Dynamics Project - Moein Razavi

8/18/2019 Dynamics Project - Moein Razavi

1/34

PROJECT :

DYNAMICS OF

MACHINERY


2/34

DYNAMICS PROJECT: CRANK-SHAPER

PROVIDED BY : MOEIN RAZAVI - SAJJAD DEHGHANI – KASRA ALION ST. IDS.: 9226081 – 92266077 – 9226033 1

Project: Crank-Shaper Mechanism

Professor: Dr. A. Taghvaei pour

Teacher Assistant: Eng. M. Asadi Khanooki

Provided by: Moein Razavi – Sajjad Dehghani – Kasra Alion

St.IDs: 9226081 – 9226077 - 9226033

Course: 2nd Semester 2014-2015


3/34



Abstract: ........................................................................................................................................................ 3

Applications: ................................................................................................................................................. 3

1. INTRODUCTION ............................................................................................................................... 3

2. CRANK AND SLOTTED LEVER QUICK RETURN MECHANISM ........................................................ 5

A. Kinematics Analysis ............................................................................................................................... 6

A-1. ASSUMPTIONS ..................................................................................................................................... 6

A-2. SOLVING THE PROBLEM ........................................................................................................................ 7

A-3. OBTAINING DESIRED VALUES AND PLOT DIAGRAMS FOR SECTION 1 ...................................................... 12

A-3-1. “MATLAB” CODES ...................................................................................................................... 12

A-3-2. TABLE ........................................................................................................................................ 13

A-3-3. RESULTS (DIAGRAMS) ................................................................................................................. 13

A-4. OBTAINING DESIRED VALUES AND PLOT DIAGRAMS FOR SECTION 2 ...................................................... 15

A-4-1. “MATLAB” CODES ...................................................................................................................... 16

A-4-2. TABLE FOR INPUT PARAMETERS ................................................................................................... 18

A-4-3. RESULTS…………………………………………………………………………………………………………...…………19

A-4-3-1. (“MATLAB” DIAGRAMS) .................................................................................................... 19

A-4-3-2. (“ADAMS” DIAGRAMS) ...................................................................................................... 22

A-5. CHART ANALYSIS (ANALYZING Q UICK-RETURN) ................................................................................... 24

A-5-1. VELOCITY DIAGRAMS .................................................................................................................. 24

A-5-2. Acceleration Diagrams ....................................................................................................24

B. Kinetics Analysis .................................................................................................................................... 25

B-1. ASSUMPTIONS AND SOLVING THE PROBLEM ........................................................................................ 25

B-2. TABLES FOR INPUT VALUES .................................................................................................................. 30

B-3. CHART ANALYSIS (ANALYZING THE TORQUE ON LINK 2) ....................................................................... 33


4/34



Abstract:

We’re going to analyze a kind of quick-return mechanism, Crank-Shaper which

employs and inversion of slider-crank linkage. This project has two phases. The

first phase is the kinematics analysis of the mechanism and the second one is the

kinetics analysis of it.

Applications:

INTRODUCTION

A quick return mechanism is a mechanism that converts rotary motion into

reciprocating motion at different rate for its two strokes, i.e. working stroke and

return stroke. When the time required for the working stroke is greater than that

of the return stroke, it is a quick return mechanism. It yields a significant

improvement in machining productivity. Currently, it is widely used in machine

tools, for instance, shaping machines, power-driven saws, and other applicationsrequiring a working stroke with intensive loading, and a return stroke with non-

intensive loading. Several quick return mechanisms can be found including the

offset crank slider mechanism, the crank-shaper mechanisms, the double crank

mechanisms, crank rocker

mechanism and Whitworth

mechanism. In mechanical

design, the designer often

has need of a linkage thatprovides a certain type of

motion for the application in

designing.

