Influence line (IL) for continuous beams

Statement

Muller-Breslau principle: the influence line for a force-type response (a support reaction, shear force or bending moment) at a point in a structure has the same shape as the deflected shape of the structure obtained when the restraint that carries that response is removed and a unit displacement is imposed in its direction at that point. The ordinate of the ILD at any point is the displacement of the structure at that point, measured in the direction of the unit load.

Basis (virtual work): after removing the restraint, the virtual work of the real forces Q (the response) and the unit load P = 1 moving through the displaced shape must be zero, so $Q\cdot\Delta_Q = 1\cdot y(x)$. With $\Delta_Q = 1$ the displacement $y(x)$ equals the influence ordinate.

Procedure

1. Remove the restraint related to the response and apply the corresponding unit displacement.
2. Reaction: remove the support and displace it by 1 vertically. Shear: cut the beam at the section and give a unit relative vertical displacement (the two parts stay parallel). Bending moment: insert a hinge at the section and give a unit relative rotation.
3. Draw the deflected shape; the ordinate at any point is the influence value.

Determinate and indeterminate structures

• In a statically determinate beam the displaced shape consists of straight lines, so the ordinates are exact.
• In a continuous (indeterminate) beam the rest of the beam remains supported, so the shape is curved and is found from the elastic curve (moment-area/compatibility). The principle gives the correct qualitative shape immediately, so the critical loading can be decided (for example, load alternate spans for maximum positive moment).

Example: two equal spans of 6 m (A hinged, B and C rollers), ILD for the support moment $M_B$

Insert a hinge at B (so the beam cannot carry moment there) and give the two sides a unit relative rotation in the hogging sense (the beam kinks into an inverted V at B). A, B and C stay on the axis, so each span bends into a curve below the axis, with a total kink of 1 radian at B. The deflected shape is the ILD: both spans lie below the axis, i.e. a load on either span produces a hogging moment at B (negative in the sagging-positive convention).

Ordinates from the elastic curve (equal spans, constant $EI$): $M_B = -\dfrac{x(L^2 - x^2)}{4L^2} = -\dfrac{x(36 - x^2)}{144}$ for a load at $x$ from A in span AB (and symmetric in BC). Sample: $x = 3$: $M_B = -\dfrac{3(36-9)}{144} = -0.5625$.

**----------------------*----------------------**
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^                       ^                       ^

Thus a load on either span produces hogging at B; for the maximum hogging moment at B load both spans. Similarly, the ILD of the reaction $R_B$ is the deflected shape when the support B is removed and B is lifted by 1 (positive in both spans, equal to 1 at B):

The largest hogging ordinate of $M_B$ is about 0.56 m (kN·m per kN), near mid-span of each span; the $R_B$ ordinate is 1.0 under B. A UDL $w$ over both spans gives $R_B = w\times$ (area of the ILD).

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method

Propped cantilever AB, $L = 8$ m, A fixed, B roller, section 2-2 at 4 m from A. Redundant: $R_B = \dfrac{x^2(3L-x)}{2L^3}$. Shear at the section (left part): load left of the section, $V = -R_B$; load right, $V = 1 - R_B$.

Sample: $x = 6$: $R_B = \dfrac{36\times18}{2\times512} = 0.633$, $V = 0.367$. At the section: $R_B(4) = \dfrac{16\times20}{1024} = 0.3125$, giving $V = -0.3125$ just left and $+0.6875$ just right.

Ordinates (2 m interval)

ILD shape

                        |***
                        |   ******
                        |         ******
                        |               ******
*************-----------|---------------------***
             *********  |
                      **|


^                                               ^

Jump of 1.0 at the section (from $-0.3125$ to $+0.6875$); the ILD is zero at A and B.

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method

Propped cantilever AB, $L = 10$ m, A fixed, B roller. The reaction moment at A is the moment at the fixed support (hogging, so negative in the above convention). Muller-Breslau: replace the fixed support by a hinge and give it a unit rotation; the deflected shape of the beam gives the ILD.

Analytically, with a unit load at $x$ from A, $a = x$, $b = L - x$: fixed-end moment at A is $ab^2/L^2$ and the fixed-end moment at B is $a^2b/L^2$. Releasing B (roller) carries half of the latter to A:

$$M_A = \frac{ab^2}{L^2} + \frac{a^2 b}{2L^2} = \frac{x(L-x)(2L-x)}{2L^2}\quad(\text{hogging})$$

Sample: $x = 4$: $M_A = \dfrac{4\times6\times16}{2\times100} = 1.92$ (hogging).

