---
title: "Moment capacity of unrestrained ISMB 500"
subject: "Design of Steel and Timber Structure"
chapter: 7
key: "semester_6/design_of_steel_and_timber_structure:moment-capacity-unrestrained-ismb500"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/moment-capacity-unrestrained-ismb500
site: BCT Engineering
---

# Moment capacity of unrestrained ISMB 500

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

- Asked in: 2077 Chaitra, 10 marks
- Video: [DESIGN LATERALLY UNSUPPORTED BEAM QUESTION BY AMIT SINGH](https://www.youtube.com/watch?v=xWgvqtBXmYA) — Study with Civil buddy
- 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/moment-capacity-unrestrained-ismb500
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7

## Question

Calculate the moment carrying capacity of laterally unrestrained beam made of ISMB500 and length of member is equal to 6m. Assume necessary data suitably.

## Short answer

Elastic critical moment M_cr = 278.7 kN·m; non-dimensional slenderness λ_LT = 1.355; reduction factor χ_LT = 0.440; design bending strength f_bd = 100.1 N/mm². Moment capacity M_d = 205 kN·m.

## Answer

**Approach.** Moment capacity of a laterally unrestrained beam, IS 800:2007 cl. 8.2.2: $M_d = \beta_bZ_pf_{bd}$, $f_{bd} = \chi_{LT}f_y/\gamma_{m0}$, with $M_{cr}$ from Annex E.

#### Assumed data
- ISMB 500 @ 86.9 kg/m (IS 808): $D = 500$, $b_f = 180$, $t_w = 10.2$, $t_f = 17.2$ mm, $I_{yy} = 1370$ cm⁴, $Z_e = 1810$ cm³.
- Fe 410: $f_y = 250$ N/mm². Simply supported over 6 m, ends restrained against torsion with the flanges free to warp, so $L_{LT} = 1.0L = 6000$ mm (Table 15). No intermediate lateral restraint. Load applied at the shear centre level (normal loading, the usual assumption).

#### Step 1: Section properties
Plastic modulus $Z_p = b_ft_f(D - t_f) + t_w(D - 2t_f)^2/4 = 2048$ cm³ (plastic section, $\beta_b = 1$; $b/t_f = 5.23$, $d/t_w = 42.3$).

$$
\begin{aligned} I_t &= \frac{2\times180\times17.2^3 + (500 - 34.4)\times10.2^3}{3} = 77.5\ \text{cm}^4 \\ I_w &= \frac{I_y(D - t_f)^2}{4} = \frac{1370\times10^4\times482.8^2}{4} = 79.8\times10^{10}\ \text{mm}^6 \end{aligned}
$$

#### Step 2: Elastic critical moment
$$
M_{cr} = \sqrt{\frac{\pi^2EI_y}{L_{LT}^2}\left(GI_t + \frac{\pi^2EI_w}{L_{LT}^2}\right)} = 278.7\ \text{kN·m}
$$
($E = 2\times10^5$, $G = 0.769\times10^5$ N/mm²).

#### Step 3: Reduction factor
Rolled section: $\alpha_{LT} = 0.21$.
$$
\begin{aligned} \lambda_{LT} &= \sqrt{\frac{\beta_bZ_pf_y}{M_{cr}}} = \sqrt{\frac{2047546\times250}{278.7\times10^6}} = 1.355 \\ \phi_{LT} &= 0.5[1 + 0.21(\lambda_{LT} - 0.2) + \lambda_{LT}^2] = 1.540 \\ \chi_{LT} &= \frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \lambda_{LT}^2}} = 0.440 \\ f_{bd} &= \frac{0.440\times250}{1.1} = 100.1\ \text{N/mm}^2 \end{aligned}
$$

#### Step 4: Moment capacity
$$
M_d = \beta_bZ_pf_{bd} = 2047546\times100.1/10^6 = 205.0\ \text{kN·m}
$$
(limit $1.2Z_ef_y/\gamma_{m0} = 494$ kN·m, not governing).

**Answer: moment carrying capacity $M_d \approx 205$ kN·m**, about 44% of the fully restrained value $Z_pf_y/\gamma_{m0} = 465$ kN·m. For a UDL this corresponds to a factored load $w = 8M_d/L^2 = 45.5$ kN/m including self-weight.

## Check yourself

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

1. How is M_d for a laterally unrestrained beam found?
   A. M_d = beta_b Z_p f_bd with f_bd = chi_LT f_y / gamma_m0
   B. M_d = Z_p f_y / gamma_m0
   C. M_d = M_cr
   D. M_d = Z_e f_y
   Answer: A. The bending stress is reduced by chi_LT, which depends on the LTB slenderness derived from M_cr. (Key point: LTB design strength)

2. Why is L_LT = 1.0 L = 6000 mm taken?
   A. Simply supported, ends restrained against torsion with flanges free to warp (Table 15), no intermediate restraint
   B. The beam is fixed at both ends
   C. A cantilever is assumed
   D. The load is on the tension flange
   Answer: A. For simply supported ends with torsional restraint and free warping, the effective length for LTB equals the span. (Key point: Effective length)

3. What expression gives lambda_LT?
   A. sqrt(beta_b Z_p f_y / M_cr)
   B. sqrt(M_cr / (Z_p f_y))
   C. L / r_y
   D. Z_p / Z_e
   Answer: A. The ratio of plastic moment to elastic critical moment, under a root; here 1.355. (Key point: LTB slenderness)

4. What is the outcome compared with full lateral restraint?
   A. About 205 kN·m, roughly 44% of the restrained 465 kN·m
   B. About 465 kN·m, unchanged
   C. About 494 kN·m, set by the 1.2 Z_e limit
   D. About 100 kN·m, which is 20%
   Answer: A. chi_LT = 0.440 gives f_bd = 100.1 N/mm^2 and M_d = 205 kN·m; the 1.2 Z_e limit of 494 kN·m does not govern. (Key point: Result comparison)

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