Top chord $U_1U_2$. Section just right of $U_1$ and take moments about $L_3$ (where the two web members meet): $M_{L_3}=45(8)-36(4)=216$; with lever arm $3$ m, $U_1U_2=-216/3=\boxed{72\ \text{kN (C)}}$.
Diagonals meeting at $L_3$. Vertical equilibrium of the cut and geometry (each diagonal rises $3$ over $4$, length $5$) give $U_1L_3=\boxed{15\ \text{kN (T)}}$ and $L_3U_3=\boxed{45\ \text{kN (T)}}$.
Member
Force
$U_1U_2$
72 kN C
$U_1L_3$
15 kN T
$L_3U_3$
45 kN T
Part (b)
King-post truss: 120 kN horizontal at the apex $U_1$, 60 kN at $L_1$ and $L_2$; pin at $L_3$, roller at $L_5$.
Given. Apex $U_1$ at $(8,6)$, mid-height joints $M_1(4,3),M_2(12,3)$, bottom chord at $4$ m spacing; loads $120$ kN horizontal (right) at $U_1$ and $60$ kN down at $L_1$ and $L_2$; pin at $L_3$, roller at $L_5$. Find. $L_2L_3,\;L_3L_4,\;M_1L_3$.
Reactions. $\sum M_{L_3}=0:\;8\,R_{L_5}=60(4)+60(8)-120(6)=0\Rightarrow R_{L_5}=0$. Then $R_{L_3}=120$ kN ↑ and the pin supplies $120$ kN horizontally (leftward) to balance the applied $120$ kN.
$L_2L_3$. Isolating the left of a vertical section through the left panel and taking moments about $M_1$ gives $L_2L_3=\boxed{80\ \text{kN (C)}}$.
$L_3L_4$. The right half carries no net vertical load ($R_{L_5}=0$), so $L_3L_4=\boxed{0}$.
$M_1L_3$. Vertical equilibrium at the left web (diagonal rises $3$ over $4$, length $5$) gives $M_1L_3=\boxed{50\ \text{kN (C)}}$.
