Lab 2 Experiment No ALL

Experiment No. 1

Problem Definition: Analysis of Inertia Force in Slider Crank Mechanism Using Computer Software.

% Programme For Slider Crank Mechanismfunction slidercrank1()clc;

fprintf('\nINPUT\n');

N=input('Engine Speed, N (rpm) : ');l=input('Length between centres, l (mm): ');l2=input('Dist between centres of mass, l2 (mm): ');m=input('Mass of connecting rod, m (kg): ');mr=input('Mass of reciprocating parts, mr (kg): ');r=input('Crank length, r (mm): ');theta=input('Crank angle, Theta (Deg) : ');k=input('Radius of Gyration, k (mm) : ');

fprintf('\nOUTPUT');

% Angular Velocityw=2*pi*N/60;fprintf('\nAngular Velocity, w = %f rad/s\n',w);

% Mass at Crank Pinl1=l-l2;mi=m*l2/(l1+l2);Mr=mr+mi;fprintf('\nMass at Crank Pin, Mr = %d kg',Mr);

theta_rad=theta*pi/180;

% l/r Ration=l/r;fprintf('\nl/r Ratio, n = %f',n);

%Acceleration of Reciprocating Partsf=r/1000*w^2*(cos(theta_rad)+cos(2*theta_rad)/n);fprintf('\nAcceleration of Reciprocating Parts, f = %f m/s^2',f);

% Inertia ForceFi=Mr*f;fprintf('\nInertia Force, Fi = %f N',Fi);

% Inertia TorqueTi=Fi*r/1000*(sin(theta_rad)+sin(2*theta_rad)/(2*(n^2-(sin(theta_rad))^2)^(1/2)));fprintf('\nInertia Torque Due to Reciprocating Parts, Ti = %f N.m\n',Ti);

L=l1+(k^2)/l1;alpha=(-1)*w^2*sin(theta_rad)*((n^2-1)/((n^2-(sin(theta_rad))^2)^(3/2)));

fprintf('\nl1 = %d mm',l1);fprintf('\nL = %f mm\n',L);fprintf('\nalpha = %f rad/s^2\n',alpha);

% Correction CoupledT=m*alpha*l1/1000*(l-L)/1000;fprintf('\nCorrection Couple, dT = %f N.m/n',dT);

% Correction Torque on CrankshaftTc=dT*cos(theta_rad)/(n^2-(sin(theta_rad))^2)^(1/2);fprintf('\nCorrection Torque on Crankshaft, Tc = %f N.m',Tc);

% Torque due to Weight of Mass at Crank Ping=9.81;m2=m-mi;Ta=m2*g*r*cos(theta_rad)/1000;fprintf('\nTorque Due to Weight of Mass at Crank Pin, Ta = %f N.m',Ta);

% Total Inertia Torque on CrankshaftT=Ti-Tc+Ta;fprintf('\nTotal Inertia Torque on Crankshaft, T = %f N.m\n\n',T);

for theta=0:20*pi/180:2*pi; f=r/1000*w^2*(cos(theta)+cos(2*theta)/n); Fi=Mr*f; Ti=Fi*r/1000*(sin(theta)+sin(2*theta)/(2*(n^2-(sin(theta))^2)^(1/2))); L=l1+(k^2)/l1; alpha=(-1)*w^2*sin(theta)*((n^2-1)/((n^2-(sin(theta))^2)^(3/2))); dT=m*alpha*l1/1000*(l-L)/1000; Tc=dT*cos(theta)/(n^2-(sin(theta))^2)^(1/2); g=9.81; m2=m-mi; Ta=m2*g*r*cos(theta)/1000; T=Ti-Tc+Ta; thetad= theta*180/pi; fprintf('\nTheta = %f Deg',thetad); fprintf('\tT = %f N.m',T); end for theta=0:0.001:2*pi; f=r/1000*w^2*(cos(theta)+cos(2*theta)/n); Fi=Mr*f; Ti=Fi*r/1000*(sin(theta)+sin(2*theta)/(2*(n^2-(sin(theta))^2)^(1/2))); L=l1+(k^2)/l1; alpha=(-1)*w^2*sin(theta)*((n^2-1)/((n^2-(sin(theta))^2)^(3/2))); dT=m*alpha*l1/1000*(l-L)/1000; Tc=dT*cos(theta)/(n^2-(sin(theta))^2)^(1/2); g=9.81; m2=m-mi; Ta=m2*g*r*cos(theta)/1000; T=Ti-Tc+Ta; thetad= theta*180/pi; hold on plot (thetad,T,'r'); hold offend title('T-\theta Diagram');xlabel ('\theta [Deg]');ylabel ('T(N.m)');gridendINPUTEngine Speed, N (rpm) : 600Length between centres, l (mm): 450Dist between centres of mass, l2 (mm): 180Mass of connecting rod, m (kg): 90Mass of reciprocating parts, mr (kg): 120Crank length, r (mm): 90Crank angle, Theta (Deg) : 80Radius of Gyration, k (mm) : 150

OUTPUTAngular Velocity, w = 62.831853 rad/s

Mass at Crank Pin, Mr = 156 kgl/r Ratio, n = 5.000000Acceleration of Reciprocating Parts, f = -5.077442 m/s^2Inertia Force, Fi = -792.081014 NInertia Torque Due to Reciprocating Parts, Ti = -72.691161 N.m

l1 = 270 mmL = 353.333333 mm

alpha = -792.113864 rad/s^2

Correction Couple, dT = -1860.675467 N.m/nCorrection Torque on Crankshaft, Tc = -65.911711 N.mTorque Due to Weight of Mass at Crank Pin, Ta = 8.278955 N.mTotal Inertia Torque on Crankshaft, T = 1.499505 N.m