Fig.: App.-1 - Shaping Machine


5/34


6/34



1. CRANK AND SLOTTED LEVER QUICK RETURN MECHANISM

This mechanism is mostly used in shaping machines (Fig.: App.-1), slotting

machines (Figs.: App.-2) and in rotary internal combustion engines (Fig.: App.-3). In this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, as

shown in (Fig.: App.-4). The link 3 corresponds to the connecting rod of a

reciprocating steam engine. The driving crank CB revolves with uniform angular

speed about the fixed centre C. A sliding block is attached to the crank pin at B

slides along the slotted bar AP and thus causes AP to oscillate about the pivoted

point A. A short link PR transmits the motion from AP to the ram which carries the

tool and reciprocates along the line of stroke R1R2. The line of stroke of the ram

(i.e. R1R2) is perpendicular to AC produced. In the extreme positions, AP1 and

AP2 are tangential to the circle and the cutting tool is at the end of the stroke.

The forward or cutting stroke occurs when the crank rotates from the position

CB1 to CB2 (or through an angle β) in the clockwise direction. The return stroke

occurs when the crank rotates from the position CB2 to CB1 (or through angle α)

in the clockwise direction.

Fig.: App.-4


7/34


8/34


9/34



Fig. P-2: Inversion of Slider-Crank Mechanism (First Part)

Fig. P-3: Slider-Crank Mechanism


10/34



In this part, in order to make our calculations simpler, we solve the slider-crank

mechanism. It means that we first obtain the angular velocity and acceleration of

ground and link 2 to relative to link 4. (Like the solution of the problem shown in

Fig. P-3.)

We choose a secondary coordinate for this part as shown in Fig.P-2. The

equations for this part of problem are as follows:

θ1’=2π+π/2-θ4 (P-1)θ2’=θ2-θ4 (P-2)

L1eiθ1’+L2eiθ2’=R3 (P-3)

By differentiating Equation (P-1), we get:

L1iω1/4eiθ1’+ L2iω2/4eiθ2’=V3 (P-4)

We know that in the secondary coordinate (X’-Y’), link 3 has velocity only in the X’

direction. So the imaginary part of the equation (P-2) equals zero. By solving this

equation we obtain the angular velocity of link 4:

=

((’))/((’)) ( )

Similarly by differentiating equation (P-4), knowing that the imaginary part of the

acceleration of link 4 is zero, we’ll find the angular acceleration of link 4:

L1 eiθ1’ (iα1/4-ω1/42) – L2 eiθ2’ (iα2/4-ω2/42) =A3 (P-6)

=−(^∗(’)+(−)^∗(’)))

∗(’)∗(+(−)/) (P-7)


11/34



2. Second part Includes Links 4, 5 and 6 (Fig. P-4). This part is again a mechanism

like the mechanism shown in Fig. P-3. Since we have obtained the angular velocity

and acceleration of link 4 (Equations P-3 and P-4, respectively) with respect to the

angular velocity and acceleration of the input link (link 2), we can now obtain the

angular velocity and acceleration of link 5 using a solution similar to which

described for the calculation of the angular velocity and acceleration of link 4. The

solution for this part is as follows:

Fig. P-4: Slider-Crank Mechanism (Second Part)


12/34



L4eiθ4+L5eiθ5=R6 (P-8)

By differentiating Equation (P-6) two times, knowing that imaginary parts of the

linear velocity and acceleration of link 6 is zero, we obtain the angular velocity

and acceleration of link 5:

L4iω4eiθ4+ L5iω5eiθ5=V6 (P-9)

L4eiθ4 (iα4-ω42) – L5ei5 (iα5-ω52) =A6 (P-10)

= ()

() (P 11)

= (^()+^())

() (P-12)


