---
title: "ILD and BM at mid span for triangular load"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:ild-triangular-load-mid-span-c"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ild-triangular-load-mid-span-c
site: BCT Engineering
---

# ILD and BM at mid span for triangular load

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

- Asked in: 2065 Chaitra, 8 marks
- Video: [Lecture - 008 Influence Lines for Determinate Structures (Part 2)](https://www.youtube.com/watch?v=8xTw2dUb7-Y) — Pinoy Ipsum 
- 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/ild-triangular-load-mid-span-c
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw influence line diagram and calculate the bending moment at mid span C for the beam shown in fig-7. [Figure: simply supported beam AB of span L, C at mid span; a triangular load increasing from zero at C to W kN/m at B over the right half.]

## Short answer

The ILD for M_C is a triangle with peak L/4 at C. Integrating the triangular load from C to B yields M_C = WL^2/48 (sagging).

## Answer

**Data:** simply supported beam $AB$ of span $L$; $C$ at mid-span. A triangular load acts on the right half $CB$, zero at $C$ and rising to $W$ kN/m at $B$.

#### ILD for bending moment at C
A triangle with peak $\dfrac{ab}{L} = \dfrac{(L/2)(L/2)}{L} = \dfrac{L}{4}$ at $C$, 0 at $A$ and $B$. At a distance $u$ from $B$ (in the right half) the ordinate is

$$
y = \frac{L/4}{L/2}\,u = \frac{u}{2}
$$

```text
   A o--------------C--------------o B
                    /\  L/4
   ILD M_C:      0 /    \ 0
   Load on CB:   0 at C ........ W at B
```

#### Load intensity
At distance $u$ from $B$: $w(u) = W\left(1 - \dfrac{2u}{L}\right)$, so $w = W$ at $B$ ($u=0$) and 0 at $C$ ($u = L/2$).

#### Bending moment
$M_C = \int w\,y\,du$ over the loaded length:

$$
\begin{aligned} M_C &= \int_0^{L/2} W\left(1-\frac{2u}{L}\right)\frac{u}{2}\,du = \frac{W}{2}\left[\frac{u^2}{2} - \frac{2u^3}{3L}\right]_0^{L/2} \\ &= \frac{W}{2}\left[\frac{L^2}{8} - \frac{L^2}{12}\right] = \frac{W}{2}\times\frac{L^2}{24} = \frac{WL^2}{48} \end{aligned}
$$

#### Check by statics
Total load $=\tfrac12\cdot W\cdot\dfrac{L}{2} = \dfrac{WL}{4}$, acting at $\dfrac{2}{3}\cdot\dfrac{L}{2} = \dfrac{L}{3}$ from $C$, i.e. $\dfrac{5L}{6}$ from $A$. $R_A = \dfrac{WL}{4}\times\dfrac{L/6}{L} = \dfrac{WL}{24}$.

$$
M_C = R_A\times\frac{L}{2} = \frac{WL}{24}\times\frac{L}{2} = \frac{WL^2}{48}
$$

**Answer: $M_C = \dfrac{WL^2}{48}$ (sagging).**

## Check yourself

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

1. What is the peak ordinate of the M_C influence line at mid-span of span L?
   A. L/2
   B. L/4
   C. L/8
   D. L
   Answer: B. ab/L = (L/2)(L/2)/L = L/4. (Key point: Peak of ILD)

2. How is M_C found from the ILD for a distributed load?
   A. Sum of load intensity times ILD ordinate integrated over the loaded length
   B. Peak ordinate times total load
   C. W x L/4
   D. Only the area of the ILD
   Answer: A. M_C = integral of w(u) y(u) du over the loaded length, since each element w du acts like a point load. (Key point: Method of ILD integration)

3. What is the result for M_C?
   A. WL^2/24
   B. WL^2/16
   C. WL^2/48
   D. WL^2/12
   Answer: C. The integral gives (W/2)(L^2/8 - L^2/12) = WL^2/48. (Key point: Result)

4. In the check, the triangular load WL/4 acts at 5L/6 from A. What is R_A?
   A. WL/8
   B. WL/12
   C. WL/24
   D. WL/4
   Answer: C. R_A = (WL/4)(L/6)/L = WL/24, and M_C = R_A x L/2 = WL^2/48. (Key point: Statics check)

## Related in Theory of Structures I

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- [Moment area: mid-span deflection under UDL](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-ss-udl-midspan) — Theory of Structures I, chapter 4
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- [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
- [Suspension 100 m: UDL on right half, SF and BM at 20 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-100m-5knm2-10knm2-right-half) — Theory of Structures I, chapter 7, asked 2×
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## Related in other subjects

- [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7
- [ILD: bending moment at D, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-d-ab-bc-2) — Theory of Structures II, chapter 4
- [ILD: bending moment at B, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-b-two-span-8) — Theory of Structures II, chapter 4
- [Mid-span deflection of beam, 10 kN load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ass-beam-deflection-mid-span-4m) — Strength of Materials, chapter 7