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) Simple span with an overhang

Given. Pin at A, roller at B, span $AB = 6$ m; the beam continues 2 m past B to a free end C. A uniformly distributed load of $12\ \text{kN}\,\text{m}^{-1}$ covers the 6 m span only (the load block on the figure stops at the roller); a concentrated $30$ kN acts down at C.

Find. The two reactions, and the shear and bending moment diagrams with their maximum positive and negative ordinates.

Approach. Take moments about A for the whole beam to get the roller reaction, then vertical equilibrium; build the shear diagram from the left and integrate it for the moment diagram.

1. Roller reaction from moments about A. The distributed load resultant is $wL = 12(6) = 72$ kN acting at mid-span, 3 m from A, and the tip load acts $6+2 = 8$ m from A: $$\sum M_A = 0:\qquad R_B(6) = 72(3) + 30(8) = 216 + 240 = 456$$ $$R_B = \frac{456}{6} = \boxed{76.0\ \text{kN}\ \uparrow}$$
2. Pin reaction from vertical equilibrium. With the total downward load $72 + 30 = 102$ kN, $$R_A = 102 - 76 = \boxed{26.0\ \text{kN}\ \uparrow}$$ There is no horizontal load, so the horizontal component at the pin is zero.
3. Shear diagram. Working from the left, $V(x) = R_A - wx = 26 - 12x$ over the span, which falls from $+26$ kN at A to $26 - 72 = -46$ kN just left of B. The roller lifts the diagram by 76 kN, so just right of B the shear is $-46 + 76 = +30$ kN and stays constant along the unloaded overhang until the 30 kN tip load closes it. The maximum ordinates are therefore $+30$ kN (overhang, and $+26$ kN at A) and $-46$ kN just left of B.
4. Point of zero shear and the peak sagging moment. Setting $V = 0$ in the span, $x_0 = R_A/w = 26/12 = 2.167$ m, and the moment there is the area of the shear diagram to its left: $$M_{\max}^{+} = \frac{R_A^{\,2}}{2w} = \frac{26^2}{2(12)} = \frac{676}{24} = \boxed{28.17\ \text{kN}\,\text{m}}$$
5. Moment over the support. The overhang is a cantilever loaded only by its tip force, so the hogging moment at B follows in one line from the right-hand free body: $$M_B = -\,30(2) = \boxed{-60.0\ \text{kN}\,\text{m}}$$ which the span-side expression reproduces: $M_B = 26(6) - 12(6)^2/2 = 156-216 = -60$ kN·m. The moment diagram is therefore positive (sagging) from A to the contraflexure point and negative (hogging) from there to B, closing linearly to zero at C.
6. Contraflexure point. Setting $26x - 6x^2 = 0$ gives $x = 26/6 = 4.333$ m from A; the sagging region is $0 \le x \le 4.333$ m and the hogging region runs from there to C.

2(b) L-frame with an internal hinge

Given. A 3 m column, built in at its base A, carries a uniformly distributed horizontal pressure load of $4\ \text{kN}\,\text{m}^{-1}$ over its full height. Rigidly attached at the top of the column (joint D) is a horizontal member running 7.5 m to a roller at E, carrying $8\ \text{kN}\,\text{m}^{-1}$ downward over its whole length. An internal hinge sits 1.5 m from D, i.e. 6 m from the roller.

Find. The four support reaction components and the shear and bending moment diagrams for both members.

Approach. The hinge supplies the fourth equation. Isolate the 6 m piece between the hinge and the roller, take moments about the hinge to get the roller reaction, then work back through the hinge into the column.

1. Check the count. Three reaction components at the fixed base plus one at the roller, less three equilibrium equations, less one condition equation at the hinge, gives $4 - 3 - 1 = 0$: determinate.
2. Roller reaction from the hinge sub-structure. The 6 m piece from the hinge to E carries $8(6) = 48$ kN whose resultant sits 3 m from the hinge, and the moment at the hinge is zero by definition: $$\sum M_{\text{hinge}}^{\,\text{right}} = 0:\qquad R_E(6) = 48(3) = 144 \;\Rightarrow\; R_E = \boxed{24.0\ \text{kN}\ \uparrow}$$
3. Force carried through the hinge. Vertical equilibrium of the same piece gives the shear delivered to the left-hand part: $$V_{\text{hinge}} = 48 - 24 = 24.0\ \text{kN}\ \text{(downward on the left part)}$$
4. Base reactions. The left-hand free body is the column plus the 1.5 m stub, carrying $8(1.5) = 12$ kN of its own, the 24 kN from the hinge, and $4(3) = 12$ kN of horizontal pressure. Hence $$V_A = 12 + 24 = \boxed{36.0\ \text{kN}\ \uparrow}, \qquad H_A = 12.0\ \text{kN}\ \leftarrow$$ and taking moments about the base, with the stub loads at 0.75 m and 1.5 m from the column line and the pressure resultant 1.5 m up, $$M_A = 12(0.75) + 24(1.5) + 12(1.5) = 9 + 36 + 18 = \boxed{63.0\ \text{kN}\,\text{m}}$$
5. Shear in the horizontal member. The column delivers 36 kN upward into joint D, so $V(x) = 36 - 8x$ measured from D: $+36$ kN at D, $+24$ kN at the hinge (matching step 3), zero at $x = 4.5$ m and $-24$ kN at the roller.
6. Moments in the horizontal member. Working from the roller end, $M(x) = 24(7.5-x) - 4(7.5-x)^2$ gives $$M_D = -45.0\ \text{kN}\,\text{m}, \qquad M_{\text{hinge}} = 0, \qquad M_{\max}^{+} = M(4.5) = \boxed{+36.0\ \text{kN}\,\text{m}}$$ so the member hogs 45 kN·m at the column, passes through zero at the hinge as it must, and sags to a peak of 36 kN·m under the point of zero shear 4.5 m from D.
7. Column actions. The column carries a constant axial compression of 36 kN. Its transverse shear runs linearly from $12$ kN at the base to zero at the top (the horizontal member has no axial force because the roller cannot push horizontally), and its bending moment is $$M(y) = 63 - 12y + 2y^2$$ measured up from the base: $63.0$ kN·m at A, falling monotonically to $63 - 36 + 18 = 45.0$ kN·m at D, which closes on the beam value. The whole column diagram is one sign — tension on the outer (left) face.

