cive702 midterm report combined
TRANSCRIPT
Table of Contents INTRODUCTION .......................................................................................................... 1
FLEXIBILITY MATRIX COMPUTATION .................................................................... 2
Beam length ............................................................................................................... 2
Geometric centroid .................................................................................................... 2
Principal Second Moments of Area ........................................................................... 2
Polar Second Moment of Area .................................................................................. 2
Shear Center ............................................................................................................... 3
Material Properties .................................................................................................... 3
Flexibility Matrix ........................................................................................................ 3
SECTION PROPERTY COMPARISON......................................................................... 3
Discussion .................................................................................................................. 4
STRESS COMPUTATION ............................................................................................. 4
Bending ....................................................................................................................... 4
Shear Stress ................................................................................................................ 5
Shear Area .................................................................................................................. 6
Torsion ........................................................................................................................ 7
DISPLACEMENT COMPUTATION ............................................................................. 7
Comparison .............................................................................................................. 10
CONCLUSION ............................................................................................................ 10
INTRODUCTION The cantilevered beam shown in Figure 1, an L4x4 angle with length of 84 inches, was analyzed in the following steps:
1. Section properties, specifically, principal second moments of area, geometric centroid and shear center, were computed by hand. Hand results were compared with results obtained from SolidWorks.
2. Stress distributions due to shear, bending and torsion were computed by hand. Also, the principle of virtual forces was leveraged to compute the shear areas of the cross section, and this result was compared with values published for this cross section1.
3. The three displacements (๐๐1,๐๐2,๐๐3) and three rotations (๐๐4,๐๐5,๐๐6) of the cantilever tip, induced by a vertical force applied to the free end of the beam and through the shear center, were computed using the generalized beam element formulation. MATLAB was used for matrix computation and results were compared to those obtained from a 3-D beam element model in Abaqus.
In all steps, percentage errors were computed from the comparison and values were reported in table format.
Figure 1. L4x4x1/4 Aluminum Alloy, Grade 6061-76, cantilevered beam
1 Zureick, A., Emkin, L., Gwangseok, N., Structural Engineering, Mechanics and Materials Research Report No. 07-1, February 2007
PAGE 1
Figure 2. L4x4 Cross Section Dimensions
FLEXIBILITY MATRIX COMPUTATION All computations for the property values presented hereafter can be found in Appendix A.
BEAM LENGTH
Beam length, ๐ฟ๐ฟ, was given in the problem statement as 84 inches.
GEOMETRIC CENTROID
Geometric centroid was computed to be ๐ฆ๐ฆ๏ฟฝ = ๐ง๐งฬ = 1.093 in., relative to bottom left corner of section.
PRINCIPAL SECOND MOMENTS OF AREA
The angle of rotation that relates the coordinate system shown in Figure 1 to the principal coordinate system was computed to be ๐๐ = 45ยฐ. Using this value, principal second area moments for the yโ and zโ axis were computed to be ๐ผ๐ผ๐ฆ๐ฆโฒ =4.855 in4 and ๐ผ๐ผ๐ฆ๐ฆโฒ = 1.225 in4.
POLAR SECOND MOMENT OF AREA
๐ฝ๐ฝ = .0403 in4
Y
Z
PAGE 2
SHEAR CENTER
The following values for location of the shear center are given in the principal coordinate system with origin located at geometric centroid
๐ฆ๐ฆ๐ ๐ ๐ ๐ = โ1.369 in
๐ง๐ง๐ ๐ ๐ ๐ = 0 in
MATERIAL PROPERTIES
Modulus of Elasticity, Shear Modulus, and Poissonโs ratio for Grade 6061-T6 aluminum obtained from asm.matweb.com
๐ธ๐ธ = 10E + 07 psi
๐บ๐บ = 3.77E + 06 psi
๐๐ = 0.33
FLEXIBILITY MATRIX
A MATLAB script was used, see Appendix B, to generate the following flexibility matrix for a general beam element with the properties presented above
4.34E-06 0 0 0 0 0
0 0.016155 0 0 0 0.000288
[๐น๐น] = 0 0 0.004096 0 -7.3E-05 0
0 0 0 0.000553 0 0
0 0 -7.3E-05 0 1.73E-06 0
0 0.000288 0 0 0 6.86E-06
SECTION PROPERTY COMPARISON As stated in the introduction, the section property values for L4x4 angle were obtained by hand and compared to those obtained from modeling the section in SolidWorks and online sources. The comparison, including percentage errors, is presented in Table 1.
