---
title: "Foundation base for ISHB 350 column"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:foundation-base-ishb350-120kn"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/foundation-base-ishb350-120kn
site: BCT Engineering
---

# Foundation base for ISHB 350 column

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

- Asked in: 2070 Magh, 10 marks
- Video: [DESIGN COLUMN BASES / DESIGN BASE PLATE](https://www.youtube.com/watch?v=tTEZpElnIfg) — 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/foundation-base-ishb350-120kn
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design the foundation base for an ISHB 350 column to carry factored load of 120[?] KN. Assume steel and M20 concrete.

## Short answer

Assuming the doubtful load is 1200 kN, provide base plate 450 x 350 x 14 mm with 4 M20 anchor bolts and 6 mm fillet weld. If the load is actually 120 kN, minimum 50 mm projection and flange thickness govern, giving 350 x 450 x 12 mm plate.

## Answer

**Reading of the doubtful value:** the load is printed as "120[?] kN"; it is read as **1200 kN** (factored), consistent with the similar ISHB 350 base-plate questions. If the load is really 120 kN, the minimum 50 mm projection and thickness $\ge t_f$ govern: a plate $350\times450\times12$ mm would do. Data: ISHB 350: $h=350$, $b_f=250$, $t_f=11.6$ mm; steel E250; concrete M20.

#### 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): $L=350+2\times50=450$ mm, $B=250+2\times50=350$ mm. Area provided $=157500$ mm² $>A_{req}$ ✓
$$
w=\frac{P}{BL}=\frac{1200\times10^3}{450\times350}=7.62\ \text{N/mm}^2<9.00\ \text{N/mm}^2
$$

#### 3. Thickness (cl. 7.4.3.1)
Projections $a=50$ mm and $b=50$ mm (larger and smaller):
$$
t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times7.62\times(50^2-0.3\times50^2)\times1.10}{250}}=12.11\ \text{mm}
$$
It must also be not less than the flange thickness $t_f=11.6$ mm. **Provide $t_s=14$ 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 450 × 350 × 14 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. For 1200 kN on a 450 x 350 mm plate with M20 concrete, which result is correct?
   A. w = 7.62 N/mm^2 < 9.00 N/mm^2, safe
   B. w = 7.62 N/mm^2 > 6 N/mm^2, unsafe
   C. w = 9.00 N/mm^2 exactly
   D. w = 15.2 N/mm^2, unsafe
   Answer: A. w = 1200x10^3/(450x350) = 7.62 N/mm^2, which is less than the permissible 0.45 x 20 = 9.00 N/mm^2. (Key point: Plate area and pressure)

2. In t_s = sqrt(2.5 w (a^2 - 0.3 b^2) gamma_m0 / f_y), what are a and b?
   A. a is the larger projection and b the smaller projection beyond the column
   B. a is the plate length and b the plate breadth
   C. a is the flange thickness and b the web thickness
   D. a is the bolt spacing and b the edge distance
   Answer: A. a and b are the larger and smaller projections of the plate beyond the column. Here both equal 50 mm, giving 12.11 mm. (Key point: Thickness formula inputs)

## Related in Design of Steel and Timber Structure

- [Base plate for ISHB 350, 1200 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abase-plate-ishb350-1200kn) — Design of Steel and Timber Structure, chapter 6, asked 2×
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