1. ME 302 DYNAMICS OF MACHINERY
STATIC FORCE ANALYSIS
Dr. Sadettin KAPUCU
© 2007 Sadettin Kapucu

2. StaticEquilibrium
From Newton’s first law, a body is in static equilibrium if the resultant of all the forces acting on a rigid body is zero. This condition can be expressed mathematicaly as:
In space, these two vector equation yields six scalar equations:
In planar, there are three scalar equations:

3. Force Systems
In coplanar force system under which the body acted on is in equilibrium, we can have following force systems:
Two force member : If there are only two forces acting on a body, it is called a “ two force member”.
To satisfy sum of forces equal to zero, the two forces should be equal and opposite. F1 F2
To satisfy sum of torque is equal to zero distance between the forces must be zero. This means that the forces are collinear.

4. Force Systems
Three force member : If there are only three forces acting on a body, it is called a “ three force member”.
To satisfy sum of forces equal to zero, the vectors must form a closed polygon and coplanar. F3 F1 F2
To satisfy sum of torque is equal to zero, the lines of application of all the three forces intersects at one single point. This point is called the point of concurrency.

5. Solving static force problems
Separate the mechanism into its links, considering each a free body with all the acting external and constraint forces on it.
Apply the rules of statics each free body which are
Solution of vector equations can be by arithmetical and or graphical.

6. Solving static force problems
Graphical Approach : We draw straight lines to represent vectors which are in proper directions and lengths proportional to the magnitudes of the vectors and in an articulated manner as depicted in Figure. Vectors form a closed polygon called a “vector loop” F3 F1 F2

7. Solving static force problems
Arithmetical Approach : The simplest arithmetical approach is to separate vector equation into components. q3 F3 F1 F2 q1 q2
These two component equations are not no longer vector equations. They are scalar and can be simultanously solved to find max two of the following;

8. Example
An external force of 10 N is acting horizontally on the rocker link, 30 mm from the point D. Find the amount of torque to be applied to the crank AB to keep the mechanism in static equilibrium.
a1 = 80 mm
a2 = 30 mm
a3 = 70 mm
a4 = 50 mm

9. Example
First step is a position analysis to find the angles of the crank CD and the coupler links.
The simplest way is to draw the mechanism to scale and measure the required angles by a protractor directly from the figure.
Or we can take a purely arithmetical approach.
and

10. Example FCx Arithmetical method:
Separate the mechanism into free bodies of links,
Put all the acting and interacting forces,
Then, apply the law of statics for each free body.
FCy
FBx
FCy
FCx
FBy
FBy
FBx
FDx
FAx
FDy
FAy

11. Static equations for link 4;
FDx
FDy
FCy
FCx

12. Static equations for link 3;
FCy
FCx
FBx
FBy
Static equations for link 3;

13. Static equations for link 2;
FBx
FBy
FAx
FAy
Static equations for link 2;

14. FBx FCy FCx FBy FDx FAx FDy FAy
We have 9 equations to solve simultaneously; FAx, FAy, FBx, FBy, FCx, FCy, FDx, FDy, and T.
ANS.

15. Example Graphical method: Draw the mechanism in scale,
Measure the unknown quantities directly from the scaled drawing,
Separate the mechanism into free bodies of links (scaled drawing),
State whether the link is two force - three force member and then put all the acting and interacting forces,
Apply the law of statics for each free body.

16. F43 F23 F34 F32 F14 F14 F34 F12 Link 3 is two force member,
Link 4 is three force member
F23
F34
F32
10 N
F14
F14
F34
F12

17. F14 & F34 are measured directly from the scaled force polygon.
10 N stands for 50 mm
F14 stands for 22.5 mm
F34 stands for 32.5 mm