# ENGG 202 Lecture Notes - Lecture 1: Wind Engineering, Ferrari 250

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Published on 26 Jan 2015

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ENGG 202 Tutorial 10 March 19/20, 2013

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300 mm 250 mm

3D Equilibrium of Rigid Bodies

1) The bent rod ABEF is supported in the position

shown by bearings at C and D and by wire AH.

Neither bearing is capable of exerting couples. The

bearing at D is not capable of exerting axial thrust,

while the bearing at C is. The force P = 400 N and

is parallel to the y- axis. Assume the weight of the

rod is negligible. Portion AB of the rod is 250 mm

long. a) Write the one equation of equilibrium that

could be used to find the tension in the cable AH

directly? ANS: ΣMCD = 0 b) Determine the support

reactions at bearings C and D and the tension in the

wire AH. Express your answers in Cartesian vector

format.

ANS:T=230.9 j + -400.0 k N, D = 505.2 j + -66.7 k N, C = -336.1 j + 466.7 k

N

Solution Strategy: Draw FBD of the bar ABCDEF. There are 2 force reactions at D (y and z), 3

force reactions at C, and one unknown cable tension for a total of 6 unknowns. A) By summing

moments about line CD, the force in the cable could be found directly (1 eqn 1 unknown). B)

Using vectors, sum the moments about C = 0 to obtain 3 eqns of equil and solve for reactions at

D and the cable force. Then write the force equil eqns to solve for reactions at C.

2) The lever AB is welded to the bent rod BCD which is

supported by bearings at E and F and by cable DG. The

cable DG lies in a y-z plane. The force P is 330 N and is

parallel to the x-axis. Knowing that both bearings are

properly aligned (and therefore cannot exert

couples/moments) and only the bearing at E does not exert

any axial thrust, determine (a) the tension in cable DG (b)

the reactions at bearings E and F

ANS: a) 561 N b) Ez=-228.4 N, Ex=157 N, Fx=173.1 N, Fy=264.3N, Fz=723.6N

Solution Strategy: Draw FBD of the assembly

ABEFCD. There are 2 force reactions at E (x and z), 3

force reactions at F, and one unknown cable tension for

a total of 6 unknowns. Using vectors, sum the moments about F = 0 to obtain 3 eqns of equil and

solve for reactions at E and the cable force. Then write the force equil eqns to solve for reactions

at F.

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