13.3 Influence Lines for Determinate and Indeterminate Structures

13.3 Influence Lines for Determinate and Indeterminate Structures

Bridge girders, crane runways, and industrial highway structures are subjected to transient vehicular moving loads. Unlike buildings carrying stationary dead loads, bridge design requires structural engineers to determine which vehicle positions produce governing maximum internal reactions, shears, and moments. The primary analytical tool for this task is the Influence Line (IL).

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1. Concept of the Influence Line

An Influence Line is a graph showing the variation of a specific structural reaction or internal action (shear force, bending moment, axial force, or deflection) at one specific fixed cross-section as a single unit vertical concentrated load ($1\text{ kN}$) traverses across the span.

Influence Lines vs. Shear and Moment Diagrams

A frequent source of confusion among candidates is the distinction between Shear Force / Bending Moment Diagrams (SFD/BMD) and Influence Lines:

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2. Influence Lines for Statically Determinate Beams

Consider a simply supported beam $AB$ of span $L$. A unit load ($1\text{ kN}$) moves from $x = 0$ (support $A$) to $x = L$ (support $B$).

Support Reactions

• Reaction $R_A$: When the load is at $x$, summing moments about $B$ gives $R_A(L) - 1(L - x) = 0$: $IL(R_A) = 1 - \frac{x}{L}$ Shape: Linear triangle, starting at $+1.0$ at $A$ ($x=0$) and decreasing to $0.0$ at $B$ ($x=L$).
• Reaction $R_B$: Summing moments about $A$ gives $R_B(L) - 1(x) = 0$: $IL(R_B) = \frac{x}{L}$ Shape: Linear triangle, starting at $0.0$ at $A$ and rising to $+1.0$ at $B$.

Shear at Section $C$ ($x = a$ from left support, $b = L - a$ from right support)

• When the unit load is to the left of $C$ ($0 \le x < a$): $V_C = -R_B = -\frac{x}{L}$ At $x = 0$, $V_C = 0$. As $x \to a^-$, $V_C = -\frac{a}{L}$.
• When the unit load is to the right of $C$ ($a < x \le L$): $V_C = +R_A = 1 - \frac{x}{L} = +\frac{b}{L}$ At $x \to a^+$, $V_C = +\frac{b}{L}$. At $x = L$, $V_C = 0$.
• Discontinuity: At section $C$, the influence line drops by exactly $1.0$ unit: $\Delta V = +\frac{b}{L} - \left(-\frac{a}{L}\right) = \frac{a+b}{L} = 1.0$

Bending Moment at Section $C$ ($x = a$)

• When the unit load is to the left of $C$ ($0 \le x \le a$): $M_C = R_B \cdot b = \left(\frac{x}{L}\right) b = \frac{b x}{L}$
• When the unit load is to the right of $C$ ($a \le x \le L$): $M_C = R_A \cdot a = \left(1 - \frac{x}{L}\right) a = \frac{a (L - x)}{L}$
• Peak Ordinate: At $x = a$, both equations yield the maximum moment ordinate: $y_{\text{max}} = \frac{a b}{L}$ Shape: A triangle with apex ordinate $\frac{ab}{L}$ at section $C$, sloping linearly to zero at supports $A$ and $B$.

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3. The Müller-Breslau Principle

Heinrich Müller-Breslau formulated in 1886 a powerful geometric theorem for constructing qualitative and quantitative influence lines:

Müller-Breslau Principle: The influence line for any reaction or internal force of a structure is proportional to the deflected shape of the structure obtained by removing the restraint corresponding to that action and introducing a corresponding unit virtual displacement (or rotation) in the direction of the action.

Application to Determinate vs. Indeterminate Structures

• Statically Determinate Structures: Removing a restraint converts the structure into a kinematically determinate mechanism (rigid bodies connected by hinges). Because rigid segments cannot bend, the resulting deflected shape consists entirely of straight line segments.
• Statically Indeterminate Structures: Removing one constraint leaves the remaining primary structure statically stable. Applying a unit displacement induces elastic bending. Therefore, the influence line for any action in an indeterminate structure consists of smooth continuous elastic curves.

Qualitative IL for Middle Reaction R_B of a Two-Span Continuous Beam:
[A]▲================[B]▲================[C]▲

Step 1: Remove vertical restraint at support B.
Step 2: Push joint B upward by unit displacement Δ_B = 1.0.

