Liquor Lock Final Design

Introduction:
As incidents involving underage drinking continue to increase, the need for a reliable way
to secure liquor bottles has become necessary. In a recent survey, 65% of teens reported that they
acquired their alcohol from friends and family, a portion of which was taken without permission.
Everyone knows how teenagers sneak alcohol: just take a barely noticeable amount from a few
bottles lying around the house. The creative ones add water or iced tea to refill the small amount
that was taken. For other uses, a liquor bottle lock can remove the temptation to drink for anyone
trying to lay off of alcohol for a little bit. Lastly, a scotch connoisseur will always be looking for
a new way to protect his or her precious single malts.
The typical solution to secure liquor bottles has been the classic liquor cabinet. A liquor
cabinet is simple to use and can lock multiple bottles at once. The issue with a liquor cabinet is
generally a high cost and size. Nice liquor cabinets can be pretty pricey, and the larger the
cabinet, the more expensive. In addition, what if you only need to secure a couple bottles? If you
have a large collection of liquor bottles, is it worth shelling out a few hundred dollars for a large
enough cabinet? The prospect of an individual liquor bottle lock aims to provide users with this
alternative.
Project Definition & Functional Requirements:
The design objective is as follows:
Design a lock that will secure onto the top of a liquor, wine, or champagne bottle
and prevent undesired access.
While the project definition is quite simple, the original functional requirements with testable
conditions were carefully thought out to ensure adequate lock design and are listed below.
Table 1: Final Functional Requirements
Functional Requirement Testable Condition
Very Secure
-Must resist the strength of a strong human
-Smashing of the bottle is the only way to remove the lock
Cheap -Approximately $20
Easy to Use -Takes 5-10 seconds to secure or unlock a bottle
Lightweight
-Approximately 0.5-1.0 pounds
-Cannot make the bottle & lock assembly “top heavy”
Versatile -Can adjust to fit different sized different sized cylindrical necks
Aesthetically Pleasing -A classy bottle of liquor should still look classy with the lock on

Technical Feasibility Calculations:
Since this will be an individual bottle lock, the lock will slide over the top of the bottle
and then secure into place. This is different from typical locks however, where the object being
secured is completely bound by the lock. In this particular case, forces must be present to prevent
someone from pulling the lock off even while it is secured. Liquor bottles have irregularities in
the cap for threading purposes such as those shown in Figure 1, which should be taken advantage
of to prevent the lock from moving upward.
Figure 1: Typical Geometric Irregularities in Liquor Bottles
However, the lock itself should provide sufficient clamping force to ensure that the lock
is well secured. Friction between the lock and the bottle, initiated by a clamping force from a
compression spring, will resist any pulling motion. There are a few limiting factors to consider
however:
1. How much force can the bottle withstand?
2. Can reasonable spring constants generate enough frictional force?
3. The lock must be easy to use. How much grip strength can an average person provide to
compress the springs to secure the lock?
Compressive Strength Analysis of a Glass Bottle
When clamping down the lock, the spring will compress against one side of the bottle,
and by Newton’s third law, that same force will be felt on the other side of the cylinder.
However, the spring force will not be acting at a point, it will be applied across a surface,
resulting in a spring pressure (Ps). While, the two forces cancel out in the radial direction, the
walls of the cylinder will experience forces similar to a beam in bending. The cylinder can be cut
in half and be analyzed as a curved beam in bending, subjected to Ps. The length L is the length
of application, and the two ends can be treated as fixed, as they are attached to surrounding glass
and cannot move or bend. Figure 2 shows the free body diagrams of the pressure on the liquor
bottle.

