Numerical of Bolt design, punch design and standard sizes and speed of shaft. Different stresses acting on bolt and punch. Failure of bolt and punch under stresses. Standardization, Preferred numbers, Basic and derived series. NOTE: In the numerical no 1 of punching on slide no 10. The Answer of shear force is obtained considering thickness of plate = 5 mm. (If 6 mm is considered than the answer according to formula will be 395,841 N)

1. Design of Machine Elements
Numerical of Design of Bolt, Punch and
Standard sizes and speeds of shaft using
Basic series.
Gaurav Mistry
AssistantProfessor
Diwaliba Polytechnic, UTU.

2. Gaurav Mistry 2
 Consideration and Requirements: Design of MachineElements
 Factors affecting Machine Design / General considerations in Machine
Design.
 Geometry and Kinematics of the machine.
 Type of load and the stresses produced because of the load.
 Selection of materials.
 Climatic and Chemical Influence.
 Production Facilities.
 Use of Standard Parts.
 Limit, Fit and Tolerance.
 Frictional Resistance and Lubrication.
 Cost Consideration.
 Consumer/ Customer/ User’s Requirements.
(a) Main Requirements: (b) Additional
Requirement:
i. Output Capacity i. Transportability
ii. Reliability ii. Noiseless operation
iii. Expected service life iii. Simple and easy control
iv. Weight and space limitations iv. Easy maintenance
v. Appearance v. Safety of operations

3. Gaurav Mistry 3
 Bolt Design Design of MachineElements
After substituting the area
equation in
we can find
the core diameter 𝑑 𝑐 .
Later the nominal diameter
d can be obtained by,
𝑑 𝑐 = 0.84 d Area =
𝜋
4
x (𝑑 𝑐)2
Area = 𝜋 x d x t
Area =
𝜋
4
x (D2 – d2)
𝜎𝑡 =
𝑃
𝐴
c
If we say M16 bolt, then here 16 stands for 16 mm nominal diameter (d) of the bolt.
𝑑 𝑐
D d

4. c
Gaurav Mistry 4
 Numerical of Bolt Design Design of MachineElements
1. Determine the tensile load carrying capacity of M24 bolt having
allowable tensile stress = 60 (N/mm2) MPa, core diameter of bolt 𝒅 𝒄 =
0.84 times the nominal diameter d.
Solution: Figure shows the loading condition and area under tension of
the bolt.
Here given Nominal diameter d = 24 mm, Allowable stress 𝜎𝑡 = 60 N/mm2
Now, the Maximum tensile load required can be calculated from
𝜎𝑡 =
𝑃
𝐴
where A , and 𝑑 𝑐 = 0.84 d
Therefore 𝑑 𝑐 = 0.84 x 24 mm = 20.16 mm
Substituting 𝑑 𝑐 = 20.16 mm and 𝜎𝑡 = 60 N/mm2 in P = 𝜎𝑡 x
𝜋
4
x (𝑑 𝑐)2
P = 60 x
𝜋
4
x (20.16)2
P = 19152.33 N (Max. Tensile Load) ……………………….Answer

5. c
Gaurav Mistry 5
Design of MachineElements
2. Determine the tensile load carrying capacity of M16 bolt having
allowable tensile stress = 80 (N/mm2) MPa, core diameter of bolt 𝒅 𝒄 =
0.84 times the nominal diameter d.
Solution: Figure shows the loading condition and area under tension of
the bolt.
Here given Nominal diameter d = 16 mm, Allowable stress 𝜎𝑡 = 80 N/mm2
Now, the Maximum tensile load required can be calculated from
𝜎𝑡 =
𝑃
𝐴
where A , and 𝑑 𝑐 = 0.84 d
Therefore 𝑑 𝑐 = 0.84 x 16 mm = 13.44 mm
Substituting 𝑑 𝑐 = 13.44 mm and 𝜎𝑡 = 80 N/mm2 in P = 𝜎𝑡 x
𝜋
4
x (𝑑 𝑐)2
P = 80 x
𝜋
4
x (13.44)2
P = 13500 N (Max. Tensile Load) ……………………….Answer
 Numerical of Bolt Design

