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 β_R times the total relative rotation of the spring or M_C = (β_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:
\quad\quad\quad\quad{ P}_{cr1}=\frac{M_{C}}{2\Delta_{A}}=\frac{\beta_{R}{\Bigg\lgroup}2\frac{\Delta_{A}}{ L}{\Bigg\rgroup}}{2\Delta_{A}}=\frac{\beta_{R}}{ L}=\beta{ L}\quad\quad(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
\quad\quad\quad\quad H_B = -(\beta)(3\Delta _A)\quad\quad(b)
3. Analyze: Sum moments about D in Fig. 11-5d to find the critical load:
\quad\quad\quad\quad P_{cr2}={\frac{-H_{B}\left({\frac{3L}{2}}\right)}{4\Delta_{A}}}={\frac{9}{8}}\beta L\quad\quad (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 β_{R1} ~ and ~ β_{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 b 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 = β\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
\quad\quad\quad\quad \theta_{D}={\frac{\Delta_{D}-\Delta_{C}}{L}}\ \ \ \ \ \ \theta_{B}={\frac{2}{L}}\Delta_{C}\quad\quad (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
\quad\quad\quad\quad 2{P}\Delta_{C}-\beta\Delta_{D}\frac{L}{2}+\beta_{R1}(\theta_{D}-\theta_{B})=0
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:
\quad\quad\quad\quad{\Bigg\lgroup}2P – {\frac{3\beta_{R1}}{L}}{\Bigg\rgroup}\Delta_{C}+{\Bigg\lgroup}\frac{\beta_{R1}}{L} – \,\frac{L\beta}{2} {\Bigg\rgroup}\Delta_{D} = 0 \quad\quad (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
\quad\quad\quad\quad P(\Delta_{D}\,-\,\Delta_{C})-\beta_{R1}(\theta_{D}\,-\,\theta_{B})-\beta_{R2}\theta_{D}\,-\,\beta\Delta_{D}L=0
which can be rewritten as
\quad\quad {\Bigg\lgroup}\frac{3\beta_{R1}}{L}-P+\frac{\beta_{R2}}{L}{\Bigg\rgroup}\Delta_{C}+{\Bigg\lgroup}{P}-\frac{\beta_{R1}}{L}-\frac{\beta_{R2}}{L} – {L}\beta{\Bigg\rgroup}\Delta_{D} = 0 \quad\quad(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

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, \quad\quad\quad\quad \beta_{R1} = \beta_{R2} = \beta L^2 , and Eq. (g) becomes

So the determinant of the coefficient matrix in Eq. (h) (known as the characteristic equation) is
\quad\quad\quad\quad P^{2}\,-\,{\Bigg\lgroup}{\frac{17}{4}}\beta{L}{\Bigg\rgroup} P+{\frac{7}{2}}(\beta{L})^{2}\,=\,0\quad\quad (i)
Solving Eq. (i) using the quadratic formula results in two values of the critical load:
\quad\quad\quad\quad P_{cr1} = \beta{L}{\Bigg\lgroup}\frac{17-\sqrt{65}}{8}{\Bigg\rgroup} = 1.117 ~\beta{L} \\ \quad\quad\quad\quad P_{cr2} = \beta{L}{\Bigg\lgroup}\frac{17+\sqrt{65}}{8}{\Bigg\rgroup} = 3.13 ~\beta{L}
4. Finalize: These are the eigenvalues of the combined 2DOF system when elastic connection stiffnesses are defined as \beta_{R1} = \beta_{R2} = \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:
\quad\quad\quad\quad \Phi_{1}={\Bigg\lgroup}\begin{array}{l}{{0.653}}\\ {{1}}\end{array}{\Bigg\rgroup}~~\Phi_{2}={\Bigg\lgroup}\begin{array}{l}{{-0.153}}\\ {{1}}\end{array}{\Bigg\rgroup}
Application of combined model to Columns 1 and 2.
If the rotational spring stiffness \beta_{R2} goes to infinity and translational spring stiffness \beta 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).