Figure 2: Liquor Bottle Free Body Diagram
The equations listed below will not yield truly accurate results because many assumptions
were made. For example, the entire curved beam will not deform, only a small portion
underneath the pressure application, as the rest of it, not just the ends, are fixed. In addition, it is
difficult to gauge the area of application of the pressure, since it is a flat surface to a cylinder. It
is expected that the force required to crush the bottle will be calculated as larger than it will
actually be. This calculation is performed to just get a general gauge of the strength of a bottle. If
the bottle is determined to be safe with a large factor of safety, this simple analysis was
successful. Otherwise, an FEA analysis will be performed. The worst-case scenario will be
analyzed and all relationships are taken from Shigley’s Mechanical Engineering Design.
The max moment in a beam with two fixed supports and a center load (provides higher
moment than a pressure load) is:
𝑀 𝑚𝑎𝑥 = 𝐹𝑠 ∗
𝐿
8
(1)
The second moment of area for the semicircle is:
𝐼 =
𝜋
8
∗ (𝑟2
4
− 𝑟1
4
) (2)
With the neutral axis in the middle of the wall, the max distance on the semicircle from the
neutral axis is 𝑐 =
𝑡
2
.
The stress in the cylinder is:
𝜎 =
𝑀 𝑚𝑎𝑥∗𝑐
𝐼
(3)
The neck of a standard 1.75 liter bottle of Jameson whiskey has an outer diameter of 1.5” and a
wall thickness of 1/16”. The neck length is 4”. Glass has an ultimate strength of approximately
33 MPa or 4,780 psi. This gives the following parameters to apply to equations 1 through 3.

Table 2: Parameters for Calculating Compressive Strength on Bottle
Parameter Value
R2 0.75”
R1 0.6875”
t 0.0625”
L 4.0”
Applying these parameters, and setting the stress in equation 3 equal to the ultimate strength of
glass with a factor of safety of 10 applied, the spring force required to damage the glass is found
to be 560 lb. Again, note this is with a factor of safety with 10 applied to the ultimate strength of
the glass. So, while the model is not extremely accurate due to its simplicity, it can be stated with
confidence that the glass bottle will not break under the force of a compressive spring.
Frictional Forces from Compression Springs
The purpose of this feasibility calculation is to see if the frictional force required to resist
a pulling force, can be generated by relatively cheap springs with reasonable spring constants.
Figure 3 shows the free body diagram model of the frictional and pulling forces.
Figure 3: Free Body Diagram of Forces on Secured Liquor Bottle
FP represents someone attempting to rip off the secured lock. Fs and FN represent the spring and
corresponding normal forces. Ff will represent the frictional forces. Notice that the effect is
mirrored on the opposite side.
A simple force balance for the vertical and horizontal directions yields:
𝐹𝑆 = 𝐹 𝑁 and 𝐹𝑃 = 2 ∗ 𝐹𝑓

A simple model of static friction is given by:
𝐹𝑓 = 𝜇 𝑠 ∗ 𝐹 𝑁, where 𝜇 𝑠 is the coefficient of static friction
Lastly, Hooke’s Law states that:
𝐹𝑠 = 𝑘 ∗ 𝑥, where k is the spring constant and x is the distance the spring is compressed.
Glass generally has very high coefficients of static friction, and the irregularities in
geometry will only add to that. 0.70 is a rather conservative estimate of this coefficient. If we
assume the person pulling is generating a force of 100 lb, and the spring compresses 0.50”, we
find that the desired spring constant would be approximately 140 lb/in. This would generate a
compressive force of 70 lb on the bottle. Both of these numbers are very reasonable. Note that
springs can be added in parallel to create the desired spring constant.
Grip Strength of Average Person
Thus far, it’s been concluded that a bottle is plenty strong enough to withstand the
clamping force of a compression spring, and the frictional force generated by the spring can
withstand a pesky teenager. Although at this point it would be nice to design something that
clamps down at say, 500 lb, on the bottle, the person assembling the lock must be kept in mind.
The lock should generate enough force to secure the bottle, but should also be able to be
compressed simply enough by the user. No need to have worried parents attempting to stand on
the lock to generate enough compressive force on the spring.
Helen C. Roberts et al completed a detailed measurement of grip strength in clinical
trials. The study was performed using a handgrip dynamometer, shown below, that was squeezed
for three seconds to determine the force generated.
Figure 4: Handgrip Dynamometer for Measuring Grip Strength
The results from the study are tallied below. All units are in pounds of force.
Table 3: Results of Study on Grip Strength

Rating Males Females
Excellent > 141 > 84
Very good 123-141 75-84
Above Average 114-122 66-74
Average 105-113 57-65
Below Average 96-104 49-56
Poor 88-95 44-48
Very Poor < 88 < 44
Keep in mind that utilizing two hands to close the lock is not an issue and thus, the average
person will be able to generate more force than what is listed in table 3. Still, the user should not
be struggling with the lock, so staying on the low end of the table, a compressive force anywhere
from 50 – 75 lb should be adequate. Coincidentally, the compressive force calculated in the
previous section (70 lb) falls within this range, which is a good sign that this lock is feasible.
Solid Model
The final design of the liquor lock is shown below fully securing a bottle of wine.

