07-Str-A1 · December 2015

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Reactions, shear and bending moment 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 plane structures, dimensions and loads read from the examination figures.

Find. The reactions, the shear force and bending moment diagrams for each structure, with the maximum and minimum ordinates and the sign of every segment.

2(a) — Gerber beam with an internal hinge

Approach. The hinge supplies the fourth equation: take moments about it for the length to its left, then apply global vertical equilibrium and a global moment equation for the two remaining reactions.

1. Reaction at the pin, from the hinge condition. The length from the pin to the hinge carries only the first 9 m of the uniform load, $6 \times 9 = 54$ kN acting at 4.5 m from the pin, and the moment of everything on that side about the hinge must vanish: $$ \sum M_{\text{hinge, left}} = 0 : \qquad R_A(9) - 54(4.5) = 0 \quad\Longrightarrow\quad \boxed{R_A = 27\ \text{kN} \uparrow} $$
2. Global vertical equilibrium. The total load is the uniform load plus the two point loads, $$ \sum F_y = 0 : \qquad R_A + R_B + R_C = 6(12) + 18 + 18 = 108\ \text{kN}, $$ so $R_B + R_C = 81$ kN.
3. Global moment about the pin. Taking moments about the support at 0 m, with the uniform load resultant 72 kN at 6 m, $$ 12 R_B + 21 R_C = 72(6) + 18(15) + 18(18) = 1026\ \text{kN}\cdot\text{m}. $$ Solving with $R_B + R_C = 81$ gives $$ \boxed{R_B = 75\ \text{kN} \uparrow, \qquad R_C = 6\ \text{kN} \uparrow} $$ The small reaction at the far support is the signature of this arrangement: the right-hand span carries its own two point loads, but the 3 m overhang beyond the hinge levers most of that load back onto the 12 m support.
4. Shear force. Working from the left, the shear starts at $+27$ kN and falls at 6 kN per metre, passing through zero at $x = 27/6 = 4.5$ m and reaching $$ V(12^-) = 27 - 6(12) = -45\ \text{kN} $$ just left of the roller. The 75 kN reaction lifts it to $+30$ kN, each 18 kN load steps it down, and the final $-6$ kN is closed by $R_C$: $$ V = +30 \to +12 \to -6\ \text{kN}. $$ The maximum ordinate is $+30$ kN and the minimum is $-45$ kN.
5. Bending moment. The moment is zero at the pin, rises to its peak where the shear vanishes, $$ M_{\max} = 27(4.5) - \tfrac{1}{2}(6)(4.5)^2 = \boxed{+60.75\ \text{kN}\cdot\text{m}} $$ and returns to zero at the hinge, as it must. Beyond the hinge the beam hogs over the roller, $$ M(12) = 27(12) - \tfrac{1}{2}(6)(12)^2 = 324 - 432 = \boxed{-108\ \text{kN}\cdot\text{m}} $$ which is the minimum ordinate. In the right-hand span the moment climbs at 30 kN per metre to $-18$ kN·m under the first 18 kN load, crosses zero at $x = 16.5$ m, reaches $+18$ kN·m under the second load and falls to zero at the end support.
6. Sign of each segment. The beam sags (positive) from the pin to the hinge and again from 16.5 m to the right-hand support; it hogs (negative) from the hinge to 16.5 m. Both changes of sign are points of contraflexure and are marked on the diagram.

2(b) — Bent with a horizontal load at the corner

Approach. Take moments about the pin at the foot of the leg — the 12 kN horizontal load then contributes through its 3 m lever arm — and plot the two members on one developed abscissa so that the continuity of moment round the rigid corner is visible.

