Design mini-project for TY mechanical students

DOMC Project 2
Topic: Lifting Table
Batch: G2
Team members
1. Michael Susngi (111910138)
2. Pranav Jadhav (111910139)
3. Rahul Jadhav (111910140)
4. Ravindra Shinde (111910141)

1.Problem Statement:
“Design a suitable mechanical system for changing the height of study table
through 600mm”
Specifications:
1. Load acting on system = 1000N
2. Space constraint = area of table is 1000x700mm
3. Thickness of wooden table is 25mm
4. Standard height of study table = 750mm

2. Solution:
a. Given Data:
i. Application: Height adjustment drawing table
ii. Load capacity: 1000N
iii. Space constraints: area of table is 1000x700mm
iv. Working conditions: Normal working conditions
b.General assumptions of design:
i. Material is assumed to be isotropic and homogeneous.
ii. Load acting on the component is assumed to be static.
iii. If dynamic loading is occurring, then it is converted into quasi-static loading.
c. Specific assumptions of design:
i. Load is Distributing equally on each leg of table.

3.Theory:
a)Introduction:
Adjust height of table according to Height of students. Easy and fast lifting of
table with the help of motor.
b)Different Possible Mechanisms:
i. Scissor Lift
It is very simple and effective
mechanism for lifting, but it
cannot withstand large amount
of weight.

ii. Hydraulic lift 1.
Smooth operation with high lifting
capacity and precise movements. 2.
It is very costly, and operation is also
noisy.
iii. Worm Drive Mechanism:
1. Large speed reduction.
2. Large increase in torque.
3. Self-Locking Mechanism.
4. Less efficient
c) Most Suitable Mechanism: The most suitable
mechanism as per us is worm drive mechanism because:
1. It is more precise.
2. More efficient.
3. Makes less vibration when operating.

4. Is self-locking.
5. Operates silently
d)Working:
....Lifting-Table-v12.mp4
WhatsApp Video 2021-10-11 at 8.46.05 AM.mp4

e) Power flow Diagram:
Motor
Muff Coupling
Worm
Worm Gear
Shaft
Fork Arm
Table

f) Advantages:
1. The table can cater the needs of people of each height thus providing each
one with a perfect writing or drawing position
g) Disadvantages:
1. Cost is more than ordinary table.
2. Periodic maintenance is required
h)Necessity:
Many people around the world find it difficult to find a table with height
suitable to them and making a custom table for each person is costly so to
meet their demands of one-size-fits-all we have come up with this idea of
varying height table.

i) Orthographic views with notations:

j) List of components:
Part No. Part Name Type
9 Movable legs Custom
2 Worm & Worm gear Custom
8 Frame Custom
11 Shaft Custom
3 Bearing Standard
12 Muff Coupling Custom
10 Fork Arm Custom
13 Keys Custom
5 Nut Standard
4 Bolt Standard
6 Screws Standard
7 Table top Standard
1 Motor Standard

5.Design Procedure:
a) FBD of whole system:

b)Basic forces and moments calculations:
i. The load acting on the table is 𝐏 = 𝟏𝟎𝟎𝟎𝐍
ii. As we assumed that, the load is equally distributed on each leg. Thus, load
on each leg
∴
𝐏
𝟒
= 𝟐𝟓𝟎𝐍 … (1)
iii. Now, torque acting on system is T,
T = Force on each leg X Length of fork arm … (2)
iv. Length of fork arm is given by,

v. Length of fork arm = (3502
+ 3002
)0.5
= 461mm … (3)
vi. From equation (1), (2) & (3)
Torque = 250 × 0.460 × 4 = 𝟒𝟔𝟎𝐍𝐦
vii. Axial thrust on gear tooth is Pa, obtained from FBD of shaft

viii. Radial force on gear is obtained from FBD of worm,
Pr X d = Mt, after knowing d1 (diameter of worm) , one can easily get Pr
thus, Pr =
16000
40
= 400N