Figure 5: Full View of the Locked Liquor Bottle and Assembly

Operation
To better illustrate the operation of the lock, a simpler model is shown below in figures 6
through 11. In addition, all materials have been made clear to ensure that the internal parts can be
seen. To utilize the lock, the lock is fully opened and placed around the neck of the liquor bottle.
Figure 6: Lock fully Opened around Bottle
The following shows a zoom-in on the spring assembly. The lock is able to adjust to different
sized bottles by use of the sliding pin. The sliding pin allows the locking portion to slide back
and forth. For smaller sized bottles, the lock will slide farther to allow it to accommodate the
smaller neck. For larger sized bottles, the locking portion doesn’t have to slide as far down. The
compression pad transfers the force from the springs to the bottle.

Figure 7: Close-up View of Spring Assembly
The lock is then closed until the pad is about to make contact with the liquor bottle.
Figure 8: Springs about to Compress as Bottle Makes Contact
Next, the user fully compresses the springs against the bottle. The compression pad forces itself
tightly against the bottle, providing the frictional force necessary secure the lock.
Figure 9: Fully Compressed Lock on Liquor Bottle
The digits are aligned to the correct combination, and the key is inserted.

Figure 10: Correct Combination Selected
The key is fully inserted and the combination is jumbled to secure the lock in place.
Figure 11: Secured Lock
And now, the lock is completely secured!
Technical Analysis of the Compressive Springs

As can be seen from the solid model images above, the clamping force is provided by a
rectangular “compression pad” which protrudes from the spring housing which holds the springs.
When the pad makes contact with the bottle, the pins protruding from underneath the rectangular
section compress the springs. To properly design these springs the following requirements were
placed on the design.
 A small force (~5 lb) should be exerted on the pad when it is fully protruded. This forces
it to stay in place, and prevents any wobbly components.
 The springs will be squared and grounded
 The springs will be placed in a hole to prevent stability issues
 The springs must generate 50 to 75 lb of compression force through 0.50” of deflection
 Must be more than one spring along the length of the pad to give more support as the pad
is compressed
The process to properly design a spring is iterative. The general method used in this case is to
select a guess for a wire diameter, number of springs used, and spring index, and then using the
above requirements to determine the proper dimensions for the spring. From there, similar
springs were searched for through McMaster Carr. A part number was selected, and then
reanalyzed through the same process to ensure it still met the requirements. Lastly, price was
compared and the process was repeated for other part numbers to determine the best choice.
Note the following parameters utilized in the design process:
d - wire diameter
D - mean diameter
OD - outside diameter
ID - inside diameter
k - spring rate
A - parameter for calculating ultimate strength
m - parameter for calculating ultimate strength
SUT - ultimate tensile strength
Ssy - shear strength
G - shear modulus
C - spring index
KB - direct shear factor
Nt - total number of coils
Na - number of active coils
Ls - shut length
Lf - free length
τ - shear stress
y - deflection
The procedure will be demonstrated for using three hard drawn springs, with a wire diameter of
0.105” and a spring index of 8.
1. 𝐹𝑚𝑎𝑥 =
𝑀𝑎𝑥 𝐿𝑜𝑎𝑑
# 𝑜𝑓 𝑆𝑝𝑟𝑖𝑛𝑔𝑠
=
75
3
= 25 𝑙𝑏, 𝐹 𝑚𝑖𝑛 =
𝑀𝑖𝑛 𝐿𝑜𝑎𝑑
# 𝑜𝑓 𝑆𝑝𝑟𝑖𝑛𝑔𝑠
=
5
3
= 1.67 𝑙𝑏

