The document discusses the design of a pneumatic robot for costic recovery in a textile factory. It outlines several objectives for the project, including designing each component mathematically and determining stress analysis. It also discusses limitations like a lack of internet access and cost constraints. Several chapters analyze the design concepts, including two models - one powered by an engine and the other by an electric motor. Design considerations like material selection, working mechanisms, and input parameters are evaluated. Specific component designs are analyzed, such as the column support, shaft, and connecting base and colemun. The document provides details on the problem, objectives, methodology, conceptual designs, and structural analysis of key parts for the pneumatic robot.

1. Chapter one
Project (case study)
1. Costic recoveryusing pneumatic robot in processing department.
1.1 Introduction
The textile industry is one of the most complicated industry among the manufacturing
industry. Costic recovery is one of the major problems faced in Almeda textile factory. a
study of processing department that there are many complicated processes and chemicals
used throughout the production. for the purposes of continuous blanching in processing
department they are used costic soda(NaoH).this chemical is strong and hazard .
1.2. Problem ofthe statement
In every stages of almeda textile factory various types of chemicals are used. those chemical are
hazards to human being. they can damage seriously to different parts of body such us blindness
of eye, skin,mantality retarded.evnthogh this said effect those chemicals are being performed by
human labor in this factory in which is not suitable manner by caring through their body. their
for this costic is very dangerous which causes dieses to human being do to their strength.
The conventional treatment of this hazard representation of this machine is pneumatic robot. This
works instead of human energy. This type of machine is very easy to operate, time sever, which
working fluid is air. Punmatic robot machines are used to left in to large height to transport from
one elevation to another. those type of machines are used to transport goods from one room to
another room ,but in some department there is not qualified machines to be functional like in
costic recovery. For those purposes I am starting to design simple pneumatic robot which is
operated with one person.
1.3. Objective of the Project
1.3.1. General Objective
The main objective of this project is designing and modeling of pneumatic robot machine.
1.3.2. Specific Objective
 To design mathematically for each components of pneumatic robot machine.
 To determine and calculate stress analysis, detail drawing of pneumatic robot machine
and assembly drawing of the machine by solid work software.
1.4 Scope of the Study

2. This project will be designed and modeled based on studies conducted by referring the existing
reference books and mechanical software. The main scope of this project is starting from
conducting literature review, design analysis of the components up to conducting modeling and
manufacturing drawing pneumatic robot machine.
1.5. Limitation of study
There is not internet assess
There is not enough cost to manufacture
There is not enough information that given from the librarian.
CHAPTERTHREE
2.3. Literature review
Textile industry is one of the most complicated industries among manufacturing industry (seicuk,
2005).costic recovery is the process of addition water to costic soda for the purpose of
continuous bleaching. in almeda textile factory the process of costic recovery is performed in
which is not suitable manner.

3. CHAPTRE THREE
METHODOLOGY
To conduct the design of the pneumatic robot machine the processes follows the product design
development (PDD) process.
Collecting data is done on existing parts of Tigray and adwa areas; this information was gathered from
adigrat university mechanical department and almeda textile factory. The final design is the output of data
inputs, conceptual design, and analytic design. Data collection was through informal interviews, different
design books and from the internet. The final drawing is to be prepared using solid work software.
Farther more, the necessary thing is examine the cost estimation of the pneumatic robot machine by
taking some information standard markets to know the cost of pneumatic robot component and depending
on the manufacturing process, assembly process,labor cost etc…After taking and collecting some
necessarily information for this project, process and analysis of that’s information has done. Data
collection was done through the two basic ways
Primary Data: -
 Direct observation,
 Formal & informal interview with different professionals and from written data.
Secondary Data: -
 Different research works and pneumatic robot manufacturers
 From internet.
Generally, after collecting the data by primary and secondary source, the process follows through PDD
sequence and procedure from identifying customer need, design analysis and cost estimation. The
methodology process follows PDD procedure as:

4. Identifycustomerneed
Establishtargetspecification
Generate product concept
Selectproductconcept
Designanalysis
Cost estimation

