mod 2

CONNECTIONS IN STEEL STRUCTURES

Bolted connection

1 and 2 are axial connections, 3 and 4 are eccentric connections.

1. Lap joint (1 problem)
2. Double cover butt joint (1 problem)
3. Eccentric connection (Design)
4. Eccentric connection (Analysis)

Welded connection

1. Lap weld connection (1 sum)
2. Butt weld connection (1 sum)
3. Eccentric connection (Design)
4. Eccentric connection (Analysis)

So for this unit its enough that you study 8 problems. If time is constrained study first 6.

Bolted connections

These are the things you should know by-heart so that you can solve the problems during exam with the necessary speed:

1. Diameter of hole for a diameter of bolt
2. Basic stresses for bolts

During exams in the questions we are given the diameter of bolt like 16mm, 20mm, 24mm etc. Know that we drill a hole in the plate to insert the bolt. The hole shouldn't be the same as the diameter of the bold. We should give it extra clearance.

So look at this table in page number 73:

We see that the diameter of the hole should be the diameter of bolt/fastener + clearances.

If the question doesn't mention the diameter of the bolt, just take it as 20mm and just know the clearance will be 2mm and therefore the diameter of the hole should be 22mm.

next

focus on grade 4.6

Given bolt grade = 4.6

4 * 6 = 24

Therefore, fyb (yield stress of the bolt) = 240MPa

Remmeber that for grade 4.6 bolt, the fub (ultimate stress of bolt) is 400Mpa.

For 4.8, the fyb is 320. There was .2 increment from 4.6. Multiply .2 by 100. You get 20. Add it to 400. You get 420. This pattern keeps on for the rest of entries too.

So just remember. For grade 4.6 the fyb is 240 and the fub is 400. For 4.8 its 320 and 400 + .2 * 100 = 420.

for steel

Remember these 4 values:

1. 4.6 grade = 240Mpa fyb, and 400Mpa fub
2. yield stress = fy = 250 MPa
3. ultimate stress = fu = 410 MPa

Anb = Area at root
Asb = Area at shank
Area at root < Area at shank.
Anb = 0.78 * Asb

Take pitch as 60 and edge as 40.

Design procedure for bolted connection

Strength of one bolt = Bolt value

i) Shear capacity of bolt ii) Bearing capacity of bolt

Shearing capacity of bolt

Bearing capcity of bolt

Finding number of bolts

bolt value will be lesser of these two: the shearing capacity and the bearing capacity.

We will find out the strength of one bolt. Based on that we will find out the strength of the connection.

Question will say: "100kN force is acting", and we find the bolt value (the lesser of shearing capacity and bearing capacity) and we divide the load by the bolt value to get the number of bolts

Design details

• Lap joint
• Butt joint

Checking of design

Efficiency of joint

The strength of joint

Note we got the strength of joint here in earlier steps.

KTU Krishnendu

BV = Strength of the bolt = least of its two failure strengths:

1. shear
2. bearing

Design strength of bolt Vdb = Bolt value = Least of shear strength of bolt and bearing strength of bolt.

Butt joint

When given the working load, you should convert it into the design load by multiplying it with 1.5. But when you are given the factored load, you can take it as it is and start the design process.

When designing, you are probably going to chose M16 bolt. Then. Take the pitch as 2.5d and end/edge distance as 1.5do. M16, 2.5d = 2.5*(16+2) and 1.5do = 1.5 (16+2).

Single cover butt joint, when we get to the point of calculating the bearing capacity of the bolt, we need to know the value of t. Know that it will be the smaller of three things:

1. plate 1
2. plate 2
3. coverplate

It is OK for the diagram to look like this when we are drawing single cover butt joint:

Also note, we are multiplying Vdsb by packing factor Bpk:

And, that Bpk was found from the thickness of the packing plate.

Note how we are finding the t for bearing capacity:

plate 1 thickness vs plate 2 thickness vs sum of two packing plate thickness.

Note how the edge distances were also computed:

Comments