Question 1 of 9: Statical indeterminacy and slope-deflection degrees of freedom

Question 1: Statical indeterminacy and slope-deflection degrees of freedom (9 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. Three plane structures, all with rigid (moment-resisting) joints unless a hinge is drawn as an open circle. Structure (i) is a two-bay frame that is two storeys high over the left bay and one storey high over the right bay, with all three column bases pinned. Structure (ii) is a single-bay portal with the left base built in and the right base pinned, carrying a uniformly distributed horizontal load down the left column. Structure (iii) is a continuous beam on a roller at each end, propped by two columns whose bases are built in, with an internal hinge in the first span.

Find. For each structure, the degree of statical indeterminacy $n_s$ and the minimum number $k$ of independent joint displacements needed to set up the slope-deflection equations.

Approach. Count static indeterminacy from the member/joint/reaction tally $n_s = 3m + r - 3j - c$ and confirm it independently by the closed-loop rule $n_s = 3(\text{number of closed loops, counting the ground as a member}) - (\text{released equations})$; then count kinematic unknowns as (independent joint rotations) + (independent joint translations), dropping every rotation that the modified stiffness of a moment-free end already eliminates.

1. Part (a) — state the two counting rules. For a plane structure with $m$ members, $j$ joints (supports included), $r$ independent reaction components and $c$ released equations (one per internal hinge between two members), $$n_s = 3m + r - 3j - c$$ The independent check treats the ground as one extra rigid member: each closed loop supplies three redundants and every release removes one, so $$n_s = 3\,n_{\text{loops}} - \sum(\text{releases})$$ A pinned support is one release relative to a built-in one; a roller is two.
2. Structure (i) — tally the members and joints. The left and middle columns each run through the mid-level joint, so each counts as two members; the right column is one; the top beam is one and the mid-level beam is two spans. Hence $m = 2 + 2 + 1 + 1 + 2 = 8$. The joints are the three bases, the three mid-level joints and the two top joints, so $j = 8$, and three pinned bases give $r = 3 \times 2 = 6$ with $c = 0$: $$n_s = 3(8) + 6 - 3(8) = \boxed{6}$$
3. Structure (i) — confirm by loops, then count $k$. Three closed loops exist (the upper left bay, and the two ground-closed loops under each bay), and the three pinned bases are three releases, so $n_s = 3(3) - 3 = 6$, which agrees. For the kinematics the beams are inextensible, so every joint on one floor shares a single horizontal displacement: there are two independent sways. The pinned bases carry zero moment, so their rotations are absorbed by the modified stiffness $3EI/L$ and are not unknowns. That leaves the five rigid joints (three at mid level, two at the top): $$k = 5 + 2 = \boxed{7}$$
4. Structure (ii) — a plain portal. Here $m = 3$, $j = 4$, and $r = 3$ (built-in base) $+\,2$ (pinned base) $= 5$, with no internal release: $$n_s = 3(3) + 5 - 3(4) = \boxed{2}$$ The loop check gives the same answer: one ground-closed loop, $3(1) = 3$, less one release for the pinned base, $3 - 1 = 2$. Kinematically the two column tops rotate and the frame sways once, while the built-in base is fixed and the pinned base is moment-free, so $$k = 2 + 1 = \boxed{3}$$
5. Structure (iii) — work along the beam. The beam is cut into four members by the hinge and the two column heads, and the columns add two more, so $m = 6$; the joints are the two beam ends, the hinge, the two column heads and the two column bases, $j = 7$; the reactions are one per roller and three per built-in base, $r = 1 + 1 + 3 + 3 = 8$; and the hinge releases one equation, $c = 1$: $$n_s = 3(6) + 8 - 3(7) - 1 = \boxed{4}$$ The loop check: three closed loops give 9, and the releases are two per roller, one per roller, plus the hinge, $2 + 2 + 1 = 5$, so $9 - 5 = 4$ — the same.
6. Structure (iii) — find the minimum $k$. The span between the end roller and the hinge has zero moment at both ends, so it is a simply supported span that simply delivers $wL/2$ to the hinge; the short piece between the hinge and the first column is then a determinate cantilever off that column, carrying a known tip force. Neither adds an unknown. The far roller is moment-free, so its rotation goes into a modified stiffness. Because both end supports are rollers, the beam line can translate horizontally against the flexure of the two built-in columns, which is one sway. The unknowns are therefore the two column-head rotations plus that sway: $$k = 2 + 1 = \boxed{3}$$