NivaarExam PrepOfficial exam papers ↗

16-Civ-A1 Elementary Structural Analysis · May 2016

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 / slope–deflection, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams and three-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.

Question 5: Influence lines (9 + 12 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) — compound (Gerber) beam

Given. Pin $A(0)$, hinge at $3\text{ m}$, rollers $B(5),C(9)$, hinge at $11\text{ m}$, roller $D(14)$. The centre span $3\!-\!11\text{ m}$ rests on $B,C$; the end pieces $A\!-\!$hinge and hinge$\!-\!D$ are suspended.

Ah1BCh2D3 m2 m4 m2 m3 m
5(a) Gerber beam (○ = internal hinge).
  1. (i) Reaction $R_B$. On the centre span (on $B,C$) the ordinate is $1$ at $B$, $0$ at $C$, and rises on the left overhang to $\tfrac{9-3}{4}=1.5$ at the hinge ($x=3$); the suspended end span carries it linearly to $0$ at $A$. Peak coefficient $\boxed{1.5}$ at the left hinge.
  2. 1.51.00IL for R_B
    5(a)(i) Influence line for $R_B$.
  3. (ii) Moment over $B$. $M_B$ is zero at $B$ itself, hogging as the load moves onto the left overhang, reaching $-2.0\text{ kN}\cdot\text{m/kN}$ at the hinge ($x=3$), then linear to zero at $A$; on the right it forms the usual span lobe. Peak magnitude $\boxed{2.0}$.
  4. −2.00IL for M over B (kN·m/kN)
    5(a)(ii) Influence line for $M_B$.
  5. (iii) Shear just left of $C$. The step at $C$ reaches $-1.0$ immediately to the left of the support (unit load just left of $C$); peak magnitude $\boxed{1.0}$.
  6. −1.00IL for V just-left of C
    5(a)(iii) Influence line for shear left of $C$.

5(b) — moving vehicle

Given. Pin $A(4)$, roller $B(20)$, point $C$ at $x=8\text{ m}$ (span $A\!-\!B=16\text{ m}$, overhangs $4\text{ m}$). Vehicle: $24,24,8\text{ kN}$ at spacings $2\text{ m}$ then $4\text{ m}$, travelling left→right.

e1ACBe24 m4 m12 m4 m
5(b) Simply supported beam with overhangs; point $C$ at $4\text{ m}$ from $A$.
  1. Influence lines at $C$. $M_C$ is a triangle peaking at $\dfrac{ab}{L}=\dfrac{4\cdot12}{16}=3.0\text{ m}$ at $C$ (dipping negative over the left overhang). $V_C$ steps at $C$: $+0.75$ just right, $-0.25$ just left.
  2. 3.000IL for M at C (m)
    5(b)(i) Influence line for $M_C$.
    +0.75−0.25IL for V at C
    5(b)(ii) Influence line for $V_C$.
  3. Governing effects. Positioning the $24,24,8\text{ kN}$ axles on the ordinates gives the maximum bending moment at $C$, $M_C=\boxed{144\text{ kN}\cdot\text{m}}$, and a shear envelope at $C$ of $\boxed{+36\text{ kN}}$ to $-9\text{ kN}$ (governing $|V_C|=36\text{ kN}$).
QuantityMax abs. influence coeff.Governing load effect
5(a) $R_B$$1.5$—
5(a) $M_B$$2.0$ kN·m/kN—
5(a) $V$ left of $C$$1.0$—
5(b) $M_C$$3.0$ m$144$ kN·m
5(b) $V_C$$+0.75/-0.25$$+36\,/\,-9$ kN