Question 11.1: Two idealized columns are shown in Fig. 11-5. Both columns a...

Structure 1. We begin by considering the equilibrium of Structure 1 in a dis-turbed position caused by some external load and defined by small rotation angle \theta_{D} (Fig. 11-5a). Summing moments about D, we get the following equilibrium equation:

\Sigma M_{D}=0 \quad P \Delta_{A}=H_{B}\left(\frac{3 L}{2}\right)   (a)

where    \Delta_{A}=\theta_{D}\left(L+2 \frac{L}{2}\right)=\theta_{D}(2 L)  (b)

and  H_{B}=\beta \Delta_{B}=\beta\left[\theta_{D}\left(\frac{3 L}{2}\right)\right]  (c)

Since the angle \theta_{D} is small, lateral displacement \Delta_{A} is obtained using Eq. (b). The force H_{B} in the translational spring at B is the product of spring constant β and small horizontal displacement \Delta_{B}. Substituting the expression for \Delta_{A} from Eq. (b) and the expression for H_{B} from Eq (c) into Eq. (a), and solving for P, we find that the critical load P_{\mathrm{cr}} for Structure 1 is

P_{\mathrm{cr}}=\frac{H_{B}}{\Delta_{A}}\left(\frac{3 L}{2}\right)=\frac{\beta \theta_{D}\left(\frac{3 L}{2}\right)}{\theta_{D}(2 L)}\left(\frac{3 L}{2}\right)=\frac{9}{8} \beta L    (d)

The buckled mode shape for Structure 1 is the disturbed position shown in Fig. 11-5a.

Structure 2. The translational spring at B is now replaced by a roller sup-port, and the structure is assembled using two rigid bars (ABC and CD) joined by a rotational spring having stiffness \beta_{R}. If we sum moments about D for the undisturbed structure, we conclude that horizontal reaction H_{B} is zero. Next, we consider the equilibrium of Structure 2 in a disturbed position, once again defined by small rotation angle \theta_{D} (Fig. 11-5b). Using a free-body dia-gram of the upper bar ABC (Fig. 11-5c) and noting that the moment M_{c} is equal to rotational stiffness \beta_{R} times the total relative rotation of the spring, we have

M_{C}=\beta_{R}\left(\theta_{C}+\theta_{D}\right)=\beta_{R}\left(2 \theta_{D}+\theta_{D}\right)=\beta_{R}\left(3 \theta_{D}\right)      (e)

We see that equilibrium of bar ABC requires that

\Sigma M_{C}=0 \quad M_{C}-P\left(\Delta_{A}+\Delta_{C}\right)=0    (f)

Substituting expressions for M_{c} , \Delta_{A} , and \Delta_{c} into Eq. (f), we obtain

So the critical load P_{\mathrm{cr}} for Structure 2 is

P_{\mathrm{cr}}=\frac{3 \beta_{R}}{2 L} \text { or } P_{\mathrm{cr}}=\frac{3}{2 L}\left(\frac{2}{5} \beta L^{2}\right)=\frac{3}{5} \beta L    (g)

The buckled mode shape for Structure 2 is the disturbed position shown in Fig. 11-5b.
Combined Model and Analysis. We can create a more advanced or complex structure model by combining the features of Structure 1 and Structure 2 into a single structure, as shown in Fig. 11-5d. This idealized structure is shown in its disturbed position and now has both translational spring β at B and rotational elastic connection \beta_{R} at joint C where rigid bars ABC and CD are joined. Note that two rotation angles, \theta_{C} and \theta_{D} , are now required to uniquely describe any arbitrary position of the disturbed structure (alternatively, we could use translations \Delta_{B} and \Delta_{C} , for example, instead of \theta_{C} and \theta_{D}). We will refer to position angles \theta_{C} and \theta_{D} as degrees of freedom. Hence, the combined structure has two degrees of freedom and, therefore, has two possible buckled mode shapes and two different critical loads, each of which causes the associated buckling mode. In con-trast, we see now that Structures 1 and 2 are single degree of freedom struc-tures, because only \theta_{D} is needed (or alternatively, \Delta_{C}) to define the buckled shape of each structure depicted in Figs. 11-5a and b.

