CIVE702 Midterm Report Combined

CIVE 702
MIDTERM
WINTER 2015
Cyle Teal
CIVE 702
February 12, 2015

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 (𝑈𝑈1, 𝑈𝑈2, 𝑈𝑈3) and three rotations (𝑈𝑈4, 𝑈𝑈5, 𝑈𝑈6) 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, 𝐿𝐿, was given in the problem statement as 84 inches.
GEOMETRIC CENTROID
Geometric centroid was computed to be 𝑦𝑦� = 𝑧𝑧̅ = 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 𝜗𝜗 = 45°. Using this value,
principal second area moments for the y’ and z’ axis were computed to be 𝐼𝐼𝑦𝑦′ =
4.855 in4
and 𝐼𝐼𝑦𝑦′ = 1.225 in4
.
POLAR SECOND MOMENT OF AREA
𝐽𝐽 = .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
𝑦𝑦𝑠𝑠𝑠𝑠 = −1.369 in
𝑧𝑧𝑠𝑠𝑠𝑠 = 0 in
MATERIAL PROPERTIES
Modulus of Elasticity, Shear Modulus, and Poisson’s ratio for Grade 6061-T6
aluminum obtained from asm.matweb.com
𝐸𝐸 = 10E + 07 psi
𝐺𝐺 = 3.77E + 06 psi
𝜈𝜈 = 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
[𝐹𝐹] = 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
(%)
𝑦𝑦� 1.093 in 1.0723 in 1.93
𝑧𝑧̅ 1.093 in 1.0723 in 1.93
𝐼𝐼𝑦𝑦′ 4.855 in4 4.7333 in4 2.57
𝐼𝐼𝑧𝑧′ 1.225 in4 1.2245 in4 4.08E-02
𝐽𝐽 .0403 in4 .0404 in4 (1) 0.25
1. SolidWorks’ 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 𝑅𝑅 = 0.375 in at the edge formed by the joining of the two legs and 𝑅𝑅 =
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
𝜎𝜎𝑥𝑥 = 688.78𝑧𝑧′
+ 2511.8𝑦𝑦′
psi
Table 2 presents values for axial stress at the locations shown in Figure 3.
Table 2. Axial Stress at Select Locations
Point 𝒚𝒚 𝒛𝒛 𝒚𝒚′ 𝒛𝒛′ 𝝈𝝈𝒙𝒙 (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
𝜏𝜏𝜏𝜏 =
𝑉𝑉𝑦𝑦′ 𝑄𝑄𝑧𝑧′
𝐼𝐼𝑧𝑧′
+
𝑉𝑉𝑧𝑧′ 𝑄𝑄𝑦𝑦′
𝐼𝐼𝑦𝑦′
The equations for 𝑄𝑄𝑦𝑦′ and 𝑄𝑄𝑧𝑧′ were computed for each rectangle of the cross
section in MATLAB, see Appendix C.
Vertical leg
𝑄𝑄𝑦𝑦′ = �
𝑧𝑧̅
4
−
2907
4000
� �
121√2
250
−
√2�𝑧𝑧̿ + 2907
1000� �
4
� , −.843 ≤ 𝑧𝑧̅ ≤ 2.907
𝑄𝑄𝑧𝑧′ = − �
𝑧𝑧̅
4
−
2907
4000
� �
121√2
250
+
√2�𝑧𝑧̿ + 2907
1000� �
4
� , −.843 ≤ 𝑧𝑧̅ ≤ 2.907
Horizontal leg
𝑄𝑄𝑦𝑦′ = �
𝑦𝑦�
4
−
2907
4000
� �
121√2
250
−
√2�𝑦𝑦� + 2907
1000� �
4
� , −1.093 ≤ 𝑦𝑦� ≤ 2.907
A
C
B
D
E
𝒛𝒛
𝒚𝒚
PAGE 5

