Check (reconstructed figure). Adopted geometry: pin at A$(0,0)$, crown hinge C$(8,6)$, pin at E$(18,1)$ (right base 1 m above the left); an $80$ kN vertical load at the crown; a wind/pressure load on the right rafter taken as $24$ kN/m over its 5 m rise, i.e. a $120$ kN horizontal resultant ($\leftarrow$). The three-hinge statics is exact for this model.
Given. Determinate three-hinged frame ($r=4$, one crown condition).
6 (adopted): three-hinged gable; 80 kN at crown, horizontal wind on the right rafter.
Approach. Four reaction components from three global equations plus the crown condition $M_C=0$ (moment of the forces on one side of C vanishes).
Crown condition (right side). Taking moments about C of the E-reaction and the wind gives one equation relating $E_x,E_y$.
Global equilibrium. Solving with $\sum F_x,\sum F_y,\sum M_A$ yields $A=(A_x,A_y)=(\boxed{88,\,66})$ kN and $E=(E_x,E_y)=(\boxed{32,\,14})$ kN; resultants $|A|=110$ kN, $|E|=34.9$ kN.
Left rafter is a two-force member. The support reaction at A acts along A–C (direction $0.8,0.6$), so the left rafter carries pure axial force — $\boxed{M=0}$ along its whole length.
Right rafter. Moment is zero at the crown hinge and at the pin E; the wind produces a mid-rafter peak $M_{\max}\approx\boxed{+75\text{ kN}\cdot\text{m}}$ (at $x\approx13$ m). This is the governing bending ordinate of the frame.