6. Gaurav Mistry 6
Design of MachineElements
3. The Bolt with circular head is assembled to a thick plate. An axial tensile load of 30kN is
acting on the lower side of the bolt. If the permissible stresses for the bolt material in
tensile, crushing and shear are 60 N/mm2, 80 N/mm2 and 40 N/mm2 . Determine the bolt
diameter and thickness of the head. If diameter of bolt head is D = 1.6 d (nominal diameter
of bolt) check the bolt for crushing failure.
Solution:
Given P = 30kN = 30000 N, 𝜎𝑡 = 60 N/mm2 , 𝜎𝑐= 80 N/mm2 and 𝜏 = 40 N/mm2
D = 1.6 d where
(i) For finding the bolt diameter d, considering the tensile failure of the bolt through threaded
portion (weak portion) under tension.
Now we know that Tensile strength of bolt P = 𝜎𝑡 x
𝜋
4
x (𝑑 𝑐)2
Substituting the values 30000 = 60 x
𝜋
4
x (𝑑 𝑐)2
Therefore, 𝑑 𝑐 = 25.23 mm and as we know 𝑑 𝑐 = 0.84 d we can write d =
𝑑 𝑐
0.84
,
d =
25.23
0.84
= 30 mm. Standard diameter of bolt d = 30 mm ……….Answer.
𝑑 𝑐
 Numerical of Bolt Design

7. Gaurav Mistry 7
Design of MachineElements
(ii) For finding the thickness t and diameter D of bolt head, considering the shear failure of
the bolt head portion under tension.
Now we know that Strength of bolt head under shearing (area of circumference, internal part of
ring), P = 𝜏 x 𝜋 x d x t
Substituting the values 30000 = 40 x 𝜋 x 30 x t
Therefore, 𝑡 = 7.95 = 8 mm
Thickness of bolt t = 8 mm ……….Answer.
Now, D = 1.6 d, D = 1.6 (30) = 48 mm.
(iii) For checking the bolt under crushing failure at bolt head section.
Strength of bolt at crushing area will be P =
𝜋
4
x (D2 – d2) x 𝜎𝑐
𝜎𝑐 = P /
𝜋
4
x (D2 – d2)
Substituting the values, 𝜎𝑐 = 30000/
𝜋
4
x (D2 – d2) = 30000/
𝜋
4
x (482 – 302)
𝜎𝑐 = 27.2 N/mm2 which is very less than allowable crushing stress 80 N/mm2.
Therefore the bolt is safe against crushing failure.
D d
t
 Numerical of Bolt Design

8. Gaurav Mistry 8
Design of MachineElements
 Punching Design
𝝅
𝟒
x (d)2
(Cross sectional
Area of Punch
under crushing)
𝝅 x d x t
(Cross sectional
Area of Hole
under shear)
d
t
d
 For Punching the plate, it is required to create fracture in the plate and hence
required punching force will be based on plate’s ultimate shear strength.(𝜏 𝑢)
 Punch is used to cut the hole in the plate but should not deform, hence the
stresses induced in the punch due to crushing should not exceed the permissible
crushing stress. (𝜎𝑐)
 i.e. Resisting forcefor the punch (strength of Punch) > Maximum shear force for
punching then only the plate can be punched.
𝝈 𝒄 x (cross sectional area of punch under crushing) > 𝝉 𝒖 x (cross section area of
hole under shear)
(Circumference x thickness)

9. Gaurav Mistry 9
Design of MachineElements
 Punching Design
 Resisting force forthe punch (strength of Punch) > Maximumshear force for
punching then only the plate can be punched.
𝝈 𝒄 x (cross sectional area of punch under crushing) > 𝝉 𝒖 x (cross section area of
hole under shear)
Similarly,
For punching Square hole in a plate.
(Circumference x thickness)
𝟒 𝒂 x t
(Cross sectional
Area of Hole
under shear)
a
a x a = a2
(Cross sectional
Area of Punch
under crushing)

10. Gaurav Mistry 10
 Numerical of Punching Design Design of MachineElements
1. Determine the force required to cut the 60 mm diameter blank from 6 mm thick
plate. Ultimate shear stress for the plate material, 𝝉 𝒖 = 350 N/mm2
Solution:
Here given 𝜏 𝑢 = 350 N/mm2, Plate thickness t = 6 mm and Blank diameter d = 60 mm
Now, The (Shear) force required to cut the blank, F = ?
We know that
F = 𝜏 𝑢 x (cross section area of hole under shear)
F = 𝜏 𝑢 x 𝜋 x d x t
F = 350 x 𝜋 x 60 x 6
F = 329,867.229 N = 330 kN ……………….(Answer)
(Circumference x thickness)
d
t
𝝅 x d x t
(Cross sectional
Area of Hole
under shear)

