WINTER 14 EXAMINATIONS Subject Code: 12310 Model …msbte.engg-info.website/sites/default/files/12310_winter_2014... · WINTER – 14 EXAMINATIONS Subject Code: 12310 Model Answer

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

(ISO/IEC - 27001 - 2005 Certified)

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WINTER – 14 EXAMINATIONS Subject Code: 12310 Model Answer Page No: ____/ N Important Instructions to examiners: 1) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. 2) The model answer and the answer written by candidate may vary but the examiner may try to assess the understanding level of the candidate. 3) The language errors such as grammatical, spelling errors should not be given more importance. (Not applicable for subject English and Communication Skills) 4) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn. 5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate’s answers and model answer. 6) In case of some questions credit may be given by judgment on part of examiner of relevant answer based on candidate’s understanding. 7) For programming language papers, credit may be given to any other program based on equivalent concept.



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Q. NO.

MODEL ANSWER

MARKS TOTAL MARKS

1.

(A) Attempt any THREE of the following: 3Χ4=12

(a)

Force acting on sunk key. Types of Keys The following types of keys are important from the Subject point of view: 1. Sunk keys, 2. Saddle keys, 3. Tangent key . 4. Round keys, and 5. Splines

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b

•When the component is subjected to several types of loads simultaneously, it i necessary to determine the state of stresses under such conditions. For example, a transmission shaft is subjected to bending moment as well as twisting moment (torque) at the same time. • The plane, on which, only normal stresses acts and no shear stress, is called as principal plane. The magnitude of normal stress acting on the principal plane is called principal stress. • Consider an element of a plate subjected to two dimensional stresses as shown in Fig. (b). • In this analysis, the stresses are classified into two groups : (a) Normal stress,(b) Shear stress. Normal stress is perpendicular to area under consideration, while shear stress acts over the area. Refer Fig. (c), showing the stresses acting on oblique plane. • "Major principal stress is the maximum value of normal stress acting on the principal plane, whereas, the minimum value of normal stress acting on

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principal plane is called asminor principal stress". Maximum principal stress, Ϭt1 ={(ϭx + ϭy)/2}+ ½ √{[ ϭx - ϭy]2 + 4 2ד} Minimum Principle stress, Ϭt2 ={(ϭx + ϭy)/2}- ½ √{[ ϭx - ϭy]2 + 4 2ד} Also maximum shear stress, {2ד 2 + 4[ϭx - ϭy ]}√ ½ = max={(ϭx - ϭy)/2} ד

c 1.Lap Joint: It is a joint between two overlapping components. It consists of fillet welds. Fillet weld ishaving triangular cross-section joining two surfaces at right angles to each other. The examples of lap joint are single transverse fillet, double transverse fillet, parallel fillet welds. Based on the relative positions of the load axis with respect to the fillet axis, the fillet welds are classified intotwo types. (i) Parallel fillet weld: "If the load axis is parallel to the axis of the fillet, it is known parallel fillet weld". (ii) Transverse fillet weld : "If the load axis is perpendicular to the axis of the fillet, it is known as transverse fillet weld". The transverse fillet weld can be a single transverse fillet weld or a double transverse fillet weld.

2. Butt Joint: (i) The butt weld as shown in Fig. is obtained by placing the plates to be joined side by side with their edges nearly touching each other (ii) The small gap is maintained between the edges for the filler material. (iii) The examples of butt joints are square butt, single V-butt, single If-butt, double V-butt and double U-butt. (i) Square butt weld : If the thickness of the plates is less than 5 mm, the edge of the plates do not require bevelling and hence the joint used is known as square butt weld. (ii) Single V-butt weld or single U-butt weld: If the thickness of the plates between 5 and 12.5 mm, the edges are bevelled to V or V groove and accordingly single V-butr or single V-butt weld may be used. (iii) Double V-butt weld or Double If-butt weld : If the thickness of the plates is more than 12.5 mm, it is necessary to bevel and weld the plates from both sides. In such cases, double V -butt or double U'-butt welds are used

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Butt weld

d

Important Terms Used in Riveted Joints The following terms in connection with the riveted joints are important from the subject of view : (i)Pitch. It is the distance from the centre of one rivet to the centre of the next rivet measured parallel to the seam as shown in Fig. It is usually denoted by p. (ii)Back pitch. It is the perpendicular distance between the centre lines of the successive rows as shown in Fig. It is usually denoted by Pb. (iii)Diagonal pitch. It is the distance between the centres of the rivets in adjacent rows of zig-zagriveted joint as shown in Fig. It is usually denoted by Pd. (iv)Margin or marginal pitch. It is the distance between the centre of rivet hole to the nearest edge of the plate as shown in Fig. It is usually denoted by m.

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(B) Attempt any ONE of the following: 1Χ6=06

(a) Bolt of uniform strength 02 6m



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If the shank of the bolt is turned down to a diameter equal or even slightly less than the core diameter of the thread (D) as shown in Fig. (b), then shank of the bolt will undergo higher stress. This means that a shank will absorb a large portion of the energy, thus relieving the material at the sections near the thread. The bolt, in this way, becomes stronger and lighter and it increase shock absorbing capacity of the bolt because of an increased modulus of resilience. This gives us bolts of uniform strength. The resilience of a bolt may also be increased by increasing its length.

