---
title: "Effect of lateral restraint on moment capacity"
subject: "Design of Steel and Timber Structure"
chapter: 7
key: "semester_6/design_of_steel_and_timber_structure:lateral-restraint-effect-moment-capacity"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/lateral-restraint-effect-moment-capacity
site: BCT Engineering
---

# Effect of lateral restraint on moment capacity

Design of Steel and Timber Structure · Chapter 7: Flexure Members

- Asked in: 2073 Magh, 4 marks
- Video: [Open Beams Have a Serious Weakness](https://www.youtube.com/watch?v=XSl7ZntK94E) — The Engineering Hub
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/lateral-restraint-effect-moment-capacity
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7

## Question

What is the effect of laterally restrained and unrestrained compression flange in bending moment carrying capacity of beam?

## Short answer

Restrained compression flange achieves full plastic capacity M_d = β_b Z_p f_y / γ_m0. Unrestrained length causes lateral-torsional buckling, reducing capacity to M_d = β_b Z_p f_bd, where f_bd = χ_LT f_y/γ_m0 and χ_LT < 1 decreases as span increases.

## Answer

The bending moment capacity of a beam depends on whether its compression flange is held against sideways movement.

#### Laterally restrained compression flange
The flange cannot buckle sideways, so the beam fails only when the section yields. The full plastic (or elastic) section strength is available (IS 800:2007 cl. 8.2.1):
$$
M_d = \frac{\beta_bZ_pf_y}{\gamma_{m0}} \le \frac{1.2Z_ef_y}{\gamma_{m0}}
$$
where $\beta_b = 1$ for plastic and compact sections, and $Z_e/Z_p$ for semi-compact ones.

#### Laterally unrestrained compression flange
The compression flange behaves like a column and tends to buckle sideways with a twist (lateral-torsional buckling, cl. 8.2.2). The capacity is reduced:
$$
M_d = \beta_bZ_pf_{bd},\quad f_{bd} = \chi_{LT}\frac{f_y}{\gamma_{m0}},\quad \chi_{LT} = \frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \lambda_{LT}^2}} \le 1
$$
with $\lambda_{LT} = \sqrt{\beta_bZ_pf_y/M_{cr}}$. The longer the unrestrained length and the smaller $I_y$, $I_t$, $I_w$, the smaller $M_{cr}$ and the lower the capacity.

#### Example: ISMB 400, Fe 410 (computed with the IS 800 formulas)
Plastic modulus $Z_p = 1161$ cm³, restrained capacity $M_d = 264$ kN·m.

| Unrestrained length (m) | $M_{cr}$ (kN·m) | $\chi_{LT}$ | $M_d$ (kN·m) | Fraction of restrained $M_d$ |
|---|---|---|---|---|
| 1 | 2449 | 0.97 | 255 | 97% |
| 2 | 677 | 0.87 | 229 | 87% |
| 3 | 343 | 0.72 | 190 | 72% |
| 4 | 222 | 0.57 | 150 | 57% |
| 6 | 129 | 0.37 | 98 | 37% |
| 8 | 91 | 0.27 | 72 | 27% |

#### Practical effect
- Short unrestrained length (about $\lambda_{LT} \le 0.4$): no loss of capacity.
- A long unbraced length can cut the capacity to less than half; the beam then needs a larger section or lateral bracing.
- Restraint is obtained from a connected slab, cross beams, bracing or end torsional restraint, and a larger $I_y$ (wider flange) improves the unrestrained strength.

## Check yourself

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

1. For a laterally restrained compression flange, the moment capacity is:
   A. Md = betab Zp fy/gamma_m0, not more than 1.2 Ze fy/gamma_m0
   B. Md = Mcr
   C. Md = chi_LT Zp fy/gamma_m0
   D. Md = Ze fy only
   Answer: A. The flange cannot buckle sideways, so the full section strength is available (cl. 8.2.1). (Key point: Restrained flange capacity)

2. What happens when the compression flange is unrestrained?
   A. Capacity is unchanged
   B. Capacity rises because of twisting
   C. Only shear capacity changes
   D. The beam can buckle laterally with twist (LTB), so capacity falls with fbd = chi_LT fy/gamma_m0
   Answer: D. The compression flange behaves like a column and buckles sideways, reducing Md by chi_LT. (Key point: Unrestrained flange)

3. For the ISMB 400 example, how does capacity change as the unrestrained length grows from 1 m to 8 m?
   A. It stays about 97%
   B. It falls from about 97% to about 27% of the restrained capacity
   C. It rises to 127%
   D. It falls only to 87%
   Answer: B. Mcr drops with length, so chi_LT falls (0.97 to 0.27) and the beam may need bracing or a larger section. (Key point: Effect of unrestrained length)

## Related in Design of Steel and Timber Structure

- [Safe UDL on ISLB 600 beam, restrained vs unrestrained](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-moment-islb600) — Design of Steel and Timber Structure, chapter 7
- [Moment capacity of unrestrained ISMB 500](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Amoment-capacity-unrestrained-ismb500) — Design of Steel and Timber Structure, chapter 7
- [UDL capacity of MB 300 with flange plates](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-beam-mb300-plates) — Design of Steel and Timber Structure, chapter 7
- [Beam with flange plates, 30 kN/m + 16 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-additional-plates-30knm-16kn) — Design of Steel and Timber Structure, chapter 7
- [Compression capacity of strut of two L75x75x8](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Astrut-two-l75x75x8) — Design of Steel and Timber Structure, chapter 6
- [Lateral-torsional buckling; flange curtailment](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Altb-curtailment-plate-girder) — Design of Steel and Timber Structure, chapter 7

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