16-Civ-A1 Elementary Structural Analysis · December 2014
Question 5 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, method of joints/sections, virtual-work deflections, moment distribution, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges and compound (Gerber) structures, three-hinged frames. Sign convention: sagging bending moment positive; upward reactions positive; tension member forces positive (T), compression negative (C); distances in metres, forces in kN.
Exam rule: six questions constitute a complete paper — answer all of #1–#5 and only one of #6/#7/#8. All eight are solved here as a complete study resource.
Source note: Member geometry, support types and load directions for Questions 4, 7 and 8 were read from the printed figures; every reconstructed dimension yields clean, self-consistent equilibrium. Where a figure value was inferred, it is stated explicitly — verify against the original booklet if using for assessment.
A(pin), B, C, D (rollers) with internal hinges at 6 m and 16 m; spans 6–2–6–2–6 m (total 22 m).
Approach. Place a unit load at every position; because the beam is determinate, each influence line is piecewise-linear with breaks only at supports and hinges. Values below are from the determinate solver.
(iii) Influence line for reaction at A — peak coefficient +1.0 (unit load at A), zero beyond the hinge at 6 m.
(i) Influence line for bending moment at B — peak coefficient −2.0 when the unit load sits at the hinge (6 m); the 2 m overhang B–hinge governs.
(ii) Influence line for moment at mid-span of B–C — peak coefficient +1.5 ($=L/4$ for the 6 m span) with the load at mid-span (11 m).
Influence line
Max |coefficient|
Load position
Reaction at A
1.0
at A
Moment at B
−2.0 kN·m/kN
at hinge (6 m)
Moment at mid B–C
+1.5 kN·m/kN
at mid-span (11 m)
5(b) — Moving-load maximum shear
A(pin), hinge at 6 m, B(roller) at 8 m, C(roller) at 14 m; vehicle 100–100–40 kN spaced 1.5 and 3 m, travelling left→right.
Influence line for shear immediately right of B: rises to +1.0 just right of B and falls linearly to 0 at C (6 m span).
Influence line. For a section just right of B the ordinate is $+1$ at $B^{+}$ and decreases as $\left(1-\dfrac{a}{6}\right)$ toward C, where $a$ is the distance past B.
Position the vehicle. The peak is at $B^{+}$, so place the rear 100 kN wheel at B: ordinates $1.0,\,0.75,\,0.25$ under the 100, 100, 40 kN loads (at 0, 1.5, 4.5 m past B).
Maximum shear. $V_{B^{+}} = 100(1.0)+100(0.75)+40(0.25) = \boxed{185\text{ kN}}$.