13/34



A-3. OBTAINING DESIRED VALUES AND PLOT DIAGRAMS FOR SECTION 1

A-3-1. “MATLAB” CODES

The Codes in “Matlab” for the first section are as follows:

theta2=input('Please Enter the Amount of Theta2(Deg):')*pi/180; w2=input('Please Enter the Amount of Omega2 (Rev/min) :')*pi/30; alpha2=input('Please Enter the Amount of alpha2 (Rev/min^2):')*2*pi/3600; l1=457; l2=152; l2p=0:0.1:l2; l3=762; l3p=0:0.1:l3; l4=508; l4p=0:0.1:l4; h=816.917; r=sqrt(l1^2+l2^2+2*l1*l2*sin(theta2));

theta3=acos(l2*cos(theta2)/r); theta4=asin((h-l3*sin(theta3))/l4); thetap1=2*pi+pi/2-theta3; thetap2=theta2-theta3;

w3=w2/(1+(l1*cos(thetap1))/(l2*cos(thetap2))); alpha3=(alpha2-(l1*w3^2*sin(thetap1)+l2*(w2-

w3)^2*sin(thetap2)))/(l2*cos(thetap2)*(1+(w2-w3)/w3)); w4=-l3*cos(theta3)*w3/(l4*cos(theta4));

alpha4=(w4/w3)*alpha3+(l3*w3^2*sin(theta3)+l4*w4^2*sin(theta4))/(l4*cos(theta4)); rp3=r*exp(1i*theta3); r45=l3p*exp(1i*theta3); r6=l3*exp(1i*theta3)+l4p*exp(1i*theta4); vp3=l2*1i*w2*exp(1i*theta2); ap3=l2*exp(1i*theta2)*(1i*alpha2-w2^2); v45=l3p*1i*w3*exp(1i*theta3); v45p=abs(v45); a45=l3p*exp(1i*theta3)*(1i*alpha3-w3^2); a45p=abs(a45); v6=-(l3*w3*sin(theta3)+l4p*w4*sin(theta4)); v6p=abs(v6); a6=-l3*(alpha3*sin(theta3)+w3^2*cos(theta3))-l4p*(alpha4*sin(theta4)+w4^2*cos(theta4)); a6p=abs(a6); subplot(3,2,1); plot(r45); subplot(3,2,2);

plot(r6); subplot(3,2,3); plot(l3p,v45p); subplot(3,2,4); plot(l4p,real(v6)); subplot(3,2,5); plot(l3p,a45p); subplot(3,2,6); plot(l4p,a6p); Published with MATLAB® R2014a

http://www.mathworks.com/products/matlabhttp://www.mathworks.com/products/matlab


14/34



A-3-2. TABLE

Section 1 1st Exam. 2nd Exam. 3rd Exam.

Initial Angular Position

of Link 2 (θ2 – Degrees) 167 200 70Initial Angular Velocity

of Link 2 (ω2 – Rev/min) 9.5 15 5Initial Angular

Acceleration of Link 2

(α2 – Rev/min2)

0 5 -7.5

A-3-3. RESULTS (DIAGRAMS)

Tab. P- 1: Input Parameters for Section 1

Fig. P-5: Sec.1 - Exm.1


15/34






16/34



A-4. OBTAINING DESIRED VALUES AND PLOT DIAGRAMS FOR SECTION 2

In the second section, the user enters the initial values for angular position andvelocity of link 2, then enters the time function of its angular acceleration. We can

obtain the time functions for angular velocity and position of link 2 by integrating

the function of angular acceleration 2 times, knowing the initial values of angular

velocity and position. Since we want to plot our diagrams only in a period of time

(not more or less), we have to find the time period of a complete cycle. In order

to do that, we set our obtained equation for angular position of link 2 equal to 2π.

There may be a couple of values for the time period, so we’ll choose the minimum

of positive values. Then we plot our desired diagrams in the period of onecomplete cycle.