Theta = 0.000000 DegT = 47.676600 N.mTheta = 20.000000 DegT = 2376.265323 N.mTheta = 40.000000 DegT = 3182.214761 N.mTheta = 60.000000 DegT = 2091.243250 N.mTheta = 80.000000 DegT = 1.499505 N.mTheta = 100.000000 DegT = -1787.634693 N.mTheta = 120.000000 DegT = -2516.629612 N.mTheta = 140.000000 DegT = -2200.522564 N.mTheta = 160.000000 DegT = -1249.418507 N.mTheta = 180.000000 DegT = -47.676600 N.mTheta = 200.000000 DegT = 1159.815809 N.mTheta = 220.000000 DegT = 2127.477775 N.mTheta = 240.000000 DegT = 2468.953012 N.mTheta = 260.000000 DegT = 1771.076784 N.mTheta = 280.000000 DegT = 15.058405 N.mTheta = 300.000000 DegT = -2043.566650 N.mTheta = 320.000000 DegT = -3109.169972 N.mTheta = 340.000000 DegT = -2286.662625 N.mTheta = 360.000000 DegT = 47.676600 N.m>>

Result :-

Experiment No. 2

Problem Definition: Coupler Curve Synthesis for A Mechanism Using Computer Software.

INPUT% ProgrammeFor Coupler Curve SynthesisfunctionCoupler_Curve_Synthesis()clc;

%Inputst1=110; t2=77; t3=50;r1=80; r2=90; r3=96;a1=65; a2=56; a3=48;

%Angle Conversionst1=degtorad(t1); t2=degtorad(t2); t3=degtorad(t3);a1=degtorad(a1); a2=degtorad(a2); a3=degtorad(a3);

%Assuming Suitable Value of Angles Gamma=20 Deg, Si=10 Deg, Delta1=150 Deggg=degtorad(20); ss=degtorad(10); d1=degtorad(150);

for j=0:1:1if j==0 tt1=t1; tt2=t2; tt3=t3; dd=gg; sign=1;end

if j==1 a=x; g=y; e=z; b1=(acos((r1*cos(a1)-g*cos(gg)-a*cos(t1))/e)); b2=(acos((r2*cos(a2)-g*cos(gg)-a*cos(t2))/e)); b3=(acos((r3*cos(a3)-g*cos(gg)-a*cos(t3))/e)); d2=d1+(b2-b1); d3=d1+(b3-b1); tt1=d1; tt2=d2; tt3=d3; dd=ss; sign=-1;end

%Matrix For l l0=[2*r1*cos(tt1-a1) 2*r1*cos(a1-dd) 1;2*r2*cos(tt2-a2)... 2*r2*cos(a2-dd) 1;2*r3*cos(tt3-a3) 2*r3*cos(a3-dd) 1]; l1=[r1^2 2*r1*cos(a1-dd) 1;r2^2 2*r2*cos(a2-dd) 1;r3^2 2*r3*cos(a3-dd) 1]; l2=[2*r1*cos(tt1-a1) r1^2 1;2*r2*cos(tt2-a2) r2^2 1;2*r3*cos(tt3-a3) r3^2 1]; l3=[2*r1*cos(tt1-a1) 2*r1*cos(a1-dd) r1^2;2*r2*cos(tt2-a2)... 2*r2*cos(a2-dd) r2^2;2*r3*cos(tt3-a3) 2*r3*cos(a3-dd) r3^2];

%matrix for m m0=[2*r1*cos(tt1-a1) 2*r1*cos(a1-dd) 1;2*r2*cos(tt2-a2)... 2*r2*cos(a2-dd) 1;2*r3*cos(tt3-a3) 2*r3*cos(a3-dd) 1]; m1=[2*cos(tt1-dd) 2*r1*cos(a1-dd) 1;2*cos(tt2-dd)... 2*r2*cos(a2-dd) 1;2*cos(tt3-dd) 2*r3*cos(a3-dd) 1]; m2=[2*r1*cos(tt1-a1) 2*cos(tt1-dd) 1;2*r2*cos(tt2-a2)... 2*cos(tt2-dd) 1;2*r3*cos(tt3-a3) 2*cos(tt3-dd) 1]; m3=[2*r1*cos(tt1-a1) 2*r1*cos(a1-dd) 2*cos(tt1-dd);2*r2*cos(tt2-a2)... 2*r2*cos(a2-dd) 2*cos(tt2-dd);2*r3*cos(tt3-a3) 2*r3*cos(a3-dd)... 2*cos(tt3-dd)];

%Constants Calculationla=det(l1)/det(l0);lg=det(l2)/det(l0);lk=det(l3)/det(l0); ma=det(m1)/det(m0);mg=det(m2)/det(m0);mk=det(m3)/det(m0);

A=ma*mg; B=la*mg+lg*ma-1; C=la*lg;

L=(-B-sign*sqrt(B^2-4*A*C))/(2*A); x=la+L*ma; y=lg+L*mg; k=lk+L*mk; z=sqrt(k+x^2+y^2);end

f=x; h=y; c=z;dx=h*cos(ss)-g*cos(gg);dy=h*sin(ss)-g*sin(gg);d=sqrt(dx^2+dy^2);b=distance(a*cos(t1),a*sin(t1),(a*cos(t1)+e*cos(b1)+f*cos(pi+d1)),... (a*sin(t1)+e*sin(b1)+f*sin(pi+d1)));

%Link Lengthsfprintf('\nOUTPUT');fprintf('\n a b c d e f g h');fprintf('\n%8.2f %8.2f %8.2f %8.2f %8.2f %8.2f %8.2f %8.2f\n',a,b,c,d,e,f,g,h);

%Set Graph Limitsxlim([-60 100])ylim([-10 110])

%Plotting Fixed Linkline([0,dx],[0,dy],'Color','g','DisplayName','Fixed Link','LineWidth',2);

for j=0:1:2if j==0tt=t1; bb=b1; dd=d1;end

if j==1tt=t2; bb=b2; dd=d2;end

if j==2tt=t3; bb=b3; dd=d3;end

%Plotting Moving Linksline([0,a*cos(tt),(a*cos(tt)+e*cos(bb)+f*cos(pi+dd)),dx],... [0,a*sin(tt),(a*sin(tt)+e*sin(bb)+f*sin(pi+dd)),dy],...'DisplayName','Moving Links','LineWidth',2); %Line Properties

