MECH3460
Executive Summary
For the given problem, it became evident that the most appropriate and desirable results
according to the given criteria are as follows:
For the Bolt

The bolt grade that will be used will be 8.8

A bolt diameter of 6.8mm will be used, this gives rise to an M7 bolt

To ensure that the flange is safe but also minimal spacing occurs between the
truncated hollow cylinders, 22 bolts will be used

The maximum allowable number of bolts to be used is 23.49 bolts

A preload of 0.75 will be chosen for the typical reusable connection.

The Bolts will have a required grip length of 14.3mm this will ensure a desirable
flange width and in turn will produce desirable results for the flange stiffness.
Flange Dimensions
According to the given diagram in the criteria which can be seen on page 12, the
flange will have the following dimensions:

A = 3.84mm the thickness of the cylinder, determined using the thin wall theory

B = 11.13mm distance from flange width to the edge of cylinder wall

C = 18.42mm flange width

D = 7.15mm which is determined according to the grip length

E = 7.15mmn also determined according to the grip length
Cost

The overall cost will be $72.45 this includes:
 $8.05 Flange costs
 $10.40 Bolt costs
 $54 Drilling and handling costs.
1|Page
MECH3460
Given Criteria

A cylinder is to be attached to a crankcase

The base of each cylinder is to have a flange through which it will be bolted

The pressure in the cylinder is considered to cycle between 16MPa and atmospheric
pressure every second revolution

