Influence Lines - Structural Analysis - Civil Engineering (CE) PDF Download

Influence Line Diagram and Rolling Loads

An influence line diagram (ILD) shows the variation of a structural response (reaction, shear force, bending moment, axial force, slope or deflection) at a specific point of a structure as a unit concentrated load moves across the structure from one end to the other. The ordinate of an ILD at a given position equals the value of the chosen response when a unit load is placed at that position.

Simply Supported Beam

The ordinate of the influence line at a section (denote ordinate by y₁) represents the magnitude of that stress function when a unit concentrated load is placed at that section.

1. Meaning of ordinate. The ordinate (y₁) at a point on the ILD is the value of the required response (for example reaction, shear, moment) caused by a unit load placed at the abscissa corresponding to that ordinate.
2. Total vertical reaction due to a system of loads. For a given loading system consisting of concentrated loads P₁, P₂ and a uniformly distributed load (UDL) of intensity w over a length A of the span, the resultant vertical reaction at B is obtained by superposition as
   R_B = P₁ Y₁ + P₂ Y₂ + wA where Y₁ and Y₂ are the ordinates of the ILD for reaction at B at the positions of P₁ and P₂ respectively, and wA represents the contribution of the distributed load equivalent to the area under the ILD over the loaded region.
3. Influence line for shear at a section C. The ILD for shear at C shows the variation of shear just to the right (or just to the left) of the section as a unit load moves. It is piecewise linear with a jump of magnitude 1 at the section when the unit load crosses the section.
4. Influence line for bending moment at a section C. The ILD for bending moment at C is typically triangular (or piecewise linear) with maximum ordinate at the position where the unit load is at C.
5. Maximum bending moment at C due to moving loads. For a system of loads including concentrated loads and a UDL, the bending moment maximum at C can be obtained by superposition. For example, for a concentrated load P at some position and a distributed load component, a form of the general result is
   BM_max at C = P·a·b / l + wA where a and b are distances partitioning the span l and wA denotes contribution of distributed loading obtained from the area of ILD under the loaded region.
6. Other illustrative ILD shapes and examples. Influence lines for various responses on a simply supported beam (reactions, shear at a point, moment at a point) are standard and are either straight lines or piecewise linear polygons depending on the location of supports, internal hinges and load positions.
7. Position of moving load for maximum shear at C. The positions of a moving uniform load (or moving concentrated loads) that produce maximum positive or negative shear at C may be found by sliding the load and comparing the algebraic integrals (areas) of the ILD under the loaded portion. The maximum positive and negative shear values can be written in terms of w and the appropriate loaded-area quantities:
   Maximum positive SF at C = w·A₁
   Maximum negative SF at C = w·A₂

Position of load for maximum positive SF at C 

Position of load for maximum negative SF at C 

Position of load for maximum positive SF at C

Constructing Influence Lines - basic procedures

• Unit-load (static) method. Place a unit load at a position x along the span. Compute the response (reaction, shear, moment, axial force) at the required point by static equilibrium. Repeat for different x to obtain the ILD.
• Superposition. For multiple moving loads or distributed moving loads, evaluate the contribution of each load by multiplying the load value by the IL ordinate at its position and sum (or integrate) to get the total response.
• Müller-Breslau principle (graphical method). To get the qualitative shape of an ILD for a function (reaction, shear, moment), remove the restraint associated with that function and apply a unit displacement or rotation corresponding to the function. The deflected shape (subject to sign convention) is the influence line. This principle is useful to sketch ILDs quickly and to determine where ordinates are positive or negative.

Worked derivation: Influence line for reaction at B, shear and moment at a section C on a simply supported span

Consider a simply supported beam AB of span L with a unit load at a distance x from A. Let section C be at distance a from A (0 ≤ a ≤ L).

Reaction at B when unit load is at x:

The reaction at B is obtained by vertical equilibrium:

R_B = x / L

This is the IL ordinate for reaction at B when the unit load is at x (so the IL for RB is a straight line varying from 0 at A to 1 at B).

