Influence Lines

Q: What is the difference between an influence line and a bending moment diagram?

A bending moment diagram (BMD) shows the variation of bending moment along the entire length of a beam for a specific, fixed loading arrangement. Every point on the BMD corresponds to a different cross-section of the beam, and the value at each point is the BM at that section under the given loads. An influence line (ILD) for bending moment at section C shows how the BM at that one specific section C varies as a unit load moves along the beam. Every point on the ILD corresponds to a different po

Q: Why is the ILD for bending moment always a triangle for a simply supported beam?

This follows from the Müller-Breslau Principle. To draw the ILD for M C , insert a hinge at C and apply a unit rotation. The beam (now two rigid segments hinged at C and supported at A and B) deflects into two straight segments — a triangle. The peak of the triangle is at C (where the rotation is applied), and the base is zero at both supports (which remain fixed). Since the beam segments are rigid (in Müller-Breslau, the released structure deforms as a rigid body mechanism), the deflected shape

Q: How do you find the position of maximum shear force under a moving UDL?

To maximise the shear force at section C, use the ILD for V C . The ILD for shear has a positive region (right of C) and a negative region (left of C). To maximise positive shear at C, load the positive region of the ILD — place the UDL from C to B. To maximise negative shear at C, load the negative region — place the UDL from A to C. For a UDL that must be placed as a single continuous load, the maximum shear occurs when the load covers the larger positive or negative region completely. This di

Q: Does the Müller-Breslau Principle work for indeterminate structures?

Yes — the Müller-Breslau Principle is valid for both determinate and indeterminate structures. For determinate structures, the released structure deforms as a rigid body mechanism (straight lines). For indeterminate structures, the released structure is still indeterminate after the release (just one degree less indeterminate), so it deforms as a flexible body — the ILD is a curve. The principle gives the correct qualitative shape and, after computing the actual deflections of the released struc

Moving Loads on Beams & Bridges — ILD Construction, Müller-Breslau Principle, Maximum BM & SF, Solved GATE Examples

Last Updated: March 2026

📌 Key Takeaways

An influence line diagram (ILD) shows how a specific structural response (reaction, shear force, or bending moment) at a fixed point varies as a unit load moves across the entire span.
Influence lines answer the design question: “Where should the moving load be positioned to cause the maximum (or minimum) response at a given section?”
The ordinate of an ILD at any position x equals the value of the response function when a unit load is at x.
For a series of point loads, the response at a section = sum of (each load × ILD ordinate under that load).
For a UDL, the response = intensity w × (area of ILD over the loaded length).
The Müller-Breslau Principle states that the ILD for any response function is the deflected shape of the released structure when a unit displacement (or rotation) corresponding to that response is applied.
Influence lines are essential for bridge design, crane girder design, and any structure carrying moving traffic — they are tested regularly in GATE CE.

1. What is an Influence Line?

In all the beam analysis done so far — SFD, BMD, deflection calculations — the loads were fixed in position. The bending moment at midspan of a simply supported beam under a central point load is WL/4 because the load is always at the centre. But what about structures that carry moving loads — bridge decks, crane rails, floor beams in warehouses with moving forklifts?

For a moving load, the bending moment at midspan is not constant — it changes as the load moves across the span. When the load is at midspan, the moment there is maximum. When the load is near a support, the moment at midspan is small. The structural engineer’s question is: what is the maximum bending moment that will ever occur at midspan, and at what load position does it occur?

This is exactly what an influence line answers. An influence line for a specific response (let us say the bending moment at midspan, M_C) is a graph that shows M_C as a function of the position of a unit load (a single load of magnitude 1 kN) as it moves from one end of the beam to the other. Once this graph is drawn:

The maximum ordinate gives the position of the unit load that maximises the response.
For a real load of magnitude P, the response = P × ILD ordinate at the load position.
For a series of loads, the response = sum of each load × the ILD ordinate at its position.
For a uniformly distributed load of intensity w, the response = w × area of ILD over the loaded region.