2(c) Bent member with loads normal to the rafter

Given. A single bent member: a pin at A, rising 6 m over a horizontal run of 8 m to the apex B, then falling to a roller at C, 4.5 m beyond the apex on the same level as A. Two loads act perpendicular to the rising member — 15 kN at 3 m from A along the member, and 16 kN a further 2 m along (5 m from A).

Find. The reactions, and the developed shear and bending moment diagrams with their extreme ordinates.

Approach. The rising member is a 3–4–5 triangle, so its unit vectors are exact; resolve the two perpendicular loads into components, take moments about A for the roller reaction, and then read the internal actions from free bodies cut along the member axis.

1. Member geometry. With $A = (0,0)$ and $B = (8,6)$, $$L_{AB} = \sqrt{8^2 + 6^2} = 10\ \text{m}, \qquad \hat{u} = (0.8,\ 0.6), \qquad \hat{n} = (0.6,\ -0.8)$$ where $\hat n$ is the inward normal, the direction the load arrows point. The falling member $BC$ has $\sqrt{4.5^2+6^2} = 7.5$ m, a 3–4–5 triangle again.
2. Load components and positions. Along the rafter, $$15\hat n = (9.0,\ -12.0)\ \text{kN at } (2.4,\ 1.8), \qquad 16\hat n = (9.6,\ -12.8)\ \text{kN at } (4.0,\ 3.0)$$ so the total applied force is $(18.6,\ -24.8)$ kN.
3. Roller reaction from moments about A. Using $M = x F_y - y F_x$ for each load, $$2.4(-12.0) - 1.8(9.0) = -45.0, \qquad 4.0(-12.8) - 3.0(9.6) = -80.0$$ $$\sum M_A = 0:\qquad -45.0 - 80.0 + 12.5\,C_y = 0 \;\Rightarrow\; C_y = \frac{125}{12.5} = \boxed{10.0\ \text{kN}\ \uparrow}$$ The exactly round result is the confirmation that the loads really are normal to the rafter and that the roller acts vertically.
4. Pin reaction. From the two force equations, $$A_y = 24.8 - 10.0 = \boxed{14.8\ \text{kN}\ \uparrow}, \qquad A_x = -18.6 \;\Rightarrow\; \boxed{18.6\ \text{kN}\ \leftarrow}$$
5. Shear and axial force along AB. Resolving the resultant of everything below a section at distance $s$ from A onto the member axes, the axial force is $6.0$ kN tension everywhere in AB (the perpendicular loads contribute nothing along the axis), while the shear steps down at each load: $$V = +23.0\ \text{kN}\ (0 \le s < 3), \quad +8.0\ \text{kN}\ (3 < s < 5), \quad -8.0\ \text{kN}\ (5 < s \le 10)$$ The steps of 15 kN and 16 kN are exactly the applied loads, which checks the resolution.
6. Moments along AB. Integrating the shear from A, $$M(3) = 23.0(3) = 69.0, \qquad M(5) = 69.0 + 8.0(2) = \boxed{85.0\ \text{kN}\,\text{m}}, \qquad M(10) = 85.0 - 8.0(5) = 45.0\ \text{kN}\,\text{m}$$ The peak sits at the 16 kN load, where the shear changes sign.
7. Member BC. The falling member carries no load, so the actions are constant along it. Resolving the roller reaction $(0,\,10.0)$ onto the axes of $BC$, whose unit vector from B to C is $(0.6,\,-0.8)$, $$N_{BC} = 8.0\ \text{kN compression}, \qquad V_{BC} = 6.0\ \text{kN}$$ and the moment falls linearly from $45.0$ kN·m at the apex to zero at the roller — consistent with $10.0 \times 4.5 = 45.0$ kN·m taken directly about B from the C side.
8. Signs. Every ordinate of the bending moment diagram is positive on the convention used here: the loads press on the rafter from outside, so the whole member sags with tension on its outer face, and there is no point of contraflexure anywhere. The only sign change in the shear diagram is at the 16 kN load.