Showing posts with label Mechanical Engineering. Show all posts
Showing posts with label Mechanical Engineering. Show all posts

Thursday 31 January 2019

Automotive Computational Fluid Dynamics (CFD) Analysis

     This post is about the numerical analysis of an Ahmed body. It was a new experience because of the area between the car's floor and the road which is different from most of the numerical analysis performed in open channel aeronautics and turbo-machinery.

     The numerical analysis was performed using the commercial software, SolidWorks Flow Simulation. The software employs κ − ε turbulence model with damping functions, SIMPLE-R (modified), as the numerical algorithm and second-order upwind and central approximations as the spatial discretization schemes for the convective fluxes and diffusive terms. The time derivatives are approximated with an implicit first-order Euler scheme. Flow simulation solves the Navier–Stokes equations, which are formulations of mass, momentum, and energy conservation laws for fluid flows. To predict turbulent flows, the Favre-averaged Navier–Stokes equations are used.

     The software generates Cartesian mesh using immersed boundary method. The mesh had a cell size of 0.035 m in the far field regions within the computational domain. A fine mesh was need between the road the the car's floor to make sure the interaction of the car's floor with the road was captured accurately. Therefore the mesh between the car's floor and the road was refined to have a cell size of 0.00875 m. Another mesh control was applied around the body to refine the mesh with a cells size of 0.0175 m to capture the trailing vortices. The resulting mesh had 209,580 total cells, among those cells, 31,783 cells were at the solid fluid boundary. The computational domain size was ~1L x 1.12L x 3L where L being the vehicle's length. The computational domain along with the computational mesh is shown in Fig. 1.



Fig. 1 Mesh, computational domain and the boundary conditions.

     The red arrows within the Fig. 1 represents the inlet boundary condition of ambient (free-stream) velocity and the blue arrows represent the outlet boundary condition of the ambient pressure. The green arrows represents the co-ordinates axes direction.

     The results from the numerical analysis were compared with [1-3]. The results are within 10% of the experimental results. The velocity (superimposed by the velocity streamlines) and pressure profiles around the car body at various free-stream velocities is shown in Fig. 2.


Fig. 3 Velocity and pressure plots. From the top, Row 1, L-R; ambient velocity of 30 and 40 m/s. Row 2, L-R; ambient velocity of 60 and 80 m/s. Row 3, ambient velocity of 105 m/s.

     It was a good experience learning about automotive CFD after spending a long time in aeronautic/turbo-machinery CFD. Thank you for reading. Please share my work. If you would like to collaborate on a project please reach out.


[1] F.J.Bello-Millán, T.Mäkelä, L.Parras, C.delPino, C.Ferrera, "Experimental study on Ahmed's body drag coefficient for different yaw angles", Journal of Wind Engineering and Industrial Aerodynamics, Volume 157, October 2016, Pages 140-144.

[2] Guilmineau E., Deng G.B., Queutey P., Visonneau M. (2018) Assessment of Hybrid LES Formulations for Flow Simulation Around the Ahmed Body. In: Deville M. et al. (eds) Turbulence and Interactions. TI 2015. Notes on Numerical Fluid Mechanics and Multidisciplinary Design, vol 135. Springer, Cham.

[3] A. Thacker, S.Aubrun, A.Leroy, P.Devinant, "Effects of suppressing the 3D separation on the rear slant on the flow structures around an Ahmed body", Journal of Wind Engineering and Industrial Aerodynamics, Volumes 107–108, August–September 2012, Pages 237-243.

Sunday 7 October 2018

High Camber Wing CFD Simulation

     This post is about the numerical simulation of a high camber, large aspect ratio wing. The wing had an aspect ratio of 5:1. The Reynolds number of flow was 500,000. The wing was at an angle of attack of zero degree. The aero-foil employed had a cross section of NACA 9410.

     The software employed was Flow Simulation Premium. A Cartesian mesh was created using the immersed boundary method. The mesh had 581,005 cells. Among those 581,005 cells, 55,882 were at the solid-fluid boundary. A time step of ~0.00528167 s was employed*. The domain was large enough to accurately trace the flow around the wing without any numerical or reversed flow errors. The software employs κ-ε turbulence model with damping functions, SIMPLE-R (modified) as the numerical algorithm and second order upwind and central approximations as the spatial discretization schemes for the convective fluxes and diffusive terms. The time derivatives are approximated with an implicit first-order Euler scheme.

     The mesh is shown in Fig. 1. The four layers of different mesh density are also visible in Fig. 1, the mesh is refined near the wing surface using a mesh control. The velocity around the wing section is shown in Fig. 2, using a cut plot at  the center of the wing. In Fig. 2, the wing body is super imposed by pressure plot. The velocity vectors showing the direction of flow are superimposed on both the wing body and the velocity cut plot.


Fig. 1, The computational domain.


Fig. 2, The velocity and pressure plots.

     The results of the simulation was validated against the results from XFLR5 software. XFLR5 predicted slightly higher lift and slightly less drag on the wing for same boundary conditions because the XFLR5 simulations were inviscid.

     Thank you for reading. If you would like to contribute to the research, both financially and scientifically, please feel free to reach out.

     *Time step is averaged because of the fact that a smaller time step was employed at the start of the numerical simulation.

Saturday 29 September 2018

Improvement of the Volume Flow Rate Through a Blower Fan

     In this post, an improvement in the volume flow rate through the blower fan assembly made are presented. The only thing changed in the blower fan was the cross section of the fan blades. In the previous version, the blade cross section resembled a flat plate with fillets at the leading edge. The trailing edge in the previous design was blunt. In the modified design, there aero-foils were selected, namely NACA 9410, NACA 9420 and the NACA 9430. All the other parameters were kept the same to the previous case. The CAD models of the modified fan blades are shown in Fig. 1.

Fig. 1, Fan blade geometries.

     The velocity contours are shown in Fig. 2 while the pressure contours are shown in Fig. 3, super imposed with velocity vectors and the computational mesh. The volume flow rate was the most for the fan with blades having cross-section of NACA 9410 aero-foil, followed by the fan with blades having cross-section of NACA 9420 and the NACA 9430 cross sections, respectively.

Fig. 2, Pressure contours. Row 1, L-R; fan with the NACA 9430 and NACA 9420 cross sections. Row 2, fan with NACA 9410 cross sections.

