NAME- ANURAG KUMAR

CLASS- B.TECH MECHATRONICS 5TH SEM

ROLL NO. – 41822002

MACHINE DESIGN PRACTICALS

 PRACTICAL – 3

AIM - Design of boiler riveted joint.

Question:
Design a double-riveted lap joint for a boiler shell
to sustain a steam pressure of 1.5 MPa 1.5MPa.
The shell has an internal diameter of 1.2 m 1.2m
and is made of steel with the following allowable
stresses:
 Allowable tensile stress ( 𝜎 𝑡 σ t ): 80 MPa

 Allowable shear stress ( 𝜏 τ): 60 MPa 60MPa

80MPa

 Allowable crushing stress ( 𝜎 𝑐 σ c ): 120 MPa

120MPa
Determine:
 The thickness of the boiler shell ( 𝑡 t).
 Dimensions of the rivets and their pitch.

Solution:
Step 1: Determine the thickness of the boiler shell ( 𝑡 t)
The hoop stress due to internal pressure is given by:
𝜎𝑡
𝑝⋅𝑑2⋅𝑡,σt
2⋅t p⋅d , where:
𝑝
1.5 MPa = 1.5 × 1 0 6 Pa p=1.5MPa=1.5×10 6 Pa, 𝑑 =
1.2 m = 1200 mm d=1.2m=1200mm, 𝜎 𝑡 = 80 MPa σ t
=80MPa. Rearranging for 𝑡 t:
𝑡
𝑝⋅𝑑2⋅𝜎𝑡
1.5 × 1 0 6 ⋅ 1200 2 ⋅ 80 × 1 0 6
1.8 × 1 0 9 160 × 1 0 6
11.25 mm . t= 2⋅σ t
p⋅d
2⋅80×10 6
1.5×10 6 ⋅1200
160×10 6
1.8×10 9
=11.25mm. Rounding up for safety, 𝑡
12 mm t=12mm.
Step 2: Dimensions of the rivets Assume standard
proportions for a double-riveted lap joint:
Diameter of rivet ( 𝑑 𝑟 d r ): 𝑑 𝑟
6𝑡
6 12
6 ⋅ 3.46
20.8 mm . d r =6 t =6 12 =6⋅3.46=20.8mm. Adopt 𝑑 𝑟
= 22 mm d r =22mm (nearest standard size).
Step 3: Pitch of rivets ( 𝑝 p) To ensure the joint is
strong in all failure modes (tensile, shear, crushing), the
pitch is determined by the weakest mode.
Tensile failure of the plate: The tensile strength of the
plate between two rivets must equal the strength of
the rivets in shear. Effective area of plate per pitch:
𝐴 plate
( 𝑝 − 𝑑 𝑟 ) ⋅ 𝑡 . A plate =(p−d r )⋅t. Force resisted by
plate:
𝐹 plate
𝐴 plate ⋅ 𝜎 𝑡
( 𝑝 − 22 ) ⋅ 12 ⋅ 80. F plate =A plate ⋅σ t

double shear: 𝐴 rivet

=(p−22)⋅12⋅80. Shear failure of rivets: Each rivet is in

2⋅𝜋⋅𝑑𝑟24
𝜋⋅2224
380 mm 2 . A rivet =2⋅ 4 π⋅d r 2
=π⋅ 4 22 2
=380mm 2 . Force resisted by rivets:
𝐹 rivet
𝑛 ⋅ 𝐴 rivet ⋅ 𝜏
2 ⋅ 380 ⋅ 60
45 , 600 N . F rivet =n⋅A rivet ⋅τ=2⋅380⋅60=45,600N.
Equating forces:
( 𝑝 − 22 ) ⋅ 12 ⋅ 80
45 , 600. (p−22)⋅12⋅80=45,600. 𝑝 − 22
45 , 600 960
47.5 mm . p−22= 960 45,600 =47.5mm. 𝑝 = 69.5 mm .
p=69.5mm. Adopt 𝑝 = 70 mm p=70mm (nearest
practical value).
Step 4: Verify crushing failure Crushing force per rivet:
𝐹 crush
𝑑𝑟⋅𝑡⋅𝜎𝑐
22 ⋅ 12 ⋅ 120
31 , 680 N . F crush =d r ⋅t⋅σ c =22⋅12⋅120=31,680N.
For two rivets:
𝐹 crush, total
2 ⋅ 31 , 680
63 , 360 N . F crush, total =2⋅31,680=63,360N. Since 𝐹
crush, total
𝐹 rivet F crush, total
F rivet , crushing is safe.
Final Design: Shell thickness: 12 mm 12mm. Rivet
diameter: 22 mm 22mm. Pitch: 70 mm 70mm.