c) Selection of prime mover – Stepper motor
i. Reasons for using stepper motor
1. Light weight compared to Similar AC or DC motors
2. Silent and Vibrationless operation
3. Provides higher torque at lower speeds

ii. The torque needed on the shaft to lift the table 460,000Nmm, but as we
assuming the speed reduction of 30 from worm-to-worm gear
iii. Thus, required torque on worm should be
𝟒𝟔𝟎
𝟑𝟎
= 15.33Nm
iv. The standard value of motor available is of 16Nm torque, whose
specification is given by
“ Nema 42 ,3-phase, 50Hz, 1.1KW(1.475HP) 5A/220V/16Nm”
v.
Electrical specification Physical specification
Number of Phase: 3 Frame Size: 110 x 110 mm
Rated Current/phase: 5 A Shaft Diameter: Φ19 mm
Torque: 16 Nm Body Length: 162.5 mm
Operating voltage: 220V Weight: 10.5 kg
vi. Reference: https://www.omc-stepperonline.com/nema-42-3-phase-
stepper-motor-bipolar-16-nm-2266-24oz-in-5-0a-110x110x162-5mm.html

vii. Now let us find the time (t) required for lifting of table
for 1.1KW power of stepper motor, RPM is
𝑃 = 𝑇 × 𝜔 ∴ 𝑵 =
𝑃×60
2𝜋𝑇
=
1.1×103×60
2𝜋×16
= 𝟔𝟓𝟔. 𝟓𝟏𝟒 𝒓𝒑𝒎
viii. But due to speed reduction in worm drive, the actual rpm of shaft is
𝑵′
=
𝑁
30
= 𝟐𝟏. 𝟖𝟖 𝒓𝒑𝒎
Here, N & N’ denotes RPM of motor and shaft respectively.
ix. Tangential velocity of fork arm
∴ 𝑉 = 𝑟𝜔 = 45 × 0.5 × 10−3
×
2𝜋 × 21.88
60
= 0.05156 𝑚/𝑠
x. The total lift of the table is 600mm
𝑡 =
𝑑
𝑣
=
0.6
0.05156
= 11.63 𝑠𝑒𝑐
As the time obtained is nearly 12 sec which is quite
acceptable since we must move table slowly.
This states that the selection of motor is appropriate

6.Part by part design:
a) Part No. is 12 and Part Name is Muff coupling.
i. Function: Transmission of power from Motor to worm shaft

ii. Free Body diagram for Muff Coupling based on Torque:
iii. Most critical mode of failure under application of torque is,
torsional shear failure in muff
Worm Shaft
Muff
Motor Shaft

iv. The martial select for Muff is grey cast iron with grade FG200
The Muff is usually made of grey cast iron of Grade FG 200.
v. Sut = 200N/mm2
vi. Factor of safety (fs) = 2, because
vii. Allowable shear stress is,
Ʈ =
𝟐𝟎𝟎
2
= 100N/mm2
viii. Standard empirical relations available for design of Muff are,
d = shaft diameter = 19mm … (From motor specification)
D = 2d+13 = 2(19) +13 = 51mm
L = 3.5d = 66.5mm ≈ 68mm
ix. The torsional shear stress in the sleeve is calculated by treating it as a
hollow cylinder.
J =
𝜋(𝐷4− 𝑑4)
32
=
𝜋(514− 194)
32
= 694803 mm4
&
r =
𝐷
2
= 25.5mm
Ʈ =
𝑇×𝑟
𝐽
=
460000×25.5
694803
= 16.88 < 100N/mm2
i.e., Design is safe

x.
b)Part No. is 13 and Part Name is Keys for motor and worm shaft.
i. Function: Prevents a relative rotation between the two parts and may
enable torque transmission to occur.
ii. 2 orthographic views of key
Notation Meaning Value(mm)
d Shaft diameter 19
D Muff diameter 51
L Length of muff coupling 68

