mm mm F1 F2

Answer 1

solution:- Stept:- Draw FBD of joystick and calculate reaction forces at A and c:- fi=55KN F2=135KN FABT 250 mm Now equilibrium equations.- allo t Сх 6 5 ť Fi cy loo ij EM C=0;- 4F K7 F_ x100 + 5F, X175-FAR X250-0 => 135x100+*x55x175- FAR X250 =0 => FAB = 84.8 KN 10) EFXD Fast Coc - 36=0 cx = 2X55-8468 Cx = - 51.8KNY i Elyson Cy - F - 56=0 cy= 135+2x55 07 (44 = 179KN

Net force at RE roztcy? (-57-8)2 +(1792 [RE 186-34 KN Step2:- Smallest diameter required for pin at point A:- : at pin a double shear FAB will act. TA < Tallowable * A da TA Tallowable FAbla FAbla safety coefficient → FAbla 80x106 - AA S -> 84.8x103 80x106 &XFxdk lol da = 0.02477 m da = 24.76 mm 25 mm / Ary Step3: - similarly at pin at c :- łc Tallowable Safety coefficient Rc 80X106 Ill da? 11 => 186.34X103 80x106 JA Xd? dc = 51.92mm 52mm Any dc = Real Rc => case: Single Shear
Thanks