four bar mechanism - kts...four bar mechanism ground - any link or links that are fixed with respect...

Four bar mechanism

Mechanism Design

Four bar mechanism

Ground - Any link or links that are fixed with respect to the reference frame

Crank - A link which makes a complete revolution and is pivoted to ground

A four-bar mechanism consists of3 moving bodies and 4 joints (2 of thebodies are connected to the ground). The mechanism has 1 DoF.

Four bar mechanismCoupler - A link which has complex motion and is not pivoted to ground

Rocker - A link which has oscillatory (back and forth) rotation betweentwo limiting angles and pivoted to ground

Four bar mechanismLaw of Grashof (Grashof Condition):

The shortest link of a four-bar mechanism can make a fullrevolution relative to the other links if the sum of the shortestand the largest link is less then or equal to the sum of thelengths of the other two links.

S+L ≤ P+QS = length of shorter linkL=length of longest linkP=length of one remaining linkQ=length of the other remaining link

Four bar mechanism

• The shortest link is pin-jointed to the frame (crank and rocker mechanism),

• The shortest link is the coupler (double rocker mechanism),

• The shortest link is the frame. In this case all other links make a full revolution relative tothe frame (double crank mechanism)

When the Grashof condition is true (S+L < P+Q), there are still three possibilities (Class I case)

Four bar mechanism

• All inversions will be rockers in which no link can fully

When the Grashof condition is not true (S+L > P+Q), there are still one possibilities (Class II case)

Four bar mechanism

Parallelogram form Antiparallelogram form

When the Grashof condition is (S+L = P+Q), there are still one possibilities (Class III case)

Task 1: Double crank four bar mechanism

Grashof condition is true (S+L < P+Q)

The frame is the shortest link of fourbar mechanism.

All other links make a full revolution relative to the frame

Individual Project

Mechanism design SM-N Mechanism Analysis 2019/2020

1) Solve the velocity of point A

2) Determine the maximum value of the angular acceleration on the rocker (part 3)

Part 1 – crank

Dimensionlength = L1height = H1width = W1

Part 2 – coupler (ternary part)Location point ADistance from crank = Vwidth = W2

Part 3 – rocker

Dimensionlength = L3height = H3width = W3

GroundDimension H, V

Angular velocity of crank:ω= 10 s-1

No. L1 H1 W1 V W2 L3 H3 W3 H L1 40 10 10 50 10 200 10 10 100 1402 40 10 10 50 10 200 10 10 100 1603 40 10 10 50 10 200 10 10 100 1804 60 10 10 50 10 200 10 10 140 1605 60 10 10 50 10 200 10 10 140 1606 60 10 10 50 10 200 10 10 140 1607 80 10 10 50 10 300 10 10 180 1408 80 10 10 50 10 300 10 10 180 1609 80 10 10 50 10 300 10 10 180 180

10 100 10 10 50 10 300 10 10 200 14011 100 10 10 50 10 300 10 10 200 16012 100 10 10 50 10 300 10 10 200 180

Individual Project



Information: name of BODY/PARTname of MARKER

No. L1 H1 W1 V W2 L3 H3 W3 H L1 40 10 10 50 10 200 10 10 100 1402 40 10 10 50 10 200 10 10 100 1603 40 10 10 50 10 200 10 10 100 1804 60 10 10 50 10 200 10 10 140 1605 60 10 10 50 10 200 10 10 140 1606 60 10 10 50 10 200 10 10 140 1607 80 10 10 50 10 300 10 10 180 1408 80 10 10 50 10 300 10 10 180 1609 80 10 10 50 10 300 10 10 180 180

10 100 10 10 50 10 300 10 10 200 14011 100 10 10 50 10 300 10 10 200 16012 100 10 10 50 10 300 10 10 200 180

A

Individual Project



1. Select body: PART_4marker: MARKER_7

2. Select characteristic: Translational_Velocity

3. Select Component: X(for the left plot)

4. Select Component: Y(for the right plot)

Individual Project


A


Information: name of JOINT

Individual Project


2. Select constraint: JOINT_1 3. Select characteristic: Angular_Acceleration

4. Select Component: Z


1. Insert new page

Trajectory of point A

1. Insert new page

1. Select body: PART_4marker: MARKER_7

2. Select characteristic: Translational_Displacement

4. Select Component: Y(for the Dependent Axis)

3. Select Component: X(for the Independent

Axis)

Literature: ADAMS_Full_Simulation_Guide_2005.pdf

If a user of Adams creates a contact between two bodies, a model must be chosen for the calculation of the normal force. A choice is given between the IMPACT function model and the POISSON model (Restitution). It is also possible to define a custom function.

After choosing the IMPACT function model, four variables must be defined; stiffness, force exponent, damping and penetration depth.

stiffness - stiffness of the boundary surface interactionforce exponent - A positive real variable that specifies the exponent of the force deformation characteristic. For a stiffening spring characteristic, e > 1.0. For a softening spring characteristic, 0 < e < 1.0.damping – maximum damping coefficient.penetration depth - boundary penetration at which Adamsapplies full damping

The second built-in method for calculating the normal forces is called Restitution in Adams. Most important in this method is the so-called restitution coefficient (COR - Coefficient OfRestitution). It defines a continuum between a perfectly elastic (COR = 1.0) and perfectly inelastic (COR = 0.0) collision. The difference between both limits is that in an elastic collision the kinetic energy is conserved and in an inelastic collision the kinetic energy is not conserved. In a perfectly inelastic collision the reduction of kinetic energy equals the total kinetic energy before the collision in a center-of-momentum frame. Even though the behaviour of kinetic energy differs in these cases, in all collisions the total momentum is conserved.

The coefficient (e) is defined as the ratio of relative speeds after and before an impact, taken along the line of the impact:

how much of the kinetic energy remains for the objects to rebound from one another vs. how much is lost as heat, or work done deforming the objects

four bar mechanism - kts...four bar mechanism ground - any link or links that are fixed with respect...

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