Published on 26 Jan 2015

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ENGG 202 Tutorial 12 April 2/3, 2013

Frames and Machines

1) The 80 kg ventilation door OD with mass center at G is

held in the equilibrium position shown (open) by means of a

moment M applied at A to the linkage mechanism. Member

AB is parallel to the door for the 30˚ position shown. All

joints are pin connections.

a) Determine the forces acting on member OCGD at O and C.

b) Determine M

ANS: CB=840.9 N (C), Ox= -528.6N, Oy=130.8N, M = 706.3 Nm

Solution Strategy: a) draw the FBD of the gate OCGD, there will be 3 unknowns

(2 at O and CB is a 2 force member), write 3 eqns of equil to solve for these

unknowns. You will need to apply some knowledge of geometry to find the angle

of member CB. B) Next draw an FBD of the member AB, there will be 2 new

unknowns at A as well as the unknown M. sum moments about A and solve for M.

2) The clamp presses two blocks of wood

together. Determine the horizontal and vertical

reactions at pin C if the blocks are pressed

together with a vertical normal force of 200 N.

ANS: Cx = 500 N, Cy = -200 N (on ABC)

Solution Strategy: draw the FBD of the upper

jaw plus upper block, there are only 3

unknowns. (force acting upward at block of

200 N, BE is a 2 force member, 2 unknowns at

C), solve.

3) The loaded bucket of the bulldozer weighs 11 kN with a centre of mass at H. The weight of

arm AD weighs 2.5 kN and acts at B, while the weight of arm DG is 1.5 kN and acts at E. The

weights of the hydraulic pistons (CJ, EI, and BF) can be neglected. Find the force P in piston CJ

and the forces in pistons BF and EI for static equilibrium. Pistons CJ and EI are horizontal.

ANS: CJ = 20.3 kN (T), BF = 11.6 kN

(C), EI = 3.18 kN (T)

Solution Strategy: draw the FBD of the

bucket, there are only 3 unknowns (2 at

G and 1 at I), write 1 eqns of equil (sum

moments about G) and solve for EI.

Next draw an FBD of arm DEFG and

the bucket H (do not detach EI), there

are 3 unknowns (2 at D and 1 at F),

write 1 eqns of equil (sum moments