2. 𝐷 = 𝐶 ∗ 𝑑 = 8 ∗ 0.105 = 0.84"
3. 𝐾 𝐵 =
4∗𝐶+2
4∗𝐶−3
=
4∗8+2
4∗8−3
= 1.17
4. 𝜏 = 𝐾 𝐵 ∗
8∗𝐹 𝑚𝑎𝑥∗𝐷
𝜋∗𝑑3
= 54.2 𝑘𝑝𝑠𝑖 *This is the shear stress at the max force
5. 𝐴 = 140 𝑘𝑝𝑠𝑖 ∗ 𝑖𝑛 𝑚
, 𝑚 = 0.190 (From Table 10-4 in Shigley’s)
6. 𝑆 𝑈𝑇 =
𝐴
𝑑 𝑚 = 215 𝑘𝑝𝑠𝑖, %𝑆 𝑈𝑇 = 0.45 (From Table 10-6 in Shigley’s)
7. 𝑆𝑠𝑦 = (%𝑆 𝑈𝑇) ∗ 𝑆 𝑈𝑡 = 215 ∗ 0.45 = 96.7 𝑘𝑝𝑠𝑖
8. 𝑛 𝑦 =
𝑆 𝑠𝑦
𝜏
=
96.7
54.2
= 1.78 *Spring will not yield
9. 𝑘 =
𝐹 𝑚𝑎𝑥−𝐹 𝑚𝑖𝑛
𝑥
=
25−1.67
0.50
= 46.7 𝑙𝑏/𝑖𝑛
10. 𝐺 = 11.5 𝑀𝑝𝑠𝑖 (From Table 10-5 in Shigley’s)
11. 𝑁𝑎 =
𝑑4∗𝐺
8∗𝐷3∗𝑘
= 6.32 𝑐𝑜𝑖𝑙𝑠 = 7 𝑐𝑜𝑖𝑙𝑠 *Round-up for ease of manufacturing
12. 𝑘 =
𝑑4∗𝐺
8∗𝐷3∗𝑁 𝑎
= 42.1 𝑙𝑏/𝑖𝑛 *Close enough to original k
13. 𝑁𝑡 = 𝑁𝑎 + 2 = 9 𝑐𝑜𝑖𝑙𝑠
14. 𝐿 𝑠 = 𝑑 ∗ 𝑁𝑡 = 0.945"
15. 𝑦𝑖 =
𝐹 𝑚𝑖𝑛
𝑘
=
1.67
42.1
= 0.04"
16. 𝑦𝑐𝑙𝑎𝑠ℎ = 0.15 ∗ 𝑥 = 0.15 ∗ 0.50 = 0.075 *Accounts for non Hooke’s Law behavior
17. 𝐿 𝑓 = 𝑥 + 𝐿 𝑠 + 𝑦𝑖 + 𝑦𝑐𝑙𝑎𝑠ℎ = 1.56"
18. 𝑦𝑠 = 𝐿 𝑓 − 𝐿 𝑠 = 0.61" *Deflection to shut length
19. 𝐹𝑠 = 𝑦𝑠 ∗ 𝑘 = 25.7 𝑁
20. 𝜏 𝑠 = 𝐾 𝐵 ∗
8∗𝐹𝑠∗𝐷
𝜋∗𝑑3 = 56.1 𝑘𝑝𝑠𝑖
21. 𝑛 𝑠 =
𝑆 𝑠𝑦
𝜏 𝑠
= 1.72 *Spring will not fail at shut length
22. 𝐹𝑡𝑜𝑡𝑎𝑙 = 𝑘 ∗ 𝑥 ∗ (# 𝑜𝑓 𝑠𝑝𝑟𝑖𝑛𝑔𝑠) = 63 𝑁 *Total Force is within range

23. 𝐿 𝑐𝑟𝑖𝑡 = 5.26 ∗ 𝐷 = 4.4" > 𝐿 𝑓 *Spring will not buckle
24. 𝑆𝑠𝑢 = 0.67 ∗ 𝑆 𝑢𝑡 = 144 𝑘𝑝𝑠𝑖
25. 𝑆𝑠𝑒 = 0.5 ∗ 𝑆𝑠𝑢 = 72 𝑘𝑝𝑠𝑖
26. 𝜏 𝑎𝑙𝑡 = 𝐾𝑏 ∗
8∗(𝐹 𝑚𝑎𝑥−𝐹 𝑚𝑖𝑛)∗𝐷
2∗𝜋∗𝑑3
= 25.2 𝑘𝑝𝑠𝑖
27. 𝜏 𝑚𝑒𝑎𝑛 = 𝐾𝑏 ∗
8∗(𝐹 𝑚𝑎𝑥+𝐹 𝑚𝑖𝑛)∗𝐷
2∗𝑝𝑖∗𝑑3 = 28.8 𝑘𝑝𝑠𝑖
28. 𝑛𝑓𝑎𝑡𝑖𝑔𝑢𝑒 =
1
(
𝜏 𝑎𝑙𝑡
𝑆 𝑠𝑒
)+(
𝜏 𝑚𝑒𝑎𝑛
𝑆 𝑠𝑢
)
= 1.82  *Spring is designed for infinite life
29. 𝐹𝑖𝑛𝑎𝑙 𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠:
 𝑂𝐷 = 𝐷 + 𝑑 = 0.945"
 𝑑 = 0.105"
 𝑘 = 42.1 𝑙𝑏/𝑖𝑛
 𝐿 𝑓 = 1.56"
 𝑁𝑡 = 9 𝑐𝑜𝑖𝑙𝑠
Through this process, an adequate design for a three-spring arrangement has been
achieved. However, there is a very minimal chance that an exact spring is commonly
manufactured with the specified dimensions. To avoid custom manufacturing costs, a similar
spring will be selected and reevaluated through the process above. There are a couple of
important parameters to pay attention to here. The first is the factor of safety at max deflection.
The pins will have stoppers on them, preventing the springs from unnecessarily compressing. So
in this case, the factor of safety does not have to be greater than 1 at the shut length, provided
that the deflection to the shut length is greater than the maximum 0.50” deflection. The other
important factors are obviously cost, and the total compressive force applied to the bottle.
Overall, this process was repeated multiple times for three and two spring configurations
and different springs from McMaster-Carr were analyzed. The best results of each configuration
are shown below. The green cells highlight which design was better for the parameters that are
important.

