---
title: "Section modulus of fixed beam at collapse"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:fixed-beam-section-modulus-collapse-60kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/fixed-beam-section-modulus-collapse-60kn
site: BCT Engineering
---

# Section modulus of fixed beam at collapse

Theory of Structures II · Chapter 5: Introduction to plastic analysis

- Asked in: 2066 Bhadra, 10 marks
- Video: [Plastic Analysis Problem No - 2 ( Finding Collapse Load in a Non Prismatic Fixed beam )](https://www.youtube.com/watch?v=Ovrp_aoxcnM) — Stan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/fixed-beam-section-modulus-collapse-60kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

A single spanned fixed beam of length 9 m has two concentrated forces applied vertically downwards at 3 m distance from each end. The left and right forces are 60 kN and 120 kN respectively. Calculate the section modulus required to render the system into collapse condition, if the yield stress and load factor for the materials used are 250 MPa and 1.15 respectively.

## Short answer

Factored loads are 69 and 138 kN; governing mechanism has hinges at A, D and B, giving Mp = 172.5 kNm. Required plastic section modulus Zp = Mp/sigma_y = 690 cm3.

## Answer

#### Given data
Fixed beam AB, $L = 9$ m, loads 60 kN at 3 m from A and 120 kN at 3 m from B (6 m from A). $\sigma_y = 250$ MPa, load factor 1.15.

Factored (collapse) loads: $W_1 = 1.15\times60 = 69$ kN at C (3 m from A), $W_2 = 1.15\times120 = 138$ kN at D (6 m from A).

The required plastic section modulus $Z_p$ follows from $M_p = \sigma_yZ_p$ (the section modulus needed for the beam to just collapse under the factored loads).

#### Possible mechanisms (hinges at the fixed ends and under one load)
The beam is indeterminate to the second degree (3 hinges for a mechanism).

**Mechanism 1: hinges at A, C and B** (under the 69 kN load). Part AC (3 m) rotates $\theta_1$, part CB (6 m) rotates $\theta_2$ with $3\theta_1 = 6\theta_2$, i.e. $\theta_1 = 2\theta_2$. The hinge rotations are A: $2\theta_2$, C: $3\theta_2$, B: $\theta_2$; the deflection at C is $6\theta_2$ and at D (3 m from B) it is $3\theta_2$.
- Internal work $= M_p(2 + 3 + 1)\theta_2 = 6M_p\theta_2$
- External work $= 69\times6\theta_2 + 138\times3\theta_2 = 828\theta_2$
$$
M_p = \frac{828}{6} = 138\ \text{kN·m}
$$

**Mechanism 2: hinges at A, D and B** (under the 138 kN load). Part AD (6 m) rotates $\theta_1$, part DB (3 m) rotates $\theta_2$ with $6\theta_1 = 3\theta_2$, i.e. $\theta_2 = 2\theta_1$. Hinge rotations: A $\theta_1$, D $3\theta_1$, B $2\theta_1$; deflection at D $= 6\theta_1$, at C $= 3\theta_1$.
- Internal work $= M_p(1 + 3 + 2)\theta_1 = 6M_p\theta_1$
- External work $= 69\times3\theta_1 + 138\times6\theta_1 = 1035\theta_1$
$$
M_p = \frac{1035}{6} = 172.5\ \text{kN·m}
$$

#### Collapse mechanism and $M_p$
The larger requirement governs (the beam must resist the worst mechanism): $M_p = 172.5$ kN·m, with hinges at A, D and B.

**Check (equilibrium):** free reactions: $R_A = (69\times6 + 138\times3)/9 = 92$ kN, $R_B = 115$ kN. Free moments: $M_C = 92\times3 = 276$, $M_D = 115\times3 = 345$ kN·m. With the end moments $-M_p$ at A and B, $M_D = 345 - M_p = +M_p$ gives $M_p = 172.5$, and $M_C = 276 - 172.5 = 103.5 \le M_p$, so no section exceeds $M_p$.

#### Section modulus
$$
Z_p = \frac{M_p}{\sigma_y} = \frac{172.5\times10^6\ \text{N·mm}}{250\ \text{N/mm}^2} = 6.9\times10^5\ \text{mm}^3 = 690\ \text{cm}^3
$$

**Answer: required (plastic) section modulus $Z_p = 6.9\times10^5$ mm³ = 690 cm³** (corresponding to $M_p = 172.5$ kN·m). Since the shape factor of the chosen section is not given, this is the plastic modulus; if the elastic modulus is wanted, use $Z_e = Z_p/S$ for the section selected.

## Check yourself

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

1. With load factor 1.15, what are the loads used in collapse analysis?
   A. 60 kN and 120 kN
   B. 69 kN and 138 kN
   C. 51.7 kN and 104.3 kN
   D. 75 kN and 150 kN
   Answer: B. Multiply working loads by the load factor: 60x1.15 = 69, 120x1.15 = 138. (Key point: Factored loads)

2. For hinges at A, D and B (under the 138 kN load), what Mp results?
   A. 138 kN·m
   B. 172.5 kN·m
   C. 103.5 kN·m
   D. 345 kN·m
   Answer: B. Internal work 6Mp theta; external 69x3 + 138x6 = 1035 theta; Mp = 172.5 kN·m. (Key point: Mechanism with hinge under 138 kN)

3. Mechanisms give 138 and 172.5 kN·m. What does the beam need?
   A. 138, the smaller
   B. 172.5, the larger, so the beam resists the worst mechanism
   C. 155.25, the average
   D. 310.5, the sum
   Answer: B. The larger Mp governs; it is also confirmed by the equilibrium check, where Mc = 103.5 is less than Mp. (Key point: Which mechanism governs)

4. How is the required section modulus found and what is it?
   A. Zp = Mp/sigma_y = 172.5x10^6/250 = 6.9x10^5 mm3
   B. Zp = Mp x sigma_y = 4.3x10^10 mm3
   C. Zp = Mp/sigma_y = 138x10^6/250 = 5.5x10^5 mm3
   D. Zp = Mp/(1.15 sigma_y) = 6.0x10^5 mm3
   Answer: A. Mp = sigma_y Zp, so Zp = 172.5x10^6 N·mm / 250 N/mm2 = 6.9x10^5 mm3 = 690 cm3. (Key point: Section modulus)

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