07-Str-A4 · December 2014

Question 2 of 9: Influence lines for a truss loaded through a mid-height floor system (8 marks)

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, December 2014 — 3 hours, closed book (approved Sharp or Casio calculator permitted). Nine questions over six pages: Questions 1 and 2 are compulsory (12 and 8 marks), then any two of Questions 3, 4, 5 (18 marks each) and any two of Questions 6, 7, 8, 9 (22 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. 6 (influence lines), Ch. 9 (deflections by energy methods), Ch. 10 (force/flexibility method), Ch. 11 (slope-deflection).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 3 (determinacy and stability), Ch. 8 (influence lines for floor-beam systems), Ch. 13 (least work), Ch. 15–16 (slope-deflection with and without sidesway).
A. Ghali, A. M. Neville and T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. — Ch. 4 (force method), 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).
CSA S6:19 Canadian Highway Bridge Design Code and NBCC 2020 — the Canadian codes that these analysis methods feed; no code check is required by this paper.

Sign convention used throughout. End moments are counter-clockwise positive, the convention that matches the standard six-degree-of-freedom stiffness element. In it the slope-deflection equation is

$$M_{ij}=\frac{2EI}{L}\bigl(2\theta_i+\theta_j-3\psi_{ij}\bigr)+\mathrm{FEM}_{ij}$$

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 $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$. Every translation degree of freedom is closed with the virtual-work statement

$$\sum_{\text{members}}\bigl(M_{ij}+M_{ji}\bigr)\,\psi^{*}_{ij}\;+\;\sum \mathbf{F}\cdot\mathbf{d}^{*}=0$$

in which $\mathbf{F}$ are the equivalent nodal loads and $\mathbf{d}^{*}$, $\psi^{*}$ belong to the unit sway pattern. Mixing this with the clockwise-positive convention of some textbooks produces clean-looking but wrong answers, which is the single most common failure on this subject.

All work below uses the drawing.

Question 2: Influence lines for a truss loaded through a mid-height floor system (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. A 24 m truss in four 6 m panels, pinned at $M_1$ and on a roller at $M_5$, with the moving load delivered by stringers spanning between $M_1\ldots M_5$ at mid height.

Given data — truss geometry
Quantity	Value
Panel length	6 m (four panels, 24 m overall)
Bottom chord to mid-height level	2.5 m
Mid-height level to top chord	2.5 m (overall depth 5 m)
Supports	pin at $M_1$, roller at $M_5$ (both at mid height)
Load path	stringers at mid height, panel points $M_1$ to $M_5$

Find. The influence lines for the axial forces in $M_1\!-\!L_1$ and $U_1\!-\!L_2$, with the maximum absolute ordinate on each labelled tension or compression.

Question 2: the truss. $M_2$, $M_3$ and $M_4$ are the mid-points of the verticals where the stringers land; the long diagonals cross the stringer line without connecting to it.

Approach. Establish that the truss is determinate once the stringers are recognised as a load path rather than a chord, then place a unit load at each panel point in turn and read the two member forces by a joint and by a section.

