# Exam 2 Review Oct 13, 2014 Dr. Koen – 3D Rigid Body, Trusses ```Exam 2 Review
Oct 13, 2014
Dr. Koen – 3D Rigid Body, Trusses
5 - 76 The boom AC is supported at A by
a ball-and-socket joint and by two cables
BDC and CE. Cable BDC is continuous and
passes over a pulley at D. Calculate the
tension in the cables and the x,y,z
components of reaction at A if a crate has
a weight of 80 lb.
Thought process…
3d Rigid Body Equilibrium Problem
Draw FBD to identify unknowns
Ax
Ball &amp; Socket at A
Two cables (always pull)
Assume all unknowns as positive
Decide on equations that can be used
Express each cable as a Cartesian Vector
Apply equations
Az
BD
Ay
CE
CD
CD = CD { -.231i - .923j + .308k}
r CD = &lt; -3, -12, 4 &gt;
CE = CE { .218i - .873j + .436k}
r CE = &lt; 3, -12, 6 &gt;
BD = BD { -.468i - .625j + .625k}
r BD = &lt; -3, -4, 4 &gt;
ƩFx = 0; Ax - .231CD + .218CE - .468BD = 0
ƩFy = 0; Ay - .923CD - .873CE - .625BD = 0
ƩFz = 0; Az + .308CD + .436CE+ .625 BD – 80 = 0
ƩMx = 0; .625BD(4) – 80(12) + .436CE(12) + .308CD(12) = 0
ƩMz = 0; -.218(CE12) + .231(CD)12) +.468BD(4)=0
Also… BD = CD
Using Solver….
Ax = 19.4 lb
Ay = 192 lb
Az = -25.8 lb
CD = BD = 62lb
CE = 110 lb
5 – 80 The bent rod is supported at A,
B, and C by smooth journal bearings.
Determine the magnitude of F2 which
will cause the reaction Cy at the bearing
C to equal to zero. The bearings are in
proper alignment and exert only force
reactions on the rod. Set F1 = 300 lb
Thought process…
3d Rigid Body Equilibrium Problem
Draw FBD to identify unknowns
Proper Alignment means no couple
moment reactions at supports… just
forces
Cz
Ay
Bz
Ax
Assume all unknowns as positive
6 unknowns, so need all 6 equations
Express F1 and F2 as Cartesian Vectors
Apply all six equations
Bx
Cy=0
F1 = {0i – 300 cos45j - 300sin45k} = {0 i – 212j – 212k} lb
F2 = {F2cos45sin30i + F2cos45cos30j - F2sin45k} = {.354(F2) i + .612F2j – .707(F2)k} lb
ƩFx = 0;
Ax + Bx + .354F2 = 0
ƩFy = 0;
Ay – 212 + .612F2 = 0
ƩFz = 0;
Bz + Cz – 212 - .707F2 = 0
ƩMx = 0; -Ay(4) – Bz(3) + 212(5) + 212(5) = 0
ƩMy = 0;
Ax(4) + Cz(5)= 0
ƩMz = 0;
Ax(5) + Bx(3)= 0
Using Solver….
Ax = 357 lb
Ay = -200 lb
Cz = -286 lb
Bx = -595 lb
F2= 674 lb
Bz = 974 lb
Prob 6 - 9
Determine the force in each member of the
truss and state if the members are in tension
or compression. Hint: The vertical component
of force at C must equal zero.
Thought process…
Truss by Joints problem
Find starting point – without supports?
Draw FBD, can assume tension for all, but
THINK!
Decide on path.
AB
3
Find by inspection.
4
AE
Joint A
6 kN
Joint B
3
BC
4
AB
DB
EB
ƩFy = 0;
ƩFx = 0;
4/5 AB - 6 = 0
3/5 (7.5) - AE = 0
AB = 7.5 kN (T)
Joint A
AE = 4.5 kN (C)
AB
3
4
AE
EB = 8kN (T)
Joint E
6 kN
AE
ED = 4.5 kN(C)
8 kN
Joint B
By inspection….
3
ƩFy = 0;
-4/5 (7.5) – 8 + DBsin45 = 0
ƩFx = 0;
-3/5 (7.5) + BC – DB cos 45 = 0
DB = 19.8 kN (C)
BC = 18.5 kN (T)
BC
4
AB
DB
EB
Prob 6 – 42
Determine the force in members LK, LC, and
BC of the truss, and state if the members are
in tension or compression.
Thought process…
Truss by Section
Decide where to cut, which side to use
Find support reactions
Draw FBD, assume tension for all
Look for points to sum moments about
Clockwise moments called positive…..
ƩMA = 0; 3000(4) + (3/5)LC(8)= 0
LC = -2500 lb….. 2500 lb (C)
LK
LC
ƩMC = 0; 7500(8) + (4/5) LK(6) – 3000(4) = 0
LK = -10,000 lb…. = 10,000 (C)
BC
ƩML = 0;
7500(4) – BC(3)= 0
7500 lb
BC = 10,000 lb (T)
Ms. Campbell
2D Rigid Body Equilibrium
Frames &amp; Machines
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