Influence Lines for Simple Structures

A UDL shorter than the span gives maximum bending moment at a section when the load is placed so that the section divides the span and the load in the same ratio.

ILD for bending moment at section C

Simply supported beam $AB$, span $L$; section $C$ at $a$ from $A$, $b = L-a$. The ILD for $M_C$ is a triangle with peak $y_C = ab/L$ under $C$.

        A      c    C   (l-c)    B
        o------+----+-------+----o
  ILD:         y1  ab/L     y2

Ordinate left of $C$ at distance $x$ from $A$: $y = \dfrac{xb}{L}$. Right of $C$ at distance $x$ from $B$: $y = \dfrac{xa}{L}$.

Proof

Let a UDL of intensity $w$ and length $l$ (< $L$) have a length $c$ on the left of $C$ and $(l-c)$ on the right. Ordinates under its ends are

$$y_1 = \frac{(a-c)\,b}{L}, \qquad y_2 = \frac{\{b-(l-c)\}\,a}{L}$$

$M_C = w\times(\text{area under ILD below the load})$. Shift the load by a small distance $dx$ to the right. The left end loses a strip of height $y_1$ and the right end gains a strip of height $y_2$, so

$$\frac{dM_C}{dx} = w\,(y_2 - y_1)$$

$M_C$ is maximum when $dM_C/dx = 0$, i.e. $y_1 = y_2$:

$$\begin{aligned} (a-c)\,b &= (b-l+c)\,a \\ ab - cb &= ab - la + ca \\ c(a+b) &= la \Rightarrow \frac{c}{l} = \frac{a}{L} \end{aligned}$$

Hence the part of the load on the left of $C$ is to the whole load as the left part of the span is to the whole span:

$$\frac{c}{l} = \frac{a}{L}$$

Maximum moment

With $c = la/L$ the end ordinates are equal: $y_1 = y_2 = \dfrac{ab}{L}\left(1-\dfrac{l}{L}\right)$. The area under the load is two trapezoids:

$$M_{max} = w\,\frac{l}{2}\left(\frac{ab}{L} + y_1\right) = w\,l\,\frac{ab}{L}\left(1 - \frac{l}{2L}\right)$$

Result: place the UDL so that the section divides the load in the ratio $a : b$, the same as it divides the span.

Definition

An influence line diagram (ILD) is a graph that shows how a particular structural quantity (reaction, shear force, bending moment, axial force in a member, or deflection) at one fixed location varies as a unit load moves across the structure. The ordinate at any point gives the value of that quantity at the fixed location when the unit load is at that point.

Uses

1. Critical load position: it shows where to place live loads (trucks, train wheels, UDL) to produce the maximum value of a quantity.
2. Maximum values: for concentrated loads $F = \sum P\,y$; for UDL $F = w\times\text{area}$. This gives the design reaction, shear, moment or member force.
3. Design of bridges, girders and crane girders: moving loads decide the sections of bridge girders, trusses, floor beams and gantry girders.
4. Truss design: ILDs for chord, web and counter members show whether a member changes sign (stress reversal) and how much counter bracing is needed.
5. Absolute maximum BM and SF under a train of wheel loads can be found quickly.
6. Used in Civil Engineering: bridge and highway loading (IRC/AASHTO truck, IRC class loads), railway bridges, overhead cranes, and as the basis of Müller-Breslau's principle for indeterminate structures.

Difference from BMD / SFD

Let $F$ be a structural quantity (reaction, shear, moment or member force) at a fixed section. Let $y$ be its influence ordinate (value of $F$ for a unit load) at the load position. By the principle of superposition (linear elastic structure), the effect of several loads is the sum of their separate effects.

1. Concentrated loads

A single load $P$ at a point where the ILD ordinate is $y$ gives $F = P\,y$, because a unit load gives $y$ and the effect is proportional to the load. For loads $P_1, P_2, \dots, P_n$ at ordinates $y_1, y_2, \dots, y_n$:

$$F = P_1y_1 + P_2y_2 + \dots + P_ny_n = \sum P_i\,y_i$$

2. Uniformly distributed load

Consider a UDL of intensity $w$ per unit length between $x_1$ and $x_2$. An elementary strip $dx$ carries load $w\,dx$, acting as a concentrated load at ordinate $y$ (the ILD ordinate at $x$):

$$dF = w\,dx\;y \Rightarrow F = w\int_{x_1}^{x_2} y\,dx = w\times(\text{area of ILD between }x_1\text{ and }x_2)$$

Areas above the base line are positive and those below are negative. For maximum positive effect, load only the positive portions.

