Question 2 of 8: Reactions, shear force and bending moment diagrams for three structures

Question 2: Reactions, shear force and bending moment diagrams for three 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.

Given. Three determinate structures, each of which contains one release or one change of direction that must be exploited before equilibrium can be written.

Find. All reaction components, and complete shear force and bending moment diagrams for each structure, labelled with the maximum positive and negative ordinates and with the sign of every segment.

Approach. Use the hinge first: the length of beam beyond it is a simply supported span in its own right, so solving that piece gives the force the hinge hands back to the rest of the structure, after which the remaining reactions follow from ordinary statics and the diagrams are drawn by integrating the load.

1. Part (a) — isolate the beam to the right of the hinge. The 8 m length from the hinge $H$ at $x = 12$ m to the roller $C$ at $x = 20$ m carries $wL = 2(8) = 16$ kN acting at its own mid-length. Taking moments about $H$, $$R_C(8) - 2(8)(4) = 0 \;\Rightarrow\; R_C = \frac{64}{8} = 8.0\ \text{kN} \ (\uparrow)$$ Vertical equilibrium of the same piece gives the hinge force, $V_H = 16 - 8.0 = 8.0$ kN, which the right-hand piece pushes down on the left-hand piece.
2. Transfer that force and solve the left-hand piece. The remaining structure runs from the free end at $x = 0$ to the hinge, carries 9.6 kN down at $x = 0$ and 8.0 kN down at $x = 12$ m, and is supported by the pin $A$ at $x = 2$ m and the roller $B$ at $x = 10$ m. Moments about $A$: $$9.6(2) + R_B(8) - 8.0(10) = 0 \;\Rightarrow\; R_B = \frac{80.0 - 19.2}{8} = 7.6\ \text{kN}\ (\uparrow)$$ $$\sum F_y = 0:\quad R_A = 9.6 + 8.0 - 7.6 = \boxed{10.0\ \text{kN}\ (\uparrow)}$$ with $R_B = 7.6$ kN and $R_C = 8.0$ kN, all upward. The three reactions sum to 25.6 kN, which equals the 9.6 kN point load plus the 16 kN of distributed load — the arithmetic check that should always be made.
3. Build the shear diagram by walking along the beam. Starting from the free end, the 9.6 kN load drops the shear to $-9.6$ kN, which it holds until $R_A$ lifts it to $-9.6 + 10.0 = +0.4$ kN. It stays there for the 8 m to $B$, where $R_B$ raises it to $+8.0$ kN, and it holds that value across the 2 m to the hinge because nothing is applied in between. Over the last 8 m the uniform load takes it down linearly at 2 kN per metre, so $V(x) = 8.0 - 2(x - 12)$, crossing zero at $x = 16$ m and reaching $-8.0$ kN at $C$, where $R_C$ closes the diagram. The largest ordinates are $\boxed{V_{max}^{+} = +8.0\ \text{kN}\ \text{and}\ V_{max}^{-} = -9.6\ \text{kN}}$
4. Integrate the shear to obtain the moments. Areas under the shear diagram accumulate as $$M(2) = -9.6(2) = -19.2\ \text{kN}\cdot\text{m}, \quad M(10) = -19.2 + 0.4(8) = -16.0\ \text{kN}\cdot\text{m}$$ $$M(12) = -16.0 + 8.0(2) = 0$$ which is the proof that the hinge has been located and used correctly: the moment there must vanish. Beyond the hinge $M(x) = 8(x - 12) - (x - 12)^2$, whose peak sits at the point of zero shear: $$M_{max}^{+} = 8(4) - 4^2 = \boxed{+16.0\ \text{kN}\cdot\text{m at } x = 16\ \text{m}}$$ The whole of the beam from the free end to the hinge is in hogging (negative, tension on top), with $M_{max}^{-} = -19.2$ kN.m over the pin, and the final 8 m span is entirely in sagging (positive, tension on the underside).

