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

Load deflection at plastic stage

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

Figure 1.31(b) shows load-deflection curve.