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

Question 2: Reactions, shear 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, all dimensions and loads read from the drawings.

Find. All support reactions, and shear and bending moment diagrams labelled with the maximum positive and negative ordinates.

Approach. Each structure is determinate, so global equilibrium alone fixes the reactions — with the hinge in part (c) supplying the fourth equation as $\sum M = 0$ for the segment on one side of it — after which the diagrams follow by integrating the load.

(a) Beam with a tip load and a full-length distributed load

1. Total load and reactions. The distributed load acts over the entire 14 m, so its resultant is $W = wL = 8(14) = 112$ kN acting at mid-length, $x = 7$ m, and the tip load adds 20 kN at $x = 0$. Taking moments about the pin at $x = 2$ m, $$R_{\text{roller}} = \frac{20(0 - 2) + 112(7 - 2)}{10 - 2} = \frac{-40 + 560}{8} = 65\ \text{kN}$$ and vertical equilibrium gives $$R_{\text{pin}} = 20 + 112 - 65 = \boxed{67\ \text{kN}}, \qquad R_{\text{roller}} = \boxed{65\ \text{kN}}$$ Both act upwards. The clean values are a useful check that the load really does cover the full span.
2. Shear force. Starting at the free tip and accumulating upward forces, $V = -20 - 8x$ on the first 2 m, so $V$ falls from $-20$ kN to $-36$ kN at the pin, jumps to $-36 + 67 = +31$ kN, falls again to $31 - 8(8) = -33$ kN just left of the roller, jumps to $+32$ kN and runs out to zero at the right tip. The largest ordinates are $$V_{\max}^{+} = +32\ \text{kN}, \qquad V_{\max}^{-} = -36\ \text{kN}$$ Shear passes through zero at $x = 2 + 31/8 = 5.875$ m.
3. Bending moment. Integrating the shear, $M = -20x - 4x^{2}$ up to the pin and $M = -20x - 4x^{2} + 67(x - 2)$ between the supports. At the pin $M = -56\ \text{kN}\cdot\text{m}$; at the point of zero shear $M = +4.06\ \text{kN}\cdot\text{m}$; at the roller $M = -64\ \text{kN}\cdot\text{m}$; and the moment closes to zero at the free right tip, as it must. Hence $$M_{\max}^{+} = \boxed{+4.06\ \text{kN}\cdot\text{m}}, \qquad M_{\max}^{-} = \boxed{-64\ \text{kN}\cdot\text{m}}$$ The diagram is negative (hogging, tension on top) over almost the whole beam, with a very shallow positive lobe between the two points of contraflexure. This is characteristic of a beam whose overhangs are long relative to the span: the cantilever moments very nearly swallow the sagging moment of the interior span.