5. CHAPTER FOUR
DESIGNE CONSEPTS AND SELCTION
4.1. Concept Development Process
4.1.1. Identifying Opportunity
The opportunity is the problem, which is lack of pneumatic robot machine, so the almeda textail labors
tend to work by their energy and hands by consuming a lot of time, capital and money.
4.1.2. Identifying Company Need
In Adwa tigray region most of the fabrics use traditional tools and ways to mix water and acid in order to
continuous blanching , so most of the Ethiopian fabrics need a machine that can minimize the total cost,
time and labor, which have an overall profit.
4.1.3. Organizing Raw Data
The needs that obtained by direct observation and interview. The needs are organized as follows:
 Less number of components
 Design simplicity
 Cost
 Strength
 Availability
 Manufacturability
 Durability
 Complexity
 Weight
 Operability
 Reliability
 Maintainability

6. Preparing relative importance of needs
Table1.1. Preparing relative importance of needs
Product design needs Level of importance
Less number of components 3
Light weight 10
Low cost 1
Easy to operate and maintain 2
Safe and reliable 4
Strength 9
Design simplicity 6
Durability 8
Availability 5
Manufacturability 7

7. 4.2. CONCEPT DESIGN
Concept generation is when a product development team comes up with the ideas, is the most critical step
in the engineering design process,without it there is no design. It is the process of creating, developing,
communicating ideas, which are abstract,concrete or visual. The process includes constructing through
the idea innovating the concept, developing the process and bringing a concept to real worled.we will see
sDifferent alternatives of robotic machine design. Hence,the following are some of the competitive
mechanisms anticipated to meet the quality and feature needed.
4.2.1.Model A:Designof manually operated crop cutting machine with engine as power source to the
machine and that uses chain sprocket and motorcycle chain with collecting plates welded on it as
collecting mechanism.
Figure4.1.Manually operated crop-cuttingmachinewith engineas powersourcetothe
machine
Working Mechanism
This machine consists of two mechanisms one is a four bar mechanism for reciprocation of
cutter blade over stationary cutter blade and this mechanism is used to convert rotary motion
into linear motion. Second is collecting mechanism which consist chain sprocket and motorcycle

8. chain.Thismachine ispoweredbypetrol engine. By using V-Belt power is transmitted to bevel gear
box. Bevel gear box is used to change direction of drive by 900 in the gear system. One end of
this output shaft is connected to four bar mechanism which converts rotary motion of shaft into
reciprocating motion of cutter blade. Reciprocating cutter blade slides over fixed blade and
creates scissoring action responsible for cutting the crop.
Table4.2. Material selection forModel A
Number Components Material
1 Frame Mild Steel
2 V-Belt rubber fabric
3 Pulley cast iron
4 Shaft Mild steel
5 Key mild steel
6 Bevel gear AISI 1050 heat treated (hot rolled) steel
7 Blade /Cutter Stainless Steel
10 Bolts Carbon steel with grade 40C8
11 Nut Bronze
12 Handles Mild Steel
13 Chain High carbon steel
4.2.2.Model B:
Designof manuallyoperatedcropcuttingmachine with electric motor as power source to the machine
and that uses flat belt with collecting plates bolted on it as collecting mechanism.

9. Figure4.2.Manually operated crop-cuttingmachinepowered with electricmotor
Working Mechanism
This machine is powered by electric motor. With the help of V-belt, drive power is transmitted to
gearbox. To reduce the speed of electric motor (to have suitable speed of the cutter), a spur gearbox
used.One endof thisoutputshaftis connectedtoslidercrankmechanismwhichconvertsrotarymotion
of shaftintoreciprocatingmotionof cutterblade.Reciprocatingcutterblade slidesoverfixed blade and
createsscissoringactionresponsibleforcuttingthe crops. Collectingmechanismconsistof flatbeltwith
collecting plates bolted on it.
Table4.3.Material selection for Model B
Number Components Material
1 Frame Mild Steel
2 V-Belt rubber fabric
3 Pulley cast iron
4 Shaft Mild steel
5 Key mild steel
6 Spur gear Carbon steel
7 Blade /Cutter Stainless Steel
8 Bolts Carbon steel with grade 40C8
9 Nut Bronze
10 Handles Mild Steel
11 Flat Belt for collecting mechanism Leather
12 Crop divider G.I. sheet
13 Roller Aluminium bronze
CHAPTER FIVE
DESIGN ANALYSIS
Factors affecting performance of costic recovery pneumatic robotic machines:
1. Chemical factors

