The horizontal bar as shown in Figure 8.11(a) is supported by two columns AB and CD. Each column is pin-connected at top with the horizontal bar. Both columns are solid steel bars (E = 200 GPa) with square cross-section of each side 15 mm. If the distance ‘a’ as shown is varied from 0 to 1 m, what

Question

The horizontal bar as shown in Figure 8.11(a) is supported by two columns AB and CD. Each column is pin-connected at top with the horizontal bar. Both columns are solid steel bars (E = 200 GPa) with square cross-section of each side 15 mm. If the distance ‘a’ as shown is varied from 0 to 1 m, what is the maximum value of load Q? What is the corresponding value of a?

Accepted Answer

Column AB is having one end fixed and one end pinned. In such case, it can be shown that effective length is 0.7 times the length. Refer to the result of Example 8.6. Therefore,

[latex] \left(P_{ Cr } ight)_{ AB }=\frac{\pi^2 E(I)_{ AB }}{\left(L_{ c } ight)_{ AB }^2} [/latex]

[latex] \begin{aligned} & =\frac{\pi^2 imes\left(200 imes 10^9 ight) imes \frac{1}{12}\left(15 imes 10^{-3} ight)^4}{(0.7 imes 1)^2} \quad\left( ext { as } L_{ c }=0.7 L ight) \ & =16.995 kN \end{aligned} [/latex]

Similarly,

[latex] \begin{aligned} \left(P_{ Cr } ight)_{ CD } & =\frac{\pi^2 E(I)_{ CD }}{\left(L_{ e } ight)_{ CD }^2} \ & =\frac{\pi^2 imes\left(200 imes 10^9 ight) imes \frac{1}{12}\left(15 imes 10^{-3} ight)^4}{(1.2)^2} \quad\left( ext { as } L_{ e }=L ight) \ & =5.783 kN \end{aligned} [/latex]

Now, if we consider free-body diagram of the horizontal member EBC as shown in Figure 8.11(b) above, we get for equilibrium,

[latex] \sum M_{ B }=0 \Rightarrow C_y imes 1-Q \cdot a=0 [/latex]

or [latex] C_y=Q a [/latex]

and [latex] \sum F_y=0 \Rightarrow B_y+C_y-Q=0 [/latex]

or [latex] B_y=Q-C_y=Q-Q a=Q(1-a) [/latex]

If we take [latex] C_y=\left(P_{ Cr } ight)_{ CD } [/latex], we get

Q × a = 5.783 (1)

and [latex] B_y=\left(P_{ Cr } ight)_{ AB } [/latex] gives

Q × (1 - a) = 16.995 (2)

Dividing Eq. (1) by Eq. (2), we can write

[latex] \frac{a}{1-a}=\frac{5.783}{16.995} [/latex]

or [latex] \frac{1}{a}-1=\frac{16.995}{5.783} [/latex]

which gives a = 0.254 m.
Now, maximum value of Q will be obtained when both the compression members AB and CD reach their criticality simultaneously. So, a = 0.254 m at that condition. The corresponding Q is

[latex] Q=Q_{\max }=\frac{5.783}{0.254}=22.778 kN [/latex]

Therefore, [latex] Q_{\max }=22.778 kN ext { and } a=0.254 m [/latex]

Question 8.4: The horizontal bar as shown in Figure 8.11(a) is supported b...

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