Question 2 of 9: Influence lines for a pin-jointed truss
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Civ-B1 Advanced Structural Analysis, National Examinations, May 2015 — three hours, closed book (an approved Sharp or Casio calculator is the only aid). Questions 1 and 2 are compulsory; the candidate then answers two of Questions 3, 4, 5, one of Questions 6, 7 and one of Questions 8, 9, so six questions totalling 100 marks constitute a complete paper. Marks are printed in the left margin. All nine questions are solved here, because the set is a study resource rather than an examination script.
Reference texts. R. C. Hibbeler, Structural Analysis (Pearson) — force/flexibility method, slope-deflection, moment distribution and influence lines; A. Kassimali, Structural Analysis (Cengage) — matrix stiffness formulation and support-settlement effects; J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods (Wiley); CSA S16:19 Design of Steel Structures and CSA S6:19 Canadian Highway Bridge Design Code for the Canadian design setting in which these analyses are used.
Sign convention used throughout. Bending moments are reported as sagging positive and are plotted on the tension face of each member. Slope-deflection end moments follow the counter-clockwise-positive convention stated in Question 9: $$M_{ij}=\frac{2EI}{L}\left(2\theta_i+\theta_j-3\psi_{ij}\right)+M^{F}_{ij},\qquad \psi_{ij}=\frac{(\mathbf{D}_j-\mathbf{D}_i)\cdot \mathbf{e}_2}{L}$$ where $\mathbf{e}_2$ is the member axis rotated 90 degrees counter-clockwise, and the fixed-end moments of a downward uniform load are $M^{F}_{ij}=+wL^2/12$, $M^{F}_{ji}=-wL^2/12$. The sagging moment at the ends of a member is then $M(i)=-M_{ij}$ and $M(j)=+M_{ji}$.
Question 2: Influence lines for a pin-jointed truss (9 marks)
Given. Span $6L$ between a pin at U1 and a roller at U7;
panel length $L$; depth $d=3L/4$; the unit load travels on stringers along the top chord,
so it is delivered to the truss only at the top panel points U1 to U7.
Members: 6 top chords, 4 bottom chords (L1L2 to L4L5),
5 verticals and 6 diagonals.
Find. The influence lines for the diagonal U2L2, the
vertical U3L2 and the bottom chord L2L3.
Question 2: the six-panel truss; the three members whose influence lines are required are shown heavy.
Approach. Confirm the truss is determinate, then place the unit load at each
top panel point in turn and use one section cut (for the diagonal and the chord) and one joint
equilibrium (for the vertical); the ordinates between panel points are straight lines because the
stringers deliver the load only at the joints.
Check determinacy. With $m=21$ members, $j=12$ joints and $r=3$ reaction
components,
$$m+r-2j=21+3-24=\boxed{0}$$ so the truss is statically determinate and every influence line can
be found from statics alone.
Reactions. For a unit load at distance $x$ from U1,
$$R_{U1}=1-\frac{x}{6L},\qquad R_{U7}=\frac{x}{6L}$$
Diagonal U2L2 — vertical equilibrium of a cut.
A vertical section between U2 and U3 cuts the top chord
U2U3, the diagonal U2L2 and the bottom chord
L1L2; only the diagonal has a vertical component. Its length is
$$\ell=\sqrt{L^{2}+\left(\tfrac{3L}{4}\right)^{2}}=1.25L
\;\Rightarrow\; \frac{d}{\ell}=\frac{0.75}{1.25}=0.6$$
so vertical equilibrium of the left free body gives $F=V_{\text{panel}}/0.6$. With the load at
U3 (to the right of the cut) $V=R_{U1}=2/3$ and
$$F_{U2L2}=\frac{2/3}{0.6}=\boxed{+1.1111}$$ while with the load at U2 the left free
body carries $R_{U1}-1=-1/6$, giving $-0.2778$. The influence line is a straight line from zero at
U1 to $-0.2778$ at U2, a transition line across the loaded panel that
crosses zero at $x=1.2L$, then a straight line from $+1.1111$ at U3 back to zero at
U7.
Vertical U3L2 — joint equilibrium. Joint
L2 connects two collinear bottom chords, the diagonal U2L2 and
the vertical, and carries no load (the load runs on the top chord). Resolving vertically,
$$F_{U3L2}=-0.6\,F_{U2L2}$$ so this influence line is simply the diagonal’s influence line
scaled by $-0.6$, with its peak $\boxed{-0.6667}$ (compression) at U3.
Bottom chord L2L3 — moments about U3.
A section between U3 and U4 cuts U3U4,
U3L3 and L2L3; the first two intersect at
U3, so moments about U3 isolate the chord force:
$$F_{L2L3}=\frac{M_{U3}}{d}=\frac{M_{U3}}{0.75L}$$
The influence line for $M$ at $x=2L$ on a $6L$ simple span is a triangle peaking at
$2L\cdot 4L/6L=1.3333L$, so
$$F_{L2L3,\max}=\frac{1.3333L}{0.75L}=\boxed{+1.7778}$$ under U3, falling linearly
to zero at each support. The whole influence line is positive: this chord is always in
tension.
Question 2: influence lines for the diagonal, the vertical and the bottom chord (tension positive).
Influence-line ordinates (unit load at each top panel point; tension positive)