RESULT – We designed the boiler riveted joint

 PRACTICAL – 4
AIM - Design of welded joint with eccentric loads.

Question:
A horizontal steel bracket is welded to a vertical column using

eccentric load 𝑃
a fillet weld along its edges. The bracket is subjected to an

10 kN P=10kN, which acts at a distance 𝑒 = 200 mm

e=200mm from the vertical column. The bracket dimensions

uniform on all sides of the bracket with a throat thickness 𝑡 =

are 300 mm × 200 mm 300mm×200mm, and the weld is

6 mm t=6mm.
Determine the following:
 The resultant force per unit length on the weld.

𝜏 allow 90 MPa τ allow =90MPa.

 Whether the weld is safe if the allowable shear stress is

Solution:
Step 1: Analyze the forces The eccentric load creates:
Direct Shear Force ( 𝐹 𝑠 F s ): The load 𝑃 P causes a uniform
shear stress across the weld. Torsional Moment ( 𝑀 M): The
eccentricity 𝑒 e generates a torsional moment about the
center of gravity of the weld. Step 2: Weld layout For
simplicity, assume the weld is a rectangle with total length:
𝐿 total
2 ⋅ ( 300 + 200 )
1000 mm . L total =2⋅(300+200)=1000mm. Step 3: Direct
shear force The force per unit length of weld due to direct
shear is:
𝑞𝑠
𝑃 𝐿 total
10 × 1 0 3 1000
10 N/mm . q s = L total
P
1000 10×10 3
=10N/mm. Step 4: Torsional moment effect The torsional
moment generates a varying force per unit length, depending

rectangular weld, the polar moment of inertia 𝐽 J is:

on the distance from the weld’s center of gravity. For a

𝐽
1 3 ⋅ ( 𝑎 ⋅ 𝑏 ⋅ ( 𝑎 2 + 𝑏 2 ) ) , J= 3 1 ⋅(a⋅b⋅(a 2 +b 2 )), where 𝑎
300 mm a=300mm, 𝑏 = 200 mm b=200mm:
𝐽
1 3 ⋅ ( 300 ⋅ 200 ⋅ ( 30 0 2 + 20 0 2 ) )
1 3 ⋅ 300 ⋅ 200 ⋅ ( 90 , 000 + 40 , 000 )
8.4 × 1 0 7 mm 4 . J= 3 1 ⋅(300⋅200⋅(300 2 +200 2 ))= 3 1
⋅300⋅200⋅(90,000+40,000)=8.4×10 7 mm 4 . The force per
unit length due to the torsional moment is:
𝑞𝑚
𝑀⋅𝑟𝐽,qm
J M⋅r , where 𝑀
𝑃⋅𝑒
10 kN ⋅ 200 mm = 2 × 1 0 6 N-mm M=P⋅e=10kN⋅200mm=2×10
6 N-mm, and 𝑟 r is the distance from the center of gravity of
the weld.
At the farthest points ( 𝑟 max r max ):
𝑟 max
( 300 2 ) 2 + ( 200 2 ) 2
15 0 2 + 10 0 2
22 , 500 + 10 , 000
180 mm . r max = ( 2 300 ) 2 +( 2 200 ) 2