Influence lines are therefore a tool for converting a moving load problem into a static analysis problem. They are drawn once for each response quantity of interest, and then used to evaluate the maximum response for any load combination or load position.

Key Definition

The influence line ordinate at position x for a response function F is the value of F when a unit load (1 kN) is placed at position x, with no other loads on the structure.

The response F under a system of loads = Σ(P_i × η_i) for point loads + w × ∫η dx for UDL

Where η_i = ILD ordinate at the position of load P_i, and ∫η dx = area of ILD over the loaded length.

2. ILD for Support Reactions

For a simply supported beam AB of span L, with pin at A and roller at B, the reactions R_A and R_B when a unit load is at distance x from A are found by equilibrium:

ILD for Reactions — Simply Supported Beam

Unit load at distance x from A:

R_A(x) = (L − x)/L = 1 − x/L

R_B(x) = x/L

ILD for R_A: Straight line from 1.0 at A (x=0) to 0 at B (x=L).

ILD for R_B: Straight line from 0 at A (x=0) to 1.0 at B (x=L).

Both ILDs are straight lines — the ILD for a reaction in a simply supported beam is always a straight line.

Maximum R_A occurs when unit load is at A (ordinate = 1.0). Maximum R_B occurs when unit load is at B (ordinate = 1.0).

Physical interpretation: When the unit load is at A, all of it goes directly into support A — R_A = 1, R_B = 0. As the load moves toward B, more of it transfers to B and less to A. When the load reaches B, R_A = 0 and R_B = 1. The linear variation reflects the linear lever-arm effect.

3. ILD for Shear Force

The ILD for shear force at a section C (at distance a from A) of a simply supported beam has a characteristic shape — a straight line that jumps by 1.0 at the section itself.

ILD for Shear Force at Section C (distance a from A, b = L−a from B)

When unit load is to the right of C (a < x ≤ L):

V_C = −R_B = −x/L (considering right portion; shear is negative here by convention)

Wait — let us use the consistent definition: V_C = sum of forces to the LEFT of C.

When unit load is to the left of C (0 ≤ x < a): R_A = (L−x)/L, so V_C = R_A = (L−x)/L

At x = 0: V_C = 1.0 | At x = a (just left of C): V_C = (L−a)/L = b/L

When unit load is to the right of C (a < x ≤ L): V_C = R_A − 0 = (L−x)/L (no load to left of C except R_A)

At x = a (just right of C): V_C = (L−a)/L − 1 = b/L − 1 = −a/L

At x = L: V_C = 0 − 0… = 0

Correction: When unit load is just to the right of C: V_C = R_A (no loads to left of C — the unit load is to the right) = (L−x)/L.

Just right of C (x = a⁺): V_C = (L−a)/L − 1 = −a/L (the −1 comes from the unit load which is at x = a, i.e., at C itself, contributing to the left side count)

Summary:

ILD ordinate at A (x=0): +b/L … No. Let us state the standard result:

For unit load between A and C (left of section): η_V = −(x/L) — negative, varying from 0 at A to −a/L just left of C

For unit load between C and B (right of section): η_V = +(L−x)/L — positive, varying from +b/L just right of C to 0 at B

ILD shape: Two straight lines. Left portion: 0 at A sloping to −a/L at C. Right portion: +b/L at C sloping to 0 at B. Jump of 1.0 (= a/L + b/L) at C.

Maximum positive shear at C = b/L (unit load just right of C)

Maximum negative shear at C = −a/L (unit load just left of C)

4. ILD for Bending Moment

The ILD for bending moment at section C is a triangle with a peak directly under C. This triangular shape is one of the most important results in influence line theory.

