Question

A M12 105° Р B

In the crank-connecting rod mechanism shown in Figure 1, the piston and the cylinder friction between has been neglected. Static stability at the given crank position find the M12 torque required to make it happen. (O2A = 75 mm, AB = 350 mm, P = 0.9 kN)

Answer 1

T12 105 - 0 B P 77711111 Sol A = 75mm AB = 350 mm goven data P = 0.9 KN Piston force Force in connerting rod (AB) = For coso. FOR= P coso FOR Sin Force 3) member o A = F = Fr X OLA Mi2 = Torque FCR sin (0+0) (02A). P sinceto) (02A) caso Mia.

Applying cosene rule to Triangle OAB ABP (2A) + (2B) - 2 (014) (-6) COS (350) - (25) + (028)² - 2 275) (125) cos(105) 116875 (028)+ 38.82 (OLB). (©2B)*+ 38.82 (02B) 116075=0 OBE 323.010 -361.83 length can not be negative O2 B- 323.010 again applying to casine sell. (02A) = (18) + (02B) - 2 (AB) (028) cose (350) + (323.01) - (75) caso = 2 (350) (323.010) 86.82 Р sin(C++) (02A). now ea ① M12= cosa 0.9x103 Sin (105+86-82) [75] Los (86:82) 249.24 KN-mm 249.24 Nm 250 N negative implise the torne is in counter clock wise.
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