Table 4: Spring Design Comparison
Parameter 3 Springs 2 Springs Units
Part #: 9657K402 9657K379 -
wire d: 0.100 0.105 inches
OD: 0.845 0.97 inches
L0: 2.00 1.75 inches
k: 55.40 68 lb/in
Nt: 8 6.5
Material:
Music
Wire
Music
Wire -
Compressive Force: 83 68 lb
Cost per spring: $1.36 $1.34
Total Cost $4.09 $2.67
F.O.S @ max defl: 1.93 1.88
F.O.S for infinite life: 1.10 1.07
Ultimately, the 2 and 3 spring configurations performed quite similarly. The two spring
configuration takes up less space in the design however and is approximately half the price. The
factors of safety are extremely similar so not much strength is being given up by going with two
springs. There is less compressive force on the bottle, but it still falls within the range required
by the dynamometer study.

Technical Analysis of the Sliding Pin
The sliding pin allows for the lock to adjust to different sized bottles. It is essentially a
shaft and thus, will be designed as one that can withstand the forces subjected to it with the
smallest required diameter. The figures below shows the forces and stresses that the pin will
undergo.
Figure 13: FBD of Assembly as the Spring is Compressed
As can be seen from the above figure, as the spring is compressed by the user, the pin
receives an equal force in the opposite direction. If we look at a top view of the pin in this
situation, it can be seen how the lock is attempting to shear the pin.
Figure 14: Shearing of Sliding Pin
Instead of solving for the minimum diameter required, the selection of miniature shafts from
McMaster Carr was first surveyed. It is expected that the stresses will not be too large, so it is
probably best to start with the smallest size available. The smallest miniature shaft they carry is a
3” long, 3/16” OD miniature steel shaft for $3.13. The following parameters were utilized during
the design process.

F - Force
d - Outside Diameter of Shaft
SUT - Ultimate Tensile Strength
SY - Yield Strength
t - Thickness (of plastic contact area)
𝜏 - Shear Stress
𝜎 - Bearing Stress
𝜎′ - Equivalent Stress
n - Factor of Safety
k - Fatigue Modification Factors
Kt - Stress Concentration Factor
q - Notch Sensitivity
Kf - Shear Stress Concentration Factor
The given values are shown below in the following table. Note that a conservative estimate was
made for the strength of steel, the notch sensitivity, and the stress concentration factors. The
values for notch sensitivity and stress concentration factors were taken from Shigley’s.
Table 5: Given Parameters for Sliding Pin Design
Parameter Value Units
Maximum Force 75 Lb
d 3/16 inches
SUT 48 kpsi
Sy 41 kpsi
t 0.2 inches
Kt 2.7 -
q 0.7
The approach was to find the shearing and bearing stresses acting on the pin and ensure that the
pin would not yield. Then, the modified goodman approach was utilized to ensure the pin is
designed for infinite life. The design process is outlined as follows:
1. 𝐹𝑠ℎ𝑒𝑎𝑟 =
𝐹 𝑠𝑝𝑟𝑖𝑛𝑔
2
= 37.5 𝑙𝑏
2. 𝜏 =
𝐹 𝑠ℎ𝑒𝑎𝑟
(
𝜋
4
)∗𝑑2
= 1358 𝑝𝑠𝑖
3. 𝜎 =
𝐹 𝑠ℎ𝑒𝑎𝑟
𝑑∗𝑡
= 1000 𝑝𝑠𝑖
4. 𝜎′
= √(𝐾𝑡 ∗ 𝜎)2 + 3 ∗ (𝐾𝑡 ∗ 𝜏)^2 = 6560 𝑝𝑠𝑖
5. 𝑛 𝑦 =
𝑆 𝑦
𝜎′
= 6.2 *Pin Will Not Yield
6. 𝑆 𝑒
′
=
𝑆 𝑈𝑇
2
= 24 𝑘𝑝𝑠𝑖

