For the following concept application problem, how is it that P$1=0,$ P$2=$ P$3=600$ kN $,$ and P$4=900$kN$?$ I would like drawings and the work shown so that I can understand the balances of forces. The value of P$2$ and P$3$ is particularly confusing. I understand the rest. Its the superposition method and cut and force balances that are confusing. Need it to be solved with superposition.

Concept Application $2\mathrm{.}4$

Determine the reactions at A and B for the steel bar and loading

shown in Fig. $2\mathrm{.}23$a$,$ assuming a close fit at both supports before

the loads are applied.

We consider the reaction at B as redundant and release the bar

from that support. The reaction R$\_($B$)$ is considered to be an unknown

load and is determined from the condition that the deformation $\backslash$delta of

the bar equals zero.

The solution is carried out by considering the deformation $\backslash$delta $\_($L$)$

caused by the given loads and the deformation $\backslash$delta $\_($R$)$ due to the redun$-$

dant reaction R$\_($B$)($Fig$.2\mathrm{.}23$b$).$

The deformation $\backslash$delta $\_($L$)$ is obtained from Eq$.(2\mathrm{.}10)$ after the bar has

been divided into four portions, as shown in Fig. $2\mathrm{.}23$c$.$ Follow the

same procedure as in Concept Application $2\mathrm{.}1$:

P$\_(1)=0,$P$\_(2)=$P$\_(3)=600\backslash$times $10^(3)$N$,$P$\_(4)=900\backslash$times $10^(3)$N

A$\_(1)=$A$\_(2)=400\backslash$times $10^(-6)$m$^(2),$A$\_(3)=$A$\_(4)=250\backslash$times $10^(-6)$m$^(2)$

L$\_(1)=$L$\_(2)=$L$\_(3)=$L$\_(4)=0\mathrm{.}150$m

Substituting these values into Eq$.(2\mathrm{.}10),$