150 2 +100 2

22,500+10,000 =180mm. Thus:

𝑞 𝑚 , max
2 × 1 0 6 ⋅ 180 8.4 × 1 0 7
4.29 N/mm . q m,max = 8.4×10 7
2×10 6 ⋅180 =4.29N/mm. Step 5: Resultant force per unit
length The resultant force per unit length at any point is:
𝑞 res
𝑞 𝑠 2 + 𝑞 𝑚 2 . q res
q s 2 +q m 2
. At the farthest point:
𝑞 res,max
1 0 2 + 4.2 9 2
100 + 18.4
118.4
10.9 N/mm . q res,max = 10 2 +4.29 2

100+18.4
118.4 =10.9N/mm. Step 6: Check weld safety The shear
stress in the weld is:
𝜏
𝑞 res,max 𝑡
10.9 6
1.82 MPa . τ= t q res,max

stress 𝜏 allow 90 MPa τ allow =90MPa. Therefore, the weld is

6 10.9 =1.82MPa. This is well below the allowable shear
RESULT – We designed the welded joint with eccentric loads.
 PRACTICAL – 5
AIM - Design of bolted joint with eccentric loads.

Question:
A steel plate is connected to a rigid support using a
bolted joint with four bolts arranged in a rectangular
pattern. The bolts are positioned at the corners of a

150mm×100mm. The joint is subjected to a load 𝑃 =

rectangle with dimensions 150 mm × 100 mm

10 kN P=10kN that acts at a distance 𝑒 = 200 mm

e=200mm from the center of the bolt group.
If the allowable shear stress for the bolts is 𝜏 allow
80 MPa τ allow =80MPa
determine:
 The diameter of the bolts to ensure safety.
 The maximum shear force on any bolt.

Solution:
Step 1: Analyze the forces The applied load 𝑃 P creates
two effects on the bolts:
Direct Shear Force ( 𝐹 𝑠 F s ): Uniformly distributed
across all bolts. Moment ( 𝑀 M): Eccentricity 𝑒 e
creates a moment that generates additional shear
forces on the bolts. Step 2: Calculate the direct shear
force The direct shear force is evenly distributed across
the four bolts:
𝐹𝑠
𝑃𝑛
10 , 000 4
2 , 500 N . F s = n P = 4 10,000 =2,500N. Step 3:
Calculate the torsional shear force due to the moment
The moment generated by the eccentric load is:
𝑀
𝑃⋅𝑒
10 , 000 ⋅ 200
2 , 000 , 000 N-mm . M=P⋅e=10,000⋅200=2,000,000N-
mm. The bolts are arranged in a rectangular pattern, so
we calculate the polar moment of inertia of the bolt
group:
𝐽
∑ 𝑟 2 , J=∑r 2 , where 𝑟 r is the distance of each bolt
from the center of the bolt group.
For a 150 mm × 100 mm 150mm×100mm rectangle:
𝑟1
( 150 2 ) 2 + ( 100 2 ) 2
752+502
5 , 625 + 2 , 500
8 , 125
90.14 mm . r 1 = ( 2 150 ) 2 +( 2 100 ) 2

75 2 +50 2

5,625+2,500
8,125 =90.14mm. Since there are four bolts:
𝐽
4⋅𝑟12
4 ⋅ 8 , 125
32 , 500 mm 2 . J=4⋅r 1 2 =4⋅8,125=32,500mm 2 . The
torsional shear force on a bolt is:
𝐹𝑚
𝑀⋅𝑟𝐽.Fm
J M⋅r . At the farthest bolt ( 𝑟
𝑟 1 r=r 1 ):
𝐹𝑚
2 , 000 , 000 ⋅ 90.14 32 , 500
180 , 280 , 000 32 , 500
5 , 544.92 N . F m = 32,500 2,000,000⋅90.14 = 32,500
180,280,000 =5,544.92N. Step 4: Resultant shear force
on the most stressed bolt The resultant shear force is
the vector sum of the direct and torsional components:
𝐹 res
𝐹 𝑠 2 + 𝐹 𝑚 2 . F res
F s 2 +F m 2
. Substitute the values:
𝐹 res
2 , 50 0 2 + 5 , 544.9 2 2
6 , 250 , 000 + 30 , 755 , 601.5
37 , 005 , 601.5
6 , 083.2 N . F res = 2,500 2 +5,544.92 2

