Reactions, shear and bending-moment diagrams for three determinate structures (18 marks): 24-Bld-A1 Elementary Structural Analysis, December 2016, Q2

Question 2: Reactions, shear and bending-moment diagrams for three determinate structures (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.

(a) Simple beam with overhang

Given. Pin at A ($x=0$), roller at B ($x=12\text{ m}$), free tip C ($x=16\text{ m}$); a 60 kN point load at $x=6\text{ m}$ and a 6 kN/m UDL from $x=6\text{ m}$ to the tip.

Find. $R_A$, $R_B$, and the shear/moment diagrams with every segment's max/min ordinate.

Fig. Q2(a) — beam with 4 m overhang past roller B.

Approach. Two global equilibrium equations solve the two unknown reactions; shear and moment then follow segment by segment from the left.

Reactions. UDL resultant $=6\times10=60\text{ kN}$ at $x=11\text{ m}$. $\sum M_A=0$: $R_B(12)=60(6)+60(11)=1020 \Rightarrow R_B=85\text{ kN}$. $\sum F_y=0$: $R_A=60+60-85=\boxed{35\text{ kN}}$, $R_B=\boxed{85\text{ kN}}$ (both upward).
Shear diagram. $V=35\text{ kN}$ for $0
Bending moment diagram. $M(6)=35(6)=\boxed{+210\text{ kN}\cdot\text{m}}$ (peak sagging, under the point load). $M(12)=210+(-25)(6)-3(6)^2=\boxed{-48\text{ kN}\cdot\text{m}}$ (hogging, over support B). $M(16)=-48+24(4)-3(4)^2=0$, confirming the free tip.

Q2(a) – reactions and diagram extremes
Quantity	Value
$R_A$	35 kN ↑
$R_B$	85 kN ↑
$V_{\max}$ (span AB)	+35 kN (0–6 m)
$V_{\min}$ (span AB)	−61 kN (at B−)
$V_{\max}$ (overhang BC)	+24 kN (at B+)
$M_{\max}$ (sagging)	+210 kN·m at $x=6$ m
$M_{\min}$ (hogging)	−48 kN·m at B

(b) Propped cantilever with an internal hinge

Given. Fixed at A ($x=0$), internal hinge H at $x=4\text{ m}$, roller at $x=10\text{ m}$, free tip at $x=14\text{ m}$; UDL 6 kN/m over the full 14 m.

Find. $R_A$, $M_A$, and shear/moment diagrams for both spans.

Fig. Q2(b) — cantilever A–H carrying a suspended span H–roller–tip.

Approach. Split at the hinge (zero moment, shear-only transfer): solve the right sub-beam H–roller–tip on its own, then apply the hinge's reaction as a downward point load on the left cantilever A–H.

Right sub-beam (H to tip, 10 m, own UDL 60 kN, centroid at 5 m). $\sum M_H=0$: $R_{\text{roller}}(6)=60(5) \Rightarrow R_{\text{roller}}=50\text{ kN}$. $\sum F_y=0$: $R_H=60-50=\boxed{10\text{ kN}}$ — this is the downward force the hinge transmits onto the left cantilever's tip.
Left cantilever (A to H, 4 m, own UDL 24 kN plus the 10 kN hinge load at the tip). $R_A=24+10=\boxed{34\text{ kN}}$. $M_A=24(2)+10(4)=48+40=\boxed{88\text{ kN}\cdot\text{m}}$ (hogging).
Shear/moment by segment. A–H: $V$ falls linearly $34\to10$ kN; $M$ rises monotonically $-88\to0$ kN·m (zero at the hinge, as required). H–roller: $V$ falls $10\to-26$ kN, crossing zero at $x=4+1.67=5.67\text{ m}$ where $M=\boxed{+8.33\text{ kN}\cdot\text{m}}$ (local sagging peak); $M$ reaches $-48$ kN·m at the roller. Roller–tip: $V$ falls $24\to0$ kN; $M$ rises $-48\to0$ kN·m at the free tip.