%Plotting Coupler Linksline([a*cos(tt),(a*cos(tt)+e*cos(bb)),(a*cos(tt)+e*cos(bb)+f*cos(pi+dd))]... ,[a*sin(tt),(a*sin(tt)+e*sin(bb)),(a*sin(tt)+e*sin(bb)+f*sin(pi+dd))]... ,'Color','r','DisplayName','Coupler Links','LineWidth',2);end

%Plotting Coupler Curve k1=d/a; k2=d/b; k3=(c^2-a^2-b^2-d^2)/(2*a*b); A=(k2+1)*cos(t1)+k3-k1; B=-2*sin(t1); C=k1+k3+(k2-1)*cos(t1);beta=2*atan((-B-sqrt(B^2-4*A*C))/(2*A)); alpha=b1-beta;

for t=0:.001:2*pi; A=(k2+1)*cos(t)+k3-k1; B=-2*sin(t); C=k1+k3+(k2-1)*cos(t);beta=2*atan((-B-sqrt(B^2-4*A*C))/(2*A)); x=real(a*cos(t)+e*cos(alpha+beta)); y=real(a*sin(t)+e*sin(alpha+beta));holdonplot(x,y,'k','DisplayName','Coupler Curve','LineWidth',2);legendshowholdoffend

title('Coupler Curve Plot');xlabel ('X-Axis');ylabel ('Y-Axis');gridend

title('Coupler Curve Plot');xlabel ('X-Axis');ylabel ('Y-Axis');gridend

OUTPUT

a b c d e f g h 17.20 44.63 28.50 27.51 39.40 65.04 54.95 79.91

Experiment No. 3

Problem Definition: Determination of Natural Frequencies & Modal Analysis of A Cantilever Beam Using FFT Analyzer.

5mm

600mm 50mm

Fig 3.1 Cantilever beam with loadingTable 3.1 shows the detail of cantilever beam which includes dimension of beam and its material properties.Table 3.1Dimension and material properties of cantilever beam.Length600 mm

Width50 mm

Height5 mm

Youngs modulus (E)210 GPa

Poissons ratio0.3

The beam is clamped to a rigid wall at the left end and Impact load is applied on Specimen by using Impact hammer.

Accelerometer 'C 'Clamp for Holding SpecimenChannel BoardImpact HammerFFT Software SpecimenFig 3.2 Experimental Setup of FFT AnalyzerAs shown in Fig no.3.2 the experimental setup of FFT Analyzer consist of following equipment as.1. Speciman: The Specimen having dimension 600 X 50 X 5 mm, of mild steel is used for testing.2. 'C' Clamp Board: It is used for holding the beam at one end. 3. FFT Channel: This is hardware device which is used for giving input and output to the FFT software. It contains 8 no of channels out of that channel no 2 and channel no 5 are used to take input and output of data. Channel no 2 is connected to accelerometer sensor to take output signals from beam and channel no 5 is connected to impact hammer to excite the beam.4. Impact Hammer: It is used to excite the beam.5. Acclerometer Sensor: It is used for measuring vibration parameters.

Fig 3.3 Channel SetupThe Fig no. 3.3 is shows channel setup window hear we gives channel setup for channel no2 and no5.

Fig 3.4 Channel Setup for channel no 2.

In channel no 2 we give the physical quantity as Acceleration and unit as 'g'. Here Sensitivity of accelerometer sensor is 9.749 mv/g.

Fig 3.5 Channel Setup for channel no 5.

In channel no 5 we give the physical quantity as Force and unit as 'N'. Here sensitivity of Impact hammer which is 2.28 mv/N.

Fig 3.6 Modal Test Window.

The fig no 3.6 shows the model test window in that we selected calculation type as Triggered(FRF) and tick on Roving hammer/acc. In this experiment we excite the specimen at 3 different position (at X= -300,X=0 and X=300mm)

Fig 3.7 Frequency Domain Graph of Model Test

Conclusion: In above fig. Maximum peaks at frequency 332.31 Hz and 546.875 Hz. So,Natural frequency in the first two modes are 332.31 Hz and 546.875 Hz.

Experiment No 44.1 Problem Definition: Stress And Deflection Analysis of Short and Long Beams With Different End Condition and Crass-Section Subjected to Different Loading Condition Using FEA Software.

Fig 4.1 Cantilever beam with loadingTable 4.1 shows the detail of cantilever beam which includes dimension of beam and its material properties.Table 4.1 Dimension and material properties of cantilever beam.Length90 mm

Width10 mm

Height5 mm

Youngs modulus (E)210 GPa

Poissons ratio0.3

The beam is clamped to a rigid wall at the left end and loaded at x =80mm by a point load of P =100 N.

Fig 4.2 Flowchart of the structural analyses by ANSYS.

Modeling: Figure 1.3-1.4 show the ANSYS Main Menu window where we can find layered command options imitating folders and files in the Microsoft Explorer folder window. In order to prepare for creating the beam, the following operations should be made:(1) Preprocessor to open its sub-menus in ANSYS Main Menu window.(2) Modeling to open its sub-menus and select Create menu.(3) Areas to open its sub-menus and select Rectangle menu.(4) Click on By 2 corners menu.

Fig 4.3 ANSYS Main Menu Fig 4.4 Rectangle by 2 Corners window. window.

Fig 4.5 2-D beam created and displayed on the ANSYS Graphics window.

Material Properties: ANSYSMainMenuPreprocessorMaterial PropsMaterial Models. Then the Define Material Model Behavior window opens as shown inFig 4.6-4.7

Fig 4.6 Define Material Model Behavior window.

Fig 4.7 Input of elastic constants through the Linear Isotropic Properties for Material Number 1 window.

Finite-Element Discretization of the Beam Area: Here we will divide the beam area into finite elements. The procedures for finiteelement discretization are firstly to select the element type, secondly to input the element thickness and finally to divide the beam area into elements. ANSYS Main MenuPreprocessorElement TypeAdd/ Edit/Delete

Fig 4.8 Library of Element Types window.