17/34



A-4-1. “MATLAB” CODES

The Codes in “Matlab” for the second section are as follows:

syms t; theta2i=input('Please Enter the Amount of Theta2(Deg):')*pi/180; w2i=input('Please Enter the Amount of Omega2 (Rev/min):')*pi/30; alpha2=input('Please Enter the Time Function of alpha2:')*2*pi/3600; l1=457; l2=152; l2p=0:0.1:l2; l3=762; l3p=0:0.1:l3; l4=508; l4p=0:0.1:l4; h=816.917;

w2=int(alpha2,t)+w2i; theta2=int(w2,t)+theta2i; r=sqrt(l1^2+l2^2+2*l1*l2*sin(theta2)); theta3=acos(l2*cos(theta2)./r); theta4=asin((h-l3*sin(theta3))/l4); thetap1=2*pi+pi/2-theta3; thetap2=theta2-theta3;

w3=w2./(1+(l1*cos(thetap1))./(l2*cos(thetap2))); alpha3=(alpha2-(l1*w3.^2.*sin(thetap1)+l2*(w2-

w3).^2.*sin(thetap2)))./(l2*cos(thetap2).*(1+(w2-w3)./w3)); w4=-l3*cos(theta3).*w3./(l4*cos(theta4));

alpha4=(w4./w3).*alpha3+(l3*w3.^2.*sin(theta3)+l4*w4.^2.*sin(theta4))./(l4*cos(theta4)); rp3=r.*exp(1i*theta3); r45=l3*exp(1i*theta3); r6=l3*exp(1i*theta3)+l4*exp(1i*theta4); vp3=l2*1i*w2.*exp(1i*theta2); v3=abs(vp3); ap3=l2*exp(1i*theta2).*(1i*alpha2-w2.^2); a3=abs(ap3); vp45=l3*1i.*w3.*exp(1i*theta3); v45=abs(vp45); ap45=l3*exp(1i*theta3).*(1i*alpha3-w3.^2); a45=abs(ap45); vp6=-(l3*w3.*sin(theta3)+l4*w4.*sin(theta4)); v6=abs(vp6); ap6=-l3*(alpha3.*sin(theta3)+w3.^2.*cos(theta3))-

l4*(alpha4.*sin(theta4)+w4.^2.*cos(theta4)); a6=abs(ap6); T=solve(int(w2,t)-2*pi==0); T=double(T); TP=min(T(T>0)); t=0:.01:TP; subplot(3,4,1); plot(eval(theta2),eval(v3)); title('V_3-\theta_2'); xlabel('\theta_2');


18/34



ylabel('V_3'); subplot(3,4,2); plot(eval(theta2),eval(a3)); title('A_3-\theta_2'); xlabel('\theta_2'); ylabel('A_3');

subplot(3,4,3); plot(eval(theta2),eval(v45)); title('V_4_5-\theta_2'); xlabel('\theta_2'); ylabel('V_4_5'); subplot(3,4,4); plot(eval(theta2),eval(a45)); title('A_4_5-\theta_2'); xlabel('\theta_2'); ylabel('A_4_5'); subplot(3,4,5); plot(eval(theta2),eval(v6)); title('V_6-\theta_2'); xlabel('\theta_2'); ylabel('V_6'); subplot(3,4,6); plot(eval(theta2),eval(a6)); title('A_6-\theta_2'); xlabel('\theta_2'); ylabel('A_6'); subplot(3,4,7); plot(eval(theta2),eval(w3)); title('\omega_4-\theta_2'); xlabel('\theta_2'); ylabel('\omega_4'); subplot(3,4,8); plot(eval(theta2),eval(alpha3));

title('\alpha_4-\theta_2'); xlabel('\theta_2'); ylabel('\alpha_4'); subplot(3,4,9); plot(eval(theta2),eval(w4)); title('\omega_5-\theta_2'); xlabel('\theta_2'); ylabel('\omega_5'); subplot(3,4,10); plot(eval(theta2),eval(alpha4)); title('\alpha_5-\theta_2'); xlabel('\theta_2'); ylabel('\alpha_5'); figure; polar(eval(theta2),eval(abs(rp3))); title('R_3-\theta_2'); figure; polar(eval(theta2),eval(abs(r45))); title('R_4_5-\theta_2'); figure; polar(eval(theta2),eval(abs(r6))); title('R_6-\theta_2');

Published with MATLAB® R2014a

http://www.mathworks.com/products/matlabhttp://www.mathworks.com/products/matlab


19/34



A-4-2. TABLE FOR INPUT PARAMETERS

Section 2 1st Exam. 2nd Exam. 3rd Exam.

