Truss: max forces under long UDL

Assumed truss (figure not available): Pratt truss, 5 panels of 4 m (span 20 m), height 4 m (end posts $L_0U_1$ and $L_5U_4$ at 45°). Verticals $U_1L_1\ldots U_4L_4$; diagonals $U_1L_2$, $U_2L_3$ and $U_4L_3$. Members: 2 = $U_1L_2$ (diagonal), 3 = $U_2L_2$ (vertical), 4 = $U_2L_3$ (diagonal). The UDL ($w = 10$ kN/m) is longer than the span and acts on the bottom chord, so it can cover any part of the span.

Reactions: $R_{L_0} = \dfrac{20-x}{20}$, $R_{L_5} = \dfrac{x}{20}$.

Method

Member forces are found by sections in the panel:

• Diagonals: $F = \pm V_{panel}/\sin 45^\circ$ with the panel shear from the reactions: $F_{U_1L_2}=\sqrt2\,V_{L_1L_2}$ and $F_{U_2L_3} = \sqrt2\,V_{L_2L_3}$.
• Vertical $U_2L_2$: at the unloaded joint $U_2$, $F_{U_2L_2} = -F_{U_2L_3}/\sqrt2$.

ILD ordinates for the unit load on the bottom chord (+ tension, − compression):

For a UDL longer than the span, the maximum effect of one kind is obtained by loading only the portions of the ILD of that sign. Force $= w\times$ (area of the ILD of that sign).

Maximum forces for 10 kN/m

• Member 2 ($U_1L_2$): maximum tension $=+63.64$ kN (load from the zero point at $x = 5.00$ m to the right support); maximum compression $=-7.07$ kN (load on $L_0$ to the zero point).
• Member 3 ($U_2L_2$): maximum tension $=+20.00$ kN (load from $L_0$ to the zero point at $x = 10$ m); maximum compression $=-20.00$ kN (load from the zero point to $L_5$). Here the sign changes at $x = 10$ m.
• Member 4 ($U_2L_3$): maximum tension $=+28.28$ kN (load from the zero point at $x = 10$ m, mid-span, to $L_5$); maximum compression $=-28.28$ kN (load from $L_0$ to mid-span).

Answer: member 2: 63.64 kN (T), 7.07 kN (C); member 3: 20.00 kN (T), 20.00 kN (C); member 4: 28.28 kN (T), 28.28 kN (C) (for 10 kN/m over the length that gives the maximum).