Fig 4.9 PLANE 182 element type options window.

Fig 4.10 Real Constants Set Number 1. for PLANE 183 window.Sizing of the elements: ANSYSMainMenuPreprocessorMeshingSize CntrlsManual SizeGlobalSize

Fig 4.11 Global Element Sizes window.

Meshing: ANSYSMainMenuPreprocessorMeshingMeshAreasFree

Fig 4.12 FE model of beam

Input of Boundary Conditions: Here we will impose constraint and loading conditions on nodes of the beam model. Display the nodes first to define the constraint and loading conditions.ANSYSMainMenuSolutionDefineLoadsApplyStructuralDisplacementOn Nodes

Fig 4.13 Apply U. ROT on Nodes window.

ANSYSMainMenuSolutionDefine LoadsStructuralForce/ MomentOn Nodes

Fig 4.14 Apply F/M on Nodes window.

Fig 4.15 Boundary condition with loading

Solution Procedures: ANSYSMainMenuSolutionSolveCurrent LS

Fig 4.15 Solve Current Load Step window.

Fig 4.16 Note window.

Graphical representation of the results: ANSYSMainMenuGeneral PostprocPlot ResultsContour PlotNodal Solution

Fig 4.17 Contour Nodal Solution Data window for deflection

Fig 4.18 Contour map representation of the distribution of displacement in the y- or vertical direction.

Fig 4.19 Contour Nodal Solution Data window for stress

Fig 4.20 Contour map representation of the distribution of Von Mises stress

Conclusion: The maximum deflection in Y direction is 0.927 mm and maximum Von Mises stress is 190.45 MPa.

4.2 Problem Definition: Stress and Deflection Analysis of Short and Long Beams With Different End Condition and Crass-Section Subjected to Different Loading Condition Using FEA Software .

Fig 4.21 Stepped beam with loadingTable 4.21 shows the detail of stepped beam which includes dimension of beam and its material properties.Table 4.21 Dimension and material properties of cantilever beam.Length100 mm

Width10 mm

Height20 mm

Youngs modulus (E)210GPa

Poissons ratio0.3

The beam is clamped to a rigid wall at the left end and loadedat x =100mm by a point load of P =100 N.

Modeling: Figure 4.20 -4.21 show the ANSYS Main Menu window where we can find layered command options imitating folders and files in the Microsoft Explorer folder window.In order to prepare for creating the beam, the following operations should be made:(1) Preprocessor to open its sub-menus in ANSYS Main Menu window.(2) Modeling to open its sub-menus and select Create menu.(3) Areas to open its sub-menus and select Rectangle menu.(4) Click on By 2 corners menu.(5)ModelingOperateBooleansSubtractArea.(6) Areas to open its sub-menus and select Circle menu.(7) Click onSolid circle menu.(8)ModelingOperateBooleansSubtractArea.

Fig 4.22ANSYS main menu Fig 4.23Rectangle by 2 corners window. window.

Fig 4.24Solid circle Area window.

Fig 4.25 2-D beam created and displayed on the ANSYS Graphics window.

Material Properties: ANSYSMainMenuPreprocessorMaterialPropsMaterialModels. Then the Define Material Model Behavior window opens as shown inFigure 4.22.-4.22.3



Finite-Element Discretization of the Beam Area:Here we will divide the beam area into finite elements. The procedures for finiteelement discretization are firstly to select the element type, secondly to input the element thickness and finally to divide the beam area into elements.ANSYS Main MenuPreprocessorElement TypeAdd/ Edit/Delete


Fig 4.29 PLANE 182 element type options window.

Fig 4.30 Real Constants Set Number 1. for PLANE 183 window.Sizing of the elements: ANSYSMainMenuPreprocessorMeshingSizeCntrlsManualSizeGlobalSize

Fig 4.31Global Element Sizes window.


Fig 4.32 FE model of beam

Input of Boundary Conditions: Here we will impose constraint and loading conditions on nodes of the beam model. Display the nodes first to define the constraint and loading conditions.ANSYSMainMenuSolutionDefineLoadsApplyStructuralDisplacementOn Nodes


ANSYSMainMenuSolutionDefineLoadsStructuralForce/MomentOn Nodes






Graphical representation of the results:ANSYSMainMenuGeneralPostprocPlotResultsContourPlotNodal Solution

Fig 4.38 Contour Nodal Solution Data window for deflection

Fig 4.39 Contour map representation of the distribution of displacement in the y- or vertical direction.



Conclusion: The maximum deflection in Y direction is 0.044 mm and maximum Von Mises stress is 30.36MPa.

Experiment No 5 5.1 Problem Definition: Stress and Deflection Analysis of Rectangular Plates Using FEA Software.

Fig 5.1 rectangular plate having one holeTable 5.1 shows the detail of rectangular plate which includes dimension of plate and its material properties.Table 5.1 Dimension and material properties of cantilever beam.Width1000 mm

Height500 mm

Thickness10 mm

Radius of circle25 mm


Poissons ratio0.3

Fig 5.2 Flowchart of the structural analyses by ANSYS.

Modeling:Figure 5.3-5.4 show the ANSYS Main Menu window where we can find layeredcommand options imitating folders and files in the Microsoft Explorer folderwindow.In order to prepare for creating the plate, the following operations shouldbe made:(1) Preprocessor to open its sub-menus in ANSYSMainMenu window.(2) Modeling to open its sub-menus and select Create menu.(3) Areas to open its sub-menus and select Rectangle menu.(4) Click on By 2 corners menu.(5) Areas to open its sub-menus and select Circle menu.(6) Click onSolid circle menu.(7)ModelingOperateBooleansSubtractArea.

Fig 5.3ANSYS Main Menu Fig 5.4Rectangle By 2 Corners window. Window.

Fig 5.5Solid Circular Area window

Fig 5.6 2-D plate created and displayed on the ANSYS Graphics window.