3. Couple

A couple $M_0$ is a pair of equal and opposite loads $P$ at a small distance $d$ apart, with $M_0 = P\,d$. If the ordinates at the two points are $y_1$ and $y_2$ (the second load acting opposite to the first):

$$F = P\,(y_2 - y_1) = P\,d\;\frac{y_2-y_1}{d} = M_0\,\frac{dy}{dx}$$

So the effect of a couple equals the couple magnitude multiplied by the slope of the ILD at the point of application (sign taken according to the sense of the couple, positive when clockwise on a ILD drawn with upward positive ordinates).

 Summary:   P  ->  F = P . y
            w  ->  F = w . (area of ILD under load)
            M0 ->  F = M0 . (slope of ILD)

Data: simply supported beam $0$–$1$, $L = 14$ m. $P_1 = 240$ kN at 3 m from support 0; $P_2 = 160$ kN at 4.5 m from support 1, i.e. 9.5 m from support 0.

Reaction at support 0

ILD for $R_0$: a straight line from 1 at support 0 to 0 at support 1; the ordinate at distance $s$ from 0 is $(L-s)/L$.

$$R_0 = 240\left(\frac{11}{14}\right) + 160\left(\frac{4.5}{14}\right) = 240.0\ \text{kN}$$

Reaction at support 1

Ordinate $s/L$:

$$R_1 = 240\left(\frac{3}{14}\right) + 160\left(\frac{9.5}{14}\right) = 160.0\ \text{kN}$$

(Check: $R_0 + R_1 = 400$ kN = total load.)

Bending moment beneath the loads

The ILD for $M$ at a section $x$ is a triangle with peak $x(L-x)/L$. For a load at $s$ the ordinate is $s(L-x)/L$ if $s\le x$, and $x(L-s)/L$ if $s\ge x$.

Under 240 kN ($x = 3$ m): peak ordinate $=3\times 11/14 = 2.357$; ordinate under the 160 kN load $=3\times 4.5/14 = 0.964$.

$$M_{3} = 240(2.357) + 160(0.964) = 720.0\ \text{kN m}$$

Under 160 kN ($x = 9.5$ m): peak ordinate $=9.5\times 4.5/14 = 3.054$; ordinate under 240 kN $=3\times 4.5/14 = 0.964$.

$$M_{9.5} = 240(0.964) + 160(3.054) = 720.0\ \text{kN m}$$

Shear force beneath the loads

The ILD for shear at $x$ has ordinate $-s/L$ for $s<x$ and $+(L-s)/L$ for $s>x$ (a jump of 1 at the section).

Answer: $R_0 = 240$ kN, $R_1 = 160$ kN; $M = 720$ kN m under the 240 kN load and $720$ kN m under the 160 kN load; the shear force is +240 kN from support 0 to the 240 kN load, 0 between the loads, and −160 kN from the 160 kN load to support 1.

Assumed truss (figure not available): Pratt truss, 6 panels of 5 m (span 30 m), height 5 m; verticals $U_iL_i$ and diagonals $U_1L_2$, $U_2L_3$, $U_4L_3$, $U_5L_4$; end posts $L_0U_1$ and $L_6U_5$ at 45°. $L_0$ hinge, $L_6$ roller. Reaction for unit load at $x$ from $L_0$: $R_{L_0} = (30-x)/30$.

Method

• $L_0U_1$ (end post): joint $L_0$, vertical equilibrium: $F\sin45^\circ = -R_{L_0}$, so $F_{L_0U_1} = -1.414\,R_{L_0}$ (always compression).
• $L_0L_1$: horizontal equilibrium at $L_0$: $F_{L_0L_1} = -F_{L_0U_1}\cos45^\circ = +R_{L_0}$ (always tension).
• $U_2L_2$ (vertical): the section through $U_2U_3$, $U_2L_3$ and $L_2L_3$ gives the diagonal $F_{U_2L_3} = \sqrt2\,V$, where $V$ is the shear in panel $L_2L_3$ (left forces upward positive). At the unloaded joint $U_2$ the vertical balances the diagonal: $F_{U_2L_2} = -F_{U_2L_3}/\sqrt2 = -V$. For a unit load at $L_2$: $V = \tfrac23 - 1 = -\tfrac13$, so $F_{U_2L_2} = +0.333$. For a unit load on the top chord at $U_2$, the load itself is added at the joint, which gives $-0.667$.