(b) Bent frame with a load normal to the inclined leg

1. Resolve the load on the inclined leg. The leg rises 6 m over a horizontal run of 4.5 m, so its length is $\sqrt{4.5^{2} + 6^{2}} = 7.5$ m and its direction cosines are $(0.6,\ 0.8)$. The load block is drawn parallel to the member with its arrows perpendicular to it, so the intensity is per metre of member and the resultant is $$F = 10(7.5) = 75\ \text{kN} \quad \text{acting along } (0.8,\ -0.6), \quad \text{i.e. } (60,\ -45)\ \text{kN}$$ applied at the mid-length of the leg. The vertical load on the top member is $10(7.5) = 75$ kN at its mid-length.
2. Reactions. Taking moments about the pin at the foot, with the corner 4.5 m to the right and 6 m above it and the roller a further 5.5 m along the top member, $$R_{\text{roller}} = \frac{281.25 + 618.75}{10} = \boxed{90\ \text{kN}}$$ Then horizontal and vertical equilibrium give $$H_{\text{pin}} = \boxed{60\ \text{kN} \text{ (to the left)}}, \qquad V_{\text{pin}} = 75 - 45 + \ldots = \boxed{30\ \text{kN (up)}}$$ where the vertical sum is $V_{\text{pin}} - 45 - 75 + 90 = 0$.
3. Internal forces in the inclined leg. Resolving the base reaction along and normal to the member, the axial force is a constant $12$ kN tension, while the shear falls linearly from $66$ kN at the pin to $-9$ kN at the corner, vanishing at $s = 66/10 = 6.6$ m. The bending moment therefore peaks a little below the corner: $$M_{\max} = \boxed{217.8\ \text{kN}\cdot\text{m}} \ \text{at } s = 6.6 \ \text{m}, \qquad M_{\text{corner}} = 213.75\ \text{kN}\cdot\text{m}$$ Both put the outside (upper-left) face of the leg in tension.
4. Internal forces in the horizontal member. Working from the free right-hand tip, $M = -5(7.5 - \xi)^{2}$ over the 2 m overhang, reaching $-20\ \text{kN}\cdot\text{m}$ at the roller, and $M = -5(7.5-\xi)^{2} + 90(5.5-\xi)$ inboard of it, which returns $+213.75\ \text{kN}\cdot\text{m}$ at the corner. Moment continuity at the rigid corner is the check that the two limbs have been solved consistently. The diagram crosses zero at $\xi = 5.21$ m from the corner. The shear is $-15$ kN at the corner, $-70$ kN just left of the roller, $+20$ kN just right of it and zero at the tip, so $$V_{\max}^{-} = \boxed{-70\ \text{kN}}, \qquad M_{\max}^{+} = \boxed{+213.75\ \text{kN}\cdot\text{m}}, \qquad M_{\max}^{-} = \boxed{-20\ \text{kN}\cdot\text{m}}$$

(c) Propped cantilever with an internal hinge

1. Use the hinge to break the problem in two. The built-in end supplies three components and the roller one, so four equations are needed and the hinge supplies the fourth. Isolating everything to the right of the hinge at $x = 4$ m and taking moments about the hinge, $$R_{C}(8) = 16(2) + 20(4) + 16(8) + 20(12) = 480 \ \Rightarrow \ R_{C} = \boxed{60\ \text{kN}}$$ where the four terms are, in order, the remaining 4 m of the 4 kN/m load, the 20 kN at $x = 8$ m, the 2 kN/m load over 8 m, and the 20 kN at the right tip.
2. Carry the hinge force back to the built-in end. Vertical equilibrium of the same right-hand segment gives a hinge shear of $72 - 60 = 12$ kN transmitted from the left. The left segment then carries its own 16 kN of distributed load plus that 12 kN at its tip, so $$V_{A} = 16 + 12 = \boxed{28\ \text{kN}}, \qquad M_{A} = -\left[16(2) + 12(4)\right] = \boxed{-80\ \text{kN}\cdot\text{m}}$$ with $H_{A} = 0$ because no horizontal load is applied. The negative sign means hogging — tension on the top face at the wall, as expected.
3. Shear diagram. $V$ starts at $+28$ kN and falls at 4 kN/m to $-4$ kN just left of $x = 8$ m, drops 20 kN to $-24$ kN, falls at 2 kN/m to $-32$ kN just left of the roller, jumps 60 kN to $+28$ kN, falls to $+20$ kN at the right tip and is closed by the 20 kN tip load. Hence $$V_{\max}^{+} = +28\ \text{kN}, \qquad V_{\max}^{-} = -32\ \text{kN}$$ Shear passes through zero at $x = 7$ m.
4. Bending moment diagram. Integrating, $M = 28x - 80 - 2x^{2}$ over the first 8 m. It rises from $-80\ \text{kN}\cdot\text{m}$ at the wall through exactly zero at the hinge — the check that the hinge has been located correctly — to a maximum of $+18\ \text{kN}\cdot\text{m}$ at $x = 7$ m, then falls to $-96\ \text{kN}\cdot\text{m}$ over the roller and closes to zero at the loaded right tip: $$M_{\max}^{+} = \boxed{+18\ \text{kN}\cdot\text{m}}, \qquad M_{\max}^{-} = \boxed{-96\ \text{kN}\cdot\text{m}}$$ The segments are positive (sagging) between the hinge and roughly $x = 10$ m and negative (hogging) elsewhere.