Deflected Elastic Shape (= IL for R_B):
        +1.0
         ▲
       /   \
[A]▲--/     \--▲[C]
(Continuous smooth curve, concave downward over B)

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4. Quantitative Application of Influence Lines

Once the influence line for an internal action $F$ is established, the total response under any service loading train is computed by superposition:

Concentrated Moving Loads

For a series of discrete concentrated wheel loads $P_1, P_2, \dots, P_n$ located at positions corresponding to influence ordinates $y_1, y_2, \dots, y_n$:

$F = \sum_{i=1}^n P_i y_i$

Uniform Distributed Live Load ($w_{LL}$)

For a uniform live load $w_{LL}$ extending from $x_1$ to $x_2$:

$F = \int_{x_1}^{x_2} w_{LL} y(x) dx = w_{LL} \times (\text{Net Area under IL between } x_1 \text{ and } x_2)$

• To obtain the maximum positive effect, place uniform live load only over segments where the influence line is positive.
• To obtain the maximum negative effect, place uniform live load only over segments where the influence line is negative.

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5. Moving Load Analysis: Absolute Maximum Effects

In bridge design, vehicular traffic consists of wheel load trains (such as the standard Philippine DPWH / AASHTO truck). Structural engineers must identify the absolute maximum bending moment ($M_{\text{abs,max}}$) and absolute maximum shear ($V_{\text{abs,max}}$) that can develop anywhere along the span.

Absolute Maximum Shear ($V_{\text{abs,max}}$)

For simply supported spans, absolute maximum shear always occurs at one of the end supports (where the influence ordinate is $+1.0$). The heaviest wheel load is positioned directly over the support, with trailing loads placed within the span on the steepest branch of the influence line.

Absolute Maximum Bending Moment: The Centerline Bisection Theorem

The absolute maximum bending moment does not generally occur at midspan ($L/2$), nor does it occur under the resultant $R$ of the loads. It occurs under one of the critical wheel loads near the resultant.

The Bisection Rule:

1. Compute the magnitude of the total resultant force of the wheel group: $R = \sum P_i$.
2. Determine the location of the resultant $\bar{x}_R$ by taking moments about the lead wheel: $\bar{x}_R = \frac{\sum P_i x_i}{R}$.
3. Identify the critical heavy wheel load $P_k$ located closest to the resultant $R$. Let $d$ be the distance between $P_k$ and $R$.
4. Placement for Maximum Moment: Position the load train on the beam such that the centerline of the beam bisects the distance $d$ between the critical wheel load $P_k$ and the resultant $R$:

$\text{Distance from nearest support to } P_k = \frac{L}{2} - \frac{d}{2}$ $\text{Distance from nearest support to } R = \frac{L}{2} + \frac{d}{2}$

1. Calculate the bending moment directly beneath load $P_k$ under this placement to establish $M_{\text{abs,max}}$.

Impact Factor / Dynamic Load Allowance ($IM$)

Vehicular motion, surface roughness, and engine vibration produce dynamic amplification. In classical Philippine bridge engineering (AASHTO Standard Specifications), the impact fraction $I$ is calculated as:

$I = \frac{15.24}{L + 38} \le 0.30$

where $L$ is the loaded span length in meters. In AASHTO LRFD Bridge Specifications, the dynamic load allowance is taken as a fixed percentage: $IM = 33\%$ for bridge deck and girder strength limit states, and $15\%$ for fatigue.

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6. Pattern Loading of Continuous Beams and Frames

Influence lines also explain the TOS competency "compute the maximum reactions with and without pattern loading." For a continuous beam, the influence line for positive moment in a span is positive over that span and over alternate spans, and negative over the adjacent spans. Live load should therefore be placed only where the influence ordinates have the desired sign.

ACI 318-14 (Section 6.4.2), which NSCP 2015 follows, allows the live load arrangement to be limited to two cases:

The same logic gives the maximum reaction at an interior support: load the two spans that meet at that support, then every other span beyond them.