Fig. 3, Velocity contours. Row 1, L-R; fan with the NACA 9430 and NACA 9420 cross sections. Row 2, fan with NACA 9410 cross sections.

     Thank you for reading. If you would like to collaborate, both scientifically and financially, on research projects, please reach out.

Computational Fluid Dynamics Analysis of a Blower/Centrifugal Fan: Update 01

     In this post the results from a CFD analysis of a blower fan are presented. The fan had a diameter of 66 mm and a height of 12.57 mm. The fan's rotational velocity was at 10,000 rpm. The CAD model is shown in Fig. 1.


Fig. 1, CAD Assembly of the Blower Fan.

     The simulations were completed in SolidWorks Flow Simulation Premium code. The code employs immersed boundary method to create a Cartesian mesh. The sliding mesh feature was employed to simulate the rotation of the fan at atmospheric conditions. The code employs κ-ε model with Two-Scales Wall Functions approach as the turbulence model. The numerical algorithm implemented is the SIMPLE-R, modified. The second-order upwind discretization scheme is used to approximate the convective fluxes while the diffusive terms are approximated using the central differencing scheme. The time derivatives are approximated with an implicit first-order Euler scheme.

     The numerical model for the fan had 816,994 cells of which 209,421 cells were at the solid-fluid interface. Two mesh controls were employed to refine the mesh near the blades of the fan and at the boundary of the stationery and the rotating domains. The results were indeed, mesh independent. Due to the fact that this was an internal flow problem, domain independence test was not applicable. The mesh and the computational domain is shown in Fig. 2. The curved teal arrow represents the direction of rotation of the fan. The blue arrows represent the pressure boundary conditions at the inlet and at the outlet of the fan assembly. The straight teal arrow represents the force of gravity (the arrow is inverted).


Fig. 2, The mesh and the computational domain.

     The pressure and velocity plots are shown in Fig. 3-4.

Fig. 3, Pressure contours.

Fig. 4, Velocity contour

     Thank you very much for reading. If you would like to collaborate on research projects or want a tutorial for the setup of the numerical simulations such as this one, please reach out.

Update 01

     CAD files are available here.

Thursday 16 June 2016

Finite Element Method: 1D Elasticity (MATLAB)


Code


%code solves 1D elasticity problems with up to 4 elements (5 nodes)

clc;

n=input('Press one for one, two for two, three for three and four for four elements in the system '); %elements in the system

clc;

if n==1

    A1=input('Enter the area of the element in mm^2 '); %area input

    E1=input('Enter the Young''s modulus for the element in MPa '); %Young's modulus input

    L1=input('Enter the length of the element in mm '); %length input

    DT=input('Enter the change in temperature in degree Celsius ');

    TA1=input('Enter the coefficient of thermal expansion for the element in per degree Celsius ');

    Px1i=input('Enter the pressure acting on the face at node i '); Px1j=input('Enter the pressure acting on the face at node j '); %pressure input

    X=input('Enter the body force acting on the part under examination in N/mm^3 '); %body force input

    P1i=input('Enter the nodal force on node i '); P1j=input('Enter the nodal force on node j '); %nodal force input

    clc;

    I=[1 -1;-1 1]; kg=((A1*E1)/L1)*I; %global stiffness matrix calculation

    FT=[-1 1]; ft1=E1*A1*TA1*DT*FT.'; %thermal force vector calculation

    FPi=[1 0]; fp1i=A1*Px1i*FPi.'; FPj=[0 1]; fp1j=A1*Px1j*FPj.'; %pressure force vector

    FB=[1 1]; bf1=(X*A1*L1/2)*FB.'; %body force vector

    FN1=[P1i P1j].'; %nodal force vector

    fg=ft1+fp1i+fp1j+bf1+FN1; %global force vector calculation

    u=input('Press one for u1=0, two for u2=0 '); %boundary conditions

    clc;

    if u==1

        fgu=zeros(1,1); fgu(1,1)=fg(2,1); %global force vector when u1=0

        kgu=zeros(1,1); kgu(1,1)=kg(2,2); %global stiffness matrix when u1=0

        U=kgu\fgu; %displacement calculation

        e1=(U(1,1)-0)/L1; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; %strain due to forces

        S1=ef1*E1; %stress calculation

        R1=(kg(1,2)*U(1,1))-fg(1,1); %reaction force calculation

        uU=['Displacement at node j is ',num2str(U),' mm']; disp(uU); %displacement output

        rR=['Reaction force at node i is ',num2str(R1),' N']; disp(rR); %reaction force output

        eE=['Strain in the element is ',num2str(ef1)]; disp(eE); %strain output

        sS=['Stress in the element is ',num2str(S1),' MPa']; disp(sS); %stress output

    end

    if u==2

        fgu=zeros(1,1); fgu(1,1)=fg(1,1); %global force vector when u1=0

        kgu=zeros(1,1); kgu(1,1)=kg(1,1); %global stiffness matrix when u1=0

        U=kgu\fgu; %displacement calculation

        e1=(0-U(1,1))/L1; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; %strain due to forces

        S1=ef1*E1; %stress calculation

        R1=(kg(2,1)*U(1,1))-fg(2,1); %reaction force calculation

        uU=['Displacement at node i is ',num2str(U),' mm']; disp(uU); %displacement output

        rR=['Reaction force at node j is ',num2str(R1),' N']; disp(rR); %reaction force output

        eE=['Strain in the element is ',num2str(ef1)]; disp(eE); %strain output

        sS=['Stress in the element is ',num2str(S1),' MPa']; disp(sS); %stress output

    end

end

if n==2

    A1=input('Enter the area of element number one in mm^2 '); A2=input('Enter the area of element number two in mm^2 '); %area input

    E1=input('Enter the Young''s modulus for element number one in MPa '); E2=input('Enter the Young''s modulus for element number two in MPa '); %Young's modulus input

    L1=input('Enter the length of element one in mm '); L2=input('Enter the length of element two in mm '); %length input

    DT=input('Enter the change in temperature in degree Celsius ');

    TA1=input('Enter the coefficient of thermal expansion for element number one in per degree Celsius '); TA2=input('Enter the coefficient of thermal expansion for element number two in per degree Celsius '); %thermal expansion coefficient input

    Px1i=input('Enter the pressure acting on the face at node i for element number one '); Px2j=input('Enter the pressure acting on the face at node j for element number two '); %pressure input