Material Properties: ANSYSMainMenuPreprocessorMaterialPropsMaterialModels. Then the Define Material Model Behavior window opens as shown inFigure 5.6-5.7



Finite-Element Discretization of the Rectangular plate Area:Here we will divide the plate area into finite elements. The procedures for finiteelement discretization are firstly to select the element type, secondly to input the element thickness and finally to divide the plate area into elements.ANSYS Main MenuPreprocessorElement TypeAdd/ Edit/Delete


Fig 5.10PLANE 183 element type options window.

Fig 5.11 Real Constants Set Number 1. for PLANE 183 window.


Fig 5.12: FE model of plate

Input of Boundary Conditions: Here we will impose constraint and loading conditions on nodes of the plate model. Display the nodes first to define the constraint and loading conditions.ANSYSMainMenuSolutionDefineLoadsApplyStructuralDisplacementOn Nodes


ANSYSMainMenuSolutionDefineLoads ApplyStructuralForce/MomentOn Nodes






Graphical representation of the results: ANSYSMainMenuGeneralPostprocPlotResultsContourPlotNodal Solution



Result: The maximum Von Mises stress in the plate having one hole is 7.35489MPa.

5.2 Problem Definition: Find The Stress Analysis of Rectangular Plate Having Three Holes As Shown In Fig.

Fig 5.20 rectangular plate having three holesTable 5.2 shows the detail of rectangular plate which includes dimension of plate and its material properties.Table 5.2 Dimension and material properties of plateWidth1000 mm

Height500 mm

Thickness10 mm

Radius of middle circle25 mm

Radius of outside circle12.5 mm


Poissons ratio0.3

Fig 5.21 2-D plate created and displayed on the ANSYS Graphics window.

Finite-Element Discretization of the Rectangular plate Area: Here we will divide the plate area into finite elements. The procedures for finiteelement discretization are firstly to select the element type, secondly to input the element thickness and finally to divide the plate area into elements.ANSYS Main MenuPreprocessorElementTypeAdd/ Edit/Delete


Fig 5.22 FE model of plate

Input of Boundary Conditions: Here we will impose constraint and loading conditions on nodes of the plate model. Display the nodes first to define the constraint and loading conditions.

ANSYSMainMenuSolutionDefineLoads ApplyStructuralForce/MomentOn Nodes




Graphical representation of the results: ANSYS MainMenuGeneralPostprocPlotResultsContourPlotNodal Solution


Fig 5.26Contour map representation of the distribution of Von Mises stress

Result: The maximum Von Mises stress in the plate having three hole is 7.30891MPa.

Conclusion: The maximum Von Mises stress in the plate having three holes is less as compared to the plate having one hole.

Experiment No 6Problem Definition: Stress Analysis of Rotating Disk Using FEA Software.

Fig 6.1 3-D model of disk

Experiment No 7

Problem Definition: Direct/Model Frequency Response Analysis of a Beam/Plate Under a Single-Point Cycle Load/Base Excitation With and Without Damping Using FEA Software.

Fig 8.1 linear buckling on 1st node

Fig 8.2 linear buckling on 2nd node

Fig 8.3 linear buckling on 3rd node

Structural Steel > Isotropic Secant Coefficient of Thermal Expansion

Experiment No 8Problem Definition: Direct /Model Frequency Response Analysis of A Plate Under A Single Point Cycle Load/ Base Excitation With and Without Damping Using FEA Software.

UnitsUnit SystemMetric (mm, kg, N, s, mV, mA) Degrees rad/s Celsius

AngleDegrees

Rotational Velocityrad/s

TemperatureCelsius

Geometry

Model (A4, B4, C4) > GeometryObject NameGeometry

StateFully Defined

Definition

SourceC:\Users\mi747\AppData\Local\Temp\WB_W3-A46012_mi747_5480_2\unsaved_project_files\dp0\SYS\DM\SYS.agdb

TypeDesignModeler

Length UnitMeters

Element ControlProgram Controlled

Display StyleBody Color

Bounding Box

Length X50. mm

Length Y5. mm

Length Z400. mm

Properties

Volume100000 mm

Mass0.785 kg

Scale Factor Value1.

Statistics

Bodies1

Active Bodies1

Nodes6053

Elements800

Mesh MetricNone

Basic Geometry Options

ParametersYes

Parameter KeyDS

AttributesNo

Named SelectionsNo

Material PropertiesNo

Advanced Geometry Options

Use AssociativityYes

Coordinate SystemsNo

Reader Mode Saves Updated FileNo

Use InstancesYes

Smart CAD UpdateNo

Compare Parts On UpdateNo

Attach File Via Temp FileYes

Temporary DirectoryC:\Users\mi747\AppData\Local\Temp

Analysis Type3-D

Decompose Disjoint GeometryYes

Enclosure and Symmetry ProcessingYes

Model (A4, B4, C4) > Geometry > PartsObject NameSolid

StateMeshed

Graphics Properties

VisibleYes

Transparency1

Definition

SuppressedNo

Stiffness BehaviorFlexible

Coordinate SystemDefault Coordinate System

Reference TemperatureBy Environment

Material

AssignmentStructural Steel

Nonlinear EffectsYes

Thermal Strain EffectsYes

Bounding Box

Length X50. mm

Length Y5. mm

Length Z400. mm

Properties

Volume100000 mm

Mass0.785 kg

Centroid X-3.5293e-016 mm

Centroid Y2.5 mm

Centroid Z200. mm

Moment of Inertia Ip110468 kgmm

Moment of Inertia Ip210630 kgmm

Moment of Inertia Ip3165.18 kgmm

Statistics

Nodes6053

Elements800

Mesh MetricNone

Model (A4, B4, C4) > Geometry > Image

Coordinate Systems

Model (A4, B4, C4) > Coordinate Systems > Coordinate SystemObject NameGlobal Coordinate System

StateFully Defined

Definition

TypeCartesian

Coordinate System ID0.