10.  Chemical change
 Ambient temperature
 Chemical moisture
 Chemical density
2. Machine factor
 Shape and size of pneumatic robot machine.
 Real position and speed
 Cutting blade shape and speed
 Arm speed
 Machine settings
3. Operational factor
 Height of cut
 Operational speed
In put parameters
 Weight to left =25 kg
 Minimum height to left =200mm
 Maximum height to left =950mm
5.1. Design of column support (trapezoidal support)
Assumptions
 Height(h)=300mm
 Width(b)=150mm
 Width(a)=75mm b1
 Thickness of each top and bottom(t)=12mm
 Mass applied =250kg h
 Acceleration due to gravity=10 𝑚
𝑠2⁄
 Factor of safety= 5
 Length of trapezoidal beam support =1800mm b2
 Elastic Modules (E)=210Gpa
Material
AISI steel 1030

11.  Ultimate tensile strength (𝜎ut) =848Mpa
 Yielding strength (𝜎y) =648Mpa
Analysis
First let’s determine the weight applied on it
𝑊 = 𝑚𝑔 = 25𝑘𝑔 ∗ 10 𝑚
𝑠2⁄ = 250𝑁
Then let’s determine the allowable stress and allowable shear stress of the material
𝜎𝑎𝑙𝑙 =
𝜎𝑦
𝐹. 𝑆
𝜎𝑎𝑙𝑙 =
648
5
𝜎𝑎𝑙𝑙 = 129.6𝑀𝑝𝑎
𝜏 𝑎𝑙𝑙𝑜𝑤 =
𝜎𝑦
2𝐹. 𝑆
𝜏 𝑎𝑙𝑙𝑜𝑤 =
648
2 ∗ 5
𝜏 𝑎𝑙𝑙𝑜𝑤 = 64.8𝑀𝑝𝑎
Then let’s find the reaction force formed in the leg support
125N 125N
RA RB
Then summation of force in the vertical plane is zero. That is
∑ 𝐹𝑌 = 0
𝑅 𝐴 + 𝑅 𝐵 = 250𝑁
And summation of moment at point A is zero. That is
∑ 𝑀𝐴 = 0
1800 ∗ 𝑅 𝐵 − 125 ∗ 300 − 125 ∗ 1500 = 0
𝑅 𝐵 = 125𝑁

12. Substituting
𝑅 𝐴 + 𝑅 𝐵 = 250𝑁
125+RA = 250N
RA=125N
Then let’s determine the maximum bending moment formed in the support beam.
𝜎𝑏 =
𝑀𝐶
𝐼
But moment is given by
𝑀 =
𝑅 𝐴 ∗ 𝐿
2
𝑀 =
250𝑁 ∗ 600𝑚𝑚
4
= 9.4 ∗ 103
𝑁𝑚𝑚
Also the moment of inertia is given by
Ixx =
ℎ2( 𝑎2
+4𝑎𝑏+𝑏2)
36( 𝑎+𝑏)
Ixx =
3002 (752
+4∗75∗150+1502 )
36(75+150)
Ixx = 9.03𝑚𝑚4
And the center of axis is given by
𝑐 𝑚𝑎𝑥=
𝑎+2𝑏
3( 𝑎+𝑏)
*h =
75+2∗150
3(75+150)
*300 = 166.7mm
Then the maximum bending moment formed in the support beam will be
𝜎𝑏=
𝑀∗𝐶 𝑚𝑎𝑥
𝐼
𝜎𝑏 =
9.4∗103
𝑁𝑚𝑚 ∗166.7mm
9.03𝑚𝑚4 = 173534
𝑁
𝑚𝑚2
5.2 DESIGN OF SHAFT SUPPORT
A shaft is a rotating member, usually of circular cross section, used to transmit power or motion. It
provides the axis of rotation, or oscillation, of elements such as gears, pulleys, flywheels, cranks,
sprockets, and the like and controls the geometry of their motion. However, a shaft can have a noncircular
cross section and need not be rotating. An axle, a nonrotating member that carries no torque, is used to
support rotating members.

