16-Civ-A1 Elementary Structural Analysis · December 2016
Question 7 of 8: Influence lines
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy (Ch. 2), method of joints/sections (Ch. 3), shear & moment diagrams (Ch. 4), virtual-work deflections (Ch. 8–9), influence lines (Ch. 6), moment distribution / slope–deflection (Ch. 11–12); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, three-hinged and two-hinged frames. Sign convention: sagging bending moment positive; upward reactions positive; member tension positive (T), compression negative (C).
The paper directs candidates to answer Q1–Q5 and then one of Q6/Q7/Q8. For completeness all eight questions are worked here. Because every structure is defined entirely by its figure, each drawing has been read from the printed figure and redrawn to scale below.
Given. Top chord $U_1\ldots U_7$ at $y=3\text{ m}$, six panels at $4\text{ m}$ ($24\text{ m}$); bottom nodes $L_1(4),L_2(12),L_3(20)$. Supports: rollers at $U_1$ and $U_7$, pin at $L_1$, roller at $L_3$ (four supports; the count is determinate, $m{+}r=2n=20$). Unit load travels the top chord.
Find. Influence lines and extreme coefficients for $U_2\!-\!U_3$, $L_1\!-\!U_3$, $L_2\!-\!L_3$.
7(a) — the truss and its four supports; load moves along the top chord.
Approach. Place the unit load successively at each top panel point $U_1\ldots U_7$ and record the three member forces; the influence line is the piecewise-linear locus of those ordinates.
$U_2\!-\!U_3$ is a zero-force member. Joint $U_1$ (roller, only the horizontal chord $U_1\!-\!U_2$) forces $U_1\!-\!U_2=0$; then joint $U_2$ (two collinear chords plus the vertical $U_2\!-\!L_1$) forces $U_2\!-\!U_3=0$ for every load position. Its influence line is $\boxed{\equiv 0}$.
$L_1\!-\!U_3$ (diagonal). Non-zero only for loads over the interior; ordinates $-1.25,\,-0.83,\,-0.42$ at $U_3,U_4,U_5$. Maximum $\boxed{1.25\text{ (compression)}}$, no tension.
$L_2\!-\!L_3$ (bottom chord). Ordinates $+0.33,\,+0.67,\,+1.00$ at $U_3,U_4,U_5$. Maximum $\boxed{1.00\text{ (tension)}}$, no compression.
7(a) — the three influence lines (coefficients per unit moving load).
7(b) — moving vehicle, shear just left of $B$
Given. Beam $A(0)$ pin, $B(8)$ roller, internal hinge at $x=10$, $C(16)$ roller. Vehicle: $64,\,64,\,20\text{ kN}$ at spacings $2\text{ m}$ and $3.2\text{ m}$, travelling right.
Find. Influence line for $V$ just left of $B$, and the greatest $|V_{B^-}|$ as the vehicle crosses.
7(b) — compound beam (hinge 2 m right of $B$) and the idealised vehicle.
Approach. For a unit load at position $p$ solve the determinate beam (hinge gives $M=0$ at $x=10$), read $V_{B^-}=R_A-[\text{load left of }B]$; then position the three wheels on the ordinates that maximise the (negative) shear.
Influence line. $V_{B^-}$ is $0$ at $A$, falls linearly to $-1.0$ just left of $B$ (a unit jump to $0$ just right of $B$), dips to $-0.25$ at the hinge, then returns to $0$ at $C$ — the entire line is non-positive.
Critical position. Put the heavier wheels on the steep left branch: a $64\text{ kN}$ wheel just left of $B$ (ordinate $-1.0$), the second $64\text{ kN}$ $2\text{ m}$ back at $x=6$ (ordinate $-0.75$), the $20\text{ kN}$ wheel at $x=11.2$ (ordinate $-0.20$).
Maximum shear. $V_{B^-}=64(1.0)+64(0.75)+20(0.20)=64+48+4=\boxed{116\text{ kN}}$ (negative shear, i.e. downward to the left of $B$).
