CONNECT IONS

• Connections are the devices used to join elements

of a structure together at a point such that forces
can be transferred between them safely.
• Connection design is more critical than the design
of members.
• The failure of connection usually means collapse
of a greater part or whole of the structure.
• In general, relatively more factor of safety is
provided in the design of connections.
• The rigid connection should provide
sufficient strength and ductility.
• The ductility is very useful for redistribution
of stresses and dissipation of extra energy in
case of earthquakes, etc.
TYPES O F CONNECT IONS
Based On Means Of Connection

A. Welded connections
B. Riveted connections
C. Bolted connections
Based On Forces To Be Transferred

A. Truss connections
B. Moment connections
– i) Fully rigid connections
– ii) Semi-rigid connections
C. Simple/shear connections
D. Splices
E. Brackets
 Moment Connections
• Moment connections are also referred to as rigid,
continuous frame or FR connections.
• Knee joints are the typical example.
• They are assumed to be sufficiently rigid keeping
the original angles between members practically
unchanged after application of loads.
• Greater than 90 percent moment may be
transferred with respect to ideally rigid connection
besides the full transfer of shear and other forces.
 Semi-Rigid / Partially
Restrained Connections
• Type PR connections have rigidity less than 90
percent compared with ideally rigid connections.
• Although the relative rotation between the
joining members is not freely allowed, the
original angles between members may change
within certain limits.
• They transfer some percentage of moment less
than 90 percent and full shear between the
members.
• Semi-rigid connections provide rigidity in-
between fully restrained and simple
connections.
• Approximately 20 to 90 percent moment
compared with ideal rigid joint may be
transferred.
• End moments may develop in the beams and
the maximum beam moment may be
significantly reduced.
• Usually no advantage is taken of this
reduction and beams are designed as simply
supported because of various reasons.
 Shear Connections
• Simple or shear connections have less than
20 percent rigidity.
• They are considerably flexible and the
beams become simply supported due to the
possibility of the large available rotation.
• Moment may not be transferred in larger
magnitudes with the requirement that the
shear force is fully transferred.
• In these connections, primarily the web is to
be connected because most of the shear
stresses are concentrated in it.
• Connections of beams, girders, or trusses
shall be designed as flexible joints to resist
only the reaction shears except otherwise
required.
• Flexible beam connections shall
accommodate end rotations of unrestrained
beams.
 Bearing J o i n t s
• There shall be sufficient connectors to hold
all parts of the section securely in place
when columns rest on bearing plates.
• All compression joints shall be designed to
provide resistance against uplift and tension
developed during the uplift load
combination.
SPLICES
These are used to extend the length of a
particular member.
The two sides of the member may have same or
different cross-sections.
Splice joint is a connection between two parts of
the same member whereas a regular joint is the
connection of more than one members of the
structure.
BRACKETS

Brackets are the connections used to transfer

torque besides other types of forces.
The term bracket is generally used for an extra
plate projecting out of the column and acting
like a seat for the beam.
 Types of Joints Based On Placement
Of Parts To Be Joined
The types of joint depends on factors such as the
size and shape of the members coming into the
joint, the type of loading, the amount of joint area
available for welding, and the relative costs for
various types of welds.
Butt joints
The butt joint is used mainly to join the ends of flat
plates of the same or nearly the same thickness.
A gap or groove is left between abutting members,
which is later on filled with weld (Figure 8.1).
 The principal advantage of this type of joint is to
eliminate the eccentricity developed in single lap
joints.

Groove filled with

weld

Welded Butt Bolted Butt

Joint Joint
Lap joints
The members are either overlapped with each
other or with some connecting plates like gusset
plates, splice plates, etc, as shown in Figure 8.2.
Eccentricity of load and hence moment may be
produced in these joints.
In welded lap joints, the minimum amount of lap is
to be five times the thickness of the thinner part
joined, but not less than 25 mm.
 Welded Lap Joint Bolted Lap Joint

Advantages of Lap Joints

a.The plates of different thickness can easily be

joined such as in a truss connection (Figures
8.3 and 8.4).
b.Ease of Filling: Pieces being joined do not
require the preciseness in fabrication, as do the
other types of joints.
 Lapped plate Lapped plate