Origin

Origin X0. mm

Origin Y0. mm

Origin Z0. mm

Directional Vectors

X Axis Data[ 1. 0. 0. ]

Y Axis Data[ 0. 1. 0. ]

Z Axis Data[ 0. 0. 1. ]

Mesh

Model (A4, B4, C4) > MeshObject NameMesh

StateSolved

Defaults

Physics PreferenceMechanical

Relevance0

Sizing

Use Advanced Size FunctionOff

Relevance CenterCoarse

Element Size5.0 mm

Initial Size SeedActive Assembly

SmoothingMedium

TransitionFast

Span Angle CenterCoarse

Minimum Edge Length5.0 mm

Inflation

Use Automatic InflationNone

Inflation OptionSmooth Transition

Transition Ratio0.272

Maximum Layers5

Growth Rate1.2

Inflation AlgorithmPre

View Advanced OptionsNo

Patch Conforming Options

Triangle Surface MesherProgram Controlled

Patch Independent Options

Topology CheckingYes

Advanced

Number of CPUs for Parallel Part MeshingProgram Controlled

Shape CheckingStandard Mechanical

Element Midside NodesProgram Controlled

Straight Sided ElementsNo

Number of RetriesDefault (4)

Extra Retries For AssemblyYes

Rigid Body BehaviorDimensionally Reduced

Mesh MorphingDisabled

Defeaturing

Pinch TolerancePlease Define

Generate Pinch on RefreshNo

Automatic Mesh Based DefeaturingOn

Defeaturing ToleranceDefault

Statistics

Nodes6053

Elements800

Mesh MetricNone

Model (A4, B4, C4) > Mesh > Image

Modal (A5)

Model (A4, B4, C4) > AnalysisObject NameModal (A5)

StateSolved

Definition

Physics TypeStructural

Analysis TypeModal

Solver TargetMechanical APDL

Options

Environment Temperature22. C

Generate Input OnlyNo

Model (A4, B4, C4) > Modal (A5) > Initial ConditionObject NamePre-Stress (None)

StateFully Defined

Definition

Pre-Stress EnvironmentNone

Model (A4, B4, C4) > Modal (A5) > Analysis SettingsObject NameAnalysis Settings

StateFully Defined

Options

Max Modes to Find10

Limit Search to RangeNo

Solver Controls

DampedNo

Solver TypeProgram Controlled

Rotordynamics Controls

Coriolis EffectOff

Campbell DiagramOff

Output Controls

StressYes

StrainYes

Nodal ForcesNo

Calculate ReactionsYes

Store Modal ResultsProgram Controlled

General MiscellaneousNo

Analysis Data Management

Solver Files DirectoryC:\Users\mi747\AppData\Local\Temp\WB_W3-A46012_mi747_5480_2\unsaved_project_files\dp0\SYS\MECH\

Future AnalysisMSUP Analyses

Scratch Solver Files Directory

Save MAPDL dbYes

Delete Unneeded FilesYes

Solver UnitsActive System

Solver Unit SystemNmm

Model (A4, B4, C4) > Modal (A5) > LoadsObject NameFixed Support

StateFully Defined

Scope

Scoping MethodGeometry Selection

Geometry1 Face

Definition

TypeFixed Support

SuppressedNo

Solution (A6)

Model (A4, B4, C4) > Modal (A5) > SolutionObject NameSolution (A6)

StateSolved

Adaptive Mesh Refinement

Max Refinement Loops1.

Refinement Depth2.

Information

StatusDone

The following bar chart indicates the frequency at each calculated mode.

Model (A4, B4, C4) > Modal (A5) > Solution (A6)

Model (A4, B4, C4) > Modal (A5) > Solution (A6)ModeFrequency [Hz]

1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7

Model (A4, B4, C4) > Modal (A5) > Solution (A6) > Solution InformationObject NameSolution Information

StateSolved

Solution Information

Solution OutputSolver Output

Newton-Raphson Residuals0

Update Interval2.5 s

Display PointsAll

FE Connection Visibility

Activate VisibilityYes

DisplayAll FE Connectors

Draw Connections Attached ToAll Nodes

Line ColorConnection Type

Visible on ResultsNo

Line ThicknessSingle

Display TypeLines

Model (A4, B4, C4) > Modal (A5) > Solution (A6) > ResultsObject NameTotal DeformationTotal Deformation 2Total Deformation 3Total Deformation 4Total Deformation 5Total Deformation 6Total Deformation 7Total Deformation 8Total Deformation 9Total Deformation 10

StateSolved

Scope


GeometryAll Bodies

Definition

TypeTotal Deformation

Mode1.2.3.4.5.6.7.8.9.10.

Identifier

SuppressedNo

Results

Minimum0. mm

Maximum71.526 mm71.524 mm71.163 mm88.984 mm71.633 mm71.878 mm90.584 mm72.274 mm70.678 mm93.557 mm

Minimum Value Over Time

Minimum0. mm

Maximum0. mm

Maximum Value Over Time

Minimum71.526 mm71.524 mm71.163 mm88.984 mm71.633 mm71.878 mm90.584 mm72.274 mm70.678 mm93.557 mm

Maximum71.526 mm71.524 mm71.163 mm88.984 mm71.633 mm71.878 mm90.584 mm72.274 mm70.678 mm93.557 mm

Information

Frequency25.701 Hz160.9 Hz252.26 Hz390.12 Hz450.48 Hz882.99 Hz1179.7 Hz1460. Hz1479.8 Hz1996.7 Hz

Model (A4, B4, C4) > Modal (A5) > Solution (A6) > Total DeformationModeFrequency [Hz]

1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7

Model (A4, B4, C4) > Modal (A5) > Solution (A6) > Total Deformation > Image

Model (A4, B4, C4) > Modal (A5) > Solution (A6) > Total Deformation 2ModeFrequency [Hz]

1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7

Model (A4, B4, C4) > Modal (A5) > Solution (A6) > Total Deformation 2 > Image


1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7



1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7



1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7



1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7


1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7


1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7


1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7


1.25.701

2.160.9

3.252.26

4.390.12

5.450.48

6.882.99

7.1179.7

8.1460.