6,250,000+30,755,601.5
37,005,601.5 =6,083.2N. Step 5: Determine the bolt
diameter The shear stress on the bolt is:
𝜏
𝐹 res 𝐴
4 ⋅ 𝐹 res 𝜋 ⋅ 𝑑 2 , τ= A F res

π⋅d 2
4⋅F res
, where 𝐴
𝜋 ⋅ 𝑑 2 4 A= 4 π⋅d 2
is the bolt's cross-sectional area.
Rearrange for 𝑑 d:
𝑑
4 ⋅ 𝐹 res 𝜋 ⋅ 𝜏 allow . d= π⋅τ allow
4⋅F res

. Substitute the values:

𝑑
4 ⋅ 6 , 083.2 𝜋 ⋅ 80
24 , 332.8 251.33
96.81
9.84 mm . d= π⋅80 4⋅6,083.2

251.33 24,332.8
96.81 =9.84mm. Rounding up, adopt 𝑑
12 mm d=12mm.
Final Design: Bolt diameter: 12 mm 12mm. Maximum
shear force on a bolt: 6 , 083.2 N 6,083.2N.

RESULT – We designed the bolted joint with eccentric loads.

 PRACTICAL – 6

AIM - Design of keys.

Question:

𝑑=50mm to transmit a torque of 𝑇 = 1200 N-m. The

Design a rectangular key for a shaft of diameter

key is made of steel with the following allowable

stresses:
 Allowable shear stress ( 𝜏 allow τ allow ): 60 MPa
 Allowable crushing stress ( 𝜎 𝑐 σ c ): 90 MPa .
Assume the standard proportions for a rectangular key:
 Width of key ( 𝑤 w) = 𝑑/4,
 Height of key ( ℎ h) = 𝑑/6.
Determine the length of the key ( 𝑙 l) required to
ensure safety.

Solution:

shaft diameter 𝑑
Step 1: Determine the dimensions of the key Given the
50 mm d=50mm, the dimensions of the rectangular key
are:
𝑤
𝑑4
50 4
12.5 mm , ℎ = 𝑑 6 = 50 6 = 8.33 mm . w= 4 d = 4 50
=12.5mm,h= 6 d = 6 50 =8.33mm. Rounding to
standard values:
𝑤
12 mm w=12mm, ℎ = 8 mm h=8mm. Step 2: Calculate
the torque transmitted The torque transmitted by the
key is shared between shear and crushing effects.

in shear is: 𝑇 shear

Shear stress in the key: The torque resisted by the key

𝜏 ⋅ 𝐴 shear ⋅ 𝑟 , T shear =τ⋅A shear ⋅r, where:

𝜏

𝐴 shear = 𝑤 ⋅ 𝑙 A shear =w⋅l, 𝑟 = 𝑑 2 = 25 mm = 0.025

60 MPa = 60 × 1 0 6 N/m 2 τ=60MPa=60×10 6 N/m 2 ,

m r= 2 d =25mm=0.025m. Rearranging for 𝑙 l:

𝑇
𝜏 ⋅ 𝑤 ⋅ 𝑙 ⋅ 𝑟 , T=τ⋅w⋅l⋅r, 𝑙
𝑇 𝜏 ⋅ 𝑤 ⋅ 𝑟 . l= τ⋅w⋅r T . Substitute the values:
𝑙
1200 60 ⋅ 12 ⋅ 0.025 . l= 60⋅12⋅0.025 1200 . 𝑙
1200 18

the key: The torque resisted by the key in crushing is: 𝑇

66.67 mm . l= 18 1200 =66.67mm. Crushing stress in

crush = 𝜎 𝑐 ⋅ 𝐴 crush ⋅ 𝑟 , T crush =σ c ⋅A crush ⋅r,

where:
𝜎𝑐

, 𝐴 crush = ℎ ⋅ 𝑙 . A crush =h⋅l. Rearranging for 𝑙 l:

90 MPa = 90 × 1 0 6 N/m 2 σ c =90MPa=90×10 6 N/m 2

𝑙
𝑇 𝜎 𝑐 ⋅ ℎ ⋅ 𝑟 . l= σ c ⋅h⋅r T . Substitute the values:
𝑙
1200 90 ⋅ 8 ⋅ 0.025 . l= 90⋅8⋅0.025 1200 . 𝑙
1200 18
66.67 mm . l= 18 1200 =66.67mm. Step 3: Final length
of the key Since both shear and crushing conditions
give the same result, the required length of the key is:
𝑙
67 mm ( rounded to the nearest whole number ) .
l=67mm(rounded to the nearest whole number). Final
Design: Key dimensions: 𝑤 = 12 mm , ℎ = 8 mm
w=12mm,h=8mm. Key length: 𝑙 = 67 mm l=67mm.

RESULT – We designed the keys.

 PRACTICAL – 7

AIM - Design of coupling.

Question:
Design a rigid flange coupling to transmit 𝑇

shafts have a diameter of 𝑑 = 50 mm d=50mm and the

500 N-m T=500N-m torque between two shafts. The

following permissible stresses:

 Allowable shear stress for shaft and bolts: 𝜏 allow

 Allowable crushing stress for bolts: 𝜎 𝑐 = 80 MPa

40 MPa τ allow =40MPa,

 Allowable shear stress for the flange material: 𝜏

σ c =80MPa,

flange = 20 MPa τ flange =20MPa.

Assume:

distributed on a pitch circle diameter (PCD) of 𝐷 𝑝

 Use 4 bolts in the flange connection, evenly

150 mm D p =150mm.
  Key dimensions are 𝑤 = 12 mm , ℎ = 8 mm
w=12mm,h=8mm.
Determine:
 The diameter of the bolts.
 Check the strength of the flange.
 Verify the design of the key.

Solution:
Step 1: Torque transmitted by the bolts The torque is
shared equally by all bolts. The force on each bolt due
to the torque is:
𝐹𝑏
𝑇𝑛⋅𝑅,Fb
n⋅R T , where:
𝑛
4 n=4 (number of bolts), 𝑅
𝐷𝑝2
150 2
75 mm = 0.075 m . R= 2 D p

2 150 =75mm=0.075m. Substitute the values:

𝐹𝑏
500 4 ⋅ 0.075
500 0.3
1666.67 N . F b = 4⋅0.075 500 = 0.3 500 =1666.67N.
Step 2: Diameter of the bolts The shear stress in the
bolts is:
𝜏
𝐹 𝑏 𝐴 , τ= A F b
, where 𝐴
𝜋 ⋅ 𝑑 𝑏 2 4 A= 4 π⋅d b 2

𝑑𝑏db :
is the cross-sectional area of the bolt. Rearranging for

𝑑𝑏
4 ⋅ 𝐹 𝑏 𝜋 ⋅ 𝜏 allow . d b
π⋅τ allow
4⋅F b

. Substitute the values:

𝑑𝑏
4 ⋅ 1666.67 𝜋 ⋅ 40 × 1 0 6
6666.67 125663.7
0.0531
7.29 mm . d b = π⋅40×10 6
4⋅1666.67

