STATIC FORCE ANALYSIS
WEEK 1
COUPLE
MOMENT OF COUPLE
Couple: Two equal and opposite forces (F & F’)
Moment of Couple: A vector normal to the plane
of the couple (M=RBAX F)
Conditions of Equilibrium
A system of bodies is in equilibrium, if, and only if:
In 2-D (planar) systems:
Two-Force Members
Not in equilibrium
F0, M0
Not in equilibrium
F=0, M0
In equilibrium
F=0, M=0
Condition of equilibrium, for any two-force member with no applied
torque: The forces are equal, opposite and have the same line of action.
Three-Force Members
O
Not in equilibrium
M0
In equilibrium
F=0, M=0
Condition of equilibrium, for any three-force member with no applied torque:
Forces should be coplanar
Four-Force Members: The problem is reduced to one of three-force member.
Then the approach above is applied.
Example (Graphical Solution)
Link 3 is a two-force member
b)
1 st approach:
c)
2 nd approach: Since link 4 is is a three-force member; lines of action of forces P, F
34
F
F
should
intersect
at
a
point
.
Therefore
direction
of
is
obtained.
14
14
d) Force triangle is used.
If 1st approach was used, this triangle gives direction and magnitude of F14
If 2nd approach was used, this triangle gives magnitudes of F34
,
Since, graphically
e) Action and reaction forces are equal:
f) F=0
and
F
14
(Analytical Solution)
= 5(cos68.4i+sin68.4j) X 120(-cos40i-sin40j)
+12(cos68.4i+sin68.4j) X F34 (cos 22.4i+sin 22.4j) = 0
F34 = 33.1 lb
Problem (Analytical Solution)
(without friction)
O2A=75 mm P=0.9 kN
AB=350 mm M12 =?
F=0
F34+P+F14=0
F34(cos11,95i-sin11,95j)-900i+F14j=0
i: 0.978 F34 = F34x= 900  F34 = 920.25 N
j: 0.207 F34 = F34y= F14  F14 = 190.5 N
For the moment balance,
all of the force vectors
should pass through
point B.
F34 = 920,25 /-11,95 N = 900i-190.5j N
F14 = 190.5j N
Two-force member
F=0
F23=-F43=F34
F=0
F12=-F32=F23=F34
MO2=0
M12 + O2A x F32 = 0
M12k + 0.075(cos105i+sin105j) x (-900i+190.5j) = 0
M12k – 3.69k + 65.2k = 0
M12 = -61.51 N.m M12 = -61.51k N.m
Problem (Analytical Solution)
(with friction)
=0.2 (Between piston and cylinder)
0.2N
F=0 F34+P+F14=0
F34+P+(0.2Ni+Nj)=0
F34(cos11.95i-sin11.95j)-900i+0.2Ni+Nj=0
i: 0.978 F34 -900+0.2N = 0
j: 0.207 F34 + N = 0  N = 0.207 F34
0.978 F34 -900+0.2(0.207F34)= 0  F34 = 882.61 Newton
N=182.7 Newton F14 = 36.54i+182.7j Newton
F34 = 882.61 /-11,95 Newton = 863.48i-182.75j Newton
F=0 F12=-F32=F23=F34
MO2=0 M12 + O2A x F32 = 0
M12k + 0.075(cos105i+sin105j) x (-863.48i+182.75j) = 0
M12k – 3.547k + 62.55k = 0
M12 = -59.003 N.m M12 = -59.003k N.m CONCLUSION ?