PAGE 3
Table 1. Section Property Value Comparison
Property Hand SolidWorks Percentage Error (%)
๐ฆ๐ฆ๏ฟฝ 1.093 in 1.0723 in 1.93
๐ง๐งฬ 1.093 in 1.0723 in 1.93
๐ผ๐ผ๐ฆ๐ฆโฒ 4.855 in4 4.7333 in4 2.57
๐ผ๐ผ๐ง๐งโฒ 1.225 in4 1.2245 in4 4.08E-02
๐ฝ๐ฝ .0403 in4 .0404 in4 (1) 0.25
1. SolidWorksโ value differed greatly, so obtained value of 0.0404 from Zureick et al., 2007
DISCUSSION
The percentage errors for all section properties were acceptable. The main reason for the observed differences is the cross section was modeled in SolidWorks with fillets of ๐ ๐ = 0.375 in at the edge formed by the joining of the two legs and ๐ ๐ =0.25 in at the end of the legs. The addition of material in the region were the edges meet and the subtraction of material at the end of the legs causes a change in section properties.
STRESS COMPUTATION Derivation of the equations presented in this section can be found in Appendix A
BENDING
Axial stress due to bending in the principle coordinate system is given by
๐๐๐ฅ๐ฅ = 688.78๐ง๐งโฒ + 2511.8๐ฆ๐ฆโฒpsi
Table 2 presents values for axial stress at the locations shown in Figure 3.
Table 2. Axial Stress at Select Locations
Point ๐๐ ๐๐ ๐๐โฒ ๐๐โฒ ๐๐๐๐ (psi)
A -0.968 2.907 1.37108 2.740039 5180.461
B -0.968 0.907 -0.04313 1.325825 731.9387
C -0.968 -0.968 -1.36896 0 -3438.55
D 0.907 -0.968 -0.04313 -1.32583 -948.624
E 2.907 -0.968 1.37108 -2.74004 1707.297
PAGE 4
Figure 3. Locations for Stress Computation
SHEAR STRESS
For a principal coordinate system, shear flow is given by
๐๐๐๐ =๐๐๐ฆ๐ฆโฒ๐๐๐ง๐งโฒ๐ผ๐ผ๐ง๐งโฒ
+๐๐๐ง๐งโฒ๐๐๐ฆ๐ฆโฒ๐ผ๐ผ๐ฆ๐ฆโฒ
The equations for ๐๐๐ฆ๐ฆโฒ and ๐๐๐ง๐งโฒ were computed for each rectangle of the cross section in MATLAB, see Appendix C.
Vertical leg
๐๐๐ฆ๐ฆโฒ = ๏ฟฝ๐ง๐งฬ 4โ
29074000
๏ฟฝ ๏ฟฝ121โ2
250โโ2๏ฟฝ๐ง๐งฬฟ + 2907
1000๏ฟฝ ๏ฟฝ4
๏ฟฝ ,โ.843 โค ๐ง๐งฬ โค 2.907
๐๐๐ง๐งโฒ = โ๏ฟฝ๐ง๐งฬ 4โ
29074000
๏ฟฝ ๏ฟฝ121โ2
250+โ2๏ฟฝ๐ง๐งฬฟ + 2907
1000๏ฟฝ ๏ฟฝ4
๏ฟฝ ,โ.843 โค ๐ง๐งฬ โค 2.907
Horizontal leg
๐๐๐ฆ๐ฆโฒ = ๏ฟฝ๐ฆ๐ฆ๏ฟฝ4โ
29074000
๏ฟฝ ๏ฟฝ121โ2
250โโ2๏ฟฝ๐ฆ๐ฆ๏ฟฝ + 2907
1000๏ฟฝ ๏ฟฝ4
๏ฟฝ ,โ1.093 โค ๐ฆ๐ฆ๏ฟฝ โค 2.907
A
C
B
D E
๐๐
๐๐
PAGE 5
๐๐๐ง๐งโฒ = ๏ฟฝ๐ฆ๐ฆ๏ฟฝ4โ
29074000
๏ฟฝ ๏ฟฝ121โ2
250+โ2๏ฟฝ๐ฆ๐ฆ๏ฟฝ + 2907
1000๏ฟฝ ๏ฟฝ4
๏ฟฝ ,โ1.093 โค ๐ฆ๐ฆ๏ฟฝ โค 2.907
Shear stress was computed for eight locations in the cross section and is presented in the following table.
