---
title: "Flexural rigidity; T-beam bending stresses (60 kN loads)"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:flexural-rigidity-t-beam-60kn"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/flexural-rigidity-t-beam-60kn
site: BCT Engineering
---

# Flexural rigidity; T-beam bending stresses (60 kN loads)

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2081 Bhadra, 2+6 marks
- Video: [Mechanics of Materials: Lesson 31 - How to Use the Flexure Formula](https://www.youtube.com/watch?v=9Y448EOR2vk) — Jeff Hanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/flexural-rigidity-t-beam-60kn
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

Explain flexural rigidity and its importance. A beam is loaded as shown in figure. Determine maximum tensile and compressive bending stress developed in the beam. [Figure: simply supported beam A-B-C-D with a hinge at A and a roller at D; two 60 kN downward loads at B and C; A to B 1 m, B to C 1.5 m, C to D 1 m. Cross-section: inverted T-shape with a top flange 7.5 cm + 7.5 cm either side of the web (web width 5 cm), web height 15 cm and flange thickness 5 cm as marked.]

## Short answer

Flexural rigidity EI resists bending. M_max=60 kN·m. Section centroid 13.21 cm from bottom, I=5900.3 cm⁴. Maximum tensile stress (bottom) 134.4 MPa, compressive stress (top) 69.0 MPa.

## Answer

#### Flexural rigidity
**Flexural rigidity** is the product $EI$ of Young's modulus and the moment of inertia of the section about the neutral axis. It measures the resistance of a beam to bending: from $\dfrac{M}{EI} = \dfrac{1}{R} = \dfrac{d^2y}{dx^2}$, the curvature and deflection of a beam are inversely proportional to $EI$.

**Importance:** a larger $EI$ gives smaller deflection and curvature for the same loading. It is used to check deflection limits, to compare sections (deep sections have large $I$), and it appears in the deflection and column formulas ($\delta \propto \dfrac{WL^3}{EI}$, $P_{cr} = \dfrac{\pi^2EI}{L_e^2}$).

#### Bending stresses in the beam
**Reading of the figure:** simply supported beam with span $AD = 1 + 1.5 + 1 = 3.5$ m and two 60 kN loads at B (1 m from A) and C (1 m from D). The T-section has a flange $7.5 + 5 + 7.5 = 20$ cm wide and 5 cm thick, and a web 5 cm wide and 15 cm deep (total depth 20 cm). The flange is taken at the top (if the flange is at the bottom, the tensile and compressive values below are interchanged).

**Reactions and maximum moment:** by symmetry $R_A = R_D = 60$ kN. The maximum moment is constant between B and C:

$$
M_{max} = 60\times1 = 60\ \text{kN·m} = 60\times10^6\ \text{N·mm}
$$

**Centroid** (from the bottom of the web):

$$
\bar y = \frac{(20\times5)(17.5) + (5\times15)(7.5)}{100 + 75} = 13.214\ \text{cm}
$$

**Moment of inertia about the neutral axis:**

$$
I = \left[\frac{20(5)^3}{12} + 100(17.5-13.214)^2\right] + \left[\frac{5(15)^3}{12} + 75(13.214-7.5)^2\right] = 5900.30\ \text{cm}^4
$$

**Stresses** ($\sigma = My/I$, $M = 60\times10^5$ N·cm):

$$
\sigma_{tension}\ (\text{bottom},\ y = 13.214\ \text{cm}) = \frac{60\times10^5 \times 13.214}{5900.30} = 13437.6\ \text{N/cm}^2 = 134.38\ \text{N/mm}^2
$$

$$
\sigma_{compression}\ (\text{top},\ y = 6.786\ \text{cm}) = \frac{60\times10^5 \times 6.786}{5900.30} = 6900.4\ \text{N/cm}^2 = 69.00\ \text{N/mm}^2
$$

**Answer:** maximum tensile stress $= 134.38$ N/mm² (bottom of the web); maximum compressive stress $= 69.00$ N/mm² (top of the flange).

## Check yourself

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

1. What is flexural rigidity?
   A. Section modulus Z
   B. E divided by I
   C. E times I of the section about the neutral axis
   D. M divided by I
   Answer: C. Flexural rigidity is EI, a measure of resistance to bending. (Key point: Flexural rigidity)

2. What does a larger EI do for the same loading?
   A. Reduces curvature and deflection
   B. Increases bending moment
   C. Increases deflection
   D. Has no effect
   Answer: A. Curvature 1/R = M/EI, so larger EI means less curvature and deflection. (Key point: Meaning of large EI)

3. What is the maximum bending moment between loads B and C (R_A = 60 kN, 1 m from A)?
   A. 90 kN·m
   B. 60 kN·m
   C. 30 kN·m
   D. 120 kN·m
   Answer: B. M = 60 x 1 = 60 kN·m, constant between B and C. (Key point: Maximum moment)

4. With the flange on top, why is the tensile stress larger than the compressive stress?
   A. Concrete is weaker in tension
   B. Load acts at the bottom
   C. Tension fibre is farther from the NA (13.21 cm vs 6.79 cm)
   D. Flange is thicker
   Answer: C. sigma = My/I, so the larger y at the bottom gives 134.4 vs 69.0 N/mm^2. (Key point: Tension vs compression)

## Related in Strength of Materials

- [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7
- [T-beam: maximum tensile and compressive stress](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-udl-40kn-30cm-tension) — Strength of Materials, chapter 7
- [Simply supported I-beam: stress at flange-web junction](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ai-beam-flange-web-junction-stress) — Strength of Materials, chapter 7
- [T-beam with overhang: max bending stress](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-4kn-m-10kn-overhang) — Strength of Materials, chapter 7
- [T-beam: bending stress 15 cm above base](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-udl-5m-bending-stress) — Strength of Materials, chapter 7
- [I-section: bending stress 80 mm above base](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ai-section-bending-stress-80mm-above) — Strength of Materials, chapter 7

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