---
title: "ILD: reaction at A and BM at C"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:ild-reaction-a-bm-c-80kn-20knm"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ild-reaction-a-bm-c-80kn-20knm
site: BCT Engineering
---

# ILD: reaction at A and BM at C

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

- Asked in: 2075 Bhadra, 6 marks
- Video: [1st Numerical of ILD for Support Reaction - Influence Line Diagram - GATE Structural Analysis](https://www.youtube.com/watch?v=X437yt6E3J8) — Ekeeda GATE & ESE
- 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-reaction-a-bm-c-80kn-20knm
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Using ILD, calculate reaction at A and BM at C. [Figure: beam ABCD; A hinge, D roller; AB = 2 m with 80 kN at B, BC = 1 m, CD = 5 m carrying 20 kN/m UDL.]

## Short answer

Using influence lines: RA = 80 kN × 0.75 (at B) + 20 kN/m × 1.5625 m (area under UDL) = 91.25 kN. MC = 80 kN × 1.25 m + 20 kN/m × 4.6875 m² = 193.75 kN m.

## Answer

**Data:** simply supported beam $AD$ ($A$ hinge, $D$ roller). $AB = 2$ m with 80 kN at $B$; $BC = 1$ m; $CD = 5$ m with 20 kN/m UDL. So $L = 8$ m and $C$ is 3 m from $A$.

#### Reaction at A
ILD for $R_A$: ordinate $\dfrac{8-x}{8}$, i.e. 1 at $A$ and 0 at $D$.

- Ordinate at $B$ ($x=2$): $\dfrac{6}{8} = 0.75$.
- Ordinates under the UDL: $0.625$ at $C$ and $0$ at $D$, so the area $=\tfrac12\times 5\times 0.625 = 1.5625$ m.

$$
R_A = 80(0.75) + 20(1.5625) = 60 + 31.25 = 91.25\ \text{kN}
$$

#### Bending moment at C
ILD for $M_C$: a triangle with peak $\dfrac{ab}{L} = \dfrac{3\times 5}{8} = 1.875$ m under $C$; at $B$ the ordinate is $\dfrac{2\times 5}{8} = 1.25$ m.

```text
  A o--B--C-----------o D
  ILD M_C:   /\ 1.875
           1.25 at B
```

Area of the ILD under the UDL (triangle from $C$ to $D$) $=\tfrac12\times 5\times 1.875 = 4.6875$ m².

$$
M_C = 80(1.25) + 20(4.6875) = 100 + 93.75 = 193.75\ \text{kN m}
$$

Check: $M_C = R_A\times 3 - 80\times 1 = 273.75 - 80 = 193.75$ kN m.

**Answer: $R_A = 91.25$ kN, $M_C = 193.75$ kN m (sagging).**

## Check yourself

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

1. How is R_A found using its ILD?
   A. Sum of load x ordinate for point loads plus w x area under the UDL
   B. Divide total load by 2
   C. Take the maximum ordinate only
   D. Multiply the UDL by the span
   Answer: A. Effect = sum of P x y plus w x area of the ILD under the UDL. (Key point: Using ILD for reaction)

2. What is R_A?
   A. 60 kN
   B. 91.25 kN
   C. 31.25 kN
   D. 180 kN
   Answer: B. 80(0.75) + 20(1.5625) = 60 + 31.25 = 91.25 kN. (Key point: R_A value)

3. Section C is 3 m from A on an 8 m span. What is the peak of the M_C ILD?
   A. 3 m
   B. 1.875 m
   C. 2.67 m
   D. 1.25 m
   Answer: B. ab/L = 3 x 5 / 8 = 1.875 m. (Key point: Peak of M_C ILD)

4. What is M_C and how can it be checked?
   A. 193.75 kN m; R_A x 3 - 80 x 1
   B. 273.75 kN m; R_A x 3
   C. 100 kN m; 80 x 1.25
   D. 93.75 kN m; 20 x 4.6875
   Answer: A. 80(1.25) + 20(4.6875) = 193.75, matching 91.25 x 3 - 80 x 1 = 193.75 kN m. (Key point: M_C result)

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