The operating speed is to be 2100rpm

The bore of the cylinder is given to be 120mm
Problem

Propose a number, grade, preload, and sizes of bolts

Propose appropriate dimensions (A...E) for the flanges shown on the diagram.
Designers comment
It is important to understand that with a problem such as the one proposed here, there is
no right or wrong answer; there is simply a safe or unsafe answer. Even with the safe
answer, there are many possibilities that can be put forth, the one that will be more
desirable than other answers will be highly affected by the cost, grade, simplicity and
availability of the bolts rather than the potential.
As an engineer, it then becomes your responsibility to determine the safer or rather
cheaper approach of such a problem and this may take hours even days to determine
mainly because many approaches are possible each giving a different answer and different
‘correct’ values. The question then becomes which is the more desirable design.
2|Page
MECH3460
Calculation of Proof, Gerber and Goodman lines
In the given criteria the bolts may be of grades 6.8, 8.8, 9.8, 10.9 or other. For
our problem we will first begin by allowing the solutions to initially rotate about
a bolt grade of 8.8.
Hence the Su, Sy, Se and Sp can be determined and calculated for bolt grade of
8.8
Min Tensile strength (N/mm2)
Min yield strength (N/mm2)
Endurance Strength (N/mm2)
Min Proof strength (N/mm2)
Su
Sy
Se
Sp
830
660
111.6682
564.4
Lecture notes
Lecture notes
0.155*rel*Su
Su*0.8*0.85
Lecture notes
Table 1
The Proof Line
The proof line is a straight line which joins Sp on the mean stress axis to Sp on
the fatigue axis.
Sa (alternating stress)
0
564.4
Sm (mean stress)
564.4
0
Lecture Notes
Equation 24
The Goodman Line
The Goodman line is straight and joins (0,Se) to (Su,0) on the fatigue diagram.
Sa (alternating stress)
0
111.7
Sm (mean stress)
830
0
Lecture Notes
Equation 26
The Gerber Line
The Gerber line is a curve given by equation 25 in lecture notes. For a factor of
safety (FS) of 1, the Gerber line represents a reasonable mean fit through the
data, indicating a 50% likelihood of fatigue failure over which the fatigue
strength Se has been evaluated.
Lecture Notes
Equation 25
3|Page
MECH3460
Calculate Gerber Line
Gerber Line
Sm
830
787.4071
742.3746
694.4278
642.9152
586.8986
524.9381
454.6097
371.1873
262.469
Se/FS =
0
Sa
0
10.47211
20.94422
31.41633
41.88844
52.36055
62.83266
73.30477
83.77688
94.24899
104.7211
Where Sm =(SQRT(1-Sa/Se))*(Su)
Calculation of Safe Gerber Line
The ‘safe’ Gerber line is proportional to the original line but its intercepts
at both axes are reduced by the factor of safety as the FS is above 1. A FS of
1.1 was chosen.
Calculate Safe
Gerber Line
Se/FS
Sm
754.5454545
713.8171831
670.6199361
624.4415318
574.5636209
519.9225734
458.8198015
388.2160972
301.5093133
176.3632442
34.67070564
0
Sa
0
10
20
30
40
50
60
70
80
90
95
95.201
Where Sm =(SQRT(1-FS*Sa/Se))*Su/FS
Shigley p337
rel = 0.868
for 95%
The Safe line, Goodman line, Gerber line and safe Gerber line can now all be
plotted to obtain a graph. In order to ensure that the bolts will take the load
should one fail, a reliability factor that predicts 95% or more for any one bolt to
not fail.
f
4|Page
MECH3460
Fatigue diagram-8.8
200
Alternating Stress (MPa)
Proof
Goodman
100
Gerber
Safe Gerber
0
0
200
400
600
800
1000
Mean Stress(MPa)
Through the use of the above Graph a rough determination of whether the
bolts will fail can be derived.
If the point of a specific preload, diameter, and grip length of a bolt lies to the
left of the proof line and also beneath or on the Safe Gerber then the bolt is
deemed safe and desirable.
Calculating mean and alternating stresses
In order to calculate the mean bolt stress and alternating bolt stress it is vital to
first obtain the max and min bolt stress and also the max and min force per
bolt.
This is done as follows:
Bolt preload force N
Fi
Max Force Per bolt
Min Force Per bolt
Max Bolt Stress
Min Bolt Stress
Mean Bolt Stress
alt Bolt Stress
Fmax
Fmin
smax
smin
smean
salt
4644.530579
5636.477119
4644.530579
560.6707469
462
511.3353735
49.33537347
pr*Sy*As
Fi+P*rat/n
Fi
Fmax/As
Fmin/As
(smax+smin)/2
(smax-smin)/2
Where the Bolt preload force is dependent on the preload (0.75) and also the
thread stress area.
5|Page
MECH3460
The solution for this problem can be determined in one of two ways, a simple approach
whereby only the bolt diameter is varied and all the other parameters are kept constant and
are selected by the designer, making this problem a simple mathematical problem. The
second approach is a more complicated approach and requires the designer to take into
consideration all aspects of the design as all parameters have the ability to change and
nothing is kept constant.
The Simple Approach
In this approach it is the designer’s job to predetermine all parameters of the bolted joint
except for the bolt diameter and so in turn also the flange width.
The bolt diameter can be determined by graphing the appropriate parameters of the bolt on
the Goodman graph and from there we can see what a recommended diameter is as it
would remain below the Safe Gerber curve and to the left of the proof line.
For this approach the parameters which will be chosen to remain constant are as follows.
Grip Length = 8mm
Grade of bolt = 8.8
Preload = 0.7
Number of bolts = 6 bolts
GL = 8mm
Bolt Grade = 8.8
Preload = 0.7
# of bolts = 6
These parameters are simply designer’s choice and are used to obtain a required safe
diameter for a bolt. Once these have been determined then using these values, the
corresponding mean and alternating bolts stress can be drawn and graphed on the
Goodman Graph for a range of different bolt diameters.
Grip
Length
8mm
8mm
8mm
8mm
8mm
8mm
8mm
8mm
8mm
8mm
8mm
Preload
No Of Bolts
Diameter
0.7
0.7
0.7
0.7
0.7
0.7
0.7
0.7
0.7
0.7
0.7
6
6
6
6
6
6
6
6
6
6
6
5mm
6mm
7mm
8mm
9mm
10mm
11mm
12mm
13mm
14mm
15mm
S(mean)
MPa
676.2663
619.85
583.29
558.204
540.2135
526.86
516.67
508.73
502.4
497.28
493.0789
S(alt)
MPa
214.266
157.85
121.297
96.204
78.2135
64.86
54.68
46.73
40.4
35.28
31.08
6|Page
MECH3460
Fatigue diagram
200
Proof
Alternating Stress (MPa)
Goodman
Gerber
Safe Gerber
100
Diameter effect
0
0
200
400
600
Mean Stress (MPa)
800
1000
Desired Diameter for 6 Bolts is >=12mm
It can be seen from the above graph that all bolts that have a diameter of 12 or more lay
below the safe Gerber and to the left of the Proof Line, hence proving to be safe.
This exact same approach can be repeated for a different variable whereby the diameter is
set to be constant along with other parameters and the grip length for example is taken to
vary.
Grip Length = ?
Number of bolts = 6 bolts
Preload
0.7
0.7
0.7
0.7
0.7
0.7
0.7
No of
Bolts
6
6
6
6
6
6
6
Preload = 0.7
Bolt Diameter = 12mm
Grade of bolt = 8.8
Bolt Diameter
Grip Length
12mm
12mm
12mm
12mm
12mm
12mm
12mm
4mm
6mm
8mm
10mm
12mm
14mm
16mm
S(mean)
MPa
514.77
512.566
510.56
508.72
507.044
505.5
504.05
Diam =12mm
Bolt Grade = 8.8
Preload = 0.7
# of bolts = 6
S(alt)
MPa
52.77
50.566
48.56
46.72
45.044
43.49
42.05
7|Page
MECH3460
Desired Grip Length >= 8mm
It can be seen from the above graph that all bolts that have a grip length of 8mm or more
lay below the safe Gerber and to the left of the Proof Line, hence proving to be safe.
And hence this approach, although simple can prove to be quite reliant however it cannot
solve more complicated problems where more than one variable are changing.
8|Page
MECH3460
The Complicated approach
The complicated approach involves one whereby all variables change, this includes the
number of bolts, their diameter, grade and grip length along with flange dimensions.
In order to calculate the appropriate bolt diameters, many variables are to begin changing
and others are to remain constant.
The first approach is to set a point on the safe Gerber line at smn, this will be our “saFS”
SaFS = (1-((FS*smn/Su))*(Se/FS)
We also need to obtain our excess fatigue strength, and this will be denoted as “eSf”
eSf = saFS - salt
Now we select a grade to be examined, our first grade to be examined will be 8.8, for grade
8.8 our Su, Sy, Sp and Se will all change accordingly.
For grade 8.8 we will examine the effect of changing one variable, then two variables, then
three variables for 2 bolts, 3 bolts ... 10 bolts, which means we first select to examine the
Refer to
appendix 2
effect of the variables on two bolts, these variables will be the diameter, grip length and
Preload, by changing them one by one, then two by two, then all three using solver.
After this process was complete it showed that the solutions obtained by solver were highly
dependent upon the variables and their values placed into the program. Hence this means
that the solution of the problem is highly dependent on an educated guess.
For a grade of 8.8 and for 5+ bolts solver solutions gave us a grip length of 8mm and a
diameter of 12mm which seemed very biased as only the preload varied. Hence this way of
solving the problem was deemed unfit by the designer.
A much more reliable hence would be the one that is highlighted in appendix 3.
A comparison of grades was performed with different preloads. Using solver we were able
to obtain the optimum diameter, grip length and number of bolts for each preload at each
grade. The way this was performed was simply by inserting the value 1 for the diameter and
grip length and hence this allowed the program to have a larger scale to work with and so
the output would be desirable and this was clearly apparent in our results.
A quick snap shot of the solver variables is illustrated below and reflects the value for just
one value.
9|Page
Refer to
appendix 3
MECH3460
Here:

