Course on Design of Steel Structures

Professor Damodar Maity

Department of Civil Engineering

Indian Institute of Technology Kharagpur

Lecture 63

Module 12

Design of Gusset Base

In this lecture I will be discussing about a design example of the gusset base. So in last

lecture we have discussed the design procedure of the gusset base and which includes the

determination of the gusset plate dimensions and the base plate dimensions and the number of

bolts. So how to find out step by step that will be clear if we go through this example.

(Refer Slide Time: 0:53)

Example. A column section ISHB 350 @ 710.2 N/m carries a factor axial compressive load

of 1700 kN and factored bending moment of 85 kN-m. Design the base plate and its

connections. Assume concrete pedestal of M-20 grade.

(Refer Slide Time: 2:30)

Solution:

Assume Fe 410 grade of steel: f u=410 MPa, f y=250 MPa

For M20 grade of concrete:

Bearing strength of concrete ¿0.45 f ck=0.45×20=9 N/mm2

Partial safety factor: (Table 5, IS 800: 2007)

γm0=1.1 γmb=1.25

Properties of ISHB 350 @ 710.2 N/m: [table 1, SP-6(1)-1964]

t f=11.6 mm tw=10.1 mm

D=350 mm b f=250 mm

A=9221 mm2

Design compressive load, P=1700 kN

Design bending moment, M=85 kN-m

Eccentricity, e=MP

=85×106

1700×103=50 mm

(Refer Slide Time: 4:42)

Let us provide 16 mm thick gusset plate, one on each side of the column flanges and two

cleat angles ISA 200×150×15 mm.

Provide a length of base plate, L=550 mm.

Now,eL=

50550

=111

<16

Hence, the base plate is in compression throughout.

(Refer Slide Time: 6:42)

To limit the bearing pressure from concrete to 0.45 f ck , the width required,

B=2P

L×0.45 f ck

=2×1700×103

550×0.45×20=686.87 mm ≃690 mm

(Refer Slide Time: 9:10)

Projection of base plate beyond leg angle toe

¿690−(350+2×16+2×150 )

2=4 mm

Let us provide a base plate 690×550 mm in size.

Area provided, A=690×550=379.5×103 mm2

Section modulus of the base plate,

Ze=550×6902

6=43.64×106 mm3

(Refer Slide Time: 11:48)

Maximum pressure,

f max=PA

+MZe

=1700×103

379.5×103+85×106

43.64×106

¿6.43 N/mm2 < 9 N/mm2

(Refer Slide Time: 13:13)

Minimum pressure,

350

551

690

2.536.43

x

x

5.64

 

f min=PA

−MZe

=1700×103

379.5×103−85×106

43.64 ×106

¿2.53 N/mm2 < 9 N/mm2

Calculation for the thickness of base plate:

Base pressure at section x-x,

¿2.53+(6.43−2.53 )×551690

=5.64 N/mm2

Moment at critical section x-x,

¿5.64×139×1392

+12×139× (6.43−5.64 )×( 23 ×139)

¿59573.1 N-mm

(Refer Slide Time: 18:48)

Moment capacity of base plate,

M d=1.2f y

γm0

Ze

¿1.2×2501.1

×( 16 ×1×t a2)=45.45 t a

2 N-mm

⇒ 45.45 t a2=59573.1

⇒ t a=36.2 mm (aggregate thickness of base plate and cleat angle)

Therefore, thickness of base plate,

t b=36.2−15=21.2 mm ≃22 mm > 11.6 mm

Provide a base plate 690×550×22 mm in size.

Bolted connection:

Provide 24-mm diameter bolts of grade 4.6.

The bolts will be in single shear.

Strength bolt in single shear (cl. 10.3.3, IS 800:2007)

¿

Anb( f ub

√3 )γmb

=(0.78×

π ×242

4 )×( 400√3 )1.25

×10−3= 65.2 kN

(Refer Slide Time: 23:09)

Strength of bolt in bearing ¿2.5kbdt f u /γmb (cl. 10.3.4, IS 800:2007)

For 24 mm diameter bolts the minimum edge distance,

e=1.5×d0=1.5× (24+2 )=3 9 mm

The minimum pitch, p=2.5×24=60 mm

Let us provide an edge distance of 40 mm and pitch of 65 mm.

kb is smaller of

( e3d0

=40

3×26=0.51) ,( p

3d0

−0.25=65

3×26−0.25=0.58) ,

( f ub

f u

=400410

=0.98) and 1.0

Hence kb=0.51

∴ Strength in bearing ¿2.5×0.51×24×11.6×4101.25

×10−3

¿116.43 kN

Hence, the strength of bolt ¿65.2 kN

Assuming column end and gusset material to have complete bearing, 50% of the load will be

assumed to pass directly and 50% of the load will pass through the connections.

Number of bolts required to connect column flanges with gusset plates,

n1=0.5×1700

65.2=13.03≃16

Provide 8, 24 mm diameter bolts on each flange in two rows as shown in the figure.

The number of bolts required to connect the cleat angle with gusset plate will be the same.

(Refer Slide Time: 27:29)

Dimension of gusset plate:

Height of gusset plate ¿200+2×40+65=345 mm

Length of gusset plate ¿ length of base plate ¿550 mm

Provide gusset plate 550×345×16 mm in size.

(Refer Slide Time: 29:36)

250

550

Gusset plate550×345×16 mm

Cleat angle200×150×15 mm

Fig. Gusseted base with bolted connections

So this is how one can design a gusset plate when the column is under compression and

moment, okay. So this is all about the todays lecture and before going to finish this entire

lecture I would like to suggest one thing that please go through the IS code 800-2007 IS: 800-

2007, this is not a mathematical course or some structural analysis course that everything you

will be understanding through logic, here logic is definitely there but numerical equations are

there expressions are there which is derived from some experimental observations like some

coefficients are obtained from experimental observations, right.

So and those are given in tabulated form which will be to remember few of them or we have

to go through that code and then we can find out. so when you will be going through my

lecture I would suggest you take a code and try to follow that code with this lecture then it

will be easier and you do not have to remember all the expressions many big expressions are

given where many numerical expressions means are given which is difficult to remember its

coefficients are difficult to remember so those things you can follow from the code, okay.

So whenever you will go through the lecture please sit with the code as well and revise the

things along with code and try to find out in which page or which clause these codal

provisions are there so that you can remember, right. And through this lecture you will not

understand everything unless you go through some workout example, okay and if you go

through workout example then it will be clear and when you will be going through workout

example please refer the code so that how the coefficients are obtaining, how the equations

we are going to get that you can find out and accordingly you can design the thing.

So through this lecture means entire lecture means I hope that you will get a concept of

design of limit state method for steel structure. So how to design a steel member using limit

state method that concepts I hope it will be clear however it will be much clear if you work

out some example then the concept will be much clear, I thank for your patients and I hope

you will enjoy this lecture, thank you.