Count the truss correctly. The mid-height line is the floor system: it carries the moving load to the panel points $M_1\ldots M_5$ and is not a chord (the question says the diagonals are not connected there). Removing it leaves 13 bars and 8 joints with three reactions, so $$m+r-2n=13+3-2(8)=0$$ and the truss is statically determinate. Counting the stringers as chord members instead makes the truss read one degree indeterminate and the question unanswerable by statics — this is the trap.
Recognise that the influence lines are piecewise linear. Between two panel points the stringer is a simple beam, so a load at fraction $s$ of a panel splits as $(1-s)$ and $s$ to the two adjacent panel points. Every influence ordinate is therefore a straight line between the five values computed at $M_1\ldots M_5$, and a load applied directly at either support goes straight into it, giving zero in all bars.
Influence line for $M_1\!-\!L_1$, by joint $M_1$. Only the two end diagonals and the reaction meet there. Horizontal equilibrium gives $F_{M_1U_1}=-F_{M_1L_1}$, and vertical equilibrium then gives $$R_{M_1}-2F_{M_1L_1}\frac{2.5}{6.5}=0\;\Longrightarrow\;F_{M_1L_1}=R_{M_1}\frac{6.5}{5}=1.3\,R_{M_1}$$ The member force is simply 1.3 times the left reaction, so its influence line has the shape of the reaction influence line.
Evaluate at the five panel points. With $R_{M_1}=1-x/24$, $$\boxed{\eta_{M_1L_1}=0,\;+0.975,\;+0.650,\;+0.325,\;0 \text{ at } M_1\ldots M_5}$$ $$\boxed{\text{maximum } +0.975 \text{ (TENSION), load at } M_2}$$
Influence line for $U_1\!-\!L_2$, by a section. Cut the second panel through $U_1U_2$, $U_1L_2$ and $L_1L_2$. The diagonal runs from $(6,5)$ to $(12,0)$, length $\sqrt{6^{2}+5^{2}}=\sqrt{61}=7.8102$ m, so its vertical component is $5/\sqrt{61}$ of the axial force. Vertical equilibrium of the free body left or right of the cut gives $$F_{U_1L_2}=\pm\,\frac{\sqrt{61}}{5}\,V_{\text{panel}}$$
Evaluate at the five panel points. With the load at $M_3$ the panel shear is the right-hand reaction $0.5$ and the left free body needs tension; with the load at $M_2$ the load sits on the other side of the cut and the sign reverses: $$\boxed{\eta_{U_1L_2}=0,\;-0.3905,\;+0.7810,\;+0.3905,\;0 \text{ at } M_1\ldots M_5}$$ $$\boxed{\text{maximum } +0.7810=\sqrt{61}/10 \text{ (TENSION), load at } M_3}$$
Cross-check the peak by hand. Unit load at $M_3$: reactions $0.5$ each, left free body has no applied load, so $0.5=F\,(5/\sqrt{61})$ and $F=0.5\sqrt{61}/5=0.78102$, tension — exactly the computed ordinate.

Both influence lines are triangles with the apex at the panel point nearest the member, which is the signature of a floor-beam load path: the load can only enter the truss at $M_1\ldots M_5$, so nothing curved or discontinuous can appear between them.

Question 2(a): influence line for $M_1\!-\!L_1$. Wholly tensile, peaking at 0.975 with the load over $M_2$.

Question 2(b): influence line for $U_1\!-\!L_2$. It reverses sign as the load crosses the cut panel; the compressive lobe reaches only half the tensile peak.

Question 2 — influence ordinates (per unit moving load)
Load at	$M_1$	$M_2$	$M_3$	$M_4$	$M_5$	Maximum absolute ordinate
$M_1-L_1$	0	+0.975	+0.650	+0.325	0	0.975 tension (load at $M_2$)
$U_1-L_2$	0	−0.3905	+0.7810	+0.3905	0	0.7810 tension (load at $M_3$)

Topic. Influence lines for truss members when the moving load reaches the truss through a floor system rather than directly at the joints.

Key relations. Determinacy $m+r-2n=0$; panel distribution $(1-s)$ and $s$ to adjacent panel points; diagonal force $=\pm(L_{\text{diag}}/d)\,V_{\text{panel}}$ where $d$ is the truss depth resolved perpendicular to the chords.

Why this works. A floor system makes the influence line a straight line between panel points, so five numbers describe the whole function. Each of those five follows from one joint equation or one section cut, so an eight-mark question needs no matrix work at all.

Common pitfall. Treating the drawn mid-height line as a chord. It has bars along it in the drawing, but the sentence "diagonals are not connected at the mid-height level" plus the phrase "loads move ... on beams" identifies it as a stringer line; counting it as a chord gives $m+r-2n=1$ and blocks the whole solution.

Source. Kassimali, Structural Analysis, 6th ed., Ch. 8 (influence lines for floor-beam and truss systems); Hibbeler, Structural Analysis, 10th ed., §6.3–6.4.

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