We can now observe that if rotational spring \beta_{R} becomes infinitely stiff in the combined structure (Fig. 11-5d) (but β remains finite), the two degree of freedom (2DOF) combined model reduces to the single degree of free-dom (SDOF) model of Fig. 11-5a. Similarly, if translational spring β becomes infinitely stiff in Fig. 11-5d (while \beta_{R} remains finite), the elastic support at B becomes a roller support. We conclude that the solutions for P_{\mathrm{cr}} for Structures 1 and 2 in Eqs. (d) and (g) are simply two special-case solutions of the general combined model in Fig. 11-5d.

Our goal now is to find a general solution for the 2DOF model in Fig. 11-5d and then show that solutions for P_{\mathrm{cr}}  for Structures 1 and 2 can be obtained from this general solution.
First, we consider the equilibrium of the entire 2DOF model in the dis-turbed position shown in Fig. 11-5d. Summing moments about D, we get

where  \Delta_{A}=\left(\theta_{C}+\theta_{D}\right) L

and    H_{B}=\beta \Delta_{B}=\beta\left(\theta_{C} \frac{L}{2}+\theta_{D} L\right)

Combining these expressions, we obtain the following equation in terms of the two unknown position angles (\theta_{C} and \theta_{D}) as

\theta_{C}\left(P-\frac{3}{4} \beta L\right)+\theta_{D}\left(P-\frac{3}{2} \beta L\right)=0  (h)

We can obtain a second equation which describes the equilibrium of the disturbed structure from the free-body diagram of bar ABC alone (Fig. 11-5e). The moment at C is equal to rotational spring stiffness \beta_{R} times the relative rotation at C, and the spring force H_{B} is equal to the spring con-stant β times the total translational displacement at B

M_{C}=\beta_{R}\left(\theta_{C}-\theta_{D}\right)  (i)

and    H_{B}=\beta \Delta_{B}=\beta\left(\theta_{C} \frac{L}{2}+\theta_{D} L\right)  (j)

Summing moments about C in Fig. 11-5e, we get the second equilibrium equation for the combined model as

\Sigma M_{C}=0 \quad P\left(\theta_{C} L\right)-M_{C}-H_{B} \frac{L}{2}=0  (k)

Inserting expressions for M_{C} using Eq. (i) and H_{B} using Eq. (j) into Eq. (k) and simplifying gives

\theta_{C}\left(P-\frac{1}{4} \beta L-\frac{\beta_{R}}{L}\right)+\theta_{D}\left(\frac{\beta_{R}}{L}-\frac{1}{2} \beta L\right)=0  (l)

We now have two algebraic equations in Eqs. (h) and (l) and two unknowns (\theta_{C} ,\theta_{D}). These equations can have nonzero (i.e., nontrivial) solu-tions only if the determinant of the coefficients of \theta_{C} and \theta_{D} is equal to zero. Substituting the assumed expression for \beta_{R}=(2 / 5) \beta L^{2} and then evalu – ating the determinant produces the following characteristic equation for the system:

P^{2}-\left(\frac{41}{20} \beta L\right) P+\frac{9}{10}(\beta L)^{2}=0  (m)

Solving Eq. (m) using the quadratic formula results in two possible val-ues of the critical load:

These are the eigenvalues of the combined 2DOF system. Usually, the lower value of the critical load is of more interest, because the structure will buckle first at this lower load value. If we substitute P_{\mathrm{cr1}} and P_{\mathrm{cr2}} back into Eqs. (h) and (l), we can find the buckled mode shape (i.e., eigenvector) asso-ciated with each critical load.

Application of combined model to Structures 1 and 2. If the rotational spring stiffness \beta_{R} goes to infinity while the translational spring stiffness β remains finite, the combined model (Fig. 11-5d) reduces to Structure 1 because the rotation angles \theta_{C} and \theta_{D} are equal, as shown in Fig. 11-5a. Equating \theta_{C} and \theta_{D} in Eq. (h) and solving for P results in P_{\mathrm{cr}} = (9/8)βL , which is the critical load for Structure 1 [see Eq. (d)].

If the rotational spring stiffness \beta_{R} remains finite while the transla-tional spring stiffness β goes to infinity, the combined model (Fig. 11-5d) reduces to Structure 2. The translational spring becomes a roller support, so \Delta_{B}=0 (i.e., H_{B}=0 ) while rotation angle \theta_{C}=-2 \theta_{D} (i.e., \theta_{C} is clockwise, so negative, as shown in Fig. 11-5b). Inserting β = 0 and \theta_{C}=-2 \theta_{D} into Eq. (l) gives the critical load for Structure 2 [see Eq. (g)].