Kinetic analysis has to be performed to design the shape and thickness of the links, the joints and to determine the required input torque/force and the resulting output torque/force.

B

A

Force analysis of a 4-Bar

Known:

4

ω2 α2 • The length of all links; r1, r2, r3, and r4

3

2

O4

O2

• Input information; θ2, ω2, α2, and T2 (torque)

• The results of position, velocity, and acceleration analysis; θ3, θ4, ω3, ω4, and α3, α4

• All external loads acting on the links

Unknown:

• Forces acting on all joints; FO2, FA, FB, and FO4

• Inertia forces acting on all links; FI2, FI3, FI4,

• Inertia torques acting on all links; TI2, TI3, TI4,

Ken Youssefi

• The input torque (or force) needed to obtain the

Mechanical Engineering Dept. required output torque (or force)

1

Force Analysis – Static Equilibrium

Static equilibrium – 2D case

∑ Fx = 0, ∑ Fy = 0, ∑ M = 0

Two force members – only two forces acting, one at each joint.

A rigid link acted on by two forces is in static equilibrium only if the two forces are collinear and have the same magnitude in opposite direction

X

F1 = - F2

Ken Youssefi

Mechanical Engineering Dept.

2

Force Analysis – Static Equilibrium

Three force members – only three forces act on a link. Either three joint forces, or two joint forces and an external or inertia or gravitational force.

The link is in static equilibrium only if the force-vectors add up to zero and they pass through the same point

(summation of moment about that point is zero).

Ken Youssefi

Mechanical Engineering Dept.

3

Force Analysis – Graphical Method

B

P = 120 lb

C

4

3

A

2

O2A = 3, AB = O4B = 6, O4C = 4.5,

O2O4 = 2.4

O2

O4

Find the input torque, T2 , to maintain static equilibrium at the position shown or what should be the input torque to exert 120 lb of force for the position shown. Use graphical method, construct force polygon.

F34

B

Free-Body Diagram of link 4

Parallel to link3

AB is a two force member, F34

(force of link3 on link 4) has to be collinear to link 3.

M

Direction of F14 has to pass through point M

Ken Youssefi

P = 120 lb

Mechanical Engineering Dept.

O4

F14

4

Force Analysis – Graphical Method

Force polygon

F34

B

1 - Draw and scale force P

2 - Draw two lines parallel to F14 and F34

P = 120 lb

directions from the two ends of vector P

3 - Locate the intersection, complete the

polygon and scale the force F34

M

P = 120 lb

O4

F14

F34 = 113.52 lb.

Parallel to F14 direction Parallel to F34 direction Ken Youssefi

Mechanical Engineering Dept.

5

Force Analysis – Graphical Method

Free-Body Diagram of link 3

F43

B

Free-Body Diagram of link 2

Draw and scale link 2, O2A = 3

Scale the perpendicular distance from O2 to the force F32

F32 = 113.52

d = 2.428

3

A

A

2

O2

F12 = 113.52 lb

F23 = F43 = 113.52

T2 = (F32)(d) = 113.52 x 2.428 = 275.6 in-lb (ccw)

Ken Youssefi

Mechanical Engineering Dept.

6

Force Analysis – Graphical Method

Principle of Superposition

In linear systems the output (response) is directly proportional to the input, linear relationship between the output and the input exist.

Principle of superposition can be used to solve problems involving linear systems by considering each of the inputs to the system separately. And then, combine the individual results to obtain the total response of the system.

B

B

P

Q

P

C

C

D

=

4

3

A

2

Ken Youssefi

O2

B

Q

O4

A

+

4

3

2

O2

O4

A

D

4

3

2

T2 = (T2) Due to force P + (T2) Due to force Q

Mechanical Engineering Dept.

O2

O4

7

Dynamic Force Analysis

Inertia forces and D’Alembert’s Principle

=

An unbalanced set of forces acting on a rigid body

The translation and rotation caused by the unbalanced forces

Resultant force R…