---
title: "Moment area: girder with 120 kN and 80 kN"
subject: "Theory of Structures I"
chapter: 4
key: "semester_4/theory_of_structures_i:moment-area-14m-girder-two-loads"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-14m-girder-two-loads
site: BCT Engineering
---

# Moment area: girder with 120 kN and 80 kN

Theory of Structures I · Chapter 4: Deflection of Beams

- Asked in: 2069 Poush, 16 marks
- Video: [Moment Area Method Example 1 - Structural Analysis](https://www.youtube.com/watch?v=x8AdLwAGplE) — structurefree
- 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/moment-area-14m-girder-two-loads
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4

## Question

A horizontal girder of steel having uniform section 14 m long is simply supported at its end. It carries concentrated loads of 120 kN and 80 kN at two points 3 m and 4.5 m from the two end supports respectively. Calculate the deflection and slopes of the girder at the point under the loads using moment area method. Take $I = 16\times10^8\ \text{mm}^4$ and $E = 210\ \text{kN/mm}^2$. Verify the results using conjugate beam method. Also find magnitude and location of the maximum deflection in the beam.

## Short answer

Moment area results: at 120 kN load, slope 0.00414 rad clockwise, deflection 15.64 mm down; at 80 kN load, slope 0.00282 rad anticlockwise, deflection 19.93 mm down. Maximum deflection 23.65 mm down at 6.87 m from left support.

## Answer

#### Given data
Simply supported girder, span 14 m. 120 kN at 3 m from the left support A; 80 kN at 4.5 m from the right support B (i.e. 9.5 m from A). $I = 16\times10^8$ mm$^4 = 1.6\times10^{-3}$ m$^4$, $E = 210$ kN/mm$^2 = 2.1\times10^8$ kN/m$^2$, so

$$
EI = 2.1\times10^8\times1.6\times10^{-3} = 3.36\times10^5\ \text{kNm}^2
$$

#### Part 1: Moment-area method

#### Step 1: Reactions
$R_{A} = 120.00$ kN, $R_{B} = 80.00$ kN (positive = upward; check: $\sum F_y = 200.00$ kN equals the total load 200.00 kN).

#### Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces
| Portion (m) | $M$ (kNm) | stiffness | Area of $M/EI$ | $\bar x$ from A (m) |
|---|---|---|---|---|
| 0 to 3 | $120 x$ | EI | 540.00 | 2.000 |
| 3 to 9.5 | $360$ | EI | 2340.00 | 6.250 |
| 9.5 to 14 | $1120 - 80 x$ | EI | 810.00 | 11.000 |

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

#### Step 3: Tangent at A
Deviation of the support at $x = 14$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$
t = (540.00)(12.000) + (2340.00)(7.750) + (810.00)(3.000) = \frac{27045.00}{EI}
$$

That support does not deflect, so $\theta_A = -t/14$:

$$
\theta_A = \frac{-1931.786}{EI}\ \text{(clockwise)}
$$

**At the 120 kN load ($x = 3$ m):**

$$
\theta = \theta_A + \sum A = -1931.786 + [(540.00)] = \frac{-1391.786}{EI}\ \text{(clockwise)}
$$

$$
y = \theta_A\,x + t_{P/A} = -5795.357 + [(540.00)(1.000)] = \frac{-5255.357}{EI}\ \text{(downward)}
$$

**At the 80 kN load ($x = 9.5$ m):**

$$
\theta = \theta_A + \sum A = -1931.786 + [(540.00) + (2340.00)] = \frac{948.214}{EI}\ \text{(anticlockwise)}
$$

$$
y = \theta_A\,x + t_{P/A} = -18351.964 + [(540.00)(7.500) + (2340.00)(3.250)] = \frac{-6696.964}{EI}\ \text{(downward)}
$$

**At the point of maximum deflection ($x = 6.866$ m):**

$$
\theta = \theta_A + \sum A = -1931.786 + [(540.00) + (1391.76)] \approx 0\ \text{(horizontal tangent)}
$$

$$
y = \theta_A\,x + t_{P/A} = -13263.641 + [(540.00)(4.866) + (1391.76)(1.933)] = \frac{-7945.729}{EI}\ \text{(downward)}
$$

#### Maximum deflection
Between the loads the bending moment is constant, $M = 360$ kNm, so the slope changes by $360/EI$ per metre. The slope is $-1391.79/EI$ under the 120 kN load ($x = 3$ m), so it is zero at

$$
-1391.79 + 360\,(x - 3) = 0 \;\Rightarrow\; x = 6.866\ \text{m from A}
$$

where the deflection is $y_{max} = \dfrac{7945.7}{EI}$ downward (same formula as above).

