Beam with two 10 kN/m end UDLs, a central 60 kN point load, and supports A and B.
Given. Simple beam with supports A ($x=2$ m) and B ($x=12$ m); UDL $10$ kN/m over the left overhang ($0\!-\!2$ m) and right overhang ($12\!-\!16$ m); a $60$ kN downward point load at mid-span ($x=7$ m). Find. Reactions and SFD/BMD.
Resultants. Left UDL $=10(2)=20$ kN at $x=1$; right UDL $=10(4)=40$ kN at $x=14$; total load $=20+60+40=120$ kN.
Moment about A. $\sum M_A=0:\;10\,R_B=60(5)+40(12)-20(1)=760\Rightarrow R_B=\boxed{76\ \text{kN}\uparrow}$.
Shear. Starting from the free left end: $-20$ kN just left of A, jumps to $+24$ at A, holds to the load, drops to $-36$, holds to B, jumps to $+40$, and grades to $0$ at the right tip. $V_{\max}=+40,\;V_{\min}=-36$ kN.
Moment. $M=0$ at both free ends; $M_A=-20$ and $M_B=-80$ kN·m (hogging over the supports); the peak sagging value under the 60 kN load is $M(7)=44(5)-20(6)=+100$ kN·m.
Shear-force diagram (kN).
Bending-moment diagram (kN·m), plotted on the tension side.
Quantity
Value
$R_A,\ R_B$
44 kN ↑, 76 kN ↑
$V_{\max},\ V_{\min}$
+40, −36 kN
$M$ max sag (at load)
+100 kN·m
$M$ max hog (at B)
−80 kN·m
Part (b) — compound (Gerber) beam
Continuous beam with two internal hinges: rollers at A ($x=0$) and B ($x=10$), fixed end at $x=22$, hinges at $x=12$ and $x=20$, UDL 2 kN/m throughout.
Given. $2$ kN/m over the full $22$ m; rollers at A ($x=0$) and B ($x=10$), a fixed end at $x=22$, and internal hinges at $x=12$ and $x=20$. With $r=5$ and $c=2$ the beam is determinate. Find. Reactions and SFD/BMD.
Suspended span first. The floating span between the hinges (12–20 m, length 8 m) is simply supported on the two hinges; each hinge carries $2(8)/2=8$ kN. It therefore delivers $8$ kN down to the beam on each side.
Left segment (0–12 m). Loads: UDL $2(12)=24$ kN at $x=6$ plus $8$ kN at the hinge ($x=12$). $\sum M_A=0:\;10\,R_B=24(6)+8(12)=240\Rightarrow R_B=\boxed{24\ \text{kN}\uparrow}$, and $R_A=32-24=\boxed{8\ \text{kN}\uparrow}$.
Right (cantilever) segment (20–22 m). Loads: $8$ kN at $x=20$ plus UDL $2(2)=4$ kN at $x=21$. Vertical reaction $R_F=12$ kN ↑; fixing moment $M_F=8(2)+4(1)=\boxed{20\ \text{kN}\cdot ext{m (hogging)}}$.
Diagrams. Shear is limited to $\pm12$ kN. The bending moment sags to $+16$ kN·m at the low points of the two supported spans, is exactly zero at both hinges, and hogs to $-20$ kN·m over support B and at the fixed end.
Shear-force diagram (kN).
Bending-moment diagram (kN·m); note $M=0$ at both hinges.
Quantity
Value
$R_A,\ R_B,\ R_{fixed}$
8, 24, 12 kN ↑
Fixing moment $M_F$
20 kN·m (hog)
$V_{\max}/V_{\min}$
±12 kN
$M$ max sag / max hog
+16 / −20 kN·m
Part (c) — cross (T) frame
Horizontal beam A–J–B with a vertical stub at J carrying two opposed 10 kN horizontal loads.
Given. A horizontal beam, pin at A ($x=0$) and roller at B ($x=10$), with a vertical stub at the junction J ($x=6$) rising 2 m and dropping 3 m; a $10$ kN load acts left at the top of the stub and $10$ kN acts right at the bottom. Find. Reactions and SFD/BMD.
The two loads form a couple. They are equal, opposite and 5 m apart, so the net force is zero and the net moment is $M_J=10(5)=50$ kN·m (counter-clockwise). Hence $A_x=0$.
Reactions. With only a pure couple applied, $\sum M_A=0:\;10\,B_y+50=0\Rightarrow B_y=\boxed{5\ \text{kN}\downarrow}$ and $A_y=\boxed{5\ \text{kN}\uparrow}$.
Beam moments. From A, $M$ rises linearly to $+30$ kN·m just left of J ($5\times6$), then the applied couple steps it down to $-20$ kN·m just right of J, returning to zero at B.
Stub moments. The upper arm delivers $10(2)=20$ kN·m and the lower arm $10(3)=30$ kN·m at J; these balance the beam step ($30+20=20+30$). The largest moment anywhere is $\boxed{30\ \text{kN}\cdot ext{m}}$.
Check: $B_y$ comes out downward, so at B the beam tends to lift; the roller must be able to resist uplift (or be replaced by a pin/tie). This is flagged rather than hidden.