ILD for Bending Moment at Section C (distance a from A, b = L−a from B)

When unit load is at position x:

For x ≤ a (load to left of or at C): M_C = R_B × b = (x/L) × b = bx/L

At x = 0: M_C = 0 | At x = a: M_C = ab/L (maximum)

For x ≥ a (load to right of or at C): M_C = R_A × a = ((L−x)/L) × a = a(L−x)/L

At x = a: M_C = ab/L (maximum) | At x = L: M_C = 0

ILD shape: Triangle. Zero at A, rises linearly to peak of ab/L directly under C, then falls linearly back to zero at B.

Peak ordinate = ab/L (at position C itself)

For midspan C (a = b = L/2): Peak = L/4. Maximum midspan moment under unit load = L/4 (when load is at midspan) ✓

Key result for UDL using ILD: The maximum bending moment at C under a full-span UDL w = w × area of triangular ILD = w × (1/2 × L × ab/L) = wab/2. For midspan of a simply supported beam: M_max = w × L/2 × L/4 = wL²/8 ✓ — this confirms the standard formula.

5. Müller-Breslau Principle

The Müller-Breslau Principle provides a powerful, intuitive way to draw influence lines without any calculation — by observing the shape of the deflected structure under a specific imposed deformation.

Müller-Breslau Principle — Statement

The influence line for any response function (reaction, shear force, or bending moment) at a section is the deflected shape of the structure when the structure is “released” at that section and a unit deformation corresponding to the response function is imposed.

For a reaction R_A: Remove the support at A (release the reaction). Apply a unit vertical displacement at A (in the direction of R_A). The resulting deflected shape of the beam is the ILD for R_A.

For a shear force V_C: Insert a shear release at C (allow relative vertical movement between the two parts). Apply a unit relative vertical displacement at C (keeping both sides horizontal). The deflected shape is the ILD for V_C.

For a bending moment M_C: Insert a hinge at C (allow relative rotation). Apply a unit relative rotation at C. The deflected shape is the ILD for M_C.

Why is this useful? For determinate structures, the ILD shapes are always straight lines — which makes them easy to sketch quickly. For indeterminate structures (continuous beams, frames), the Müller-Breslau deflected shapes are curved — they give the correct qualitative shape and can be quantified using any deflection calculation method.

Practical significance — loading for maximum response: The Müller-Breslau Principle immediately tells you where to place a UDL to maximise a response. Simply cover all portions of the ILD that have the same sign as the desired response. For maximum positive bending moment at C, load all portions of the beam where the ILD for M_C is positive. For maximum negative bending moment (hogging), load where the ILD is negative.

Response Function	Release Type	Unit Deformation	ILD Shape (Determinate SS Beam)
Reaction R_A	Remove support at A	Unit upward displacement at A	Straight line: 1.0 at A, 0 at B
Shear V_C	Shear release at C	Unit relative vertical displacement at C	Two straight lines with unit jump at C
Bending Moment M_C	Hinge at C	Unit relative rotation at C	Triangle with peak ab/L at C

6. Using ILDs — Point Loads and UDL

Response Under a System of Point Loads

When loads P₁, P₂, …, P_n are at positions x₁, x₂, …, x_n:

F = P₁η₁ + P₂η₂ + … + P_nη_n = Σ P_iη_i

Where η_i = ILD ordinate at position x_i.

If η_i is positive, that load contributes positively to F (increases the response). If η_i is negative, that load reduces F.

Response Under a UDL

For a UDL of intensity w (kN/m) applied over a length from x₁ to x₂:

F = w × ∫_x1^x2 η(x) dx = w × (area of ILD between x₁ and x₂)

For maximum positive F: load all regions where η > 0.

For maximum negative F: load all regions where η < 0.

Area of a triangle = (1/2) × base × height

Area of a trapezoid = (1/2) × (sum of parallel sides) × height

7. Maximum BM Under a Moving Load System

When a series of concentrated loads (a train of loads) moves across a beam, the bending moment at any section changes as the loads move. The critical question for design is: what is the maximum bending moment that will occur at section C for any position of the load train?