    X=input('Enter the body force acting on the part under examination in N/mm^3 '); %body force input

    P1i=input('Enter the nodal force on node i for element number one '); P1j=input('Enter the nodal force on node j for element number one '); P2i=P1j; P2j=input('Enter the nodal force on node j for element number two '); %nodal force input

    clc;

    I=[1 -1;-1 1]; ke1=((A1*E1)/L1)*I; ke2=((A2*E2)/L2)*I; %element stiffness matrix calculation

    kg=zeros(n+1,n+1); %initialize global stiffness matrix

    kg(1,1)=ke1(1,1); kg(1,2)=ke1(1,2); kg(1,3)=0; kg(2,1)=ke1(2,1); kg(2,2)=ke1(1,1)+ke2(2,2); kg(2,3)=ke2(2,1); kg(3,1)=0; kg(3,2)=ke2(2,1); kg(3,3)=ke2(2,2); %global stiffness matrix formulation

    FT=[-1 1]; ft1=E1*A1*TA1*DT*FT.'; ft2=E2*A2*TA2*DT*FT.'; %thermal force vector

    FPi=[1 0]; fp1i=A1*Px1i*FPi.'; FPj=[0 1]; fp2j=A2*Px2j*FPj.'; %pressure force vector

    FB=[1 1]; bf1=(X*A1*L1/2)*FB.'; bf2=(X*A2*L2/2)*FB.'; %body force vector

    FN1=[P1i P1j].'; FN2=[P2i P2j].'; %nodal force vector

    f1=ft1+fp1i+bf1+FN1; f2=ft2+fp2j+bf2+FN2; %force vector calculation

    fg=zeros(3,1); %global force vector initialize

    fg(1,1)=f1(1,1); fg(2,1)=f1(2,1)+f2(1,1); fg(3,1)=f2(2,1); %global force vector calculation

    u=input('Press one for u1=0, two for u3=0, three for u1=u3=0 '); %boundary conditions

    clc;

    if u==1

        fgu=zeros(2,1); fgu(1,1)=fg(2,1); fgu(2,1)=fg(3,1); %global force vector when u1=0

        kgu=zeros(2,2); kgu(1,1)=kg(2,2); kgu(1,2)=kg(2,3); kgu(2,1)=kg(3,2); kgu(2,2)=kg(3,3); %global stiffness matrix when u1=0

        U=kgu\fgu; %displacement calculation

        e1=(U(1,1)-0)/L1; e2=(U(2,1)-U(1,1))/L2; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; ef2=e2-e0; %strain due to forces

        S1=ef1*E1; S2=ef2*E2; %stress calculation

        R1=(kg(2,1)*U(1,1))-fg(1,1); %reaction force calculation

        uU=['Displacement at node two and three is ',num2str(U(1,1)),' mm and ',num2str(U(2,1)),' mm']; disp(uU); %displacement output

        rR=['Reaction force at element one, node i is ',num2str(R1),' N']; disp(rR); %reaction force output

        eE=['Strain in the element one and two is ',num2str(ef1),' and ',num2str(ef2)]; disp(eE); %strain output

        sS=['Stress in the element one and two is ',num2str(S1),' MPa and ',num2str(S2),' MPa']; disp(sS); %stress output

    end

    if u==2

        fgu=zeros(2,1); fgu(1,1)=fg(1,1); fgu(2,1)=fg(2,1); %global force vector when u3=0

        kgu=zeros(2,2); kgu(1,1)=kg(1,1); kgu(1,2)=kg(1,2); kgu(2,1)=kg(2,1); kgu(2,2)=kg(2,2); %global stiffness matrix when u3=0

        U=kgu\fgu; %displacement calculation

        R1=(kg(3,2)*U(2,1))-fg(3,1); %reaction force calculation

        e1=(U(2,1)-U(1,1))/L1; e2=(0-U(2,1))/L2; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; ef2=e2-e0; %strain due to forces

        S1=ef1*E1; S2=ef2*E2; %stress calculation

        uU=['Displacement at node one and two is ',num2str(U(1,1)),' mm and ',num2str(U(2,1)),' mm']; disp(uU); %displacement output

        rR=['Reaction force at element two, node j is ',num2str(R1),' N']; disp(rR); %reaction force output

        eE=['Strain in the element one and two is ',num2str(ef1),' and ',num2str(ef2)]; disp(eE); %strain output

        sS=['Stress in the element one and two is ',num2str(S1),' MPa and ',num2str(S2),' MPa']; disp(sS); %stress output

    end

    if u==3

        fgu=zeros(1,1); fgu(1,1)=fg(2,1); %global force vector when u3=0

        kgu=zeros(1,1); kgu(1,1)=kg(2,2); %global stiffness matrix when u1=u3=0

        U=kgu\fgu; %displacement calculation

        R1=(kg(2,1)*U(1,1))-fg(1,1); %reaction force calculation

        R3=(kg(3,2)*U(1,1))-fg(3,1); %reaction force calculation

        e1=(U(1,1)-0)/L1; e2=(0-U(1,1))/L2; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; ef2=e2-e0; %strain due to forces

        S1=ef1*E1; S2=ef2*E2;%stress calculation

        uU=['Displacement at node two is ',num2str(U(1,1)),' mm']; disp(uU); %displacement output

        rR1=['Reaction force at element one, node i is ',num2str(R1),' N']; disp(rR1); %reaction force output

        rR3=['Reaction force at element two, node j is ',num2str(R3),' N']; disp(rR3); %reaction force output

        eE=['Strain in the element one and two is ',num2str(ef1), ' and ',num2str(ef2)]; disp(eE); %strain output

        sS=['Stress in the element one and two is ',num2str(S1),' MPa and ',num2str(S2),' MPa']; disp(sS); %stress output

    end

end

if n==3

    A1=input('Enter the area of element number one in mm^2 '); A2=input('Enter the area of element number two in mm^2 '); A3=input('Enter the area of element number three in mm^2 '); %area input

    E1=input('Enter the Young''s modulus for element number one in MPa '); E2=input('Enter the Young''s modulus for element number two in MPa '); E3=input('Enter the Young''s modulus for element number three in MPa '); %Young's modulus input

    L1=input('Enter the length of element one in mm '); L2=input('Enter the length of element two in mm '); L3=input('Enter the length of element three in mm '); %length input