Diameter is to be greater than or equal to zero

Grip length also greater than or equal to zero

N being number of bolts was set to 10 for this specific example

Preload also set to 0.7 for this example

eFs was set to zero
Reason eFs is set to zero is simple to set up the analysis in a way whereby the bolt
diameter varies to arrive at the diameter for which the excess fatigue strength is
minimised to practically zero.
Hence for a reliability factor of 95% grades 8.8, 9.8 and 10.8 were all compared.
For each grade a preload of 0.7, 0.75, 0.8, and 0.85 were all compared using solver
For each preload bolts ranging from 6 to 34 bolts were all compared
Finally the grip length and diameter for each of these bolts, at each preload and grade were
all taken into account and the best was chosen.
Refer to appendix 3 for a large comparison of all the bolts and their grades and preloads.
10 | P a g e
CHOSEN
VARIABLES
Diameter
=6.8mm
Grip Length
=14.3mm
# of Bolts =22
Preload =0.75
Grade =8.8
MECH3460
Reason for Design Selection
The above values for the bolt were selected above others for the following reasons:

Grade 0.75 was selected above 0.7 because in the criteria it is suggested that a
preload of 75% of proof stress be used for the typical reusable connection and up to
90% for those joints that will remain permanent connected or rebuilt very few times.

Some specific bolts passed that had a grip length smaller than the bolt diameter this
is seen unreasonable as the grip length will soon determine the flange width and so
a larger grip length is seen more desirable.

Grade 8.8 was selected over 9.8 and 10.8 mainly because for high grades it was seen
that the diameter and griplength began to have minimal effect on the bolt mean
stress, hence the points remained to the right of the proof line but would at times
come below the safer gerber curve which wasn’t good enough.

Cost was a major consideration in selecting the required design along with all the
criteria mentioned above. In appendix 3, the cost of all bolts that passed were
calculated and the one that was cheapest was selected however the cheapest was
for a preload of 0.7, hence the second cheapest was selected with a preload of 0.75.

The maximum number of bolts was also taken into consideration, this number was
also calculated and in appendix 3 was also placed near all the bolts that passed,
however some bolts that required 34 bolts had a max bolt of 27 and so deemed unfit
for the design, hence this also played a major role in selection criteria.
Flange Dimensions
11 | P a g e
Refer to
appendix 3
MECH3460
For the above Dimensions
1. The grip length that has already been found represents D + E
As the designer the optimal strength will occur at the point where D = E
GL = D + E
and
D=E
GL = 14.3mm
2D = 14.3mm
D = 7.15mm
therefore
E = 7.15mm
2. The Flange width (C)
In order to calculate the flange width we need to use trigonometry and sketch the
hollow cylindrical tube and the compare this value to the one derived through our
spreadsheet Appendix 4.
The Flange width hence is equal to:
Flange width = washer diameter + 2x(7.12tan30)
Washer Diameter = 1.5 x 6.8 = 10.2mm
Flange width = 10.2 + 2x(7.12tan30)
Flange Width = 18.42mm
Hence C = 18.42
12 | P a g e
MECH3460
3. The thickness A can be determined as follows:
(for a cylinder)
Hence the thickness will be:
𝑡=
𝑃𝑟
𝜎
And it is known that the pressure in the cylinder is 16MPa = P
And the radius is given as 60mm = r
Finally the stress is 250MPa = 𝜎 (for steel)
Hence 𝑡 =
16𝑥60
A = 3.84mm
250
B = 11.13mm
t = 3.84mm
C = 18.42mm
hence A = 3.84mm
D = 7.15mm
4. By observation:
E = 7.15mm
B = (C + A)/2
= (18.42 + 3.84)/2
=11.13mm
HENCE:
13 | P a g e
MECH3460
The figure below shows sections across the flange and along the flange through the bolt
centrelines.
The truncated hollow cones (frustums), are a means whereby they can be used to estimate
flange stiffness. It is also important that a decision be made, that determines how close
these bolts are allowed to come close to each other, and also allow the variable S to be
greater than 0. These truncated cones along with the grip length GL determine the
minimum width and thickness of the flanges.
Calculation Of S

From the above tructated hollow cylinders it is vital to obtain the value for S, which
is the distance that separates the edges of frustums.