13. Figure 5.2. Shaft
Material Selection
Mild steelis general-purpose steel bars for machining, suitable for lightly stressed components including
studs, bolts, gears, handles and shafts.
It is less costly.
Mechanical Properties
Ultimate tensile strength σut = 430Mpa
Yield strength σy = 220Mpa
Ultimate shear stress = 360 MPa
Factor of safety =5
Modulus of elasticity E =210Gpa
Density, ρ = 7860
kg
m3
Given Parameters and Assumptions
Design power,𝐏 = 𝟎. 𝟖𝟓𝐤𝐰
Shaft speed, 𝐍 𝟐 = 𝟑𝟎𝟎𝐫𝐩𝐦
Shaft length =10mm
Required
 Diameter of shaft, ds
Analysis

14. The shaft is subjected to torque of, T =
60P
2πN
=
60×850
2π×300
= 27.056Nm
𝜏 =
𝜏𝑢
𝑓.𝑠
𝜏 =
360
5
= 72
𝑁
𝑚𝑚2
Diameter of the solid shaft
Let d = Diameter of the solid shaft.
We know that torque transmitted by the shaft,
T =
𝑃∗60
2𝜋𝑁
,T=27.056Nmm
We also know that torque transmitted by the solid shaft (T),T=
𝜋
16
∗ 𝜏 ∗ 𝑑3
27.056*103
Nmm =
𝜋
16
∗ 72 ∗ 𝑑3
, 1913.82𝑚𝑚3
=𝑑3
,d=√1913.82
3
= 12.42𝑚𝑚
5.3. DESIGN OF SHAFT CONECTING BASE AND COLEMUN
Material specification selected for the shaft is plain carbon steel to British Standard specification
BS 970 080M30, Hardened and Tempered, whose properties are as shown in Appendix B and
the material yield strength is 700MPa both in tension and pure compression and 450MPa in
shear.
Mechanical Properties
allowable compression=700Mpa
Allowable tensile stress σt = 700Mpa
Yield strength σy = 700Mpa
Allowable shear stress(𝜏) = 450MPa
Factor of safety =5
Given Parameters
Weight (w)=mg=250N
Analysis
𝒘 = 𝝈𝒄 ∗ 𝑨𝒄 , Where 𝝈𝒄=compressive stress
Ac=cross sectional area

15. 𝒘
𝝈𝒄
=Ac, Ac=
𝝅
𝟒
𝒅 𝟐 ,
𝟐𝟓𝟎𝑵
𝟕𝟎𝟎
𝑵
𝑴𝑴 𝟐
=
𝝅
𝟒
𝒅 𝟐,𝟎. 𝟑𝟓𝟕 =
𝝅
𝟒
𝒅 𝟐,𝟒 × 𝟎. 𝟑𝟓𝟕 = 𝝅𝒅 𝟐
1.4285= 𝝅𝒅 𝟐,
𝟏.𝟒𝟐𝟖𝟓
𝝅
= 𝒅 𝟐, 𝟎. 𝟒𝟓𝟒𝟕 = 𝒅 𝟐,𝒅 = √ 𝟎. 𝟒𝟓𝟒𝟕 = 𝟔𝟕𝒎𝒎
5.3.1 Shaft stress
Compressive stresses duo to axial loads using the new diameter is
𝜎𝑐 =
𝑤
𝜋
4
𝑑2
=
250𝑁
𝜋
4
67𝑚𝑚2
= 4.75
𝑁
𝑚𝑚
2
5.3.2 Torque required to rotate the shaft
𝑇1 = 𝑝 ×
𝑑𝑚
2
Where p is effort, dm is mean diameter
T1=200N×
67
2
𝑚𝑚 = 6700𝑁𝑚𝑚
The shear stress due to this torque using the new diameter is given
𝜏 =
𝑇1×𝑑
2𝑗
, 𝑗 =
𝜋
32
𝑑4 =
6700𝑁𝑚𝑚×67𝑚𝑚×32
𝜋×674 𝑚𝑚4
= 0.224
𝑁
𝑚𝑚2
5.3.3 Principal stresses
Maximum principal stress is as follows:
𝜎𝑐𝑚𝑎𝑥 =
1
2
⌈ 𝜎𝑐 + (√𝜎𝑐2 + √4𝜏2⌉ =
1
2
⌈4.75 + (√4.752 + √4 × 0.2242⌉ = 4.974
𝑁
𝑀𝑀2
Design value of 𝜎𝑐 =
𝜎𝑎𝑙𝑙
𝑓.𝑠
=
700
𝑁
𝑚𝑚2
5
= 140
𝑁
𝑚𝑚2
𝜎𝑐𝑚𝑎𝑥 ≤ 𝜎𝑐, 4.974
𝑁
𝑀𝑀2
≤ 140
𝑁
𝑚𝑚2
, 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑠𝑎𝑓𝑒
Maximum shear stresses is as follows
𝜏𝑚𝑎𝑥 =
1
2
(√𝜎𝑐2 + 4𝜏2) =
1
2
(√4.752 + 4 × 0.2242) = 2.38
𝑁
𝑚𝑚2