Criterion for Maximum BM at Section C Under a Moving Load Train

The bending moment at section C is maximum when the load train is positioned such that:

“The average load per unit length to the left of C equals the average load per unit length to the right of C”

Equivalently: The resultant of all loads on the beam is equidistant from the load nearest to C on each side.

In practice, the critical position is found by trial: move the load train until this condition is approximately satisfied, checking with the load immediately to the left and right of C. The maximum BM occurs when the resultant of loads crosses C.

Practical Method — Resultant Position Criterion

For maximum BM at section C:

Find the resultant of all loads in the train (magnitude and position).
Position the load train so that section C and the resultant are equidistant from the midspan (i.e., C and the resultant straddle the midspan symmetrically).
Calculate the BM at C for this position.
Repeat with each load placed at C (one load at a time) and check which gives the largest BM.

8. Absolute Maximum Bending Moment

The absolute maximum bending moment in a simply supported beam under a moving load train is the largest bending moment that occurs anywhere in the beam for any position of the load train. Both the location of the critical section and the position of the load train are unknown — this is the most general case.

Absolute Maximum BM — Key Theorem

The absolute maximum bending moment in a simply supported beam under a moving load train occurs under one of the concentrated loads.

Position of load train for absolute maximum BM:

The absolute maximum BM occurs when the load train is positioned such that the midspan of the beam bisects the distance between the resultant of all loads and the load directly under which the BM is being calculated.

In other words: Place the load such that the midpoint of the beam lies halfway between that load and the resultant of all loads on the beam.

Condition: x̄ = (R + x_i)/2 where x̄ is the midspan position, R is the position of the resultant, and x_i is the position of the load under consideration.

Equivalently: The load and the resultant are placed symmetrically about the midspan.

Procedure for absolute maximum BM:

Find the resultant of all loads — magnitude and position (centroid of the load system).
For each load P_i (candidate for position of absolute max), find the distance d between P_i and the resultant.
Position the load train so that P_i and the resultant are equidistant from midspan (each at d/2 from midspan, on opposite sides).
Calculate the BM under P_i for this position.
Repeat for all candidate loads. The largest value is the absolute maximum BM.

9. Worked Example 1 — ILD Construction for Simply Supported Beam

Problem: A simply supported beam AB has span 10 m. Draw the ILD for: (a) reaction R_A, (b) shear force at C (4 m from A), and (c) bending moment at C (4 m from A).

(a) ILD for R_A

R_A(x) = (10 − x)/10 = 1 − x/10

Key ordinates: x = 0 (at A): η = 1.0 | x = 4 m (at C): η = 0.6 | x = 10 m (at B): η = 0.0

Shape: Straight line from 1.0 at A to 0 at B. Entirely positive.

(b) ILD for Shear Force at C (a = 4 m, b = 6 m)

For unit load between A and C (0 ≤ x ≤ 4):

η_V(x) = −x/L = −x/10

x = 0: η = 0 | x = 4⁻: η = −4/10 = −0.4

For unit load between C and B (4 ≤ x ≤ 10):

η_V(x) = (L−x)/L = (10−x)/10

x = 4⁺: η = 6/10 = +0.6 | x = 10: η = 0

Shape: From 0 at A, negative slope to −0.4 just left of C; jumps to +0.6 just right of C; positive slope back to 0 at B. Jump = 0.4 + 0.6 = 1.0 ✓

Maximum positive V_C = 0.6 (unit load just right of C)

Maximum negative V_C = −0.4 (unit load just left of C)

(c) ILD for Bending Moment at C (a = 4 m, b = 6 m)

Peak ordinate = ab/L = 4×6/10 = 2.4 m (at position C, x = 4 m)

For unit load between A and C: η_M(x) = bx/L = 6x/10 = 0.6x

x = 0: η = 0 | x = 4: η = 2.4

For unit load between C and B: η_M(x) = a(L−x)/L = 4(10−x)/10 = 0.4(10−x)

x = 4: η = 0.4×6 = 2.4 ✓ | x = 10: η = 0

Shape: Triangle. Zero at A, rises to peak 2.4 m at C, falls back to zero at B. Entirely positive.

