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

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Column 1.
1. Conceptualize [hypothesize, sketch]: Begin by considering the equilibrium of Column 1 in a displaced position caused by some external load P applied at A causing small lateral displacement \Delta_{A} (Fig. 11-5a). Sum lateral forces for the entire column to find that reaction H_{B} = 0. Next, draw the free-body diagram of bar ABC (Fig. 11-5c). Note that \Delta_{C} = \Delta_{A} and that bar ABC rotation is \Delta_{A} /(L/2) = 2\Delta_{A} /L.

2. Categorize [simplify, classify]: The rotation of bar ABC can be expressed as 2\Delta_{A} /L because the rotation is small, so the elastic connection resisting moment at C is equal to rotational stiffness \beta_{R} times the total relative rotation of the spring or M_{C} = (\beta_{R} )(2\Delta_{A} /L).

3. Analyze [evaluate; select relevant equations; carry out mathematical solution]: Sum moments about C in Fig. 11-5c and solve for P to obtain the critical load P_{cr1} for Column 1:

P_{ cr 1}=\frac{M_{C}}{2 \Delta_{A}}=\frac{\beta_{R}\left(\frac{2 \Delta_{A}}{L}\right)}{2 \Delta_{A}}=\frac{\beta_{R}}{L}=\beta L            (a)

4. Finalize [conclude; examine answer—Does it make sense? Are units correct?How does it compare to similar problem solutions?]: The buckled mode shape for Column 1 is the displaced position shown in Fig. 11-5a.

Part (a): Column 2.

1, 2. Conceptualize, Categorize: Investigate the equilibrium of Column 2 in a displaced position, once again defined by lateral displacement \Delta_{A}, as shown in Fig. 11-5b. Use a free-body diagram of the entire column (Fig. 11-5d) and sum lateral forces to find that

H_{B}=-(\beta)\left(3 \Delta_{A}\right)            (b)

3. Analyze: Sum moments about D in Fig. 11-5d to find the critical load:

P_{ cr 2}=\frac{-H_{B}\left(\frac{3 L}{2}\right)}{4 \Delta_{A}}=\frac{9}{8} \beta L             (c)

4. Finalize: The buckled mode shape for Column 2 is the displaced position shown in Fig. 11-5b.

Combined model and analysis.

1. Conceptualize: Now create a more advanced structure model by combining the features of Column 1 and Column 2 into a single column as shown in Fig. 11-5e. This idealized structure has elastic rotational springs at both C and D with rotational stiffnesses \beta_{R1}  and  \beta_{R2}, respectively. The roller support remains at B, and the sliding support at D is now restrained by an elastic translational spring with stiffness β. Small lateral displacements \Delta_{C}  and  \Delta_{D} are selected as the degrees of freedom that define the possible displaced positions of the column. (Alternatively, rotation angles such as \theta_{B}  and  \theta_{D} could be used as degrees of freedom to uniquely describe any arbitrary position of the displaced structure.) Hence, the combined structure has two degrees of freedom; therefore, it has two possible buckled mode shapes and two different critical loads—each of which causes the associated buckling mode. In contrast, Columns 1 and 2 are single degree-of-freedom structures because only \Delta_{A} is needed to define the buckled shape of each structure depicted in Figs. 11-5a and b.

2. Categorize: Observe that if rotational spring \beta_{R2} becomes infinitely stiff and translational spring stiffness \beta = 0 in the combined structure (Fig. 11-5e) (while \beta_{R1} remains finite), the two degree of freedom (2DOF) combined model reduces to the single degree-of-freedom (SDOF) model of Fig. 11-5a. Similarly, if rotational spring stiffness \beta_{R1} goes to infinity and spring \beta_{R2} = 0 in Fig. 11-5e (while translational spring stiffness β remains finite), the model becomes that shown in Fig. 11-5b. It follows that the solutions for P_{cr} for Columns 1 and 2 [Eqs. (a) and (c)] are simply two special case solutions of the general combined model in Fig. 11-5e.