𝑄𝑄𝑧𝑧′ = �
𝑦𝑦�
4
−
2907
4000
� �
121√2
250
+
√2�𝑦𝑦� + 2907
1000� �
4
� , −1.093 ≤ 𝑦𝑦� ≤ 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 𝒚𝒚�
(in)
𝒛𝒛�
(in)
𝒍𝒍
(in)
𝑨𝑨
(in2)
𝒚𝒚′
(in)
𝒛𝒛′
(in)
𝑸𝑸𝒚𝒚′
(in3)
𝑸𝑸𝒛𝒛′
(in3)
𝝉𝝉
(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
𝐴𝐴𝑦𝑦′ =
𝑡𝑡𝐼𝐼𝑦𝑦′
2
∫ 𝑄𝑄𝑦𝑦′
2𝑏𝑏
𝑎𝑎
𝐴𝐴𝑧𝑧′ =
𝑡𝑡𝐼𝐼𝑧𝑧′
2
∫ 𝑄𝑄𝑧𝑧′
2𝑏𝑏
𝑎𝑎
MATLAB was used to perform the computations for shear area and results are as
follows
𝐴𝐴𝑦𝑦′ = 0.821 in2
𝐴𝐴𝑧𝑧′ = 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
(%)
𝐴𝐴𝑦𝑦′ 0.821 0.820 0.1
𝐴𝐴𝑧𝑧′ 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., 𝑀𝑀1 = 0). Hence, no torsional stress
due to torsion should be present in the beam, theoretically.
DISPLACEMENT COMPUTATION
Translational displacements (𝑈𝑈1, 𝑈𝑈2, 𝑈𝑈3) and rotational displacements (𝑈𝑈4, 𝑈𝑈5, 𝑈𝑈6)
of the free end of the beam were computed in MATLAB using the following
equation
{𝑢𝑢} = [𝐹𝐹]{𝑓𝑓}
where [𝐹𝐹] is the flexibility matrix, presented previously, {𝑓𝑓} the force vector,
{𝑓𝑓} =
⎩
⎪
⎨
⎪
⎧
0
−36.63
−36.63
0
3077
−3077⎭
⎪
⎬
⎪
⎫
and {𝑢𝑢} the displacement vector.
The displacements resulting from the computation are given as
{𝑢𝑢} =
⎩
⎪
⎨
⎪
⎧
0
−1.4778
−.3736
0
. 008
−.0316 ⎭
⎪
⎬
⎪
⎫
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 – Abaqus
PAGE 8

The displacements U2 and U3 in Abaqus can be related to 𝑈𝑈𝑈𝑈′ and 𝑈𝑈𝑈𝑈′ in the
following way
𝑈𝑈𝑦𝑦′ = −𝑈𝑈3 cos 45° + 𝑈𝑈2 sin 45°
𝑈𝑈𝑧𝑧′ = 𝑈𝑈2 cos 45° + 𝑈𝑈3 sin 45°
Figure 7 and Figure 8 show displacement as a function of distance from the
clamped end in the y’ and z’ 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 (%)
𝑈𝑈𝑈𝑈 0 -1.46E-02 100
𝑈𝑈𝑈𝑈′ -1.4778 -0.597 149
𝑈𝑈𝑈𝑈′ -.3736 -0.150 147
𝜗𝜗𝑥𝑥 0 -3.2E-18 100
𝜗𝜗𝑦𝑦′ 0.008 9.422E-03 15.1
𝜗𝜗𝑧𝑧′ -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

CIVE702 Midterm Report Combined

1
Table of Contents
Parameters ......................................................................................................................... 1
Force vector ....................................................................................................................... 1
Flexibility matrix ................................................................................................................ 2
Displacement vector ............................................................................................................ 2
Parameters
P = 51.8; % [lb]
L = 7*12; % [in]
% Material Properties
E = 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 center
Zsc = 0;
%Ysc = -1.36;
Ysc = 0;
% Angle to principal axes
theta = pi/4;
Force vector
F1-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 matrix
F = 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 0
0 0.0162 0 0 0 0.0003
0 0 0.0041 0 -0.0001 0
0 0 0 0.0006 0 0
0 0 -0.0001 0 0.0000 0
0 0.0003 0 0 0 0.0000
Displacement vector
u = F*f
u =
0
-1.4778
-0.3736
0
0.0080
-0.0316

1
Table of Contents
Symbolic variables .............................................................................................................. 1
Constants ........................................................................................................................... 1
Shear Area ......................................................................................................................... 1
Shear stress ........................................................................................................................ 2
Symbolic variables
syms y z
Constants
based 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 Area
Computation Ay' and Az'
% Shear area for y', Ay_prime
Qy_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_prime
Qz_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 stress
Computation 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*(
Tau2 =
- (3663*(y/4 - 721/992)*((15*2^(1/2))/31 - (2^(1/2)*(y + 721/248))/4))/490 - (3663*
Published with MATLAB® R2014b

CIVE702 Midterm Report Combined

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