    DT=input('Enter the change in temperature in degree Celsius ');

    TA1=input('Enter the coefficient of thermal expansion for element number one in per degree Celsius '); TA2=input('Enter the coefficient of thermal expansion for element number two in per degree Celsius '); TA3=input('Enter the coefficient of thermal expansion for element number three in per degree Celsius '); %thermal expansion coefficient input

    Px1i=input('Enter the pressure acting on the face at node i for element number one '); Px3j=input('Enter the pressure acting on the face at node j for element number three '); %pressure input

    X=input('Enter the body force acting on the part under examination in N/mm^3 '); %body force input

    P1i=input('Enter the nodal force on node i for element number one '); P1j=input('Enter the nodal force on node j for element number one '); P2i=P1j; P2j=input('Enter the nodal force on node j for element number two '); P3i=P2j; P3j=input('Enter the nodal force on node j for element number three '); %nodal force input

    clc;

    I=[1 -1;-1 1]; ke1=((A1*E1)/L1)*I; ke2=((A2*E2)/L2)*I; ke3=((A3*E3)/L3)*I; %element stiffness matrix calculation

    kg=zeros(n+1,n+1); %initialize global stiffness matrix

    kg(1,1)=ke1(1,1); kg(1,2)=ke1(1,2); kg(2,1)=ke1(2,1); kg(2,2)=ke1(2,2)+ke2(1,1); kg(2,3)=ke2(1,2); kg(3,2)=ke2(2,1); kg(3,3)=ke2(2,2)+ke3(1,1); kg(3,4)=ke3(1,2); kg(4,3)=ke3(2,1); kg(4,4)=ke3(2,2); %global stiffness matrix calculation

    FT=[-1 1]; ft1=E1*A1*TA1*DT*FT.'; ft2=E2*A2*TA2*DT*FT.'; ft3=E3*A3*TA3*DT*FT.'; %thermal force vector

    FPi=[1 0]; fp1i=A1*Px1i*FPi.'; FPj=[0 1]; fp3j=A3*Px3j*FPj.'; %pressure force vector

    FB=[1 1]; bf1=(X*A1*L1/2)*FB.'; bf2=(X*A2*L2/2)*FB.'; bf3=(X*A3*L3/2)*FB.'; %body force vector

    FN1=[P1i P1j].'; FN2=[P2i P2j].';  FN3=[P3i P3j].'; %nodal force vector

    f1=ft1+fp1i+bf1+FN1; f2=ft2+bf2+FN2; f3=ft3+fp3j+bf3+FN3; %force vector calculation

    fg=zeros(4,1); %global force vector initialize

    fg(1,1)=f1(1,1); fg(2,1)=f1(2,1)+f2(1,1); fg(3,1)=f2(2,1)+f3(1,1); fg(4,1)=f3(2,1); %global force vector formulation

    u=input('Press one for u1=0, two for u4=0, three for u1=u4=0 '); %boundary conditions

    clc;

    if u==1

        fgu=zeros(3,1); fgu(1,1)=fg(2,1); fgu(2,1)=fg(3,1); fgu(3,1)=fg(4,1); %global force vector when u1=0

        kgu=zeros(3,3); kgu(1,1)=kg(2,2); kgu(1,2)=kg(2,3); kgu(2,1)=kg(3,2); kgu(2,2)=kg(3,3); kgu(2,3)=kg(3,4); kgu(3,2)=kg(4,3); kgu(3,3)=kg(4,4); %global stiffness matrix when u1=0

        U=kgu\fgu; %displacement calculation

        R1=(kg(1,2)*U(1,1))-fg(1,1); %reaction force calculation

        e1=(U(1,1)-0)/L1; e2=(U(2,1)-U(1,1))/L2; e3=(U(3,1)-U(2,1))/L3; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; ef2=e2-e0; ef3=e3-e0; %strain due to forces

        S1=ef1*E1; S2=ef2*E2; S3=ef3*E3; %stress calculation

        uU=['Displacement at node two, three and four is ',num2str(U(1,1)),' mm, ',num2str(U(2,1)),' mm and ',num2str(U(3,1)),' mm']; disp(uU); %displacement output

        rR=['Reaction force at element one, node i is ',num2str(R1),' N']; disp(rR); %reaction force output

        eE=['Strain in the element one, two and three is ',num2str(ef1),', ',num2str(ef2),' and ',num2str(ef3)]; disp(eE); %strain output

        sS=['Stress in the element one, two and three is ',num2str(S1),' MPa, ',num2str(S2),' MPa and ',num2str(S3),' MPa']; disp(sS); %stress output

    end

    if u==2

        fgu=zeros(3,1); fgu(1,1)=fg(1,1); fgu(2,1)=fg(2,1); fgu(3,1)=fg(3,1); %global force vector when u4=0

        kgu=zeros(3,3); kgu(1,1)=kg(1,1); kgu(1,2)=kg(1,2); kgu(2,1)=kg(2,1); kgu(2,2)=kg(2,2); kgu(2,3)=kg(2,3); kgu(3,2)=kg(3,2); kgu(3,3)=kg(3,3); %global stiffness matrix when u4=0

        U=kgu\fgu; %displacement calculation

        R1=(kg(4,3)*U(3,1))-fg(4,1); %reaction force calculation

        e1=(0-U(3,1))/L1; e2=(U(3,1)-U(2,1))/L2; e3=(U(2,1)-U(1,1))/L3; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; ef2=e2-e0; ef3=e3-e0; %strain due to forces

        S1=ef1*E1; S2=ef2*E2; S3=ef3*E3; %stress calculation

        uU=['Displacement at node one, two and three is ',num2str(U(1,1)),' mm, ',num2str(U(2,1)),' mm and ',num2str(U(3,1)),' mm']; disp(uU); %displacement output

        rR=['Reaction force at element three, node j is ',num2str(R1),' N']; disp(rR); %reaction force output

        eE=['Strain in the element one, two and three is ',num2str(ef1),', ',num2str(ef2),' and ',num2str(ef3)]; disp(eE); %strain output

        sS=['Stress in the element one, two and three is ',num2str(S1),' MPa, ',num2str(S2),' MPa and ',num2str(S3),' MPa']; disp(sS); %stress output

    end

    if u==3

        fgu=zeros(2,1); fgu(1,1)=fg(2,1); fgu(2,1)=fg(3,1); %global force vector when u1=u4=0

        kgu=zeros(2,2); kgu(1,1)=kg(2,2); kgu(1,2)=kg(2,3); kgu(2,1)=kg(3,2); kgu(2,2)=kg(3,3); %global stiffness matrix when u4=u1=0