At these specific points the pressure is equal to zero

They should be minimal but also greater than zero
Firstly the circumference of the major cylinder:
𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 2𝜋r
= 2 x 60 x 3.14
= 376.9911mm
Now we chose to have 22bolts which means 22 times the flange width
=373.21mm
And this means that there should be 22 spaces between the bolts, or 22S
14 | P a g e
MECH3460
Hence:
22S = 376.9911-373.21
S = 0.17mm
S = 0.17186mm
Which is extremely small, hence is a reasonable distance to have between the
frustums. The pressure at these points is zero and so there is a minimum amount of
space where the pressure is zero meaning the pressure I quite distributed.
Cost and max number of bolts
The cost and max number of bolts can only be determined once the bolt diameter, grip
length, preload and grade have all be obtained. For our specific results we can
Refer to
appendix 4
determine these results.
Flange cost = volume of flange x 0.07 = $8.05
Bolt cost = Vb*0.14*(Su*Su/100000-Su/100+3.6) = $10.40
Drilling and handling cost = 10 + 2n = $54
Total Cost = sum of above 3 = $72.45
The max number of bolts
This was determined by using the equation:
Max number of bolts = Lb/w = length of flange/width of flange = 23.49
Bolt Conclusion
Hence the final bolt parameters were as follows:
Number Of bolts (N) = 22Bolts
Max Number of Bolts = 23.49
Bolt Diameter (Db) = 6.8mm
Cost = $72.50
Grip Length (GL) = 14.3mm
Bolt of M7
Preload (pr) = 0.75
15 | P a g e
MECH3460
Examination Of a Range of Preloads on Gerber Parabola
When examining a range of preloads there is again a number of ways by which this process
can be performed. The simpler way is for the designer to set specific values for the bolt
parameters and then allow the preload to have many values and compare the results.
No of Bolts
Bolt
Diameter
12
12
12
12
12
12
12
12
12
6
6
6
6
6
6
6
6
6
Grip
Length
8
8
8
8
8
8
8
8
8
Preload
S(mean)
S(alt)
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
180.56
246.56
312.56
378.56
444.56
510.56
576.56
642.56
708.56
48.56
48.56
48.56
48.56
48.56
48.56
48.56
48.56
48.56
From the above table it is evident that the diameter, grip length, # of bolts and grade are set
to be constants throughout.
Fatigue diagram Grade 8.8
200
Alternating Stress (MPa)
Proof
Goodman
Gerber
100
0
0
100
200
300
400
500
Mean Stress (MPa)
600
700
800
900
16 | P a g e
MECH3460
From the above diagram it can be clearly seen that:

For the given conditions a preload of <=0.7 is required

Preloads of 0.8+ fail as they are to the right of proof line and above safe Gerber.

Preload slightly affects the Bolt alternating stress

Preload greatly affects the bolt Mean Stress.
The second comparison which is slightly more difficult is to plot the preload chosen for our
solution (0.75) along with one below and one above and compare.
The graph on the next page is an illustration of a comparison of 3 different preloads (0.7,
0.75 and 0.8) for grade 8.8 bolts with the desired dimensions given
From the graph we can see that:

An increase in the preload increases the bolt mean stress

An increase in the preloads has almost no affect on the bolt alt stress

In order to be safe an increase in preload would require a dramatic increase in the
diameter and a decrease in the grip length

These would be unreasonable.
17 | P a g e
MECH3460
Examination Of a Range of Preloads on Deflection Diagram
In order to draw a force deflection diagram there are certain values that must be obtained,
these include:

Bolt stiffness

Flange stiffness

Preload force

Flange force

Bolt force
According to our calculations we have the following values:
For a preload of 0.75
Bolt force
Flange force
Bolt Preload force
Bolt stiffness (kb)
Flange stiffness (km)
Fmax
16495.23317
-8269.972406
14681.80509
5.39E+05
1.91E+06
18834.19
All of the following deflections depend heavily on the above values
𝜕𝑝 =
𝜕𝐵𝑖 =
𝐵𝑜𝑙𝑡 𝑃𝑟𝑒𝑙𝑜𝑎𝑑 𝐹𝑜𝑟𝑐𝑒
𝐵𝑜𝑙𝑡 𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠
𝜕𝑚𝑖 =
𝐵𝑜𝑙𝑡 𝑃𝑟𝑒𝑙𝑜𝑎𝑑 𝐹𝑜𝑟𝑐𝑒
𝐹𝑙𝑎𝑛𝑔𝑒 𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠
𝑁𝑒𝑤 𝐵𝑜𝑙𝑡 𝐹𝑜𝑟𝑐𝑒 − 𝐵𝑜𝑙𝑡 𝑃𝑟𝑒𝑙𝑜𝑎𝑑 𝐹𝑜𝑟𝑐𝑒
𝐵𝑜𝑙𝑡 𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠
From the above deflection equations it can be clearly seen that all deflections rely on the
bolt preload force, and also the Bolt preload force DEPENDS on the preload (pr).
𝐵𝑜𝑙𝑡 𝑃𝑟𝑒𝑙𝑜𝑎𝑑 𝐹𝑜𝑟𝑐𝑒 = 𝑝𝑟 × 𝑆𝑦 × 𝐴𝑠
18 | P a g e
MECH3460
Hence a change in the preload will result in a change in all deflections.
To draw our force deflection diagram it is vital to calculate our deflections first, as all our
required forces have been obtained.
Our chosen preload is 0.75, hence our deflections become:
For preload of 0.75
𝝏𝑩𝒊
𝝏𝒎𝒊
𝝏𝑩𝒊 + 𝝏𝒎𝒊
𝝏𝒑
2.72E-02mm
7.70E-03mm
3.49E-02mm
3.36E-03mm
Now that all values are obtained the following graph can be drawn
Now a comparison of different preloads, it is expected that a decrease in a preload for
example from 0.75 to 0.7 will result in:

A decrease in the bolt preload force and hence decreasing 𝜕𝑚𝑖 and also 𝜕𝐵𝑖 making
the entire graph shift to the left.

