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.

(a) Simply supported beam with an overhang

Given. A straight beam 16 m long measured from the left-hand pin. The concentrated load and the distributed load are read from the figure as follows.

The extent of the load block is the reading that decides the whole answer. Its left edge sits exactly on the 60 kN arrow and its right edge on the end of the member line, so the ruling covers 10 m and not the 6 m bay between the load and the roller. The reactions that follow are integers, which is the usual confirmation that a hand-drawn extent has been read correctly.

Find. The two reactions, and the shear and bending moment diagrams with the maximum and minimum ordinates labelled.

Approach. Take moments about the pin to find the roller reaction, then build the shear diagram from the left by accumulating the loads and read the moment diagram off it, the moment being the running area under the shear curve.

1. Resolve the loading into two resultants. The distributed load acts over $16 - 6 = 10$ m, so $$W = w L = (6\ \text{kN/m})(10\ \text{m}) = 60\ \text{kN}$$ acting at the centroid of the block, $x = 11$ m. Together with the 60 kN concentrated load at $x = 6$ m the total applied load is 120 kN.
2. Take moments about A to find the roller reaction. With $R_B$ acting at $x = 12$ m, $$\sum M_A = 0: \quad 12 R_B = (60)(6) + (60)(11) = 360 + 660 = 1020$$ $$R_B = \boxed{85\ \text{kN} \uparrow}$$
3. Vertical equilibrium gives the pin reaction. $$R_A = 120 - 85 = \boxed{35\ \text{kN} \uparrow}$$ Both reactions are integers, which is the arithmetic check on the load extent read in the Given. block.
4. Build the shear diagram from the left. Over $0 \le x \lt 6$ m no load acts, so $V = +35$ kN, constant. At $x = 6$ m the 60 kN load drops the shear to $35 - 60 = -25$ kN, after which the ruling removes 6 kN per metre: $$V(x) = -25 - 6(x - 6)\ \text{kN}, \qquad 6 \le x \lt 12\ \text{m}$$ so that just left of the roller $V = -25 - 36 = -61$ kN. The roller adds 85 kN, lifting the shear to $+24$ kN, and it then falls linearly to zero at the free end, as it must.
5. Locate the maximum sagging moment. The shear does not pass through zero inside a span here; it jumps straight from $+35$ kN to $-25$ kN at the concentrated load, so the peak sagging moment is at $x = 6$ m: $$M_{\max} = R_A x = (35)(6) = \boxed{+210\ \text{kN}\cdot\text{m}}$$
6. Compute the hogging moment over the roller. The tidiest route is from the right, using the overhang alone: $$M_B = -\frac{w a^{2}}{2} = -\frac{(6)(4)^{2}}{2} = \boxed{-48\ \text{kN}\cdot\text{m}}$$ Working from the left gives the same figure, $M = (35)(12) - (60)(6) - 3(6)^{2} = 420 - 360 - 108 = -48\ \text{kN}\cdot\text{m}$, which is the check that the shear diagram and the reactions agree.
7. Find the point of contraflexure. Between the load and the roller the moment is $M = 210 - 25u - 3u^{2}$ with $u = x - 6$; setting this to zero gives $3u^{2} + 25u - 210 = 0$, so $u = 5.18$ m and the moment changes sign at $x = 11.18$ m. The diagram is therefore positive (sagging) from the pin to $x = 11.18$ m and negative (hogging) from there to the free end.

(b) Propped cantilever with an internal hinge and an overhang

Given. A beam 14 m long built into a wall at the left-hand end, with an internal hinge 4 m from the wall, a roller 6 m beyond the hinge (at x = 10 m) and a free end 4 m beyond the roller. A uniformly distributed load of 6 kN/m covers the whole 14 m — the load block on the figure runs from the wall face to the free end.

Find. The reactions at the built-in end and at the roller, and the shear and bending moment diagrams with maximum and minimum ordinates.

Approach. The hinge makes the beam determinate, so start with the free body to the right of the hinge, which contains only one unknown reaction; the force it transmits back through the hinge then loads the cantilever portion.