11. Gaurav Mistry 11
Design of MachineElements
2. A 30 mm diameter punch has an allowable stress of 150 N/mm2 is
used to punch the hole. Determine the thickness of the plate
having ultimate shear stress for the plate material, 𝝉 𝒖 = 450 N/mm2
Solution:
Here given 𝜏 𝑢 = 450 N/mm2, 𝜎𝑐 = 150 N/mm2 and Blank diameter d = 30 mm
Plate thickness t = ?
Now for shearing the plate,
Strength of punch > Shearing force for punching hole
𝜎𝑐 x (cross sectionalarea of punch under crushing) >
𝜏 𝑢 x (cross section area of hole under shear)
𝜎𝑐 x
𝝅
𝟒
x (d)2 > 𝜏 𝑢 x 𝜋 x d x t
150 x
𝝅
𝟒
x (30)2 > 450 x 𝜋 x 30 x t
Solving we get
Plate Thickness t = 2.5 mm ………..(Answer)
(Circumference x thickness)
d
t
𝝅 x d x t
(Cross sectional
Area of Hole
under shear)
𝝅
𝟒
x (d)2
(Cross sectional
Area of Punch
under crushing)
d
 Numerical of Punching Design

12. Gaurav Mistry 12
Design of MachineElements
3. Determine the smallest size of the hole that can be punched in a 9 mm
Thick plate. Allowable crushing stress for the punch is twice the ultimate
shearing stress of the plate.
Solution:
Here given 𝜎𝑐 = 2𝜏 𝑢 Plate thickness t = 9 mm and
Mini diameter of hole d = ?
Now for shearing the plate,
Strength of punch > Shearing force for punching hole
𝜎𝑐 x (cross sectionalarea of punch under crushing) >
𝜏 𝑢 x (cross section area of hole under shear)
𝜎𝑐 x
𝝅
𝟒
x (d)2 > 𝜏 𝑢 x 𝜋 x d x t
Substituting 𝜎𝑐 = 2𝜏 𝑢 and t = 9
2𝜏 𝑢 x
𝝅
𝟒
x (d)2 > 𝜏 𝑢 x 𝜋 x d x 9
2 x
𝟏
𝟒
x (d) > 9
d > 9 x 2 = 18 mm
Therefore mini. dia. of hole, d = 18 mm ………..(Answer)
(Circumference x thickness)
d
t
𝝅 x d x t
(Cross sectional
Area of Hole
under shear)
𝝅
𝟒
x (d)2
(Cross sectional
Area of Punch
under crushing)
d
 Numerical of Punching Design

13. Gaurav Mistry 13
Design of MachineElements
4. Determine the smallest size of a square hole that can be punched in a
10 mm thick plate. Allowable crushing stress for the punch is 200 N/mm2
and the ultimate shearing stress of the plate is 𝝉 𝒖 = 300 N/mm2
Solution:
Here given 𝜎𝑐 = 200 N/mm2, 𝜏 𝑢 = 300 N/mm2 and
Plate thickness t = 10 mm. Mini size of square hole, each side of a = ?
Now for shearing the plate,
Strength of punch > Shearing force for punching hole
𝜎𝑐 x (cross sectionalarea of punch under crushing) >
𝜏 𝑢 x (cross section area of hole under shear)
𝜎𝑐 x (a)2 > 𝜏 𝑢 x 4 𝑎 x t
200 x (a)2 > 300 x 4 𝑎 x 10
200 x (a) > 300 x 4 x 10
a > 60 mm i.e. each side of sq. hole is 60 mm
Therefore mini. size of sq. hole, a x a = 60 mm x 60 mm..(Answer)
(Circumference x thickness)
a
a x a = a2
(Cross sectional
Area of Punch
under crushing)
𝟒 𝒂 x t
(Cross sectional
Area of Hole
under shear)
 Numerical of Punching Design

14. Gaurav Mistry 14
 Standard sizes using PR and Basic
series
Design of MachineElements
 Standardization is the process of making things of the same type all having
the same basic / Characteristics.
 For mechanical engineering design, shapes and dimensions of commonly
used machine elements such as bolts, screws and nuts, rivets, bearings,
keys, etc., the relevant standards are used.
 The Preferred numbers contains the powers of 10.
 The intermediate are obtained by dividing each decimal range into 5, 10,
20, 40 and 80 geometrically equal steps.
 The series thus obtained are the basic series and designated as R5, R10,
R20, R40 and R80 respectively. (Here R stands for French engineer, Renard who
first introduced this series)
 Each of this 5 basic series has one step ratios i.e. seriesfactors

