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Bolts

Chapter 20 – add-on

Material taken from Shigley, 1986, Standard Handbook of Machine Design, and Mott, 2003, Machine Elements in Mechanical Design

Bolted JointsJoints 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 JointsShear 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.

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Shigley, 1986, Standard Handbook of Machine Design

Types of Shear Joints con’tOnly 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 ProcedureIn 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.

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Shigley, 1986, Standard Handbook of Machine Design

Shigley, 1986, Standard Handbook of Machine Design

Bearing-Type JointsA 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:

rbmAF

=τ

Where F = total force, b = # of shear planes, m = # of bolts and Ar = shank area

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Bearing-Type Joints con’tThe shear stress within each bolt in the joint will be:

A bolt can have different cross-sectional areas. If the plane passes through the unthreaded body of the bolt, the area is simply:

TAF

=τ

4

2dABπ

=

AT = total area of m bolts

Bearing-Type Joints con’tIf 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:Unified:

Metric:

29743.04

⎟⎠⎞

⎜⎝⎛ −

π=

ndAs

( )2

9382.04

PdAs −π

=

n = threads per inch

Bearing-Type Joints con’tAn 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:

)1133(757.12

9743.075.04

55

)1425(209.24

)75.0(55

222

222

mminA

mminA

s

B

=⎟⎠⎞

⎜⎝⎛ −

π=

=π

=

5

Shigley, 1986, Standard Handbook of Machine Design

Shigley, 1986, Standard Handbook of Machine Design

Bearing-Type Joints con’tThe shear stress in each bolt will be:

which is well within the shear stress allowed for A325 steel bolts.

)5.66(9646757.1209.2

38250 MPapsiAF

T=

+==τ )5.66(9646

757.1209.238250 MPapsi

AF

T=

+==τ

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Bearing-Type Joints con’tTensile Stress in the Plate: to compute the tensile stress in the plates, first compute the cross-sectional area of a row containing the most bolts.With references to Figs. 23-2 and 23-3, that area will be: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:

GB

mdlF

=σ

)3.29(42502)5.4(

38250 MPapsiAF

===σ

Bearing-Type Joints con’tThese 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

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Bearing-Type Joints con’twhere lG = 2.25 inch , m = 5, and d = 0.75 in.

Then,

Note that the allowable bearing stresses listed in Table 23-1 are greater than the allowable shear stresses for the same plate material.

GB

ldmF

**=σ

)3.31(4533)25.2)(75.0(5

38250 MPapsiB ==σ

Bearing-Type Joints con’tTearout 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:

)6.76(11111)2)(75.0(6

100000 MPapsi==τ

Shigley, 1986, Standard Handbook of Machine Design

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Bearing-Type Joints con’tFriction-Type 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.

Bearing-Type Joints con’tComputing Slip Resistance. To compute the slip resistance of the joint under a shear load, use the following expression:

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.

mbFR PASS ***µ=Where, µs = slip coefficient, and FPA = average pre-load

Shigley, 1986, Standard Handbook of Machine Design

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Bearing-Type Joints con’tTo 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:

)373(83810)5)(2)(17000(493.0

kNlbbmFR PASS

==µ=

Bearing-Type Joints con’tComparing 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’tThese 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.

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Eccentrically Loaded Bolted JointsThe 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.

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Mott, 2003, Machine Elements in Mechanical Design

Mott, 2003, Machine Elements in Mechanical Design

Mott, 2003, Machine Elements in Mechanical Design

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Mott, 2003, Machine Elements in Mechanical Design

Mott, 2003, Machine Elements in Mechanical Design