Steel strucure lec # (18)

Prof. Zahid Ahmad Siddiqi
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.
RIVETED AND BOLTED TRUSS
CONNECTIONS

T
T
TT
Failure Plane
Shaft of Fastener
Grip
Head of Fastener
Bearing Stresses
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
(mm)
A325M Bolts
Pretension (kN)
A490M Bolts
Pretension (kN)
Standard Hole
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
Strength
(MPa)
ft
Shearing
Strength in
Bearing Type
Connections
(MPa)
fv
1- A502, grade 1, hot driven
rivets.
310 0.75 172 0.75
2- A502, grade 2 or 3, hot driven
rivets.
414 0.75 228 0.75
3- A307 bolts. 310 0.75 165 0.75
4- A325M bolts (Fu = 825 MPa)
when threads are not excluded
from shear planes.
0.75 Fu
= 620
0.75 0.40 Fu
= 330
0.75
5- A490M bolts (Fu = 1035
MPa) when threads are not
excluded from shear planes.
0.75 Fu
= 780
0.75 0.40 Fu
= 414
0.75

TYPE OF STRESSES IN FASTENERS
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 p/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
f = 0.75 and W = 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
f = 0.75 and W = 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
T
Figure 8.28. Rivet Under Single Shear.
T
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/2
T/2
Figure 8.30. Rivet Under 4-Times Shear.
T/3
T/3
T/3

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 ³ 2d
otherwise Rn = 1.2 Fu ´ Lc ´ t
(In case deformation at service load is a
design consideration)

where f = 0.75 (LRFD)
and W = 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 fRns and fRn

Rivet/bolt shear strength,
f Rns = resistance factor ´ rivet shear
strength ´ area in shear ´ number
of shear planes
= f ´ rivet shear strength ´p/4 d2 ´ n
Rivet/bolt plate bearing strength,
fRn = resistance factor ´ bearing strength ´
area in bearing
= 0.75´ 2.4Fu´ d´t when Lc ³ 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:

tg
ta
ta £ tg
T
T
Ru = lesser of 1) shear strength of rivet
in single shear
fRIS = 0.65´330´p/4 d2´1 / 1000
= 0.168 d2 (kN)

2) strength of rivet
based upon its bearing on
plate.
fRn = 0.75 ´ 2.4 ´ 400 ´ d ´ ta
= 0.72 d ta (kN) when Lc ³ 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) fR2s = 0.65´330´p/4 d2 ´2 / 1000
= 0.337d2 (kN)
2) fRn = 0.75 ´ 2.4 ´ 400 ´d ´ tg
= 0.72 d tg (kN) when Lc ³ 2d

T/2
T/2
T
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 = t6
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 ´ p/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.

ADVANTAGES OF BOLTS OVER RIVETS
1. Smaller working crews are required as
compared with riveting. The efficiency of
construction is approximately double per
person.
2. Less high strength bolts may be needed as
compared with rivets.
3. The experience required to properly install
bolts is significantly lesser than is necessary
for welded and riveted connections.
4. Less noise is produced during construction as
compared with riveting.

5. Cheaper equipment is used to make bolted
connections compared with welded or
riveted connections.
6. Fire in case of welding and hot material in
case of rivets is avoided.
7. Fatigue strength of fully tight high strength
bolts is greater than that of rivets.
8. Future changes are very easy with bolts.

ROCEDURE FOR DESIGN OF
RIVETED TRUSS CONNECTIONS
1. Find design capacity of the member, ftTn or
fcPn.
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 ft Tn or fcPn

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:
N = (rounded to higher whole number)
u
u
R
F

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
= distance between the centroid of element
and the interface surface.
x
x

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.
45°45°
L 76 ´ 64 ´ 9.5
L 76 ´ 64 ´ 9.5
2Ls 102 ´ 102 ´ 9.5
F2 = 350 kN(T) F1 = 168 kN(T)
F3 = 70 kN(C)
Length = 1.0 m
F4 = 210 kN(T)

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.
zr
Kl
3.13
10001´
A = 1240 mm2
= @ 76
fcFcr = 165.66 MPa
0.5fcPn = 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 mm2
0.5ft 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:
0.5ftTn = 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 0.5ftTn)
– 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
t6 106d = =
= 18.97 mm say 18 mm
Rivet Values
on the next slide

L76´64´9.5, assuming Lc ³ 2d,
Ru = lesser of
fR1S = 0.75 ´ 228 ´ p/4(18)2 ´
1/1000
= 43.51 kN
fRn = 0.75 ´ 2.4 ´ 400 ´18 ´
9.5/1000
= 123.12 kN
= 43.51 kN
2Ls 102´102´9.5, assuming Lc ³ 2d,
Ru = lesser of
fR2S = 0.75 ´ 228 ´ p/4 (18)2 ´
2/1000
= 87.03 kN
fRn = 0.75 ´ 2.4 ´ 400 ´ 18 ´
10.0/1000 = 129.60 kN
= 87.03 kN

Number of rivets for members 1and 2 =
03.87
1.221
= 2.54 say 3
Number of rivets for member 3 =
51.43
7.102
= 2.36 say 3
Number of rivets for member 4 =
51.43
210
= 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
= 114 mLet edge 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
= 133 mm

x
x
Let pitch= 60 mm for diagonal members and 130
mm for bottom chord members.
for L76´64´9.5 = 17.9 mm
For U = 0.80, lpref = 5.0
= 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.

