FORCE ANALYSIS OF MECHANISMS

Methods of Linkage Force Analysis

Superposition Method (sequential solution of links/subsystems equations). Two approaches can be used o Graphical layout o Hand calculation

Matrix Method: computer solution (simultaneous solution of system of equations) The steps of solving a mechanism under static loading is reviewed first, then D'Alembert principle and inertia forces are introduced, which converts the dynamic problem into kinetostatic problem, where the superposition method and the graphical approach can be utilized. Newton's Equations:

Statics: ∑ 𝐹 = 0 ∑𝑀 = 0

In planar mechanism the vector force equation is equivalent to two scalar equations in the x and y directions, and the moment equation is just a single equation in the z-direction

STATIC FORCE ANALYSIS SUPERPOSITION - GRAPHICAL METHOD

Static Force Systems

In the graphical approach it is useful to recognize the following static force system to

determine the directions (line of action but not the sense) of the internal forces.

1. Two Force Member: a rigid body acted upon by two

forces is in static equilibrium only if the two forces are

collinear, equal in magnitude, and opposite in

direction.

2. Three Force Member: a rigid body acted upon by three

forces is in static equilibrium only if the three forces

are concurrent at some point and their sum is zero.

3. A rigid body acted upon by a couple is in static

equilibrium only if acted upon by another coplanar

couple, which is equal in magnitude, and opposite in

direction.

Internal Forces at cut joints:

Revolute A revolute joining link-1 and link-2 has two constraints

(internal force vector) that prevent the relative translation

motion between the two links and has a DOF in the

rotation direction.

Prismatic A prismatic joining link-1 and link-2 has two constraints

(internal force and moment) that prevent the relative

translation motion in the tangent direction and the relative

rotation. The prismatic has a DOF in the tangent

direction.

Higher Pairs A higher pair joint such as between meshing gears, can

and follower, or pin and slot has one constraint (internal

force) in the normal direction and two DOF in the tangent

and rotational directions.

Graphical Static Force Analysis (Using Force Polygon Method)

Following are the Steps for graphically solving static force problem in mechanisms

1. Graphical position analysis: draw the space diagram (mechanism configuration) at the required

position using a suitable scale.

2. On the space diagram draw the external forces/moments acting on the mechanism in their

corresponding locations.

3. Draw the free body diagram for each subsystem:

a. Introduce cuts at joints or links' sections, draw an exploded drawing of the mechanism (break the

mechanism into a suitable number of subsystems-links/group of links),

b. Draw the external forces on the links and internal constraining forces at the introduced cuts.

4. Identify the type of static force system for each subsystem and apply the conditions of equilibrium

to get information about the unknown internal forces.

5. Analyze links sequentially by drawing the force polygon for each subsystem/link

a. Start with the link which can be analyzed immediately (subject to known external force)

b. Use the results from the previous step to construct remaining force polygons by

examining forces transmitted between subsystems.

Youtube Videos

a) Four bar mechanism: http://www.youtube.com/watch?v=x2zzkIu8uAo b) Slider crank: http://www.youtube.com/watch?v=AVyfRqqyUR0 c) Complex mechanism: http://www.youtube.com/watch?v=s08Mde4aflE

Example: Four bar mechanism (same as video-a above)

Required to determine the magnitude and direction of M12 in equilibrium with the

external force P

1. The space diagram (mechanism configuration) is drawn at the required position using

a suitable scale.

2. On the space diagram the external force is drawn acting on link-4 in its position.

3. Introduce cuts, identify the internal forces at the cuts, and draw the free body

diagram for each subsystem.

a. Introduce cuts at joints or links' sections, draw an exploded drawing of the

mechanism (break the mechanism into a suitable number of subsystems-

links/group of links),

b. Draw the external forces on the links and internal constraining forces at the

introduced cuts.

Note the notation of the

internal forces. For example F23

is the force from link-2 on link-

3. Also note the Newton third

law is used. For example F23 is

equal to F32 and opposite in

direction.

At this stage neither the

directions nor magnitudes of

the internal forces are known.

4. Identify the type of static force system for each subsystem and apply the conditions

of equilibrium to get information about the directions of the unknown internal

forces.

As link-3 is two force

member hence the

directions of F23 and F43 are

known. Hence the directions

of F32 and F34 are known.

As link 4 is three force

member with the direction

of the external force P and

F34 known, the direction of

F14 is defined.

As link-2 is subject to moment M12, hence for static equilibrium F12 is parallel to F32 to form

a couple which is equal in magnitude and opposite in direction to M12.

5. Analyze links sequentially by drawing the force polygon for each subsystem/link

a. Start with the link which can be analyzed immediately (subject to known

external force)

As the external force P is known in

magnitude and direction we start

drawing the force polygon of link-4.

Starting with the external force P

which is known in magnitude and

sense (arrow direction), the force P

is first drawn to scale.

Then the intersection of the two lines

which are parallel to F14 and F34 are drawn at the ends of P define the polygon (triangle in

this case). The senses of F14 and F34 (their arrows directions) are drawn such that the

polygon closes to satisfy ∑ 𝐹 = 0, or in this case: P+F14+F34 = 0

As shown in the figure the force polygon may be drawn in two ways, however the result,

i.e., the magnitude and sense of direction of F34 and F14 are the same.

b. Use the results from the previous step to construct remaining force polygons by

examining forces transmitted between subsystems.

With F34 is known from the force polygon of link-4, hence F43, F23, F32, and F12 are all

known in magnitude and sense of action.

F32, and F12 form a couple which is equal and opposite to M12. Hence M12 = F12 X d, where

d is the normal distance between F32, and F12. The distance d is measured from the space

diagram and space scale is utilized.

Note 1:

All the five steps of graphical force analysis are done on the same page as shown below.

Note 2:

One can save time by careful draw

the directions of the internal forces

and the force polygons directly on

the space diagram without an

exploded drawing and FBD

Problems

In solving the problems do the following:

Print the problem page and consider the space diagram is drawn according to the scale given in

the problem. Hence steps 1 and 2 are completed.

On the printed page directly solve the problem starting from step 3.

You will need your drawing tools.

Ether use:

A drawing board, hence fix the printed page on the board and start solving the problem.

or

Just use triangles (45x45 and 30x60) to draw parallel lines.

Of course in both cases you will need panicle, eraser, scaled straight edge, etc.

1. Determine the shaft torque Ts at O2 in static

equilibrium with the shown external force

(S.S. 1:5)

2. Determine the shaft torque Ts at O2 in

static equilibrium with the shown

external force (S.S. 1:5)

3. Determine the shaft torque Ts at O2 in static equilibrium with the shown external force

(S.S. 1:5)

4. Determine the shaft torque Ts at O2 in static equilibrium with the shown external force

(S.S. 1:5)