Influence lines for the shear immediately left of support B: 07-Str-A4, December 2018, Q2

Question 2: Influence lines for the shear immediately left of support B (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. Two three-support beams of the same overall layout. In (a) an internal hinge is drawn between A and B, which makes the beam determinate; (b) has no hinge and is one degree indeterminate. No dimensions are printed; for the ordinates below the drawn proportions are taken as span AB = 10 m with the hinge 7.0 m from A, and span BC = 10 m.

Find. The influence line for the shear force on the section immediately to the left of support B in each beam, and the ordinate of largest absolute value on each.

Structure (a) — roller, internal hinge, roller, pin. The hinge is what makes the beam determinate.

Approach. Work from first principles rather than from a remembered picture: the shear just left of B equals the sum of the upward forces acting to the left of that section, which for both beams is $V = R_{A} - 1$ when the unit load stands left of the section and $V = R_{A}$ when it stands right of it. Everything then follows from the influence line for $R_{A}$.

(a) Split the determinate beam at the hinge. Segment A–H is a simple span carried by the roller at A and by the hinge force at H; segment H–B–C is a beam on B and C with the overhang H–B. The decisive consequence is that a load anywhere to the right of the hinge delivers nothing to A: with no load on A–H the hinge force is zero, so $R_{A} = 0$.
(a) Unit load between A and the hinge. The simple span gives $R_{A} = (x_{h} - x)/x_{h}$, so $$\eta(x) = R_{A} - 1 = -\dfrac{x}{x_{h}}$$ a straight line from 0 at A to $\boxed{-1.000}$ at the hinge.
(a) Unit load between the hinge and B. Now $R_{A} = 0$ and the load is still left of the section, so $\eta = 0 - 1 = -1.000$ for every position: the influence line runs flat at −1.000 from the hinge to B. This plateau is the feature that distinguishes the determinate answer, and it is the part most often drawn as a continuing ramp.
(a) Unit load in span BC. The load is now right of the section and $R_{A}$ is still zero, so $\eta \equiv 0$ throughout BC. The influence line jumps by exactly 1.000 across support B, as it must for a section adjacent to a support.

Influence line (a). Maximum absolute ordinate 1.000, held constant from the hinge to the section; identically zero in span BC.

Structure (b) — the same supports without the hinge, so one degree indeterminate.

(b) Choose the redundant. Take the interior reaction $R_{B}$ as the redundant on a simple beam of span $S = L_{1} + L_{2}$. With $\delta_{BB} = L_{1}^{2}L_{2}^{2}/(3EIS)$ and the standard mid-point deflection formula for $\delta_{Bx}$, the reaction influence line comes out in closed form. For the load in span AB, with $\xi = x/L$ and equal spans, $$R_{A}(\xi) = 1 - \tfrac{5}{4}\xi + \tfrac{1}{4}\xi^{3}$$
(b) Convert to shear. Left of the section $\eta = R_{A} - 1 = -\tfrac{5}{4}\xi + \tfrac{1}{4}\xi^{3}$, a cubic that falls monotonically from 0 at A to $$\eta(B^{-}) = \boxed{-1.000}$$ That value is exact for any span ratio: with the unit load standing directly over B the reaction at A vanishes, so the shear just left of the section is $0 - 1$.
(b) The far span reverses. Loading BC lifts the far end, so $R_{A}$ — and with it $\eta$ — goes negative again. Writing $s$ for the distance from C and $\sigma = s/L_{2}$, $$\eta = \dfrac{L_{2}^{2}}{2L_{1}S}\left(\sigma^{3} - \sigma\right)$$ which for equal spans reduces to $\tfrac{1}{4}(\sigma^{3}-\sigma)$ and reaches its extremum at $\sigma = 1/\sqrt{3}$: $$\eta_{\min,BC} = -0.0962 \ \text{at}\ 0.577L_{2} = 5.77\ \text{m from C}$$
(b) Compare the two answers. Both beams give the same governing ordinate, −1.000 immediately left of B, because that value is a statement of statics at the section and cannot depend on the redundancy. What the redundancy changes is the shape: (a) is piecewise straight with a plateau and a dead span, (b) is cubic everywhere and spills a small reversed lobe into the far span. Reporting only the number without that distinction throws away most of the eight marks.

Influence line (b). The cubic reaches −1.000 at the section and carries a small secondary lobe of −0.0962 into span BC.

Question 2 — influence-line ordinates
Beam	Feature	Ordinate
(a) determinate	At A	0
	Hinge to B (constant plateau)	−1.000 (maximum absolute)
	Throughout span BC	0
(b) indeterminate	At A	0
	Immediately left of B	−1.000 (maximum absolute)
	Extremum in span BC	−0.0962 at 0.577L from C

Topic. Influence lines for shear at a section adjacent to a support, determinate and indeterminate.

Key relations. $V_{B^{-}} = R_{A} - 1$ (load left of the section), $V_{B^{-}} = R_{A}$ (load right of it); Müller-Breslau: the influence line for an action is the deflected shape produced by releasing that action and imposing a unit relative displacement; far-span reaction lobe $R_{A} = L_{2}^{2}(\sigma^{3}-\sigma)/(2L_{1}S)$.

Why this works. Naming the free body before writing anything is what makes both parts short. In (a) the hinge severs the load path: the piece to the left of the hinge can only feel loads standing on itself, so $R_{A}$ is identically zero for every load position beyond it, and the influence line is forced flat. In (b) the beam is continuous, so a load anywhere still reaches A, but only through the redundant — and when it stands in the far span it does so with the opposite sign, which is the physical origin of the reversed lobe and of the hold-down force real two-span beams need at their end supports. The unit ordinate at the section itself is a matter of statics alone: it survives any change of span ratio, any change of EI and the presence or absence of the hinge.

Common pitfall. Drawing the textbook two-triangle shear influence line for a section at mid-span instead of one adjacent to a support — here there is no beam to the right of the section within span AB, so the whole diagram is one-signed. The second pitfall is sketching (a) as a single straight ramp from A to B, which quietly ignores the hinge and loses the plateau that the eight marks are actually testing.

Source. Hibbeler, Structural Analysis, 10th ed., §6.1–6.3 (determinate) and §10.10 (Müller-Breslau for indeterminate beams); Kassimali, 6th ed., §8.1 and §13.7.

Question 2 of 9: Influence lines for the shear immediately left of support B

Question 2: Influence lines for the shear immediately left of support B (8 marks)