44 44
32 32
50
60
60
60
50
60
60
50
38
64
130 130
50
Between
50 and
114
Between
50 and
114
60
44 44
32 32
50
60
60
60
50
60
60
50
38
64
130 130
50
Between
50 and
114
Between
50 and
114
60

Block shear may be checked by a procedure
discussed earlier. The readers are required to
perform this check.
To verify that An ³ 0.85 Ag and U = 0.80 are left
as exercise for the readers.
DESIGN OF LOADED TRUSS JOINTS
Spliced joint is a joint where the upper or lower
chord is discontinuous and it is designed like an
ordinary joint even if load is acting on it.
The loaded joint with continuous upper or lower
chord member is called un-spliced loaded joint.

This is also designed as for an unloaded joint.
However, the transverse component of the load is
to be transferred through the member to the
gusset plate, and this force is not considered in
the truss analysis for the axial forces (Figure
8.36).
Transverse load component (V) acts on top chord
at the joint of Figure and is to be transferred to
gusset plate through the rivets.
Hence, resultant force on rivets R¢u is to be
calculated as follows:

F1
F2
V
P
F4
F5
F3
q4
q3
q
F1
F2
V
P
F4
F5
F3
q4
q3
q

R¢u = resultant force on each rivet
N = number of rivets provided for the top
chord
Fu = design force for the connection
When R¢u =
22
÷
ø
ö
ç
è
æ
+÷
ø
ö
ç
è
æ
N
V
N
Fu
If R¢u > Ru:
1. The number of rivets is increased.
2. Diameter of the rivets is increased.
3. Lug angle may also be used.
£ Ru OK

BOLTS SUBJECTED TO ECCENTRIC SHEAR
Eccentrically loaded bolt/rivet groups are
subjected to direct shears and torque as shown in
Figure 8.54.
In a truss, if the center of gravity of a member is
not in line with the center of gravity of the bolts at
its end connections, moments are produced.
Eccentricity is quite obvious in a bracket and is
also present in shear connection of beam with a
column.

In Figure 8.55, a typical group of four bolts is
shown with the load acting at some eccentricity,
e, from the centroid of the bolt system.
This load may be transferred to the centroid but
it will then be accompanied by a torque (T = P ´
e), as shown in Figure 8.56.
Both Figures 8.55 and 8.56 are equivalent as far
as the structural behavior is concerned.

a)
Bracket
b) Shear
Connection of
Beam With Column
a)
Bracket
b) Shear
Connection of
Beam With Column

c.g. c.g.
e P P
Figure 8.55.
Typical Group of Bolts/Rivets Subjected
to Eccentric Shear.
Figure 8.56.
Eccentric Shear On a Bolt/Rivet
Group Converted to Shear and
Torque at the Centroid.
12
3 4
Tu = P ´ eGroup of
Bolts c.g. c.g.
e P P
Figure 8.55.
Typical Group of Bolts/Rivets Subjected
to Eccentric Shear.
Figure 8.56.
Eccentric Shear On a Bolt/Rivet
Group Converted to Shear and
Torque at the Centroid.
12
3 4
Tu = P ´ eGroup of
Bolts

The bolts shown in Figure 8.56 are subjected to
a downward force (P / N) in each fastener called
direct shear force plus the shear force due to
the torque called eccentric shear.
The force in each fastener (Fi) due to the torque
and distance of each fastener (di) from the
centroid are shown in Figure 8.57.

F1
F2
F3 F4
e P
d1
d2
d3
d4
c.g
.
F1
F2
F3 F4
e P
d1
d2
d3
d4
c.g
.

The torque causes the plate to rotate about the
centroid of the bolt connection keeping the amount of
rotation or strain at a particular bolt being
proportional to its distance from the centroid.
These strains produce stresses and forces Fi in the
bolts. Greater is the distance of a fastener from the
centroid, more will be the force.
Further, force in each fastener will be perpendicular
to the distance vector between the centroid and that
fastener.
These eccentric forces will produce same turning
effect as that of the applied torque (clockwise or
counterclockwise).

1
1
d
F
2
2
d
F
3
3
d
F
4
4
d
F
Magnitude Of Eccentric Shear
From Figure 8.57,
Tu =P ´ e = F1 d1 + F2 d2 + F3 d3 + F4 d4 I
For force to be proportional to the distance, we can
write:
= = =
1
2
11
d
dF
2
2
22
d
dF
3
2
33
d
dF
4
2
44
d
dF
Equation I can be written as:
Tu = +++
= ( )2
4
2
3
2
2
2
1
1
1 dddd
d
F
+++ II

Þ Tu = å 2
1
1 d
d
F
Þ F1 =
å 2
1
d
dTu
III
å 2
2
d
dTu
å 2
3
d
dTu
å 2
4
d
dTu
Similarly F2 = , F3 = , F4 =
IV
Each force Fi is perpendicular to the line drawn
from the centroid to the particular bolt.
It is usually more convenient to break these down
into vertical and horizontal components (Figure
8.58).