Ordinates (2 m interval; values shown as moments in the sagging-positive convention)

ILD shape

**--------------------------------------------***
  **                                     *****
    ****                            *****
        *****                *******
             ****************
^                                               ^

The ILD is entirely hogging, zero at A and B, with a maximum $-1.92$ at about $x = 4.2$ m ($0.42L$).

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method (Muller-Breslau)

Propped cantilever AB of span $L = 10$ m: A fixed, B roller, constant $EI$. Remove the support at B and give B a unit upward displacement. The ILD for $R_B$ is the deflected shape of the cantilever AB (fixed at A) scaled so that the ordinate at B equals 1.

By Maxwell's reciprocal theorem, the deflection at B due to a unit load at $x$ from A equals the deflection at $x$ due to a unit load at B. So

$$R_B(x) = \frac{\delta_{Bx}}{f_{BB}} = \frac{x^2(3L-x)/6EI}{L^3/3EI} = \frac{x^2(3L-x)}{2L^3}$$

Sample: $x = 8$ m: $R_B = \dfrac{8^2(3\times10 - 8)}{2\times10^3} = 0.704$.

Ordinates (2 m interval)

ILD shape

                                            *****
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               ************
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^                                               ^

The ordinate is 0 at A (a load at the fixed support is carried by A), increases with zero slope at A and reaches 1 at B. It is positive throughout, so a downward load always increases $R_B$. A UDL over the whole span gives $R_B = w\times$ (area of the ILD) $= \dfrac{3wL}{8}$.

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method

Propped cantilever AB, $L = 5$ m, A fixed, B roller, constant $EI$. The structure is indeterminate to the first degree; take $R_B$ as the redundant. For a unit load at $x$ from A, the deflection at B of the released cantilever is $\delta_{Bx} = \dfrac{x^2(3L-x)}{6EI}$, and $f_{BB} = \dfrac{L^3}{3EI}$, so

$$R_B = \frac{x^2(3L-x)}{2L^3},\qquad R_A = 1 - R_B,\qquad M_A = x - R_BL\ \text{(hogging)} = \frac{x(L-x)(2L-x)}{2L^2}$$

Sample: $x = 2$: $R_B = \dfrac{4\times13}{250} = 0.208$, so $R_A = 0.792$; $M_A = 2 - 0.208\times5 = 0.96$ (hogging).

ILD for the vertical reaction $R_A$ at the fixed end (1 m interval)

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^                                               ^

Muller-Breslau: release the vertical restraint at A (keeping the fixed rotation) and displace A upward by 1; the shape equals $1 - R_B(x)$.

ILD for the reaction moment $M_A$ at the fixed end (1 m interval; hogging shown negative)

The moment ILD is negative (hogging) throughout, with a maximum of about $-0.96$ at $x \approx 0.42L = 2.1$ m, and it is zero at A and B.

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method

Propped cantilever, span $L = 7$ m, fixed at the left end and on a roller at the right end. Section A is 1 m from the right end, i.e. 6 m from the fixed end. Redundant: $R_B$.

$$R_B = \frac{x^2(3L-x)}{2L^3} = \frac{x^2(21-x)}{686}$$

Shear at A (left part): load to the left of A ($x < 6$): $V_A = R_{fixed} - 1 = -R_B$; load to the right of A ($x > 6$): $V_A = R_{fixed} = 1 - R_B$.

Sample: $x = 3$: $R_B = \dfrac{9\times18}{686} = 0.236$, so $V_A = -0.236$. At the section ($x = 6$): $R_B = \dfrac{36\times15}{686} = 0.787$, so $V_A = -0.787$ (just left) and $+0.213$ (just right).

Ordinates (1 m interval)

ILD shape

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             ***********                 |
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                                ******   |
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^                                               ^

The ILD is negative from the fixed end up to the section, with a maximum of $-0.787$ just left of A, jumps by +1 at the section to $+0.213$, and falls to zero at the roller.

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method (Muller-Breslau / equilibrium)

Beam AB: A fixed, B roller, $L = 10$ m, section 1-1 at 4 m from A. The beam is singly indeterminate; take $R_B$ as the redundant.

