07-Str-A1 · December 2015

Question 5 of 8: Propped continuous frame by moment distribution

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

Notes on this paper

Paper format. National Examinations, December 2015, 07-Str-A1 — Elementary Structural Analysis. Three hours, closed book, approved Sharp or Casio calculator only. Six questions constitute a complete paper: answer ALL of Questions 1 to 4, then ONLY TWO of Questions 5, 6, 7 or 8. Marks are shown in the left margin (6 + 18 + 18 + 18 + 20 + 20 = 100). All eight questions are worked below, because the four alternatives exercise four different methods — moment distribution, three-hinged-frame statics, influence lines and virtual work — and the complete set is the more useful study resource.

Reference texts.

R. C. Hibbeler, Structural Analysis, 10th ed. — Ch. 2 (determinacy and stability), Ch. 3–5 (trusses, internal loadings, frames), Ch. 6 (influence lines), Ch. 8–9 (deflections and virtual work), Ch. 11–12 (slope-deflection and moment distribution).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 3 (equilibrium and determinacy), Ch. 4 (plane trusses), Ch. 5 (beams and frames), Ch. 7 (virtual work), Ch. 8–9 (influence lines), Ch. 16–17 (slope-deflection and moment distribution).
K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis, 5th ed. — a parallel treatment of the same syllabus.
C. H. Norris, J. B. Wilbur and S. Utku, Elementary Structural Analysis, 4th ed. — the classical text this exam code is named after.

Check: the roller in Question 2(c). The roller under the inclined member of Question 2(c) is drawn on a bed ruled parallel to the member, so its reaction is taken normal to the member axis. A vertical-roller reading is also arithmetically self-consistent. The distinction changes only the reaction components and the axial force: the shear and bending-moment diagrams the question asks for are identical under either reading, and both sets of reactions are reported.

Question 5: Propped continuous frame by moment distribution (20 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 52 m beam built in at both ends, propped at two points by 4 m columns with pinned bases, and released by two internal hinges.

Given data for Question 5
Quantity	Value
Beam, joint 1 to joint 4	16 m + 2 m + 16 m + 2 m + 16 m = 52 m overall
Column positions and length	at 16 m and at 36 m; each 4 m long, pinned at the base (joints 5 and 6)
Internal hinges	joint 2 at 18 m and joint 3 at 34 m
Uniform load	12 kN/m over the whole 52 m of beam
End supports	Built in (fixed) at joints 1 and 4
Members	Same $EI$ throughout, inextensible; column tops rigidly framed into the beam

Find. The shear force and bending moment diagrams for the beam and for the columns, with the maximum and minimum ordinates labelled.

Figure 8 — structure 5 with its shear force and bending moment diagrams for the beam. The structure and the loading are symmetric about mid-span, so only the left half needs to be distributed. Sagging moment is plotted positive.

Approach. Cut at the two hinges. The 16 m length between them is a simply supported span whose end reactions load the two 2 m overhangs; each overhang then delivers a known moment to its column joint, so only one rotation is unknown on each half and a single moment-distribution cycle suffices.