        U=kgu\fgu; %displacement calculation

        R1=(kg(1,2)*U(1,1))-fg(1,1); R2=(kg(4,3)*U(2,1))-fg(4,1); %reaction force calculation

        e1=(U(1,1)-0)/L1; e2=(U(2,1)-U(1,1))/L2; e3=(0-U(2,1))/L3; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; ef2=e2-e0; ef3=e3-e0; %strain due to forces

        S1=ef1*E1; S2=ef2*E2; S3=ef3*E3; %stress calculation

        uU=['Displacement at node two and three is ',num2str(U(1,1)),' mm and ',num2str(U(2,1)),' mm']; disp(uU); %displacement output

        rR=['Reaction force at element one, node i and element three, node j is ',num2str(R1),' N and ',num2str(R2),' N']; disp(rR); %reaction force output

        eE=['Strain in the element one, two and three is ',num2str(ef1),' , ',num2str(ef2),' and ',num2str(ef3)]; disp(eE); %strain output

        sS=['Stress in the element one, two and three is ',num2str(S1),' MPa, ',num2str(S2),' MPa and ',num2str(S3),' MPa']; disp(sS); %stress output

    end

end

if n==4

    A1=input('Enter the area of element number one in mm^2 '); A2=input('Enter the area of element number two in mm^2 '); A3=input('Enter the area of element number three in mm^2 '); A4=input('Enter the area of element number four in mm^2 '); %area input

    E1=input('Enter the Young''s modulus for element number one in MPa '); E2=input('Enter the Young''s modulus for element number two in MPa '); E3=input('Enter the Young''s modulus for element number three in MPa '); E4=input('Enter the Young''s modulus for element number four in MPa '); %Young's modulus input

    L1=input('Enter the length of element one in mm '); L2=input('Enter the length of element two in mm '); L3=input('Enter the length of element three in mm '); L4=input('Enter the length of element four in mm '); %length input

    DT=input('Enter the change in temperature in degree Celsius ');

    TA1=input('Enter the coefficient of thermal expansion for element number one in per degree Celsius '); TA2=input('Enter the coefficient of thermal expansion for element number two in per degree Celsius '); TA3=input('Enter the coefficient of thermal expansion for element number three in per degree Celsius '); TA4=input('Enter the coefficient of thermal expansion for element number four in per degree Celsius '); %thermal expansion coefficient input

    Px1i=input('Enter the pressure acting on the face at node i for element number one '); Px4j=input('Enter the pressure acting on the face at node j for element number four '); %pressure input

    X=input('Enter the body force acting on the part under examination in N/mm^3 '); %body force input

    P1i=input('Enter the nodal force on node i for element number one '); P1j=input('Enter the nodal force on node j for element number one '); P2i=P1j; P2j=input('Enter the nodal force on node j for element number two '); P3i=P2j; P3j=input('Enter the nodal force on node j for element number three '); P4i=P3j; P4j=input('Enter the nodal force on node j for element number four '); %nodal force input

    clc;

    I=[1 -1;-1 1]; ke1=((A1*E1)/L1)*I; ke2=((A2*E2)/L2)*I; ke3=((A3*E3)/L3)*I; ke4=((A4*E4)/L4)*I; %element stiffness matrix calculation

    kg=zeros(n+1,n+1); %initialize global stiffness matrix

    kg(1,1)=ke1(1,1); kg(1,2)=ke1(1,2); kg(2,1)=ke1(2,1); kg(2,2)=ke1(2,2)+ke2(1,1); kg(2,3)=ke2(1,2); kg(3,2)=ke2(2,1); kg(3,3)=ke2(2,2)+ke3(1,1); kg(3,4)=ke3(1,2); kg(4,3)=ke3(2,1); kg(4,4)=ke3(2,2)+ke4(1,1); kg(4,5)=ke4(1,2); kg(5,4)=ke4(2,1); kg(5,5)=ke4(2,2); %global stiffness matrix calculation

    FT=[-1 1]; ft1=E1*A1*TA1*DT*FT.'; ft2=E2*A2*TA2*DT*FT.'; ft3=E3*A3*TA3*DT*FT.'; ft4=E4*A4*TA4*DT*FT.'; %thermal force vector

    FPi=[1 0]; fp1i=A1*Px1i*FPi.'; FPj=[0 1]; fp4j=A4*Px4j*FPj.'; %pressure force vector

    FB=[1 1]; bf1=(X*A1*L1/2)*FB.'; bf2=(X*A2*L2/2)*FB.'; bf3=(X*A3*L3/2)*FB.'; bf4=(X*A4*L4/2)*FB.'; %body force vector

    FN1=[P1i P1j].'; FN2=[P2i P2j].';  FN3=[P3i P3j].'; FN4=[P4i P4j].';%nodal force vector

    f1=ft1+fp1i+bf1+FN1; f2=ft2+bf2+FN2; f3=ft3+bf3+FN3; f4=ft4+fp4j+bf4+FN4; %force vector calculation

    fg=zeros(5,1); %global force vector initialize

    fg(1,1)=f1(1,1); fg(2,1)=f1(2,1)+f2(1,1); fg(3,1)=f2(2,1)+f3(1,1); fg(4,1)=f3(2,1)+f4(1,1); fg(5,1)=f4(2,1); %global force vector formulation

    u=input('Press one for u1=0, two for u5=0, three for u1=u5=0 '); %boundary conditions0

    clc;

    if u==1

        fgu=zeros(4,1); fgu(1,1)=fg(2,1); fgu(2,1)=fg(3,1); fgu(3,1)=fg(4,1); fgu(4,1)=fg(5,1); %global force vector when u1=0

        kgu=zeros(4,4);

        kgu(1,1)=kg(2,2); kgu(1,2)=kg(2,3); kgu(2,1)=kg(3,2); kgu(2,2)=kg(3,3); kgu(2,3)=kg(3,4); kgu(3,2)=kg(4,3); kgu(3,3)=kg(4,4); kgu(3,4)=kg(4,5); kgu(4,3)=kg(5,4); kgu(4,4)=kg(5,5);%global stiffness matrix when u1=0

