Bolts
Chapter 20 – add-on
Material taken from Shigley, 1986, Standard
Handbook of Machine Design, and Mott, 2003,
Machine Elements in Mechanical Design
Bolted Joints
™ Joints
are an extremely important part of any
structure.
™ Whether held together by bolts or rivets or
weldmelts or adhesives or something else, joints
make complex structures and machines possible.
™ Bolted joints, at least, also make disassembly and
reassembly possible.
Types of Shear Joints
Shear joints are found almost exclusively in structural
steel work.
™ Such joints can be assembled with either rivets or bolts.
™ Since the early 1950s, bolts have steadily gained in
popularity.
™ Two basic types of joint are used, lap and butt, shown in
the next slide.
™ These are further defined as being either (1) friction-type
joints, where the fasteners create a significant clamping
force on the joint and the resulting friction between joint
members prevents joint slip, or (2) bearing-type joints,
where the fasteners act as points to prevent slip.
™
1
Shigley, 1986, Standard Handbook of Machine Design
Types of Shear Joints con’t
™ Only
bolts can be used in friction-type joints,
because only bolts can be counted on to develop
the high clamping forces required to produce the
necessary frictional resistance to slip.
™ Rivets or bolts can be used in bearing-type joints.
Allowable-Stress Design Procedure
In the allowable-stress design procedure, all fasteners in
the joint are assumed to see an equal share of the applied
loads.
™ Empirical means have been used to determine the
maximum working stresses which can be allowed in the
fasteners and joint members under these assumptions.
™ A typical allowable shear stress might be 20% of the
ultimate shear strength of the material.
™ A factor of safety (5:1) has been incorporated into the
selection of allowable stress.
™
2
Shigley, 1986, Standard Handbook of Machine Design
Shigley, 1986, Standard Handbook of Machine Design
Bearing-Type Joints
™A
successful bearing-type joint must size the parts
so that the fasteners will not shear, the joint plates
will not fail in tension nor be deformed by bearing
stresses, and the fasteners will not tear loose from
the plates.
™ Stresses (τ) within a rivet are:
τ=
F
bmAr
Where F = total force, b = # of shear planes, m = # of bolts and Ar = shank area
3
Bearing-Type Joints con’t
™ The
shear stress within each bolt in the joint will
be:
τ=
F
AT
AT = total area of m bolts
™A
bolt can have different cross-sectional areas. If
the plane passes through the unthreaded body of
the bolt, the area is simply:
AB =
πd
4
2
Bearing-Type Joints con’t
™ If
the shear plane passes through the threaded
portion of the bolt, the cross-sectional area is
considered to be the tensile-stress area of the
threads and can be found by:
2
n = threads per inch
0.9743 ⎞
π⎛
™ Unified:
As = ⎜ d −
⎟
4⎝
n ⎠
2
π
As = (d − 0.9382P )
4
™ Metric:
Bearing-Type Joints con’t
™ An
example based on Fig. 23-2: the bolts are
ASTM A325 steel, m = 5 bolts, F = 38 250 lb
(170.1 kN), d = ¾ in (19.1 mm), n = 2 (one
through the body of each bolt, one through the
thread).
™ The total cross-sectional area through the bodies of
all 5 bolts and then the threads is:
2
5π(0.75)
= 2.209in 2(1425mm2)
4
2
5π ⎛
0.9743 ⎞
2
2
5As =
⎜ 0.75 −
⎟ = 1.757in (1133mm )
4 ⎝
2 ⎠
5AB =
4
Shigley, 1986, Standard Handbook of Machine Design
Shigley, 1986, Standard Handbook of Machine Design
Bearing-Type Joints con’t
™ The
τ=
shear stress in each bolt will be:
F
38250
=
= 9646psi(66.5MPa)
AT 2.209 + 1.757
τ=
F
38250
=
= 9646psi(66.5MPa)
AT 2.209 + 1.757
which is well within the shear stress allowed
for A325 steel bolts.
5
Bearing-Type Joints con’t
Stress in the Plate: to compute the tensile
stress in the plates, first compute the crosssectional area of a row containing the most bolts.