16. Design value of 𝜏 =
𝜏𝑦
𝑓.𝑠
=
450
𝑁
𝑚𝑚2
5
= 90
𝑁
𝑀𝑀2
𝜏𝑚𝑎𝑥 ≤ 𝜏, 2.38
𝑁
𝑚𝑚2 ≤ 90
𝑁
𝑀𝑀2 , 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑠𝑎𝑓𝑒
Cheek; those maximum shear and compressive stresses are less than the permissible stresses which is safe
design.
5.4. DESIGN OF LEVER
A lever is a rigid rod or bar capable of turning about a fixed point called fulcrum. It is used as a machine
to lift a load by the application of a small effort. The ratio of load lifted to the effort applied is called
mechanical advantage. Sometimes, a lever is merely used to facilitate the application of force in a desired
direction. A lever may be straight or curved and the forces applied on the lever (or by the lever) may be
parallel or inclined to one another. The principle on which the lever works is same as that of moments.
According to the principle of moments,
Material Selection
Mild steel is general-purpose steel bars for machining, suitable for lightly stressed components including
studs, bolts, gears, leveres, handles and shafts.
Mechanical Properties
Allowable tensile stress σt = 100Mpa
Yield strength σy = 220Mpa
Allowable shear stress = 55 MPa
Factor of safety =5
Modulus of elasticity E =210Gpa
Given Parameters and Assumptions
Width of the lever =200mm
Effort (p) = 200 N
Height of the lever= 2800mm

17. Weight (w) =250N
Thickens=30mm
Length of the lever from the furculum to the weight is 200 mm, L1
Length of the lever from furculum to effort is 250 mm, L2
Required
 Furculum, (FR)
Analysis
The principle on which the lever works is same as that of moments.
According to the principle of moments,
W*L1=P*L2 or
𝑤
𝑝
=
𝑙2
𝑙1
Mechanical advantage,(M.A)=
𝑤
𝑝
=
𝑙2
𝑙1
P L2 FR L1 W
The forces acting on the lever are Fig. 5.3. Type of lever.
1. Load (W), L1
2. Effort (P), L2
3. Reaction at the fulcrum F (RF)
M.A=
𝑤
𝑝
=
𝑙2
𝑙1
, M.A=
𝑤
𝑝
=
250𝑁
200𝑁
=1.25
M.A=
𝑙2
𝑙1
= 1.25=
𝐿2
200𝑚𝑚
= 𝐿2 = 200𝑚𝑚 ∗ 1.25 = 250𝑚𝑚
Then summation of force in the vertical plane is zero. That is
∑ 𝐹𝑌 = 0
𝑃 + 𝑊 = 𝐹𝑅
200N+250N=FR

18. FR=450N
5.5. Design of Stabilizer h
Assumption
Length =145mm L
Height=96mm
Thickens=10mm

19. 5.6. Design of base
Eccentric Loaded Riveted Joint
When the line of action of the load does not pass through the centroid of the rivet system and thus all
rivets are not equally loaded, then the joint is said to be an eccentric loaded riveted joint. The eccentric
loading results in secondary shear caused by the tendency of force to twist the joint about the centre of
gravity in addition to direct shear or primary shear.
Let P = Eccentric load on the joint, and e = Eccentricity of the load i.e. the distance between the line of
action of the load and the centroid of the rivet system i.e. G.
The following procedure is adopted for the design of an eccentrically loaded riveted joint
1. First of all, find the centre of gravity G of the rivet system.
Let A = Cross-sectional area of each rivet
x1, x2, x3 etc. = Distances of rivets from OY, and
y1, y2, y3 etc. = Distances of rivets from OX.
We know that 𝑋̅=
𝐴1×𝑋1+𝐴2×𝑋2+𝐴3𝑋3+....
𝐴1+𝐴2+𝐴3+⋯
=
𝐴×𝑋1+𝐴×𝑋2+𝐴𝑋3+....
𝑛𝐴
=
𝑋1+×𝑋2+𝑋3+....
𝑛
(where n = Number of rivets)
Similarly, 𝑦̅ =
𝑦1+𝑦2+𝑦3
𝑛