Initial Angular Position

of Link 2 (θ2 – Degrees) 167 85 230Initial Angular Velocity

of Link 2 (ω2 – Rev/min) 9.5 79 0Time Function for

Angular Acceleration of

Link 2 (α2 – Rev/min2)

0 1.5t 5t 3+7t

Tab. P- 2: Input Parameters for Section 2


20/34



A-4-3. RESULTS

A-4-3-1. (“MATLAB” DIAGRAMS)



21/34





22/34



Fig. P-10: Sec.2 – Exm3.


23/34



A-4-3-2. (“ADAMS” DIAGRAMS)

Figs. P-11: Sec.2 – Exm1.


24/34



Figs. P-12: Sec.2 – Exm2.


25/34



A-5. CHART ANALYSIS (ANALYZING Q UICK-RETURN)

A-5-1. VELOCITY DIAGRAMS

We see in the charts that the each of the velocity diagrams for Link 6 in “X”

Direction (Figs. P-11 and Figs. P-12) has local 2 peaks per a period, one of them is

much higher than the other one. It shows that the “return” part (Positive Peak) of

the motion would be done with much more velocity, hence it takes less time to be

done. This is due to the fact that in “go” part (Negative Peak), the angular

displacement of Link 2 is much bigger than it in the “return” (Positive Peak) part

of the motion.

A-5-2. ACCELERATION DIAGRAMS

As we see in the acceleration diagrams of Link 6 in “X” direction (Figs. P-11 andFigs. P-12), again there are two peaks in each period with one of them positive

and the other one, negative. These peaks show the “start” and the “end” of the

quick-return motion. The “positive” one is for the beginning and the “negative”

on is for the end of the quick-return motion. We also see that the magnitudes of

these negative and positive peaks are the same; it means that the time for

“acceleration” equals the time for “deceleration”.


26/34



B. Kinetics Analysis

B-1. ASSUMPTIONS AND SOLVING THE PROBLEM

Let us consider the mechanism in (Fig. P-1) where Link 2 is a driver and rotates

with a known constant angular velocity ω2. The motion is opposed by a known

force P acting on slider 6 as shown. We wish to find the forces at all the pairs or

joints and the torque which the shaft at O2 must exert on crank 2 to drive the

mechanism. It’ll be assumed that the forces of the gravity on the links are

negligible. The acceleration polygon appears in (Fig. P-13-A and Fig. P-13-B). In

(Figs. P-14) each link is shown isolated and the magnitude, direction, and location

of inertia are also shown.

Fig. P-13-A: Acceleration Polygon


27/34



In order to have much more accuracy, we drew our acceleration diagram using

“SolidWorks 2015TM”.

Fig. P-13-B: Acceleration Polygon (by SolidWorks 2015)


28/34



(A)

(B)

(C)(D)

(E)

(F)

(G)

Figs. P-14


29/34



The force analysis is started with link 6, which is shown in (Fig. P-14-A). The

unknowns are the magnitude of F16 and the magnitude and direction of F56. The

horizontal component of F56 is F56H, and its magnitude can be found from thesummation of horizontal forces on link 6.