Under a full-span UDL w: M_C = w × area = w × (1/2 × 10 × 2.4) = 12w kN·m

Check: M_C = R_A×a = (w×10/2)×4 = 20w. Wrong? Wait: M_C = w×(R_A×4 − w×4×2) = w × (5×4 − 4×2×w/w)… Let us use the formula: M_C = wab/2 = w×4×6/2 = 12w ✓

10. Worked Example 2 — Maximum Response Under a System of Point Loads

Problem: A simply supported beam AB of span 12 m. Section C is at 5 m from A. A train of three loads moves across: P₁ = 20 kN, P₂ = 30 kN (2 m behind P₁), P₃ = 25 kN (2 m behind P₂). Find the maximum bending moment at C.

ILD for M_C (a = 5 m, b = 7 m, L = 12 m)

Peak ordinate at C = ab/L = 5×7/12 = 35/12 = 2.917 m

ILD equations:

Left of C (0 ≤ x ≤ 5): η = bx/L = 7x/12

Right of C (5 ≤ x ≤ 12): η = a(L−x)/L = 5(12−x)/12

Strategy — Critical Load Positions

Maximum M_C occurs when the load train is positioned to maximise Σ(P_i × η_i). Try placing each load in turn directly at C and compute M_C.

Case 1 — P₂ (30 kN) at C (x = 5 m):

P₁ is 2 m ahead of P₂, so P₁ is at x = 3 m (left of C).

P₃ is 2 m behind P₂, so P₃ is at x = 7 m (right of C).

η at x=3: 7×3/12 = 1.75 m | η at x=5: 2.917 m | η at x=7: 5(12−7)/12 = 25/12 = 2.083 m

M_C = 20×1.75 + 30×2.917 + 25×2.083 = 35 + 87.5 + 52.08 = 174.58 kN·m

Case 2 — P₁ (20 kN) at C (x = 5 m):

P₂ at x = 7 m, P₃ at x = 9 m.

η at x=5: 2.917 m | η at x=7: 2.083 m | η at x=9: 5(12−9)/12 = 1.25 m

M_C = 20×2.917 + 30×2.083 + 25×1.25 = 58.33 + 62.5 + 31.25 = 152.08 kN·m

Case 3 — P₃ (25 kN) at C (x = 5 m):

P₂ at x = 3 m, P₁ at x = 1 m.

η at x=1: 7×1/12 = 0.583 m | η at x=3: 1.75 m | η at x=5: 2.917 m

M_C = 20×0.583 + 30×1.75 + 25×2.917 = 11.67 + 52.5 + 72.92 = 137.08 kN·m

Maximum M_C = 174.58 kN·m (when P₂ = 30 kN is at C)

11. Worked Example 3 — Maximum BM Under UDL

Problem: A simply supported beam AB of span 10 m. A moving UDL of 15 kN/m of length 4 m can be placed anywhere on the beam. Find the maximum bending moment at midspan C (x = 5 m).

Solution

ILD for M_C (midspan, a = b = 5 m): Triangle, peak = 5×5/10 = 2.5 m at midspan.

Left half: η = x/2 (rises from 0 at A to 2.5 at C)

Right half: η = (10−x)/2 (falls from 2.5 at C to 0 at B)

To maximise M_C: Place the UDL where the ILD ordinate is maximum — centred on C (peak of ILD triangle).

Place UDL from x = 3 m to x = 7 m (4 m long, centred at midspan).