Truss Connection

Beam
bracket
Splice joint
The pieces can be slightly shifted to accommodate
minor errors in fabrication or to make adjustments
in length.
c. Ease of Joining: The edges of the pieces
being joined do not need special preparation and
are usually sheared or flame cut.
Occasionally the pieces are positioned by a small
number of erection bolts, which may be either left
in place or removed after the welding is
completed.
Tee joint
In a tee joint, one member
meets the other member at
right angles, as shown in
Figure 8.4.
Corner joint
A typical example of corner
joint is shown in Figure 8.5.
Edge joint
The parts to be joined come
parallel to each other from one
side and are joined at their edge.
WELDING
Welding is a process in which metallic parts are
connected together by heating their surfaces to a
fluid state and allowing the parts to flow together
and join with or without the addition of other
molten metal.
General Types Of Welding
Gas welding
In gas welding a mixture of oxygen and
acetylene is burned at the tip of a torch or
blowpipe held in the welder’s hand.
Additional metal is introduced by a metal rod
known as filler or welding rod.
Gas welding is a rather slow process as
compared to other means of welding and is
normally used for repair and maintenance work
and not for the fabrication and erection of large
steel structures.
Electric arc welding
In arc welding an electric arc is formed between
the pieces being welded connected to negative
terminal of battery and an electrode held in the
operator’s hand with some type of holder
connected to positive terminal of battery.
The arc is a continuous spark which upon contact
brings the electrode and the piece being welded to
the melting point.
The resistance of the air or gas between the
electrode and the piece being welded changes the
electrical energy into heat.
A temperature of somewhere between 3100 and
5500 oC is produced in the arc.
In electric-arc welding the metallic rod, which is
used as the electrode, melts off in to the joint as it
is being made.
Advantages Of Welding

1- Welded structures allow the elimination

of a large percentage of the gusset and
splice plates necessary for riveted or bolted
structures along with the elimination of rivet
or bolt heads.
In some bridge trusses it may be
possible to save up to 15% or more of the
steel weight by using welding making the
structure economical.
2- Welding requires appreciably less labor
than does riveting because one welder can
replace the standard four person riveting crew.
However, skilled and experienced welders are
needed for better quality.
3Welding has a much wide range of
application than riveting or bolting. Consider a
steel pipe column and the difficulties of
connecting it to other steel members by riveting
or bolting.
4Welded structures are more rigid because the
members are often welded directly to each
other.
5- Welding changes and repairs are quick and
easy.
6- Welding has relative silence of operation.

7- Fewer pieces are used and as a result time

is saved in detailing, fabrication and field
erection.
8- Welded connections are not recommended
for temporary connections, where bolts are
preferred.
9- Welding gives truly continuous structures
with smooth and clean surfaces.
 Types Of Welds Depending
Upon Weld Shape

Groove welds
This type of weld is used in approximately 15% of
construction. A groove of one of the shapes
shown in Figure 8.8 is formed between the
adjoining surfaces, which is then filled with weld.
 Name Symbol Use
t  10mm
t1 t2

1. Square
t  12mm
weld

2. Single - V
3. Double - V t > 12mm
4. Single - bevel t  12mm
5. Double - bevel t > 12mm
6. Single - U t  12mm
7. Double - U t > 12mm
8. Single - J t  12mm
9. Double - J t > 12mm
Fillet Welds
Fillet welds owing to their overall economy, ease
of fabricating and adaptability are the most widely
used (in approximately 80% of construction).
It is actually triangular filling of weld around the
overlapping edges.
Slot and Plug Welds
In this type of welding, the pieces to be joined are
placed one above the other and a hole or slot is
drilled in the top plate.
This hole or slot is then filled with the weld
material (Figure 8.9).
 Symbol :
A A

Slot weld
(Called plug weld
if circular)

Section AA
Intermittent Welds
The effective length of any segment of
intermittent fillet welding shall be not less than 4
times the weld size, with a minimum of 38mm.
Minimum effective length of one weld segment
should be 4 tw, but not less than 38 mm. In lap
joints, the minimum amount of lap shall be five
times the thickness of the thinner part joined,
but not less than 25 mm.

1 3 5 7 1 3 5 7
2 4 6 8 2 4 6
 Other Welding Symbols
Some other common symbols are shown in Figure.

= weld all around

= field weld

= flush contour

= convex contour

= concave contour
 Standard Welding Symbol
A standard weld symbol is used on the
drawings and it gives complete information
about the referenced weld.
A typical standard weld symbol is shown in
Figure 8.11 and the terms used in it are
explained below:
 6 150

The symbol indicates fillet weld on near or

arrow side. Size of weld is 6 mm and length
of weld is 150 mm.
50@150 or
12 50 - 150

The symbol shows 12 mm thick fillet weld on far

or opposite-to-arrow side. The weld is
intermittent with length of each segment equal to
50 mm and pitch equal to 150 mm.
 6 150

6mm fillet weld, 150mm long is present on both

sides. As indicated, if weld dimensions are same
on both sides, write only once. Further, it is field
weld.
10 50 - 150

A staggered, intermittent, 10mm fillet weld,

50mm long, 150 on centers, is provided on
both sides.
Minimum Weld Size For Fillet Welds
The minimum fillet weld sizes for
various thicknesses of thinner tp1

parts joined are given by AWS tp2

D1.1 (American Welding Society)
and are reproduced in Table 8.1.
Table 8.1. Minimum Fillet Weld Sizes.
Base metal thickness of thinner Minimum leg size of fillet weld
part joined (tp2) (tw)min.
mm mm
0 < tp1  6 3
6 < tp1  13 5
13 < tp1  19 6
19 < tp1 8
Maximum Fillet Weld Size

1- Along edges of material less than 6

mm thick,
(tw)max. = tp1 where tp1 =
thickness of thinner plate joined.
2- Along edges of material 6 mm or
more in thickness,
(tw1)max. = tp1 − 2
Practical Weld Size
The smallest practical weld size is about
3mm and the most economical size is probably
about 8mm giving the best efficiency of welder.
This 8mm weld is the largest size that can be
made in one pass with the shielded arc welding
process.
Optimum weld size (tw)opt = 8mm
Minimum Length Of Fillet Weld
There is always a slight tapering off in the
region where the fillet weld is started and where
it ends.
Therefore, if the length is very small, large
percentage difference is created between actual
and expected strengths.
Hence, the minimum effective length of a fillet
weld is specified as four times its nominal size.