With support B removed, the cantilever deflection at B due to a unit load at $x$ from A is $\delta_{Bx} = \dfrac{x^2(3L-x)}{6EI}$, and due to a unit load at B it is $f_{BB} = \dfrac{L^3}{3EI}$. Compatibility (zero deflection at B) gives

$$R_B = \frac{\delta_{Bx}}{f_{BB}} = \frac{x^2(3L-x)}{2L^3}$$

Shear at the section (left part):

• load to the left of the section ($x < 4$): $V = R_A - 1 = -R_B$
• load to the right of the section ($x > 4$): $V = R_A = 1 - R_B$

So the ILD is the $R_B$ curve with a downward-reversed part on the left and a unit jump at the section.

Sample: $x = 6$: $R_B = \dfrac{36(30-6)}{2000} = 0.432$, so $V = 1 - 0.432 = 0.568$. At the section: $R_B(4) = \dfrac{16\times26}{2000} = 0.208$, so $V = -0.208$ just left and $+0.792$ just right.

Ordinates

ILD shape

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^                                               ^

Maximum negative ordinate $-0.208$ and maximum positive ordinate $+0.792$ at the section (a jump of 1 unit). The ordinates are 0 at A and at B.

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method

Two-span beam ABC (A hinged, B and C rollers), $L_1 = 8$ m, $L_2 = 10$ m, constant $EI$. Take $M_B$ as the redundant; by Muller-Breslau the ILD for $M_B$ is the deflected shape of the beam with a hinge inserted at B and a unit rotation (kink) applied there. Ordinates follow from the three-moment equation with zero moments at A and C.

For a unit load at $x$ from A in span AB:

$$M_B = -\frac{x\,(L_1^2 - x^2)}{2L_1(L_1+L_2)}$$

For a unit load at distance $y$ from C in span BC:

$$M_B = -\frac{y\,(L_2^2 - y^2)}{2L_2(L_1+L_2)}$$

Sample: $x = 4$ m: $M_B = -\dfrac{4(64-16)}{2\times8\times18} = -0.667$. $y = 4$ m (x = 14): $M_B = -\dfrac{4(100-16)}{2\times10\times18} = -0.933$.

Ordinates of $M_B$

ILD shape

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  ****             **  *                     **
      *****   *****     **                ***
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^                    ^                          ^

The ILD is negative (hogging) throughout and zero at A, B and C. The largest ordinate in AB is about $-0.68$ near $x = 4.6$ m, and in BC about $-1.07$ near $x = 12.2$ m (5.8 m from C); the tabulated value at 12 m is $-1.067$.

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method

Two-span beam: AB = 10 m, BC = 20 m (A hinged, B and C rollers), D at 5 m from A. The ILD for $M_D$ is the simple-span ILD for $M_D$ plus the contribution of the continuity moment at B:

$$\text{ILD}(M_D) = \text{ILD}_{\text{simple}}(M_D) + \frac{5}{10}\,\text{ILD}(M_B)$$

with

$$M_B = -\frac{x(L_1^2-x^2)}{2L_1(L_1+L_2)}\ \ (\text{load in AB}),\qquad M_B = -\frac{y(L_2^2-y^2)}{2L_2(L_1+L_2)}\ \ (\text{load in BC},\ y \text{ from C}).$$

The simple-span ordinate is $x/2$ for $x \le 5$ and $(10-x)/2$ for $x \ge 5$.

Sample: $x = 5$: simple ordinate $= 2.5$; $M_B = -\dfrac{5(100-25)}{2\times10\times30} = -0.625$; so $M_D = 2.5 + 0.5(-0.625) = 2.1875$. Sample: $x = 15$ ($y = 15$): $M_B = -\dfrac{15(400-225)}{2\times20\times30} = -2.1875$; $M_D = 0.5\times(-2.1875) = -1.094$.

Ordinates of $M_D$ (2.5 m in AB, 5 m in BC)

ILD shape

       ***
      *   *
    **     **
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**-------------***----------------------------***
                  ***                   ******
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^               ^                               ^

Maximum positive ordinate $+2.19$ at D; the ordinate is zero at B (x = 10 m) and the negative loop in BC reaches about $-1.28$ near x = 18.5 m (the tabulated value at the middle of BC, x = 20 m, is $-1.25$).