The horizontal and vertical components of the
distance d1 are represented by h1 and n1,
respectively, and the horizontal and vertical
components of force F1 are represented by H1
and V1, respectively.
q1
V1
H1
h1
v1
d1
c.g
.
q1F1
q1
V1
H1
h1
v1
d1
c.g
.
q1F1

1
1
1
1
d
h
F
V
=
1
1
d
h
From Figure 8.58:
or V1 = F1
Substituting the value of F1 from above:
å 2
1
d
hTu
1
1
1
1
d
v
F
H
=
1
1
d
v
å 2
1
d
vTu
V1 =
Similarly, Þ H1 = F1
Þ H1 = VI
V

å 2
d
vT iu
å 2
d
hT iu
In general: Hi = and Vi = VII
2
1d 2
2d 2
nd 2
ih
2
in
Sd2 for a rivet system may be calculated as
S + + - - - + or as S + S
.
The total shear for each fastener is calculated as
the sum of eccentric shear and direct shear. The
magnitude of this total shear will vary for each
fastener.
The fastener which has the maximum shear is
called the most heavily stressed fastener and is
to be located before further calculations.

One way of locating this critical rivet is to find
resultant shear for each fastener and then finding
the maximum value.
However, this procedure may require lengthy
calculations.
A visual observation is used to reduce such extra
work.
The fasteners having the maximum distance from
the centroid (corner fasteners) are expected to be
more critical.

Further, out of these, whichever has both
horizontal and vertical components of direct and
eccentric shears in the same direction is expected
to be the most critical.
If there is any doubt about the condition of this
fastener by visual observation, all the competing
fasteners must by investigated by calculating the
resultant shear.
The procedure is clear in Example 8.6.

Required Number of Fasteners
When the eccentricity of the load on a bolt group
is less than about 60 mm, it is neglected.
For a shear force of Pu (kN) and a torque of Tu
(kN-m), the number of A502 Grade 2 rivets (N) of
15 mm diameter may be approximately
determined as under:
N =
000,1570
uu TP
+

Example 8.6: Find the diameter of rivets for
the bracket shown in Figure 8. 59.
P=800kN (factored force)
10050 100 250
50
75
100
125
10
7
4
1
11
8
5
2
12
6
9
3
20°
c.g.
P=800kN (factored force)
10050 100 250
50
75
100
125
10
7
4
1
11
8
5
2
12
6
9
3
20°
c.g.

Solution:
Let y = distance from top of bracket to
centroid of the rivet system,
and A = area of a single rivet.
Taking moment of the resultant rivet area about the
top edge of bracket (12A ´ y) and equating it to the
sum of moments of the individual areas about the
same edge, y may be evaluated.
y =
( ) ( ) ( ) ( )
A12
350A3225A3125A350A3 +++
= 187.5 mm

N
Px
12
8.751
Rivet 10 is expected to be critical as horizontal and
vertical forces due to moment and applied load add
into one another. However, rivet-1 may also be
investigated.
Horizontal component of load = Px = 800 ´ cos20°
= 751.8 kN
Vertical component of load = Py = 800 ´ sin20°
=273.6 kN
Horizontal force on each rivet = =
= 62.65 kN

N
Py
12
6.273
Vertical force on each rivet = =
Torque at centroid Tu
= 751.8 ´ 187.5 (counterclockwise)
- 273.6 ´ 350 (clockwise)
= 45,202.5 kN-mm (counterclockwise)
= 22.80 kN
d1 = 22
5.162100 +
= 190.8 mm = d3
d2 = 162.5 mm

22
5.37100 +
22
5.62100 +
22
5.137100 +
d4 =
= 106.8 mm = d6
d5 = 37.5 mm
d7 =
= 117.9 mm = d9
d8 = 62.5 mm
d10 =
= 170.0 mm = d12
d11 = 137.5 mm
Sd2 = 231,848 mm2

Rivet Value
Ru = 0.75 ´ p/4d2 ´ 228 / 1000
= 0.1343 d2
å 2
10
d
vTu
848231
5137520245
,
.., ´
å 2
10
d
hTu
848231
100520245
,
., ´
Shear For Rivet 10
Eccentric Shear
H10 = = = 26.81 kN
V10 = = = 19.50 kN

Total horizontal force = 62.65 + 26.81
= 89.46 kN
Total vertical force = 22.80 + 19.50
= 42.30 kN
Resultant force = 22
30424689 .. +
Rivet Diameter
The rivet diameter may be found by equating the
rivet value to the shear force in the most heavily
stressed rivet, as under:
= 98.96 kN

0.1343 d2 = 98.96
d = 27.15 mm
Use 28 mm diameter rivets
Check Fastener Bearing Strength
This check is performed if the plate thicknesses
are given. The bolt bearing value must be
greater than its shear value.

Steel strucure lec # (18)

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