1. Part (b) — again start beyond the hinge. The 6 m length from the hinge at $x = 2$ m to the free end at $x = 8$ m carries $2(6) = 12$ kN at its centroid $x = 5$ m and is propped by the roller at $x = 6$ m. Moments about the hinge give $$R(4) - 12(3) = 0 \;\Rightarrow\; R = \boxed{9.0\ \text{kN}\ (\uparrow)}$$ and vertical equilibrium leaves $V_H = 12 - 9.0 = 3.0$ kN to be carried by the hinge, acting downward on the 2 m cantilever stub behind it.
2. Solve the stub to get the fixed-end reactions. The stub from the wall to the hinge carries nothing but that 3.0 kN at its tip, so $$\begin{aligned} V_{wall} &= 3.0\ \text{kN}\ (\uparrow), \\ M_{wall} &= 3.0(2) = \boxed{6.0\ \text{kN}\cdot\text{m (hogging)}} \end{aligned}$$ The wall therefore holds the beam with an upward force of 3.0 kN and a restraining moment of 6.0 kN.m; together with the 9.0 kN roller reaction these balance the 12 kN of applied load.
3. Shear. The shear is a constant $+3.0$ kN over the first 2 m, then falls linearly under the uniform load, $V(x) = 3.0 - 2(x - 2)$, reaching $-5.0$ kN just to the left of the roller and crossing zero at $x = 3.5$ m. The roller lifts it to $+4.0$ kN, and the last 2 m of uniform load brings it back to zero at the free end, as it must. The extreme ordinates are $+4.0$ kN and $-5.0$ kN.
4. Moment. Integrating, the moment rises from $-6.0$ kN.m at the wall to zero at the hinge, then $M(x) = 3.0(x - 2) - (x - 2)^2$ up to the roller: $$M_{max}^{+} = 3.0(1.5) - 1.5^2 = \boxed{+2.25\ \text{kN}\cdot\text{m at } x = 3.5\ \text{m}}$$ $$M(6) = 3.0(4) - 4^2 = -4.0\ \text{kN}\cdot\text{m}$$ The last value is confirmed independently by the overhang, $-2(2)(1) = -4.0$ kN.m. The greatest negative ordinate is the fixed-end value, $M_{max}^{-} = -6.0$ kN.m. The beam hogs from the wall to the hinge, sags between the hinge and $x = 5.0$ m (where the moment returns to zero), and hogs again from there to the free end.

1. Part (c) — resolve the bent structure. Place the origin at the pin $D$, with the top chord $BC$ at 4 m and the 20 kN horizontal load applied to the right leg 2 m above $D$. Horizontal equilibrium is immediate, because the roller at $A$ cannot supply a horizontal component: $$\sum F_x = 0:\quad 20 + D_x = 0 \;\Rightarrow\; D_x = \boxed{20\ \text{kN acting to the left}}$$
2. Take moments about the pin. The uniform load resultant is $2(10) = 20$ kN acting downward at mid-span, and the 20 kN horizontal load acts 2 m above $D$: $$\sum M_D = 0:\quad -10 A_y + 20(5) - 20(2) = 0 \;\Rightarrow\; A_y = \frac{100 - 40}{10} = \boxed{6.0\ \text{kN}\ (\uparrow)}$$ $$\sum F_y = 0:\quad D_y = 20 - 6.0 = 14.0\ \text{kN}\ (\uparrow)$$
3. Left leg AB. Nothing acts on this 2 m leg except the reaction that passes along it, so it carries a pure axial thrust of 6.0 kN compression with zero shear and zero bending moment everywhere. Candidates lose marks by forgetting to say so; a member with no diagram still needs the statement.
4. Top chord BC. The leg delivers 6.0 kN upward and no horizontal force to $B$, so measuring $s$ from $B$, $$\begin{aligned} V(s) &= 6.0 - 2s, \\ M(s) &= 6.0 s - s^2 \end{aligned}$$ The shear runs from $+6.0$ kN at $B$ to $-14.0$ kN at $C$ and vanishes at $s = 3.0$ m, where $$M_{max}^{+} = 6.0(3) - 3^2 = \boxed{+9.0\ \text{kN}\cdot\text{m}}$$ At the far end $M(10) = 60 - 100 = -40.0$ kN.m, the largest negative ordinate on this member. The chord sags over the first 6 m and hogs over the last 4 m, the change of sign occurring at $s = 6.0$ m.
5. Right leg CD. Working up from the pin, the 20 kN horizontal reaction is the only transverse force below the load point, so the shear there is a constant 20 kN and the moment grows linearly to $20(2) = 40.0$ kN.m at the load level. Above the load the two horizontal forces cancel, so the shear is zero and the moment stays at 40.0 kN.m all the way to $C$. That value matches the $-40.0$ kN.m carried in from the chord, which is the joint-equilibrium check at the knee. The leg also carries 14.0 kN of axial compression.