Image two is the given example,how if the question 5.6 change to below?

Please please explain more detail in every process

5.6 Calculate the shear flow in the web of rib B. Exercise 5.6 Exercise 5.6 At the right is the skin/stringer structure surrounding the rib at section B. Since Pz=0, and, by symmetry, Ixz=0, Equation 5.2.4 reduces to P~x(i)=IzPyyiAi. From the data on the figure, we obtain P~x(1)=P~x(4)=50lb/in.P~x(2)=P~xr(3)=50lb/in. Applying Equation 5.2.3 to stringers 2, 3 and 4 yields q~(2)=q~(1)+P~x(2)q~(2)=q~(1)+50q~(3)=q~(2)+P~x(3)q~(3)=(q~(1)+50)+50q~(3)=q~(1)+100q~(4)=q~(3)+P~x(4)q~(4)=(q~(1)+100)50q(4)=q~(1)+50 Moment equivalence around flange 1 requires (10q~(2))10+(10q~(3))10=100010+150010 or 2q~(1)+150=250q~(1)=50 This together with Equations (a) means q~(1)=50lb/in.q~(3)=150lb/in.q~(2)=100lb/in.q~(4)=100lb/in. The directions of these shear flows are as shown in the figure above. They are reversed in direction on the free-body diagram of the rib at the right. To find the shear flow qw in the rib web, we draw a free-body diagram at the left, of rib stiffener 4-1. The net 87