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.

(a) Beam with two loaded overhangs and a central point load

Given. A 16 m beam supported at $x = 2$ m (pin, A) and $x = 12$ m (roller, B), carrying 10 kN/m over the left overhang (0 to 2 m), a 60 kN point load at $x = 7$ m, and 10 kN/m over the right overhang (12 to 16 m).

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

Approach. Replace each distributed block by its resultant, take moments about A for the roller reaction, then integrate the load to build the shear diagram and integrate the shear to build the moment diagram.

1. Reduce the distributed loads. The left block is $W_1 = 10 \times 2 = 20$ kN acting at $x = 1$ m and the right block is $W_2 = 10 \times 4 = 40$ kN acting at $x = 14$ m. Together with the 60 kN load the total downward force is 120 kN.
2. Take moments about A ($x = 2$ m). With $R_B$ acting 10 m to the right of A, $$ 10\,R_B = 20(1-2) + 60(7-2) + 40(14-2) = -20 + 300 + 480 = 760 $$ so that $$ \boxed{R_B = 76.0\ \text{kN}\ (\uparrow)} $$
3. Vertical equilibrium. Summing forces, $R_A = 120 - 76 = \boxed{44.0\ \text{kN}\ (\uparrow)}$. As a check, moments about B give $10 R_A = 20(12-1) + 60(12-7) - 40(14-12) = 220 + 300 - 80 = 440$, the same answer.
4. Shear force. Working from the left free end, the shear falls linearly to $-20$ kN just left of A, jumps by $+44$ kN to $+24$ kN, stays constant to the 60 kN load, drops to $-36$ kN, stays constant to B, jumps by $+76$ kN to $+40$ kN, and falls linearly to zero at the free right end. The governing ordinate is $$ \boxed{V_{\max} = +40\ \text{kN just right of B}} $$ with $-36$ kN the largest negative value.
5. Bending moment. Integrating the shear, the moment is zero at both free ends. Over the left overhang it is hogging, $M = -10x^{2}/2$, reaching $-20$ kN·m at A. It then rises linearly at 24 kN per metre to the point load, $$ M(7) = -20 + 24(5) = \boxed{+100\ \text{kN}\!\cdot\!\text{m}} , $$ falls at 36 kN per metre to $M(12) = 100 - 36(5) = \boxed{-80\ \text{kN}\!\cdot\!\text{m}}$ at B, and closes to zero at the free end. The right-hand overhang value checks independently as $-10(4)^{2}/2 = -80$ kN·m.
6. Read off the sign changes. The moment is negative (hogging) from 0 to 2 m and again from about 9.8 m to 16 m, and positive (sagging) in between; the contraflexure point in the interior span is where the linear branch from $+100$ crosses zero, at $x = 7 + 100/36 = 9.78$ m.

(b) Continuous beam with two internal hinges and a built-in end

Given. A 22 m beam carrying 2 kN/m over its whole length, supported on rollers at $x = 0$ (A) and $x = 10$ m (B), with internal hinges at $x = 12$ m (C) and $x = 20$ m (D) and a built-in end at $x = 22$ m (E).

Find. All reactions, including the fixing moment, and the two diagrams with their extreme ordinates.

Approach. The two hinges cut the beam into three rigid lengths. Analyse the suspended length C–D first, because it is simply supported by the hinge forces; then carry those forces into the cantilever D–E and into the two-span length A–C.