No change in the flange stiffness and bolt stiffness hence the gradients Km and Kb
will remain the same and the graph will shift along these gradients.

A decrease in all the forces (Bolt preload, Bolt and Flange)
19 | P a g e
MECH3460
An increase in the preload will result in:

Increase in the bolt preload force

Increase in the forces

Increase in the deflections

No change in bolt stiffness or flange stiffness (gradients)
20 | P a g e
MECH3460
Now to ensure that the flanges do not separate with lower preloads or even higher preloads
we must analyse the following image:
Q
P
-Fi
mi
Fi
p
Fb
Fm= Fb=0
From
Lecture
Notes

-Fm
Fi =-Fm=Fb
bi
P
Fb=Fi+rP
Fb
p
Q
Fm=0


Fm=-Fi+(1-r)P
Here it can be seen that the flanges begin to separate at the point where Fm=0 which is
directly linked to a dramatic increase in the bolt force. 0.33
In our case a decrease in preload from 0.75 to 0.7 will cause the bolt force and flange force
to change from:
Bolt force
Flange force
16495.23317N
-8269.972406N
Bolt force
Flange force
15516.44617
-7291.1854
To
Which means that the flanges will not separate as the flange is -7291.1854 which is negative
because it is in compression. Hence the flange force is not 0, and pushes against the bolt
force.
For the flanges to start separating the flange force must be >=0 and the Bolt force must also
exist in the positive direction.
Through trial and error on the excel spread sheet it became evident that the flanges will
begin to separate when the preload falls below 0.33
Hence all preloads >0.33 are deemed safe, including 0.7, 0.75 and 0.8.
21 | P a g e
MECH3460
Bolt Failure
In order to calculate the loads that are transported to other bolts as a result of a bolt failing
it is vital to obtain all the values for the forces per bolt.
For 22 bolts there exists 8225.3N in each bolt which means an overall force of 180956.6N, if
one bolt were to fail then 391.7N will be distributed amongst the rest of the bolts, if two fail
then 430.85N will be distributed and so on.
Number
of bolts
force per bolt
(N)
Extra Force Per
Bolt Due to failure (N)
% increase
Due to failure
22
21
20
19
18
17
16
15
14
13
12
11
10
8225.260766
8616.93985
9047.786842
9523.98615
10053.09649
10644.45511
11309.73355
12063.71579
12925.40977
13919.67207
15079.64474
16450.52153
18095.57368
0
391.6790841
430.8469925
476.1993075
529.1103417
591.3586171
665.2784443
753.9822369
861.693985
994.2622904
1159.972672
1370.876794
1645.052153
0%
4.761904762
5
5.263157895
5.555555556
5.882352941
6.25
6.666666667
7.142857143
7.692307692
8.333333333
9.090909091
10
%
%
%
%
%
%
%
%
%
%
%
%
%
Hence we can see an exponential growth in the failure up till 10 bolts failing, the graph
below illustrates this idea.
Force (N)
Force Distribution due to Bolt Failure
1800
1600
1400
1200
1000
800
600
400
200
0
0
2
4
6
8
10
12
14
Number Of Bolts
22 | P a g e
MECH3460
References
1. Efunda-thin walled pressure vessels accessed on the 2/10/11 from
http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vess
el.cfm
2. Andrei Lozzi Lecture notes- University of Sydney accessed on the 1/10/11
3. J. E. Shigley and C. R. Mischke, Mechanical Engineering Design, 5th ed., McGraw-Hill
(1989).
4. W.J. Orvis, Excel for Scientists and Engineers, 2nd ed., Symbex (1995).
5. Andrei Lozzi, and Paul Briozzo-The Practical Optimisation of Machine Components,
Int. J. Engng Ed. Vol. 16, No. 1, pp. 39±49, 2000
6. P. Orlov, Fundamentals of Machine Design, Vol. 5, MIR Moscow (1980).
23 | P a g e
MECH3460
APPENDIX 1
SELECTED VALUES
Symbol
Bore of cylinder (radius-mm)
Y modulus of flange (N/mm^2)
Y modulus of bolt
Bolt preload force N
Bolt Su (N/sq mm)
Bolt Sy (N/sq mm)
Min Proof Strength
Endurance limit (N/sq mm)
Cylinder yield strength (Mpa)
Reliability
Cone angle (degrees)
Loss of area due to thread
External Force N
r
Em
Eb
Fi
Su
Sy
Sp
Se
Csy
kc
Ac
Al
Ptot
Reliability factor
rel
PROPERTIES:
Bolt stem area
Thread stress area (sq mm)
Bolt stiffness N/mm
Flange stiffness N/mm
Stiffness ratio
Force/bolt N
Max Force Per bolt
Min Force Per bolt
Max Bolt Stress
Min Bolt Stress