Q2(b) – reactions and diagram extremes
Quantity	Value
$R_A$	34 kN ↑
$M_A$	88 kN·m (hogging)
$R_H$ (hinge shear transfer)	10 kN
$R_{\text{roller}}$	50 kN ↑
$M_{\min}$	−88 kN·m at A
local $M_{\max}$ (sagging)	+8.33 kN·m at $x=5.67$ m
$M$ at roller	−48 kN·m

(c) Three-hinge-type gable frame (pin–roller, rigid peak)

Given. Pin at A $(0,0)$, rafter to a rigid peak at $(12,9)$, rafter down to a roller at D $(15,5)$ (D sits 5 m above A). UDL 10 kN/m on the horizontal projection, full 15 m.

Find. $R_A$, $R_D$, and the shear/moment diagram along each rafter.

Fig. Q2(c) — asymmetric gable frame; UDL is per horizontal metre.

Approach. Resolve the horizontal-projection UDL into an equivalent load per metre of each inclined rafter ($w_{\text{eff}}=w\cos\theta$), then take global equilibrium (only vertical reactions arise since the total load is vertical and there is no other horizontal action) followed by a section cut along each rafter for shear and moment.

Equivalent rafter loads. Rafter A–Peak: length $15\text{ m}$ (9,12,15 triple), horizontal run 12 m $\Rightarrow$ total load $=10(12)=120\text{ kN}$, centroid at $x=6$. Rafter Peak–D: length 5 m, horizontal run 3 m $\Rightarrow$ total load $=10(3)=30\text{ kN}$, centroid at $x=13.5$.
Reactions. Total $W=150\text{ kN}$ at $\bar x=\dfrac{120(6)+30(13.5)}{150}=7.5\text{ m}$. $\sum M_A=0$: $R_D(15)=150(7.5)\Rightarrow R_D=\boxed{75\text{ kN}}$. $\sum F_y=0$: $R_A=\boxed{75\text{ kN}}$ (both vertical; $R_{Ax}=0$ since the load has no horizontal component and only one support resists horizontal thrust—confirmed by the frame solve).
Rafter A–Peak (local $x$ from A, $w_{\text{trans}}=8\times12/15=6.4\text{ kN/m}$ transverse component). $V(0)=+60\text{ kN}$, falling to $V(15)=-36\text{ kN}$ at the peak, crossing zero at $x=9.375\text{ m}$ where $M=\boxed{+281.25\text{ kN}\cdot\text{m}}$ — the governing design moment for the whole frame. $M(\text{peak})=+180\text{ kN}\cdot\text{m}$.
Rafter Peak–D. $V$ runs $-27\to-45\text{ kN}$ (one sign throughout, so no interior extremum): $M$ falls monotonically from $+180$ kN·m at the peak to $0$ at the roller D.

Q2(c) – reactions and diagram extremes
Quantity	Value
$R_A$	75 kN ↑ (vertical only)
$R_D$	75 kN ↑ (vertical only)
$M_{\max}$	+281.25 kN·m, 9.375 m up rafter A–Peak
$M$ at peak	+180 kN·m
$V$ range, A–Peak	+60 kN (A) to −36 kN (peak)
$V$ range, Peak–D	−27 kN (peak) to −45 kN (D)

Topic: Reactions and internal-force diagrams for statically determinate beams and frames.

Key relations: $\sum F_y=0$, $\sum M=0$ for reactions; $\dfrac{dV}{dx}=-w$, $\dfrac{dM}{dx}=V$ along any straight member; at a hinge $M=0$ and shear alone is transmitted; for an inclined member under a load specified per horizontal metre, resolve $w_{\text{eff}}=w\cos\theta$ along the member before integrating.

Why this works: Every additional support beyond the minimum three reactions of a simple beam is handled by the same two global equations as long as an internal release (hinge) supplies the missing condition; splitting the structure at that release turns one indeterminate-looking sketch into two ordinary free bodies solved in sequence.

Common pitfall: applying the horizontal-projection UDL directly as a per-metre-of-rafter load without the $\cos\theta$ correction, which overstates both the total load and the peak moment; and forgetting the shear sign change inside a segment, which is exactly where the true (interior) maximum moment sits — not at the segment ends.

Source: Hibbeler §4.2–4.3, §6.1–6.2 (shear/moment diagrams); Kassimali §5.1–5.3.

Question 2 of 8: Reactions, shear and bending-moment diagrams for three determinate structures (18 marks)

Question 2: Reactions, shear and bending-moment diagrams for three determinate structures (18 marks)

(a) Simple beam with overhang

(b) Propped cantilever with an internal hinge

(c) Three-hinge-type gable frame (pin–roller, rigid peak)