---
title: "ILD: shear at C, propped cantilever 14 m"
subject: "Theory of Structures II"
chapter: 4
key: "semester_5/theory_of_structures_ii:ild-shear-c-propped-14m"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-shear-c-propped-14m
site: BCT Engineering
---

# ILD: shear at C, propped cantilever 14 m

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

- Asked in: 2075 Asoj, 6 marks
- Video: [Influence Line Diagrams for Propped Cantilever Beams - Problem No 2 (Shear Force at C)](https://www.youtube.com/watch?v=p24SaLdKj3k) — 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-shear-c-propped-14m
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4

## Question

Draw ILD for S.F. at point C of the propped cantilever beam shown in figure below. [Figure: beam AB, A fixed, B roller, span 14 m, C at 6 m from B.]

## Short answer

Shear ILD at C: left part V = -R_B, right part V = 1 - R_B. At C (8 m from A), ordinates are -0.40 just left and +0.60 just right; zero at ends.

## 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, $L = 14$ m, A fixed, B roller, C at 6 m from B (8 m from A). Redundant: $R_B$.
With support B removed, the cantilever deflection at B due to a unit load at $x$ from A is $\delta_{Bx} = \dfrac{x^2(3L-x)}{6EI}$, and due to a unit load at B it is $f_{BB} = \dfrac{L^3}{3EI}$. Compatibility (zero deflection at B) gives
$$
R_B = \frac{\delta_{Bx}}{f_{BB}} = \frac{x^2(3L-x)}{2L^3}
$$
Shear at C (left part): load left of C: $V_C = -R_B$; load right of C: $V_C = 1 - R_B$.

Sample: $x = 10$: $R_B = \dfrac{100\times32}{2\times2744} = 0.583$, so $V_C = 0.417$. At C: $R_B(8) = \dfrac{64\times34}{5488} = 0.397$, giving $-0.397$ (just left) and $+0.603$ (just right).

#### Ordinates (2 m interval)
| x from A (m) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 |
|---|---|---|---|---|---|---|---|---|
| $V_C$ | 0 | -0.0292 | -0.1108 | -0.2362 | -0.3965 / 0.6035 | 0.4169 | 0.2128 | 0 |

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

^                                               ^
```
The ordinates are 0 at A and B; the jump at C is 1.0 (from $-0.397$ to $+0.603$).

## Check yourself

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

1. Which redundant is used to solve the propped cantilever?
   A. M_A
   B. R_B
   C. V_C
   D. R_A
   Answer: B. Removing the prop at B leaves a cantilever; R_B is found from zero deflection at B. (Key point: Redundant)

2. For load to the right of C, the shear at C (left part) is
   A. -R_B
   B. R_B
   C. 1 - R_B
   D. 1 + R_B
   Answer: C. With the load right of C, the unit load is not on the left part, so the shear there is the upward unit minus R_B; for loads left of C it is -R_B. (Key point: Shear rule)

3. What is the jump in the ILD at section C?
   A. 0.5
   B. R_B at C
   C. 0.603
   D. 1.0
   Answer: D. Shear ILD jumps by unity at the section, from -0.397 to +0.603. (Key point: Jump)

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