Table 3. Shear Stress at Select Locations
Point ๐๐๏ฟฝ (in)
๐๐๏ฟฝ (in)
๐๐ (in)
๐จ๐จ (in2)
๐๐โฒ (in)
๐๐โฒ (in)
๐ธ๐ธ๐๐โฒ (in3)
๐ธ๐ธ๐๐โฒ (in3)
๐๐ (psi)
A -0.968 2.907 0.000 0.000 1.371 2.740 0.000 0.000 0.000
B -0.968 1.907 1.000 0.250 0.664 2.033 0.166 0.508 -4.112
C -0.968 0.907 2.000 0.500 -0.043 1.326 -0.022 0.663 -4.915
D -0.968 0.000 2.907 0.727 -0.684 0.684 -0.497 0.497 -2.780
E -0.968 -0.968 3.875 0.969 -1.369 0.000 -1.326 0.000 2.501
F 0.000 -0.968 2.907 0.727 -0.684 -0.684 -0.497 -0.497 4.657
G 0.907 -0.968 2.000 0.500 -0.043 -1.326 -0.022 -0.663 4.996
H 1.907 -0.968 1.000 0.250 0.664 -2.033 0.166 -0.508 3.486
J 2.907 -0.968 0.000 0.000 1.371 -2.740 0.000 0.000 0.000
SHEAR AREA
Shear area was determined for the principal directions using the following equations
๐ด๐ด๐ฆ๐ฆโฒ =๐๐๐ผ๐ผ๐ฆ๐ฆโฒ2
โซ ๐๐๐ฆ๐ฆโฒ2๐๐๐๐
๐ด๐ด๐ง๐งโฒ =๐๐๐ผ๐ผ๐ง๐งโฒ2
โซ ๐๐๐ง๐งโฒ2๐๐๐๐
MATLAB was used to perform the computations for shear area and results are as follows
๐ด๐ด๐ฆ๐ฆโฒ = 0.821 in2
๐ด๐ด๐ง๐งโฒ = 0.809 in2
PAGE 6
A comparison was made to published values for shear area of an L4 x 4 x 1/4 angle section and is presented in Table 4.
Table 4. Shear Area Comparison
Hand (in2)
Published (in2)
Percentage Error (%)
๐ด๐ด๐ฆ๐ฆโฒ 0.821 0.820 0.1
๐ด๐ด๐ง๐งโฒ 0.809 0.833 2.88
TORSION
Since end load was applied through the shear center of the cross section, no moment about the x-axis was generated (i.e., ๐๐1 = 0). Hence, no torsional stress due to torsion should be present in the beam, theoretically.
DISPLACEMENT COMPUTATION Translational displacements (๐๐1,๐๐2,๐๐3) and rotational displacements (๐๐4,๐๐5,๐๐6) of the free end of the beam were computed in MATLAB using the following equation
{๐ข๐ข} = [๐น๐น]{๐๐}
where [๐น๐น] is the flexibility matrix, presented previously, {๐๐} the force vector,
{๐๐} =
โฉโชโจ
โชโง
0โ36.63โ36.63
03077โ3077โญ
โชโฌ
โชโซ
and {๐ข๐ข} the displacement vector.
