Tension Joints: Worked Example

Tension joints
Worked example
Tension joints
Clamping force is called the pretension or bolt preload.

It exists in the connection after the nut has been
properly tightened no matter whether the external
tensile load P is exerted or not.
The clamping force produces tension in the bolt

inducing compression in the members.
The grip l of a connection is the total thickness of the

clamped material, including the washers.
Tension joints
The stiffness of the portion of a bolt or screw within the
clamped zone can be treated as two springs in series:
1 1 1 𝑘1 𝑘2
= + or 𝑘=
𝑘 𝑘1 𝑘2 𝑘1 +𝑘2
The spring rates of the threaded and unthreaded portions

of the bolt in the clamped zone are, respectively,
𝐴𝑡 𝐸 𝐴𝑑 𝐸
𝑘𝑡 = 𝑘𝑑 =
𝑙𝑡 𝑙𝑑
Where At = tensile-stress area (see Table)

lt = length of threaded portion of grip
Ad = major-diameter area of fastener
ld = length of unthreaded portion in grip
Tension joints
• Then the estimated effective stiffness kb of the bolt or cap screw in
the clamped zone is
𝐴𝑑 𝐴𝑡 𝐸
𝑘𝑏 =
𝐴 𝑑 𝑙𝑡 + 𝐴 𝑡 𝑙𝑑
• The members of the clamped can also be treated as compressive

springs in series
1 1 1 1
= + + ⋯+
𝑘𝑚 𝑘1 𝑘2 𝑘𝑛
Tension joints
The pressure distribution at the

member interface can be
approximated by a pressure-cone.
Osgood* reports a range of 25◦ ≤ α ≤

33◦ for most combinations.
Compression of a member with the equivalent
elastic properties represented by a frustum of a
Using α = 30° covers most cases hollow cone. Here, l represents the grip length.
*C. C. Osgood, “Saving Weight on Bolted Joints,” Machine Design, Oct. 25, 1979
Tension joints
The spring rate or stiffness and contraction 𝛿 due to a compressive force P are related by
𝑃 𝜋 𝐸 𝑑 𝑡𝑎𝑛𝛼
𝑘= =
𝛿 2𝑡 𝑡𝑎𝑛𝛼 + 𝐷 − 𝑑 𝐷 + 𝑑
𝑙𝑛
2𝑡 𝑡𝑎𝑛𝛼 + 𝐷 + 𝑑 𝐷 − 𝑑
With α = 30
0.5774𝜋𝐸𝑑
𝑘=
1.155𝑡 + 𝐷 − 𝑑 𝐷 + 𝑑
𝑙𝑛
1.155𝑡 + 𝐷 + 𝑑 𝐷 − 𝑑
Use these equations to separately solve for each frustum in the joint. Then individual stiffnesses
are assembled to obtain km
An exponential curve-fit of the form, 𝑘𝑚 = 𝐸 𝑑 𝐴 𝑒𝑥𝑝 𝐵𝑑/𝑙

Tension joints
To compute the external load P, let
Fi = preload
P = external tensile load
Pb = portion of P taken by bolt
Pm = portion of P taken by members
Fb = Pb + Fi = resultant bolt load
Fm = Pm − Fi = resultant load on members
C = stiffness constant of the joint
Tension joints
The load P is tension, and it causes the connection to
stretch, or elongate, through some distance δ
Recalling that k is the force divided by the deflection,

then
𝑃𝑏 𝑃𝑚
𝛿= 𝑎𝑛𝑑 𝛿=
𝑘𝑏 𝑘𝑚
𝐾𝑚
Or 𝑃𝑚 = 𝑃𝑏
𝑘𝑏
Tension joints
Since P = Pb + Pm
𝑘𝑏 𝑃
𝑃𝑏 = = 𝐶𝑃
𝑘𝑏 + 𝑘𝑚
And
𝑃𝑚 = 𝑃 − 𝑃𝑏 = 1 − 𝐶 𝑃
Where
𝑘𝑏
𝐶=
𝑘𝑏 + 𝑘𝑚
Resultant bolt load is
Fb = Pb + Fi = C P + Fi Fm < 0
Resultant connected members load is