Ordinates when the unit load moves on the bottom chord (+ tension)

Ordinates when the unit load moves on the top chord

Shapes (bottom-chord loading)

• $L_0U_1$: always compression. 0 at $L_0$ (load directly on the support), peak $-1.179$ at $L_1$, then a straight line decreasing to 0 at $L_6$. Beyond $L_1$ the ordinate is $-1.414\,R_{L_0}$.
• $L_0L_1$: always tension. 0 at $L_0$, peak $+0.833$ at $L_1$, then falling linearly to 0 at $L_6$ (ordinate $=R_{L_0}$).
• $U_2L_2$: $+0.167$ at $L_1$, $+0.333$ at $L_2$, $-0.5$ at $L_3$, then $-0.333$, $-0.167$ and 0. The sign changes between $L_2$ and $L_3$, at $x = 10 + 5\times\frac{0.333}{0.833} = 12$ m from $L_0$, so the vertical is stressed in both tension and compression (the part of the ILD with a negative area is larger).
• With top-chord loading, the ordinates of $L_0U_1$ and $L_0L_1$ are the same linear functions of $R_{L_0}$ (taken at the top joints), while $U_2L_2$ takes $-0.667$ at $U_2$, since the load acts directly on the vertical.

Data: simply supported beam $AB$, $L = 20$ m. Section $C$ at $a = 5$ m from $A$, $b = 15$ m. Let a unit load be at a distance $x$ from $A$.

ILD for reaction $R_A$

$R_A = \dfrac{20-x}{20}$: a straight line from 1 at $A$ to 0 at $B$.

ILD for reaction $R_B$

$R_B = \dfrac{x}{20}$: a straight line from 0 at $A$ to 1 at $B$.

ILD for shear force at C

• Load to the left of $C$ ($x<5$): from the right part, $V_C = -R_B = -\dfrac{x}{20}$. At $C$ the ordinate is $-\dfrac{5}{20} = -0.25$.
• Load to the right of $C$ ($x>5$): $V_C = R_A = \dfrac{20-x}{20}$. At $C$ the ordinate is $+\dfrac{15}{20} = +0.75$.

So the ILD is two parallel lines: from 0 at $A$ to $-0.25$ just left of $C$, a jump of 1 at $C$ to $+0.75$, and then down to 0 at $B$.

ILD for bending moment at C

• Load left of $C$: $M_C = R_B\times 15 = \dfrac{15x}{20}$.
• Load right of $C$: $M_C = R_A\times 5 = \dfrac{5(20-x)}{20}$.
• Peak at $C$: $\dfrac{ab}{L} = \dfrac{5\times 15}{20} = 3.75$ m.

 R_A:  1 \
          \_____ 0 at B
 SF_C: 0 \-0.25 | +0.75 \ 0
 BM_C: 0 /\ 3.75 (at C) \ 0

The positive area of the SF ILD is $\tfrac12\times 15\times 0.75 = 5.625$ m and the negative area is $\tfrac12\times 5\times 0.25 = 0.625$ m. The area of the BM ILD is $\tfrac12\times 20\times 3.75 = 37.5$ m².

Data: girder $AB$, $L = 100$ m; section $C$ at $a = 35$ m from $A$, $b = 65$ m. Wheel loads from the rear: 100, 160, 180, 120 kN at spacings 4 m, 3 m, 2 m; the train moves left to right (assumed as drawn, so the 120 kN load leads). Total $W = 560$ kN.

(i) Maximum BM at C

ILD for $M_C$: peak $\dfrac{ab}{L} = \dfrac{35\times 65}{100} = 22.75$ m; ordinate $0.65x$ for $x\le 35$ and $0.35(100-x)$ for $x\ge 35$.

