chapter 10 - simple connections.pdf

7/26/2019 Chapter 10 - Simple Connections.pdf

http://slidepdf.com/reader/full/chapter-10-simple-connectionspdf 1/32

CE134

-

STRUCTURAL STEEL

DESIGN

J. Berlin P. Juanzon PhD



Connections of structural steel members areof critical importance. An inadequateconnection, which can be the “weak link” ina structure, has been the cause of numerousfailures. Failure of structural members israre; most structural failures are the result of

poorly designed or detailed connections.



Modern steel structures are connected bywelding or bolting (either high-strength or“common” bolts) or by a combination of both.Until fairly recently, connections were either

welded or riveted.



Failure of the fastener:

where P is the load acting on an individual fastener, A isthe cross-sectional area of the fastener, and d is itsdiameter. The load can then be written as:



Although the loading in this case is notperfectly concentric, the eccentricity is smalland can be neglected



Failure of the connected part because ofbearing exerted by the fasteners:



Bearing stress would be computed as:

whereP = force applied to the fastenerd = fastener diameter, and

t = thickness of the part subjectedto the bearing.



Shear tear-out at the end of a connectedelement:

where0.6Fu = shear fracture stress of theconnected partlc = distance from edge of hole to edgeof connected part

t = thickness of connected part



To prevent excessive elongation of the hole,

an upper limit is placed on the bearing. Thisupper limit is proportional to the projectedbearing area times the fracture stress, or:

whereC = a constant

d = bolt diametert = thickness of the connectedpart

If excessive deformation at service load is a concern,

and it usually is, C is taken as 2.4



The nominal bearing strength of a single bolttherefore can be expressed as:

Where:lc = clear distance, in the direction parallel to the

applied load, from the edge of

the bolt hole to the edge of the adjacenthole or to the edge of the material

t = thickness of the connected partFu = ultimate tensile stress of the connected part

(not the bolt)



When computing the bearing strength for a bolt, use thedistance from that bolt to the adjacent bolt or edge in thedirection of the bearing load on the connected part.



For load and resistance factor design, theresistance factor is f = 0.75, and the designstrength is:

When computing the distance lc, use the actual hole diameter(which is 1⁄ 16-inch larger than the bolt diameter), and do notadd the 1⁄16 inch as required in AISC B4.3b for computing thenet area for tension and shear. In other words, use a holediameter of:



To maintain clearances between bolt nuts and to provideroom for wrench sockets, AISC J3.3 requires that center-to-center spacing of fasteners (in any direction) be no less than 2-

2 ⁄3d and preferably no less than 3d, where d is the fastenerdiameter.

Minimum edge distances (in any direction), measuredfrom the center of the hole, are given in AISC Table J3.4 as afunction of bolt size. The spacing and edge distance to beconsidered, denoted s and le,



Check bolt spacing, edge distances, andbearing for the connection shown in Figure.



A plate with width of 200mm and thickness of16mm is to be connected to two plates of thesame width with half the thickness by 19mm

diameter rivets. The rivet holes have a diameter2mm larger than the rivet diameter. The plateis A36 steel with yield strength Fy=248MPa,allowable tensile stress is 0.60Fy and allowable

bearing stress of 1.35Fy. The rivets are A502,Grade 2, hot driven rivets with allowable shearstress of 150MPa.



a. What is the maximum load that can beapplied without exceeding the allowabletensile stress in the plates

b. What is the maximum load that can beapplied without exceeding the allowable shearstress in the rivets.

c. What is the maximum load that can beapplied without exceeding the allowablebearing stress between plates and the rivets.



Design of fillet welded connections

Fillet welds are most common and used widely Weld sizes are specified in 1/16 in. increments

Fillet welds are usually fail in shear, where the shearfailure occurs along a plane through the throat of theweld

Shear stress in fillet weld of length L subjected to load P

fv =

a

a

Throat = a x cos45o

= 0.707 a

a

a

Throat = a x cos45o

= 0.707 a

Failure Plane

L

wLa707.0

P



Limitations on weld dimensions

Minimum size (amin) Weld size need not exceed the thickness of the thinner part joined.

amin depends on the thickness of the thicker part joined

If the thickness of the thicker part joined (T) is less than or equal to¾ in. amin = ¼ in.

If T is greater than ¾ in.

amin = 5/16 in. Maximum size (amax)

Maximum size of fillet weld along edges of connected parts

for material with thickness < 0.25 in., amax = thickness of thematerial

for plates with thickness 0.25 in., amax = thickness of material -1/16 in.

Minimum length (Lw) Minimum effective length of fillet weld = 4 x size of fillet weld

Effective length of fillet weld > 1.5 in.



Example 1 Design the fillet welded connection

system for a double angle tension member 2L 5x 3½ x 1/2 made from A36 steel to carry afactored ultimate load of 250 kips.

Step I. Design the welded connectionConsidering only the thickness of the angles; amin = 1/4 in.

Considering only the thickness of the angles; amax = 1/2 - 1/16 in. = 7/16in.

Design, a = 3/8 in. = 0.375 in.

§ Shear strength of weld metal = f Rn = 0.80 x 0.60 x FEXX x 0 . 7 0 7 x a x Lw

= 8 . 9 x Lw kips

Strength of the base metal in shear = f Rn =1 .0x0 .58xFy x t x Lw

= 10.44 Lw kips

§ Shear strength of weld metal governs, f Rn = 8 . 9 Lw kips



Design strength f Rn > 250 kips Therefore, 8.9 Lw > 250 kips

Therefore, Lw > 28.1 in.

Design length of 3/8 in. E70XX fillet weld =30.0 in.

Shear strength of fillet weld = 267 kips

Connection layout

Connection must be designed to minimizeeccentricity of loading. Therefore, the center orgravity of the welded connection must coincidewith the center of gravity of the member.

Tu

f L2

f L1

L1

L2

3.4 in.T

u

f L2

f L1

L1

L2

3.4 in.



Connection layout Connection must be designed to minimize eccentricity

of loading.

The c.g. of the welded connection must coincide withc.g. of the member

Total length of weld required = 30 in.

Two anglesassume each angle will have weldlength of 15 in.

Tu

f L2

f L1

L1

L2

3.4 in.T

u

f L2

f L1

L1

L2

3.4 in.



The tension force Tu acts along the c.g. of themember, which is 1.65 in. from the top and 3.35 in.from the bottom (AISC manual). Let, f be the strength of the fillet weld per unit length.

Therefore, fL1 + fL2 = Tu

And fL2 x 3.35 - fL1 x 1.65 = 0 - taking moments about themember c.g.

Therefore, L1 = 2 . 0 L2

But, L1 + L2 = 15.0 in. Therefore, L1 = 10 in. and L2 =5in .

Design: L1 = 10.0 in. and L2 = 5.0 in.



Consider another layout

Tu

f L2

L1

L2

f L1

5f 3.4 inT

u

f L2

L1

L2

f L1

5f 3.4 in

fL1 + fL2 + 5f = Tu

fL2 x 3.5 + 5f x 0.85 - fL1 x 1.65 = 0 - Moment about member c.g.

Additionally, L1 + L2 + 5 = 15.0 in.

Therefore, L1 = 7.6 in. and L2 = 2.4 in.

Design: L1

= 8.0 in. and L2

= 3.0 in.

chapter 10 - simple connections.pdf

Documents