---
title: "Base plate for ISHB 450, 1400 kN"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:base-plate-ishb450-1400kn"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/base-plate-ishb450-1400kn
site: BCT Engineering
---

# Base plate for ISHB 450, 1400 kN

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

- Asked in: 2070 Bhadra, 8 marks
- Video: [Design of slab base (base slab)](https://www.youtube.com/watch?v=Q1GjZpCOih4) — Simpleciviltamil
- 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/base-plate-ishb450-1400kn
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design the base plate for the column ISHB 450 to carry a factored load of 1400 kN. Take E250 grade of steel and M20 grade of concrete.

## Short answer

Bearing strength 9 N/mm2; required area 155556 mm2; use 550 x 350 mm plate, pressure 7.27 N/mm2. Thickness 11.83 mm but not less than flange 13.7 mm, so provide 14 mm plate with 4 M20 anchor bolts and 6 mm fillet weld.

## Answer

**Data:** ISHB 450 @ 87.2 kg/m: $h=450$, $b_f=250$, $t_f=13.7$, $t_w=9.8$ mm. Factored load $P=1400$ kN; 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
$$
A_{req}=\frac{P}{0.45f_{ck}}=\frac{1400\times10^3}{9.00}=155556\ \text{mm}^2
$$
Projection of about 50 mm all round the column (not less than 50 mm): $L=450+2\times50=550$ mm, $B=250+2\times50=350$ mm. Area provided $=192500$ mm² $>A_{req}$ ✓
$$
w=\frac{P}{BL}=\frac{1400\times10^3}{550\times350}=7.27\ \text{N/mm}^2<9.00\ \text{N/mm}^2
$$

#### 3. Thickness (cl. 7.4.3.1)
Projections $a=50$ mm and $b=50$ mm (larger and smaller):
$$
t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times7.27\times(50^2-0.3\times50^2)\times1.10}{250}}=11.83\ \text{mm}
$$
It must also be not less than the flange thickness $t_f=13.7$ 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 550 × 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. What is the required area for P = 1400 kN on M20 concrete?
   A. 155556 mm^2
   B. 70000 mm^2
   C. 192500 mm^2
   D. 233333 mm^2
   Answer: A. A_req = 1400x10^3/(0.45x20) = 1400x10^3/9 = 155556 mm^2. The value 192500 mm^2 is the area provided (550 x 350). (Key point: Bearing area)

2. ISHB 450 has h = 450 and b_f = 250. With 50 mm projection all round, what is the plate size?
   A. 550 x 350 mm
   B. 500 x 300 mm
   C. 450 x 250 mm
   D. 650 x 450 mm
   Answer: A. L = 450 + 100 = 550 mm and B = 250 + 100 = 350 mm. (Key point: Plate size from ISHB 450)

3. The computed t_s is 11.83 mm but the flange thickness is 13.7 mm. What thickness is provided and why?
   A. 12 mm, from the formula
   B. 14 mm, because t_s must not be less than t_f
   C. 10 mm, because the plate sits on grout
   D. 20 mm, to carry anchor bolt tension
   Answer: B. The plate must be at least as thick as the column flange (13.7 mm), so 14 mm is adopted. (Key point: Thickness decision)

## Close variants set in other exams

- [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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