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

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,

\left(P_{ Cr }\right)_{ AB }=\frac{\pi^2 E(I)_{ AB }}{\left(L_{ e }\right)_{ AB }^2}

=\frac{\pi^2 \times\left(200 \times 10^9\right) \times \frac{1}{12}\left(15 \times 10^{-3}\right)^4}{(0.7 \times 1)^2}           \text { (as } L_{ e }=0.7 L \text { ) }

=16.995 kN

Similarly,

\left(P_{ Cr }\right)_{ CD }=\frac{\pi^2 E(I)_{ CD }}{\left(L_{ e }\right)_{ CD }^2}

=\frac{\pi^2 \times\left(200 \times 10^9\right) \times \frac{1}{12}\left(15 \times 10^{-3}\right)^4}{(1.2)^2} \quad\left(\text { as } L_e=L\right)

=5.783 kN

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

\sum M_{ B }=0 \Rightarrow C_y \times 1-Q \cdot a=0

or                C_y=Q a

and              \sum F_y=0 \Rightarrow B_y+C_y-Q=0

or                  B_y=Q-C_y=Q-Q a=Q(1-a)

If we take C_y=\left(P_{ Cr }\right)_{ CD } , we get

Q × a = 5.783                   (1)

and B_y=\left(P_{ Cr }\right)_{ AB } gives

Q × (1 – a) = 16.995             (2)

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

\frac{a}{1-a}=\frac{5.783}{16.995}

or            \frac{1}{a}-1=\frac{16.995}{5.783}

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

Q=Q_{\max }=\frac{5.783}{0.254}=22.778  kN

Therefore, Q_{\max } = 22.778 kN and a = 0.254 m.