Area of ILD under UDL (from 3 m to 7 m):

The ILD is a triangle. The ordinates at x=3 and x=7 are:

η(3) = 3/2 = 1.5 m | η(7) = (10−7)/2 = 1.5 m | η(5) = 2.5 m (peak)

Area under ILD from 3 to 5 (left half): trapezoid = (1/2)(1.5 + 2.5)(2) = 4.0 m²

Area under ILD from 5 to 7 (right half, symmetric): 4.0 m²

Total area = 8.0 m²

Maximum M_C = w × area = 15 × 8.0 = 120 kN·m

Verification: With UDL from 3 to 7 m, R_A = 15×4×(10−5)/10 = 30 kN. M_C = 30×5 − 15×2×1 = 150 − 30 = 120 kN·m ✓

12. Worked Example 4 — Absolute Maximum BM

Problem: A simply supported beam of span 20 m carries a moving load system: P₁ = 100 kN at the front, P₂ = 200 kN at 4 m behind P₁, P₃ = 150 kN at 3 m behind P₂. Find the absolute maximum bending moment and its location.

Step 1 — Resultant of Load System

Total load R = 100 + 200 + 150 = 450 kN

Taking moments about P₁ (front load) to find position of resultant:

R × ȳ = 200×4 + 150×(4+3) = 800 + 1050 = 1850

ȳ = 1850/450 = 4.11 m behind P₁

Resultant is 4.11 m behind P₁ (i.e., between P₂ at 4 m and P₃ at 7 m — closer to P₂).

Distance between P₂ and resultant = 4.11 − 4.0 = 0.11 m (resultant is 0.11 m behind P₂).

Step 2 — Critical Load Position (P₂ likely governs)

For absolute maximum BM under P₂:

Midpoint of beam = 10 m from either end.

Place P₂ and resultant symmetrically about midspan:

Midpoint between P₂ and resultant = 0.11/2 = 0.055 m from P₂.

Place this midpoint at the beam’s midspan (10 m from A).

P₂ position from A = 10 − 0.055 = 9.945 m from A ≈ 9.95 m from A

Load positions: P₁ = 9.95 − 4 = 5.95 m from A | P₂ = 9.95 m | P₃ = 9.95 + 3 = 12.95 m from A

Step 3 — BM Under P₂

R_A = (100×5.95 + 200×9.95 + 150×12.95)/20

= (595 + 1990 + 1942.5)/20 = 4527.5/20 = 226.375 kN

M under P₂ = R_A × 9.95 − 100×(9.95−5.95)

= 226.375×9.95 − 100×4

= 2252.4 − 400 = 1852.4 kN·m

(This would be compared against BM under P₁ and P₃ for the absolute maximum — P₂ being the heaviest load typically governs.)

13. ILDs for Indeterminate Structures

For statically indeterminate structures (continuous beams, propped cantilevers, fixed beams), the ILDs are curved lines rather than straight lines. This is because the reactions of an indeterminate structure depend on the compatibility conditions (which involve EI), not just equilibrium — and the resulting response functions are not linear in the load position.

The Müller-Breslau Principle still applies — the ILD shape is the deflected shape of the released structure under a unit deformation. But now, computing this deflected shape requires a full deflection analysis (double integration, moment-area method, or MDM) rather than simple geometry.

Structure Type	ILD Shape	Analysis Method
Simply supported beam	Straight lines (triangles)	Equilibrium only
Propped cantilever	Curved (cubic)	Force method + deflection
Fixed beam	Curved (cubic)	Force method + deflection
Continuous beam	Curved, changes sign	MDM or Slope-Deflection

Qualitative ILD for continuous beams (important for GATE MCQs): For a two-span continuous beam ABC (pin at A, roller at B, roller at C), the ILD for the intermediate reaction R_B has positive ordinates in both spans and is curved. The ILD for M at an interior section of a continuous beam can have both positive and negative regions — meaning some load positions cause sagging and others cause hogging at that section. This has direct design implications: reinforcement must be provided at both top and bottom of continuous beam sections.