7. 𝑘 𝑎 = 2.7 ∗ (𝑆 𝑢𝑡)−0.265
= 0.968
8. 𝑘 𝑏 = (
𝑑
3
)
0.107
= 1.35 *This was set equal to 1 to be conservative
9. 𝑘 𝑐 = 𝑘 𝑑 = 𝑘 𝑒 = 1
10. 𝑆 𝑒 = 𝑘 𝑎 ∗ 𝑘 𝑏 ∗ 𝑘 𝑐 ∗ 𝑘 𝑑 ∗ 𝑘 𝑒 ∗ 𝑆 𝑒
′
= 23.2 𝑘𝑝𝑠𝑖
11. 𝐾𝑓 = 1 + 𝑞 ∗ (𝐾𝑡 − 1) = 2.2
12. 𝜏 𝑎𝑙𝑡 = 𝜏 𝑚𝑒𝑎𝑛 =
𝜏
2
= 680 𝑝𝑠𝑖
13. 𝜎 𝑚𝑒𝑎𝑛 = 𝜎 𝑎𝑙𝑡 =
𝜎
2
= 500 𝑝𝑠𝑖
14. 𝜎 𝑎𝑙𝑡
′
= √(𝐾𝑓 ∗ 𝜎 𝑎𝑙𝑡)
2
+ 3 ∗ (𝐾𝑓 ∗ 𝜏 𝑎𝑙𝑡)
2
= 2800 𝑝𝑠𝑖
15. 𝜎 𝑚𝑒𝑎𝑛
′
= √(𝐾𝑓 ∗ 𝜎 𝑚𝑒𝑎𝑛)
2
+ 3 ∗ (𝐾𝑓 ∗ 𝜏 𝑚𝑒𝑎𝑛)
2
= 2800 𝑝𝑠𝑖
16. 𝑛𝑓 =
1
(
𝜎 𝑎𝑙𝑡
′
𝑆 𝑒
)+(
𝜎 𝑚𝑒𝑎𝑛
′
𝑆 𝑈𝑇
)
= 5.3 *Pin is designed for infinite life
It was found that the miniature 3/16” shaft from McMaster Carr will easily handle the
maximum stresses that will be induced in the pin. The table below summarizes the part
information.
Table 5: Sliding Pin Values
Part Number 1327K103 -
Diameter 3/16 inches
Length 3 inches
Material 2L14 Steel -
Price $3.13 -

Technical Analysis of Bolted Joints
In order to hold the locked assembly together, a set of screws were needed to secure
everything in place. Since this is a lock, all screws must be internal once the device is fully
locked, otherwise someone could just take it apart as they please. The screws will be tightly
secured, but after that, there will be very little alternating loads. Thus, fatigue is not a concern for
the screw design. The material being screwed into is plastic, so specially designed screws must
be considered. Size is also an issue, as the screws cannot take up too much space. McMaster
Carr has different lengths of #0 18-8 Stainless Steel screws for plastic. #0 indicates a major
diameter of 0.060” and is the smallest available screw sizes. It would be ideal if these screws
could withstand the necessary forces so they will be analyzed first. All screws will be tapped into
the plastic, rather than utilizing a nut. The figure below shows a simple tapped hole for a screw.
Figure 15: Tapped Hole
The following table provides the parameters for the screws found from McMaster Carr.
Table 6: Stainless Steel Plastic Screw Dimensions
Length 0.25 inches
Major Diameter 0.060 inches
Minor Diameter 0.046 inches
Young’s Modulus 28 106
psi
Length of Tapped Portion 0.15 inches
Proof Strength 33 kpsi
The process involved finding the stiffness of both the bolt and the plastic joint and
finding what preload will be placed on each screw. The proof strength of the screw was assumed
to be a low value since it was never specified. Once the preload was found, the applicable bolting
factors of safeties were found to see what the screws could withstand.
1.) 𝑘 𝑏 = 𝐸 𝑏 ∗
𝐴 𝑡
𝑙 𝑡
= 186.1 𝑘𝑖𝑝/𝑖𝑛𝑐ℎ