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method

Beam ABC: A fixed, B and C supports, $AB = 2L$, $BC = L$, constant $EI$. Use the moment-distribution (stiffness) form: rotational stiffness at B of AB is $4EI/2L = 2EI/L$ (far end fixed) and of BC is $3EI/L$ (far end C is a roller, moment zero). So the distribution factors at B are $DF_{BA} = 2/5 = 0.4$ and $DF_{BC} = 3/5 = 0.6$.

Unit load in AB at $x$ from A: $M_{BA}^F = \dfrac{x^2(2L-x)}{(2L)^2}$; support moment $M_B = (1 - 0.4)M_{BA}^F$:

$$M_B = -0.6\,\frac{x^2(2L-x)}{4L^2}$$

Unit load in BC at $y$ from B: the fixed-end moment at B with C released is $\dfrac{y(L-y)(2L-y)}{2L^2}$, and $M_B = (1-0.6)$ of it:

$$M_B = -0.4\,\frac{y(L-y)(2L-y)}{2L^2}$$

Sample: $x = L$: $M_B = -0.6\times\dfrac{L^2\cdot L}{4L^2} = -0.15L$. $y = 0.5L$: $M_B = -0.4\times\dfrac{0.5\cdot0.5\cdot1.5}{2}L = -0.075L$.

Ordinates (at 0.25 and 0.5 times each span length)

ILD shape

*****---------------------------*--------------**
     ****                     ** ****     *****
         ****                *       *****
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^                               ^               ^

The ILD is hogging throughout; the largest ordinate in AB is about $-0.178L$ near $x = 1.33L$, and in BC about $-0.077L$ near $0.42L$ from B (mid-span value $-0.075L$).

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method

Propped cantilever AB, $L = 12$ m, A fixed, B roller. By the compatibility (or carry-over) argument, a unit load at $x$ from A gives the hogging fixed-end moment

$$M_A = \frac{x(L-x)(2L-x)}{2L^2}$$

(fixed-end moment $ab^2/L^2$ plus half of the far-end moment $a^2b/L^2$ carried over when the roller is released). The ILD is the rotated shape of the beam when A is changed to a hinge and given a unit rotation.

Sample: $x = 6$: $M_A = \dfrac{6\times6\times18}{2\times144} = 2.25$.

Ordinates (3 m interval)

ILD shape

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^                                               ^

All ordinates are hogging; the maximum is about $-2.31$ at $x = 0.42L = 5.07$ m, and the value at mid-span is $-2.25$.

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method

Propped cantilever AB, $L = 14$ m, A fixed, B roller, C at 6 m from B (8 m from A). Redundant: $R_B$. With support B removed, the cantilever deflection at B due to a unit load at $x$ from A is $\delta_{Bx} = \dfrac{x^2(3L-x)}{6EI}$, and due to a unit load at B it is $f_{BB} = \dfrac{L^3}{3EI}$. Compatibility (zero deflection at B) gives

$$R_B = \frac{\delta_{Bx}}{f_{BB}} = \frac{x^2(3L-x)}{2L^3}$$

Shear at C (left part): load left of C: $V_C = -R_B$; load right of C: $V_C = 1 - R_B$.

Sample: $x = 10$: $R_B = \dfrac{100\times32}{2\times2744} = 0.583$, so $V_C = 0.417$. At C: $R_B(8) = \dfrac{64\times34}{5488} = 0.397$, giving $-0.397$ (just left) and $+0.603$ (just right).

Ordinates (2 m interval)

ILD shape

                           |***
                           |   *****
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************---------------|------------------***
            *********      |
                     ******|
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^                                               ^

The ordinates are 0 at A and B; the jump at C is 1.0 (from $-0.397$ to $+0.603$).

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Reading of the figure

The beam has two equal spans of $4@2 = 8$ m, hinged at A and on rollers at B and C. Section 5 is taken as the middle section of span BC (3 sections per span, numbered 1-3 in AB and 4-6 in BC), that is 12 m from A (4 m from B and 4 m from C).

Method

The ILD for the moment at a section in BC is the ILD of the simple span BC plus the effect of the support moment $M_B$:

$$\text{ILD}(M_5) = \text{ILD}_{\text{simple}} + \frac{4}{8}\,\text{ILD}(M_B)$$

Here $M_B = -\dfrac{x(L_1^2 - x^2)}{2L_1(L_1+L_2)}$ for a load in AB at $x$ from A, and $M_B = -\dfrac{y(L_2^2 - y^2)}{2L_2(L_1+L_2)}$ for a load in BC at $y$ from C. With $L_1 = L_2 = 8$ m: $M_B = -\dfrac{x(64-x^2)}{256}$.

