Question 2 of 9: Influence Lines for the Shear Immediately Left of Support C
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations,
December 2016 — 07-Str-A4 Advanced Structural Analysis. Three
hours, CLOSED BOOK (an approved Sharp or Casio calculator is permitted).
Questions 1 and 2 are compulsory (12 and 8 marks); the candidate then
answers TWO of Questions 3–5 (18 marks each) and TWO of
Questions 6–9 (22 marks each), so six questions constitute a complete
paper worth 100 marks. All nine questions are solved here,
because the set is intended as a study resource rather than as an exam script.
Reference texts. R. C. Hibbeler, Structural
Analysis, 10th ed. (Ch. 6 influence lines, Ch. 9 virtual work and
Castigliano's theorems, Ch. 10 force method, Ch. 11 slope-deflection,
Ch. 12 moment distribution, Ch. 16 stiffness method for frames);
A. Kassimali, Structural Analysis, 6th ed.; A. Ghali,
A. M. Neville & T. G. Brown, Structural Analysis: A Unified
Classical and Matrix Approach, 7th ed.; J. C. McCormac,
Structural Analysis: Using Classical and Matrix Methods, 4th ed.
Canadian design context for the same structures: CSA S16, CSA A23.3 and
the National Building Code of Canada 2020.
Sign convention used throughout.
Member-end moments follow the counter-clockwise-positive slope-deflection form
$M_{ij}=\dfrac{2EI}{L}\left(2\theta_i+\theta_j-3\psi_{ij}\right)+\mathrm{FEM}_{ij}$,
with the chord rotation $\psi_{ij}=\left[(\mathbf{D}_j-\mathbf{D}_i)\cdot
\mathbf{e}_2\right]/L$ and $\mathbf{e}_2$ the member axis turned
$+90^\circ$. For a downward uniformly distributed load on a horizontal member,
$\mathrm{FEM}_{ij}=+wL^{2}/12$ and $\mathrm{FEM}_{ji}=-wL^{2}/12$. Sagging
moments are recovered as $M_{\text{sag}}(i)=-M_{ij}$ and
$M_{\text{sag}}(j)=+M_{ji}$, and every bending-moment diagram is plotted on the
tension side of the member.
Question 2: Influence Lines for the Shear Immediately Left of Support C (8 marks)
Given. A four-support beam A–B–C–D
carrying a travelling unit load. In (a) two internal hinges lie inside span BC,
making the beam a determinate Gerber (cantilever–suspended-span) system; in
(b) the hinges are removed and the beam is a two-degree indeterminate continuous
beam. The drawing carries no dimensions; scaling it gives
$AB=CD=L$, $BC=1.5L$, with the hinges about $0.2\,BC$ inside each interior
support.
Find. The influence line for the shear force at a
section immediately to the left of support C, for both beams, with the largest
absolute ordinate quoted.
Check: the exam figure is
undimensioned, so the span ratios $1:1.5:1$ and the hinge positions at
$0.2\,BC$ are scaled from the printed drawing. The governing answer
— the maximum absolute ordinate of $1.000$ at the section — is
independent of those proportions; only the secondary ordinates quoted for the
continuous beam depend on them.
Approach. Use the Müller-Breslau principle: the
influence line for an action is the deflected shape produced by releasing the
corresponding restraint and imposing a unit relative displacement across the
release. For the determinate beam the released shape is a mechanism and the
influence line is a set of straight lines; for the indeterminate beam it is an
elastic curve.
Structure (a): the Gerber beam and its influence line for the shear immediately left of C. The line is identically zero from A to the first hinge.
Identify the free body that owns the section (structure a).
The hinges cut the beam into three determinate pieces: A–B with a
cantilever tail to hinge 1, the suspended span between the two hinges, and
hinge 2–C–D. The section lies in the third piece, so only
load that reaches that piece can produce shear there.
Build the ordinates. A unit load anywhere between A and
hinge 1 is carried entirely by A and B, so the influence ordinate is
identically $\eta=0$. A unit load in the suspended span delivers a fraction
$\eta_{h}$ to hinge 2, rising linearly from $0$ at hinge 1 to $1$ at
hinge 2, and that fraction is the whole of the shear at the section. A unit
load between hinge 2 and C sits to the left of the cut with no support
between it and the section, giving $\eta=-1$; and a unit load anywhere in span
CD is beyond the section, so $\eta=0$ again.
Peak ordinate. The influence line is therefore a zero
plateau from A to hinge 1, a straight ramp to $-1$ at hinge 2, a
constant $-1$ plateau up to the section, and an abrupt return to zero across
support C. Its largest absolute ordinate is
$$\boxed{|\eta|_{\max}=1.000\ \text{immediately left of C}},$$
with $\eta=-0.500$ at the midspan of BC. The unit jump across the section is
the influence-line statement of the fact that a load placed on the section
itself changes the shear by exactly its own magnitude.
Structure (b): the same beam made continuous. The influence line is now a smooth elastic curve with small reversed lobes in the outer spans.
Apply Müller-Breslau to the continuous beam. Insert a
shear release immediately left of C and impose a unit relative transverse
displacement across it, keeping the two faces parallel. Because support C sits
essentially at the release, the right-hand face cannot move, so the
left-hand face must take the whole unit displacement. The ordinate at the
section is therefore exactly
$$\boxed{|\eta|_{\max}=1.000},$$
the same peak as the determinate beam — continuity redistributes the
shape, not the discontinuity.
Read the rest of the shape. Solving the released beam
(spans $1:1.5:1$) gives $\eta=-0.500$ at the midspan of BC, and small reversed
lobes of about $+0.071$ at the midspan of AB and $-0.071$ at the midspan of CD.
The ordinate is zero at A, at B, at D and immediately to the right of
C. The practical consequence is the one the question is really testing: for the
continuous beam a lane load must be placed over span BC and over
span AB or CD selectively, because loading the outer spans changes the sign
of the contribution.
Question 2 — influence-line ordinates for $V$ immediately left of C