Fm = Pm − Fi = (1 − C)P − Fi Fm < 0
Statically Loaded Tension Joint with Preload
• Tensile stress in the bolt can be found as
𝐶𝑛𝑃 𝐹𝑖
𝜎𝑏 = + = 𝑆𝑝
𝐴𝑡 𝐴𝑡
• The limiting value of σb is the proof strength Sp

• A load factor n or safety fact or can be introduced
• Any value of n > 1 ensures that the bolt stress is less than the proof strength
Statically Loaded Tension Joint with Preload
𝑆𝑝 𝐴𝑡
Yield factor of safety guarding against exceeding Sp 𝑛𝑝 =
𝐶𝑃+𝐹𝑖
𝑆𝑝 𝐴𝑡 −𝐹𝑖
Load factor of safety guarding against overloading 𝑛𝐿 =
𝐶𝑃
𝐹𝑖
Load factor for joint separation 𝑛0 =
𝑃(1−𝐶 ሻ
0.90 𝐴𝑡 𝑆𝑝 𝑓𝑜𝑟 𝑝𝑒𝑟𝑚𝑎𝑛𝑒𝑛𝑡 𝑗𝑜𝑖𝑛𝑡

Fi can be determined as: 𝐹𝑖 = ൝
0.75 𝐴𝑡 𝑆𝑝 𝑓𝑜𝑟 𝑛𝑜𝑛𝑝𝑒𝑟𝑚𝑎𝑛𝑒𝑛𝑡 𝑗𝑜𝑖𝑛𝑡
SP is the proof strength and can be found in tables 8 -11

Example
The figure illustrates the connection of a
cylinder head to a pressure vessel using 10
bolts and a confined-gasket seal. The
effective sealing diameter is 150 mm.
Other dimensions are: A = 100, B = 200, C =
300, D = 20, and E = 20, all in millimetres.
The cylinder is used to store gas at a static
pressure of 6 MPa.
ISO class 8.8 bolts with a diameter of 12 mm
have been selected. This provides an
acceptable bolt spacing.
What load factor n results from this
selection?
Cylinder head is steel; cylinder is
grade 30 cast iron
Example
If a full gasket is present in the joint, the

gasket pressure p is found by dividing the
force in the member by the gasket area
per bolt.
For 10 bolts, the external tensile load per
bolt is
𝐴𝜎 1 𝜋
𝑝= = 1502 6 10−3
𝑁 10 4
= 10.6 𝑘𝑁
Example
Bolt grip length, LG = D + E = 40 mm
ISO class 8.8 bolts with ∅12 mm selected; from
table for d = 12 mm, H = 10.8 mm
Recall
Then
LT = 2D + 6 = 2(12) + 6 = 30 mm
Total grip length LG + H = 40 + 10.8 = 50.8 mm
Dimensions of Hexagonal Nuts
Example
• Round up from Table A-17
Since LG + H = 50.8 mm
Bolt length L ≈ 60mm
• Length of useful unthreaded portion:

ld = L – LT = 60 – 30 = 30 mm
Table A–17
• Length of threaded portion:

lt = l – ld = 40 - 30 = 10 mm
• Area of unthreaded portion Preferred Sizes (When a choice can be made, use
one of these sizes; however, not all parts or items
𝜋𝑑 4 𝜋 12 2
𝐴𝑑 = = = 113 𝑚𝑚2 are available in all the sizes shown in the table.)
4 4
Example
• Area of unthreaded portion
𝜋𝑑 4 𝜋 12 2
𝐴𝑑 = = = 113 𝑚𝑚2
4 4
• Area of threaded portion (from Table):
𝐴𝑡 = 84.3 𝑚𝑚2
• Effective stiffness kb of the bolt:

