Truss ILDs for AB, CD, EJ, FH

Assumed truss (figure not available): top joints $A, B, C, D, E, F, G$ (left to right) directly above bottom joints $N, M, L, K, J, I, H$, 6 panels of 3 m (18 m overall), height 4 m. Verticals at all joints; diagonals $AM$, $BL$, $CK$, $DJ$, $EI$, $FH$ (all sloping down to the right). Supports as given: $N$ hinge and $I$ roller ($x = 15$ m), so the last panel $IH$ (members $FG$, $GH$, $FH$, $IH$) overhangs 3 m beyond the roller. The loads are not given in the text, so the ILDs are drawn for a unit load on the bottom chord at $x$ from $N$:

$$R_N = \frac{15-x}{15}, \qquad R_I = \frac{x}{15} \qquad (0\le x\le 18)$$

For a load at $H$: $R_N = -0.2$, $R_I = 1.2$. Diagonal length $\sqrt{3^2+4^2} = 5$ m, $\sin\theta = 0.8$.

Method

• AB (top chord, panel $NM$): section through $AB$, $AM$, $NM$; moments about $M$ ($x=3$, where $AM$ and $NM$ meet), lever arm 4 m: for a load at or right of $M$, $F_{AB} = -\dfrac{3R_N}{4} = -0.75R_N$; for a load at $N$, 0.
• CD (top chord, panel $LK$): section through $CD$, $CK$, $LK$; moments about $K$ ($x=9$): load at or right of $K$: $F_{CD} = -\dfrac{9R_N}{4}$; load at or left of $L$ (right part, only $R_I$ at 6 m from $K$): $F_{CD} = -\dfrac{6R_I}{4} = -1.5R_I$.
• EJ (vertical): at the unloaded top joint $E$, $F_{EJ} = -0.8F_{EI}$, and $F_{EI} = V/0.8$ from the shear $V$ in panel $JI$, so $F_{EJ} = -V$. For a load at or left of $J$: $V = R_N - 1$, so $F_{EJ} = 1 - R_N = \dfrac{x}{15}$; for a load at or right of $I$: $F_{EJ} = -R_N$ (0 at $I$, $+0.2$ at $H$).
• FH (diagonal, overhang panel $IH$): joint $G$ is unloaded and has only the two non-collinear members $FG$ and $GH$, so $F_{FG} = F_{GH} = 0$. At joint $H$ the remaining members are $IH$ and $FH$. For a load left of $I$, $H$ is unloaded and $F_{FH} = 0$. For the load at $H$: $0.8\,F_{FH} = 1$, so $F_{FH} = +1.25$ (tension).

Ordinates (per unit load on the bottom chord; + tension, − compression)

Shapes

• AB: 0 at $N$, compression peak $-0.6$ at $M$, decreasing linearly to 0 at the roller $I$, then tension on the overhang ($+0.15$ at $H$).
• CD: compression, a triangle with peak $-0.9$ under $K$, zero at $I$, then tension on the overhang ($+0.45$ at $H$).
• EJ: tension, rising linearly to $+0.8$ at $J$, falling to 0 at $I$, then rising again to $+0.2$ at $H$.
• FH: zero from $N$ to $I$, then rising linearly to $+1.25$ (tension) at $H$.

The force in any member for the actual loads is $F = \sum P\,y$ (concentrated loads) or $w\times$ area (UDL), using these ordinates.