3. Analyze: The goal now is to find a general solution for the 2DOF model in Fig. 11-5e and then show that solutions for P_{cr} for Columns 1 and 2 can be obtained from this general solution.
First, consider the equilibrium of the entire 2DOF model in the displaced position shown in Fig. 11-5f. Sum horizontal forces to find that H_{B} = \beta \Delta_{D}. Rotation angles \theta_{B}  and  \theta_{D} in Fig. 11-5f can be expressed in terms of translations \Delta_{C}  and  \Delta_{D} as

\theta_{D}=\frac{\Delta_{D}-\Delta_{C}}{L} \quad \theta_{B}=\frac{2}{L} \Delta_{C}              (d)

Sum moments about B in free-body diagram ABC (Fig. 11-5g), noting that the moment at C is equal to rotational spring stiffness \beta_{R1} times the relative rotation (\theta_{D} – \theta_{B} ) at C, to get

Substituting expressions for \theta_{D}  and  \theta_{B} from Eqs. (d) produces the following equation in terms of unknown displacements \Delta_{C}  and  \Delta_{D}:

\left(2 P-\frac{3 \beta_{R 1}}{L}\right) \Delta_{C}+\left(\frac{\beta_{R I}}{L}-\frac{L \beta}{2}\right) \Delta_{D}=0           (e)

Obtain a second equation that describes the equilibrium of the displaced structure from the free-body diagram of bar CD alone. Summing moments about C for bar CD alone gives

which can be rewritten as

\left(\frac{3 \beta_{R 1}}{L}-P+\frac{\beta_{R 2}}{L}\right) \Delta_{C}+\left(P-\frac{\beta_{R 1}}{L}-\frac{\beta_{R 2}}{L}-L \beta\right) \Delta_{D}=0              (f)

There are now two algebraic equations [Eqs. (e) and (f)] and two unknowns (\Delta_{C}  and  \Delta_{D}). Expressing Eqs. (e) and (f) in matrix form gives

\left(\begin{array}{cc}2 P-\frac{3 \beta_{R 1}}{L} & \frac{\beta_{R 1}}{L}-\frac{L \beta}{2} \\\frac{3 \beta_{R 1}}{L}-P+\frac{\beta_{R 2}}{L} & P-\frac{\beta_{R 1}}{L}-\frac{\beta_{R 2}}{L}-L \beta\end{array}\right)\left(\begin{array}{l}\Delta_{C} \\\Delta_{D}\end{array}\right)=\left(\begin{array}{l}0 \\0\end{array}\right)               (g)

These homogeneous equations have a nonzero (nontrivial) solution only if the determinant of the coefficient matrix in Eq. (g) is equal to zero. If both elastic connections have the same rotational stiffness, \beta_{R 1}=\beta_{R 2}=\beta L^{2} , and Eq. (g) becomes

\left(\begin{array}{cc}2 P-3 L \beta & \frac{L \beta}{2} \\4 L \beta-P & P-3 L \beta\end{array}\right)\left(\begin{array}{c}\Delta_{C} \\\Delta_{D}\end{array}\right)=\left(\begin{array}{l}0 \\0\end{array}\right)             (h)

So the determinant of the coefficient matrix in Eq. (h) (known as the characteristic equation) is

P^{2}-\left(\frac{17}{4} \beta L\right) P+\frac{7}{2}(\beta L)^{2}=0              (i)

Solving Eq. (i) using the quadratic formula results in two values of the critical load:

4. Finalize: These are the eigenvalues of the combined 2DOF system when elastic connection stiffnesses are defined as \beta_{R 1}=\beta_{R 2}=\beta L^{2} . Usually the lower value of the critical load is of more interest because the structure will buckle first at this lower load value. Substitute P_{cr1}  and  P_{cr2} back into Eq. (e) or (f) to find the buckled mode shape (eigenvector) associated with each critical load. The resulting eigenvectors are given here and are shown in Fig. 11-5h:

Application of combined model to Columns 1 and 2.
If the rotational spring stiffness \beta_{R2} goes to infinity and translational spring stiffness β goes to zero while rotational stiffness \beta_{R1} remains finite, the combined model (Fig. 11-5e) reduces to Column 1, and the critical load obtained from the solution of Eq. (g) is that given in Eq. (a). [Alternatively, equating \Delta_{C}  and  \Delta_{D} and setting \beta = 0 in Eq. (e) confirms P_{cr1} in Eq. (a)]. If, instead, rotational spring stiffness \beta_{R1} goes to infinity and rotational stiffness \beta_{R2} goes to zero while translational stiffness β remains finite, the solution of Eq. (g) gives the critical load for Column 2 in Eq. (c).