(𝑙w)min. = 4 tw
If this requirement is not met, the size of the weld
for calculating strength should be considered to
be one-fourth of the effective length provided.
The effective length of any segment of
intermittent fillet weld shall be not less than 4tw,
with a minimum of 38 mm.

Recommended Maximum Weld Length

𝑙max. = 30 tw
If the weld length is greater than this limit, it is
better to use intermittent weld at a clear
spacing of 100 - 150mm.
Strength Of Weld
Strength of weld depends upon the following factors:
1- Size of weld (tw).
2- Length of weld (𝑙1 , 𝑙2).
3- Type of electrode.

4 Type of weld.
5Type of base metal. 6-
Thickness of plates.
Table 8.3. Shielded Metal Arc Welding (SMAW) Electrodes.

Electrode Type Minimum Tensile Strength (FE)

MPa

E60 425

E70 495

E80 550

E100 690

E110 760
Area of weld = te  length of weld = 0.707 tw 𝑙w

The effective throat of the weld (te) is the shortest

distance from the root of the weld to its theoretical
face.
For the 45 or equal leg fillet, the throat
dimension is 0.707 times the leg of the weld (tw),
but it has a different value for fillet weld with
unequal legs, as shown in Figure 8.16.
Adopted Or Selected Weld Size (tw)
Three limiting weld sizes, (tw)min, (tw)max and
(tw)opt are found as explained earlier and are
arranged in ascending or descending order.
The middle value is then selected and is
rounded to the nearest whole number millimeter.

Selected Weld Length

Selected weld length at any face of the member
(𝑙1, 𝑙2, and 𝑙3) should be greater than or equal to
the calculated value but should be within (𝑙w)min
and (𝑙w)max.
 Weld Value (Rw)
It is the strength or load carrying capacity in kN of a unit
length of the weld (usually 1mm) depending on weld or
member strength, whichever is lesser.

Rw = lesser of the following two:

1) Rnw =   effective throat (te)  unit length  weld
shear strength
= 0.75  0.707  tw  1  0.6 FE / 1000
2) RBM = 0.75  0.6 Fu  Ans / 1000
= 0.75  0.6 Fu  t  1 / 1000
where t = thickness of base metal
REQUIRED LENGTH OF WELD

The total weld length required is calculated by

dividing the design force with the weld value.
This weld length is then divided into weld on three
sides of the member namely 𝑙1, 𝑙2 and 𝑙3, as shown
in Figure 8.17.
These calculations are made depending on the
basic requirement that no moment should be
generated at the connection.
Fu
𝑙w = = 𝑙1 + 𝑙2 + 𝑙 3
Rw
𝑙2
𝑙2

Fu
𝑙3

𝑙1 𝑙1

𝑙1

𝑙2
 B
P2

d−y
𝑙3 Fu
P3
Gravity axis
y
P1
A 𝑙1

𝑙1 + 𝑙2 + 𝑙3 = 𝑙 = Fu
w Rw
P1 = Rw 𝑙1
P2 = Rw 𝑙2
P3 = Rw 𝑙3
Fu = Rw 𝑙w
Taking moments about point A and equating it to
zero, following expression is obtained:
P2(d) + P3(d/2) – P  y = 0
𝑙2d + 𝑙3 d/2 − 𝑙w y =0
𝑙w y 𝑙
𝑙2 = − 3
d 2
Similarly taking moments about the point B,
length 𝑙1 may be calculated as follows:
𝑙w (d − y ) 𝑙
𝑙1 = − 3
d 2
Length of weld on that side of the member will be
greater which is closer to centroidal axis, like
towards the projecting leg of the member, etc.
If 𝑙1 is greater than 𝑙2 and 𝑙3 is first selected
equal to zero, the following procedure may be
used to check the lengths for the minimum
and the maximum limits.
Check 𝑙1:
If 𝑙1  (𝑙w)max and 𝑙1  (𝑙w)min Use 𝑙1 without any change

If 𝑙1 < (𝑙w)min Increase 𝑙1 to (𝑙w)min

If 𝑙1 > (𝑙w)max a)Provide 𝑙3 equal to length of end face of

the member and revise 𝑙1 and 𝑙2 (most
common solution)
b)Increase tw, if it is lesser than (tw)max and
revise calculations
c) Provide intermittent weld
 P R O C E D U R E F O R D E S IG N O F
WELDED T R U S S C O N N E C T I O N S

