ILD for BM at F and stress in support BD

Data and assumptions: beam $ABC$ with $AB = BC = 20$ m, supported at the ends $A$ and $C$ and at $B$ by two inclined members $BD$ and $BE$ ($D$ and $E$ are 12 m horizontally on either side of $B$ and 9 m below it). For the structure to be statically determinate an internal hinge at $B$ is assumed, with vertical supports at $A$ and $C$. Then $AB$ and $BC$ act as two simply supported beams that rest on the hinge $B$, which is held by the pair of inclined struts.

Member $BD$: horizontal 12 m, vertical 9 m, length $15$ m, $\sin\theta = 0.6$, $\cos\theta = 0.8$.

ILD for bending moment at F (5 m right of A)

Unit load at $x$ from $A$. For a load on $AB$ (which is simply supported over 20 m), $F$ is at $a = 5$ m and $b = 15$ m.

• $0\le x\le 5$: $M_F = R_B'\times 15 = 0.75x$ (where $R_B' = x/20$)
• $5\le x\le 20$: $M_F = R_A\times 5 = 0.25\,(20-x)$
• A load on $BC$ ($x>20$) is carried by $BC$, $C$ and the hinge, and does not affect $AB$: $M_F = 0$.

Peak ordinate under $F$: $\dfrac{5\times 15}{20} = 3.75$ m.

 A    F          B (hinge)         C
 o----+----------o-----------------o
 ILD M_F: 0 /\ 3.75 \_ 0 ________ 0

ILD for the force in the strut BD

A unit load on either span sends a vertical force $V_B$ to the hinge $B$: on $AB$, $V_B = \dfrac{x}{20}$; on $BC$, $V_B = \dfrac{40-x}{20}$. The ordinate of $V_B$ is 1 at $B$ and 0 at $A$ and $C$.

At the hinge $B$ no horizontal load comes from the beams, so horizontal equilibrium gives $F_{BD}\cos\theta = F_{BE}\cos\theta$, i.e. $F_{BD} = F_{BE}$. Vertical equilibrium:

$$2F_{BD}\sin\theta = V_B \Rightarrow F_{BD} = \frac{V_B}{2\times 0.6} = 0.833\,V_B$$

Both struts are in compression. The ILD for the force (and for the stress $=F_{BD}/A_{BD}$) is a triangle with the apex $0.833$ under $B$:

 ILD F_BD:   0 /\ 0.833 (at B) \ 0

Force in $BD$ for loads: $F_{BD} = 0.833\sum P\,y_B$, with $y_B$ the ordinate of the triangle ($y = x/20$ on $AB$).