---
title: "Slab base for SC 300, 1200 kN"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:column-slab-base-sc300"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/column-slab-base-sc300
site: BCT Engineering
---

# Slab base for SC 300, 1200 kN

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

- Asked in: 2077 Chaitra, 10 marks
- Video: [Slab Base Designing | Problem 2 | Column Bases | Design Of Steel Structures](https://www.youtube.com/watch?v=rWKCG-ecYOM) — Jawad Khan
- 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/column-slab-base-sc300
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design slab base for a column SC300 carrying factored axial load of 1200KN if concrete grade used is M20.

## Short answer

Bearing strength9 MPa. Base plate400×400 mm. Thickness from bending12 mm, but flange thickness16 mm, so provide16 mm.4 M20 anchor bolts.,6 mm weld.

## Answer

**Assumption:** the dimensions of "SC 300" are not in the section tables used here, so the column is taken as a stocky column section of overall size $h=300$ mm, $b_f=300$ mm and flange thickness $t_f\approx16$ mm (similar to a heavy column section); only $h$, $b_f$ and $t_f$ matter here. Factored axial load $P=1200$ kN, pedestal M20, steel E250 ($f_y=250$).

#### 1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)
$0.45f_{ck}=0.45\times20=9.00$ N/mm²

#### 2. Plate size
$$
A_{req}=\frac{P}{0.45f_{ck}}=\frac{1200\times10^3}{9.00}=133333\ \text{mm}^2
$$
Projection of about 50 mm all round the column (not less than 50 mm): $L=300+2\times50=400$ mm, $B=300+2\times50=400$ mm. Area provided $=160000$ mm² $>A_{req}$ ✓
$$
w=\frac{P}{BL}=\frac{1200\times10^3}{400\times400}=7.50\ \text{N/mm}^2<9.00\ \text{N/mm}^2
$$

#### 3. Thickness (cl. 7.4.3.1)
Projections $a=50$ mm and $b=50$ mm (larger and smaller):
$$
t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times7.50\times(50^2-0.3\times50^2)\times1.10}{250}}=12.02\ \text{mm}
$$
It must also be not less than the flange thickness $t_f=16$ mm. **Provide $t_s=16$ 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 400 × 400 × 16 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M20 concrete.**

## Check yourself

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

1. A_req for 1200 kN at 9.0 N/mm2 is
   A. 111111 mm2
   B. 160000 mm2
   C. 133333 mm2
   D. 108000 mm2
   Answer: C. 1200 x 10^3 / 9 = 133333 mm2. (Key point: Required area)

2. For 300 x 300 column with 50 mm projections, plate is
   A. 300 x 300 mm
   B. 350 x 350 mm
   C. 500 x 500 mm
   D. 400 x 400 mm
   Answer: D. 300 + 2 x 50 = 400 on each side. (Key point: Plate size)

3. Calculated ts = 12.02 mm but tf = 16 mm. What is provided?
   A. 12 mm
   B. 10 mm
   C. 16 mm
   D. 25 mm
   Answer: C. Plate thickness must be at least the column flange thickness. (Key point: Thickness requirement)

4. How is the column connected to the base plate?
   A. Nominal fillet weld, with load transferred by contact from machined end
   B. Full penetration weld designed for 1200 kN
   C. Bolts only
   D. No connection
   Answer: A. Load passes mostly by bearing; weld and 4 anchor bolts are for fixing. (Key point: Connection)

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