In (Fig. P-14-B), force F65H is equal and opposite to F56H. The magnitude of F65V can

be found by the summation of Moments about point ”45”. Then from a force

polygon for link 5, the magnitude and direction of F45 are found. Next, in (Fig. P-

14-C) F54 is known and is equal and opposite to F45. There are four unknowns in

(Fig. P-14-C): the magnitude and direction of F14 and the magnitude and location

of F34. Since there are only three equations of equilibrium, we cannot determinethese forces by considering this link alone. If consider link 3 shown in (Fig. P-14-D)

we note that here there are also four unknowns: the magnitude and direction of

F23 and magnitude and location of F43. However, for the combination of links 3

and 4 we have only six unknowns, and they can be analyzed in combination. From

the free body of link 3 we see that F23 causes no torque about the center of mass

point”3”, and thus F43 must be of such a magnitude as to balance the forces, and

its line of action must be displayed from point “3” enough to balance the inertia

torque. F43 can then be resolved into a force passing through point “3” and atorque about point 3 sufficient to balance the inertia torque. This is shown in (Fig.

P-14-E). The equal and opposite force and torque on link 4 as shown in (Fig. P-14-

F) makes link 4 a free body with three unknowns. The magnitude of F34 can be

found by setting the sum of the moments about O4 equal to zero. F14 can then be

found from a force polygon for link 4.

We replaced F43 in (Fig. P-14-D) with force F13 and T43, which are shown in (Fig. P-

14-E). Thus in (Fig. P-14-D):

F43a = T43 = I3α3 (P-13)

And thus the moment arm “a” can be determined as follows:

a =I3α3

F43 (P-14)


30/34



This locates the actual line of action of F43. In the FBD of link 3, the F43 will be

equal and opposite to F34, which as determined earlier. F23 can now be

determined from a force polygon for link 3.

The FBD of link 2 is shown in (Fig. P-14-G). F32 is equal and opposite to F23, and theF12 can be determined from a force polygon for link 2. Finally, by summing

moments about O2, the torque T2, which the shaft at O2 exerts on crank 2, can be

determined.

In tables below the results of calculations are shown (We obtained the values for

angular acceleration and velocity (Tab. P-3) by entering input parameters of (Tab.

P-4) in “Matlab”.)


31/34



B-2. TABLES FOR INPUT VALUES

Link/Joint I(kg-

mm2)

α(Rad/s2) AG(mm/s2) M(kg) F(kg.mm/s2)

2/12 3.498 0 75.21 0.147 1658

3/23 26000 -0.1796 150.42 15.6 1654

4/14 11070 -0.1796 68.91 18.56 503.3

5/45 740.83 -0.047 137.25 3.67 1078

6/56 0 0 137.72 17.83 1081

The Torque that must be exerted on Link 2 is obtained:

T2 = 514.3 N-mm

Kinetics 1st Exam. 2nd Exam.

Initial Angular Position of Link 2

(θ2 – Degrees) 167 200Initial Angular Velocity of Link 2

(ω2 – Rev/min) 9.5 15Initial Angular Acceleration ofLink 2 (α2 – Rev/min2) 0 5

Input Force P (N) 1500 1500

Tab. P- 3: Input Parameters for Kinetics Part

Tab. P- 4: Input Parameters for Kinetics Part


32/34



Figs. P-15: Adams Diagrams (Examination 1)


33/34



Figs. P-16: Adams Diagrams (Examination 2)


34/34


B-3. CHART ANALYSIS (ANALYZING THE TORQUE ON LINK 2)

As we see in the diagrams (Fig. P-15) there is one peak and one valley in each

period of the motion for the force values in joints, that show the beginning and

the end of the quick-return motion. Also it can be concluded from the diagrams

that how the forces vary in the motion.

We know that the magnitude of the torque acting on link 2 is proportional to the

force acting on it. When links 2 and 4 are placed along each other, the resultant

force that link 2 exerts on link 4, has got only component along X direction (that

should act against the force applied on slider (link 6)). In other situations (except

where links 2 and 4 are along each other) the X component of the force that link 2

exerts on link 4 should act against the forced applied on the slider, hence theresultant of the force that link 2 exerts on link 4 is larger. The valleys in the

diagram show these conditions. By the same reason (stating the force that link 2

exerts on link 4 by X and Y components) the peaks of the diagram occur when

links 2 and 4 are normal to each other.

Documents

Dynamics Project - Moein Razavi