A column of length 1 m has the cross-section shown in Fig. 8.18. If the ends of the column are pinned and free to warp, calculate its buckling load; E=70,000 N / mm ^{2}, G=30,000 N / mm ^{2}.
Repeat using the MATLAB and the calculated section properties.
In this case, the shear center S is positioned on the C x axis, so that y_{ S }=0 and Eq. (8.79) applies. The distance \bar{x} of the centroid of area C from the web of the section is found by taking first moments of area about the web:
\left|\begin{array}{cc}P-P_{ CR (x x)} & -P x_{ S } \\-P x_{ S } & \left.I_{0}-\left(P-P_{CR(\theta)}\right) / A\right)\end{array}\right|=0 (8.79)
2(100+100+100) \bar{x}=2 \times 2 \times 100 \times 50
which gives
\bar{x}=33.3 mm
The position of the shear center S is found using the method of Example 16.3; this gives x_{ S }=-76.2 mm. The remaining section properties are found by the methods specified in Example 8.4 and follow:
A=600 mm ^{2}, \quad I_{x x}=1.17 \times 10^{6} mm ^{4}, \quad I_{y y}=0.67 \times 10^{6} mm ^{4}
I_{0}=5.35 \times 10^{6} mm ^{4}, \quad J=800 mm ^{4}, \quad \Gamma=2488 \times 10^{6} mm ^{6}
From Eq. (8.77),
P_{ CR (x x)}=\frac{\pi^{2} E I_{x x}}{L^{2}} P_{ CR (y y)}=\frac{\pi^{2} E I_{y y}}{L^{2}} P_{ CR (\theta)}=\frac{A}{I_{0}}\left(G J+\frac{\pi^{2} E \Gamma}{L^{2}}\right) (8.77)
P_{ CR (y y)}=4.63 \times 10^{5} N , \quad P_{ CR (x x)}=8.08 \times 10^{5} N , \quad P_{ CR (\theta)}=1.97 \times 10^{5} N
Expanding Eq. (8.79),
\left(P-P_{ CR (x x)}\right)\left(P-P_{ CR (\theta)}\right) I_{0} / A-P^{2} x_{ S }^{2}=0 (i)
Rearranging Eq. (i),
P^{2}\left(1-A x_{ S }^{2} / I_{0}\right)-P\left(P_{ CR (x x)}+P_{ CR (\theta)}\right)+P_{ CR (x x)} P_{ CR (\theta)}=0 (ii)
Substituting the values of the constant terms in Eq. (ii), we obtain
P^{2}-29.13 \times 10^{5} P+46.14 \times 10^{10}=0 (iii)
The roots of Eq. (iii) give two values of critical load, the lowest of which is
P=1.68 \times 10^{5} N
It can be seen that this value of flexural–torsional buckling load is lower than any of the uncoupled buckling loads P_{ CR (x x)}, P_{ CR (y y)}, \text { or } P_{ CR (\theta)}; the reduction is due to the interaction of the bending and torsional buckling modes.
The value of the critical buckling load is obtained through the following MATLAB file:
% Declare any needed variables
syms P
L = 1000;
t = 2;
E = 70000;
G = 30000;
y_s = 0;
x_s = -76.2;
A = 600;
I_xx = 1.17e6;
I_yy = 0.67e6;
I_0 = 5.32e6;
J = 800;
T = 2488e6;
% Evaluate Eq.(8.77)
P_CRyy = pi^2*E*I_yy/L^2;
P_CRxx = pi^2*E*I_xx/L^2;
P_CRtheta = A*(G*J + pi^2*E*T/L^2)/I_0;
% Substitute results of Eq. (8.77) into Eq.(8.79)
eq_I = det([P-P_CRxx -P*x_s; -P*x_s I_0*(P-P_CRtheta)/A]);
% Solve eq_I for the critical buckling load (P)
P = solve(eq_I,P);
% Output the minimum value of P to the Command Window
disp([‘P =’ num2str(min(double(P))) ‘N’])
The Command Window output resulting from this MATLAB file is as follows:
P = 167785.096 N