1.Write all the known data including selected

member sections, factored member forces, etc.
2.In case of lap joints, the amount of lap shall be
five times the thickness of the thinner part
joined, but not less than 25mm.
3Decide gusset plate thickness such that it
should be:
a) same throughout the truss,
 b) comparable to greatest thickness of
members joining with it,
c) not less than 6mm, and
d)preferably kept at a minimum of 10mm.
This thickness is most commonly used.
e)Size and shape of the gusset plate are
decided during drawing as explained in
Reference-1 (in instructions to make
working drawing for a truss).
4- In case of members with reversal of forces,
only design for the greater magnitude force and
use the corresponding section capacity.
5- Find out load carrying capacity of the
member, tTn or cPn, if not known.
6.The design factored force (Fu) for a member
discontinued at the joint is taken as the greater of
applied load and 50% (any value may be specified
for effective use of the member strength up to
100%) of the section capacity.
7.If the member is double angle section, consider
Fu as half of the above force for one angle. The
weld will be designed for one angle and the same
will be provided on the other side.
8. Find d and y for the section from the table.
9. Select size of weld (tw) considering (tw)min,
(tw)max and (tw)opt.

10. Decide the type of electrode to be used.

11. Find weld value (Rw) as smaller of  Rnw and
 RBM.
Rnw =   te10.6 FE / 1000
where te = 0.707 tw :  = 0.75
RBM =   tp110.6 Fu / 1000
for base plate subjected to shear,  =0.75
12. Calculate total weld length required (lw) as
follows:
Fu
𝑙w =
Rw
13. Calculate (𝑙w)min and (𝑙w)max.
14.Divide total weld length (𝑙w) into 𝑙1 and 𝑙2, which are
weld lengths at top and bottom of the member,
considering 𝑙3 = 0 in the start.

𝑙w d−y
𝑙1 = 𝑙w  y / d and 𝑙1 =
d
Greater value is provided on that face of the
member which is closer to the centroidal axis.
 .

15- Check lengths 𝑙1 and 𝑙2 for minimum and

maximum limits and decide the side weld length 𝑙3.
a- Assuming that 𝑙1 is the greater length, first
check it against the limiting values as follows:
If 𝑙1  (𝑙w)min and 𝑙1  (𝑙w)max  OK
If 𝑙1 < (𝑙w)min  use 𝑙1 = (𝑙w)min
If 𝑙1 > (𝑙w)max 
i) Take 𝑙3 = d, 𝑙1 and 𝑙2 will be previous
values minus d / 2.
ii) If 𝑙1 is still bigger than (𝑙w)max., we can
increase tw or use intermittent weld.
b- Similar check is made for the smaller length
out of 𝑙1 and 𝑙2.

The minimum length of one segment of

intermittent weld should be larger of 4tw and
38mm.
16- The connection length for a tension member
must be such that a better shear lag factor may be
achieved. The preferred connection length may
be calculated as under:
x
U = 1−
𝑙
x
For U = 0.9 1− 𝑙 = 0.9
pref
 x
= 0.1 𝑙 pref = 10 x
𝑙pref

where x = distance between centroid of the

element and the plane of load tranfer
17- Check block shear strength, for tension
members only.
The nominal strength for block shear is the lesser
of the following two cases because only that will
cause the final separation of the block from the
member.
Rn = lesser of 0.6 Fu Anv + Ubs Fu Ant
and 0.6 Fy Agv + Ubs Fu Ant
 ,

Nominal tension rupture strength = Ubs Fu Ant

Nominal shear rupture strength = 0.6 Fu Anv
Shear yielding strength = 0.6 Fy Agv
0.6Fy  yield shear strength = y
0.6Fu  ultimate shear strength = u
 = 0.75 (LRFD) and  = 2.00 (ASD)
Agv = gross area subjected to shear
Anv = net area in shear
Ant = net area in tension
Ubs = tensile rupture strength reduction factor
(subscript ‘bs’ stands for block shear)
= 1.0 when tensile stress is uniform
 18- Show results of weld design on a neat
sketch using standard weld symbol.

Example 8.1: Design weld for the tension-

member shown in Figure 8.22 using E 70 electrode.
The thickness of gusset plate is 10 mm and the
factored tensile force is 300 kN.
 l2

d–y
Tu = 300 kN

l1 L89  76  9.5
Solution:

From tables (Reference – 1), y = 27.4 mm,

d – y = 61.6 mm and A = 1480 mm2.
t Tn = 0.9  250  1480/1000 = 333.0 kN
t Tn = 0.75  400  1.0  1480/1000
= 444.0 kN

 t Tn /2 = 166.5 kN
Design force for the connection,
Fu = greater of 166.5 kN and 300 kN
= 300 kN
tp1 = 9.5 mm ; tp2 = 10 mm
(tw)max = tp – 2 = 7.5 mm
(tw)min = 5 mm
topt = 8 mm

