Question 1 of 9: Statical indeterminacy and structural degrees of freedom
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 07-Str-A4 Advanced Structural Analysis, National Examination, May 2014 — 3 hours, closed book (approved Sharp or Casio calculator permitted). Nine questions over seven pages: Questions 1 and 2 are compulsory (8 and 12 marks), then any two of Questions 3, 4, 5 (16 marks each) and any two of Questions 6, 7, 8, 9 (24 marks each). Six questions constitute a complete paper and total 100 marks. Marks are printed in the left margin. All nine questions are solved here, because the set is a study resource rather than a three-hour sitting.
Reference texts.
R. C. Hibbeler, Structural Analysis, 10th ed. — Ch. 9 (deflections by energy methods), Ch. 10 (force/flexibility method), Ch. 11–12 (slope-deflection and moment distribution).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 3 (determinacy and stability), Ch. 7 (deflections by work-energy), Ch. 16 (slope-deflection with sidesway).
A. Ghali, A. M. Neville and T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. — Ch. 3–4 (force and displacement methods), Ch. 5 (lack of fit and support movement).
W. McGuire, R. H. Gallagher and R. D. Ziemian, Matrix Structural Analysis, 2nd ed. — Ch. 4–5 (member stiffness, assembly, kinematic constraints).
Throughout this paper end moments are quoted with the counter-clockwise-positive sign convention, the one that matches the standard six-degree-of-freedom stiffness element. In that convention the slope-deflection equation is
the chord rotation is $\psi_{ij}=\bigl[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_2\bigr]/L$ with $\mathbf{e}_2$ the member axis turned $+90^\circ$, the fixed-end moment of a downward UDL is $\mathrm{FEM}_{i}=+wL^{2}/12$ and $\mathrm{FEM}_{j}=-wL^{2}/12$, and the ordinary sagging bending moment is recovered as $M_{\text{sag}}(i)=-M_{ij}$, $M_{\text{sag}}(j)=+M_{ji}$. Mixing this with the clockwise-positive convention of some textbooks is the single most common source of clean-looking but wrong answers on this paper.
Question 1: Statical indeterminacy and structural degrees of freedom (8 marks)
Given. Three plane frames, all members inextensible: (a) a stepped frame whose upper panel spans the left bay and whose lower beam runs on to a third column, all three columns pin-based, with a UDL on the upper beam and on the right lower bay; (b) a symmetric beam under a full-length UDL, roller-supported at both ends, carried on two pin-based columns, with an internal hinge just inboard of each end support; (c) a frame encastré at both ends made of a horizontal length $L$, a $45^\circ$ rise of horizontal run $a$ and vertical rise $a$, and a second horizontal length $L$, with a UDL over the rise.
Find. For each structure the degree of statical indeterminacy $r$, and the minimum number $k$ of joint rotations and independent joint translations that a slope-deflection analysis must carry.
Structure (a): eight members, eight joints, three pinned bases.
Structure (b): symmetric Gerber beam with two internal hinges, roller ends and two pin-based columns.
Structure (c): encastré at both ends, single 45° rise of run a.
Approach. Count indeterminacy from the plane-frame identity $r=3m+r_{\text{sup}}-3j-(\text{releases})$, then count the kinematic unknowns as (independent joint rotations) + (independent joint translations), suppressing rotations at joints known to carry zero moment and suppressing whole modes that symmetry or anti-symmetry forbids.
Set up the counting identity. For a plane frame with $m$ members, $j$ joints (supports included), $r_{\text{sup}}$ reaction components and $n_r$ internal moment releases,
Structure (a): eight members close one panel. The joints are A, B (upper beam), C, D, E (lower beam) and the three pinned bases F, G, H, so $j=8$; the members are AB, AC, CF, CD, BD, DG, DE and EH, so $m=8$; three pins give $r_{\text{sup}}=6$ and there are no releases. Hence
Cross-check (a) by counting closed loops. The frame contains one closed panel A-B-D-C, worth three redundants, and the three pinned bases give three reaction components more than the three needed for equilibrium, so $3+3=6$ — the same answer, which is the cheapest check available on a counting question.
Structure (b): the hinges make it almost determinate. Members: the five beam segments plus two columns, $m=7$; joints: the two roller ends, the two hinge points, the two beam-column joints and the two pinned column bases, $j=8$; reactions $1+1+2+2=6$; two hinges each release one moment. Then
Structure (c): only the supports are redundant. With $m=3$, $j=4$ and two encastré ends, $r_{\text{sup}}=6$, so $r=9+6-12=3$: the frame is an ordinary three-times-redundant fixed-ended member with two kinks.
Count the rotations for (a). Rigid joints occur at A, B, C, D and E — five unknown rotations. The three bases are pins, so the moment there is known to be zero; using the modified stiffness $3EI/L$ for a member with a pinned far end removes those three rotations from the unknown list, which is exactly what the question means by "joints that are known to have zero moments".
Count the translations for (a). The columns are inextensible, so A, B, C, D and E cannot move vertically; the beams are inextensible, so all joints on one floor share a single horizontal translation. There are two floors, hence two sway degrees of freedom, and $k=5+2=7$.
Count for (b), using symmetry. Each end assembly is statically determinate: the span between the roller and the hinge is a simple span delivering a known shear $wa/2$ into the hinge, and the piece from the hinge to the column is then a determinate cantilever whose end moment is a known applied load on that joint. Only the two column-head rotations survive, and because the structure and the load are both symmetric $\theta_{C_2}=-\theta_{C_1}$ and the sway vanishes, so $k=1$.
Count for (c), using inextensibility. Rotations are unknown at the two kinks B and C, giving two. For the translations, member AB is horizontal and A is fixed so $u_B=0$; member CD is horizontal and D is fixed so $u_C=0$; the inclined member then forces $(v_C-v_B)\sin 45^\circ=0$, i.e. $v_C=v_B$. One independent vertical translation survives and $k=2+1=3$.
Written out, the three counts are
$$r=3m+r_{\text{sup}}-3j-n_r$$
which gives, in turn, $r_{(a)}=3(8)+6-3(8)-0=\boxed{6}$, $r_{(b)}=3(7)+6-3(8)-2=\boxed{1}$ and $r_{(c)}=3(3)+6-3(4)-0=\boxed{3}$.
The arrows asked for in part (b) are then: on (a) a rotation arrow at each of A, B, C, D and E plus one horizontal translation arrow at each of the two floor levels; on (b) a single rotation arrow at one column head (its mirror image is equal and opposite, and no translation arrow is needed); on (c) a rotation arrow at B and at C plus one vertical translation arrow shared by B and C.