Sample: load at the section ($y = 4$): simple ordinate $= \dfrac{4\times4}{8} = 2.0$; $M_B = -\dfrac{4(64-16)}{256} = -0.75$; $M_5 = 2.0 + 0.5(-0.75) = 1.625$. Sample: load at 4 m from A: $M_B = -0.75$, $M_5 = 0.5(-0.75) = -0.375$.

Ordinates (2 m interval)

ILD shape

                                   ***
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******----------------*****--------------------**
      ****************



^                       ^                       ^

Positive (sagging) in BC with a peak of $+1.625$ under the section, and a small negative loop in AB (about $-0.385$ near x = 4.6 m; $-0.375$ at x = 4 m), zero at A, B and C.

Focal point

When a continuous beam is loaded in one span only, the bending moment diagram in every unloaded span is a straight line. For a given unloaded span this line always passes through the same point of the beam axis, whatever the load position, and this point is called the focal point of the span.

Left and right focal ratios

• Left focal ratio ($k_n$ or $k^L_n$) of span $n$: the ratio of the moment at the left (far) end to the moment at the right (near) end of an unloaded span, when the load lies on spans to its right; taken with a positive sign for the (opposite-sign) pair of moments, $k_n = -M_{n-1}/M_n$. The left focal point of that span lies $L_n/(1+k_n)$ from its right support.
• Right focal ratio ($k'_n$ or $k^R_n$) of span $n$: the ratio of the moment at the right (far) end to the moment at the left (near) end of an unloaded span, when the load lies on spans to its left; $k'_n = -M_n/M_{n-1}$. The right focal point lies $L_n/(1+k'_n)$ from its left support.

Expressions

For span $n$ between supports $n-1$ and $n$, with $L_n/I_n$ the flexibility ratio of the span:

$$k_{n+1} = \frac{L_{n+1}/I_{n+1}}{(2-k_n)\,L_n/I_n + 2\,L_{n+1}/I_{n+1}}\qquad(\text{left ratios, from the left end})$$ $$k'_{n-1} = \frac{L_{n-1}/I_{n-1}}{(2-k'_n)\,L_n/I_n + 2\,L_{n-1}/I_{n-1}}\qquad(\text{right ratios, from the right end})$$

For constant $EI$: $k_{n+1} = \dfrac{1}{4-k_n}$, $k'_{n-1} = \dfrac{1}{4-k'_n}$.

Starting values: hinged end $k = 0$; fixed end $k = \tfrac12$. Example (constant $EI$, hinged end): $k_1 = 0$, $k_2 = 0.25$, $k_3 = 0.2667$, $k_4 = 0.2679$.

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Reading of the figure

Two-span beam ABC: A hinged, B and C rollers, $AB = 6$ m, $BC = 8$ m, constant $EI$. Section x-x is 4 m from A (2 m from B).

Method

$$\text{ILD}(M_{x}) = \text{ILD}_{\text{simple}} + \frac{4}{6}\,\text{ILD}(M_B)$$

Simple-span ordinate for the section at $a = 4$, $L_1 = 6$: $x(6-4)/6 = x/3$ for $x \le 4$ and $4(6-x)/6$ for $x \ge 4$.

$$M_B = -\frac{x(L_1^2-x^2)}{2L_1(L_1+L_2)} = -\frac{x(36-x^2)}{168}\ (\text{load in AB}),\qquad M_B = -\frac{y(64-y^2)}{224}\ (\text{load in BC},\ y\text{ from C})$$

Sample: load at the section ($x = 4$): $\tfrac{4\times2}{6} = 1.333$; $M_B = -\dfrac{4(36-16)}{168} = -0.476$; $M = 1.333 + \tfrac46(-0.476) = 1.016$. Sample: load at the middle of BC ($y = 4$, x = 10): $M_B = -\dfrac{4(64-16)}{224} = -0.857$; $M = \tfrac46(-0.857) = -0.571$.

Ordinates (2 m interval)

ILD shape

             **
          ***  **
       ***       *
   ****           **
***-----------------**------------------------***
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^                    ^                          ^

Peak $+1.016$ under the section (x = 4 m). The ILD is zero at A and B, becomes negative in BC with a largest value of about $-0.59$ near x = 9.4 m, and is zero at C.