 (tw)adopted = 7.5 mm  8 mm

Rnw = 0.75  0.707  8  1  0.6  495 / 1000

= 1.26 kN/m
RBM = 0.75  0.6  400  9.5  1 / 1000
= 1.71 kN/m
 Rw = 1.26 kN/m
 Length of weld 𝑙w = 300 / 1.26
= 238 mm
𝑙w  y 238 (27.4 )
𝑙2 = = 89
d
= 73 mm (say 75 mm)
𝑙1 = 𝑙w – 𝑙2 = 165 mm

For efficiency factor (U) of 0.85,

preferred length of connection = 6.7 (x )
= 6.7 (21.1)
 145 mm
 𝑙1 = 165 mm
 (𝑙w)min = 4 tw = 32 mm
(𝑙w)min = 30 tw = 240mm
32 mm  𝑙1, 𝑙2  240 mm OK
Block Shear Strength
Perform the check as done in tension
member design
75 8

The results are

shown in Figure
8.23.
165 8
Example 8.2: Design welded connection for the
truss compression member shown in Figure 8.26
using E70 electrode. The weld length on any face
should not exceed 150 mm.
10mm Thick Gusset Plate

𝑙2

𝑙3
Pu = 600 kN

𝑙1 2Ls 1021029.5
A = 1850 mm2
L = 1.5 m
Solution:

A = 1850 mm2 for one angle

y = 29 mm
rx = 31.2 mm
Iy = 2(181  104 + 1850  342)
= 790  104 mm4
790104
ry = = 46.2 mm
21850

K𝑙 11500
R = =  48 :
rmin 31.2
Fcr = 199.13 MPa
 ½ cPn = ½  199.13  2  1850 / 1000
 368.4 kN
 Fu for 2 angles = larger of 600 and 368.4
= 680 kN
and Fu for one angle = 300kN

tp1 = 9.5 mm ; tp2 = 10 mm

(tw)max = tp – 2 = 7.5 mm
(tw)min = 5 mm
topt = 8 mm
(tw)adopted = 7.5 mm  8 mm
Rnw = 0.75  0.707  8  1  0.6  495 /
1000 = 1.26 kN/m
 .

RBM = 0.75  0.6  400  9.5  1 / 1000

= 1.71 kN/m
 Rw = 1.26 kN/m
300
𝑙w = = 238 mm
1.26
𝑙w (d − y ) 238 (102 − 29)
𝑙1 = = = 170 mm
d 102
𝑙2 = 238 – 170 = 68 mm

The joint efficiency and block shear checks are not

required here because it is a compression member.
As 𝑙1 > 150 mm, let 𝑙3 = 102 mm
102
𝑙1 = 170 – = 119 mm (say 120 mm)
2
102
𝑙2 = 68 – = 17 mm
2
(𝑙w)min = 4tw = 32
(𝑙w)max = 30 tw = 240 mm

𝑙1 is between (𝑙w)min and (𝑙w)max OK

𝑙2 < (𝑙w)min 𝑙2 = 32 mm (say 35 mm)
Final Result

𝑙1= 120 mm
𝑙2= 35 mm
𝑙3= 102 mm

To show the results on a neat sketch are left as exercise

for the reader.
 RIVETED AND BOLTED TRUSS
CONNECTIONS
Consider the example of a lap joint made by
installing a fastener and subjected to tensile or
compressive load as shown in Figure 8.27.
The fastener is placed in already drilled hole
through the parts to be joined.
The fastener has a head on one side of its shaft
for anchorage.
The other end is also worked into a head in case
of rivets and a nut is tightened at other end in
case of a bolt.
T T

Bearing Stresses
Head of Fastener

T Grip

Failure Plane
Shaft of Fastener

Figure 8.27.Lap Joint Using a Single Rivet.

The bolts may be arbitrarily tightened called snug
tight bolts.
Or they may be subjected to a predefined torque
producing pre-tension in the bolts and compression
on the joining plates known as high strength
bolts.
The distance between the two heads after placing
of the fasteners is called grip of the fastener.
A bolted joint in which the slip resistance of the
connection is also utilized is called Slip Critical
Joint.
The minimum bolt pretension for high strength bolts
is given in Table 8.4.
The pretension is measured by the turn-of-nut
method, direct tension indicator, calibrated wrench
or alternative design bolt.
Bolt Size, d A325M Bolts A490M Bolts Standard Hole
(mm) Pretension (kN) Pretension (kN) Dia (mm)
M15 80 100 17
M18 115 145 20
M20 142 179 22
M22 176 221 24
M25 225 282 28
M28 286 358 31
M30 326 408 33
M35 448 562 38
> M35 − − d+3
The rivets used for structural purposes are driven
and installed in red hot state and are therefore
known as hot driven rivets.
Once the head is made on both sides of the rivet in
red-hot state and the rivet is then allowed to cool,
compression on the parts to be joined is produced.
This is required for close packing of members at
the joint and to avoid chattering of joints.
Further, by using hot rivets, it becomes easy to
make head by hammering.
ASTM Specification A502 deals with these types
of rivets and the qualities of these rivets are
defined as Grade1, Grade 2 and Grade 3 rivets.
Grade 1 rivets are having lesser strength and
their corrosion resistance is also of ordinary
level.
Grades 2 and 3 rivets are used for higher
strength and better corrosion resistance.
The tensile and shear strengths of some
common types of rivets and bolts are given in
Table 8.5.
 .
Table 8.5. LRFD Nominal Tensile and Shearing Strengths for Rivets and Bolts.