The suspended span. The length between the hinges, 34 m − 18 m = 16 m, carries only its own uniform load and is simply supported at both hinges, so $$ R_{\text{hinge}} = \frac{wL}{2} = \frac{12(16)}{2} = 96\ \text{kN} $$ at each end, and its mid-span sagging moment is $$ M_{\text{mid}} = \frac{wL^2}{8} = \frac{12(16)^2}{8} = \boxed{+384\ \text{kN}\cdot\text{m}} $$ which is the largest sagging ordinate anywhere in the structure.
The overhang moment at the column joint. The 2 m from the column at 16 m out to the hinge at 18 m carries its own uniform load, $12 \times 2 = 24$ kN at 1 m, plus the 96 kN passed through the hinge at 2 m. Taking moments about the joint, $$ M_{\text{overhang}} = 24(1) + 96(2) = 216\ \text{kN}\cdot\text{m} $$ a hogging moment applied to the joint. The overhang has no far-end stiffness, so it must be given a distribution factor of zero — it delivers a moment, it never takes a share of the balance.
Stiffness and distribution factors at the column joint. The 16 m beam length runs to a built-in end, so its stiffness is $4EI/L$; the column runs to a pinned base, so its modified stiffness is $3EI/L$: $$ k_{\text{beam}} = \frac{4EI}{16} = 0.25\,EI, \qquad k_{\text{col}} = \frac{3EI}{4} = 0.75\,EI, $$ $$ \mathrm{DF}_{\text{beam}} = \frac{0.25}{1.00} = 0.25, \qquad \mathrm{DF}_{\text{col}} = \frac{0.75}{1.00} = 0.75 . $$ The short stiff column takes three quarters of any out-of-balance moment.
Fixed-end moments and the single balancing cycle. For the 16 m beam length under 12 kN/m, $$ \mathrm{FEM} = \frac{wL^2}{12} = \frac{12(16)^2}{12} = 256\ \text{kN}\cdot\text{m}, $$ so $M_{12} = -256$ and $M_{21} = +256$ in the usual anticlockwise-positive convention. At the joint the moments now sum to $256 + 0 - 216 = +40$ kN·m, so $-40$ kN·m must be distributed: $$ \Delta M_{\text{beam}} = -40(0.25) = -10, \qquad \Delta M_{\text{col}} = -40(0.75) = -30 . $$ Carrying over half of the beam share to the built-in end and none to the pinned base leaves $$ \boxed{M_{1} = -261\ \text{kN}\cdot\text{m}}, \qquad M_{21} = +246, \qquad \boxed{M_{\text{col, top}} = 30\ \text{kN}\cdot\text{m}} $$ and the joint checks: $246 - 30 - 216 = 0$.
Shear in the end span. With the sagging moment $-261$ kN·m at the built-in end and $-246$ kN·m just left of the column joint, $$ V_1 = \frac{M(16) - M(0) + \tfrac{1}{2}wL^2}{L} = \frac{-246 + 261 + \tfrac{1}{2}(12)(16)^2}{16} = \boxed{96.94\ \text{kN}} $$ falling at 12 kN per metre to $$ V(16^-) = 96.94 - 192 = \boxed{-95.06\ \text{kN}} $$ Just past the joint the 2 m overhang must carry $24 + 96 = 120$ kN, so $V(16^+) = +120$ kN, and the jump of $120 - (-95.06) = 215.06$ kN is the axial force delivered to the column.
Bending moment in the end span. Zero shear occurs at $x = 96.94/12 = 8.078$ m from the built-in end, where the sagging moment reaches its span maximum: $$ M_{\max} = M_1 + \frac{V_1^{\,2}}{2w} = -261 + \frac{96.9375^2}{24} = \boxed{+130.54\ \text{kN}\cdot\text{m}} $$ The moment is hogging (negative) from the built-in end to the first point of contraflexure at $x = 3.41$ m, sagging from there to $x = 12.74$ m, and hogging again over the column and out to the hinge, where it closes to zero as a hinge requires.
The columns. Each column carries 30 kN·m at its top and nothing at its pinned base, so it bends in single curvature with a constant shear of $$ V_{\text{col}} = \frac{30}{4} = 7.5\ \text{kN}, $$ and an axial compression of 215.06 kN. The two columns' shears are equal and opposite by symmetry, so they cancel and the built-in ends carry no net horizontal force; the beam simply passes the 7.5 kN thrust from one column across to the other as axial force. That is the practical content of the phrase "inextensible" in the question: it fixes every joint horizontally and vertically, leaving only joint rotations unknown, so there is no sidesway to consider.
Symmetry, and the check that it is real. The structure, the supports and the load are symmetric about mid-span at 26 m, so the right half is the mirror image of the left: $M_4 = -261$ kN·m, $V_4 = -96.94$ kN, and the second column carries the same 30 kN·m and 215.06 kN. The arithmetic check is that the left half must carry half the total load: $$ 96.94 + 215.06 = 312\ \text{kN} = 12 \times 26\ \text{m} \ \checkmark $$ A direct-stiffness solution of the same frame, run independently, reproduces $-261$, $-246$, $30$ kN·m and $96.94$ kN exactly.

Question 5 — extreme ordinates
Location	Shear force	Bending moment
Built-in end (joints 1 and 4)	96.94 kN	−261 kN·m (hogging, maximum)
End span, at 8.078 m from the fixed end	0	+130.54 kN·m (sagging)
Just inside the column joint (16 m)	−95.06 kN	−246 kN·m
Just outside the column joint (16 m)	+120 kN (maximum)	−216 kN·m
Hinges (18 m and 34 m)	±96 kN	0
Mid-span of the suspended length (26 m)	0	+384 kN·m (sagging, maximum)
Each column	7.5 kN constant; 215.06 kN axial compression	0 at the pinned base to 30 kN·m at the top

Topic. Moment distribution applied to a symmetric propped frame containing internal hinges and an overhang.

Key relations. Fixed-end moment for a uniform load, $wL^2/12$; modified stiffness $3EI/L$ for a member whose far end is pinned, and $4EI/L$ with a carry-over of one half for a member whose far end is fixed; distribution factor $\mathrm{DF}_i = k_i / \sum k$; end shear $V = \left[M_{\text{far}} - M_{\text{near}} + \tfrac{1}{2}wL^2\right]/L$; span maximum $M_{\max} = M_{\text{near}} + V^2/2w$.

Why this works. The hinges decompose an apparently formidable frame into three pieces that can be handled in turn. The middle length has moment released at both ends, so it is simply supported and its analysis is one line. Its end reactions then become known point loads on the two overhangs, and an overhang is statically determinate, so it delivers a known moment to the joint rather than an unknown one. What is left is a single joint with one unknown rotation and two stiff members, which is a single cycle of distribution — no iteration, no sidesway, because the inextensible members hold every joint in place. The 3:1 split of the out-of-balance moment simply reflects that a 4 m column pinned at its base is three times as stiff in rotation as a 16 m beam length built in at its far end.

Common pitfall. Giving the overhang a distribution factor. It has no stiffness at all, so any share allocated to it is lost from the joint and the joint no longer balances; its correct role is a fixed applied moment of 216 kN·m. The second pitfall is using $4EI/L$ for the pinned-base column, which would give distribution factors of 0.2 and 0.8 and, with the carry-over, a non-zero moment at a pin.

Source. Hibbeler, Structural Analysis, 10th ed., §12.1–12.4 (moment distribution, modified stiffness, symmetric structures); Kassimali, Structural Analysis, 6th ed., Ch. 17.

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