20. Fig.5.2 Eccentric loaded riveted joint.
2. Introduce two forces P1 and P2 at the centre of gravity ‘G’ of the rivet system. These forces
are equal and opposite to P as shown in Fig. 5.2 (b).
3. Assuming that all the rivets are of the same size, the effect of P1 = P is to produce direct shear
load on each rivet of equal magnitude. Therefore,direct shear load on each rivet,
Ps =
𝑝
𝑛
, acting parallel to the load P.
4. The effect of P2 = P is to produce a turning moment of magnitude P × e which tends to rotate
the joint about the centre of gravity ‘G’ of the rivet system in a clockwise direction. Due to the turning
moment, secondary shear load on each rivet is produced. In order to find the secondary shear load,
the following two assumptions are made :
(a) The secondary shear load is proportional to the radial distance of the rivet under consideration from
the centre of gravity of the rivet system.
(b) The direction of secondary shear load is perpendicular to the line joining the centre of the rivet to the
centre of gravity of the rivet system..
Let F1, F2, F3 ... = Secondary shear loads on the rivets 1, 2, 3...etc.
l1, l2, l3 ... = Radial distance of the rivets 1, 2, 3 ...etc. from the centre of
gravity ‘G’ of the rivet system.

21. From assumption (a),
F1 α l1 ; F2 α l2 and so on
𝐹1
𝐿1
=
𝐹2
𝐿2
=
𝐹3
𝐿3
= ⋯
F2=
𝐿2
𝐿1
× 𝐹1
F3=
𝑙3
𝑙1
× 𝐹1
We know that the sum of the external turning moment due to the eccentric load and of internal
resisting moment of the rivets must be equal to zero.
P.e=F1*L1+F2*L2+F3*L3
= F1*L1+𝐿2 ×
𝐿2
𝐿1
× 𝐹1 + 𝐿3 ×
𝑙3
𝑙1
× 𝐹1
=
𝐹1
𝑙1
[(𝑙1)2 + (𝑙2)2 + (𝑙3)2 + ⋯]
From the above expression, the value of F1 may be calculated and hence F2 and F3 etc. are
known. The direction of these forces are at right angles to the lines joining the centre of rivet to the
centre of gravity of the rivet system, as shown in Fig. 5.2 (b),and should produce the moment in the
same direction (i.e. clockwise or anticlockwise) about the centre of gravity, as the turning moment
(P × e).
5. The primary (or direct) and secondary shear load may be added vectorially to determine the
resultant shear load (R) on each rivet as shown in Fig. 5.2 (c). It may also be obtained by using the
relation
When the secondary shear load on each rivet is equal, then the heavily loaded rivet will be one
in which the included angle between the direct shear load and secondary shear load is minimum. The
maximum loaded rivet becomes the critical one for determining the strength of the riveted joint.
Knowing the permissible shear stress (𝜏),the diameter of the rivet hole may be obtained by using the
relation,
Maximum resultant shear load (R)=
𝜋
4
× 𝑑2 × 𝜏,
From Table 9.7, the standard diameter of the rivet hole ( d ) and the rivet diameter may be
specified, according to IS : 1929 – 1982 (Reaffirmed 1996).(GUPTA)

22. Assumptions
Base (b) =400mm b=L squire cross section
Length(L)=400mm
Thickens ( t) =50mm ;
Load on the bracket( P) =250kN =25 ×
103
N ;
Eccentricity (e) =400 mm;
Number of rivet (n)=7 ;
Rivet spacing (C) =100mm
Crushing stress (σc) =120MPa
Permissible shear stress (τ) = 65MPa = 65
N/mm2;
Required Fig. 5.2
Size of the rivets
Analysis
First of all, let us find the centre of gravity (G) of the rivet system.
Let 𝑥̅ = Distance of centre of gravity from OY,
𝑦̅ = Distance of centre of gravity from OX,
𝑋̅=
𝑋1+𝑋2+𝑋3+𝑋4+𝑋5+𝑋6+𝑋7
𝑛
=
100+200+200+200
7
=100mm(x1=x6 = x7 = 0)
𝑦̅=
𝑦1+𝑦2+𝑦3+𝑦4+𝑦5+𝑦6+𝑦7
𝑛
=
200+200+200+100+10
7
= 114.7𝑚𝑚(y5 = y6 = 0)