Place the 160 kN load at $C$. Load on the left of $C$ including it: $100 + 160 = 260$ kN.

$$\frac{260}{35} = 7.43 > \frac{560}{100} = 5.6 > \frac{100}{35} = 2.86$$

The condition for maximum is satisfied. Positions from $A$: 100 kN at 31 m, 160 kN at 35 m (at $C$), 180 kN at 38 m, 120 kN at 40 m.

$$M_{C,max} = 12081.0\ \text{kN m}$$

(If the train moves the other way, with the 100 kN load leading, the same method gives 12171.0 kN m.)

(ii) Absolute maximum BM

Resultant from the 120 kN (leading) load: $\bar x = \dfrac{100(9)+160(5)+180(2)}{560} = 3.679$ m behind it. For each load, place it and the resultant symmetrically about mid-span and find the moment under the load:

The largest is under the 180 kN load. With this load at 50.84 m from $A$, the loads are at 100 kN: 43.84 m, 160 kN: 47.84 m, 180 kN: 50.84 m, 120 kN: 52.84 m (all on the girder).

$$R_A = 284.7\ \text{kN}, \qquad M_{max} = R_A\times 50.84 - (\text{moments of the loads on its left})= 13293.9\ \text{kN m}$$

Answer: (i) $M_C = 12081.0$ kN m; (ii) absolute maximum BM = 13293.9 kN m under the 180 kN load.

Data: girder $AB$, $L = 60$ m; section $C$ at $a = 20$ m from $A$, $b = 40$ m. Loads from the rear: 170, 215, 150, 120 kN at spacings 3 m, 2 m, 3 m; the train moves left to right (120 kN leads, as drawn). Total $W = 655$ kN.

(i) Maximum BM at C

ILD for $M_C$: peak $\dfrac{ab}{L} = \dfrac{20\times 40}{60} = 13.33$ m; ordinate $\dfrac{2x}{3}$ for $x\le 20$ and $\dfrac{20(60-x)}{60}$ for $x\ge 20$.

Place the 215 kN load at $C$. Load on the left of $C$ including it: $170 + 215 = 385$ kN.

$$\frac{385}{20} = 19.25 > \frac{655}{60} = 10.9 > \frac{170}{20} = 8.5$$

The condition for maximum is satisfied. Positions from $A$: 170 kN at 17 m, 215 kN at 20 m (at $C$), 150 kN at 22 m, 120 kN at 25 m.

$$M_{C,max} = 8093.3\ \text{kN m}$$

(ii) Absolute maximum BM

Distance of the resultant behind the 120 kN load: $\bar x = \dfrac{170(8)+215(5)+150(3)}{655} = 4.405$ m. Trying each load with the resultant placed symmetrically about mid-span:

The largest value is under the 215 kN load, at 29.70 m from $A$. Then the loads are at: 170 kN: 26.70 m, 215 kN: 29.70 m, 150 kN: 31.70 m, 120 kN: 34.70 m (all on the girder).

$$R_A = 324.2\ \text{kN}, \qquad M_{max} = R_A\times 29.70 - (\text{moments of the loads on its left}) = 9121.0\ \text{kN m}$$

Answer: (i) $M_C = 8093.3$ kN m; (ii) absolute maximum BM = 9121.0 kN m under the 215 kN load.

Assumed truss (Pratt type): bottom joints $A$–$G$, top joints $N, M, L, K, J, I, H$ directly above them, 6 panels of 6 m (span 36 m), height 8 m; verticals at every joint; diagonals $NB$, $MC$, $LD$, $JD$, $IE$, $HF$ (sloping down towards the centre). $A$ hinge, $G$ roller. The load moves on the lower chord. Reactions: $R_A = \dfrac{36-x}{36}$, $R_G = \dfrac{x}{36}$. The diagonal $LD$ has length $10$ m, $\sin\theta = 0.8$.

Member LK (top chord, panel $CD$)

Section through $LK$, $LD$ and $CD$ (panel $CD$); $LD$ and $CD$ meet at $D$ ($x = 18$ m), so take moments about $D$ (lever arm 8 m): $F_{LK} = -\dfrac{M_D}{8}$.

• Load right of $D$: $F_{LK} = -\dfrac{18R_A}{8} = -2.25R_A$; load left of $C$: $F_{LK} = -\dfrac{18R_G}{8} = -2.25R_G$ (compression).

Member LD (diagonal)

Same section: the panel shear is carried by the diagonal, $F_{LD} = \pm\dfrac{V}{0.8}$.