The displacements resulting from the computation are given as
{๐ข๐ข} =
โฉโชโจ
โชโง
0โ1.4778โ.3736
0. 008โ.0316 โญ
โชโฌ
โชโซ
PAGE 7
Figure 4 - Figure 6 show displacement results obtained from a 3-D beam model in Abaqus, modeled with 20 elements along the length.
Figure 4. Displacement U1 โ Abaqus
Figure 5. Displacement U2 โ Abaqus
PAGE 8
Figure 6. Displacement U3 โ Abaqus
The displacements U2 and U3 in Abaqus can be related to ๐๐๐ฆ๐ฆโฒ and ๐๐๐ง๐งโฒ in the following way
๐๐๐ฆ๐ฆโฒ = โ๐๐3 cos 45ยฐ + ๐๐2 sin 45ยฐ
๐๐๐ง๐งโฒ = ๐๐2 cos 45ยฐ + ๐๐3 sin 45ยฐ
Figure 7 and Figure 8 show displacement as a function of distance from the clamped end in the yโ and zโ directions, respectively. The data was exported from Abaqus and converted using relations just mentioned.
Figure 7. Uy' vs. distance
-0.70
-0.60
-0.50
-0.40
-0.30
-0.20
-0.10
0.00
0 20 40 60 80 100
Uy'
(in)
x (in)
Uy'
PAGE 9
Figure 8. Uz' vs. distance
COMPARISON
Table 5 presents the comparison of the results obtained from MATLAB to those obtained from Abaqus.
Table 5. Displacement Comparison
MATLAB Abaqus Percentage Error (%)
๐๐๐๐ 0 -1.46E-02 100
๐๐๐ฆ๐ฆโฒ -1.4778 -0.597 149
๐๐๐ง๐งโฒ -.3736 -0.150 147
๐๐๐ฅ๐ฅ 0 -3.2E-18 100
๐๐๐ฆ๐ฆโฒ 0.008 9.422E-03 15.1
๐๐๐ง๐งโฒ -0.0316 -5.65E-03 459
CONCLUSION Hand computation of section properties is fairly straight forward and compare well with values obtained from structural analysis or CAD software; however, getting displacements computed via beam element formulation and flexibility method to agree with those obtained from said software is much more difficult. In the latter case, special care has to be taken while constructing the flexibility matrix and accounting for location of shear center of cross section.
-0.16
-0.14
-0.12
-0.10
-0.08
-0.06
-0.04
-0.02
0.00
0 20 40 60 80 100
Uz'
(in)
x (in)
Uz'
PAGE 10
1
Table of ContentsParameters ......................................................................................................................... 1Force vector ....................................................................................................................... 1Flexibility matrix ................................................................................................................ 2Displacement vector ............................................................................................................ 2
ParametersP = 51.8; % [lb]L = 7*12; % [in]% Material PropertiesE = 1*10^7; % [psi]G = 3.77*10^6; % [psi]nu = 0.33;% Areas [in^2]Ax = 1.9375;Ay = 0.821;Az = 0.833;% Second-area moments [in^4]J = .0403;Iy = 4.855;Iz = 1.225;% Shear centerZsc = 0;%Ysc = -1.36;Ysc = 0;% Angle to principal axestheta = pi/4;
Force vectorF1-F3 in [lb], M1-M3 in [in-lb]
F1 = 0;F2 = -P*cos(theta);F3 = -P*sin(theta);M1 = 0;M2 = -F2*L;M3 = F3*L;f = [F1;F2;F3;M1;M2;M3]