Method

Propped cantilever AB (A fixed, B roller) of span $L$. Muller-Breslau: remove the vertical restraint at A (slide it vertically, keeping the rotation at A fixed) and give a unit upward displacement; the deflected shape is the ILD of $R_A$. Equivalent statement: $R_A = 1 - R_B$, with $R_B = \dfrac{x^2(3L-x)}{2L^3}$ (cantilever deflection at B by a unit load at $x$, divided by the deflection $L^3/3EI$ due to a unit load at B). Therefore

$$R_A = 1 - \frac{x^2(3L-x)}{2L^3} = \frac{2L^3 - 3Lx^2 + x^3}{2L^3}$$

Sample: $x = 0.5L$: $R_A = 1 - \dfrac{0.25\times2.5}{2} = 0.6875$.

Ordinates (at 0.25L intervals; positive upward)

ILD shape

***************
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^                                               ^

The ordinate is 1 under the support A, falls smoothly (with zero slope at A) to 0 at B. It is always positive: any downward load produces an upward reaction at A.

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Reading of the figure

Two equal spans $AB = BC = 8$ m, hinged at A, rollers at B and C. Section x-x is in BC, 4 m from B (12 m from A).

Method

For a section in span BC the shear is that of the simple span BC corrected for the support moment $M_B$ (which is hogging, so it changes the end reaction at C by $M_B/L_2$):

$$V_{x} = V_{\text{simple}} - \frac{M_B}{L_2}$$

where $V_{\text{simple}} = 0$ for a load in AB, $+(L_2 - z)/L_2$ for a load right of the section ($z$ measured from B, z > 4), and $-z/L_2$ for a load left of the section in BC. With $L_1 = L_2 = 8$: $M_B = -\dfrac{x(64-x^2)}{256}$ for a load in AB, $M_B = -\dfrac{y(64-y^2)}{256}$ for a load in BC.

Sample: load at x = 4 (in AB): $M_B = -0.75$, $V = -(-0.75)/8 = +0.094$. Sample: load at the section from the left ($z = 4$): $V_{simple} = -0.5$, $M_B = -0.75$, $V = -0.5 + 0.094 = -0.406$; from the right: $+0.5 + 0.094 = +0.594$.

Ordinates (2 m interval)

ILD shape

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Jump of 1.0 at the section ($-0.406$ to $+0.594$). The ILD is positive but small in AB (maximum about 0.096) and negative in BC between B and the section.

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Data and focal ratios

Beam 1-2-3: end 1 fixed, 2 roller, 3 hinged; $L_1 = 20$ m, $L_2 = 10$ m, constant $EI$.

For span n with the far end at moment ratio $k$, the unloaded span acts as a rotational spring at the joint with stiffness $S = \dfrac{6EI}{L(2-k)}$, where $k$ is its focal ratio ($|M_{far}/M_{near}|$).

• Span 1-2 seen from joint 2 (end 1 fixed): left focal ratio $k = \tfrac12$ (focal point at $L_1/3$ from 1), so $S_{21} = \dfrac{6EI}{20(1.5)} = 0.2EI$.
• Span 2-3 seen from joint 2 (end 3 hinged): right focal ratio $k = 0$ (focal point at 3), so $S_{23} = \dfrac{6EI}{10(2)} = 0.3EI$.

Distribution of an unbalanced joint moment at 2: share to 2-1 $= 0.2/0.5 = 0.4$, share to 2-3 $= 0.3/0.5 = 0.6$.

Ordinates of $M_2$

Unit load in span 1-2 at $x$ from 1 ($a = x$, $b = 20 - x$): the fixed-end moment at 2 is $F_{21} = \dfrac{a^2b}{L_1^2}$, and the part held by span 1-2 is 0.6 of it:

$$M_2 = -0.6\,\frac{x^2(20-x)}{400}$$

Unit load in span 2-3 at $y$ from 2: with the hinge at 3 released, the end moment at 2 is $F'_{23} = \dfrac{y(L_2-y)(2L_2-y)}{2L_2^2}$, and the part held is 0.4 of it:

$$M_2 = -0.4\,\frac{y(10-y)(20-y)}{200}$$

Sample: $x = 12$: $M_2 = -0.6\times\dfrac{144\times8}{400} = -1.728$. $y = 2$ (position 22 m): $M_2 = -0.4\times\dfrac{2\times8\times18}{200} = -0.576$.

