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## Q. 1.21

A rigid horizontal bar is sup-ported by three equally spaced bars of same cross-sectional area as shown in Fig. 1.31 (a). The length of middle bar is half of the outer bars. Find the ultimate load $P_u$ on the horizontal rigid bar if the load is applied at the center. Draw load-deflection curve.

## Verified Solution

Let $T_1 and T_2$= tension in outer and inner bars, respectively

Equilibrium condition: P = $2 \times T_1 + T_2$ (i)

(1) Elastic stage:

Compatibility condition: At any stage, the elonga-tions of all the bars are equal,

or, $\frac{T_1 L_1}{A E}=\frac{T_2 L_2}{A E}$(ii)

From Eqs (i) and (ii),

$P = 2 \times T_2= 4 \times T_1$ (iii)

Since $T_2$> $T_1$ , the middle bar will reach the yield stress earlier than the outer and the corresponding load P is called the yield load $P_y$. Thus at yield stage Eq. (iii) gives,

$P_y=2 T_2=2 \sigma_y A$

And load deflection at yield stage,

$\delta_y=\frac{T_2 L_2}{A E}=\frac{\sigma_y A L / 2}{A E}=\frac{\sigma_y L}{2 E}$

(2) Plastic stage:
At plastic stage stress in all three bars are at yiel stress and the corresponding load P is called th ultimate load $P_u$.

Equilibrium condition at plastic stage gives,
$P_u=2 \times T_1+T_2=2 \sigma_y A+\sigma_y A=3 \sigma_y A$

$\delta=\frac{\sigma_y L}{E}$