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 F0, M0

Not in equilibrium F=0, M0

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

Not in equilibrium M0

In equilibrium F=0, M=0

O

Four-Force Members: The problem is reduced to one of three-force member. Then the approach above is applied.

Condition of equilibrium, for any three-force member with no applied torque: Forces should be coplanar

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

F14

should intersect at a point . Therefore direction of

d)

F14 is obtained.

Force triangle is used. If 1st approach was used, this triangle gives direction and magnitude of

If 2nd approach was used, this triangle gives magnitudes of F34 ,

F14

Since, graphically

F14

e) Action and reaction forces are equal: and f) F=0

(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 F34 = 920,25 /-11,95 N = 900i-190.5j N F14 = 190.5j N

For the moment balance, all of the force vectors should pass through point B.

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)

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 ?

0.2N