Ministry of education and science of Ukraine

NATIONAL AIRSPACE UNIVERSITY «KHAI»

Chair 202

SCREW PRESS

Explanatory note for the home task on

"Design of Machine Elements"

Executor: student __________________________

Group: _______

Advisor : assistant professor Koveza Yu. V.

Kharkov 2013

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Task blank

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 Content

1. Description of the mechanism……………………………………………. _____

2. Initial data……………………..………………………………................... _____
3. Selection of thread……………………………………………….…………_____
4. Revise calculations…………………………………………………...…… _____
5. Calculation of nut ……………………………………………….………... _____
6. Calculation of pivot……………………………………………………….. _____
7. Calculation of handle………………………………………….…..……....._____
8. Calculation of frame …..…………………..…………………….……....... _____
9. Mechanism efficiency …………………………………………...……….. _____
10. Conclusion …………..…………………………………………………... _____
The bibliography……………….………………......……............................... _____

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 1. INITIAL DATA

Load F: __________ N
Screw operating length: ________ mm
Distance L: ___________ mm
Hollowness factor α = ___
Thread ________________
Screw material: __________ (y = _______ MPa)
Nut material: ___________ (uts = _______ MPa)

2. DESCRIPTION OF THE MECHANISM

2.1. The mechanism function

The power screw mechanisms are widely used in different mechanical

engineering` constructions. Their functions are to convert rotational motion into
translational one and producing high axial forces.
The designed device is a mechanism based on power screw and called
__________________________________________________ (Fig. 1).
It is used for _________________________________________________________
___________________________________________________________________

2.2. Diagram of the mechanism

Fig. 1

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 2.3. Functions of the mechanism elements

The elements of the mechanism and their functions are:

Element
Name Function
number

2.4. Advantages and disadvantages of the mechanism

Such mechanisms are spread because of its simplicity, reliability, high transfer
ratio, self-locking, a big load-carrying capability with small dimension. Also it`s
possible to achieve high accuracy of a movement.
Main disadvantages are rapid wear, low efficiency and speed of motion.

3. THREAD SELECTION

The unsatisfactory operation of power screws is primarily due to increased

thread wear. Wear can be reduced by selecting proper materials for the screw and nut
and by decreasing unit pressure. Calculation for wear is the main method of
determination the screw diameter.
The screws should satisfy the strength condition which limits a specific bearing
pressure on the thread working surface and guarantees high wear resistance.
Strength condition has a form:
F
p  [ p],
d 2 H 1 z
hence

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 F
d2 
 h H [ p ] ,
where d 2 is thread pitch diameter; H1 is thread fillet effective height; z is number of
threads;  H  H N – nut height factor ( H  1,2...2 ,5 ), assumed value  H  ____;
d 2
HN is nut height;  h  H 1 / P is standard value for _____________ thread
 h  ____; Р is the pitch; [ p ] is allowable specific pressure depends on nut material,
in case of ___________ bronze [ p ] = ____ МPа.
Hence
F
d2   = ______mm.
 h H [ p ]
The standard _______________ thread was selected according to the parameters
with pitch diameter bigger than being calculated one:
Pitch P, The thread diameters, mm
mm Thread effective
Major Pitch Internal height H1, mm
d d2 (D2) d1 (D1)

4. REVISE CALCULATIONS

The self-locking condition 1 ° < ρ ' – ψ was checked.

P f
Here   arctg is a helix angle;    arctg is the reduced angle of
d 2 cos  
friction; f is coefficient of friction between screw and nut, f =________;   is a
working profile angle (for __________ thread it is equal _____°).
Thus
P
  arctg = = _____°,
d 2
f
   arctg = = _____°,
cos 
and self-locking condition was satisfied.
The efficiency of power screw
tan
 = = __________.
tan(    )
The combined maximum normal stress was calculated according to the third
theory of failure

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  max   2  4 2   [  ] ,
where
–   F A is the compression stress in dangerous cross-section;
– A is the cross-section area of the screw at diameter d 1 ,
–   d h / d 1 is relationship between inside diameter of the hollow cross-section of
screw and internal one, in our case α = _____ (is given);
thus

 
A  d 12 1   2 =
4
= ___________ mm2,

 F A= = _________MPa.
– the maximum shear stress
Т
 
W
d 13
– W  ( 1   4 ) is the screw polar section modulus:
16
d 3
W  1 ( 1   4 ) = = ___________ mm3,
16

– T is the total frictional torque in dangerous section

d2
Tt  F tg(    ) = = _______ Nmm;
2
Т
 
W = = _________MPa;

y
– [ ]  is an allowable stress;
[S]
– σy is yield stress of a screw material σy = ______ MPa;
– [S] is an allowable design safety factor, [S] = _____;

[ ]  y = = _________MPa;
[S]
φ is reduction factor of the basic allowable stress in a case of compressing screw. It is
chosen depending on a material of the screw and the actual screw’s flexibility
l l
  .
i min d 1

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 Here l is the maximum operation length of the screw (is given); the factor θ
depends on the screw cross-section hollowness factor: θ = ______ ;   0 ,5...2 is
factor of reducing of the effective screw length. In our case of the screw ends fixing
value ν = _______.
Thus
l
 = = _________, φ = ______.
d 1
Finally

 max   2  4 2   ______   [  ]  ________

and strength condition is satisfied.