• Load right of $D$: $F_{LD} = +\dfrac{R_A}{0.8} = 1.25R_A$ (tension).
• Load left of $C$: $F_{LD} = -1.25R_G$ (compression).

Member LC (vertical)

At the unloaded top joint $L$ the vertical balances the vertical component of the diagonal: $F_{LC} = -0.8\,F_{LD}$.

Ordinates (per unit load; + tension, − compression)

Shapes

• LK: compression only, triangle with peak $-1.125$ under $D$ (centre).
• LD: compression (peak $-0.417$ at $C$) for load left of the panel, tension (peak $+0.625$ at $D$) for load to the right; the neutral point is at $x = 12+6\times\frac{0.417}{0.417+0.625} = 14.4$ m from $A$.
• LC: tension for load on the left (peak $+0.333$ at $C$), compression to the right (peak $-0.5$ at $D$); the sign change lies between $C$ and $D$ at $x = 12 + 6\times\frac{0.333}{0.333+0.5} = 14.4$ m.

Data: simply supported beam $AB$, $L = 25$ m. Section $C$ at $a = 10$ m from $A$, $b = 15$ m. Unit load at $x$ from $A$.

Reactions

$R_A = \dfrac{25-x}{25}$, $R_B = \dfrac{x}{25}$.

ILD for bending moment at C

• Load left of $C$ ($x\le 10$): take the right part: $M_C = R_B\times 15 = 0.6x$
• Load right of $C$ ($x\ge 10$): take the left part: $M_C = R_A\times 10 = 0.4\,(25-x)$

Peak under $C$: $\dfrac{ab}{L} = \dfrac{10\times 15}{25} = 6$ m. The ILD is a triangle: 0 at $A$, $+6$ at $C$, 0 at $B$.

  A o----------C-------------------o B
               /\ 6.0
  ILD M_C:  0 /  \ 0

ILD for shear force at C

• Load left of $C$ ($x<10$): $V_C = -R_B = -\dfrac{x}{25}$; at $C^-$: $-0.4$
• Load right of $C$ ($x>10$): $V_C = R_A = \dfrac{25-x}{25}$; at $C^+$: $+0.6$

  ILD V_C:   0 \ -0.4 | +0.6 \ 0     (two parallel lines, jump of 1 at C)

Areas: $M_C$ ILD $=\tfrac12\times 25\times 6 = 75$ m²; $V_C$ ILD: positive $\tfrac12\times 15\times 0.6 = 4.5$ m, negative $\tfrac12\times 10\times 0.4 = 2$ m.

For a UDL $w$ over the whole span, $M_C = 75w$ and $V_C = 4.5w - 2w = 2.5w$; for a single load $P$ at $x$, $M_C = P\,y$ and $V_C = P\,y$ with $y$ read from the ILD.

For a train of loads $P_1,\dots,P_5$ at fixed spacings crossing a simply supported beam of span $L$, the bending moment at a section $C$ is found from the ILD of $M_C$ by trying the critical position of the train.

ILD for $M_C$

Let $C$ be at $a$ from the left support and $b = L-a$. The ILD is a triangle with peak $ab/L$ under $C$. Left slope $= b/L$, right slope $= -a/L$.

    A o-------------C-------------o B
                    ^ ab/L
    P1  P2  P3  P4  P5   ->  moving

$$M_C = \sum P_i\,y_i$$

Condition for maximum

Let $W$ be the total load on the span and $W_L$ the load on the left of $C$. Shifting the train to the right by $\delta$:

$$\Delta M_C = \frac{\delta}{L}\left(W_L\,b - W_R\,a\right) = \delta\left(W_L - \frac{W\,a}{L}\right)$$

$M_C$ is maximum when $\Delta M_C$ changes sign from positive to negative as a load $P_k$ crosses $C$:

$$\frac{W_L + P_k}{a} \ge \frac{W}{L} \ge \frac{W_L}{a}$$

where $W_L$ excludes $P_k$ (load just right of $C$) and $W_L + P_k$ includes it (just left of $C$). At such a position the average load on the left of $C$ equals the average load on the whole span, and $P_k$ stands over the section.