iii. FBD of key:
iv. Modes of failure:
1. Torsional shear failure
2. Crushing failure

v. The martial select for Key is plain carbon steel with grade C50
Because, this steel is suitable for making keys, shaft, cylinder, and
machined components requiring wear resistance
(Refer: PSG design data book, Page No. 1.9 & 1.10)
vi. Properties of C50
1. Sut = 660N/mm2
2. Syt = 380N/mm2
vii. Factor of safety chosen is 2 because,
viii. Allowable stresses:
1. c =
380
2
= 190N/mm2
2. Ʈ =
0.5 ×380
2
= 95N/mm2
ix. For square key, a =
𝑑
4
, where d is shaft diameter
∴ 𝑎 = 4.75𝑚𝑚
from table 9.3(design book by Bhandari), standard size available is 6X6mm
also, 𝑙 =
𝐿
2
, half the length of coupling
∴ 𝒍 = 𝟑𝟒𝒎𝒎
x. Hence key is specified as 6 x 6 x 34 mm

xi. Checking for induced stresses,
Ʈ =
2×𝑇
𝑎×𝑎×𝑙
=
2×16000
6×6×34
= 26.14N/mm2
< 95N/mm2
Also, c = 2 Ʈ = 52.29N/mm2
< 190N/mm2
Hence, design is safe
xii.
c) Part No. is 2 and Part Name is Worm and worm gear.
i. Functions: Use for speed reduction, also it transmits power in
perpendicular direction
ii. 2 orthographic values of worm pair,
Notation Meaning Values(mm)
a Width and height of key 6
l Length of key 34

iii. Gear Material selection (Refer: PSG Data book pg. 8.45)
Justification: 10C4, Case hardened steel is used for worm due to its
durability, high hardness, and mechanical strength.
Phosphor bronze is selected for worm wheel to reduce wear of the worms
iv. From PSG data book, we assumed sliding velocity (Vs) of 3m/s
v. Thus, Design Crushing stress = c = 159N/mm2
design bending stress = b = 64N/mm2
vi. Selection of worm gear pair:
We are going with high-speed reduction. Main reason is that with smaller
gear reduction the input torque on system is quiet high which
consequently increases the weight of motor. That’s why to reduce weight
of overall system we have gone with speed reduction up to 30
From PSG data book pg.no. 8.47
The best worm gear pair obtained is as “1/30/10/4”
where, Z1 =no. of starts on worm = 1
Z2 = no. of teeth on worm gear = 30
q = diametral quotient = 10
m = module of gear = 4

vii. Pressure angle is assumed to be 20o
. This is a standard pressure angle
used for worm gears because it avoids objectionable undercutting
regardless of lead angle.
viii. Now, we can use standard empirical relations to find out dimensions of
worm & worm gear
(Refer: PSG data book, Table 35, pg.no. 8.43)
ix.

xi.
a Distance between centers 80
d1 PCD of worm 40
da1 Tip diameter of worm 48
ɣ Lead angle 5.71o
df1 Root diameter of worm 30.48
Px Pitch of worm 12.56
d2 PCD of worm gear 120
da2 Tip diameter of worm gear 127.92
df2 Root diameter of worm gear 110.4

d)Part No. is 11 and Part Name is shaft.
i. Design of based on ASME code
ii. Function: transmission of torque and power
iii. 2 orthographic views of shaft

iv. FBD of shaft:
v. Possible modes of failure:
1. Bending failure
2. Torsional shear failure

vi. The material select for shaft is Plain carbon steel with grade 40C8
because, this steel is used for making shafts, crankshaft, automobile axle
(Refer: PSG design data book, pg.no. 1.9, 1.10)
vii. Properties of material:
1. Sut = 580N/mm2
2. Syt = 330N/mm2
viii. Allowable stress calculation:
Ʈmax = 0.30xSyt = 99N/mm2
or Ʈmax = 0.18xSut = 104.4N/mm2
hence Ʈmax = 99N/mm2
… (which is less)
Also, shaft is keyed for mounting of worm gear,
∴ Ʈmax = 0.75x99 = 74.25N/mm2
ix.

