---
title: "ILD: vertical reaction at fixed support"
subject: "Theory of Structures II"
chapter: 4
key: "semester_5/theory_of_structures_ii:ild-vertical-reaction-fixed-propped-025"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-vertical-reaction-fixed-propped-025
site: BCT Engineering
---

# ILD: vertical reaction at fixed support

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

- Asked in: 2072 Chaitra, 4 marks
- Video: [Influence Line Diagrams for Propped Cantilever Beams - Problem No 1 (Reaction at A & B and Moment A)](https://www.youtube.com/watch?v=s920Z5PGLkw) — Stan Academy
- 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-vertical-reaction-fixed-propped-025
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4

## Question

Draw influence diagram for the vertical reaction at the fixed support of a propped cantilever beam. Plot ordinates at 0.25 times span length.

## Short answer

For a propped cantilever, R_A = 1 - x^2(3L-x)/(2L^3). Ordinates at 0.25L intervals are 1, 0.9141, 0.6875, 0.3672 and 0, all positive upward, falling smoothly from the fixed support to zero at the prop.

## Answer

#### 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)
| Load at (x/L) | 0 | 0.25 | 0.50 | 0.75 | 1.00 |
|---|---|---|---|---|---|
| $R_A$ | 1 | 0.9141 | 0.6875 | 0.3672 | 0 |

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




^                                               ^
```
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.

## Check yourself

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

1. To get the ILD of R_A, Muller-Breslau says
   A. Release rotation at A
   B. Remove the prop B and rotate
   C. Cut the beam at A
   D. Remove vertical restraint at A (keep rotation fixed) and give a unit upward displacement
   Answer: D. Force-type reaction: remove that translation and displace by one unit. (Key point: Principle)

2. R_A =
   A. 1 - x^2(3L-x)/(2L^3)
   B. x^2(3L-x)/(2L^3)
   C. 1 + R_B
   D. x(L-x)/L
   Answer: A. R_A = 1 - R_B. (Key point: Formula)

3. R_A for load at mid-span is
   A. 0.5
   B. 0.6875
   C. 0.3125
   D. 0.9141
   Answer: B. 1 - 0.25 x 2.5/2 = 0.6875. (Key point: Value)

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