![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)](https://image.slidesharecdn.com/finaladu-181128091006/85/Final-adu-21-320.jpg)
Final adu
- 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