x. Therefore, resultant bending moment at the point of gear is
Mb = (1000002
+ 1250002
)0.5
= 160078.10Nmm
Mt or Torque = 16000 x 30 = 480000Nmm
xi. Take Kb = Kt =1.5 (Refer: Design book by Bhandari, Table 9.2, pg.no.334)
xii. By maximum shear stress theory,
Ʈmax =
16
𝜋𝑑𝐷𝑠3 √(𝐾𝑀2 + 𝐾𝑇2)
𝑑3
=
16
𝜋Ʈ
√((1.5 × 160078.1)2 + (1.5 × 480000)2 = 52060.23mm3
Ds = 37.34mm
But we haven’t considered the weight of worm gear while calculating
diameter of shaft, hence we consider the shaft of diameter Ds=45mm
xiii.
Ds Diameter of shaft 45

e) Part No. is 13 and Part Name is Keys for worm gear and shaft.
i. Function: Mounting Worm gear on shafts
ii. 2 orthographic views of key

v. The martial select for Key is plain carbon steel with grade C50
Because, this steel is suitable for making keys, shaft, cylinder, and
machined components requiring wear resistance
(Refer: PSG design data book, Page No. 1.9 & 1.10)
vi. Properties of C50
1. Sut = 660N/mm2
2. Syt = 380N/mm2
vii. Factor of safety chosen is 1.5 because, key failure is more economical than
gear or shaft failure in any extreme conditions.
viii. Allowable stresses:
1. c =
380
1.5
= 253.33N/mm2
2. Ʈ =
0.5 ×380
1.5
= 126.67N/mm2
ix. For rectangular sunk key & shaft diameter D= 45mm
we get bxh = 14x9 and keyway depth as 5.5mm
from table 9.3(design book by Bhandari)
and effective length is found using empirical relations available for key
∴ 𝒍 = 𝟏. 𝟓 × 𝟒𝟓 = 𝟑𝟒𝒎𝒎
x. Hence key is specified as 14 x 9 x 67.5 mm

xi. Checking for induced stresses,
Ʈ =
2×𝑇
𝑏×ℎ×𝑙
=
2×480000
14×9×67.5
= 112.87N/mm2
< 126.67N/mm2
Also, c = 2 Ʈ = 225.75N/mm2
< 253.33N/mm2
Hence, design is safe
a Width of key 14
h Height of key 9
l Length of key 67.5

f) Selection of bearing: Part No. 3 and part name is Bearing
i. From FBD of shaft, the radial and axial force acting on bearing are 200N &
250N respectively.
ii. Selected type of bearing for given application is Deep groove ball bearing,
because it has following advantages-
1. It can carry considerable axial thrust along with radial load at high
speed (Refer: PSG Data book, pg.no. 4.1)
iii. The equivalent dynamic load acting on the bearing is given by,
𝑷 = 𝑿𝑭𝒓 + 𝒀𝑭𝒂 … (1)
The value of X and Y are find using ratios,
𝐹𝑎
𝐹𝑟
=
500
400
= 1.25 < 𝑒
Thus from PSG data book, pg.no. 4.4, we get
X = 1 and Y = 0
iv. Hence equation (1) gives P = 200N
v. Expected bearing life in hrs. Lh = 6000hrs
Machines used intermittently such as lifting tackle, hand tools and

