---
title: "ILD for SF and BM at C of beam"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:ild-sf-bm-c-40-60-5knm"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ild-sf-bm-c-40-60-5knm
site: BCT Engineering
---

# ILD for SF and BM at C of beam

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

- Asked in: 2080 Chaitra, 6 marks
- Video: [SA TUTORIAL-2 INFLUENCE LINE DIAGRAM OF SHEAR FORCE AND BENDING MOMENT](https://www.youtube.com/watch?v=kd7HRaiBnqk) — Structural Analysis
- 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-sf-bm-c-40-60-5knm
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Determine the shear force and bending moment at C in the given loaded beam using ILD. Take EI constant throughout the beam. [Figure: beam with hinge A at the left and a roller support C; AB = 2 m, BC = 2 m, then D, E, F at 1 m each; 40 kN load at B, 5 kN/m UDL over the portion around C to E, 60 kN load at F at the free end. Details partly unclear.]

## Short answer

Using influence lines, bending moment at C is -190 kN m (hogging). Shear force is -67.5 kN just left of C and +70 kN just right of C.

## Answer

**Assumptions (figure partly unclear):** beam $A$–$B$–$C$ is supported by a hinge at $A$ and a roller at $C$ ($AB = BC = 2$ m, span $AC = 4$ m), with an overhang $C$–$D$–$E$–$F$ of 3 m (1 m each). Loads: 40 kN at $B$; 5 kN/m UDL from $C$ to $E$ (2 m); 60 kN at $F$ (free end). The section is taken at the support $C$.

#### Reactions (for checking)
$\Sigma M_A = 0$: $R_C\times 4 = 40\times 2 + 10\times 5 + 60\times 7 = 550 \Rightarrow R_C = 137.5$ kN; $R_A = 110-137.5 = -27.5$ kN (uplift).

#### ILD for bending moment at C
Load at distance $s$ from $A$. For $s\le 4$, $M_C = 0$ (the load is carried directly by the supports). For $s>4$, $M_C = -(s-4)$ (hogging).

```text
 A o---B-----o C---D---E---F
 ILD(M_C):  0  0   0  -1 -2 -3
```

$$
M_C = 5\times 2\times(-1) + 60\times(-3) = -10 - 180 = -190\ \text{kN m}
$$

#### ILD for shear at C
*Just left of C:* ordinate $=-s/4$ for $0\le s<4$ (so $-0.5$ at $B$ and $-1$ as the load approaches $C$); $-(s-4)/4$ for the overhang ($-0.25$ at $D$, $-0.5$ at $E$, $-0.75$ at $F$).

Using ordinates: UDL area from $C$ to $E$ $=\tfrac12(0+(-0.5))\times 2 = -0.5$, so

$$
V_{C^-} = 40(-0.5) + 5(-0.5) + 60(-0.75) = -20 - 2.5 - 45 = -67.5\ \text{kN}
$$

*Just right of C:* ordinate 0 for loads on the span, and 1 for loads on the overhang:

$$
V_{C^+} = 5\times 2\times 1 + 60\times 1 = 70\ \text{kN}
$$

Check by statics: $V_{C^-} = R_A - 40 = -27.5-40 = -67.5$ kN and $V_{C^+} = -67.5 + 137.5 = +70$ kN.

**Answer: $M_C = -190$ kN m (hogging); $V_C = -67.5$ kN just left and $+70$ kN just right of $C$.**

## Check yourself

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

1. What is the ILD ordinate for M_C when the unit load is on span AC?
   A. 0
   B. s/4
   C. -s
   D. +1
   Answer: A. Loads on the span are carried directly by supports A and C, so no moment at C is produced. (Key point: BM ILD for loads on span)

2. For a load on the overhang at distance s > 4 m from A, M_C =
   A. +(s - 4), sagging
   B. -(s - 4), hogging
   C. (s - 4)/4
   D. 0
   Answer: B. The overhanging load gives a hogging moment equal to its distance from C. (Key point: Overhang)

3. What is M_C for 5 kN/m over CE (2 m) and 60 kN at F?
   A. -180 kN m
   B. -10 kN m
   C. -190 kN m
   D. +190 kN m
   Answer: C. M_C = 5 x 2 x (-1) + 60 x (-3) = -190 kN m. (Key point: Moment value)

4. V just right of C is 70 kN and just left is -67.5 kN. The jump equals
   A. the load of 40 kN
   B. the UDL of 10 kN
   C. zero
   D. the reaction R_C = 137.5 kN
   Answer: D. Shear jumps by the support reaction at C, 137.5 kN. (Key point: Shear check)

## Related in Theory of Structures I

- [Cantilever ILD for SF and BM at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-cantilever-sf-bm-c) — 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
- [ILD at mid-span: BM and SF under given loads](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-mid-span-bm-sf-50-10-30) — Theory of Structures I, chapter 5
- [Suspension 120 m: UDL on left half, SF and BM at 30 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-120m-live-load-left-half) — Theory of Structures I, chapter 7, asked 2×
- [Overhang beam ILD for shear force at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aoverhang-ild-sf-c-2071o) — Theory of Structures I, chapter 5
- [ILD: RA, RB, SF and BM at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-ra-rb-sf-bm-c-15knm-1-5kn) — Theory of Structures I, chapter 5

## Related in other subjects

- [ILD: bending moment at fixed support, 10 m beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-fixed-support-10m-1) — Theory of Structures II, chapter 4
- [ILD: moment at support B, fixed-propped beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-b-propped-2l-l) — 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
- [ILD: moment at middle support, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-mid-support-two-span) — Theory of Structures II, chapter 4