PROJECT IN MECHANICAL ENGINEERING

1-DIMENSIONING OF SCREWS

1.1‐Load of Traction and tightening on the screw:

The diagram of a screw subjected to a tensile and tightening load is given below:

Nomenclature:
P = Axial Load (tension) [ kgf ]
Po = Tightening Load [kgf] to be used Po = 0.15 . P
external thread diameter [ mm ]
do= internal thread diameter [ mm ]
p = thread pitch [ mm ]
t = thread depth [ mm ]
a = 55° Whitworth thread
a = 60° metric thread

From the formula of tension we have: equation (I)

where:σ
allowable tensile strength [ kgf/mm² ]

A = area of the core diameter [ mm² ] equation ( II )

Substituting equation (II) into equation (I) and isolating the diameter (do) we have:

With this formula, we determine the diameter of the screw's core.

To determine the diameter of the thread (d), we consult the attached THREAD TABLE through the internal diameter.
(do) or by the formula:

d = do + 2 . t oud = do + 1.2268 . P
thread depth [ mm ]
whereP= screw step [ mm ]

See Thread Table in Attachment.

See below other formulas for calculations of other parts of the thread:

Metric triangular thread (normal and fine)

P = thread pitch
d = major diameter of the screw (normal)
d1 = smaller diameter of the screw (Ø of the core)
d2 = effective diameter of the screw (average Ø)
a = angle of the thread profile
f = clearance between the root of the nut thread and the crest of the screw thread
D = major diameter of the nut
smaller diameter of the nut
D2 = effective diameter of the nut
he = height of the screw thread
rre = rounding radius of the thread root of the screw
rri = rounding radius of the fillet of the nut

1.2-Metric Thread:

angle of the thread profile:

a = 60º.
minor diameter of the screw (Ø of the core):
d1 = d-1,2268P.
effective diameter of the screw (Ø average):
d2 = D2 = d-0.6495P.
gap between the root of the nut thread and the crest of the bolt thread:
f = 0.045P.
greater diameter of the nut:
D = d + 2f .
smaller diameter of the nut (hole):
D1 = d - 1.0825P.
effective diameter of the nut (average Ø):
D2 = d2.
height of the screw thread:
he = 0.61343P.
rounding radius of the root of the screw thread:
rre = 0.14434P.
rounding radius of the root of the nut's thread:
rri = 0.063P.

1. 3-Rosca whitworth (normal and fine triangular)

Formulas:

a = 55º
P = 1” / number of threads
hi = he = 0.6403 . P
rri= rre= 0.1373 . P
d=D
d1 = d - 2he
D2 = d2 = d-he
1.4 - Screw Strength Class:
Exercises:

1. Calculate the minor diameter of a screw (d1) for a thread with an external diameter (d) of
10 mm and a pitch (p) of 1.5 mm. Define which screw will be chosen using the table of
rings.
2. In the device in the figure below, the bushing is made of ABNT 1010 drawn steel and the screw is made of steel.
ABNT 1030 drawn. Determine the outer diameter of the bushing and the screw to support
a load of tightening of 2.5 tf (loading I). Determine which bolt should be used.
d + 1 mm

3. Dimension the screws of the support as shown in the figure below that is subjected to loading.
static. Screw material: ABNT 1020 steel rolled Load I. Determine which
screw must be used. Resp.: d = 5.05 mm;

4. Dimension the diameter of the steel screw of the tensioner in the figure below that supports a
acceptable stress of 29.87 ksi and a maximum axial load of 3307 lbf. Determine which
The screw should be used. Resp.: d = 10.23 mm; (Consider 1 ksi = 1000 psi)

5. Dimension the diameter of the same steel screw of the tensioner from the previous exercise that
it supports an admissible tension of 19.2 ksi and a maximum axial load of 1 kN. Determine
which screw should be used. Answer: d = 3.32 mm; (Consider 1 psi = 6894.757 N/m²)

6. Determine the thread of the steel screw 1040, drawn, Loading II, that supports a
Axial load of 15 kN. Answer: d = 12.9 mm;
1.5 - Tension in the Thread Flanks

Assuming that the load is uniformly distributed over the height h of the nut and that the
The threads of the bolt will fail due to tension, compression, shear or torsion at the diameter.
younger.
These are average tensions, due to the fact that the fillets distribute the load.
likewise. In view of this, strong safety factors, K > 2, must be used for purposes of
project. For a more accurate calculation of compression or tension in the thread, we will adopt the following
formula:

4°

Where:
p= thread pitch;
h = height of the nut;
d = major diameter of the thread;
d1smaller diameter of the thread;

When we need to calculate the allowable stress, we must apply both the tension or
compression, regarding shear, a safety factor k, which when dividing the stress of
flow through k, to be chosen depending on the project to be used, we find the tension
admissible. We then use the following formula:

or

Particularly for calculating the breaking or shear flow tension

we should use the data from the table below and then apply the safety factor k to find
your allowable shear stress.
Hexagonal Nuts - Metric Hexagon Nuts - Inch
Standard DIN 934 – ISO Thread 6H – DIN 13 Construction: ABSI B 18.2.2 Standard -
ANSI B 1.1 Thread - Class 2B
1.7 – Shear Stress:

A possible shear failure mode involves the tearing of thread fillets both
It's about how much of the screw. What, if one or another of these scenarios occurs, depends on the
relative resistances of the materials of the nut and bolt. If the material of the nut is weaker (like
almost always happens), its fillets can be cut along its largest diameter. If the
The screw is weaker, and its threads may be stripped along its smaller diameter.
If both materials have identical strength, the set can be torn along
of the primitive diameter. In any case, we must assume some degree of load sharing.
between the threads of the screws in order to calculate the stresses.
One way to proceed is to consider that once a complete failure requires
that all the threads of the screw are torn, these can be considered as sharing
the load equally. This hypothesis is likely valid, as long as the nut or bolt (or
both) should be ductile enough to allow each thread to tear as the assembly begins to
fail. However, if both parties are fragile (for example, high-strength steels or iron
mold) and the adjustment of the thread fillets is poor, we can imagine each fillet taking on the whole
load by shifts until there is a fracture and the work is passed on to the next filament. The reality
is included in these extremes. If we express the shear stress in terms of the number
of engaged thread fillets, a judgment must be made in each case to determine the degree
of appropriate load sharing.
The area under shearS for a thread pitch it is the area of the cylinder of its diameter
menord1:
. . .

Where:

Wi → factor that defines the percentage of the step occupied by the metal at the smaller diameter of
pig

Area factors for shear area by

thread cutting:

THREAD TYPE Wi

UNS/ISO/WITHWORTH 0.80

SQUARE 0.50

ACME (trapezoidal) 0.77

 To calculate the shear stress of the thread, we use the following formula:

. . .
1.8‐Torque

The torque on the screws is represented by:

Torque coefficient defined and is directly related to friction

what exists in the thread. is defined by the following formula:

. 0.625
2 1 This is a demonstration

The friction coefficients of threads, common screws, and nuts cover a range that goes
from 0.12 to 0.20, depending mainly on the finish, the precision of the thread, and the degree of
lubrication. On average, both µ and µ are around 0.15.
The interesting thing about the equation deké quek≈0.20 for µ = µc = 0.15, regardless of the
size of the screws used or if the threads are thick or thin. So it's more
it is convenient to write torque in the following way:
0.20

Exercises:

1. Check if the threads of an M4 screw made of SAE1030 steel with allowable stress of
1200 kgf/cm² supports a load of 150 kgf. Response: σ = 680.4 kgf/cm²;

2. Check if the threads of a steel screw with Ø ½”, with a pitch of 12 threads per inch.
The allowable tensile stress σ = 150 MPa supports a load of 350 kgf. Answer: σ = 13.9
MPa;
3. Determine the force that supports the thread of a M12 screw made of ABNT 1020 steel.
drawing, when subjected to tension, when screwed into a metal support. For this
Screw consider a safety factor k = 2.5. Response: F = 30 kN;

4. A steel support attached to a drive machine exerts a force of 22 kip, including its
own weight, over the thread fillets of the screws that hold it. How many screws Ø
¼", 20 threads per inch, are needed to secure this bracket, knowing that its tension
is the allowable equal to 290 MPa? Answer: 6 screws;

5. Check if an ISO thread with an external diameter of 10 mm and a pitch of 1.5 mm withstands
a shear stress for a load of 880 lbf, knowing that the material of
The screw is made of 8640 steel with a τ = 90 MPa. If negative, resize the screw.
specify it. Resp.: M14;

6. Verify if a 3/8" Whitworth Gas thread with an outer diameter of 19 threads per inch
resists a shear stress for a load of 350 kgf, knowing that the material
The screw is made of 1045 steel with a τ= 8 ksi. Answer: ½” Gas;