™ With references to Figs. 23-2 and 23-3, that area
will be:
™ Tensile
σB =
F
mdlG
A = 0.75(1.5) + 0.75(3) + 0.75(1.5) = 4.5 in2 (2903 mm2)
™ The
stress in 2 such cross-sections (2 splice plates)
will be:
F 38250
σ=
A
=
(4.5)2
= 4250psi(29.3MPa)
Bearing-Type Joints con’t
™ These
plates will not fail; the stress level in them is
well within the allowable tensile-stress value of
21.6 kpsi for A36 steel.
™ Bearing Stresses on the Plates. If the fasteners
exert too great a load on the plates, the latter can
be deformed; bolt holes will elongate, for example.
Shigley, 1986, Standard Handbook of Machine Design
6
Bearing-Type Joints con’t
™σB
=
F
where lG = 2.25 inch , m = 5, and
m * d * lG d = 0.75 in.
™ Then,
σB =
38250
= 4533psi(31.3MPa)
5(0.75)( 2.25)
™ Note
that the allowable bearing stresses listed in
Table 23-1 are greater than the allowable shear
stresses for the same plate material.
Bearing-Type Joints con’t
Stress. Finally, the designer should
determine whether or not the fasteners will tear out
of a joint plate, as in the lap joint shown in the next
slide.
™ In the example shown, there are 6 shear areas.
™ The shear stress in the tearout sections will be:
™ Tearout
τ=
100000
= 11111psi(76.6MPa)
6(0.75)( 2)
Shigley, 1986, Standard Handbook of Machine Design
7
Bearing-Type Joints con’t
Joints. Design a friction-type joint
using the same dimensions, materials, and bolt
pattern as in Fig 23-1, but this time preloading the
bolts high enough so that the friction forces
between joint members (between the so-called
faying surfaces) become high enough to prevent
slip under the design load.
™ Friction-Type
Bearing-Type Joints con’t
™
Computing Slip Resistance. To compute the slip
resistance of the joint under a shear load, use the
following expression:
RS = µS * FPA * b * m
Where, µs = slip coefficient, and FPA = average pre-load
Note that engineering specifications published by the
AISC and others carefully define and limit the joint
surface conditions that are permitted for structural steel
work involving friction-type joints.
™ In most cases, they are not painted or are the surfaces
polished or lubricated, since these treatments would alter
the slip coefficient.
™
Shigley, 1986, Standard Handbook of Machine Design
8
Bearing-Type Joints con’t
™ To
continue the example, assume that the joint
surfaces will be grit blasted before use, resulting in
an anticipated slip coefficient of 0.493.
™ Now, estimate the average preload in the bolts.
™ Assume that there is an average preload of 17 kip
in each of the 5 bolts in the joint.
™ Compute the slip resistance as:
RS = µSFPAbm = 0.493(17000)( 2)(5)
= 83810lb(373kN)
Bearing-Type Joints con’t
™ Comparing
Slip Resistance to Strength in Bearing.
The ultimate strength of a friction-type joint is
considered to be the lower of its slip resistance or
bearing strength.
™ To compute the bearing strength, use the same
equation used earlier.
™ This time, enter the allowable shear stress for each
material and then compute the force which would
produce that stress.
Bearing-Type Joints con’t
™ These
forces are computed separately for the
fasteners, the net section of the plates, the fasteners
bearing against the plates, and tearout.
™ The least of these forces is then compared to the
slip resistance to determine the ultimate design
strength of the joint.
9
Eccentrically Loaded Bolted Joints
™ The
next figure shows an example of an
eccentrically loaded bolted joint.
™ The motor on the extended bracket places
the bolts in shear because its weight acts
directly downward.
™ But there also exists a moment equal to
P * a that must be resisted.
™ The moment tends to rotate the bracket
and to shear the bolts.
Mott, 2003, Machine Elements in Mechanical Design
Eccentrically Loaded Bolted Joints
con’t
™ The
basic approach to the analysis and
design of eccentrically loaded joints is to
determine the forces that act on each bolt
because of all the applied loads.
™ Then, by a process of superposition, the
loads are combined vectorially to
determine which bolt carries the greatest
load.
™ That bolt is then sized.
10
Mott, 2003, Machine Elements in Mechanical Design
Mott, 2003, Machine Elements in Mechanical Design
Mott, 2003, Machine Elements in Mechanical Design
11
Mott, 2003, Machine Elements in Mechanical Design
Mott, 2003, Machine Elements in Mechanical Design
12