ACI approximate moment and shear coefficients

For continuous beams and one-way slabs with at least two spans, roughly equal spans (the larger of two adjacent spans not more than 20% longer than the shorter), uniform load, and unfactored live load not exceeding three times the dead load, ACI 318-14 Section 6.5 permits $M_u = C_m w_u \ell_n^2$ and $V_u = C_v w_u \ell_n / 2$:

Example. A three-span continuous beam has clear spans of $6.0\text{ m}$ and $w_u = 40\text{ kN/m}$. The negative moment at the exterior face of the first interior support is $w_u \ell_n^2 / 10 = 40(36)/10 = 144\text{ kN}\cdot\text{m}$. The interior-span positive moment is $40(36)/16 = 90\text{ kN}\cdot\text{m}$.

7. Comprehensive Worked Example

Worked Example: Absolute Maximum Moment Under a Three-Axle Truck

Problem: A simply supported bridge girder has a span of $L = 20.0\text{ m}$. It is traversed by a three-axle truck moving from left to right. The wheel loads are $P_1 = 50.0\text{ kN}$ (front axle), $P_2 = 150.0\text{ kN}$ (drive axle, $4.0\text{ m}$ behind $P_1$), and $P_3 = 150.0\text{ kN}$ (rear axle, $6.0\text{ m}$ behind $P_2$). Determine: (a) resultant location, (b) critical truck placement, and (c) the absolute maximum bending moment $M_{\text{abs,max}}$.

   P1 = 50 kN        P2 = 150 kN              P3 = 150 kN
       ↓                  ↓                        ↓
       |------ 4.0 m -----|--------- 6.0 m --------|

Solution:

• Step 1: Resultant Force and Location: $R = P_1 + P_2 + P_3 = 50.0 + 150.0 + 150.0 = 350.0\text{ kN}$ Taking moments about front axle $P_1$: $R \cdot \bar{x} = 50(0) + 150(4.0) + 150(4.0 + 6.0) = 600 + 1500 = 2100\text{ kN}\cdot\text{m}$ $\bar{x} = \frac{2100}{350} = 6.0\text{ m} \text{ behind } P_1$ Since $P_2$ is located $4.0\text{ m}$ behind $P_1$, the resultant $R$ is located: $d = 6.0 - 4.0 = 2.0\text{ m} \text{ behind } P_2$ (Between $P_2$ and $P_3$, closer to $P_2$).

• Step 2: Centerline Bisection Placement: The closest heavy load to the resultant is $P_2 = 150\text{ kN}$ (distance $d = 2.0\text{ m}$). The beam centerline ($L/2 = 10.0\text{ m}$) must bisect the distance between $P_2$ and $R$:

  • Distance of $P_2$ from left support $A$: $x_2 = \frac{L}{2} - \frac{d}{2} = 10.0 - \frac{2.0}{2} = 9.0\text{ m}$
  • Resultant $R$ is at $x_R = 9.0 + 2.0 = 11.0\text{ m}$ from $A$. (Centerline is at $10.0\text{ m}$, exactly halfway between $9.0\text{ m}$ and $11.0\text{ m}$!).
  • Location of all axles:
    • $P_1$ is at $x_1 = 9.0 - 4.0 = 5.0\text{ m}$ from $A$.
    • $P_2$ is at $x_2 = 9.0\text{ m}$ from $A$.
    • $P_3$ is at $x_3 = 9.0 + 6.0 = 15.0\text{ m}$ from $A$. Verification: All three axles lie within the $20.0\text{ m}$ span ($0 < 5.0, 9.0, 15.0 < 20.0$).

• Step 3: Support Reactions and Maximum Moment: Compute reaction $R_A$ using the resultant $R = 350\text{ kN}$ at $x_R = 11.0\text{ m}$: $R_A = \frac{R (L - x_R)}{L} = \frac{350 (20.0 - 11.0)}{20.0} = \frac{350 (9.0)}{20.0} = 157.5\text{ kN}$ Compute bending moment under load $P_2$ at $x_2 = 9.0\text{ m}$: $M_{\text{abs,max}} = R_A \cdot x_2 - P_1 (x_2 - x_1)$ $M_{\text{abs,max}} = 157.5 \times 9.0 - 50.0 \times (9.0 - 5.0)$ $M_{\text{abs,max}} = 1417.5 - 200.0 = 1217.5\text{ kN}\cdot\text{m}$

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8. Licensure Exam Pitfalls & Review Notes