Question 1 of 8: Determinacy and stability of six structures
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 1: Determinacy and stability of six structures (6 marks)
Given. Six plane structures. (a) A continuous beam under a
uniform load w carried on three rollers and a pin at the right-hand end,
with two internal hinges between the second and third rollers. (b) A two-bay,
three-level rigid frame carrying w on the top girder, on both bays of
the middle girder and on the right bay of the lower girder; the outer columns are
built in, the centre column is pinned at its base, and the top girder carries an
internal hinge just inside each outer column. (c) A rigid frame of two built-in
columns of unequal height joined by an inclined roof member and by a horizontal
member at mid-height, loaded by two horizontal forces P. (d) A shallow
open basket: two inclined legs built into the abutments at their upper ends and
pinned at their lower ends to a horizontal member that carries w and
rests on a roller at mid-length. (e) A hexagonal truss whose two chords and four
diagonals all meet at a central joint, pinned at the left vertex and on a roller
at the right, carrying P and 2P. (f) A triangular truss on a
pin, a roller and a second pin, loaded by two horizontal forces P.
Find. For each structure, the classification —
unstable, statically determinate, or statically indeterminate to a stated
degree.
[Figure not reproduced: Figure 1 — the six structures of Question 1, redrawn from the examination paper. A small open circle on a member is an internal hinge; hatching alone at a member end is a built-in (fixed) support. See the official exam paper.]
Approach. Count the reaction components $r$, the equations of
condition $c$ released by internal hinges, and, for the trusses, the members $m$
and joints $j$; then check the arrangement, because a favourable count
never by itself proves stability.
Three counting rules cover the whole question. For a chain of beam-type
members behaving as a set of rigid bodies,
$$ i \;=\; r - (3 + c), $$
for a rigid plane frame, in which every member end can carry a moment,
$$ i \;=\; 3m + r - 3j - c, $$
and for a plane pin-jointed truss,
$$ i \;=\; (m + r) - 2j . $$
In the frame formula it is quicker to count closed cells: treating the
foundation as one more rigid member, a plane frame is indeterminate to
$3\times$ the number of independent closed loops, less
one for every restraint released. That form is used as the check below.
(a) Gerber beam with two internal hinges. The three rollers
supply one vertical component each and the pin at the right-hand end supplies
two, so $r = 1 + 1 + 1 + 2 = 5$; the two hinges give $c = 2$. Hence
$$ i = 5 - (3 + 2) = \boxed{0} $$
and the arrangement works: the length between the two hinges is a simply
supported span carried by its neighbours, the right-hand length is held by its
own roller and pin, and the left-hand length is held vertically by its two
rollers and horizontally through the axial chain of hinges back to the pin. The
beam is statically determinate.
(b) Two-bay, three-level frame. Counting members between
joints gives two lengths of left column, three of centre column, three of right
column, one lower girder, two middle girders and two top girders, so
$m = 13$; the joints are three bases, two at the lower
girder level, three at the middle level and three at the top,
$j = 11$; the reactions are
$3 + 2 + 3 = 8$ and the two hinges give
$c = 2$. Therefore
$$ i = 3(13) + 8 - 3(11) - 2 = 39 + 8 - 33 - 2 = \boxed{12} $$
The loop check agrees: with the foundation counted as a member there are five
independent closed cells, so a wholly built-in version would be indeterminate to
degree 15; the pinned centre base removes one restraint and the two hinges remove
two more, leaving 12. The frame is statically indeterminate to the
twelfth degree.
(c) Two built-in columns with an inclined roof. Here
$m = 6$ — two lengths of each column, the horizontal member and the
inclined member — with $j = 6$, $r = 3 + 3 = 6$ and $c = 0$, so
$$ i = 3(6) + 6 - 3(6) = \boxed{6} $$
Equivalently there are two closed cells and both bases are built in, giving
$3 \times 2 = 6$. The two horizontal forces are resisted by two fixed bases that
are neither collinear nor free to rotate, so the frame is stable: it is
statically indeterminate to the sixth degree.
(d) Inclined legs pinned to a propped horizontal member. The
members are the two legs and the two lengths of the horizontal member either side
of the roller, $m = 4$; the joints are the two abutments, the two hinged corners
and the roller point, $j = 5$; the reactions are three at each built-in abutment
plus one at the roller, $r = 7$; and the two corner hinges give $c = 2$.
Therefore
$$ i = 3(4) + 7 - 3(5) - 2 = 12 + 7 - 15 - 2 = \boxed{2} $$
The two closed cells — each leg with the foundation and half the tie
— give the same number once the roller and the two hinges are debited. The
structure is statically indeterminate to the second degree.
(e) Hexagonal truss with a central joint. The members are
the two upper sloping members and the top chord, the matching three below, the
two halves of the horizontal chord through the centre and the four diagonals
radiating from that centre, giving $m = 12$; there are seven joints, and
$r = 2 + 1 = 3$, so
$$ i = (12 + 3) - 2(7) = 15 - 14 = \boxed{1} $$
Every panel is triangulated and the pin-plus-roller pair is a stable external
arrangement, so the count is trustworthy. The truss is statically
indeterminate to the first degree — internally, since externally
$r - 3 = 0$.
(f) Triangular truss on a pin, a roller and a pin. The
members are the two long sides, each split at the mid-height joint, the
horizontal member joining those two joints, the two short diagonals down to the
centre of the base and the two halves of the base, so $m = 9$; there are six
joints and $r = 2 + 1 + 2 = 5$. Hence
$$ i = (9 + 5) - 2(6) = 14 - 12 = \boxed{2} $$
The excess is external: the second pin adds one horizontal redundant and the
third support adds one vertical. The three supports are neither parallel-and-only
nor concurrent, so the assembly is stable and the truss is statically
indeterminate to the second degree.
The pattern worth carrying away is that (a) and the two trusses were settled by
arithmetic alone, while (b), (c) and (d) needed the closed-cell picture to make
the arithmetic trustworthy.