S.# Fastener Type Tensile t Shearing v

Strength Strength in
(MPa) Bearing Type
Connections
(MPa)
1- A502, grade 1, hot driven 310 0.75 172 0.75
rivets.
2- A502, grade 2 or 3, hot driven 414 0.75 228 0.75
rivets.

3- A307 bolts. 310 0.75 165 0.75

4- A325M bolts (Fu = 825 MPa) 0.75 Fu 0.75 0.40 Fu 0.75

when threads are not excluded = 620 = 330
from shear planes.
5- A490M bolts (Fu = 1035 0.75 Fu 0.75 0.40 Fu 0.75
MPa) when threads are not = 780 = 414
excluded from shear planes.
TYPE O F S T R E S S E S IN F A S T E N E R S
When the lap type connection of Figure 8.27 is
subjected to tension or compression, the fastener
is subjected to shear at a cross-section lying at the
interface of the two parts.
This cross-section at which different layers of the
fastener try to slide against each other and failure
can occur here is called a shear failure plane.
The cross-sectional area resisting shear in case of
rivets will be /4 d2 where d is the diameter of the
rivets. However, in case of a bolt, there are two
possibilities.
Bolt will have more strength if failure plane lies in
unthreaded portion and less strength if failure
plane lies within the threads.
The effective area of cross-section resisting shear
in the later case will be less, considered equal to
approximately 75% of the total area without
threads, due to grooves within the threads.
However, adjustment for this reduction is made in
the strength and then area calculated on the basis
of outer diameter is used to evaluate the strength.
Because the fastener and plate are not fully joined
with each other, the forces from the fasteners are
transferred to the plates by bearing stresses in the
plate material surrounding the bolt on one side as
shown in Figure 8.27.
Bearing Stresses
Bearing stresses are very high but local
compressive stresses produced when two surfaces
abut each other and transfer load.
If sufficient material is available around the zone of
high bearing stresses, these stresses quickly
spread over a greater region reducing the intensity.
The locally stressed material is confined in nature.
For this reason and for the reason that these local
compressive stresses cannot produce fracture and
buckling, stresses up to 3.0 times the ultimate
tensile strength of plate material (3.0 Fu) may be
allowed at nominal strength level.
According to AISC, bearing strength must be
checked for both bearing type and slip critical
connections.
(a) When deformation at bolt hole due to service
loads is a consideration:
Nominal bearing strength,
Rn = 1.2 Lc t Fu  2.4 d t Fu
where Lc = clear edge distance or clear
spacing between bolts
t = thickness of connected material
 = 0.75 and  = 2.00
 .

(b) When deformation at bolt hole due to

service loads is not a consideration:
Nominal bearing strength,
Rn = 1.5 Lc t Fu  3.0 d t Fu
 = 0.75 and  = 2.00
Shear Stresses
When two plates of a lap joint are pulled in opposite
direction as in Figure 8.28, only one failure plane is
produced and the fastener strength is determined by one
cross-section of the fastener.
T

Figure 8.28. Rivet Under Single Shear.

T/2

T/2

Figure 8.29. Rivet Under Double Shear.

This type of shear is called single shear denoted by “1s”
in calculations.
In case of the simplest half part of butt joint (Figure
8.29), three plates are trying to move relative to each
other.
Two failure planes are produced and two cross-sections
resist the applied load.
The shear strength becomes double as that of single
shear for same material and diameter of the fastener.
In other words, the applied force is divided at greater
number of cross-sections.
This type of shear is called double shear denoted by
“2s” in calculations.

In general, number of shear planes is always equal

to one less than the number of moving plates.

Number of shears = Number of moving plates – 1

For example, the fastener in Figure 8.30 is
subjected to 4- times shear because of five
moving plates.
It should be noted that any adjacent plates,
which cannot move in opposite direction, are
counted as a single unit in the above formula.
T/3

T/2

T/3

T/2

T/3

Figure 8.30. Rivet Under 4-Times Shear.

The four plates in Figure 8.31 are subjected to single
shear.

Figure 8.31. Single Shear In Rivet Joining Four Plates.

BEARING TYPE CONNECTIONS

When the loads to be transferred are larger than
the frictional resistance caused by tightening the
bolts, the members slip a little on each other
putting the fasteners in shear and the surrounding
member in bearing.
The resulting type of connections are called
bearing connections.

EFFECTIVE BEARING AREA

According to AISC, the effective bearing area of
bolts, threaded parts and rivets shall be the
diameter of such fasteners multiplied by the length
of bearing:
 Rn = 2.4 Fu  d  t when Lc  2d
otherwise Rn = 1.2 Fu  Lc  t

(In case deformation at service load is a

design consideration)
where  = 0.75 (LRFD)
and  = 2.00
t = smaller thickness of plate, subjected
to following conditions:
a) edge distance not less than 1.5 d,
b) c/c distance between fasteners not less
than 3d, and,
c) 2 or more fasteners in the line of force.
RIVET AND BOLT VALUE
The load in kN which a single rivet can
carry is called its rivet value (Ru).