household appliance, the recommended life is from 4000 to 8000 h
(Refer: design book by Bhandari, Table 15.2, pg.no.573)
vi. L10 = 60x (RPM of shaft) xL(h)/106
RPM of shaft = RPM of Motor/Velocity ratio … (1)
Power of motor = torque of motor x Angular velocity
∴ 1100 = 16 ×
2𝜋 𝑁
60
implies N(m) = 656.51rpm
Thus equation (1) yields RPM of shaft as Ns=21.88rpm
∴ 𝐿 =
60 × 21.88 × 6000
106
= 7.8768 𝑚𝑖𝑙𝑙𝑖𝑜𝑛𝑠 𝑟𝑒𝑣
vii. Dynamic load capacity is given by C = Px(L10)1/3
= 398N
viii. For series 60 and shaft diameter of 45mm
Bearing No. 6009 of extra light series the dynamic load capacity given is
1630N > 398N
This we can go with this bearing
ix. Main dimensions of bearing are,
d = inner diameter of the bearing = 45mm
D = outer diameter of the bearing = 75mm
B = axial width of the bearing = 16mm

g)Part number is 8 and part name is
Frame
i. Function: Foundation for motor
and support structure for fixed pillar.
ii. 2 orthographic views of Frame

iii. FBD
iv. Possible modes of failure: Bending Failure
v. Material selection
The material select for Frame is Carbon Steel with grade C15
Because, it is used for making lightly stressed part, easily machinable and
suitable for cold working such as bending
(Reference: PSG data book, Table 2.2, page no. 1.9, Third edition)

vi. Properties of material
1. Syt = 370N/mm2
2. Sut = 240 N/mm2
vii. Factor of safety chosen is 2, because there is no risk for human life.
viii. Design Calculations:
1) Motor Plate:
Total Force = 10.5 x 9.81 x Ftorque
= 103N + 460 x 1000/80
= 103N + 5750N
Total Force = 5853N

Bending Moment =
WL
4
= 204.855Nm
𝜎max =
𝑀𝑏 𝑌
𝐼
𝑆𝑦𝑡
𝑓𝑜𝑠
=
204.855 𝑋
𝑡
2
𝑋 12
140 𝑋 𝑡3
185 𝑋 106
=
8.7795
𝑡2
𝑡2
= 4.75 𝑋 10−5
𝑡 = 6.88𝑚𝑚
𝑡 ~ 7𝑚𝑚

2) Fixed Legs:
F = 1713.25 N x 2
𝜎max = σdirect + σbending
𝑠𝑦𝑡
𝑓𝑜𝑠
=
𝑃
𝐴
+
𝑀𝑏 . 𝑌
𝐼
370
2
=
3726.5
2 . 𝑡 . 𝑙
+
3426.5 𝑋
𝑙
2
𝑋
𝑙
2
𝑙 . 𝑙3
12
−
𝑙(𝑙−2.𝑡)3
12
185 =
3426.5
2.𝑡.30
+
3426.5 𝑋 303𝑋 3
304
12
−
30(30−2.𝑡)3
12
By solving this equation by trial and error method we get value of t
𝑡 = 0.6307𝑚𝑚
𝑡 ~ 1𝑚𝑚

3) L-Frame:
Bending Moment =
WL
4
= 512Nm
𝑌
̅ =
𝑎1.𝑌1+ 𝑎2𝑌2
𝑎1+ 𝑎2
=
30𝑡2+30𝑡− 𝑡2 𝑋 30𝑡− 𝑡2
2 𝑋 30𝑡 + 30𝑡− 𝑡2
𝐼𝑥𝑥 = 𝐼𝐺1
+ 𝑎1ℎ1
2
+ 𝐼𝐺2
+ 𝑎2ℎ2
2
𝐼𝑥𝑥 = 30. (
𝑡
2
)3
+ 30𝑡. (𝑌
̅ −
𝑡
2
)3
+ 𝑡. (30 − 𝑡)3
+ (30 − 𝑡). 𝑡. [(
30−𝑡
2
+ 𝑡) − 𝑌
̅]3