1. Free body of the beam to the right of the hinge. That piece runs from $x = 4$ m to $x = 14$ m, carries $W = (6)(10) = 60$ kN at $x = 9$ m, and is held by the roller at $x = 10$ m and by the vertical force $F$ that the hinge transmits. Moments about the roller give $$6F = (60)(1) \quad\Rightarrow\quad F = \boxed{10\ \text{kN}}$$
2. Vertical equilibrium of the same piece gives the roller reaction. $$R_2 = 60 - 10 = \boxed{50\ \text{kN} \uparrow}$$
3. Transfer the hinge force to the cantilever. The 4 m piece between the wall and the hinge carries its own $6 \times 4 = 24$ kN of ruling plus the 10 kN reaction the right-hand piece pushes down through the hinge: $$R_1 = 24 + 10 = \boxed{34\ \text{kN} \uparrow}$$ $$M_1 = -\left[ (24)(2) + (10)(4) \right] = -48 - 40 = \boxed{-88\ \text{kN}\cdot\text{m}}$$ the negative sign meaning a hogging moment, tension on the top of the beam at the wall.
4. Write the shear as one expression. Because the ruling is continuous the shear is simply $$V(x) = 34 - 6x\ \text{kN}, \qquad 0 \le x \lt 10\ \text{m}$$ falling from $+34$ kN at the wall through $+10$ kN at the hinge to $-26$ kN just left of the roller, where the 50 kN reaction lifts it to $+24$ kN. Over the overhang it falls to zero at the free end.
5. Locate and evaluate the peak sagging moment. The shear vanishes at $$x_{0} = \frac{34}{6} = 5.667\ \text{m}$$ and the moment there is $$M(x_{0}) = -88 + 34(5.667) - 3(5.667)^{2} = \boxed{+8.33\ \text{kN}\cdot\text{m}}$$
6. Check the two other controlling ordinates. At the hinge $M(4) = -88 + 136 - 48 = 0$, exactly as a hinge requires, which validates $R_1$ and $M_1$ together. Over the roller the overhang gives $M = -(6)(4)^{2}/2 = -48\ \text{kN}\cdot\text{m}$, and the moment is negative from the wall to $x = 4$ m, positive from $x = 4$ m to $x = 7.33$ m, and negative again to the free end.

(c) Bent member on a pin and a roller

Given. A single bent member. The pin is taken as the origin; the knee sits 12 m to the right of it and 9 m above it, and the member then falls 4 m over a further 3 m to a roller resting on a horizontal plane. A uniformly distributed load of 10 kN/m acts on the horizontal projection over the whole 15 m width.

Find. The reactions, and the shear and bending moment diagrams developed along the member axis with maximum and minimum ordinates.

Approach. Because the roller reaction and every applied load are vertical, the pin can carry no horizontal force; the bending moment at any section is then simply the moment of the vertical forces to its left, and the shear is the component of that same resultant taken perpendicular to whichever member the section lies on.

1. Establish that the pin carries no thrust. The only applied loading is vertical and the roller reaction is vertical, so horizontal equilibrium gives $H_1 = \boxed{0}$. This is what makes the rest of the question short.
2. Find the reactions. The total load is $W = (10)(15) = 150$ kN acting at the centroid of the projection, $x = 7.5$ m. Moments about the pin, with the roller 15 m away horizontally, give $$15 R_3 = (150)(7.5) = 1125 \quad\Rightarrow\quad R_3 = \boxed{75\ \text{kN} \uparrow}$$ $$V_1 = 150 - 75 = \boxed{75\ \text{kN} \uparrow}$$
3. Write the bending moment as a function of the horizontal coordinate. Since all forces are vertical, the height of the section does not enter and $$M(x) = 75x - \frac{10x^{2}}{2} = 75x - 5x^{2}\ \text{kN}\cdot\text{m}$$ holds over the entire structure, both members included.
4. Evaluate the peak moment. Differentiating, $\mathrm{d}M/\mathrm{d}x = 75 - 10x = 0$ at $x = 7.5$ m, which lies on the inclined member at $s = 7.5/0.8 = 9.375$ m along its axis. There $$M_{\max} = (75)(7.5) - 5(7.5)^{2} = 562.5 - 281.25 = \boxed{+281.25\ \text{kN}\cdot\text{m}}$$
5. Evaluate the moment at the knee. Substituting $x = 12$ m, $$M_{2} = (75)(12) - 5(12)^{2} = 900 - 720 = \boxed{+180\ \text{kN}\cdot\text{m}}$$ and $M = 0$ at both supports, so the whole bending moment diagram is sagging; there are no negative segments and the minimum ordinates are the zeros at the two ends.
6. Resolve the shear on member 1-2. Its axis is the unit vector $(0.8,\ 0.6)$, so the shear is $0.8$ times the net vertical force to the left of the section: $$V = 0.8\left(75 - 10x\right)$$ which runs from $+60$ kN at the pin to $0.8(75 - 120) = -36$ kN at the knee, passing through zero at $x = 7.5$ m as the moment peak requires.
7. Resolve the shear on member 2-3. Its axis is $(0.6,\ -0.8)$, so the shear is now $0.6$ times the same net vertical force: $$V = 0.6\left(75 - 10x\right)$$ giving $-27$ kN just past the knee and $-45$ kN just before the roller. The step at the knee is not an error — it is the change of direction of the member, which redistributes the same resultant between shear and axial force.