        U=kgu\fgu; %displacement calculation

        R1=(kg(1,2)*U(1,1))-fg(1,1); %reaction force calculation

        e1=(U(1,1)-0)/L1; e2=(U(2,1)-U(1,1))/L2; e3=(U(3,1)-U(2,1))/L3; e4=(U(4,1)-U(3,1))/L4; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; ef2=e2-e0; ef3=e3-e0; ef4=e4-e0; %strain due to forces

        S1=ef1*E1; S2=ef2*E2; S3=ef3*E3; S4=ef4*E4; %stress calculation

        uU=['Displacement at node two, three, four and five is ',num2str(U(1,1)),' mm, ',num2str(U(2,1)),' mm, ',num2str(U(3,1)),' mm and ',num2str(U(4,1)),' mm']; disp(uU); %displacement output

        rR=['Reaction force at element one, node i is ',num2str(R1),' N']; disp(rR); %reaction force output

        eE=['Strain in the element one, two, three and four is ',num2str(ef1),', ',num2str(ef2),', ',num2str(ef3),' and ',num2str(ef4)]; disp(eE); %strain output

        sS=['Stress in the element one, two, three and four is ',num2str(S1),' MPa, ',num2str(S2),' MPa ',num2str(S3),' MPa and ',num2str(S4),' MPa']; disp(sS); %stress output

    end

    if u==2

        fgu=zeros(4,1); fgu(1,1)=fg(1,1); fgu(2,1)=fg(2,1); fgu(3,1)=fg(3,1); fgu(4,1)=fg(4,1); %global force vector when u5=0

        kgu=zeros(4,4);

        kgu(1,1)=kg(1,1); kgu(1,2)=kg(1,2); kgu(2,1)=kg(2,1); kgu(2,2)=kg(2,2); kgu(2,3)=kg(2,3); kgu(3,2)=kg(3,2); kgu(3,3)=kg(3,3); kgu(3,4)=kg(3,4); kgu(4,3)=kg(4,3); kgu(4,4)=kg(4,4); %global stiffness matrix when u5=0

        U=kgu\fgu; %displacement calculation

        R1=(kg(5,4)*U(4,1))-fg(5,1); %reaction force calculation

        e1=(U(2,1)-U(1,1))/L1; e2=(U(3,1)-U(2,1))/L2; e3=(U(4,1)-U(3,1))/L3; e4=(0-U(4,1))/L4; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; ef2=e2-e0; ef3=e3-e0; ef4=e4-e0; %strain due to forces

        S1=ef1*E1; S2=ef2*E2; S3=ef3*E3; S4=ef4*E4; %stress calculation

        uU=['Displacement at node one, two, three and four is ',num2str(U(1,1)),' mm, ',num2str(U(2,1)),' mm, ',num2str(U(3,1)),' mm and ',num2str(U(4,1)),' mm']; disp(uU); %displacement output

        rR=['Reaction force at element four, node j is ',num2str(R1),' N']; disp(rR); %reaction force output

        eE=['Strain in the element one, two, three and four is ',num2str(ef1),', ',num2str(ef2),', ',num2str(ef3),' and ',num2str(ef4)]; disp(eE); %strain output

        sS=['Stress in the element one, two, three and four is ',num2str(S1),' MPa, ',num2str(S2),' MPa ',num2str(S3),' MPa and ',num2str(S4),' MPa']; disp(sS); %stress output

    end

    if u==3

        fgu=zeros(3,1); fgu(1,1)=fg(1,1); fgu(2,1)=fg(2,1); fgu(3,1)=fg(3,1); %global force vector when u1=u5=0

        kgu=zeros(3,3);

        kgu(1,1)=kg(2,2); kgu(1,2)=kg(2,3); kgu(2,1)=kg(3,2); kgu(2,2)=kg(3,3); kgu(2,3)=kg(3,4); kgu(3,2)=kg(4,3); kgu(3,3)=kg(4,4); %global stiffness matrix when u5=0

        U=kgu\fgu; %displacement calculation

        R1=(kg(5,4)*U(3,1))-fg(5,1); R2=(kg(1,2)*U(1,1))-fg(5,1); %reaction force calculation

        e1=(U(1,1)-0)/L1; e2=(U(2,1)-U(1,1))/L2; e3=(U(3,1)-U(2,1))/L3; e4=(0-U(3,1))/L4; %strain calculation

        e0=TA1*DT; %thermal strain

        ef1=e1-e0; ef2=e2-e0; ef3=e3-e0; ef4=e4-e0; %strain due to forces

        S1=ef1*E1; S2=ef2*E2; S3=ef3*E3; S4=ef4*E4; %stress calculation

        uU=['Displacement at node two, three and four is ',num2str(U(1,1)),' mm, ',num2str(U(2,1)),' mm, ',num2str(U(3,1)),' mm']; disp(uU); %displacement output

        rR=['Reaction force at element one, node i and element four, node j is ',num2str(R1),' N and ',num2str(R2),' N']; disp(rR); %reaction force output

        eE=['Strain in the element one, two, three and four is ',num2str(ef1),', ',num2str(ef2),', ',num2str(ef3),' and ',num2str(ef4)]; disp(eE); %strain output

        sS=['Stress in the element one, two, three and four is ',num2str(S1),' MPa, ',num2str(S2),' MPa ',num2str(S3),' MPa and ',num2str(S4),' MPa']; disp(sS); %stress output

    end

end

Sample calculations for problem with one element


With first boundary condition (u1=0)


The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 1

Enter the area of the element in mm^2 20

Enter the Young's modulus for the element in MPa 200000

Enter the length of the element in mm 100

Enter the change in temperature in degree Celsius 50

Enter the coefficient of thermal expansion for the element in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i 0

Enter the pressure acting on the face at node j 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i 0

Enter the nodal force on node j 10000

Press one for u1=0, two for u2=0 1

Result


Displacement at node j is 0.26 mm

Reaction force at node i is -10000 N

Strain in the element is 0.0025

Stress in the element is 500 MPa

With second boundary condition (u2=0)


The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 1

Enter the area of the element in mm^2 20

Enter the Young's modulus for the element in MPa 200000

Enter the length of the element in mm 100

Enter the change in temperature in degree Celsius 50

Enter the coefficient of thermal expansion for the element in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i 0