Steps

1. Place the heaviest load near $C$ and test the condition; move to the next load if it fails. Remember that loads leaving or entering the span change $W$.
2. At the critical position, read the ordinates under every load of the train from the ILD (similar triangles).
3. Compute $M_{C,max} = P_1y_1 + P_2y_2 + P_3y_3 + P_4y_4 + P_5y_5$.
4. If more than one load satisfies the test, compute for each and take the largest.
5. For the absolute maximum moment anywhere, put a load and the resultant of the loads on the span symmetric about mid-span.

Data: span $L = 25$ m, section $C$ at $a = 10$ m from the left support, $b = 15$ m. The train is a 20 kN/m UDL over 4 m (= 80 kN), then a 10 kN load 2 m beyond the UDL and a second 10 kN load another 2 m further. The train moves left to right, so the 10 kN loads lead.

ILD for $M_C$

Peak ordinate under $C$: $y_C = \dfrac{ab}{L} = \dfrac{10\times 15}{25} = 6$ m. Ordinate at $x$ from $A$: $0.6x$ (left of $C$), $0.4(25-x)$ (right of $C$).

   A o----------C---------------o B
                ^ 6.0
   ILD: 0 -> 6.0 at C -> 0

Critical position

Try positions of the leading 10 kN load (distance $p$ from $A$):

The maximum is near $p = 16$ m. Then the loads are: 10 kN at 16 m, 10 kN at 14 m, and the UDL spans 8 m to 12 m (covering $C$).

Ordinates: at 16 m: $0.4\times 9 = 3.6$; at 14 m: $0.4\times 11 = 4.4$; under the UDL: $4.8$ at 8 m, $6.0$ at 10 m, $5.2$ at 12 m.

$$\begin{aligned} M_{max} &= 10(3.6) + 10(4.4) + 20\left[\tfrac{4.8+6.0}{2}\times 2 + \tfrac{6.0+5.2}{2}\times 2\right]\\ &= 36 + 44 + 20\,(10.8 + 11.2) = 36+44+440 = 520\ \text{kN m} \end{aligned}$$

Answer: maximum bending moment at the section = 520 kN m.

Geometry: equilateral triangles of side 5 m (all inclined members at 60°), height $h = 5\sin 60^\circ = 4.33$ m, span 15 m. $L_1$ hinge, $L_4$ roller. The load travels along the bottom chord (through truss), so each ILD is linear between panel points.

Reactions for a unit load at $x$ from $L_1$: $R_{L_1} = (15-x)/15$, $R_{L_4} = x/15$.

Method of sections

• $U_1U_2$: cut through $U_1U_2$, $U_1L_2$ and $L_1L_2$ and take moments about $L_2$. Top chord is in compression: $F_{U_1U_2} = -M_{L_2}/h$.
• $U_2L_2$: pass a section through $U_1U_2$, $U_2L_2$ and $L_2L_3$. The shear $V$ (left forces upward) in this panel is carried by the inclined member: $F_{U_2L_2} = -V/\sin 60^\circ = -1.155\,V$. For a load to the left of the section $V = R_{L_1}-1$; for a load to the right $V = R_{L_1}$.
• $L_2L_3$: moments about the joint $U_2$ gives $F_{L_2L_3} = M_{U_2}/h$ (tension).

Ordinates (per unit load; + tension, − compression)

Shapes:

• $U_1U_2$: always compressive, triangle with peak $-0.770$ under $L_2$.
• $U_2L_2$: tension for load left of $L_2$ (peak $+0.385$ at $L_2$), compression for load right of $L_2$ (peak $-0.385$ at $L_3$); sign change between $L_2$ and $L_3$.
• $L_2L_3$: tension, flat $+0.577$ between $L_2$ and $L_3$.

Maximum forces for 10 kN/m over 4 m

Force $= w\times(\text{area of ILD under the load})$, with the 4 m load moved to the position giving the greatest area.

Answer: $U_1U_2 = 26.69$ kN (compression); $U_2L_2 = +11.29$ kN (tension) or $-11.29$ kN (compression); $L_2L_3 = +23.09$ kN (tension).

Assumptions (figure partly unclear): beam $A$–$B$–$C$ is supported by a hinge at $A$ and a roller at $C$ ($AB = BC = 2$ m, span $AC = 4$ m), with an overhang $C$–$D$–$E$–$F$ of 3 m (1 m each). Loads: 40 kN at $B$; 5 kN/m UDL from $C$ to $E$ (2 m); 60 kN at $F$ (free end). The section is taken at the support $C$.

