Truss ILDs and maximum forces under UDL

Geometry: equilateral triangles of side 5 m (all inclined members at 60°), height $h = 5\sin 60^\circ = 4.33$ m, span 15 m. $L_1$ hinge, $L_4$ roller. The load travels along the bottom chord (through truss), so each ILD is linear between panel points.

Reactions for a unit load at $x$ from $L_1$: $R_{L_1} = (15-x)/15$, $R_{L_4} = x/15$.

Method of sections

• $U_1U_2$: cut through $U_1U_2$, $U_1L_2$ and $L_1L_2$ and take moments about $L_2$. Top chord is in compression: $F_{U_1U_2} = -M_{L_2}/h$.
• $U_2L_2$: pass a section through $U_1U_2$, $U_2L_2$ and $L_2L_3$. The shear $V$ (left forces upward) in this panel is carried by the inclined member: $F_{U_2L_2} = -V/\sin 60^\circ = -1.155\,V$. For a load to the left of the section $V = R_{L_1}-1$; for a load to the right $V = R_{L_1}$.
• $L_2L_3$: moments about the joint $U_2$ gives $F_{L_2L_3} = M_{U_2}/h$ (tension).

Ordinates (per unit load; + tension, − compression)

Shapes:

• $U_1U_2$: always compressive, triangle with peak $-0.770$ under $L_2$.
• $U_2L_2$: tension for load left of $L_2$ (peak $+0.385$ at $L_2$), compression for load right of $L_2$ (peak $-0.385$ at $L_3$); sign change between $L_2$ and $L_3$.
• $L_2L_3$: tension, flat $+0.577$ between $L_2$ and $L_3$.

Maximum forces for 10 kN/m over 4 m

Force $= w\times(\text{area of ILD under the load})$, with the 4 m load moved to the position giving the greatest area.

Answer: $U_1U_2 = 26.69$ kN (compression); $U_2L_2 = +11.29$ kN (tension) or $-11.29$ kN (compression); $L_2L_3 = +23.09$ kN (tension).