2.) 𝑘 𝑚 = (
1
𝑘1
+
1
𝑘2
)
−1
*k1 & k2 are found through pressure cone analysis as follows
3.) 𝑘1 =
0.5774∗𝜋∗𝐸 𝑝𝑙𝑎𝑠𝑡𝑖𝑐∗𝑑
ln((1.155∗𝑡1+𝐷1−𝑑)∗(𝐷1+𝑑))
(1.155∗𝑡1+𝐷1+𝑑)∗(𝐷1−𝑑)
= 30.2 𝑘𝑖𝑝/𝑖𝑛𝑐ℎ *Eplastic ~ 28*104
psi, t1 = 0.10”
4.) 𝑘2 =
0.5774∗𝜋∗𝐸 𝑝𝑙𝑎𝑠𝑡𝑖𝑐∗𝑑
ln((1.155∗𝑡2+𝐷2−𝑑)∗(𝐷2+𝑑))
(1.155∗𝑡2+𝐷2+𝑑)∗(𝐷2−𝑑)
= 30.2 𝑘𝑖𝑝/𝑖𝑛𝑐ℎ *t2=0.15”
5.) 𝑘 𝑚 = 14.1 𝑘𝑖𝑝/𝑖𝑛𝑐ℎ
6.) 𝐶 =
𝑘 𝑏
𝑘 𝑏+𝑘 𝑚
= 0.93 *This is extremely high due to the plastic joint
7.) 𝐹𝑖 = 0.75 ∗ 𝑆 𝑝 ∗ 𝐴 𝑡 = 41 𝑙𝑏
8.) 𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝐿𝑜𝑎𝑑 𝑡ℎ𝑎𝑡 𝑒𝑥𝑐𝑒𝑒𝑑𝑠 𝑃𝑟𝑜𝑜𝑓 𝑆𝑡𝑟𝑒𝑠𝑠 =
𝑆 𝑝∗𝐴 𝑡−𝐹𝑖
𝐶
= 15 𝑙𝑏
For the selected screws, a conservative estimate gives that an external force of 15 lb on one
screw will cause the screw to exceed its proof stress. However, once the assembly is locked the
screws will never see much of an external force. The internal screws of the combination housing
shown below are the only screws that will experience a force if someone is trying to rip the lock
off as highlighted below.
Figure 16: Screws that Could Experience an External Force
However, it is difficult to assess how much force the screws will undergo since the
combination key and the sliding pin will absorb some external force. A conservative estimate is

that a total maximum force of 50 lb can be applied to the screws. This would be a 12.5 lb force
on each screw. This yields the following.
9.) 𝑛 =
(𝑆 𝑝∗𝐴 𝑡−𝐹𝑖)
𝐶∗𝑃
= 1.20 *Proof Stress will not be exceeded
It is much more likely that that only 10~15% of an external load will be applied to the screws, so
the factor of safety is much more likely to be higher. Thus, the screws outlined above will be
adequate for the design.
Assembly Instructions
1.) Take the Spring Housing and place the Springs in the designated support cylinders. Line
the Compression Pad up with the springs and place the Spring Housing top over the
Compression Pad. Take four 3/16” length screws and screw the top of the housing onto
the housing.
Figure 17: Spring Housing Assembly
2.) Line up the top of the plastic Combination Housing with the top of the Spring housing.
Take four ¼” length screws and hand tighten the four inner screws to lightly secure the
Combination Housing to the Spring Housing. Take a small Philips head screw driver to
fully tighten the screws and secure it to the housing.

Figure 18: Attaching Combination Housing to Spring Housing
3.) Take the Combination Digits and insert them into the Combination Housing.
Figure 19: Digits inserted into the Assembly
4.) Take the Bottom of the Housing and line up the digit slots. Take one 1/16” length screw
and four 3/16” length screw. The 1/16” length screw goes underneath the key hole. The
four 3/16” screws fill in the other four tapped holes. Tighten all five screws to fully
secure the bottom of the housing.

