---
title: "ILD at mid-span: BM and SF under given loads"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:ild-mid-span-bm-sf-50-10-30"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ild-mid-span-bm-sf-50-10-30
site: BCT Engineering
---

# ILD at mid-span: BM and SF under given loads

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

- Asked in: 2071 Magh, 10 marks
- Video: [Influence Line Diagrams for Simply Supported Beams](https://www.youtube.com/watch?v=P8FJdF7Jlow) — 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/ild-mid-span-bm-sf-50-10-30
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw influence line diagram for bending moment and shear force at mid span of the beam of span 20 m and determine bending moment and shear force at that section due to the loads shown in figure using the influence line diagram. [Figure: simply supported beam of span 20 m; 50 kN at 3 m from left support; 10 kN/m UDL over the middle 10 m (from 5 m to 15 m); 30 kN at 3 m from right support.]

## Short answer

For the 20 m simply supported beam, use the triangular ILD for moment (peak 5 m) and ILD for shear with jump 1 at mid-span. Compute M = 50×1.5 + 10×37.5 + 30×1.5 = 495 kN·m, V = 50(−0.15) + 10×0 + 30(0.15) = −3 kN.

## Answer

**Data:** simply supported beam $AB$, $L = 20$ m. Loads: 50 kN at 3 m from $A$; 10 kN/m UDL from 5 m to 15 m; 30 kN at 3 m from $B$ (17 m from $A$). Section at mid-span, $x = 10$ m.

#### ILD for bending moment at mid-span
A triangle with peak $\dfrac{L}{4} = 5$ m at mid-span. Ordinate at distance $x$ from the near support $= x/2$.

```text
  A o--------------------o B
              /\ 5.0
  ILD: 0 /            \ 0
```

- 50 kN at 3 m: ordinate $= 3/2 = 1.5$ m
- UDL from 5 m to 15 m: ordinates $2.5$ at 5 m, $5$ at 10 m, $2.5$ at 15 m; area $= 2\times\dfrac{(2.5+5)}{2}\times 5 = 37.5$ m²
- 30 kN at 17 m: ordinate $= 3/2 = 1.5$ m

$$
M = 50(1.5) + 10(37.5) + 30(1.5) = 75 + 375 + 45 = 495\ \text{kN m}
$$

#### ILD for shear force at mid-span
Ordinate $-\dfrac{x}{20}$ for a load left of the section and $+\dfrac{20-x}{20}$ for a load right of it; the jump at mid-span is 1 (from $-0.5$ to $+0.5$).

- 50 kN at 3 m: ordinate $= -\dfrac{3}{20} = -0.15$
- UDL from 5 to 10 m (left of the section): area $= -\tfrac12(0.25+0.5)\times 5 = -1.875$ m
- UDL from 10 to 15 m (right of the section): area $= +\tfrac12(0.5+0.25)\times 5 = +1.875$ m (net area under the UDL $=0$)
- 30 kN at 17 m: ordinate $= +\dfrac{3}{20} = +0.15$

$$
V = 50(-0.15) + 10(-1.875+1.875) + 30(0.15) = -7.5 + 0 + 4.5 = -3.0\ \text{kN}
$$

**Check:** $R_A = \dfrac{50(17)+10(10)(10)+30(3)}{20} = 97$ kN; $V = 97 - 50 - 10\times 5 = -3$ kN and $M = 97\times 10 - 50\times 7 - 10\times 5\times 2.5 = 495$ kN m.

**Answer: $M_{mid} = 495$ kN m (sagging); $V_{mid} = -3$ kN.**

## Check yourself

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

1. What is the peak of the mid-span BM ILD on a 20 m span?
   A. 10 m
   B. 5 m (L/4)
   C. 2.5 m
   D. 20 m
   Answer: B. ab/L = 10 x 10 / 20 = 5 m. (Key point: Mid-span BM ILD)

2. What is the mid-span BM from the loads?
   A. 495 kN m
   B. 395 kN m
   C. 120 kN m
   D. 375 kN m
   Answer: A. 50(1.5) + 10(37.5) + 30(1.5) = 495 kN m. (Key point: BM result)

3. What are the shear ordinates just left/right of mid-span?
   A. -0.5 and +0.5, jump of 1
   B. 0 and 1
   C. -1 and +1
   D. +0.5 and +0.5
   Answer: A. -x/20 = -0.5 on the left and (20-x)/20 = +0.5 on the right. (Key point: Shear ILD jump)

4. Why does the UDL contribute nothing to shear at mid-span?
   A. The UDL is zero
   B. Its areas on either side are equal and opposite (-1.875 and +1.875)
   C. It is outside the section
   D. Shear ILD is zero
   Answer: B. The UDL is symmetric about mid-span, so negative and positive areas cancel. V = -7.5 + 0 + 4.5 = -3 kN. (Key point: Shear UDL contribution)

## Related in Theory of Structures I

- [ILD and BM at mid span for triangular load](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-triangular-load-mid-span-c) — Theory of Structures I, chapter 5
- [ILD: reactions, BM and SF beneath loads](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-reactions-sf-bm-beneath-loads) — Theory of Structures I, chapter 5, asked 2×
- [ILDs for reactions, SF and BM, 20 m beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Abeam-ild-reactions-sf-bm-5m-section) — Theory of Structures I, chapter 5, asked 2×
- [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
- [Four loads on 50 m girder: BM at 20 m, absolute](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Afour-loads-100-150-200-150-50m) — Theory of Structures I, chapter 5
- [ILD for SF and BM at C of beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-sf-bm-c-40-60-5knm) — Theory of Structures I, chapter 5

## Related in other subjects

- [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
- [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: moment at section 5, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-section-5-two-span) — Theory of Structures II, chapter 4
- [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