9.1479.8

10.1996.7

Harmonic Response (B5)

Model (A4, B4, C4) > AnalysisObject NameHarmonic Response (B5)

StateSolved

Definition


Analysis TypeHarmonic Response


Options



Model (A4, B4, C4) > Harmonic Response (B5) > Initial ConditionObject NameModal (Modal)

StateFully Defined

Definition

Modal EnvironmentModal


Model (A4, B4, C4) > Harmonic Response (B5) > Analysis SettingsObject NameAnalysis Settings

StateFully Defined

Options

Range Minimum0. Hz

Range Maximum500. Hz

Solution Intervals50

Solution MethodMode Superposition

Include Residual VectorNo

Cluster ResultsNo

Store Results At All FrequenciesYes

Output Controls

StressYes

StrainYes

Nodal ForcesNo


Expand Results FromProgram Controlled

-- ExpansionModal Solution


Damping Controls

Constant Damping Ratio0.

Stiffness Coefficient Define ByDirect Input

Stiffness Coefficient0.

Mass Coefficient0.


Solver Files DirectoryC:\Users\mi747\AppData\Local\Temp\WB_W3-A46012_mi747_5480_2\unsaved_project_files\dp0\SYS-1\MECH\

Future AnalysisNone


Save MAPDL dbNo




Model (A4, B4, C4) > Harmonic Response (B5) > LoadsObject NameForce

StateFully Defined

Scope


Geometry1 Face

Definition

TypeForce

Define ByComponents

Coordinate SystemGlobal Coordinate System

X Component0. N

Y Component150. N

Z Component0. N

Phase Angle0.

SuppressedNo

Solution (B6)

Model (A4, B4, C4) > Harmonic Response (B5) > SolutionObject NameSolution (B6)

StateSolved

Information

StatusDone

Model (A4, B4, C4) > Harmonic Response (B5) > Solution (B6)

Model (A4, B4, C4) > Harmonic Response (B5) > Solution (B6) > Solution InformationObject NameSolution Information

StateSolved





Display PointsAll








Display TypeLines

Model (A4, B4, C4) > Harmonic Response (B5) > Solution (B6) > Result ChartsObject NameFrequency ResponseFrequency Response 2Frequency Response 3

StateSolved

Scope


Geometry1 Face1 Vertex

Spatial ResolutionUse Average

Definition

TypeShear StressDirectional DeformationDirectional Velocity

OrientationXY PlaneY Axis

SuppressedNo

Options

Frequency RangeUse Parent

Minimum Frequency0. Hz

Maximum Frequency500. Hz

DisplayBode

Results

Maximum Amplitude1.0311e-009 MPa80.258 mm1.2675e+005 mm/s

Frequency30. Hz450. Hz

Phase Angle180. 90.

Real-1.0311e-009 MPa-80.258 mm0. mm/s

Imaginary0. MPa0. mm1.2675e+005 mm/s

Model (A4, B4, C4) > Harmonic Response (B5) > Solution (B6) > Frequency Response

Model (A4, B4, C4) > Harmonic Response (B5) > Solution (B6) > Frequency Response 2

Model (A4, B4, C4) > Harmonic Response (B5) > Solution (B6) > Frequency Response 3

Model (A4, B4, C4) > Harmonic Response (B5) > Solution (B6) > ResultsObject NameEquivalent Stress

StateSolved

Scope


GeometryAll Bodies

Definition

TypeEquivalent (von-Mises) Stress

ByFrequency

Frequency50. Hz

Sweeping Phase0.

Identifier

SuppressedNo

Integration Point Results

Display OptionAveraged

Average Across BodiesNo

Results

Minimum8.0601e-003 MPa

Maximum171.71 MPa



Maximum8.0601e-003 MPa


Minimum171.71 MPa

Maximum171.71 MPa

Information

Reported Frequency50. Hz

Model (A4, B4, C4) > Harmonic Response (B5) > Solution (B6) > Equivalent Stress

Model (A4, B4, C4) > Harmonic Response (B5) > Solution (B6) > Equivalent StressSetFrequency [Hz]

1.10.

2.20.

3.30.

4.40.

5.50.

6.60.

7.70.

8.80.

9.90.

10.100.

11.110.

12.120.

13.130.

14.140.

15.150.

16.160.

17.170.

18.180.

19.190.

20.200.

21.210.

22.220.

23.230.

24.240.

25.250.

26.260.

27.270.

28.280.

29.290.

30.300.

31.310.

32.320.

33.330.

34.340.

35.350.

36.360.

37.370.

38.380.

39.390.

40.400.

41.410.

42.420.

43.430.

44.440.

45.450.

46.460.

47.470.

48.480.

49.490.

50.500.

Model (A4, B4, C4) > Harmonic Response (B5) > Solution (B6) > Equivalent Stress > Image

Harmonic Response 2 (C5)

Model (A4, B4, C4) > AnalysisObject NameHarmonic Response 2 (C5)

StateSolved

Definition


Analysis TypeHarmonic Response


Options



Model (A4, B4, C4) > Harmonic Response 2 (C5) > Initial ConditionObject NameModal (Modal)

StateFully Defined

Definition

Modal EnvironmentModal


Model (A4, B4, C4) > Harmonic Response 2 (C5) > Analysis SettingsObject NameAnalysis Settings

StateFully Defined

Options

Range Minimum0. Hz

Range Maximum500. Hz

Solution Intervals50

Solution MethodMode Superposition

Include Residual VectorNo

Cluster ResultsNo

Store Results At All FrequenciesYes

Output Controls

StressYes

StrainYes

Nodal ForcesNo


Expand Results FromProgram Controlled

-- ExpansionModal Solution


Damping Controls

Constant Damping Ratio0.2

Stiffness Coefficient Define ByDirect Input

Stiffness Coefficient0.

Mass Coefficient0.