(The values above were checked against the exact three-moment solution.)

ILD shape

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^                               ^               ^

The ILD is entirely hogging. The maximum is about $-1.78$ at x = 13.3 m in span 1-2, and about $-0.77$ at 24.2 m in span 2-3; it is zero at 1, 2 and 3.

ILD as a system-specific diagram

An influence line shows how one response (a reaction, shear or moment at one fixed location) varies as a unit load moves over the structure. It depends only on the system: the spans, support types, flexural rigidities and the chosen response, and not on the loads. So:

• the ILD of $R_B$ for a propped cantilever (zero at A, 1 at B, curved between) is quite different from that of a simply supported beam of the same length, and a beam with spans 6 m + 8 m has different ILDs from one with 8 m + 8 m;
• once the ILD for a system is drawn, it can be used for any loading on that same system: $Q = \sum P_i y_i$ for point loads, $Q = w\times(\text{area of ILD under the load})$ for a UDL. By contrast, a BMD or SFD is load-specific (it shows one response along the whole beam for one given loading), whereas an ILD is system-specific and response-specific.

Recurrence formula of the left focal ratio (loading on right spans)

Consider a continuous beam with spans numbered from the left; span $n$ lies between supports $n-1$ and $n$, with length $L_n$ and moment of inertia $I_n$. Let the load be on a span to the right of span $n+1$ (all spans from the left end up to $n+1$ unloaded), and let moments be taken with one sign convention (hogging or sagging) at all supports.

Three-moment equation at support $n$ for the unloaded spans $n$ and $n+1$ (the load terms are zero):

$$M_{n-1}\frac{L_n}{I_n} + 2M_n\left(\frac{L_n}{I_n}+\frac{L_{n+1}}{I_{n+1}}\right) + M_{n+1}\frac{L_{n+1}}{I_{n+1}} = 0$$

Because the moment in an unloaded span is linear, the moments at its two ends have opposite signs and a fixed ratio. Define the left focal ratio of span $n$:

$$k_n = -\frac{M_{n-1}}{M_n}$$

and similarly for span $n+1$: $M_n = -k_{n+1}M_{n+1}$.

Substitute $M_{n-1} = -k_nM_n$ in the equation:

$$M_n\left[(2-k_n)\frac{L_n}{I_n} + 2\frac{L_{n+1}}{I_{n+1}}\right] = -M_{n+1}\frac{L_{n+1}}{I_{n+1}}$$

so that $M_n/M_{n+1} = -k_{n+1}$ gives the recurrence formula

$$\boxed{\,k_{n+1} = \frac{L_{n+1}/I_{n+1}}{(2-k_n)\,L_n/I_n + 2\,L_{n+1}/I_{n+1}}\,}$$

For constant $EI$ this becomes $k_{n+1} = \dfrac{1}{4-k_n}$.

Starting value: if the left end of the beam is hinged, $M_0 = 0$ so $k_1 = 0$; if it is fixed, the slope there is zero, $2M_0 + M_1 = 0$, so $k_1 = \tfrac12$.

Focal point: the straight BMD of unloaded span $n$ is $M(x) = M_n\left[1 - (1+k_n)\dfrac{x}{L_n}\right]$, with $x$ measured from support $n$, so it crosses the beam axis at $x = \dfrac{L_n}{1+k_n}$ from support $n$. This point is the left focal point, and it is the same for any loading on the spans to the right of span $n$.

Muller-Breslau principle

The influence line for any force-type quantity (a reaction, shear, or bending moment) in a structure has the same shape as the deflected shape of the structure when the restraint corresponding to that quantity is removed and a unit displacement (or rotation) is given in the direction of the quantity. The ordinate of the ILD at any point is the deflection of the beam at that point. It follows from the principle of virtual work (Betti's law).

ILD for $R_C$

Beam ABC: A hinged, B and C rollers, $AB = 6$ m, $BC = 8$ m. Remove the support at C and give C a unit upward displacement: BC rotates about B as a curved shape, and AB, held by the supports A and B, bends the other way (it goes below the axis).

