Graphical reactions of three hinged arch

Principle: a three-hinged arch has the crown hinge $C$ where the bending moment is zero. If a single load $P$ acts on one half, the other half carries no load and is therefore a two-force member: the reaction at its support acts along the line joining the support to the hinge $C$.

Steps (load P on the left half)

1. Draw the arch to scale and mark the line of action of $P$.
2. The right half $CB$ is unloaded, so the reaction $R_B$ acts along the line $BC$. Extend the line $BC$ until it meets the line of action of $P$ at the point $E$.
3. The three forces $P$, $R_A$ and $R_B$ are in equilibrium, so they are concurrent. Hence $R_A$ passes through $A$ and $E$: join $AE$.
4. Draw the force triangle: draw $P$ to scale (a vertical line $ab$), then from $a$ draw a line parallel to $R_A$ ($AE$) and from $b$ a line parallel to $R_B$ ($BC$). They meet at $c$.
5. Read the reactions: $R_A = ca$ in magnitude and direction, $R_B = bc$. The horizontal components are equal and opposite, which gives the thrust $H$; the vertical components give $V_A$ and $V_B$.

   Space diagram (P on left half)        Force triangle
                  | P
        E *-------+---------           a
         /|  \                         |\
     C  / |    \                      P | \ R_B (parallel BC)
      .'  |     '.                    |  \
   A '    |       ' B                 b---c
   R_A along AE;  R_B along BC           R_A (parallel AE)

6. If $P$ lies on the right half, interchange the roles: $R_A$ acts along $AC$, extend it to meet $P$ at $E'$, and $R_B$ passes through $B$ and $E'$.
7. If several loads act, apply the method for the loads of each half separately and add the reactions vectorially.

Check: the vertical components satisfy $V_A + V_B = P$ and the moment of the reactions about any point equals that of $P$; with the numerical values, $H = \dfrac{M_C}{f}$ as in the analytical method.