---
title: "Overhang beam ILD for shear force at C"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:overhang-ild-sf-c-2071o"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/overhang-ild-sf-c-2071o
site: BCT Engineering
---

# Overhang beam ILD for shear force at C

Theory of Structures I · Chapter 5: Influence Lines for Simple Structures

- Asked in: 2071 Magh (old course), 6 marks
- Video: [Influence Line Diagrams for Overhanging beams ( Overhanging on both of the Sides )](https://www.youtube.com/watch?v=t0OrLvMDX9s) — Stan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/overhang-ild-sf-c-2071o
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw influence line diagram for shear force at C of the overhanging beam shown in figure below. [Figure: beam ABD; A hinge, B roller, AB = 20 m, overhang BD = 5 m; section C at 6 m from A.]

## Short answer

ILD for shear at C: 0 at A, decreases to -0.30 just left of C, jumps to +0.70 just right of C, then linearly to 0 at B (20 m) and further to -0.25 at free end D (25 m).

## Answer

**Data:** beam $ABD$: $A$ hinge, $B$ roller, $AB = 20$ m, overhang $BD = 5$ m. Section $C$ is 6 m from $A$. Let the unit load be at $x$ from $A$.

#### Reactions
$$
R_B = \frac{x}{20}, \qquad R_A = 1 - \frac{x}{20} = \frac{20-x}{20}
$$

These are valid for $0\le x\le 25$ (for $x>20$, $R_A$ is negative).

#### Shear at C
- **Load between $A$ and $C$ ($0\le x<6$):** take the right part: $V_C = -R_B = -\dfrac{x}{20}$. At $x = 6^-$: $-0.30$.
- **Load between $C$ and $B$ ($6<x\le 20$):** take the left part: $V_C = R_A = \dfrac{20-x}{20}$. At $x = 6^+$: $+0.70$; at $B$: 0.
- **Load on the overhang ($20\le x\le 25$):** $V_C = R_A = \dfrac{20-x}{20}$, which is negative: $-0.25$ at $D$.

| Position | $A$ (0) | $C^-$ (6) | $C^+$ (6) | $B$ (20) | $D$ (25) |
|---|---|---|---|---|---|
| Ordinate of $V_C$ | 0 | −0.30 | +0.70 | 0 | −0.25 |

```text
 A o----C--------------------o B----D
 ILD V_C:  0 \-0.30 | +0.70 \       0 .. -0.25
```

The ILD is a line from 0 at $A$ down to $-0.30$ just left of $C$, a jump of 1 to $+0.70$ just right of $C$, then a straight line falling to 0 at $B$, continuing to $-0.25$ at the free end $D$.

#### Use
- Maximum positive shear at $C$: load only the positive part, $6<x<20$; for a UDL $w$, $V = w\times\tfrac12\times 14\times 0.70 = 4.9\,w$.
- Maximum negative shear: load $A$ to $C$ and the overhang; $V = -w\,(\tfrac12\times 6\times 0.30 + \tfrac12\times 5\times 0.25) = -1.525\,w$.

## Check yourself

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

1. What are the ordinates of V_C just left and right of C (6 m from A on a 20 m span)?
   A. -0.30 and +0.70
   B. +0.30 and -0.70
   C. -0.70 and +0.30
   D. -0.30 and +0.30
   Answer: A. Left: -x/20 = -0.30; right: (20-x)/20 = 0.70; jump 1. (Key point: Shear ILD ordinates)

2. What happens to the ILD for a load on the overhang BD?
   A. It is zero
   B. The line continues from 0 at B to -0.25 at D
   C. It rises to +0.25 at D
   D. It jumps to 1
   Answer: B. V_C = R_A = (20-x)/20 stays valid and is negative: -0.25 at D. (Key point: Overhang portion)

3. For maximum negative shear at C with a UDL w, what is loaded?
   A. Only C to B
   B. A to C and the overhang, giving -1.525 w
   C. Whole beam
   D. Only the overhang, giving -0.625 w
   Answer: B. Negative areas are 0.9 (A to C) and 0.625 (overhang): sum 1.525 w. (Key point: Maximum negative shear)

## Related in Theory of Structures I

- [Overhang beam: ILD, max positive/negative moment at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aoverhang-ild-moment-d-three-loads) — Theory of Structures I, chapter 5
- [Conjugate beam: overhanging beam free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-overhang-300-150) — Theory of Structures I, chapter 4
- [ILD for BM at F and stress in support BD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-moment-f-stress-bd) — Theory of Structures I, chapter 5
- [Arch: reactions, BM, thrust and shear at D by ILD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-50m-udl-50kn-ild-d-15m) — Theory of Structures I, chapter 6
- [Strain energy and real work deflection, overhang beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Areal-work-strain-energy-overhang-5m) — Theory of Structures I, chapter 3
- [Define influence line diagram and its uses](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Adefine-ild-uses) — Theory of Structures I, chapter 5, asked 5×

## Related in other subjects

- [ILD: shear at section x-x, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-shear-section-xx-two-span) — Theory of Structures II, chapter 4
- [ILD: shear at section 2-2, propped cantilever](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-shear-section-2-2-propped) — Theory of Structures II, chapter 4, asked 3×
- [Muller-Breslau: ILD of shear force at B](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amueller-breslau-ild-shear-b-5m) — Theory of Structures II, chapter 4
- [ILD: shear at section of propped cantilever](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-shear-propped-10m-sec4) — Theory of Structures II, chapter 4