125663.7 6666.67

0.0531 =7.29mm. Adopt a standard bolt size of 𝑑 𝑏

10 mm d b =10mm.
Step 3: Check for crushing stress in the bolts The
projected area resisting crushing is:
𝐴𝑐
𝑑 𝑏 ⋅ 𝑡 , A c =d b ⋅t, where 𝑡
ℎ
8 mm t=h=8mm (thickness of flange). The crushing
stress is:
𝜎𝑐
𝐹𝑏𝐴𝑐.σc
Ac
Fb
. Substitute the values:
𝜎𝑐
1666.67 10 ⋅ 8
1666.67 80

=20.83MPa. Since 𝜎 𝑐 = 20.83 MPa < 𝜎 𝑐 , allow = 80

20.83 MPa . σ c = 10⋅8 1666.67 = 80 1666.67

MPa σ c =20.83MPa<σ c,allow =80MPa, crushing is

safe.
Step 4: Check the strength of the flange The torque
creates a shear stress in the flange material. The torque
resisted by the flange is:
𝜏 flange
𝑇 𝐴 𝑓 ⋅ 𝑅 , τ flange
A f ⋅R T , where 𝐴 𝑓
𝜋 ⋅ ( 𝑅 𝑜 2 − 𝑅 𝑖 2 ) A f =π⋅(R o 2 −R i 2 ) is the area

𝑅𝑜
of the flange resisting shear. Assume the outer radius

80 mm R o =80mm and inner radius 𝑅 𝑖 = 50 mm R i

=50mm.
𝐴𝑓
𝜋⋅(802−502)
𝜋 ⋅ ( 6400 − 2500 )
𝜋 ⋅ 3900
12252.86 mm 2 . A f =π⋅(80 2 −50
2 )=π⋅(6400−2500)=π⋅3900=12252.86mm 2 . The shear
stress in the flange is:
𝜏 flange
500 × 1 0 3 12252.86 ⋅ 75
500 × 1 0 3 919 , 000
0.544 MPa . τ flange = 12252.86⋅75 500×10 3

919,000 500×10 3
=0.544MPa. Since 𝜏 flange
0.544 MPa < 𝜏 flange,allow = 20 MPa τ flange
=0.544MPa<τ flange,allow =20MPa, the flange is safe.
Step 5: Verify the key The torque transmitted by the
key is:
𝑇
𝜏 ⋅ 𝑤 ⋅ 𝑙 ⋅ 𝑟 , T=τ⋅w⋅l⋅r, where 𝜏
40 MPa , 𝑤 = 12 mm , 𝑟 = 𝑑 2 = 25 mm
τ=40MPa,w=12mm,r= 2 d =25mm. Rearrange for 𝑙 l:
𝑙
𝑇 𝜏 ⋅ 𝑤 ⋅ 𝑟 . l= τ⋅w⋅r T . Substitute the values:
𝑙
500 × 1 0 3 40 ⋅ 12 ⋅ 25
500 × 1 0 3 12 , 000
41.67 mm . l= 40⋅12⋅25 500×10 3

12,000 500×10 3
=41.67mm. Adopt 𝑙
45 mm l=45mm for safety.
Final Design: Bolt diameter: 10 mm 10mm. Flange

50 mm 50mm. Key dimensions: 𝑤 = 12 mm , ℎ = 8

dimensions: Outer radius 80 mm 80mm, Inner radius

mm , 𝑙 = 45 mm . w=12mm,h=8mm,l=45mm.
RESULT – We designed the coupling.
 PRACTICAL – 8

AIM - Design of helical spring.

Question:
Design a helical compression spring for the following
conditions:
 Axial load, 𝑃
2000 N P=2000N,
 Maximum deflection, 𝛿 = 25 mm δ=25mm,
 Spring index, 𝐶 = 8 C=8,
 Maximum allowable shear stress, 𝜏 allow = 400

 Modulus of rigidity, 𝐺 = 80 GPa G=80GPa.