𝜎max =
𝑀𝑏 (30−𝑌
̅)
𝐼
185 =
512000.(30−𝑌
̅)
𝐼𝑥𝑥
By solving this equation using trial and error method
𝑡 = 2.9427𝑚𝑚
𝑡 ~ 3𝑚𝑚
t1 Thickness of L section 3
t2 Thickness of Plate 7
t3 Thickness of fixed pipe 1

h)Design of Nut and Bolt pair: Part number is 4,5 and Part Name is Nut
and bolt
i. Function: To attach motor to the frame.
ii. Possible modes of Failure
1. Tension failure
2. Shear failure
iii. Material selection:
The material select for shaft is Plain carbon steel with grade 40C8
because, this steel is used for making shafts, crankshaft, automobile axle
(Refer: PSG design data book, pg.no. 1.9, 1.10)
iv. Properties of material:
1. Sut = 580N/mm2
2. Syt = 380N/mm2
v. Factor of safety chosen is 3.5 because bolt is more prone to fail due to
vibrations of motor.

vi. Design calculations:
Bolt: 𝑆𝑦𝑡 = 380𝑁/𝑚𝑚2
Load per bolt = 𝑃 =
5853
4
= 1463.25𝑁
𝜎𝑚𝑎𝑥 =
𝑃
𝐴
=
1463.25
𝜋
4
.𝑑𝑐
2
𝑆𝑦𝑡
2 𝑋 𝑓𝑜𝑠
=
1463.25
𝜋
4
.𝑑𝑐
2
𝑑𝑐
2
=
1463.25 𝑋 4
𝜋 𝑋 54.28
𝑑𝑐 = 5.85𝑚𝑚
𝑑 =
𝑑𝑐
0.81
= 7.23𝑚𝑚
𝑑 ~ 8𝑚𝑚
From above dimensions we can select Bolt size as, M8
Db Diameter of Bolt 8

i) Design of movable Leg, part number is 9 and part name is movable Leg
i. Function: To hold the table in place
ii. Orthographic projections:
iii. Possible modes of failure:
1. Compression failure
iv. Material selection:

Material selected for the movable pipe is carbon steel C30 which has
undergone tempering and hardening.
v. Properties of material:
1. Sut=600 N/mm2
2. Syt=400 N/mm2
vi.
vii. Design calculation:
Frame: Factor of safety considered is 3
Therefore,
𝜎𝑚𝑎𝑥 =
𝑆𝑢𝑡
3
=
600
3
= 200 𝑁/𝑚𝑚2
𝑃 =
1000
4
= 250 𝑁
𝜎𝑚𝑎𝑥 =
𝑃
𝐴
200 =
250
2 . 28 .𝑡
𝑡 = 0.625 𝑚𝑚 ≅ 0.7 𝑚𝑚

Pin: Factor of safety considered is 2.5
Therefore,
𝜎𝑚𝑎𝑥 =
𝑆𝑢𝑡
2.5
=
600
2.5
= 240
𝑁
𝑚𝑚2
𝜎𝑚𝑎𝑥 =
𝑃
𝐴
240 =
250
2 .
𝜋
4
. 𝑑2
𝑑2
=
25
12 . 𝜋
𝑑 = 5.9 𝑚𝑚 ≅ 6 𝑚𝑚
t Thickness of pipe 1
d Diameter of supporting pin 6

j) Design of fork arm, part number is 10 and part name is Fork arm
i. Function: To transfer power from shaft to the moving leg
ii. Orthographic projections:
iii. Possible modes of failure:
1. Bending failure
Material selected for fork arm is Carbon Steel C40 which has undergone
hardening and tempering.
Sut=750 N/mm2
vi. Design calculations:

