Two idealized columns are shown in Fig. 11-5. Both columns are initially straight and vertical with load P applied at joint A. (a) The first column (Column 1, Fig. 11-5a) consists of a rigid bar ABC with a roller support at B and is connected to bar CD by an elastic connection with rotational

Question

Two idealized columns are shown in Fig. 11-5. Both columns are initially straight and vertical with load P applied at joint A.

(a) The first column (Column 1, Fig. 11-5a) consists of a rigid bar ABC with a roller support at B and is connected to bar CD by an elastic connection with rotational stiffness [latex]\beta_{R} = \beta L^{2}[/latex].

(b) The second column (Column 2, Fig. 11-5b) is a single rigid bar ABCD that has a roller support at B and a spring support at D.

Column 1 has a sliding support at D that cannot rotate, while Column 2 has a roller support at D with a translational spring with stiffness β. Find an expression for the critical load [latex]P_{cr}[/latex] for each column.

Accepted Answer

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 [latex]\Delta_{A}[/latex] (Fig. 11-5a). Sum lateral forces for the entire column to find that reaction [latex]H_{B} = 0[/latex]. Next, draw the free-body diagram of bar ABC (Fig. 11-5c). Note that [latex]\Delta_{C} = \Delta_{A}[/latex] and that bar ABC rotation is [latex]\Delta_{A} /(L/2) = 2\Delta_{A} /L[/latex].

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

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 [latex]P_{cr1}[/latex] for Column 1:

[latex]P_{ cr 1}=\frac{M_{C}}{2 \Delta_{A}}=\frac{\beta_{R}\left(\frac{2 \Delta_{A}}{L} ight)}{2 \Delta_{A}}=\frac{\beta_{R}}{L}=\beta L[/latex] (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 [latex]\Delta_{A}[/latex], 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

[latex]H_{B}=-(\beta)\left(3 \Delta_{A} ight)[/latex] (b)

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

[latex]P_{ cr 2}=\frac{-H_{B}\left(\frac{3 L}{2} ight)}{4 \Delta_{A}}=\frac{9}{8} \beta L[/latex] (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 [latex]\beta_{R1} and \beta_{R2}[/latex], 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 [latex]\Delta_{C} and \Delta_{D}[/latex] are selected as the degrees of freedom that define the possible displaced positions of the column. (Alternatively, rotation angles such as [latex] heta_{B} and heta_{D}[/latex] 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 [latex]\Delta_{A}[/latex] is needed to define the buckled shape of each structure depicted in Figs. 11-5a and b.

2. Categorize: Observe that if rotational spring [latex]\beta_{R2}[/latex] becomes infinitely stiff and translational spring stiffness [latex]\beta = 0[/latex] in the combined structure (Fig. 11-5e) (while [latex]\beta_{R1}[/latex] 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 [latex]\beta_{R1}[/latex] goes to infinity and spring [latex]\beta_{R2} = 0[/latex] 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 [latex]P_{cr}[/latex] 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 [latex]P_{cr}[/latex] 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 [latex]H_{B} = \beta \Delta_{D}[/latex]. Rotation angles [latex] heta_{B} and heta_{D}[/latex] in Fig. 11-5f can be expressed in terms of translations [latex]\Delta_{C} and \Delta_{D}[/latex] as

[latex] heta_{D}=\frac{\Delta_{D}-\Delta_{C}}{L} \quad heta_{B}=\frac{2}{L} \Delta_{C}[/latex] (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 [latex]\beta_{R1}[/latex] times the relative rotation [latex]( heta_{D} - heta_{B} )[/latex] at C, to get

[latex]2 P \Delta_{C}-\beta \Delta_{D} \frac{L}{2}+\beta_{R 1}\left( heta_{D}- heta_{B} ight)=0[/latex]

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

[latex]\left(2 P-\frac{3 \beta_{R 1}}{L} ight) \Delta_{C}+\left(\frac{\beta_{R I}}{L}-\frac{L \beta}{2} ight) \Delta_{D}=0[/latex] (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

[latex]P\left(\Delta_{D}-\Delta_{C} ight)-\beta_{R 1}\left( heta_{D}- heta_{B} ight)-\beta_{R 2} heta_{D}-\beta \Delta_{D} L=0[/latex]

which can be rewritten as

[latex]\left(\frac{3 \beta_{R 1}}{L}-P+\frac{\beta_{R 2}}{L} ight) \Delta_{C}+\left(P-\frac{\beta_{R 1}}{L}-\frac{\beta_{R 2}}{L}-L \beta ight) \Delta_{D}=0[/latex] (f)

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

[latex]\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} ight)\left(\begin{array}{l}\Delta_{C} \\Delta_{D}\end{array} ight)=\left(\begin{array}{l}0 \0\end{array} ight)[/latex] (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, [latex]\beta_{R 1}=\beta_{R 2}=\beta L^{2}[/latex] , and Eq. (g) becomes

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

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

[latex]P^{2}-\left(\frac{17}{4} \beta L ight) P+\frac{7}{2}(\beta L)^{2}=0[/latex] (i)

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

[latex]P_{ cr 1}=\beta L\left(\frac{17-\sqrt{65}}{8} ight)=1.117 \beta L[/latex]

[latex]P_{ cr 2}=\beta L\left(\frac{17+\sqrt{65}}{8} ight)=3.13 \beta L[/latex]

4. Finalize: These are the eigenvalues of the combined 2DOF system when elastic connection stiffnesses are defined as [latex]\beta_{R 1}=\beta_{R 2}=\beta L^{2}[/latex] . Usually the lower value of the critical load is of more interest because the structure will buckle first at this lower load value. Substitute [latex]P_{cr1} and P_{cr2}[/latex] 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:

[latex]\Phi_{1}=\left(\begin{array}{c}0.653 \1\end{array} ight) \quad\Phi_{2}=\left(\begin{array}{c}-0.153 \1\end{array} ight)[/latex]

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

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

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