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16-Civ-A1 Elementary Structural Analysis · December 2018

Question 5 of 8: Influence lines

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper: National Exams — December 2018, 16-Civ-A1 Elementary Structural Analysis (closed book, 3 h; approved Casio/Sharp calculator permitted). Six questions constitute a complete paper: answer Q1–Q5 and one of Q6/Q7/Q8. For completeness all eight questions are worked here.

Reference texts: R. C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy (Ch. 2), method of joints & sections (Ch. 3), shear & moment diagrams (Ch. 4), influence lines (Ch. 6), virtual-work deflections (Ch. 8–9), moment distribution (Ch. 11); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, three-hinged & indeterminate frames.

Sign convention. Upward reactions positive; sagging bending moment positive (tension on the underside), hogging negative; member axial force tension positive (T), compression negative (C).

These A1 papers are defined entirely by their figures, so every structure is redrawn to scale below.

Question 5: Influence lines (18 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.

5(a) — Determinate two-span (Gerber) beam

Given. $A(0)\!-\!B(5)\!-\!C(10)\!-\!D(12)\!-\!E(20\text{ m})$: pin at $A$, rollers at $C$ and $E$, internal hinge at $D$. Find. Influence lines for (i) $M_B$, (ii) shear just right of $C$, (iii) reaction $R_C$, each with its largest-magnitude ordinate.

hingeABCDE5 m5 m2 m8 m
5(a) — determinate beam; the hinge at $D$ separates span $A\!-\!C\!-\!D$ from the suspended $D\!-\!E$.
  1. Determinacy & method. Pin + 2 rollers ($r{=}4$) with one hinge is determinate, so every influence line is piecewise-linear; place a unit load at the control points and solve by statics.
  2. (iii) $R_C$. With the load on $A\!-\!D$, $R_C=a/10$; it reaches $1$ at $C$ and peaks $\boxed{+1.20}$ at the hinge $D$ ($a{=}12$); on the suspended span $D\!-\!E$ it falls linearly back to $0$ at $E$.
  3. (ii) Shear just right of $C$. The ordinate steps across $C$; its maximum absolute value is $\boxed{1.0}$ (just to the right of the support), tapering to zero at $A$ and beyond the hinge.
  4. (i) $M_B$ (mid-point of $A\!-\!C$). A triangular influence line peaking under $B$ with maximum ordinate $\boxed{+2.5\text{ m}}$ (the standard $ab/L=5\cdot5/10$), zero at $A,C$ and along the suspended span.

5(b) — Moving vehicle on a deck truss

Given. Deck truss, top chord $U_1(0)\!-\!U_4(24)$ at $8\text{ m}$ panels, height $3\text{ m}$; bottom nodes $L_1(4),L_2(12),L_3(20)$, pin $L_1$, roller $L_3$. A vehicle idealised as $40,40,20\text{ kN}$ at spacings $2\text{ m}$ and $4\text{ m}$ rolls along the upper chord. Find. The influence line for the diagonal $U_2\!-\!L_1$ and the maximum compression it produces.

U1U2U3U4L1L2L3
5(b) — the long diagonal $U_2\!-\!L_1$ (highlighted); load travels along the top chord.
  1. IL ordinates by sections. Placing a unit load at each top panel point and cutting the panel gives ordinates for $U_2L_1$ of $\boxed{-0.417,\,-1.25,\,-0.417,\,+0.417}$ at $U_1,U_2,U_3,U_4$ (negative = compression), linear between.
  2. Place the load train for maximum compression. Sweeping the $40/40/20\text{ kN}$ group across the ordinates, the extreme $\sum P_i\eta_i$ occurs with the leading $40\text{ kN}$ near $U_2$: $\boxed{U_2L_1=108.3\text{ kN compression}}$.

Check: The two wall supports of the 4(b) truss and the exact web of both trusses were read from the printed figure and confirmed by a full method-of-joints check; the 5(b) ordinates are confirmed by an independent section cut at each panel.

ResponseMax-magnitude influence ordinate
5(a) $R_C$$+1.20$ (at hinge $D$)
5(a) $V_{C^+}$$1.0$
5(a) $M_B$$+2.5\text{ m}$
5(b) $U_2L_1$$-1.25$ (peak); vehicle $\Rightarrow 108.3\text{ kN (C)}$