23. ∴ The centre of gravity (G) of the rivet system lies at a distance of 100 mm from OY and 114.3
mm from OX, as shown in Fig. 5.2.
We know that direct shear load on each rivet,
Ps =
𝑝
𝑛
=
250×103
𝑁
7
= 35,714.28𝑁
The direct shear load acts parallel to the direction of load P i.e. vertically downward
as shown in Fig. 5.2. Turning moment produced by the load P due to eccentricity (e)
P × e=250× 103
𝑁 × 400𝑚𝑚 =1000000Nmm
This turning moment is resisted by seven rivets as shown in Fig.5.2.
All dimensions are in millimeter (mm)
Fig. 5.2

24. Let F1, F2, F3, F4, F5, F6 and F7 be the secondary shear load on the rivets 1, 2, 3, 4, 5, 6 and 7
placed at distances l1, l2, l3, l4, l5, l6 and l7 respectively from the centre of gravity of the rivet
system as shown in Fig. 5.2
From the geometry of the figure, we find that
L1=L3=√(100)2 + (200 − 114.3)2 = 131.7𝑚𝑚
L2 = 200 – 114.3 = 85.7 mm
L4=L7 =√(100)2 + (114.3 − 100)2 = 101𝑚𝑚
L5=L6=√(100)2 + (114.3)2 = 152𝑚𝑚
Now equating the turning moment due to eccentricity of the load to the resisting moment of the
rivets, we have
250× 103
×400=
100× 106
× 131.7 = 𝐹1 × 108645 =
131.7×108
108645
= 121220.48𝑁 = 𝐹1
Since the secondary shear loads are proportional to their radial distances from the centre of
gravity, therefore F2 =F1×
𝐿2
𝐿1
=121220.48 ×
85.7
131.7
= 78880.75𝑁
F3=F1×
𝐿3
𝐿1
=121220.48𝑁, L3=L1
F4= F1×
𝐿4
𝐿1
=121220.48𝑁 ×
101
131.7
= 92963.3𝑁
F5=F1×
𝐿5
𝐿1
=121220.48𝑁 ×
152
131 .7
= 139905.1857𝑁
F6= F1×
𝐿6
𝐿1
= 121220.48𝑁 ×
152
131 .7
= 139905.1857𝑁
F7= F1×
𝐿7
𝐿1
= 𝐹4 = 92963.3𝑁

25. By drawing the direct and secondary shear loads on each rivet, we see that the rivets 3, 4 and 5
are heavily loaded. Let us now find the angles between the direct and secondary shear load for
these three rivets. From the geometry of Fig. 5.2, we find that
cosθ3=
100
𝐿3
=
100
131.7
= 0.76
cosθ4=
100
𝐿4
=
100
101
= 0.99
cosθ5=
100
𝐿5
=
100
152
= 0.6578
Now resultant shear load on rivet 3,
R3=√
(35,714.28)2 + (121220.48)2 + 2 × 35,714.28 × 121220.48× 0.76
= 150168.0853𝑁
R4=√( 𝑃𝑠)2 + ( 𝐹4)2 + 2 × 𝑃𝑠 × 𝐹4cosθ4
R4=√(35,714.28)2 + (92963.3)2 + 2 × 35,714.28 × 92963.3 × 0.99 =
R5=√( 𝑃𝑠)2 + ( 𝐹5)2 + 2 × 𝑃𝑠 × 𝐹5cosθ5
R5=√(35,714.28)2 + (139905.1857)2 + 2 × 35,714.28 × 139905.1857× 0.6578 =
The resultant shear load may be determined graphically, as shown in Fig. 5.2. From above we
see that the maximum resultant shear load is on rivet 5. If d is the diameter of rivet hole, then
maximum resultant shear load (R5),
R5=
𝜋
4
𝑑2
× 𝜏 = R5=
𝜋
4
𝑑2
× 65
5.7 DESIGN OF HAND ARM SUPPORT
ASSUMPTION b
Height (h) =200mm h
Width (b) =250mm