---
title: "ILD: fixed-end moment, 12 m beam"
subject: "Theory of Structures II"
chapter: 4
key: "semester_5/theory_of_structures_ii:ild-support-moment-fixed-12m-3m"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-support-moment-fixed-12m-3m
site: BCT Engineering
---

# ILD: fixed-end moment, 12 m beam

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

- Asked in: 2075 Chaitra, 6 marks
- Video: [SA56: Moment Influence Line for Indeterminate Beams: Part 1](https://www.youtube.com/watch?v=VcTj5OkJX-Q) — Dr. Structure
- 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-support-moment-fixed-12m-3m
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4

## Question

Draw ILD for the support moment at A by computing the ordinates at 3 meter intervals. [Figure: propped cantilever AB, A fixed, B roller, span 12 m.]

## Short answer

The ILD for M_A is entirely hogging. Ordinates every 3 m: 0, -1.97, -2.25, -1.41, 0 at 0,3,6,9,12 m. Maximum -2.31 occurs near 5.07 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.

#### 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)
| x from A (m) | 0 | 3 | 6 | 9 | 12 |
|---|---|---|---|---|---|
| $M_A$ | 0 | -1.9687 | -2.25 | -1.4062 | 0 |

#### ILD shape
```text




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

## Check yourself

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

1. The hogging M_A = x(L-x)(2L-x)/(2L^2) comes from
   A. Only ab^2/L^2
   B. Only a^2b/L^2
   C. Sum of both fixed-end moments
   D. Fixed-end moment ab^2/L^2 plus half the far-end moment a^2b/L^2
   Answer: D. Releasing the roller at B carries half of its fixed-end moment over to A. (Key point: Origin of formula)

2. M_A for a load at x = 6 m on a 12 m span is
   A. 2.25 (hogging)
   B. 3.00
   C. 1.50
   D. 1.97
   Answer: A. 6 x 6 x 18 / (2 x 144) = 2.25. (Key point: Mid-span value)

3. The ILD for the support moment at A is
   A. Sagging near B
   B. Entirely hogging, zero at A and B, with its maximum near 0.42L
   C. Maximum at mid-span exactly
   D. Hogging at A, sagging near B
   Answer: B. All ordinates keep the same sign; the maximum -2.31 is at about 0.42L. (Key point: Shape)

## Related in Theory of Structures II

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