MPa τ allow =400MPa,

Determine:
The wire diameter ( 𝑑 d),
The mean coil diameter ( 𝐷 D),


The number of active coils ( 𝑁 𝑎 N a ),



 The solid length of the spring.
Solution:
Step 1: Determine wire diameter ( 𝑑 d) using shear
stress formula The maximum shear stress in a spring is
given by:
𝜏 max
16 𝑃 𝐶 𝜋 𝑑 3 𝐾 , τ max
πd 3
16PC K, where 𝐾 K is the Wahl's correction factor:
𝐾
4 𝐶 − 1 4 𝐶 − 4 + 0.615 𝐶 . K= 4C−4 4C−1 + C 0.615 .
Substitute 𝐶
8 C=8:
𝐾
4 ⋅ 8 − 1 4 ⋅ 8 − 4 + 0.615 8
31 28 + 0.076875
1.186. K= 4⋅8−4 4⋅8−1 + 8 0.615
28 31 +0.076875=1.186. Rearrange for 𝑑 d:
𝑑
( 16 𝑃 𝐶 𝐾 𝜋 𝜏 allow ) 1 / 3 . d=( πτ allow
16PCK ) 1/3 . Substitute the values:
𝑑
( 16 ⋅ 2000 ⋅ 8 ⋅ 1.186 𝜋 ⋅ 400 × 1 0 6 ) 1 / 3 . d=(
π⋅400×10 6
16⋅2000⋅8⋅1.186 ) 1/3 . 𝑑
( 303680 1256637061.44 ) 1 / 3
( 2.417 × 1 0 − 4 ) 1 / 3

1/3 =(2.417×10 −4 ) 1/3 =0.0628m=6.28mm. Adopt 𝑑 =

0.0628 m = 6.28 mm . d=( 1256637061.44 303680 )

6.5 mm d=6.5mm (standard size).

Step 2: Mean coil diameter ( 𝐷 D) The mean coil
diameter is related to the spring index:
𝐶
𝐷 𝑑 . C= d D . Rearrange for 𝐷 D:
𝐷
𝐶⋅𝑑
8 ⋅ 6.5

coils ( 𝑁 𝑎 N a ) using deflection formula The

52 mm . D=C⋅d=8⋅6.5=52mm. Step 3: Number of active

deflection of the spring is given by:

𝛿
8 𝑃 𝐷 3 𝑁 𝑎 𝐺 𝑑 4 . δ= Gd 4
8PD 3 N a
. Rearrange for 𝑁 𝑎 N a :
𝑁𝑎
𝛿𝐺𝑑48𝑃𝐷3.Na
8PD 3
δGd 4
. Substitute the values:
𝑁𝑎
25 ⋅ 80 × 1 0 9 ⋅ ( 6.5 × 1 0 − 3 ) 4 8 ⋅ 2000 ⋅ ( 52 × 1 0 −
3)3.Na
8⋅2000⋅(52×10 −3 ) 3
25⋅80×10 9 ⋅(6.5×10 −3 ) 4
.𝑁𝑎
25 ⋅ 80 ⋅ 0.00000179 8 ⋅ 2000 ⋅ 0.0001406 . N a
8⋅2000⋅0.0001406 25⋅80⋅0.00000179 . 𝑁 𝑎
3.58 2.25
1.59. N a
2.25 3.58 =1.59. Round up to 𝑁 𝑎
2 N a =2.
Step 4: Solid length of the spring The solid length ( 𝐿 𝑠
L s ) is the length of the spring when all coils are
touching, given by:
𝐿𝑠
𝑁 𝑡 ⋅ 𝑑 , L s =N t ⋅d, where 𝑁 𝑡
𝑁 𝑎 + 2 N t =N a +2 is the total number of coils,
accounting for end coils.
𝐿𝑠
( 2 + 2 ) ⋅ 6.5

diameter: 𝑑 = 6.5 mm d=6.5mm, Mean coil diameter:

26 mm . L s =(2+2)⋅6.5=26mm. Final Design: Wire

𝐷 = 52 mm D=52mm, Number of active coils: 𝑁 𝑎 = 2

N a =2, Solid length: 𝐿 𝑠 = 26 mm L s =26mm.
RESULT – We designed the helical spring.