1. Check determinacy. There are $r = 1 + 1 + 3 = 5$ reaction components and two hinges, so $i = 5 - (3 + 2) = 0$: the beam is determinate and can be solved by statics alone.
2. The suspended length C–D. It spans 8 m between the two hinges and carries 2 kN/m, so by symmetry each hinge delivers $$ V_h = \tfrac{1}{2}(2)(8) = \boxed{8.0\ \text{kN}} $$ to the lengths on either side.
3. The cantilever D–E. This 2 m length carries its own 4 kN of distributed load plus the 8 kN handed over at the hinge D. The built-in end therefore takes $$ V_E = 4 + 8 = \boxed{12.0\ \text{kN}\ (\uparrow)} $$ and, taking moments about E, $$ M_E = -\left[8(2) + 4(1)\right] = \boxed{-20.0\ \text{kN}\!\cdot\!\text{m}} $$ (hogging, as a built-in end must be).
4. The length A–C. It runs 12 m from the left end to the hinge C, carries $2 \times 12 = 24$ kN of distributed load acting at its mid-length, and carries the 8 kN handed down at C. Taking moments about A, $$ 10\,R_B = 24(6) + 8(12) = 144 + 96 = 240 \;\Rightarrow\; \boxed{R_B = 24.0\ \text{kN}\ (\uparrow)} $$ and vertical equilibrium gives $R_A = 24 + 8 - 24 = \boxed{8.0\ \text{kN}\ (\uparrow)}$.
5. Global check. The three upward forces are $8 + 24 + 12 = 44$ kN, exactly the total load $2 \times 22 = 44$ kN.
6. Shear force. $V = 8 - 2x$ on the first span, so the shear vanishes at $x = 4$ m and reaches $-12$ kN just left of B; the roller lifts it to $+12$ kN, after which it falls at 2 kN per metre, passing 8 kN at the hinge C and zero at $x = 16$ m (the mid-point of the suspended length), reaching $-8$ kN at D and $-12$ kN just left of E. The largest ordinate either way is $\boxed{12\ \text{kN}}$.
7. Bending moment. On the first span $M = 8x - x^{2}$, giving the sagging maximum $$ M(4) = 32 - 16 = \boxed{+16.0\ \text{kN}\!\cdot\!\text{m}} $$ and $M(10) = 80 - 100 = \boxed{-20.0\ \text{kN}\!\cdot\!\text{m}}$ over the roller B. The moment returns to zero at the hinge C, rises to $2(8)^{2}/8 = +16.0$ kN·m at the middle of the suspended length, returns to zero at the hinge D, and falls to $-20.0$ kN·m at the built-in end.

The largest positive and negative ordinates therefore happen to be equal in pairs, $+16$ kN·m at $x = 4$ m and $x = 16$ m and $-20$ kN·m at $x = 10$ m and $x = 22$ m — a useful reminder that a Gerber (cantilever) beam is laid out precisely to balance the sagging and hogging peaks.

(c) Beam with a cross-piece carrying an applied couple

Given. A 10 m beam, pinned at A ($x = 0$) and on a roller at B ($x = 10$ m), with a vertical post rigidly attached at C ($x = 6$ m) extending 2 m above and 3 m below the beam. A 10 kN horizontal force acts to the left at the top of the post and a 10 kN horizontal force acts to the right at the bottom.

Find. The reactions and the two diagrams, for the beam and for the post.

Approach. Recognise that the two horizontal forces form a couple, reduce them to a single moment applied at C, then solve the beam.

1. Reduce the pair of forces. The two 10 kN forces are equal, opposite and not collinear, so their resultant force is zero and they are statically equivalent to a couple. Taking moments about C, $$ M_0 = 10(2) + 10(3) = \boxed{50\ \text{kN}\!\cdot\!\text{m}} $$ acting anticlockwise. Because the resultant force vanishes, $\Sigma F_x = 0$ gives $A_x = \boxed{0}$.
2. Reactions from the couple. A couple is the same about every point, so taking moments about A, $$ 10\,B_y + 50 = 0 \;\Rightarrow\; \boxed{B_y = 5.0\ \text{kN}\ (\downarrow)} $$ and vertical equilibrium gives $A_y = \boxed{5.0\ \text{kN}\ (\uparrow)}$. The reactions form a couple of $5 \times 10 = 50$ kN·m that balances the applied one, as it must.
3. Shear force in the beam. No vertical load acts between the supports, so the shear is constant at $\boxed{V = +5.0\ \text{kN}}$ from A to B.
4. Bending moment in the beam. Because the shear is constant, the moment varies linearly. From the left, $M = 5x$, so $$ M(6^-) = \boxed{+30.0\ \text{kN}\!\cdot\!\text{m}} , $$ and the applied couple produces a step of 50 kN·m at C. From the right, $M(6^+) = -5(10-6) = \boxed{-20.0\ \text{kN}\!\cdot\!\text{m}}$, and the difference $30 - (-20) = 50$ kN·m is exactly the applied couple, which is the check on the sign of the step.
5. The post. Each arm is a cantilever off the beam. The upper arm carries the 10 kN force at 2 m, so its moment grows linearly from zero at the free top to $10(2) = 20$ kN·m at the beam, with a constant shear of 10 kN; the lower arm carries the 10 kN force at 3 m, giving $10(3) = 30$ kN·m at the beam. The two moments add to the 50 kN·m step, and each arm carries a constant 10 kN shear.