Endurance limit (N/sq mm)
New bolt force N
New flange force N
Goodman condition
Max external force N
Ab
As
Kb
Km
rat
P
Fmax
Fmin
smax
smin
Se
Fb
Fm
Esa
Pmax
Value
60
2.07E+05
2.07E+05
14681.80509
830
660
564.4
104.7211
250
1
30
0.2
180955.7368
Equation
Sep 400
tka 0.8 surface finish
pr*Sy*As
fkc 0.897 reliability
fke 0.667 stress conc
fkd 1 temperature
Se ˆ= Sep*fka*fkb*fkc*ficd*fkf.
0.868
reliability factor for 95%
37.07526538
29.6602123
5.39E+05
1.91E+06
0.22
10644.45511
17028.59437
14681.80509
574.1224709
495
104.7211
17028.59437
-6384.139264
169.7953294
18834.19
PI*Db*Db/4
Ab*(1-Al)
Ab*Eb/I
LONG EQN
Kb/(Kb+Km)
Ptot/n
Fi+P*rat/n
Fi
Fmax/As
Fmin/As
Fi+rat*P
(1-rat)*P-Fi
ABS(salt+Se*smean/smean-Se
Fi(1+Kb/Km)
24 | P a g e
MECH3460
APPENDIX 2
FOR GRADE 8.8
GL
pr
BG
Db
FOR TWO BOLTS
FOR ONE VARIABLE
Grip Length
8
Preload
0.75
Bolt Grade
8.8
Bolt Diameter
20.38294
GL
pr
BG
Db
FOR THREE BOLTS
FOR ONE VARIABLE
Grip Length
8
Preload
0.75
Bolt Grade
8.8
Bolt Diameter
17.362
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
101.3901
0.75
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
61.68
0.75
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.0907
8.8
12
GL
pr
BG
Db
FOR TWO BOLTS
FOR TWO VARIABLE
Grip Length
9.146
Preload
0.75
Bolt Grade
8.8
Bolt Diameter
20.22779
GL
pr
BG
Db
FOR THREE BOLTS
FOR TWO VARIABLE
Grip Length
8.684427
Preload
0.75
Bolt Grade
8.8
Bolt Diameter
17.2707
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0
8.8
14.719
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0
8.8
11.8277
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
32.3885
0
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8.995
0.1395
8.8
12
GL
pr
BG
Db
FOR TWO BOLTS
FOR THREE VARIABLE
Grip Length
8.424
Preload
0
Bolt Grade
8.8
Bolt Diameter
14.66388
GL
pr
BG
Db
FOR THREE BOLTS
FOR THREE VARIABLE
Grip Length
8.424
Preload
0.342554
Bolt Grade
8.8
Bolt Diameter
12.93752
25 | P a g e
MECH3460
FOR GRADE 8.8
GL
pr
BG
Db
FOR FOUR BOLTS
FOR ONE VARIABLE
Grip Length
8
Preload
0.75
Bolt Grade
8.8
Bolt Diameter
14.8934
GL
pr
BG
Db
FOR FIVE BOLTS
FOR ONE VARIABLE
Grip Length
8
Preload
0.75
Bolt Grade
8.8
Bolt Diameter
13.21044
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
34.15
0.75
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
18.0835
0.75
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.497
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.6578
8.8
12
GL
pr
BG
Db
FOR FOUR BOLTS
FOR TWO VARIABLE
Grip Length
8.36288
Preload
0.75
Bolt Grade
8.8
Bolt Diameter
14.84567
GL
pr
BG
Db
FOR FIVE BOLTS
FOR TWO VARIABLE
Grip Length
8.14605
Preload
0.75
Bolt Grade
8.8
Bolt Diameter
13.19146
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.43
8.8
11.55
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.6115
8.8
11.527
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8.114138
0.499
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8.1119
0.659
8.8
12
GL
pr
BG
Db
FOR FOUR BOLTS
FOR THREE VARIABLE
Grip Length
8.040376
Preload
0.537907
Bolt Grade
8.8
Bolt Diameter
12.31408
GL
pr
BG
Db
FOR FIVE BOLTS
FOR THREE VARIABLE
Grip Length
8.000966
Preload
0.658564
Bolt Grade
8.8
Bolt Diameter
12.00752
26 | P a g e
MECH3460
FOR GRADE 8.8
GL
pr
BG
Db
FOR SIX BOLTS
FOR ONE VARIABLE
Grip Length
Preload
Bolt Grade
Bolt Diameter
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
7.76
0.75
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
0
0.75
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.752137
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.815
8.8
12
GL
pr
BG
Db
FOR SIX BOLTS
FOR TWO VARIABLE
Grip Length
Preload
Bolt Grade
Bolt Diameter
7.988078
0.75
8.8
11.97088
FOR SEVEN BOLTS
FOR TWO VARIABLE
GL Grip Length
pr Preload
BG Bolt Grade
Db Bolt Diameter
7.866
0.75
8.8
11.0227
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.716
8.8
11.51824
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.7855
8.8
11.5134
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8.0001
0.75311
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
7.99913
0.815869
8.8
12
GL
pr
BG
Db
FOR SIX BOLTS
