Question 1 of 9: Statical indeterminacy and the minimum number of structural degrees of freedom

Question 1: Statical indeterminacy and the minimum number of structural degrees of freedom (8 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. The three structures as drawn, all members inextensible and rigidly jointed except where an internal hinge is shown; the loading on each is as printed. No numerical dimensions are supplied — none are needed, because both counts are purely topological.

Find. For each structure, the degree of statical indeterminacy $r$ and the minimum number $k$ of joint displacements that a slope-deflection analysis must carry.

Approach. Count $r$ from $r = 3m + R - 3j - (\text{releases})$ on the rigid-jointed frame, then count $k$ as (independent joint rotations that survive after every zero-moment end is eliminated by a modified stiffness) plus (independent joint translations of the inextensible bar linkage), reducing the second group by symmetry or antisymmetry wherever the loading permits.

1. Set out the two counting rules once. For a planar rigid-jointed structure with $m$ members, $j$ joints (support joints included) and $R$ independent reaction components, $$r = 3m + R - 3j - n_{\text{rel}}$$ where $n_{\text{rel}}$ is the number of released moments (one per internal hinge between two members). For the kinematic count, the independent translations are the mobility of the linkage formed by treating every inextensible member as a rigid bar, $$n_{\Delta} = 2n_{j} - \operatorname{rank}\!\left[\mathbf{C}\right],$$ $n_j$ being the number of joints free to translate and $\mathbf{C}$ the matrix holding one compatibility row per bar (its unit-vector row) plus one row per restrained support component acting on those joints. The independent rotations are one per joint, less every end whose moment is known to be zero — a pin or roller end of a member, an internal hinge, a free tip — because such an end is condensed out with the modified stiffness $3EI/L$ rather than carried as an unknown.
2. Structure (I): the hinge makes the beam determinate, but one rotation still remains. The beam has a pin, two rollers and one internal hinge: $m = 3$, $j = 4$, $R = 2 + 1 + 1 = 4$ and $n_{\text{rel}} = 1$, so $$r = 3(3) + 4 - 3(4) - 1 = \boxed{0}$$ — statically determinate. The kinematic count is not zero, however. Rotations are eliminated at the pin support A (a member end with $M = 0$), on both faces of the hinge H, and at the free-rotating roller end C. The segment A–H then has zero moment at both ends, so it is a simple span that merely delivers a shear to H, and the only unknown left is the rotation of the interior roller B: $$k_{\text{(I)}} = \boxed{1}$$
Structure (I): pin, internal hinge and two rollers under a full-length UDL. Determinate (r = 0), yet one joint rotation survives.
3. Structure (II): a stepped frame with three independent sways. Numbering the joints 1 (left roller), 2 (left column head), 3 and 5 (the ends of the lower connecting member), 4 and 8 (the fixed bases), 6 (right column head) and 7 (right roller), there are $m = 7$ members, $j = 8$ joints and $R = 1 + 3 + 1 + 3 = 8$ reaction components with no releases: $$r = 3(7) + 8 - 3(8) = \boxed{5}$$ For the kinematics, six joints (1, 2, 3, 5, 6, 7) may translate; seven inextensible bars and the two roller components give $n_{\Delta} = 2(6) - 7 - 2 = \boxed{3}$. Physically those three are the sway of the lower storey ($u_3 = u_5$) and the two independent horizontal movements of the column heads 2 and 6, each of which drags its own overhanging beam and its roller with it. Rotations survive at joints 2, 3, 5 and 6 — the moment is zero at the roller ends 1 and 7, so those two are condensed out. Hence $$k_{\text{(II)}} = 4 + 3 = \boxed{7}$$
Structure (II): stepped two-column frame, fixed bases, roller-ended upper beams. Five times redundant, seven degrees of freedom.
4. Structure (III): symmetry kills every translation. With $m = 9$ members, $j = 10$ joints and four pinned bases ($R = 8$), $$r = 3(9) + 8 - 3(10) = \boxed{5}$$ The bare linkage has $n_{\Delta} = 3$ independent translations (the three horizontal movements $u_a = u_b$, $u_c = u_d$ and $u_e = u_f$, every vertical being locked by an inextensible column). But the structure and the loading are symmetric about the centre line, so every translation must be antisymmetric: $u_c = u_d$ together with $u_c = -u_d$ forces $u_c = u_d = 0$, and the same argument then zeroes the outer pair. No sway degree of freedom survives. The six joint rotations pair off antimetrically ($\theta_f = -\theta_a$, $\theta_e = -\theta_b$, $\theta_d = -\theta_c$), leaving three, and the four pinned bases are condensed out with $3EI/L$: $$k_{\text{(III)}} = \boxed{3}$$
Structure (III): symmetric raised-centre frame on four pinned bases under symmetric UDLs. Symmetry removes every sway.
5. Marking the answer sheet. On (I) draw one curved arrow at the interior roller. On (II) draw curved arrows at the four rigid joints 2, 3, 5 and 6 and three straight arrows: one on the lower member and one at each column head. On (III) draw curved arrows at the three joints of the left-hand half only, and write “symmetric — no sway, right half by antimetry” beside it.