f =
1.0e+03 *
0 -0.0366 -0.0366
Appendix B
2
0 3.0768 -3.0768
Flexibility matrixF = zeros(6,6);F(1,1) = L/(Ax*E);F(2,2) = L/(G*Ay)+L^3/(3*E*Iz)+L*Zsc^2/(G*J);F(2,3) = -L*Zsc*Ysc/(G*J);F(2,4) = L*Zsc/(G*J);F(2,6) = L^2/(2*E*Iz);F(3,2) = F(2,3);F(3,3) = L/(G*Az)+L^3/(3*E*Iy)+L*Ysc^2/(G*J);F(3,4) = -L*Ysc/(G*J);F(3,5) = -L^2/(2*E*Iy);F(4,2) = F(2,4);F(4,3) = F(3,4);F(4,4) = L/(G*J);F(5,3) = F(3,5);F(5,5) = L/(E*Iy);F(6,2) = F(2,6);F(6,6) = L/(E*Iz);F
F =
0.0000 0 0 0 0 00 0.0162 0 0 0 0.00030 0 0.0041 0 -0.0001 00 0 0 0.0006 0 00 0 -0.0001 0 0.0000 00 0.0003 0 0 0 0.0000
Displacement vectoru = F*f
u =
0 -1.4778 -0.3736
0 0.0080 -0.0316
1
Table of ContentsSymbolic variables .............................................................................................................. 1Constants ........................................................................................................................... 1Shear Area ......................................................................................................................... 1Shear stress ........................................................................................................................ 2
Symbolic variablessyms y z
Constantsbased on section geometry [units]
t = 0.25; % [in]h1 = 3.75; % [in]h2 = 4; % [in]A1 = h1*t; % [in^2]A2 = h2*t; % [in^2]Ycg = (A1*t/2+A2*h2/2)/(A1+A2); % [in]Zcg = (A1*(h1/2+t)+A2*t/2)/(A1+A2); % [in]
Zmax = h2-Zcg; % [in]Ymax = h2-Ycg; % [in]a1 = -Ycg+t; % [in]a2 = -Ycg; % [in]ybar = -Ycg+t/2; % [in]zbar = -Ycg+t/2; % [in]
Iy_prime = 4.855; % [in^4]Iz_prime = 1.225; % [in^4]theta = pi/4; % [rad]
Shear AreaComputation Ay' and Az'
% Shear area for y', Ay_primeQy_prime_1 = t*(Zmax-z)*(ybar*cos(theta)+(Zmax+z)*sin(theta)/2);Qy_prime_2 = t*(Ymax-y)*((Ymax+y)*cos(theta)/2+zbar*sin(theta));Ay_prime = t*Iz_prime^2/... (double(int(Qy_prime_1^2,z,a1,Zmax)+int(Qy_prime_2^2,y,a2,Zmax)))
% Shear area for z', Az_primeQz_prime_1 = t*(Zmax-z)*((Zmax+z)*cos(theta)/2-ybar*sin(theta));Qz_prime_2 = t*(Ymax-y)*(zbar*cos(theta)-(Ymax+y)*sin(theta)/2);Az_prime = t*Iy_prime^2/... (double(int(Qz_prime_1^2,z,a1,Zmax)+int(Qz_prime_2^2,y,a2,Ymax)))
Appendix C
2
Ay_prime =
0.8202
Az_prime =
0.8094
Shear stressComputation of Tau for rectangle 1 (vertical leg) and 2 (horizontal)
Vy_prime = -36.63;Vz_prime = -36.63;Tau1 = t*(Vy_prime*Qy_prime_1/Iz_prime+Vz_prime*Qz_prime_1/Iy_prime)Tau2 = t*(Vy_prime*Qy_prime_2/Iz_prime+Vz_prime*Qz_prime_2/Iy_prime)
Tau1 = (3663*(z/4 - 721/992)*((15*2^(1/2))/31 + (2^(1/2)*(z + 721/248))/4))/1942 - (3663*(z/4 - 721/992)*((15*2^(1/2))/31 - (2^(1/2)*(z + 721/248))/4))/490 Tau2 = - (3663*(y/4 - 721/992)*((15*2^(1/2))/31 - (2^(1/2)*(y + 721/248))/4))/490 - (3663*(y/4 - 721/992)*((15*2^(1/2))/31 + (2^(1/2)*(y + 721/248))/4))/1942
Published with MATLABยฎ R2014b