Figure 20: Combination Assembly fully secured.
5.) Take the plastic casing. It may be easier to put it over a liquor bottle to properly assembly
it.
Figure 21: Plastic Casing Placed over a Bottle
6.) Take the secured assembly from steps 1 through 4 and place it inside the plastic casing
and line up the slots. Insert the sliding pin through the slots leaving an equivalent amount
of length on each side.

Figure 22: Sliding Pin Inserted into Assembly
7.) Take one flanged washer and apply Loctite Instant-Bonding Adhesive to the inside of the
flanged washer. Quickly slide it over one end of the shaft. Once the glue has secured the
washer on one side of the shaft, repeat the process on the other side. Note that the glue
will begin to dry in 30 seconds, so quick assembly is necessary. The flanged washers are
to prevent the sliding pin from moving axially

Figure 23: Flanged Washers in Place to Secure Shaft
8.) Let the adhesive cure for 24 hours.
9.) The Lock is now ready for use!

Bill of Materials
A complete Bill of Materials is provided below in Table 5. Plastic Component costs were
estimated based on their size. Each important piece of the lock is separated into sections in the
table. The top listed price of each section is how much that piece costs. For example, a fully built
spring assembly costs $5.84. The total cost is the sum of the individual assembly costs.
Table 7: Bill of Materials
BOM
LEVEL
PART
NUMBER
PART NAME QTY
PROCUREMENT
TYPE
Cost
1 - Liquor Lock 1 Built $16.50
2 - Plastic Casing 1 Made to Specs $1.25
2 - Spring Assembly 1 Built $5.84
3 - Spring Housing 1 Made to Specs $1.00
3 - Spring Housing Top 1 Made to Specs $0.75
3 9657K379 Compression Spring 2 Purchased $2.67
3 - Compression Pad 1 Made to Specs $0.50
3 99461a510 3/16” Stainless Steel Screws 4 Purchased $0.92
2 - Locking Assembly 1 Built $4.33
3 - Combination Digit Housing 1 Made to Specs $1.00
3 - Combination Housing Bottom 1 Made to Specs $0.50
3 99461a515 ¼” Stainless Steel Screws 4 Purchased $0.93
3 99461a510 3/16” Stainless Steel Screws 4 Puchased $0.92
3 99461a505 1/8” Stainless Steel Screws 1 Purchased $0.23
4 - Combination Digit 3 Made to Specs $0.75
2 - Sliding Pin 1 $5.05
3 1327k103 Miniature 12L14 Steel Drive Shaft 1 Purchased $3.13
4 90097a180 Flanged Washers 2 Purchased $0.67
4 74985a57 Loctite Instant Bonding Adhesive 2 mL Purchased $1.25

Conclusion
If the functional requirements are revisited, it is seen that the functional requirements are
all met.
 Security – Lock is secure, all components cannot be broken through normal force. All
screws are secured internally to the lock so it cannot be disassembled.
 Cost – Total cost is estimated to be a little more than $16.00, less than the original $20
requirement.
 Lightweight – Solidworks analysis estimates the total weight of the lock is 0.42 pounds,
even less than the 0.50-pound minimum requirement. The lock is very lightweight and
will not cause a liquor bottle to be top-heavy.
 Easy to use – The lock can be unlocked and locked in approximately 10 seconds. It is
very easy to use, and simple to assemble. Most likely the lock would be preassembled,
but the ease of assembly could allow for users to do it themselves.
 Versatile – The lock can adjust to multiple sized liquor bottles, from skinny wine bottles
to thicker Whiskey handles.
 Aesthetically Pleasing – As you can see from figure 5, the lock looks nice on a classy
bottle of wine.
Overall, this lock design turned out to be the best overall design to accomplish the problem
statement. It is cheap, effective, versatile, and looks great.
Cheers!

Works Cited:
1. Budynas, Richard G., and J. Nisbett. Shigley's Mechanical Engineering Design. 9th ed. N.p.: McGraw-Hill
Science Engineering, 2012. Print.
2. "Handgrip Strength Test." Hand Grip Strength Test. N.p., n.d. Web. 20 Feb. 2015.
<http://www.topendsports.com/testing/tests/handgrip.htm>.
3. "McMaster – Carr.” McMaster Carr. <www.mcmaster.com>

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