For fork arm: Factor of safety considered is 3
Therefore,
𝑃 =
1000
4
= 250 𝑁
400 = 𝑙𝑐 + 𝑙𝑎
𝐵𝑦 𝑝𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑎𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚,
4002
= 3002
− 𝑙𝑎
2
𝑙𝑎 = 264.575 𝑚𝑚 ≅ 265 𝑚𝑚
∴ 𝑙𝑐 = 135 𝑚𝑚
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑎 𝑐𝑎𝑛𝑡𝑖𝑙𝑒𝑣𝑒𝑟 𝑏𝑒𝑎𝑚 𝑤𝑖𝑡ℎ 𝑝𝑜𝑖𝑛𝑡 𝑙𝑜𝑎𝑑 𝑜𝑛 𝑓𝑟𝑒𝑒 𝑒𝑛𝑑,
𝑎𝑙𝑠𝑜 𝑙𝑒𝑡 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 𝑏𝑒 3.
𝑀
𝐼
=
𝜎𝑚𝑎𝑥
𝐹𝑂𝑆 .𝑦
250 .135
1
12
.𝑡 .𝑏3
=
750
3 .
𝑏
2

𝐿𝑒𝑡 𝑡 = 10𝑚𝑚
∴ 𝑏2
= 48.769
𝑏 = 6.983 𝑚𝑚 ≅ 7 𝑚𝑚
Notation Meaning Value
lc Length of the fork prongs 135
B Height of fork prongs 7

k) Welding on the frame
i. Function: To attach all the different L shape Steel to form a frame
ii. FBD
iii. Possible modes of Failure
1. Shear failure
2. Bending failure
The material select for weld electrode is E6010 and being done by stick
metal arc welding
(Refer: PSG design data book, pg.no11.11)
1. Permissible tensile strength: 138N/mm2
2. Permissible shear strength = 70N/mm2
Design calculations:
Net weight on the plate(FN)= FG + FM

FN=5750N+130N
FN=5853N
i)Considering for the support plate
τ =
Fn
A
A =
5053
70
= 83.614 mm2
Leff =
83.614
3
= 28mm
Therefore welding on the palte should be = l1 + 15 = 29mm
Now taking the leg length =
3
0.707
≃ 5mm
Therefore since welding is on both side the weld size could be 2.5mm
ii) Considering the frame that supports the support plate of motor:
Now the frame will fail due to shear by both primary and secondary
shear
i) Considering the primary shear
τ1 =
FN
4A

τ1 =
5853
4A
N
mm2
ii)Considering secondary shear
Maximum bending moment will occur at the mid point
M = FN ×
l
8
M = 512137.5Nmm
Now taking the summation of polar moment of inertia
Jnet = J1 + J2 + J3 + J4
Jnet = 4J1
J = 4A (
l2
12
+ r1
2
)
J = 490000A mm4
Now taking the shear equation
τ2 = M ×
r
J
τ2 = 512137.5 ×
350
490000A
τ2 =
365
A
N
mm2
Now from the first consideration

τ1 =
731.625
A
τnet = τ1 + τ2
τnet =
1096.625
A
N
mm2
∴ A =
1096.625
70
mm2
considering thickness to be 3mm
l = 5.22 + 15 mm
therefore weld of lenth of 20mm and thickness of 3mm
to be welded from both the side
iii)considering the outermost structure
now calculation of centre of gravity
(30 + 30)x = 0 × 30 + 30 × 15
x = 7.5 mm
(30 + 30)y = 30 × 15 + 30 × 30
y = 22.5 mm
r1 = GG1 = √7.52 + 7.52 = 10.61mm
r2 = GG2 = √(15 − 7.5)2 + (30 − 22.5)2
since r1 = r2

J1 = J2 = J
J = A (
l2
12
+ r1
2
) = 187.36A
Jnet = 2 × J = 374.72A mm4
rmax = √(30 − 7.5)2 + (30 − 22.5)2 = 23.71mm
Now, considering the primary shear stress
τ1 =
(FN + 1000)
4A
=
(5853 + 1000)
4 × 60 × t
τ1 =
28.15
t
Considering the secondary shear
Mb =
(
FN
4
+ 500) × 235
2
Mb = 230681.875Nmm
τ2 =
M × rmax
Jnet
τ2 =
230681.875 × 23.71
824.86A
Now from the diagram
θ = tan−1
22.5
7.5