#### Numerical values with $EI = 3.36\times10^5$ kNm$^2$
| Point | Slope (rad) | Deflection (mm) |
|---|---|---|
| A | 0.00575 (clockwise) | 0 |
| Under 120 kN | 0.00414 (clockwise) | 15.64 down |
| Under 80 kN | 0.00282 (anticlockwise) | 19.93 down |
| B | 0.00523 (anticlockwise) | 0 |
| Maximum, at 6.866 m from A | 0 | 23.65 down |

#### Part 2: Verification by the conjugate beam method
#### Step 1: Reactions
$R_{A} = 120.00$ kN, $R_{B} = 80.00$ kN (positive = upward; check: $\sum F_y = 200.00$ kN equals the total load 200.00 kN).

#### Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces
| Portion (m) | $M$ (kNm) | stiffness | Area of $M/EI$ | $\bar x$ from A (m) |
|---|---|---|---|---|
| 0 to 3 | $120 x$ | EI | 540.00 | 2.000 |
| 3 to 9.5 | $360$ | EI | 2340.00 | 6.250 |
| 9.5 to 14 | $1120 - 80 x$ | EI | 810.00 | 11.000 |

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

#### Step 3: Conjugate beam
The conjugate of a simply supported beam is a simply supported beam of the same span. It carries the $M/EI$ diagram of Step 2 as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

#### Step 4: Conjugate reaction at A (slope at A)
Taking moments of the conjugate beam about the support at $x = 14$ m (conjugate moment is zero there):

$$
R'_A\times14 = (540.00)(12.000) + (2340.00)(7.750) + (810.00)(3.000) = 27045.00
$$

$$
R'_A = \frac{1931.786}{EI}\ \text{(slope at A, clockwise)}
$$

**At the 120 kN load ($x = 3$ m):**

$$
V' = R'_A - \sum A = 1931.786 - [(540.00)] = \frac{1391.786}{EI}\ \text{(slope, clockwise)}
$$

$$
M' = R'_A\,x - \sum A(x-\bar x) = 5795.357 - [(540.00)(1.000)] = \frac{5255.357}{EI}\ \text{(deflection, downward)}
$$

**At the 80 kN load ($x = 9.5$ m):**

$$
V' = R'_A - \sum A = 1931.786 - [(540.00) + (2340.00)] = \frac{-948.214}{EI}\ \text{(slope, anticlockwise)}
$$

$$
M' = R'_A\,x - \sum A(x-\bar x) = 18351.964 - [(540.00)(7.500) + (2340.00)(3.250)] = \frac{6696.964}{EI}\ \text{(deflection, downward)}
$$

The conjugate beam gives the same slopes and deflections as the moment-area method, so the results are verified.

**Answer:** slope at A $= 0.00575$ rad (clockwise); under the 120 kN load: slope $0.00414$ rad (clockwise), deflection $15.64$ mm downward; under the 80 kN load: slope $0.00282$ rad (anticlockwise), deflection $19.93$ mm downward. Maximum deflection $= 23.65$ mm downward at $6.866$ m from the left support.

## Check yourself

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

1. How is the slope at A found in the moment-area method for this simply supported girder?
   A. Theorem I applied between A and B gives it directly as zero
   B. The deviation of B from the tangent at A is found and theta_A = t/14
   C. It equals the slope at the maximum deflection point
   D. Slope at A is taken as the area of the M/EI diagram
   Answer: B. Since B does not deflect, the deviation of B from the tangent at A (moment of M/EI area about B) divided by 14 m gives theta_A. (Key point: Slope at A from Theorem II)

2. Where does the maximum deflection occur and why?
   A. At midspan, because the beam is symmetric
   B. Under the 80 kN load
   C. At 6.866 m from A, where the slope is zero in the constant-moment region
   D. At the 120 kN load because it is the larger load
   Answer: C. Between the loads M = 360 kNm is constant, so slope changes linearly; -1391.79 + 360(x-3) = 0 gives x = 6.866 m. (Key point: Max deflection location)

3. What are the reactions at the left and right supports?
   A. RA = 80 kN, RB = 120 kN
   B. RA = 120 kN, RB = 80 kN
   C. RA = 100 kN, RB = 100 kN
   D. RA = 140 kN, RB = 60 kN
   Answer: B. The 120 kN load is 3 m from A and the 80 kN load is 4.5 m from B, so RA = 120 kN and RB = 80 kN. (Key point: Reactions)

4. How does the conjugate beam check the moment-area results?
   A. Conjugate shear gives slope and conjugate moment gives deflection, matching the same numbers
   B. It gives different values that are averaged
   C. It only checks the reactions
   D. It gives deflection as conjugate shear
   Answer: A. R'A = 1931.786/EI and the conjugate shear/moment at the loads reproduce 1391.786/EI, 5255.357/EI, 948.214/EI and 6696.964/EI. (Key point: Verification by conjugate beam)

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