---
title: "ILD: shear at section 2-2, propped cantilever"
subject: "Theory of Structures II"
chapter: 4
key: "semester_5/theory_of_structures_ii:ild-shear-section-2-2-propped"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-shear-section-2-2-propped
site: BCT Engineering
---

# ILD: shear at section 2-2, propped cantilever

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

- Asked in: 2070 Chaitra, 5 marks
- 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-section-2-2-propped
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4

## Question

Draw influence line diagram for the shear at section 2-2 of the propped cantilever beam shown in figure below. Find the ordinates at 2 m interval. [Figure: beam AB, A fixed, B roller, span 8 m, section 2-2 at 4 m from A.]

## Short answer

Use redundant RB = x^2(3L-x)/(2L^3); shear is -RB for a load left of the section and 1-RB for a load right. Ordinates at 2 m intervals are 0, -0.0859, -0.3125/0.6875, 0.3672, 0, with a unit jump at the section.

## 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, $L = 8$ m, A fixed, B roller, section 2-2 at 4 m from A. Redundant: $R_B = \dfrac{x^2(3L-x)}{2L^3}$. Shear at the section (left part): load left of the section, $V = -R_B$; load right, $V = 1 - R_B$.

Sample: $x = 6$: $R_B = \dfrac{36\times18}{2\times512} = 0.633$, $V = 0.367$. At the section: $R_B(4) = \dfrac{16\times20}{1024} = 0.3125$, giving $V = -0.3125$ just left and $+0.6875$ just right.

#### Ordinates (2 m interval)
| x from A (m) | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| $V_{2-2}$ | 0 | -0.0859 | -0.3125 / 0.6875 | 0.3672 | 0 |

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


^                                               ^
```
Jump of 1.0 at the section (from $-0.3125$ to $+0.6875$); the ILD is zero at A and B.

## Check yourself

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

1. For load left of the section, V =
   A. 1 - R_B
   B. R_B
   C. -R_B
   D. 0
   Answer: C. Left-part shear: load on left gives -R_B; right gives 1 - R_B. (Key point: Shear rule)

2. R_B for L = 8 m, x = 4 m is
   A. 0.3125
   B. 0.6875
   C. 0.5
   D. 0.367
   Answer: A. 16 x 20 / 1024 = 0.3125. (Key point: Redundant)

3. Ordinates just left and right of the section
   A. -0.6875 and +0.3125
   B. -0.3125 and +0.6875
   C. 0 and 1
   D. -0.5 and +0.5
   Answer: B. The jump of 1.0 is the difference. (Key point: Jump)

## Close variants set in other exams

- [ILD: reaction at B, propped cantilever 10 m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-reaction-b-propped-10m-2m) — Theory of Structures II, chapter 4, asked 2×
- [ILD: shear at A, propped cantilever 7 m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-shear-a-propped-7m-1m) — Theory of Structures II, chapter 4, asked 2×

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