Bending moment at section C due to unit load at x (derivation).

For x ≤ a (load to the left of C):

M_C = (reaction at A)·a - (unit load)·(a - x)

reaction at A = (L - x)/L

Therefore, M_C = a·(L - x)/L - (a - x)

Simplifying, M_C = x·(L - a)/L

For x ≥ a (load to the right of C):

M_C = (reaction at A)·a

reaction at A = (L - x)/L

Therefore, M_C = a·(L - x)/L

Thus the influence line for moment at C is:

M_C = x·(L - a)/L for x ≤ a, and M_C = a·(L - x)/L for x ≥ a.

The maximum ordinate occurs when the unit load is at the section (x = a) and equals a·(L - a)/L.

Shear at section C (just to the right of the cut):

For a unit load at x ≤ a (left of the section):

V = reaction at A - (loads to left of section)

reaction at A = (L - x)/L, loads to left include the unit load ⇒ V = (L - x)/L - 1 = -x/L

For a unit load at x ≥ a (right of the section):

V = reaction at A (no loads to left of section) = (L - x)/L

So the IL for shear just to the right of C is piecewise linear with

V = -x/L (for x ≤ a) and V = (L - x)/L (for x ≥ a)

The jump across the section when the unit load crosses C equals 1 (consistent with the discontinuity of shear due to a concentrated load passing the section).

Simply Supported Beam with Overhang

When a beam is simply supported but has an overhang, influence lines remain piecewise linear but require treating the overhang region separately. Ordinates are found by placing a unit load at positions on the overhang and within the main span and computing the response by equilibrium. The ILD will exhibit linear segments corresponding to these regions and may show sign changes near the support and overhang tip.

Simply Supported Beam Carrying Internal Hinge

An internal hinge divides the span into two statically determinate segments. Influence lines for internal hinge reaction or bending at a point on either side are found by placing a unit load and solving equilibrium on the relevant segment(s). The IL shape often has zero ordinate at locations where the hinge produces no moment and piecewise linear behaviour around the hinge.

Effect of Rolling Loads (Moving UDL of finite length)

To find maximum shear force and bending moment at a section C when a uniformly distributed load (UDL) of length l′ (l′ < L) moves across a span, use the following general ideas:

Key rules for a UDL of length l′:

1. Maximum negative shear at C. Occurs when the head of the loaded region is just to the left of C so that the loaded portion on the left is AC. The maximum negative shear equals w × (area of ILD under the loaded region).
2. Maximum positive shear at C. Occurs when the tail of the loaded region is just to the right of C. The maximum positive shear equals w × (area of ILD under the loaded region) for that position.
3. Maximum bending moment at C. Occurs when the loaded region is positioned so that the average loading on the left of C equals the average loading on the right of C - equivalently, when the resultant of the loaded portion divides the span in the same ratio as the span is divided by C. Practically, the maximum moment at C is obtained by placing the l′-length UDL so as to maximise the integral of the ILD under the loaded part (i.e., align the loaded portion to cover the region where the ILD ordinates are largest).

For a moving distributed load, compute the response by integrating the product of load intensity w and the IL ordinate over the loaded length. For concentrated moving loads, sum the products of each concentrated load and the IL ordinate at its position.

Influence Line Diagram for Truss Members

Influence lines for axial force in a truss member can be constructed by placing a unit load at nodes (joints) and resolving the truss by methods such as the method of joints or sections to obtain the axial force in the member for each position of the unit load. Alternatively, apply the Müller-Breslau principle for axial force: release the member (remove it) and apply a unit displacement in the axial direction of the removed member; the resulting axial-deflection shape (projected into the member axis) is proportional to the influence line of axial force in that member.

ILD for Inclined Members

For an inclined member of a truss or frame, the IL for axial or shear/normal component is obtained by resolving the unit-load-induced forces along the member axis. Apply a unit load at each relevant joint, resolve forces by equilibrium, and plot the axial force in the inclined member as the unit load moves. The IL will generally be non-symmetric and piecewise linear depending on geometry and connectivity.