𝐴𝑑 𝐴𝑡 𝐸
𝑘𝑏 =
𝐴𝑑 𝑙𝑡 + 𝐴𝑡 𝑙𝑑
113(84.3ሻ(207ሻ
= = 504.18 𝑀𝑁/𝑚
113 10 + 84.3(30ሻ
Example
• Area of unthreaded portion
𝜋𝑑 4 𝜋 12 2
𝐴𝑑 = = = 113 𝑚𝑚2
4 4
• Area of threaded portion (from Table):
𝐴𝑡 = 84.3 𝑚𝑚2
• Effective stiffness kb of the steel bolt (E = 207 GPa):
𝐴𝑑 𝐴𝑡 𝐸 113(84.3ሻ(207ሻ
𝑘𝑏 = = = 504.18 𝑀𝑁/𝑚
𝐴𝑑 𝑙𝑡 + 𝐴𝑡 𝑙𝑑 113 10 + 84.3(30ሻ
Example
From the table of stiffness parameters of a
steel bolt
A = 0.78715, B = 0.62873 and E = 207 Gpa
From
𝑘𝑚 = 𝐸 𝑑 𝐴 𝑒𝑥𝑝 𝐵𝑑/𝑙
Stiffness Parameters of Various Member Materials
0.62873 12
= 207 12.0 0.78715 exp
40
= 2361 𝑀𝑁/𝑚
Joint members with same Young’s modulus E,
𝑘𝑠 = 2𝑘𝑚 = 4722 𝑀𝑁/𝑚
Example
From the table of stiffness parameters of the
cast iron casing
A = 0.778 71, B = 0.616 16, E = 100 GPa
From
𝑘𝑚 = 𝐸 𝑑 𝐴 𝑒𝑥𝑝 𝐵𝑑/𝑙
Stifness Parameters of Various Member Materials
061616 12
= 100 12.0 077871 exp
40
= 1124 𝑀𝑁/𝑚
Joint members with same Young’s modulus E,
𝑘𝑐𝑖 = 2𝑘𝑚 = 2248 𝑀𝑁/𝑚
Example
Total spring rate of the members,
1 1 1
= +
𝑘𝑚 𝑘𝑠 𝑘𝑐𝑖
𝑘𝑚 = 1523 𝑀𝑁/𝑚
Joint stiffness constant,

𝑘𝑏 504.18
𝐶= =
𝑘𝑏 + 𝑘𝑚 504.18 + 1523
= 0.2487
Example
𝐴𝑡 = 84.3 𝑚𝑚2
From table
𝑆𝑝 = 600 𝑀𝑃𝑎
𝐹𝑖 = 0.75 𝐴𝑡 𝑆𝑝
= 0.75 84.3 600 10−3
= 37.9 𝑘𝑁
Therefore load factor,
𝑆𝑝 𝐴𝑡 − 𝐹𝑖 600 10−3 84.3 − 37.9
𝑛𝐿 = =
𝐶𝑃 0.2487(10.6ሻ
= 4.8 ≈ 5
Example
The load P applied to a steel rod is distributed
to a timber support by an annular washer. The
diameter of the rod is 22 mm and the inner
diameter of the washer is 25 mm, which is
slightly larger than the diameter of the hole.
Determine the smallest allowable outer
diameter d of the washer, knowing that the
axial normal stress in the steel rod is 35 MPa
and that the average bearing stress between
the washer and the timber must not exceed 5
MPa
Example
Bolt tensile load
𝜋0.0222
𝑃 = 𝜎𝐴 = 35 × 106 = 13.305 × 103 𝑁
4
Required bearing area,
𝑃 13.305×103
𝐴𝑏 = = = 2.6609 × 10−3 𝑚2
𝜎𝑏 5×106
For washer do
𝜋
𝐴𝑏 = 𝑑𝑜2 − 𝑑𝑖2
4
Then,
𝑑𝑜 = 63.3 𝑚𝑚 di = 25 mm

Tension Joints: Worked Example

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Tension joints

Clamping force is called the pretension or bolt preload.

The clamping force produces tension in the bolt

The grip l of a connection is the total thickness of the

The spring rates of the threaded and unthreaded portions

Where At = tensile-stress area (see Table)

• The members of the clamped can also be treated as compressive

The pressure distribution at the

Osgood* reports a range of 25◦ ≤ α ≤

An exponential curve-fit of the form, 𝑘𝑚 = 𝐸 𝑑 𝐴 𝑒𝑥𝑝 𝐵𝑑/𝑙

Recalling that k is the force divided by the deflection,

Resultant connected members load is

• The limiting value of σb is the proof strength Sp

0.90 𝐴𝑡 𝑆𝑝 𝑓𝑜𝑟 𝑝𝑒𝑟𝑚𝑎𝑛𝑒𝑛𝑡 𝑗𝑜𝑖𝑛𝑡

SP is the proof strength and can be found in tables 8 -11

If a full gasket is present in the joint, the

• Length of useful unthreaded portion:

• Length of threaded portion:

• Effective stiffness kb of the bolt:

Joint stiffness constant,

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