θ = 71.565°
∴ x = 90 − 71565° = 18.435°
τ2x
= τ2 cos 18.435° = 6290.527 (
N
mm2
)
Similarly
τ2y
= τ2sin18.685° = 2096.85
τ2ynet
= (
2096.85
30t
) + (
28.55
t
)
τ2ynet
=
209.68
t
τnet =
231.685
t
N
mm2
Now considering the bending
Bending moment Mb =
FN × l
8 × 2
Mb = 2560.68.75 Nmm
Now
Ixx =
30t3
12
mm4
By parallel axis theorem
IG = Ixx + Ax2

IG = (
30t3
12
) + 1687.5t
IG = 1687.5t mm4
For part II
Ixx =
t × 303
12
mm4
IG =
t × 303
12
+ 30t × 7.52
mm4
IG = 2250t + 1687.5t mm4
Now adding both the moment of inertia
IGnet = 2985t mm4
Now σ =
M × ymax
IGnet
σ =
256068.75 × 22.5
5985t
N
mm2
σ =
926.66
t
N
mm2
Now by principle of maximum shear thoery
τmax = √τnet
2 + (
σ2
4
)
N
mm2

τmax =
534
t
N
mm2
t =
534
70
mm
t~8mm
Checking for failure
Case I
τ1 =
5853
4 × 3 × 56
N
mm2
τ1 = 8.40
N
mm2
∴ τ1 is less than the permissible value therefore the design is safe
Case II
τ = τ1 + τ2
τ =
731.625
A
+
365
A
N
mm2
τ = 19
N
mm2

Case III
τ = √τ1
2 +
σ2
4
N
mm2
τ = √(
231.685
6
)
2
+ (
926.66
8
× 2)
2
τ = 64.754
N
mm2
∴ The structure is safe
Weld spot Leg length (mm) Length(mm)
Supporting plate 3 29
Frame for supporting plate 5 30
Outer frame 12 30

Part No. Part Name Specifications Material Remark
1 Motor 1.1Kw,16Nm,3-phase Stepper motor Standard
2 Worm Gear Z1/Z2/q/m = 1/30/10/4
a = 80mm Phosphor bronze Standard
3 Worm Z1/Z2/q/m = 1/30/10/4
a = 80mm C10 Standard
4 Bearing Design NO. 6009 Standard
5 Bolt M8x1.25 C40 Standard
6 Shaft Nut M22x2.5 C40 Standard
7 Nut M8x1.25 C40 Standard
8 Screw M5x0.8x15 Standard
9 Table-top 1000x700x25 Wood standard
10 Frame L section = PxPxtL = 30x30x3
Motor plate = bxbxtp= 140X140x7
Fixed leg = SxSxtf = 30x30x1
C15 Custom
11 Movable leg nxnxt = 27x27x1 C30 Custom
12 Fork arm l = 334mm
lc = 135mm
ha = 7mm
C40 Custom
13 Shaft DxL = 45x760 C40 Custom
14 Muff coupling DoxDixlm = 19x51x68 FG 200 Custom
15 Keys bxhxlk = 14x9x67.5 C50 Custom

Notations Meaning
Z1 No. of starts on worm
Z2 No. of teeth on worm gear
q diametral quotient
m Module of worm gear
P Side of L section
tL Thickness of L section
b Side length of Motor Plate
tp Thickness of motor plate
S Cross section of fixed leg
tf Thickness of fixed leg
n Cross section of movable leg
t Thickness of movable leg
l Overall length of fork arm
lc Slot length of fork arm
ha Height of form arm
D Diameter of shaft
L Overall length of shaft
Do Outer diameter of Muff coupling
Di Inner diameter of Muff coupling

lm Length of muff coupling
b Width of key
h Height of key
lk Length of key

▪ REFERENCES:
o PSG Machine Design data book
o Design of Machine Elements by V.B. Bhandari (Third edition)
o Machine Drawing (New Age publishers, Third edition)

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