All Courses
All Courses
Courses by Software
Courses by Semester
Courses by Domain
Tool-focused Courses
Machine learning
POPULAR COURSES
Success Stories
Question 1 Simulate the simple crank mechanism using cylinders (diameter = 2m) in a 2D plane (XY) as shown in the figure below In the case shown above, Revolute joints are located at P0 and P1, What type of joint must be given at P2 (Joint C) so that the system doesn't have redundancy issues? NOTE: Joint C is a type of…
Kolli Surya Teja
updated on 13 Jan 2022
Question 1
Simulate the simple crank mechanism using cylinders (diameter = 2m) in a 2D plane (XY) as shown in the figure below
In the case shown above, Revolute joints are located at P0 and P1, What type of joint must be given at P2 (Joint C) so that the system doesn't have redundancy issues?
NOTE: Joint C is a type of joint that will perform translational motion only along the X-axis (Basically Joint C is the slider of the crank-slider mechanism)
Also, prove why the joint you have selected will be the ideal joint for the system using DOF calculations, also, what type of joint is this?
Ideally, after you have selected the perfect joint for Joint C to reproduce the desired motion, what will be the final DOF of this system? also, validate with calculations
Solution:
Checking units and changing the acceleration due to gravity orientation
Required points are created
Required bodies are created, and the properties will later be associated to the graphic of the cylinder respectively
cylinder graphics are created with 2 as diameter.
Required revolute joints are created.
DOF:
Joint C should be an inline joint, in order for the model to be a kinematic model.
DOF:
The inline joint (DOF=2) restricts the motion in only translation directions thus giving the net DOF of the system as O. Thus, the system will not have any redundancy issues.
For instance, if we select a revolute joint:
DOF of one rigid body = 6
DOF of two rigid bodies = 6 x 2 = 12 ........(1)
DOF removed by one revolute joint = 5
DOF removed by three revolute joints = 5x3 =15 ........ (2)
DOF of the system (1)-(2) = 12-15 = -3
Now, if we select an inline joint:
DOF of one rigid body = 6
DOF of two rigid bodies = 6 x 2 = 12 .......(1)
DOF removed by two revolute joints=5 x 2 = 10
DOF removed by one inline joint = 2
DOF removed by all joints = 10 + 2 = 12...... (2)
DOF of the system (1) - (2) = 12 - 12 = 0
Thus, the inline joint selected will be ideal for the system.
Question 2
After selecting joint C, simulate the following mechanism with initial angular velocity at the joint in P0 as 0.5 rad/sec (for mass and inertial properties obtain from associated geometry)
Now, plot a graph for velocity vs time on Joint C or Point P2 w.r.t the ground
Find the maximum velocity of Point P2 w.r.t the ground
What is the DOF of the system now and why?
Solution:
Enabling the angular velocity for the first revolute joint.
Required outputs are requested using the outputs functionality
Plot:
DOF:
The DOF of the systems remains unchanged (i.e. O) as it depends on the alterations in motions, couplers or joints. By inducing a velocity, the DOF of the system will not get affected.
The same is validated below. Go to Tools - Check Model - DOF to check the details of our model.
Question 3
Now, suppress the initial angular velocity, and apply a load of 1000 KN on P1 and simulate and answer the following questions
Now, plot a graph for the reaction torque vs time at P0 and explain your results
Compare the range of motion of the crank mechanism between this case and the previous case and explain the reason for the difference if any
Solution:
Required force is applied at the revolute joint
Required output are requested
plots:
Reaction torque is zero because it is not constrained at joint 0.
Range of motion:
case 1: with angular vel
case 2: with 1000kN force
Question 4
Now, in the same case as Question 3 create a BISTOP function at point P0 with range (-10,0) degrees
Now, again, plot a graph for the reaction torque vs time at P0 and explain your results
A screenshot of the entire interface must be attached with the plot for all the simulated cases.
Solution:
New Force is applied at point 0 which contains bitstop expression to limit the motion
Required output are requested using Vtorq function.
Plot:
the reactive torque is shown above.
Leave a comment
Thanks for choosing to leave a comment. Please keep in mind that all the comments are moderated as per our comment policy, and your email will not be published for privacy reasons. Please leave a personal & meaningful conversation.
Other comments...
Week 8 & 9 - Four Bar Mechanism
Modify either the Bistop function or the motion curve that we apply to the revolute joint to stop the trunk lid exactly when the trunk lid comes in contact with the body of the car. Solution: Model with graphics imported: Default Motion curve: Motion vs Time_s Hyperview results: at t=3.11sec. collision is detected.…
13 Jan 2022 06:28 AM IST
Week 4 Project - Simple Crank Mechanism
Question 1 Simulate the simple crank mechanism using cylinders (diameter = 2m) in a 2D plane (XY) as shown in the figure below In the case shown above, Revolute joints are located at P0 and P1, What type of joint must be given at P2 (Joint C) so that the system doesn't have redundancy issues? NOTE: Joint C is a type of…
13 Jan 2022 06:26 AM IST
Week 1 - Free Falling
OBJECTIVE: To Simulate and plot a freefalling body (point mass) of mass 1 kg and inertial properties 1 kgm2 in the y-axis with simulation time of 2sec THEORY: Free Falling Object: The term "free fall" refers to an object falling freely with no forces operating on it other than gravity. Mathematical equation for free…
08 Jan 2022 07:35 AM IST
Problem Set-1 c Week 11Challenge : Transient Handling
The following table contains information about two vehicles FWD passenger car: Code: clear all close all clc l = 2.538984 ; %Wheelbase (m) a = 0.890016; %cg to front axle (m) b = l-a; %cg to rear axle (m) m = 907.18474; %Mass (kg) jz = 1128 ; %moment of inertia (kg-m^2) cf = 61171.9437; %front cornering stiffness…
29 Dec 2021 06:59 AM IST
Related Courses
0 Hours of Content
Skill-Lync offers industry relevant advanced engineering courses for engineering students by partnering with industry experts.
© 2025 Skill-Lync Inc. All Rights Reserved.