Question 4 of 8: Truss member forces (tension / compression)

Question 4: Truss member forces (tension / compression) (18 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

4(a) — Warren truss

Given. Bottom chord $L_1\!\ldots\!L_5$ at $6\text{ m}$ spacing (pin at $L_1$, roller at $L_5$); top nodes $U_1\!\ldots\!U_4$ at panel mid-points, height $4\text{ m}$; loads $24\text{ kN}\rightarrow$ at $U_1$ and $32,40,48\text{ kN}\downarrow$ at $L_2,L_3,L_4$. Find. $U_1U_2,\,L_2U_2,\,L_2L_3$.

1. Reactions. $\Sigma M_{L_1}=0:\;24R_{L_5}=32(6)+40(12)+48(18)+24(4)\Rightarrow R_{L_5}=68\text{ kN}\uparrow$; $R_{L_1}=52\text{ kN}\uparrow$ with horizontal $24\text{ kN}\leftarrow$.
2. Section & joints. A section just right of $L_2$ and method of joints give, in the panel of interest (chord slope $4$-in-$3$, diagonal length $5\text{ m}$): $$\boxed{U_1U_2=102\text{ kN (C)},\quad L_2U_2=25\text{ kN (C)},\quad L_2L_3=117\text{ kN (T)}}$$

4(b) — Wall-mounted panel truss

Given. $2\times3\text{ m}$ bays, $4\text{ m}$ high; pin at $U_1$ and horizontal roller at $L_1$ (both on the wall); two crossing diagonals $L_1\!-\!U_2$ and $U_1\!-\!L_3$ (uncoupled at the crossing); loads $20\text{ kN}\downarrow$ at $U_2$, $36\text{ kN}\rightarrow$ at $U_3$, $60\text{ kN}\downarrow$ at $L_2$ and $L_3$. Find. $L_1U_2,\,L_1L_2,\,U_1L_3$.

1. Reactions. Vertical only at the pin $U_1$: $U_{1y}=20+60+60=140\text{ kN}$; horizontals $U_{1x}=-186\text{ kN}$, $L_{1x}=+150\text{ kN}$ (they carry the $36\text{ kN}$ and the overturning couple).
2. Method of joints. Working from the loaded joints inward (diagonals length $5\text{ m}$ and $\sqrt{52}\text{ m}$): $$\boxed{L_1U_2=100\text{ kN (C)},\quad L_1L_2=90\text{ kN (C)},\quad U_1L_3=108.2\text{ kN (T)}}$$