---
title: "Slab base for ISSC 200, 1200 kN"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:slab-base-isc200"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/slab-base-isc200
site: BCT Engineering
---

# Slab base for ISSC 200, 1200 kN

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

- Asked in: 2081 Chaitra, 8 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/slab-base-isc200
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a slab base for ISSC 200 which has to carry a design axial load of 1200 kN resting on a concrete pedestal with M 20 grade of concrete.

## Short answer

Bearing strength9 MPa. Plate370×370 mm. Thickness from bending22.08 mm, use25 mm.4 M20 anchor bolts.,6 mm weld.

## Answer

**Assumption:** the section "ISSC 200" (SC 200) is taken with overall size $h=200$ mm, $b_f=200$ mm and flange thickness $t_f\approx15$ mm (only these three values are needed for the base plate). Factored axial load $P=1200$ kN, pedestal M20, E250 steel.

#### 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 85 mm all round the column (not less than 50 mm): $L=200+2\times85=370$ mm, $B=200+2\times85=370$ mm. Area provided $=136900$ mm² $>A_{req}$ ✓
$$
w=\frac{P}{BL}=\frac{1200\times10^3}{370\times370}=8.77\ \text{N/mm}^2<9.00\ \text{N/mm}^2
$$

#### 3. Thickness (cl. 7.4.3.1)
Projections $a=85$ mm and $b=85$ mm (larger and smaller):
$$
t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times8.77\times(85^2-0.3\times85^2)\times1.10}{250}}=22.08\ \text{mm}
$$
It must also be not less than the flange thickness $t_f=15$ mm. **Provide $t_s=25$ 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 370 × 370 × 25 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. Why were 85 mm projections used instead of 50 mm?
   A. To reduce thickness
   B. To meet anchor bolt rules
   C. Because fy is low
   D. For 50 mm the plate area 300 x 300 = 90000 mm2 is below A_req 133333 mm2
   Answer: D. The small column needs a larger plate; 370 x 370 = 136900 mm2 exceeds the 133333 mm2 required. (Key point: Projection choice)

2. w = 1200000/(370 x 370) = 8.77 N/mm2. Is this acceptable?
   A. No, above 9.0
   B. No, above 0.67 fck
   C. Yes, below 9.0 N/mm2
   D. Pressure is not checked
   Answer: C. Pressure must not exceed 0.45 fck = 9.0 N/mm2. (Key point: Pressure check)

3. Calculated ts = 22.08 mm. What is provided?
   A. 22 mm
   B. 25 mm
   C. 16 mm
   D. 15 mm
   Answer: B. Round up to the next standard plate thickness, 25 mm. (Key point: Thickness)

4. Larger projections a, b lead to
   A. thicker plate because ts depends on a^2 and b^2
   B. thinner plate
   C. no change
   D. less concrete pressure only
   Answer: A. ts is proportional to sqrt(a^2 - 0.3 b^2), so projection increases thickness. (Key point: Effect of larger projection)

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