The rivet value is smaller of rivet shear

strength and the plate bearing strength.

Ru = lesser of Rns and Rn

Rivet/bolt shear strength,
 Rns = resistance factor  rivet shear
strength  area in shear  number
of shear planes
=   rivet shear strength /4 d2  n
Rivet/bolt plate bearing strength,
Rn = resistance factor  bearing strength 
area in bearing
= 0.75 2.4Fu dt when Lc  2d

where d is the outer or nominal diameter of the rivet or

bolt.
Rivet Value In Case Of Lap Joint

Typical example of a lap joint is the connection of

a single angle section (thickness = ta) with the
gusset plate (thickness = tg), as shown in Figure
8.32.
Because of two moving plates, the rivets will be
subjected to single shear (1s). Most commonly,
the angle thickness is lesser than the gusset
plate thickness.

Using A502 Grade-2 rivets, the rivet value may be

calculated as follows:
 ta
T

T tg

ta  tg

Ru = lesser of 1) shear strength of rivet

in single shear
RIS = 0.65330/4 d21 / 1000
= 0.168 d2 (kN)
 2) strength of rivet
based upon its bearing on
plate.
Rn = 0.75  2.4  400  d  ta
= 0.72 d ta (kN) when Lc  2d
for A36 steel

Rivet Value In Case Of Half Butt Joint

Typical example of a half butt joint is the

connection of double angle section with the
gusset plate (Figure 8.33).
The rivets are subjected to double shear.
For bearing, the total load is either resisted by
the thickness of gusset plate (tg) or two times the
angle thickness (2ta), usually tg is lesser than 2ta.
Using A502 Grade 2 rivets, the rivet value is
evaluated as under:
Ru =lesser of
1) R2s = 0.65330/4 d2 2 / 1000
= 0.337d2 (kN)
2) Rn = 0.75  2.4  400 d  tg
= 0.72 d tg (kN) when Lc  2d
 T/
2

T/
2

REQUIRED CLEARANCES
Minimum Edge Distance

The minimum distance from center of rivet to

the edge should preferably be not less than
1.5d.
If this distance is not maintained, detailed
formulas given in the chapter on tension
members and in another article in this chapter
are to be satisfied.
The distance should be kept equal to 2.5d + 2
(mm) to obtain bearing strength equal to 2.4 Fu.

Minimum Spacing Of Fasteners

The minimum longitudinal and transverse

spacing of fasteners (pitch or gage) shall
preferably be not less than 3d.
This is to avoid stress concentrations and to
make drilling of holes and tightening of bolts
easier.
For spacing lesser than 3d, formulas given earlier
for tension members must be checked.
Further, to improve bearing strength of bolts, it is
better to slightly increase this spacing to a value
of 3d+3 mm, which given clear spacing (Lc) equal
to 2d.
Maximum Edge Distance And Spacing
The maximum distance from the center of
fastener to the nearest edge of parts shall be
lesser of 12 t and 150 mm, where ‘t’ is the
smaller thickness of the connected parts.
The pitch of fasteners is kept lesser than the
following maximum value to prevent corrosion
of loose plates from inside of the over lap:
pmax = lesser of 305 mm and 24 t for painted
and non-corrosive steels
pmax = lesser of 180 mm and 14 t for
unpainted steels
where t = thickness of thinner part joined
DIAMETER OF FASTENER

Minimum diameter of rivets and bolts for trusses

and other building structures is 15 mm. The
usual size of rivets or bolts used for specific
purposes is as under:

In buildings: 15, 18, 20 mm

In bridges: 22, 25, 28 mm
In warehouses and towers: 30, 32, 35 mm
Preferable diameter of fastener is usually taken
by the following expression:

d= 6 t
rounded to the nearest available size
where t = thickness of thicker part, mm
Economical diameter of fastener means the
diameter of a rivet or bolt for which the shearing
strength is theoretically equal to bearing strength
of the parts to be joined. However, for most
practical cases, it becomes difficult to use this
diameter.
For A 502 Grade 2 rivets connecting double angles
section with the gusset plate both of A36 steel,
0.65  /4 d2  3302 = 0.75  2.4  400  d  tG
Economical diameter, d = 2.14 tG
(rounded to nearest available size)
The grip of a rivet shall not exceed 8 times the
diameter of the holes in any case.
 ROCEDURE FOR DESIGN OF
RIVETED TRUSS CONNECTIONS
1. Find design capacity of the member, tTn or
cPn.
2.Compare calculated factored force in the
member with the given percentage of the
member capacity (usually not less than 50%)
and select the design force for connection as
follows:
Fu = larger of 1) factored force
2) a %age of t Tn or cPn
3. For un-spliced top and bottom chord members,
the difference of forces in the adjacent panels is to
be used in a way to get the maximum possible
answer.
4.Decide the rivet diameter which should remain
same throughout the truss.
5.Find rivet value (Ru) according to single or
double shear.
6. Find number of rivets (N) as follows:
Fu
N= (rounded to higher whole number)
Ru
7. For better joint efficiency and lesser stress
concentrations, a minimum of 3 rivets is preferred.
Cost of few extra rivets is much lesser than extra
cost spent on the member for lesser joint
efficiency.
8. The length of joint should not be excessive. If
numbers of rivets are more than approximately
five, arrange them in more rows. However, the
connection length of a tension member (𝑙) must be
large enough to give better joint efficiency.
For U = 0.9, 𝑙pref = 10 x
x = distance between the centroid of element
and the interface surface.
9. Decide the spacing and edge distances of
rivets depending on minimum and maximum
requirements. In transverse direction, place the
rivets along standard gages.
10.Check block-tearing strength as discussed
earlier for welded connections.
11. Make a neat sketch to show the results.
12.Verify that net area and U, in case of tension
members, are greater than or equal to the values
taken during the member design.
 .

Example 8.3: Design rivets for the connection

shown in Figure 8.34 with the condition that each
member should be able to develop at least 50% of
the effective strength. Use A502 Grade 2 hot driven
rivets. The magnitudes of forces are all factored.
Gusset plate is 10 mm thick.
L 76  64  9.5 F3 = 70 kN(C)
Length = 1.0 m

F4 = 210 kN(T)
L 76  64  9.5

F2 = 350 kN(T) F1 = 168 kN(T)

45 45

2Ls 102  102  9.5

50% capacity of L76649.5 in compression
For compression members, double angle sections
should be preferred.
However, the capacity calculated based on the
assumption of concentric loading for single angle
section will be on safer side for the connection.
A = 1240 mm2
K𝑙
= 11000  76
rz 13.3

cFcr = 165.66 MPa

0.5cPn = ½  165.66  1240 / 1000  102.7 kN
50% capacity of L76649.5 in tension
The calculation may be based on the
assumption that An is equal to 85% of Ag in the
absence of accurate value of net area and the
value of U may be taken equal to 0.85.
Both of these assumptions are to be checked
after the connection design.
A = 1240 mm 2
0.5t Tn = lesser of
1. 1/2  0.9  250 1240/1000 = 139.5
kN
2. 1/2 0.754000.850.801240/1000 = 126.5
kN

= 126.5 kN
50% Capacity of 2Ls1021029.5 in tension
Area of one angle = 1850 mm2
Using the same assumptions as above:
0.5tTn = lesser of
1. ½
0.925021850/1000
= 416.3 kN
2. ½
0.754000.850.8021850/1000 =
377.4 kN
= 377.4 kN
Fu for members 1and 2= larger of
1) (larger of F2 and its 0.5tTn)
– F1
= 377.4 – 168 = 209.4 kN
2) 210 cos45+ 102.7 cos45
= 221.1 kN
= 221.1 kN
Fu for member 3 = larger of
1) 70 kN
2) 102.7 kN
= 102.7 kN
Fu for member 4 = larger of
1) 210 kN
2) 126.5 kN
= 210 kN
d = 6 t = 6 10
= 18.97 mm say 18 mm
Rivet Values
on the next slide
 c
Ru = lesser of
R1S = 0.75  228  /4(18)2 
1/1000
= 43.51 kN
Rn = 0.75  2.4  400 18 
9.5/1000
= 123.12 kN
= 43.51 kN
2Ls 1021029.5, assuming Lc  2d,
Ru = lesser of
R2S = 0.75  228  /4 (18)2 
2/1000
= 87.03 kN
Rn = 0.75  2.4  400  18 
10.0/1000 = 129.60 kN
 221.1
Number of rivets for members 1and 2 =
87.03
= 2.54 say 3
Number of rivets for member 3 102.7
=
43.51
= 2.36 say 3
210
Number of rivets for member 4 =
43.51
= 4.83 say 5

Minimum edge distance to provide Lc = 2d:

= 2.5 d + 2 = 2.5  18 + 2 = 47 mm
 .

Maximum edge distance = lesser of

1) 12 t = 12  9.5 = 114 mm
2) 150 mm
L=et114 medge distance
= 50 mm for inclined members
Minimum pitch = 3d = 3  18 =
54 mm
Maximum pitch, considering unpainted
surfaces,
= lesser of 1) 180 mm
2) 14 t = 133 mm
 .

Let pitch= 60 mm for diagonal members and 130

mm for bottom chord members.
x for L76649.5 = 17.9 mm
For U = 0.80, 𝑙pref = 5.0 x
= 89.5 mm
Connection length for L76649.5 in tension
= 4  60
= 240 mm > 89.5 mm OK
Provide all rivets in a single line along the
standard gage line. The rivets on bottom chord
are spread closer to maximum value of pitch to
satisfy the shape of gusset plate.
 32 32
44
44
50
60
60 50
60 60
60 60
50 50

38
64

Between 130 130 Between

50 and 50 and