---
title: "ILD for BM at F and stress in support BD"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:ild-moment-f-stress-bd"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ild-moment-f-stress-bd
site: BCT Engineering
---

# ILD for BM at F and stress in support BD

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

- Asked in: 2065 Chaitra, 8 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-moment-f-stress-bd
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw influence line diagram for bending moment at F (5 m right of A) and for the stress in the support BD of the structure shown in fig-6. [Figure: beam ABC, AB = BC = 20 m, with A and C supported at the ends; B is supported by inclined members BD and BE, D and E being 12 m on either side horizontally and 9 m below B.]

## Short answer

The beam has a hinge at B. The ILD for M_F is a triangle with peak 3.75 m at F, zero outside AB. The force in BD equals 0.833 times the vertical reaction at B, giving a triangular ILD with peak 0.833 (compression) at B.

## Answer

**Data and assumptions:** beam $ABC$ with $AB = BC = 20$ m, supported at the ends $A$ and $C$ and at $B$ by two inclined members $BD$ and $BE$ ($D$ and $E$ are 12 m horizontally on either side of $B$ and 9 m below it). For the structure to be statically determinate an internal hinge at $B$ is assumed, with vertical supports at $A$ and $C$. Then $AB$ and $BC$ act as two simply supported beams that rest on the hinge $B$, which is held by the pair of inclined struts.

Member $BD$: horizontal 12 m, vertical 9 m, length $15$ m, $\sin\theta = 0.6$, $\cos\theta = 0.8$.

#### ILD for bending moment at F (5 m right of A)
Unit load at $x$ from $A$. For a load on $AB$ (which is simply supported over 20 m), $F$ is at $a = 5$ m and $b = 15$ m.

- $0\le x\le 5$: $M_F = R_B'\times 15 = 0.75x$ (where $R_B' = x/20$)
- $5\le x\le 20$: $M_F = R_A\times 5 = 0.25\,(20-x)$
- A load on $BC$ ($x>20$) is carried by $BC$, $C$ and the hinge, and does not affect $AB$: $M_F = 0$.

Peak ordinate under $F$: $\dfrac{5\times 15}{20} = 3.75$ m.

```text
 A    F          B (hinge)         C
 o----+----------o-----------------o
 ILD M_F: 0 /\ 3.75 \_ 0 ________ 0
```

#### ILD for the force in the strut BD
A unit load on either span sends a vertical force $V_B$ to the hinge $B$: on $AB$, $V_B = \dfrac{x}{20}$; on $BC$, $V_B = \dfrac{40-x}{20}$. The ordinate of $V_B$ is 1 at $B$ and 0 at $A$ and $C$.

At the hinge $B$ no horizontal load comes from the beams, so horizontal equilibrium gives $F_{BD}\cos\theta = F_{BE}\cos\theta$, i.e. $F_{BD} = F_{BE}$. Vertical equilibrium:

$$
2F_{BD}\sin\theta = V_B \Rightarrow F_{BD} = \frac{V_B}{2\times 0.6} = 0.833\,V_B
$$

Both struts are in compression. The ILD for the force (and for the stress $=F_{BD}/A_{BD}$) is a triangle with the apex $0.833$ under $B$:

| Position | $A$ | $B$ | $C$ |
|---|---|---|---|
| $F_{BD}$ per unit load | 0 | 0.833 (C) | 0 |

```text
 ILD F_BD:   0 /\ 0.833 (at B) \ 0
```

Force in $BD$ for loads: $F_{BD} = 0.833\sum P\,y_B$, with $y_B$ the ordinate of the triangle ($y = x/20$ on $AB$).

## Check yourself

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

1. What assumption makes the structure statically determinate?
   A. A fixed support at A
   B. An internal hinge at B, with vertical supports at A and C
   C. Neglecting the struts
   D. A roller at B
   Answer: B. With a hinge at B, AB and BC act as simple spans resting on the hinge, which is held by the inclined struts. (Key point: Statical determinacy)

2. A unit load on span BC gives what value of M_F?
   A. Positive, equal to ab/L
   B. Negative
   C. Zero, because it is carried by BC, C and the hinge and does not affect AB
   D. 0.25 (20 - x)
   Answer: C. BC is a separate simple span; it sends load only to C and the hinge B, not into AB. (Key point: Load on BC and M_F)

3. What is the peak ordinate of the M_F influence line (F is 5 m from A on a 20 m span)?
   A. 5 m
   B. 3.75 m
   C. 15 m
   D. 7.5 m
   Answer: B. ab/L = 5 x 15/20 = 3.75 m. (Key point: Peak of M_F ILD)

4. How is F_BD obtained from the vertical force V_B at the hinge?
   A. F_BD = V_B
   B. F_BD = V_B/(2 sin(theta)) = 0.833 V_B, since BD = BE and sin(theta) = 0.6
   C. F_BD = 2 V_B
   D. F_BD = V_B/0.8
   Answer: B. Horizontal equilibrium gives F_BD = F_BE; vertically 2 F_BD sin(theta) = V_B, with sin(theta) = 9/15 = 0.6. (Key point: Strut force)

## 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
- [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
- [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 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
- [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×
- [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×

## 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 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