---
title: "Overhang beam: max BM and SF at D, either direction"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:overhang-max-bm-sf-d-either-direction"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/overhang-max-bm-sf-d-either-direction
site: BCT Engineering
---

# Overhang beam: max BM and SF at D, either direction

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

- Asked in: 2069 Poush, 8 marks
- Video: [Overhanging Beam: Shear Force and Bending Moment Diagram [SFD BMD Problem 3] By Shubham Kola](https://www.youtube.com/watch?v=uGqQADFkWfo) — Shubham Kola
- 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/overhang-max-bm-sf-d-either-direction
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

For the overhang beam determine the maximum positive and negative bending moment and shear force at D due to the three concentrated loads as shown in figure which moves in either directions. [Figure: overhanging beam C-A-D-B-E; CA = 2 m, AD = 3 m, DB = 5 m, BE = 2 m; supports at A and B; loads 20 kN, 25 kN, 30 kN with spacings 2 m and 3 m.]

## Short answer

Max positive BM at D = 81.875 kN m (25 kN over D), max negative = −37.50 kN m (30 kN at C). Max positive shear = 25.00 kN (20 kN leading, 30 kN just right of D), max negative shear = −11.25 kN (30 kN just left of D).

## Answer

**Data:** same overhanging beam $C$–$A$–$D$–$B$–$E$ ($CA = 2$, $AD = 3$, $DB = 5$, $BE = 2$ m; supports $A$ and $B$). Loads 20, 25 and 30 kN with spacings 2 m (20–25) and 3 m (25–30). Because the train may move in either direction, both arrangements are checked: 30 kN leading (**case I**) and 20 kN leading (**case II**). $x$ is measured from $C$.

#### ILDs for D
- $M_D$ ordinates (m): $C$: −1.25; $A$: 0; $D$: +1.875; $B$: 0; $E$: −0.75.
- $V_D$ (shear just at $D$): for a load between $C$ and $D$, $V_D = -R_B = -\dfrac{x-2}{8}$; for a load between $D$ and $E$, $V_D = R_A = \dfrac{10-x}{8}$.

| Position | $C$ (0) | $A$ (2) | $D^-$ (5) | $D^+$ (5) | $B$ (10) | $E$ (12) |
|---|---|---|---|---|---|---|
| $V_D$ ordinate | +0.25 | 0 | −0.375 | +0.625 | 0 | −0.25 |

The ILD of shear has a jump of 1 at $D$.

#### Bending moment at D
**Case I (30 kN leads):** maximum $+$: 30 kN at $x=8$, 25 kN over $D$ ($x=5$), 20 kN at $x=3$:

$$
M^{+} = 30(0.75) + 25(1.875) + 20(0.625) = 81.875\ \text{kN m}
$$

Maximum $-$: 30 kN at $C$: $M^{-} = 30(-1.25) = -37.50$ kN m.

**Case II (20 kN leads):** the best positive value is only 75.00 kN m (20 kN at $B$, 25 kN at $x = 8$, 30 kN over $D$) and the negative value -31.25 kN m, both smaller in magnitude.

#### Shear force at D
**Positive shear** (case II, 20 kN leading): 30 kN just to the right of $D$ ($0.625$), 25 kN at $x = 8$ ($0.25$), 20 kN at $B$ (0):

$$
V^{+} = 30(0.625) + 25(0.25) + 20(0) = 25.00\ \text{kN}
$$

**Negative shear** (case I): 30 kN just to the left of $D$ ($-0.375$), 25 kN at $A$ ($0$), the 20 kN not yet on the beam:

$$
V^{-} = 30(-0.375) = -11.25\ \text{kN}
$$

(Case II gives only -10.62 kN for the negative shear.)

**Answer (either direction): $M_D^{+} = 81.875$ kN m; $M_D^{-} = -37.50$ kN m; $V_D^{+} = 25.00$ kN; $V_D^{-} = -11.25$ kN.**

## Check yourself

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

1. Why are two arrangements (case I and case II) checked for the 20-25-30 kN train?
   A. The train may move in either direction, so either 30 kN or 20 kN may lead
   B. The beam has two supports so two cases are needed
   C. Shear and moment need different loads
   D. The overhangs are of different length
   Answer: A. Since the train can travel either way, the 30 kN leading and 20 kN leading arrangements are both tried and the larger effect is taken. (Key point: Both directions of travel)

2. For maximum positive M_D (case I) where is the 25 kN load placed, and what is M+?
   A. At C, giving 81.875 kN m
   B. Over D, giving 81.875 kN m
   C. Over D, giving 75.00 kN m
   D. At B, giving 37.50 kN m
   Answer: B. The largest ordinate (+1.875) is at D, so the 25 kN load goes over D with 30 kN at x = 8 and 20 kN at x = 3: 30(0.75)+25(1.875)+20(0.625) = 81.875 kN m. (Key point: Maximum positive moment at D)

3. Where does the negative ILD for M_D lie, and what is the maximum negative moment?
   A. Between A and B; -81.875 kN m
   B. Only at D; -37.50 kN m
   C. On the overhangs; 30 kN at C gives -37.50 kN m
   D. On the overhangs; 20 kN at E gives -31.25 kN m as the governing value
   Answer: C. Negative ordinates occur on the overhangs; the largest, -1.25 at C, with the 30 kN load there gives 30(-1.25) = -37.50 kN m. (Key point: Maximum negative moment at D)

4. What is special about the influence line for shear at D?
   A. It is a single triangle with peak at D
   B. It is zero for loads on the overhangs
   C. It has a jump of 1 at D, from -0.375 to +0.625
   D. It has equal ordinates on both sides of D
   Answer: C. Shear ILD is discontinuous at the section: ordinates -0.375 just left and +0.625 just right of D, a jump of 1. (Key point: Shear ILD at D)

5. Which results are obtained for maximum positive and negative shear at D?
   A. V+ = 25.00 kN (case II), V- = -11.25 kN (case I)
   B. V+ = 81.875 kN, V- = -37.50 kN
   C. V+ = 11.25 kN (case I), V- = -25.00 kN (case II)
   D. V+ = 25.00 kN and V- = -10.62 kN both from case I
   Answer: A. Positive shear uses 30 kN just right of D with 20 kN leading (case II): 30(0.625)+25(0.25) = 25 kN; negative shear uses 30 kN just left of D (case I): -11.25 kN. (Key point: Maximum shear at D)

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