Enter the pressure acting on the face at node j 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i -10000

Enter the nodal force on node j for element number one 0

Press one for u1=0, two for u2=0 2

Result


Displacement at node i is -0.26 mm

Reaction force at node j is 10000 N

Strain in the element is 0.0025

Stress in the element is 500 MPa

Sample calculations for problem with two elements


Problem 5.1a, Fagan pg. 94

With first boundary condition (u1=0)


The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 2

Enter the area of element number one in mm^2 20

Enter the area of element number two in mm^2 10

Enter the Young's modulus for element number one in MPa 200000

Enter the Young's modulus for element number two in MPa 200000

Enter the length of element one in mm 100

Enter the length of element two in mm 100

Enter the change in temperature in degree Celsius 50

Enter the coefficient of thermal expansion for element number one in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number two in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i for element number one 0

Enter the pressure acting on the face at node j for element number two 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i for element number one 0

Enter the nodal force on node j for element number one 0

Enter the nodal force on node j for element number two 10000

Press one for u1=0, two for u3=0, three for u1=u3=0 1

Result


Displacement at node two and three is 0.26 mm and 0.77 mm

Reaction force at element one, node i is -10000 N

Strain in the element one and two is 0.0025 and 0.005

Stress in the element one and two is 500 MPa and 1000 MPa

With second boundary condition (u3=0)


Problem 5.1a, Fagan pg. 94 mirrored

The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 2

Enter the area of element number one in mm^2 10

Enter the area of element number two in mm^2 20

Enter the Young's modulus for element number one in MPa 200000

Enter the Young's modulus for element number two in MPa 200000

Enter the length of element one in mm 100

Enter the length of element two in mm 100

Enter the change in temperature in degree Celsius 50

Enter the coefficient of thermal expansion for element number one in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number two in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i for element number one 0

Enter the pressure acting on the face at node j for element number two 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i for element number one -10000

Enter the nodal force on node j for element number one 0

Enter the nodal force on node j for element number two 0

Press one for u1=0, two for u3=0, three for u1=u3=0 2

Result


Displacement at node one and two is -0.77 mm and -0.26 mm

Reaction force at element two, node j is 10000 N

Strain in the element one and two is 0.005 and 0.0025

Stress in the element one and two is 1000 MPa and 500 MPa

With third boundary condition (u3=u1=0)


Problem 5.1b, Fagan pg. 94

The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 2

Enter the area of element number one in mm^2 20

Enter the area of element number two in mm^2 10

Enter the Young's modulus for element number one in MPa 200000

Enter the Young's modulus for element number two in MPa 200000

Enter the length of element one in mm 100

Enter the length of element two in mm 100

Enter the change in temperature in degree Celsius 50

Enter the coefficient of thermal expansion for element number one in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number two in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i for element number one 0

Enter the pressure acting on the face at node j for element number two 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i for element number one 0

Enter the nodal force on node j for element number one 0

Enter the nodal force on node j for element number two 0

Press one for u1=0, two for u3=0, three for u1=u3=0 3

Result


Displacement at node two is 0.0033333 mm

Reaction force at element one, node i is 266.6667 N

Reaction force at element two, node j is -266.6667 N

Strain in the element one and two is -6.6667e-05 and -0.00013333

Stress in the element one and two is -13.3333 MPa and -26.6667 MPa

Sample calculations for problem with three elements


Problem 5.1f, Fagan pg. 95

With first boundary condition (u1=0)


The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 3

Enter the area of element number one in mm^2 20

Enter the area of element number two in mm^2 20

Enter the area of element number three in mm^2 10

Enter the Young's modulus for element number one in MPa 200000

Enter the Young's modulus for element number two in MPa 200000

Enter the Young's modulus for element number three in MPa 200000

Enter the length of element one in mm 100

Enter the length of element two in mm 100

Enter the length of element three in mm 100

Enter the change in temperature in degree Celsius 0

Enter the coefficient of thermal expansion for element number one in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number two in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number three in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i for element number one 0

Enter the pressure acting on the face at node j for element number three 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i for element number one 0

Enter the nodal force on node j for element number one 20000

Enter the nodal force on node j for element number two 0

Enter the nodal force on node j for element number three 10000

Press one for u1=0, two for u4=0, three for u1=u4=0 1

Result


Displacement at node two, three and four is 1.25 mm, 1.5 mm and 2 mm

Reaction force at element one, node i is -50000 N

Strain in the element one, two and three is 0.0125, 0.0025 and 0.005

Stress in the element one, two and three is 2500 MPa, 500 MPa and 1000 MPa

With second boundary condition (u4=0)


Problem 5.1f, Fagan pg. 95 mirrored

The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 3

Enter the area of element number one in mm^2 10

Enter the area of element number two in mm^2 20

Enter the area of element number three in mm^2 20

Enter the Young's modulus for element number one in MPa 200000

Enter the Young's modulus for element number two in MPa 200000

Enter the Young's modulus for element number three in MPa 200000

Enter the length of element one in mm 100

Enter the length of element two in mm 100

Enter the length of element three in mm 100

Enter the change in temperature in degree Celsius 0

Enter the coefficient of thermal expansion for element number one in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number two in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number three in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i for element number one 0

Enter the pressure acting on the face at node j for element number three 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i for element number one -10000

Enter the nodal force on node j for element number one 0

Enter the nodal force on node j for element number two -20000

Enter the nodal force on node j for element number three 0

Press one for u1=0, two for u4=0, three for u1=u4=0 2

Result


Displacement at node one, two and three is -2 mm, -1.5 mm and -1.25 mm

Reaction force at element three, node j is 50000 N

Strain in the element one, two and three is 0.0125, 0.0025 and 0.005

Stress in the element one, two and three is 2500 MPa, 500 MPa and 1000 MPa

With third boundary condition (u4=u1=0)


Problem 5.1f, Fagan pg. 95 with temperature load only, no nodal forces

The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 3

Enter the area of element number one in mm^2 20

Enter the area of element number two in mm^2 20

Enter the area of element number three in mm^2 10

Enter the Young's modulus for element number one in MPa 200000

Enter the Young's modulus for element number two in MPa 200000

Enter the Young's modulus for element number three in MPa 200000

Enter the length of element one in mm 100

Enter the length of element two in mm 100

Enter the length of element three in mm 100

Enter the change in temperature in degree Celsius 50

Enter the coefficient of thermal expansion for element number one in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number two in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number three in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i for element number one 0

