Pin A at $x=0$; roller B at $x=12$ m; roller C at $x=21$ m
Internal hinge H at $x=9$ m
UDL $w=6$ kN/m over $0\le x\le12$ m
Point loads $18$ kN ↓ at $x=15$ and $x=18$ m
Find. Reactions and the SFD / BMD with extreme ordinates.
2(a): pin–roller–roller compound beam, one internal hinge at 9 m.
Approach. Cut at the hinge (moment there is zero); solve the left rigid link, carry the hinge shear into the right link, then finish by statics.
Left link A–H ($0\le x\le9$, carries $6\times9=54$ kN at $x=4.5$). Take moments about the hinge: $R_A(9)=54(4.5)\Rightarrow R_A=\dfrac{243}{9}=\boxed{27\text{ kN}\uparrow}$. Hinge shear $V_H=54-27=27$ kN.
Right link H–C carries the $27$ kN transferred down at H ($x=9$), the remaining UDL $6\times3=18$ kN at $x=10.5$, and $18{+}18$ kN at $x=15,18$. Moments about B ($x=12$): $R_C(9)=27(3)+18(1.5)-18(3)-18(6)$, i.e. $9R_C=54\Rightarrow R_C=\boxed{6\text{ kN}\uparrow}$.
Shear. $V=+27$ at A, falling under the UDL to $-27$ at the hinge and $-45$ just left of B; jump $+75$ to $+30$; constant to $x=15$, drop to $+12$, then $-6$ after $x=18$, closing at C. $\;V_{\max}=+30,\;V_{\min}=-45$ kN.
Moment. Sagging peak where $V=0$ at $x=4.5$: $M=27(4.5)-3(4.5)^2=\boxed{+60.75\text{ kN}\cdot\text{m}}$; zero at the hinge; hogging minimum over B, $M_B=6(9)-18(3)-18(6)=\boxed{-108\text{ kN}\cdot\text{m}}$; back to zero at C.
2(a): SFD and BMD. Sagging positive; hogging −108 kN·m over support B is the governing (minimum) ordinate.
Quantity
Value
$R_A,\,R_B,\,R_C$
27, 75, 6 kN (all ↑)
$V_{\max},\,V_{\min}$
+30 kN, −45 kN
$M_{\max}$ (sag)
+60.75 kN·m at $x=4.5$ m
$M_{\min}$ (hog)
−108 kN·m over B
2(b) — rigid Γ-bent (beam + hanging leg)
Given. Horizontal beam 0–9 m at eaves level, roller at the left end (vertical reaction), a 3 m leg dropping from the right end to a pin base; $w=6$ kN/m down on the beam; a $12$ kN horizontal force ($\rightarrow$) at the beam–column knee.
Find. Reactions, SFD/BMD of both members.
2(b): roller at A, pin at base B, 12 kN horizontal at knee J.
Approach. Only the base pin resists horizontal load; take global equilibrium.
Moments about B $(9,0)$: roller $R_A$ (at $x=0$, up), UDL $54$ kN at $x=4.5$, and the $12$ kN at height $3$ m: $-9R_A+54(4.5)-12(3)=0\Rightarrow R_A=\dfrac{207}{9}=\boxed{23\text{ kN}\uparrow}$.
Beam. $V(x)=23-6x$ ⇒ zero at $x=3.83$ m, giving $M_{\max}=23(3.83)-3(3.83)^2=\boxed{+44.1\text{ kN}\cdot\text{m}}$ (sag); at the knee $M_J=23(9)-3(9)^2=-36$ kN·m (hog).
Leg. Constant shear $12$ kN; moment $0$ at the pin base, rising linearly to $12(3)=\boxed{36\text{ kN}\cdot\text{m}}$ at the knee — equal and opposite to the beam-end moment, so the rigid joint balances ($\sum M_J=0$).
Quantity
Value
$R_A$ (roller)
23 kN ↑
$V_B,\,H_B$ (base pin)
31 kN ↑, 12 kN ←
Beam $M_{\max}$ / knee $M$
+44.1 (sag, at 3.83 m) / −36 (hog) kN·m
Leg $M$
0 at base → 36 kN·m at knee
2(c) — bent with an inclined roller
Check (reconstructed figure). The faint page-3 drawing is read as: pin at A$(0,0)$; vertical left column 7.2 m to B$(0,7.2)$; horizontal top beam B→C carrying $6$ kN/m over a 9.6 m span, C$(9.6,7.2)$; a roller on a 3:4 incline at C giving a reaction perpendicular to that surface. The method below is exact for this model; if the roller sits at a different node the numbers scale accordingly.
Given. $w=6$ kN/m over 9.6 m; pin A; inclined roller (normal $\mathbf n=(-\tfrac35,\tfrac45)$) at C.
Find. Reactions.
Resultant load. $W=6(9.6)=57.6$ kN ↓ at $x=4.8$ m.
Moments about A with roller force $R\mathbf n$ at C: $x_C(\tfrac45R)-y_C(-\tfrac35R)=W(4.8)$, i.e. $(9.6)(0.8R)+(7.2)(0.6R)=276.48\Rightarrow 12.0R=276.48\Rightarrow R=\boxed{23.0\text{ kN}}$.
Beam moment. The 9.6 m span under $6$ kN/m develops a free sagging moment up to $wL^2/8=6(9.6)^2/8=69.1$ kN·m, superposed on the axial thrust delivered by the inclined reaction.