5. THE NUT CALCULATION

The nuts are usually made from materials, which have in pair with a screw a
low coefficient of friction and good wear resistance. Such materials are tin and tinless
bronze, brass, metal ceramics.
The nut is made as a cylinder bush (Fig. 2) pressed into unmovable frame.
Nut material is _________________ (UTS =______ MPa).
In nut calculation we accept the assumption, that axial force is distributed
through the fillets uniformly, and helix
angle is so small, that we can consider
them as plane fillets.

The number of threads

4F
z
[ p ] [ d 2  d 12 ]
where [p] is an allowable pressure
([p]=______ MPа). Hence

z  _______ ,

assumed number of fillets z = ____.

The nut height should be equal Fig. 2

or more than H  zP = ______=______
n
mm.
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 Outside nut diameter was constructively determined as
Dn = d + (2...4)P = __________= ____ mm.

6. CALCULATION OF PIVOT

The rotating screw carries axial load by means of a pivot. The minimal
resistance to rotation has pivot with thrust ball bearing that is why this support type
was chosen.
To specify a standard thrust ball bearing in mechanisms with a manual drive
static load-carrying capacity should be calculated as
C0 = SF,
where S = 1,2…1,5 is a safety factor; С0 is a static load-carrying capacity, N. Thus
C0 = SF = ___________ = ______ N.
According to this value in the catalogue [3] the thrust bearing with static load-
carrying capacity more being calculated was chosen. Its parameters are:
a code ____________, internal d = _______ mm; external D =_________ mm,
height Н = _______ mm.
Friction torque in the pivot was calculated as follows
d
T F f `,
f 2
where d is bearing internal diameter, f` is reduced coefficient of friction, for thrust ball
bearing f`= _______. Thus
d
Tp  F f` = = ________ Nmm.
2

7. CALCULATION OF THE HANDLE

7.1. Handle length

In small devices, as a rule, a manual drive is utilized. Handle length was

determined from the condition of equivalent the motive moment to resisting one in the
power screw taking into account mean worker’s force Q = 200 … 250 N (Fig. 3):
Tt  T f
l  100 mm = = ________ mm.
Q
The calculated value was approximated to the nearest normal dimension:
l = _______ mm.
7.2. Handle diameter

The handle material and its allowable bending stress were chosen. Diameter of
the handle was defined from a bend strength condition like for cantilever:

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 M 32 Ql Ql
b    [  b ] , d h  3 .
W x  d h3 0 ,1 b 
Allowable stress for __________________ is [σb] = ______ MPa. Thus
Ql
dh  3 = = ________ mm.
0 ,1 b 

From ergonomic reason diameter was chosen: d = ______ mm.

Fig. 3
8. CALCULATION OF THE FRAME

The press case was made by the moulding from _____________________. On

Fig. 4 the design of the cast type case is shown.

a
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 b

Fig. 4
A T-section profile (Fig. 5) of section I–I with minimum Wx = 0,029 h3 was
chosen in relative units. The characteristic size h = hI was defined from a bend
condition
M FH FH
b   3
 [ b ] , h  3 = = ____ mm.
W x 0 ,029 h 0 ,029 [  ]
b
The nearest bigger integer size is h = ______ mm.

b2 1 b 1 = 2/ 3 h
b 2 = 1/ 8 h
h 1 = 1/ 4 h
c = 0,3h
h

A = 0,27h 2
W 1 = 0,029h 3
c
h1

2 b1 W 2 = 0,067h 3

Fig. 5
The size hII ≈ 1,1hI = ______ mm was assumed and section II–II checked on
strength taking into account stress of a bend and a tension/compression:

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 F( H  c ) F F( H  c ) F
t     =
Wt A 0 ,067 hII 0 ,27 hII3
3

= = ____  [  ] = ______

F( H  c ) F F( H  c ) F
c    3
 =
Wc A 0 ,029hII 0 ,27 hII3

= = ____  [  ] = ______

where t, С are the stresses of a tension and compression accordingly; Wt and WС are
the flexural section modulus for the tension and compressed fibres. In a case of
T-section with parameters according to Fig. 5 the flexural section modulus accordingly
are Wt = 0,067 h3, WС = 0,029 h3.

9. MECHANISM EFFICIENCY

The mechanism efficiency

FP
м  = = ________ .
2 ( Tt  T f )

10. THE CONCLUSION

The device being designed completely satisfies the initial requirements. Used
safety factors provide safe device operation under given load.
The mechanism has simple reliable design, low cost in manufacture and
service. The efficiency factor of the mechanism is equal ______%.

THE BIBLIOGRAPHY

1. Анурьев В. И. Справочник конструктора-машиностроителя. В 3х т.-

М.: Машиностроение, 1979.
2. Курмаз Л.В. Конструирование узлов и деталей машин: Справочное
учеб.-методич. пособие/ Л.В. Курмаз, О.Л. Курмаз. − М.: Высш. шк., 2007.
− 455 с.: ил.
3. Назин В. И. Проектирование подшипников и валов. Учебное
пособие.- Х.: «ХАИ», 2004.- 220 с.
4. Проектирование механизмов с передачей винт-гайка / В.И. Назин. – Учеб.
пособие. – Харьков: Нац. аэрокосм. ун-т «Харьк. авиац. ин-т», 2006. – 122 с.
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