Enter the pressure acting on the face at node j for element number three 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i for element number one 0

Enter the nodal force on node j for element number one 0

Enter the nodal force on node j for element number two 0

Enter the nodal force on node j for element number three 0

Press one for u1=0, two for u4=0, three for u1=u4=0 3

Result


Displacement at node two and three is 0.0025 mm and 0.005 mm

Reaction force at element one, node i and element three, node j is 300 N and -300 N

Strain in the element one, two and three is -7.5e-05 , -7.5e-05 and -0.00015

Stress in the element one, two and three is -15 MPa, -15 MPa and -30 MPa

Sample calculations for problem with four elements


With first boundary condition (u1=0)


The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 4

Enter the area of element number one in mm^2 20

Enter the area of element number two in mm^2 15

Enter the area of element number three in mm^2 10

Enter the area of element number four in mm^2 5

Enter the Young's modulus for element number one in MPa 200000

Enter the Young's modulus for element number two in MPa 200000

Enter the Young's modulus for element number three in MPa 200000

Enter the Young's modulus for element number four in MPa 200000

Enter the length of element one in mm 100

Enter the length of element two in mm 100

Enter the length of element three in mm 100

Enter the length of element four in mm 100

Enter the change in temperature in degree Celsius 50

Enter the coefficient of thermal expansion for element number one in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number two in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number three in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number four in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i for element number one 0

Enter the pressure acting on the face at node j for element number four 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i for element number one 400

Enter the nodal force on node j for element number one 200

Enter the nodal force on node j for element number two 400

Enter the nodal force on node j for element number three 200

Enter the nodal force on node j for element number four 400

Press one for u1=0, two for u5=0, three for u1=u5=0 1

Results


Displacement at node two, three, four and five is 0.06 mm, 0.12333 mm, 0.17333 mm and 0.22333 mm

Reaction force at element one, node i is -2400 N

Strain in the element one, two, three and four is 0.0005, 0.00053333, 0.0004 and 0.0004

Stress in the element one, two, three and four is 100 MPa, 106.6667 MPa 80 MPa and 80 MPa

With second boundary condition (u5=0)


The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 4

Enter the area of element number one in mm^2 5

Enter the area of element number two in mm^2 10

Enter the area of element number three in mm^2 15

Enter the area of element number four in mm^2 20

Enter the Young's modulus for element number one in MPa 200000

Enter the Young's modulus for element number two in MPa 200000

Enter the Young's modulus for element number three in MPa 200000

Enter the Young's modulus for element number four in MPa 200000

Enter the length of element one in mm 100

Enter the length of element two in mm 100

Enter the length of element three in mm 100

Enter the length of element four in mm 100

Enter the change in temperature in degree Celsius 50

Enter the coefficient of thermal expansion for element number one in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number two in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number three in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number four in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i for element number one 0

Enter the pressure acting on the face at node j for element number four 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i for element number one -400

Enter the nodal force on node j for element number one -200

Enter the nodal force on node j for element number two -400

Enter the nodal force on node j for element number three -200

Enter the nodal force on node j for element number four -400

Press one for u1=0, two for u5=0, three for u1=u5=0 2

Results


Displacement at node one, two, three and four is -0.22333 mm, -0.17333 mm, -0.12333 mm and -0.06 mm

Reaction force at element four, node j is 2400 N

Strain in the element one, two, three and four is 0.0004, 0.0004, 0.00053333 and 0.0005

Stress in the element one, two, three and four is 80 MPa, 80 MPa 106.6667 MPa and 100 MPa

With third boundary condition (u5=u1=0)


The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 4

Enter the area of element number one in mm^2 20

Enter the area of element number two in mm^2 15

Enter the area of element number three in mm^2 10

Enter the area of element number four in mm^2 5

Enter the Young's modulus for element number one in MPa 200000

Enter the Young's modulus for element number two in MPa 200000

Enter the Young's modulus for element number three in MPa 200000

Enter the Young's modulus for element number four in MPa 200000

Enter the length of element one in mm 100

Enter the length of element two in mm 100

Enter the length of element three in mm 100

Enter the length of element four in mm 100

Enter the change in temperature in degree Celsius 50

Enter the coefficient of thermal expansion for element number one in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number two in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number three in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number four in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i for element number one 0

Enter the pressure acting on the face at node j for element number four 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i for element number one 0

Enter the nodal force on node j for element number one 0

Enter the nodal force on node j for element number two 0

Enter the nodal force on node j for element number three 0

Enter the nodal force on node j for element number four 0

Press one for u1=0, two for u5=0, three for u1=u5=0 3

Results


Displacement at node two, three and four is -0.0058 mm, -0.0002 mm, 0.0032 mm

Reaction force at element one, node i and element four, node j is -132 N and 132 N

Strain in the element one, two, three and four is -0.000158, -4.4e-05, -6.6e-05 and -0.000132

Stress in the element one, two, three and four is -31.6 MPa, -8.8 MPa -13.2 MPa and -26.4 MPa

Sample Calculations for problems involving variable cross-sectional area


Problem 5.1d, Fagan pg. 95

The command window at various steps

Press one for one, two for two, three for three and four for four elements in the system 2

Enter the area of element number one in mm^2 25

Enter the area of element number two in mm^2 15

Enter the Young's modulus for element number one in MPa 200000

Enter the Young's modulus for element number two in MPa 200000

Enter the length of element one in mm 100

Enter the length of element two in mm 100

Enter the change in temperature in degree Celsius 50

Enter the coefficient of thermal expansion for element number one in per degree Celsius 2/1000000

Enter the coefficient of thermal expansion for element number two in per degree Celsius 2/1000000

Enter the pressure acting on the face at node i for element number one 0

Enter the pressure acting on the face at node j for element number two 0

Enter the body force acting on the part under examination in N/mm^3 0

Enter the nodal force on node i for element number one 0

Enter the nodal force on node j for element number one 0

Enter the nodal force on node j for element number two 10000

Press one for u1=0, two for u3=0, three for u1=u3=0 1

Results


Displacement at node two and three is 0.21 mm and 0.55333 mm

Reaction force at element one, node i is -10000 N

Strain in the element one and two is 0.002 and 0.0033333
Stress in the element one and two is 400 MPa and 666.6667 MPa