Reactions (for checking)

$\Sigma M_A = 0$: $R_C\times 4 = 40\times 2 + 10\times 5 + 60\times 7 = 550 \Rightarrow R_C = 137.5$ kN; $R_A = 110-137.5 = -27.5$ kN (uplift).

ILD for bending moment at C

Load at distance $s$ from $A$. For $s\le 4$, $M_C = 0$ (the load is carried directly by the supports). For $s>4$, $M_C = -(s-4)$ (hogging).

 A o---B-----o C---D---E---F
 ILD(M_C):  0  0   0  -1 -2 -3

$$M_C = 5\times 2\times(-1) + 60\times(-3) = -10 - 180 = -190\ \text{kN m}$$

ILD for shear at C

Just left of C: ordinate $=-s/4$ for $0\le s<4$ (so $-0.5$ at $B$ and $-1$ as the load approaches $C$); $-(s-4)/4$ for the overhang ($-0.25$ at $D$, $-0.5$ at $E$, $-0.75$ at $F$).

Using ordinates: UDL area from $C$ to $E$ $=\tfrac12(0+(-0.5))\times 2 = -0.5$, so

$$V_{C^-} = 40(-0.5) + 5(-0.5) + 60(-0.75) = -20 - 2.5 - 45 = -67.5\ \text{kN}$$

Just right of C: ordinate 0 for loads on the span, and 1 for loads on the overhang:

$$V_{C^+} = 5\times 2\times 1 + 60\times 1 = 70\ \text{kN}$$

Check by statics: $V_{C^-} = R_A - 40 = -27.5-40 = -67.5$ kN and $V_{C^+} = -67.5 + 137.5 = +70$ kN.

Answer: $M_C = -190$ kN m (hogging); $V_C = -67.5$ kN just left and $+70$ kN just right of $C$.

Geometry: bottom chord $A$–$B$–$C$–$D$–$E$, 4 panels of 4 m (span 16 m); top joints $H$, $G$, $F$ at 3 m height above $B$, $C$, $D$. $A$ hinge, $E$ roller. Diagonals $AH$, $HC$, $CF$, $FE$; verticals $HB$, $GC$, $FD$. The unit load moves along the bottom chord, so ordinates are needed at the panel points only and the ILD is a straight line between them (floor beam action).

Reactions: for a unit load at $x$ from $A$, $R_A = \dfrac{16-x}{16}$, $R_E = \dfrac{x}{16}$.

Method of sections (section through panel $BC$, cutting $HG$, $HC$, $BC$)

Diagonal $HC$ has $\sin\theta = 3/5$, $\cos\theta = 4/5$.

Member BC (moments about $H$, the intersection of $HG$ and $HC$... $H$ lies on both):

• Unit load right of the panel ($x\ge 8$, left part considered): $F_{BC}\times 3 = R_A\times 4 \Rightarrow F_{BC} = \tfrac43 R_A$
• Unit load left of the panel ($x\le 4$, right part considered): $F_{BC}\times 3 = R_E\times 12 \Rightarrow F_{BC} = 4R_E = x/4$

Member HG (moments about $C$): $F_{HG} = -\dfrac{8R_A}{3}$ for load right of $C$; $F_{HG} = -\dfrac{8R_E}{3}$ for load left of $B$ (top chord in compression).

Member CH (vertical equilibrium of the left part): $F_{CH}\sin\theta = R_A$ for load right of $C$, so $F_{CH} = \tfrac53 R_A$ (tension); for load left of $B$, $F_{CH} = -\tfrac53 R_E$ (compression).

Ordinates (per unit load; + tension, − compression)

Between $B$ and $C$ the ILDs are straight lines joining the ordinates at $B$ and $C$.

Shapes

• BC: tension only; triangle-like with peak $+1.00$ at $B$, $0.667$ at $C$, $0$ at $A$ and $E$.
• CH: compression ($-0.417$ at $B$) for load left of the panel, tension ($+0.833$ at $C$) for load right; a neutral point lies between $B$ and $C$ at $x = 4 + 4\times\frac{0.417}{0.417+0.833} = 5.33$ m from $A$.
• HG: compression throughout with peak $-1.333$ under $C$.