---
title: "ILD: moment at section x-x, continuous beam"
subject: "Theory of Structures II"
chapter: 4
key: "semester_5/theory_of_structures_ii:ild-moment-section-xx-two-span"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-moment-section-xx-two-span
site: BCT Engineering
---

# ILD: moment at section x-x, continuous beam

Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams

- Asked in: 2073 Shrawan, 5 marks
- Video: [EX 07 | INFLUENCE LINE DIAGRAM | TWO SPAN CONTINUOUS BEAM | INDETERMINATE STRUCTURE | ILD](https://www.youtube.com/watch?v=GZweFbmDFfw) — Structural Funda by:- KheNiSiR
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-moment-section-xx-two-span
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4

## Question

Draw influence line diagram for moment at section x-x of the continuous beam shown in figure below. Find the ordinates at 2 m intervals. [Figure: two-span beam, hinged at the left end, rollers at the middle and right supports; spans 6 m and 8 m... dimension marks 4 m and 6 m from the left end (section x-x at 4 m from the left support) and 8 m for the second span.]

## Short answer

Use ILD(M_x) = simple-span ordinate plus (4/6)ILD(M_B). Peak sagging ordinate is +1.016 at x = 4 m, zero at A, B and C, and the ILD is negative in BC with minimum about -0.59 near x = 9.4 m.

## Answer

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)
| x from A (m) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 |
|---|---|---|---|---|---|---|---|---|
| $M_{x-x}$ | 0 | 0.4127 | 1.0159 | 0 | -0.5 | -0.5714 | -0.3571 | 0 |

#### ILD shape
```text
             **
          ***  **
       ***       *
   ****           **
***-----------------**------------------------***
                      ***                *****
                         ****************


^                    ^                          ^
```
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For section x-x at 4 m in AB = 6 m, the M_B contribution is multiplied by
   A. 4/6
   B. 2/6
   C. 1/2
   D. 6/8
   Answer: A. The factor is a/L1 = 4/6. (Key point: Carry factor)

2. Maximum positive ordinate (load at the section) is
   A. 1.333
   B. 1.016
   C. 0.476
   D. 0.571
   Answer: B. 1.333 + (4/6)(-0.476) = 1.016. (Key point: Peak)

3. For a load in BC the M_x ordinate is
   A. Positive
   B. Zero
   C. Negative
   D. Equal to the simple-span ordinate
   Answer: C. Simple-span part is zero and M_B is hogging. (Key point: Load in BC)

## Related in Theory of Structures II

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