Question 2 of 8: Reactions, shear and bending-moment diagrams

Question 2: Reactions, shear and bending-moment diagrams (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.

2(a) Simply-supported beam with overhang

Given. Pin at $A$ ($x{=}0$), roller at $B$ ($x{=}6\text{ m}$), overhang to $x{=}8\text{ m}$; UDL $12\text{ kN/m}$ over $A$–$B$; point load $30\text{ kN}$ at the free tip. Find. Reactions and SFD/BMD extremes.

1. Reactions. $\sum M_A=0:\;R_B(6)=12(6)(3)+30(8)=216+240=456\Rightarrow R_B=\boxed{76\text{ kN}}$; then $R_A=12(6)+30-76=\boxed{26\text{ kN}}$.
2. Shear. $V=+26$ at $A$, falling with the UDL to $26-12(6)=-46$ just left of $B$; the reaction lifts it to $-46+76=+30$, constant across the overhang, and the $30\text{ kN}$ tip load returns it to $0$. Extremes $V_{\max}=\boxed{+30\text{ kN}}$, $V_{\min}=\boxed{-46\text{ kN}}$.
3. Moment. $V=0$ at $x=26/12=2.17\text{ m}$, giving peak sagging $M=26(2.17)-6(2.17)^2=\boxed{+28.2\text{ kN}\cdot\text{m}}$. Over the roller $M(6)=26(6)-12(6)(3)=-60\text{ kN}\cdot\text{m}$ (check from the overhang: $-30(2)=-60$), the peak hogging $\boxed{-60\text{ kN}\cdot\text{m}}$, returning to $0$ at the tip.

2(b) L-frame with internal hinge

Given. Column fixed at base, height $3\text{ m}$, horizontal UDL $4\text{ kN/m}$; rigid corner into a horizontal beam carrying $8\text{ kN/m}$ down; internal hinge $1.5\text{ m}$ from the corner; roller $6\text{ m}$ beyond the hinge (i.e. $7.5\text{ m}$ from the corner). Find. Reactions and SFD/BMD. The hinge makes the frame determinate ($r=4$, $c=1$).

1. Hinge release fixes the roller. Taking moments about the hinge for the beam to its right (length $6\text{ m}$, UDL $8$): $R_C(6)=8(6)(3)\Rightarrow R_C=\boxed{24\text{ kN}}$.
2. Base reactions. $\sum F_y:\;V_A=8(7.5)-24=\boxed{36\text{ kN}}$; $\sum F_x:\;H_A=4(3)=\boxed{12\text{ kN}}$ (leftward, resisting the wind); $\sum M_A:\;M_A=8(7.5)(3.75)-24(7.5)+4(3)(1.5)=\boxed{63\text{ kN}\cdot\text{m}}$.
3. Beam BMD. With corner moment and $M=0$ at the hinge, $M(x)=-45+36x-4x^2$ (x from corner). Corner (hogging) $\boxed{-45\text{ kN}\cdot\text{m}}$; $V=0$ at $x=4.5\text{ m}$ giving peak sag $M=-45+36(4.5)-4(4.5)^2=\boxed{+36\text{ kN}\cdot\text{m}}$; zero at hinge and roller. Beam shear runs $+36$ (corner) to $-24$ (roller).
4. Column BMD. Parabolic under the $4\text{ kN/m}$: base $\boxed{+63\text{ kN}\cdot\text{m}}$ (max), corner $-45\text{ kN}\cdot\text{m}$; column shear $12\to 0$.

2(c) Gable (bent simply-supported member)

Given. Pin at $A(0,0)$, roller at $B(12.5,0)$, apex $(8,6)$. Two loads perpendicular to the left rafter (length $10\text{ m}$): $15\text{ kN}$ at $3\text{ m}$ and $16\text{ kN}$ at $5\text{ m}$ from $A$. Find. Reactions and BMD/SFD. Each perpendicular load resolves as $(0.6,-0.8)\times P$.

1. Reactions. Load components: $15\to(9,-12)$, $16\to(9.6,-12.8)$. $\sum F_x:\;A_x=-(9+9.6)=\boxed{18.6\text{ kN}}$ (leftward). $\sum M_A=0\Rightarrow B_y=\boxed{10\text{ kN}}$; $\sum F_y:\;A_y=24.8-10=\boxed{14.8\text{ kN}}$.
2. Moments (from the load-free side). At the apex, only $B_y$ acts to the right: $M_{apex}=10(12.5-8)=\boxed{45\text{ kN}\cdot\text{m}}$. At the $16\text{ kN}$ point ($5\text{ m}$): $M=10(12.5-4.0)=\boxed{85\text{ kN}\cdot\text{m}}$ (the peak). At the $15\text{ kN}$ point ($3\text{ m}$): $M=69\text{ kN}\cdot\text{m}$; zero at both supports. All sagging (tension on the inner face).
3. Shear (normal to axis). Left rafter: $+23\text{ kN}$ ($A$ to $3\text{ m}$), $+8$ ($3$–$5\text{ m}$), $-8$ ($5\text{ m}$–apex); right rafter constant $+6\text{ kN}$.