15. Gaurav Mistry 15
 Standard sizes using PR and Basic
series
Design of MachineElements

16. Gaurav Mistry 16
 Standard sizes using PR and Basic
series
Design of MachineElements
Relation between Progression Ratio,
Range Ratio and Number of Terms in
Geometric Series.
Φ = 𝑅(
1
𝑁−1
)
Where,
Φ = Progression Ration,
𝑅 = Range Ratio,
R =
𝐿𝑎𝑠𝑡 𝑇𝑒𝑟𝑚
𝐹𝑖𝑟𝑠𝑡 𝑇𝑒𝑟𝑚
,
N = Number of Terms in Geometric Series
Meaning of progression ration
Φ = 𝑅(
1
𝑁−1
)
= 𝐴 = 10
5
= 10(
1
5
)
On comparing we see N – 1 = 5.
Therefore N = 6, total number of terms in series.
And R =
𝐿𝑎𝑠𝑡 𝑇𝑒𝑟𝑚
𝐹𝑖𝑟𝑠𝑡 𝑇𝑒𝑟𝑚
=10 so we can say
for example last term is 100 and first term is 10

17. Gaurav Mistry 17
 Derived series from Basic series Design of MachineElements
 Sometimes a required series is obtained by transforming the basic series which is
known as derived series.
 They are obtained by taking every second, third, fourth or Pth term of a basic series.
 These are designated by symbol of the corresponding basic series followed by the
solidus division - sign and the number 2, 3, 4 or p.

18. Gaurav Mistry 18
 Numerical of Standard sizes using
PR, Basic Series and Derived series
Design of MachineElements
1. Determine the standard size of 6 round bars having smallest diameter of 10 mm and
largest diameter of 32 mm. Also determine sectional modulus of each round bar.
Solution:
∅ 𝐷 = Progression Ratio for round bar diameter.
∅ 𝑍 = Progression Ratio for sectionalmodulus.
N = Number of standardround bars = 6
Here ,
∅ 𝐷 = 𝑅
(
1
𝑁−1
)
=
32
10
(
1
6−1
)
=
32
10
(
1
5
)
= 1.2619 = 1.26 (R 10 basic series)
∅ 𝐷 = 𝑅(
1
𝑁−1
)
where R =
𝐿𝑎𝑠𝑡 𝑇𝑒𝑟𝑚
𝐹𝑖𝑟𝑠𝑡 𝑇𝑒𝑟𝑚
=
𝐷 𝑚𝑎𝑥
𝐷 𝑀𝑖𝑛
=
32
10
and N = 6

19. Gaurav Mistry 19
 Numerical of Standard sizes using
PR, Basic Series and Derived series
Design of MachineElements

20. Gaurav Mistry 20
 Numerical of Standard sizes using
PR, Basic Series and Derived series
Design of MachineElements
2. Determine the six standard spindle speeds of the machine having minimum speed of
224 rpm and maximum speed of 710 rpm.
Solution:
Given, 𝑁 𝑚𝑎𝑥 = 710 rpm and 𝑁 𝑀𝑖𝑛 = 224 rpm and N = Number of speeds = 6
Here ,
Φ = 𝑅
(
1
𝑁−1
)
=
710
224
(
1
6−1
)
=
710
224
(
1
5
)
= 1.26 (R 10 basic series)
Here N1 = 224 x Φ0
= 224 x 1.260 = 224 rpm
N2 = 224 x Φ1
= 224 x 1.261 = 288 rpm
N3 = 224 x Φ2
= 224 x 1.262 = 356 rpm
N4 = 224 x Φ3
= 224 x 1.263
= 448rpm
N5 = 224 x Φ4
= 224 x 1.264 = 565 rpm
N6 = 224 x Φ5
= 224 x 1.265 = 711 rpm (approx. to 710)
Φ = 𝑅(
1
𝑁−1
)
where R =
𝐿𝑎𝑠𝑡 𝑇𝑒𝑟𝑚
𝐹𝑖𝑟𝑠𝑡 𝑇𝑒𝑟𝑚
=
𝑁 𝑚𝑎𝑥
𝑁 𝑀𝑖𝑛
=
710
224
and N = 6

21. Gaurav Mistry 21
Design of MachineElements
REFERENCES:
1. Design of Machine Elements, S. B. Soni, Atul prakashan.
2. www.google.com