14. Common Mistakes Students Make

Confusing influence lines with bending moment diagrams: A BMD shows how the bending moment varies along the beam for a fixed load position. An ILD shows how the bending moment at a fixed section varies as the load moves. These are fundamentally different diagrams — a BMD has units of kN·m while an ILD ordinate for BM has units of metres (kN·m per kN of unit load = m). Mixing up these two diagrams is the most common conceptual error in influence line problems.
Forgetting the sign of the shear force ILD: The ILD for shear force at section C is negative (below the baseline) for loads to the left of C and positive for loads to the right (using left-upward-positive convention). Students often draw it the other way around. Always verify: a unit load just to the right of C produces positive shear at C (upward force R_A to the left of C is positive shear).
Using peak ILD ordinate as the maximum response for a UDL: For a UDL, the response is w × (area of ILD over the loaded region), NOT w × (peak ordinate). The peak ordinate applies only for a single concentrated unit load at the peak location. For a UDL over the full span, you must integrate (or compute the area of) the ILD over the entire loaded length.
Not checking whether all loads are on the beam when positioning a load train: When positioning a load train for maximum BM at a section, ensure all loads are within the beam span (0 to L). If a calculated position places a load outside the span, the load is not on the beam and does not contribute — recalculate with only the loads actually on the beam.
Applying the absolute maximum BM theorem to one load only: The absolute maximum BM must be checked for each load in the train — the heaviest load does not always govern if it is far from the resultant. Always check BM under each load (for each candidate position) and take the largest.

15. Frequently Asked Questions

What is the difference between an influence line and a bending moment diagram?

A bending moment diagram (BMD) shows the variation of bending moment along the entire length of a beam for a specific, fixed loading arrangement. Every point on the BMD corresponds to a different cross-section of the beam, and the value at each point is the BM at that section under the given loads. An influence line (ILD) for bending moment at section C shows how the BM at that one specific section C varies as a unit load moves along the beam. Every point on the ILD corresponds to a different position of the moving unit load, and the ordinate at each position is the BM at C when the unit load is there. The BMD is drawn once for a given loading; the ILD is drawn once for a given section and used for any loading.

Why is the ILD for bending moment always a triangle for a simply supported beam?

This follows from the Müller-Breslau Principle. To draw the ILD for M_C, insert a hinge at C and apply a unit rotation. The beam (now two rigid segments hinged at C and supported at A and B) deflects into two straight segments — a triangle. The peak of the triangle is at C (where the rotation is applied), and the base is zero at both supports (which remain fixed). Since the beam segments are rigid (in Müller-Breslau, the released structure deforms as a rigid body mechanism), the deflected shape is always straight lines. For indeterminate beams, the segments are not rigid — they must bend to satisfy compatibility, giving curved ILDs.

How do you find the position of maximum shear force under a moving UDL?

To maximise the shear force at section C, use the ILD for V_C. The ILD for shear has a positive region (right of C) and a negative region (left of C). To maximise positive shear at C, load the positive region of the ILD — place the UDL from C to B. To maximise negative shear at C, load the negative region — place the UDL from A to C. For a UDL that must be placed as a single continuous load, the maximum shear occurs when the load covers the larger positive or negative region completely. This direct use of the ILD shape is much faster than trying multiple positions by trial.

Does the Müller-Breslau Principle work for indeterminate structures?

Yes — the Müller-Breslau Principle is valid for both determinate and indeterminate structures. For determinate structures, the released structure deforms as a rigid body mechanism (straight lines). For indeterminate structures, the released structure is still indeterminate after the release (just one degree less indeterminate), so it deforms as a flexible body — the ILD is a curve. The principle gives the correct qualitative shape and, after computing the actual deflections of the released structure, gives the exact quantitative ILD ordinates. This makes Müller-Breslau particularly valuable for indeterminate structures where deriving ILDs analytically from first principles is complex.