FOR THREE VARIABLE
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.75
8.8
11.96935
8.000044
0.752161
8.8
12.00034
FOR SEVEN BOLTS
FOR ONE VARIABLE
GL Grip Length
pr Preload
BG Bolt Grade
Db Bolt Diameter
8
0.75
8.8
11.005
FOR SEVEN BOLTS
FOR THREE VARIABLE
GL Grip Length
7.999559
pr Preload
0.814872
BG Bolt Grade
8.8
Db Bolt Diameter
11.99657
27 | P a g e
MECH3460
FOR GRADE 8.8
GL
pr
BG
Db
FOR EIGHT BOLTS
FOR ONE VARIABLE
Grip Length
Preload
Bolt Grade
Bolt Diameter
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
0
0.75
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
0
0.75
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.8603
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.8944
8.8
12
GL
pr
BG
Db
FOR EIGHT BOLTS
FOR TWO VARIABLE
Grip Length
Preload
Bolt Grade
Bolt Diameter
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.835
8.8
11.5156
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
8
0.872714
8.8
11.508
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
7.999247
0.860295
8.8
12
GL
pr
BG
Db
Grip Length
Preload
Bolt Grade
Bolt Diameter
7.999117
0.894453
8.8
12
GL
pr
BG
Db
FOR EIGHT BOLTS
FOR THREE VARIABLE
Grip Length
Preload
Bolt Grade
Bolt Diameter
7.999249
0.860016
8.8
11.99415
FOR NINE BOLTS
FOR THREE VARIABLE
GL Grip Length
7.999119
pr Preload
0.894169
BG Bolt Grade
8.8
Db Bolt Diameter
11.99314
8
0.75
8.8
10.229
7.769098
0.75
8.8
10.25874
FOR NINE BOLTS
FOR ONE VARIABLE
GL Grip Length
pr Preload
BG Bolt Grade
Db Bolt Diameter
FOR NINE BOLTS
FOR TWO VARIABLE
GL Grip Length
pr Preload
BG Bolt Grade
Db Bolt Diameter
8
0.75
8.8
9.587
7.689
0.75
8.8
9.626
28 | P a g e
MECH3460
APPENDIX 3
GRADE 8.8
Preload Of 0.7
No of Bolts
(n)
6
7
8
9
10
14
18
22
26
30
34
Diameter (Db)
9.44538
9.54438
9.72989
9.72488
9.70372
6.08345
6.42744
5.28726
4.60542
3.78565
3.1265
Grip Length (GL)
0
0
0
0
0
17.4712
5.49221
9.53978
11.334529
16.2156
21.1241
fail
fail
fail
fail
fail
pass
pass
pass
pass
pass
pass
Under Gerber & left of poof line
FAIL
FAIL
FAIL
FAIL
FAIL
PASS
FAIL
FAIL
FAIL
PASS
PASS
$53.50
MAX
NUMBER
OF BOLTS
22.76
$80.50
$91.40
28.21
25.46
COST
29 | P a g e
MECH3460
No of Bolts (n)
6
7
8
9
10
14
18
22
26
30
34
Diameter (Db)
9.55618
9.78565
9.82467
9.84664
9.77712
6.78952
6.78064
6.87064
5.6578
5.4845
3.3996
GRADE 8.8
Preload Of 0.75
Grip Length (GL)
0
fail
0
fail
0
fail
0
fail
0
fail
14.4013
pass
5.64665
pass
14.237
pass
11.3623
pass
16.1213
pass
20.6142
pass
Under Gerber & left of poof line
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
PASS
PASS
PASS
FAIL
MAX #
COST
Of bolts
$72.50
23.49
$73.90
28.2
$86.80
24.64
Fatigue diagram for grade 8.8
200
Proof
Alternating Stress (MPa)
Goodman
Gerber
Safe Gerber
100
Preload of 0.75
0
0
100
200
300
400
500
Mean Stress (MPa)
600
700
800
900
The selected design is the one that is highlighted and corresponds to 22 bolts, diameter
6.87, and a grip length of 14.237 this is denoted by the lowest point on the brown line and is
well bellow the safe gerber and well to the left of the proof line.
30 | P a g e
MECH3460
GRADE 8.8
Preload Of 0.8
No of Bolts (n)
6
7
8
9
10
14
18
22
26
30
34
Diameter (Db)
9.67735
9.90291
9.92405
9.989212
9.85339
7.71813
6.85095
6.32927
5.24749
4.55027
3.71095
Grip Length (GL)
0
0
0
0
0
10.5145
8.88536
6.90257
11.6113
14.3281
20.4257
fail
fail
fail
fail
fail
pass
pass
pass
pass
pass
pass
Under Gerber &
left of poof line
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
fail
fail
fail
fail
fail
fail
pass
pass
pass
pass
pass
Under Gerber &
left of poof line
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
Preload Of 0.85
No of Bolts (n)
6
7
8
9
10
14
18
22
26
30
34
Diameter (Db)
10.0288
10.0288
10.02816
9.980791
9.932514
9.767455
6.018207
7.289413
6.082594
4.967597
4.18159
Grip Length (GL)
0
0
0
0
0
0
23.12888
3.453169
8.231354
14.44333
19.26523
Fatigue diagram for grade 8.8
200
Alternating Stress (MPa)
Proof
Goodman
Gerber
100
Safe Gerber
Grade 8.8, preload 0.8
0
0
200
400
600
Mean Stress (MPa)
800
1000
31 | P a g e
MECH3460
No of Bolts (n)
6
7
8
9
Diameter (Db)
10.0737
10.19035
10.134