Ordinates (statics + three-moment equation): the support moment $M_B = -\dfrac{x(36-x^2)}{168}$ for a load in AB at $x$ from A, and

• load in AB: $R_C = \dfrac{M_B}{8}$ (negative, since the load in AB pulls C down),
• load in BC at $z$ from B: $R_C = \dfrac{z}{8} + \dfrac{M_B}{8}$ with $M_B = -\dfrac{y(64-y^2)}{224}$, $y = 8 - z$ measured from C.

Sample: $x = 4$ (in AB): $M_B = -\dfrac{4(36-16)}{168} = -0.476$, $R_C = -0.0595$. $z = 2$: $y = 6$, $M_B = -\dfrac{6(64-36)}{224} = -0.75$, $R_C = \dfrac28 - \dfrac{0.75}{8} = 0.156$.

Ordinates (2 m interval)

ILD shape

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^                    ^                          ^

The ordinate at C is 1; it is 0 at A and B; there is a small negative loop in AB (about $-0.06$ near x = 4 m) and the BC portion is positive and rises steeply to 1 at C.

Definition

In a continuous beam loaded on one span only, the bending moment diagram in every unloaded span is a straight line, and it cuts the beam axis at a fixed point called the focal point of that span. The focal point ratio is the ratio of the moments at the two ends of such an unloaded span (far-end moment to near-end moment, with opposite signs):

• Left focal ratio $k_n$: for an unloaded span to the left of the loaded span, $k_n = -\dfrac{M_{n-1}}{M_n}$ (moment at the left end divided by the moment at the right end).
• Right focal ratio $k'_n$: for an unloaded span to the right of the loaded span, $k'_n = -\dfrac{M_n}{M_{n-1}}$.

Derivation of the left focal ratio

Consider a continuous beam with spans numbered from the left; span $n$ lies between supports $n-1$ and $n$, with length $L_n$ and moment of inertia $I_n$. Let the load be on a span to the right of span $n+1$ (all spans from the left end up to $n+1$ unloaded), and let moments be taken with one sign convention (hogging or sagging) at all supports.

Three-moment equation at support $n$ for the unloaded spans $n$ and $n+1$ (the load terms are zero):

$$M_{n-1}\frac{L_n}{I_n} + 2M_n\left(\frac{L_n}{I_n}+\frac{L_{n+1}}{I_{n+1}}\right) + M_{n+1}\frac{L_{n+1}}{I_{n+1}} = 0$$

Because the moment in an unloaded span is linear, the moments at its two ends have opposite signs and a fixed ratio. Define the left focal ratio of span $n$:

$$k_n = -\frac{M_{n-1}}{M_n}$$

and similarly for span $n+1$: $M_n = -k_{n+1}M_{n+1}$.

Substitute $M_{n-1} = -k_nM_n$ in the equation:

$$M_n\left[(2-k_n)\frac{L_n}{I_n} + 2\frac{L_{n+1}}{I_{n+1}}\right] = -M_{n+1}\frac{L_{n+1}}{I_{n+1}}$$

so that $M_n/M_{n+1} = -k_{n+1}$ gives the recurrence formula

$$\boxed{\,k_{n+1} = \frac{L_{n+1}/I_{n+1}}{(2-k_n)\,L_n/I_n + 2\,L_{n+1}/I_{n+1}}\,}$$

For constant $EI$ this becomes $k_{n+1} = \dfrac{1}{4-k_n}$.

Starting value: if the left end of the beam is hinged, $M_0 = 0$ so $k_1 = 0$; if it is fixed, the slope there is zero, $2M_0 + M_1 = 0$, so $k_1 = \tfrac12$.

Focal point: the straight BMD of unloaded span $n$ is $M(x) = M_n\left[1 - (1+k_n)\dfrac{x}{L_n}\right]$, with $x$ measured from support $n$, so it crosses the beam axis at $x = \dfrac{L_n}{1+k_n}$ from support $n$. This point is the left focal point, and it is the same for any loading on the spans to the right of span $n$.

Numerical example

Hinged left end A, span 1: $L_1 = 6$ m, $I_1 = I$; span 2: $L_2 = 4$ m, $I_2 = 2I$.

• $k_1 = 0$ (hinged end).
• $k_2 = \dfrac{4/2}{(2-0)\times6/1 + 2\times4/2} = \dfrac{2}{12+4} = 0.125$.
• The focal point of span 2 lies $L_2/(1+k_2) = 4/1.125 = 3.56$ m from its right end (support 2), that is 0.44 m from support 1.