07-Str-A1 · May 2018

Question 6 of 8: Moment Distribution of a Symmetric Four-Column Frame

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

Question 6: Moment Distribution of a Symmetric Four-Column Frame (22 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. A 24 m continuous beam carrying 30 kN/m over its whole length on four columns, each 4 m long with a pinned base. The beam joints are at $x = 0, 8, 16$ and 24 m and two internal hinges sit 1 m either side of the interior joints, at $x = 9$ m and $x = 15$ m. All members share the same $EI$ and are inextensible.

Find. Shear-force and bending-moment diagrams for every member with the maximum and minimum ordinates marked.

Approach. Three reductions take a third-degree problem down to a two-joint table: detach the simply supported drop-in span between the hinges; treat the 1 m stub between each interior joint and its hinge as a determinate cantilever delivering a known moment of zero stiffness; and use symmetry to rule out sidesway, leaving only the rotations at joints 2 and 3 of the left-hand half.

1. Confirm the degree and the symmetry. Counting, $m = 7$, $j = 8$, $r = 8$ and $c = 2$, so $$i = (3 \times 7 + 8) - (3 \times 8 + 2) = 29 - 26 = 3.$$ All four bases are pins, the outer spans are equal at 8 m, the hinges are placed symmetrically and the load is uniform, so the structure and its loading are symmetric about $x = 12$ m. Sidesway is an antisymmetric displacement and cannot arise under symmetric loading; the inextensible drop-in span also ties the two halves together horizontally. Hence $\Delta = 0$ and no sway-correction pass is needed.
2. Detach the drop-in span. The 6 m length between the hinges is pinned at both ends, so it is simply supported: $$R = \frac{wL}{2} = \frac{30 \times 6}{2} = 90.0\ \text{kN at each hinge}, \qquad M_{\text{mid}} = \frac{wL^2}{8} = \frac{30 \times 36}{8} = 135.0\ \text{kN}\cdot\text{m}.$$
3. Reduce the 1 m stub to a known joint moment. Each stub carries its own 30 kN of distributed load plus the 90 kN hinge reaction at its tip, so it delivers to joint 3 a fixed hogging moment $$M_{\text{stub}} = 30(1)(0.5) + 90(1) = 105.0\ \text{kN}\cdot\text{m}$$ and a downward force of 120 kN. Being determinate, the stub has no rotational stiffness, so it must never be given a distribution factor — it enters the joint equation as a known applied moment.
4. Write the slope-deflection equations for the left half. The pinned-base columns take the modified stiffness $3EI/L$ with no carry-over, and the beam takes the standard form with $\mathrm{FEM} = wL^2/12 = 160.0$ kN·m: $$M_{21} = 0.75EI\theta_2, \qquad M_{34} = 0.75EI\theta_3,$$ $$M_{23} = 0.5EI\theta_2 + 0.25EI\theta_3 - 160, \qquad M_{32} = 0.5EI\theta_3 + 0.25EI\theta_2 + 160.$$
5. Enforce joint equilibrium and solve. At joint 2 the column and the beam must balance; at joint 3 the column, the beam and the stub moment of $-105$ must balance: $$1.25EI\theta_2 + 0.25EI\theta_3 = 160, \qquad 0.25EI\theta_2 + 1.25EI\theta_3 = -55.$$ Solving, $$EI\theta_2 = 142.5, \qquad EI\theta_3 = -72.5.$$
6. Back-substitute for the end moments. $$M_{21} = \boxed{+106.88\ \text{kN}\cdot\text{m}}, \quad M_{23} = -106.88, \quad M_{32} = \boxed{+159.38\ \text{kN}\cdot\text{m}}, \quad M_{34} = -54.38,$$ all in kN·m and in the clockwise-positive member convention. Joint 3 checks: $159.38 - 54.38 - 105.00 = 0$.
7. Shears and the peak sagging moment in span 2–3. With the beam-sign end moments $-106.88$ and $-159.38$ kN·m, $$V_2 = \frac{M_3 - M_2 + wL^2/2}{L} = \frac{-159.38 + 106.88 + 960}{8} = 113.44\ \text{kN},\qquad V_3 = 113.44 - 240 = -126.56\ \text{kN}.$$ The zero-shear point is at $x = 113.44/30 = 3.781$ m and $$M_{\max} = M_2 + \frac{V_2^{\,2}}{2w} = -106.88 + \frac{113.44^2}{60} = \boxed{+107.59\ \text{kN}\cdot\text{m}}.$$
8. Columns and reactions. Each column carries zero moment at its pinned base and its joint moment at the top, so its shear is that moment divided by 4 m: 26.72 kN in the outer columns and 13.59 kN in the inner ones. Their axial forces are the vertical reactions, 113.44 kN at nodes 1 and 8 and 246.56 kN at nodes 4 and 5, which sum to 720 kN, exactly $30 \times 24$. The two column shears within one half do not cancel; their sum of 13.13 kN is the axial thrust the beam transmits across the centre line, and quoting it is what shows the symmetry argument was understood rather than assumed.

The whole calculation was repeated independently with a direct-stiffness model in which the drop-in span was replaced by an axial-only bar carrying two 90 kN nodal loads. It reproduced every end moment and every reaction to five figures, which is the check worth making on any hand-worked indeterminate frame: asserting the arithmetic against your own algebra proves nothing if the formula itself is wrong.