9.4287
10
14
18
22
26
30
34
8.2422
7.0064
6.12485
5.5546
4.673
4.02253
3.43415
GRADE 9.8
Preload Of 0.7
Grip Length (GL)
0
fail
0
fail
0
fail
2.4517
pass
7.6952
14.838
8.22659
9.6389
13.3329
16.5556
14.6202
Under Gerber & left of poof line
FAIL
FAIL
FAIL
FAIL
pass
pass
pass
pass
pass
pass
pass
FAIL
PASS
PASS
PASS
PASS
PASS
FAIL
COST
$55.35
$55.95
$83.75
$73.20
$82.40
Fatigue diagram for grade 9.8
200
Proof
Alternating Stress (MPa)
Goodman
Gerber
Safe Gerber
100
preload 0.7
0
0
100
200
300
400
500
Mean Stress (MPa)
600
700
800
900
For grade 9.8 it can be seen that at a preload of 0.7 certain bolts did pass with an acceptable
bolt number, diameter and grip length however these were all rejected.
32 | P a g e
MAX # OF
BOLTS
22.91
30.1
24.48
28.7746
27.31
MECH3460
No of Bolts
(n)
6
7
8
9
10
14
18
22
26
30
34
Diameter
(Db)
10.12438
10.24
10.24526
9.941088
9.39285
6.2395
6.51126
5.3516
4.66466
3.504
3.17686
GRADE 9.8
Preload Of 0.75
Grip Length
(GL)
0
fail
0
fail
0
fail
2.5204
pass
2.6626
pass
17.09926
pass
5.717899
pass
9.8907
pass
11.64789
pass
17.5098
pass
21.4379
pass
No of Bolts
(n)
6
7
8
9
10
14
18
22
26
30
34
Diameter
(Db)
9.9243
10.4622
10.3597
10.2572
10.1788
6.9959
7.2891
6.56532
5.01596
4.3164
3.464
Preload Of 0.8
Grip Length
(GL)
0
fail
0
fail
0
fail
0
fail
0
fail
14.4532
pass
3.205
pass
3.1075
pass
11.7746
pass
14.7579
pass
21.4708
pass
Under Gerber & left of poof line
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
Diameter
(Db)
12.83744
10.23007
10.47703
10.35594
10.75029
8.06986
5.52359
6.55539
5.829402
4.74577
3.86967
Preload Of 0.85
Grip Length
(GL)
11.6737
fail
0
fail
0
fail
0
fail
2.887808
fail
10.46152
pass
26.14563
pass
7.3326
pass
8.49144
pass
14.67206
pass
21.01302
pass
Under Gerber & left of poof line
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
No of Bolts
(n)
6
7
8
9
10
14
18
22
26
30
34
Under Gerber & left of poof line
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
33 | P a g e
MECH3460
Grade Of 10.8
Preload Of 0.7
No of Bolts (n)
6
7
8
9
10
14
18
22
26
30
34
Diameter (Db)
9.2538
9.4364
10.70556
9.04266
8.44772
5.81633
6.317
5.169268
4.481661
3.66536
3.00581
Grip Length (GL)
0
0
0
7.518414
8.198683
19.77218
5.9356
10.22242
12.16345
17.24137
22.4597
fail
fail
fail
pass
pass
pass
pass
pass
pass
pass
pass
Under Gerber & left
of poof line
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
pass
fail
fail
pass
pass
pass
pass
pass
pass
pass
pass
Under Gerber & left
of poof line
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
pass
pass
fail
fail
fail
pass
pass
pass
pass
pass
pass
Under Gerber & left
of poof line
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
FAIL
Preload Of 0.75
No of Bolts (n)
6
7
8
9
10
14
18
22
26
30
34
Diameter (Db)
10.25466
9.5669
10.4474
10.9946
9.22624
6.85187
2.61963
5.98371
4.94543
4.2349
3.392638
Grip Length (GL)
25.8664
0
0
2.5204
8.260016
15.5233
7.33614
7.5568
12.22749
15.42605
22.21756
Preload Of 0.8
No of Bolts (n)
6
7
8
9
10
14
18
22
26
30
34
Diameter (Db)
13.11176
13.3925
10.23893
11.09146
11.0167
8.2129
5.666439
7.3603163
5.92359
4.817995
3.930792
Grip Length (GL)
12.17597
1.843005
0
0
0
11.11866
26.59522
2.40999
9.078244798
15.4557
21.920985
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MECH3460
APPENDIX 4
Max no of bolts
nx
23.4912261 Lb/w
OTHER DIMENSIONS:
Cylinder min thickness (mm)
Width of flange (mm)
Length of flange at bolts mm
Volume of flange (cubic cm)
Volume of bolts (cubic cm)
Total volume
Tcyl
w
Lb
Vm
Vb
Vt
2.494153163
18.52569578
435.1913082
114.7818515
34.02880269
148.8106542
sqrt(3/4)*12*60/Csy
1.5*Db‡ + l/sqrt(3)
2*PI*(60‡ + w/2)
PI*l *(w*w‡+2*w*60)/1000
(n*PI*Db*Db/4)*(l‡*4*Db)/1000
Vm‡ +
Vb
COSTS:
Flange cost ($)
Bolt cost ($)
Drilling and handling cost ($)
Total cost
SSm
SSb
SSd
SSt
8.034729607
10.42846687
54
72.46319648
Vm 0.07
Vb*0.14*(Su*Su/100000-Su/100+3.6)
10+2n
sum the above 3
Once the bolt diameter, grip length, grade,and preload are determined, the above
spreadsheet can be used in order to determine the total cost of the bolts.
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