---
title: "Slab base for ISMB 350, 850 kN service"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:slab-base-ismb350-850kn"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/slab-base-ismb350-850kn
site: BCT Engineering
---

# Slab base for ISMB 350, 850 kN service

Design of Steel and Timber Structure · Chapter 6: Compression Members

- Asked in: 2072 Magh, 8 marks
- Video: [Column Base Slab](https://www.youtube.com/watch?v=2rImEAL4ejo) — Dr. Amar Chougule
- 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/slab-base-ismb350-850kn
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a slab base for a column ISMB 350 @ 52.4 kg/m to carry a service load of 850 KN. Assume Fe410 grade steel and M25 concrete.

## Short answer

Factored load1275 kN. Bearing strength11.25 MPa. Base plate460×250 mm, pressure11.09 MPa. Thickness18 mm.4 M20 anchor bolts.,6 mm weld.

## Answer

**Data:** ISMB 350 @ 52.4 kg/m: $h=350$, $b_f=140$, $t_f=14.2$, $t_w=8.1$ mm. Service load $850$ kN → factored load $P=1.5\times850=1275$ kN. Steel Fe410 ($f_y=250$); concrete M25.

#### 1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)
$0.45f_{ck}=0.45\times25=11.25$ N/mm²

#### 2. Plate size
$$
A_{req}=\frac{P}{0.45f_{ck}}=\frac{1275\times10^3}{11.25}=113333\ \text{mm}^2
$$
Projection of about 55 mm all round the column (not less than 50 mm): $L=350+2\times55=460$ mm, $B=140+2\times55=250$ mm. Area provided $=115000$ mm² $>A_{req}$ ✓
$$
w=\frac{P}{BL}=\frac{1275\times10^3}{460\times250}=11.09\ \text{N/mm}^2<11.25\ \text{N/mm}^2
$$

#### 3. Thickness (cl. 7.4.3.1)
Projections $a=55$ mm and $b=55$ mm (larger and smaller):
$$
t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times11.09\times(55^2-0.3\times55^2)\times1.10}{250}}=16.07\ \text{mm}
$$
It must also be not less than the flange thickness $t_f=14.2$ mm. **Provide $t_s=18$ mm.**

#### 4. Connections
- The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (load is transferred mainly by contact).
- Anchor bolts: 4 nos. M20, embedded about 450 mm with a washer plate (for erection and to resist any uplift).

**Answer: base plate 460 × 250 × 18 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M25 concrete.**

## Check yourself

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

1. The question gives a service load of 850 kN. What load is used for the base plate design?
   A. 850 kN
   B. 1.1 x 850 = 935 kN
   C. 1.5 x 850 = 1275 kN
   D. 850/1.5 = 567 kN
   Answer: C. Limit state design uses factored load: 1.5 x 850 = 1275 kN. (Key point: Factored load from service load)

2. With P = 1275 kN and a 460 x 250 mm plate, what is the bearing pressure and is it safe for M25 concrete?
   A. 11.09 N/mm^2, safe because it is below 11.25 N/mm^2
   B. 11.09 N/mm^2, unsafe because it is above 9 N/mm^2
   C. 5.5 N/mm^2, safe because it is below 25 N/mm^2
   D. 11.25 N/mm^2, unsafe because it equals the limit
   Answer: A. w = 1275x10^3/(460x250) = 11.09 N/mm^2, below 0.45 x 25 = 11.25 N/mm^2. (Key point: Plate size check)

3. The formula gives about 16.07 mm and the flange is 14.2 mm thick. What is the rule behind the 18 mm plate?
   A. Plate thickness is calculated and rounded up to a practical size that is not less than t_f
   B. Plate thickness must equal the web thickness
   C. Plate thickness is 1.5 times the flange thickness
   D. The smaller value of the two is adopted
   Answer: A. The thickness must satisfy both the formula and t_s >= t_f; the larger requirement, 16.07 mm, is rounded up to 18 mm. (Key point: Thickness governing check)

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