1. Reactions. Horizontal equilibrium is settled by the pin alone, so the pin pushes back 12 kN to the left. Moments about the pin, with the 54 kN load resultant 4.5 m from the roller and the 12 kN load 3 m above the pin, give $$ 9 R_A = 54(4.5) - 12(3) = 243 - 36 = 207 \quad\Longrightarrow\quad \boxed{R_A = 23\ \text{kN} \uparrow} $$ and vertical equilibrium then gives $$ \boxed{D_y = 54 - 23 = 31\ \text{kN} \uparrow, \qquad D_x = 12\ \text{kN} \leftarrow} $$
2. Shear in the horizontal member. $V(x) = 23 - 6x$, so the shear runs from $+23$ kN at the roller to $23 - 54 = -31$ kN just left of the corner, crossing zero at $$ x_0 = \frac{23}{6} = 3.833\ \text{m}. $$
3. Bending moment in the horizontal member. At the point of zero shear, $$ M_{\max} = \frac{R_A^{\,2}}{2w} = \frac{23^2}{2(6)} = \boxed{+44.083\ \text{kN}\cdot\text{m}} $$ a sagging (positive) peak, and at the corner $$ M(9) = 23(9) - \tfrac{1}{2}(6)(9)^2 = 207 - 243 = \boxed{-36\ \text{kN}\cdot\text{m}} $$ a hogging (negative) value. The beam changes sign at $x = 2 \times 3.833 = 7.667$ m.
4. The leg. Below the corner the only forces are the two pin reactions, so the leg carries a constant shear of 12 kN and a moment that varies linearly with the height $y$ above the pin, $$ M(y) = -12\,y, $$ zero at the pin and $-36$ kN·m at the corner — equal to the moment delivered by the horizontal member, as a rigid corner with no applied couple requires. The leg also carries 31 kN of axial compression.
5. Extreme ordinates. Over the whole bent the shear ranges from $+23$ kN to $-31$ kN, and the bending moment from $+44.083$ kN·m in the span to $-36$ kN·m at the corner. The horizontal member sags over its first 7.667 m and hogs thereafter; the leg hogs throughout, with tension on its outer (right-hand) face.

2(c) — Inclined member with a load on its horizontal projection

Approach. Because the roller bed is ruled parallel to the member, its reaction is normal to the member; its lever arm about the pin is then exactly the length A to B. Take moments about the pin for the reaction, resolve the internal actions along and normal to the member, and note that the peak moment sits under the point of zero shear.

1. Geometry. The member rises 3 in 4, so its unit vector is $(0.8,\ 0.6)$ and the outward normal is $(-0.6,\ 0.8)$. The roller sits at a horizontal offset of 9.6 m, that is $$ L_{AB} = \sqrt{9.6^2 + 7.2^2} = 12.0\ \text{m}, \qquad L_{AC} = \sqrt{12^2 + 9^2} = 15.0\ \text{m}. $$ The whole 12 m projection carries the load, so the resultant is $6 \times 12 = 72$ kN acting downwards at a horizontal distance of 6 m from the pin.
2. Reaction at the roller. Because the reaction is normal to AB, its lever arm about the pin is the member length $L_{AB} = 12$ m, and $$ \sum M_A = 0 : \qquad R_B (12) = 72(6) \quad\Longrightarrow\quad \boxed{R_B = 36\ \text{kN}} $$ directed normal to the member, up and to the left. Its components are $(-21.6,\ +28.8)$ kN.
3. Reaction at the pin. Global equilibrium then gives $$ A_x = 21.6\ \text{kN} \rightarrow, \qquad A_y = 72 - 28.8 = 43.2\ \text{kN} \uparrow . $$ Read instead as a vertical roller, the same figure yields $R_B = 45$ kN vertical with $A_x = 0$ and $A_y = 27$ kN. The two readings differ by a force of $27(0.8,\,0.6)$ acting along the member, which produces neither shear nor moment anywhere — hence the identical diagrams noted at the head of this paper.
4. Shear force. Resolving the forces to the left of a section at horizontal coordinate $x$ onto the normal direction, $$ V(x) = 21.6 - 4.8\,x \quad (x < 9.6\ \text{m}), $$ so the shear starts at $+21.6$ kN, passes through zero at $x = 4.5$ m — that is 5.625 m along the member — and reaches $-24.48$ kN just below the roller. The 36 kN reaction lifts it to $+11.52$ kN, from which it decays linearly to zero at the free end. The extreme ordinates are $+21.6$ kN and $-24.48$ kN.
5. Bending moment. Below the roller the moment about a section at $x$ is $$ M(x) = 27x - 3x^2, $$ a parabola whose peak lies at the point of zero shear: $$ M_{\max} = 27(4.5) - 3(4.5)^2 = \boxed{+60.75\ \text{kN}\cdot\text{m}} $$ sagging. At the roller $$ M(9.6) = 27(9.6) - 3(9.6)^2 = 259.2 - 276.48 = \boxed{-17.28\ \text{kN}\cdot\text{m}} $$ hogging, and beyond it the moment closes to zero at the free end, as a free end must. The member sags from the pin to $x = 9$ m and hogs over the last 3 m of projection.
6. Axial force, as a check. Resolving along the member, $N(x) = 3.6x - 43.2$ kN, so the member carries 43.2 kN of compression at the pin, 8.64 kN just below the roller and exactly zero at the free end — the arithmetic closes.