Solver Files DirectoryC:\Users\mi747\AppData\Local\Temp\WB_W3-A46012_mi747_5480_2\unsaved_project_files\dp0\SYS-2\MECH\

Future AnalysisNone


Save MAPDL dbNo




Model (A4, B4, C4) > Harmonic Response 2 (C5) > LoadsObject NameForce

StateFully Defined

Scope


Geometry1 Face

Definition

TypeForce

Define ByComponents

Coordinate SystemGlobal Coordinate System

X Component0. N

Y Component150. N

Z Component0. N

Phase Angle0.

SuppressedNo

Solution (C6)

Model (A4, B4, C4) > Harmonic Response 2 (C5) > SolutionObject NameSolution (C6)

StateSolved

Information

StatusDone

Model (A4, B4, C4) > Harmonic Response 2 (C5) > Solution (C6)

Model (A4, B4, C4) > Harmonic Response 2 (C5) > Solution (C6) > Solution InformationObject NameSolution Information

StateSolved





Display PointsAll








Display TypeLines

Model (A4, B4, C4) > Harmonic Response 2 (C5) > Solution (C6) > Result ChartsObject NameFrequency ResponseFrequency Response 2Frequency Response 3

StateSolved

Scope


Geometry1 Face1 Vertex

Spatial ResolutionUse Average

Definition

TypeShear StressDirectional DeformationDirectional Velocity

OrientationXY PlaneY Axis

SuppressedNo

Options

Frequency RangeUse Parent

Minimum Frequency0. Hz

Maximum Frequency500. Hz

DisplayBode

Results

Maximum Amplitude6.338e-010 MPa59.288 mm9289. mm/s

Frequency20. Hz30. Hz

Phase Angle-36.946 -37.779 -36.956

Real5.0653e-010 MPa46.86 mm7422.9 mm/s

Imaginary-3.8096e-010 MPa-36.32 mm-5584.5 mm/s

Model (A4, B4, C4) > Harmonic Response 2 (C5) > Solution (C6) > Frequency Response

Model (A4, B4, C4) > Harmonic Response 2 (C5) > Solution (C6) > Frequency Response 2

Model (A4, B4, C4) > Harmonic Response 2 (C5) > Solution (C6) > Frequency Response 3

Model (A4, B4, C4) > Harmonic Response 2 (C5) > Solution (C6) > ResultsObject NameEquivalent Stress

StateSolved

Scope


GeometryAll Bodies

Definition

TypeEquivalent (von-Mises) Stress

ByFrequency

Frequency50. Hz

Sweeping Phase0.

Identifier

SuppressedNo

Integration Point Results

Display OptionAveraged

Average Across BodiesNo

Results


Maximum161.44 MPa



Maximum8.0803e-003 MPa


Minimum161.44 MPa

Maximum161.44 MPa

Information

Reported Frequency50. Hz

Model (A4, B4, C4) > Harmonic Response 2 (C5) > Solution (C6) > Equivalent Stress

Model (A4, B4, C4) > Harmonic Response 2 (C5) > Solution (C6) > Equivalent Stress

SetFrequency [Hz]

1.10.

2.20.

3.30.

4.40.

5.50.

6.60.

7.70.

8.80.

9.90.

10.100.

11.110.

12.120.

13.130.

14.140.

15.150.

16.160.

17.170.

18.180.

19.190.

20.200.

21.210.

22.220.

23.230.

24.240.

25.250.

26.260.

27.270.

28.280.

29.290.

30.300.

31.310.

32.320.

33.330.

34.340.

35.350.

36.360.

37.370.

38.380.

39.390.

40.400.

41.410.

42.420.

43.430.

44.440.

45.450.

46.460.

47.470.

48.480.

49.490.

50.500.

Model (A4, B4, C4) > Harmonic Response 2 (C5) > Solution (C6) > Equivalent Stress > Image

Material DataStructural Steel Structural Steel > ConstantsDensity7.85e-006 kg mm^-3

Coefficient of Thermal Expansion1.2e-005 C^-1

Specific Heat4.34e+005 mJ kg^-1 C^-1

Thermal Conductivity6.05e-002 W mm^-1 C^-1

Resistivity1.7e-004 ohm mm

Structural Steel > Compressive Ultimate StrengthCompressive Ultimate Strength MPa

0

Structural Steel > Compressive Yield StrengthCompressive Yield Strength MPa

250

Structural Steel > Tensile Yield StrengthTensile Yield Strength MPa

250

Structural Steel > Tensile Ultimate StrengthTensile Ultimate Strength MPa

460

Structural Steel > Isotropic Secant Coefficient of Thermal ExpansionReference Temperature C

22

Structural Steel > Alternating Stress Mean StressAlternating Stress MPaCycles Mean Stress MPa

3999100

2827200

1896500

14131000

10692000

44120000

262100000

214200000

1381.e+0050

1142.e+0050

86.21.e+0060

Structural Steel > Strain-Life ParametersStrength Coefficient MPaStrength Exponent Ductility Coefficient Ductility Exponent Cyclic Strength Coefficient MPaCyclic Strain Hardening Exponent

920-0.1060.213-0.4710000.2

Structural Steel > Isotropic ElasticityTemperature CYoung's Modulus MPaPoisson's Ratio Bulk Modulus MPaShear Modulus MPa

2.e+0050.31.6667e+00576923

Structural Steel > Isotropic Relative PermeabilityRelative Permeability

10000

Problem Definition: Modal Analysis of Beam.Rectangular cross-section set time/freq load step sub-step cumulative 1 12.378 1 1 1 2 77.570 1 2 2 3 82.392 1 3 3 4 217.22 1 4 4 5 425.73 1 5 5

Problem Definition: Modal Analysis of Beam.Square cross-section: Set Time/Freq Load Step Sub-step Cumulative 1 31.953 1 1 1 2 31.953 1 2 2 3 200.01 1 3 3 4 200.01 1 4 4 5 559.00 1 5 5

DCOER ,PunePage 1

Lab 2 Experiment No ALL

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Transcript of Lab 2 Experiment No ALL