Bolt of uniform strength

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b Given t= 10mm,ϭt= 400MPa, ϭc= 640MPa,320=דMPa 1. Diameter of rivet since the thickness of plate is greater than 08mm . d = 6 √ t = 6 √ 10=18.97 Select d= 20mm 2.pitch of Rivet(p) Select a double riveted lap joint,hence Tearing resistance Pt = (P-d) x t x ϭt = (p-20) x 10 x 400 Pt =4000(p-20)----------------(i) And shearing resistance Ps = n x π /4 x d2 x ד = 2 x π /4 x (10)2 x 320 = 50265.48 N --------(ii) Equating (i) aand (ii) We get 4000(p-20) = 50265.48

p=32.56= 33mm

p=33mm

3.Distant between the rows(Pb)

Pb = 0.33P + 0.67d=(0.33 x 33) + (0.67 x 20)

Pb = 24mm

4.Margine(m)

m = 1.5d= 1.5 x 20

m= 30mm

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2. Attempt any TWO of the following: 2Χ8=16

(a) Given data: P = 8 kW = 8 X 103 W N = 750rpm 's = 35 MPa = 35 Nzmmד ci = 15 N/mm2, ϭt = ϭck = 60 N/mm2דThe power transmitted by steel shafts, P= 2πNT /60 Therfore Torque = T = P X 60 /2πN = 8 x 103 x 60 / 2 x π x 750 . . T = 101.859 N-m = 101.859 x 103 N-mm We know that, torque transmitted by shaft is given by T= π/16 xד s x d3 101.859 x 103 = π/16 x35x d3

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Diameter of shaft, d = 25.56== 30 mm (say) (i)Design of hub: Usual proportions are, D = Outer diameter of hub = 2d = 2 x 30 = 60 mm L = Length of hub = 1.5 x d = 1.5 x 30 = 45 mm k = d/D = 30/60 = 0.5 Considering hub as a hollow shaft transmitting the same torque as that of shaft.Then we have T= π/16 x דci x D3(1-k4) 101.859 x 103= π/16 x דci x 603(1-0.54) ci=2.561 N/mm2דThus, the induced shear stress in the cast iron hub is less than the given permissible shear stress. Hence, the design is safe. (ii) Design of flange: Take tf=d / 2 = 30 /2 = 15 mm While transmitting the torque, the flange is under shear. The torque transmitted is T = Circumference of hub x Thickness of flange x Shear stress x Radius of hub =(π x D) x tf xד f x D/2 101.859 x 103=(π x 60) x 15 xד f x 60/2 f=1.2 N/mm2דThus, induced shear stress is less than given permissible shear stress for flange material.Hence, the design is safe.

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b Stresses in Pipes:The stresses in pipes due to the internal fluid pressure are determined by Lame's equation.According to Lame's equation, tangential stress at any radius x Ϭt ={[p (ri)2] / [(ro)2 –(ri)2] } /{1 +[(ro)2 / x2]} And Radial stress at any radius x Ϭr ={[p (ri)2] / [(ro)2 –(ri)2] } /{1 - [(ro)2 / x2]} where p = Internal fluid pressure in the pipe, ri = Inner radius of the pipe, and ro = Outer radius of the pipe The various types of pipe joints are as follows. 1. Socket or a coupler joint. The most common method of joining pipes is by means of a socket or a coupler as shown in Fig. This type of joint is mostly used for pipes carrying water at low pres ure and where the overall smallness of size is most es essential.

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& 02

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Socket or coupler joint 2. Nipplejoint. In this type of joint, a nipple which is a small piece of pipe threaded outside' screwed in the internally threaded end of each pipe, as shown in Fig. The disadvantage of this joint is that it reduces the area of flow

Nipple joint. 3.Unionjoint. In order to disengage pipes joined by a socket, it is necessary to unscrew pipe from one end. This is sometimes inconvenient when pipes are long. The union joint, as shown in Fig. provide the facility of disengaging the pipes by simply unscrewing a coupler nut.



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Union joint 4. Spigot and socket joint. A spigot and socket joint as shown in Fig. , is chiefly used for pipes which are buried in the earth.

Spigot and socket joint 5. Expansion joint. The pipes carrying steam at high pressures are usually joined by means of expansion joint. This joint is used in steam pipes to take up expansion and contraction of pipe line due to change of temperature



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Expansion bends

Expansion joints

(c) Question is vague and hence follow a general procedure as given below for any framed structure:

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3. Attempt any FOUR of the following: 4Χ4=16

(a) Stress concentration can be defined as the increase in the iintensity of stress due to various factors such as abrupt change in cross section, sharp corners,presence of holes, internal deformities, cracks,etc. The presence of stress concentration cannot be totally eliminated but it may be reduced to some extent. A device or concept that is useful in assisting a design engineer to visualize the presence of stress concentration and how it may

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be reduced is that of stress flow lines, as shown in Fig. The reduction of stress concentration means that the stress flow lines shall maintain their spacing as far as possible.

Method of reducing stress contraction in cylinder members with shoulders

Method of reducing stress contraction in cylinder members with holes

Method of reducing stress contraction in cylinder members with holes The stress concentration effects of a press fit may be reduced by making more gradual transition from the rigid to the more flexible shaft. The various ways of reducing stress concentration for such cases are shown in Fig. a,b,c

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(b) Following are the advantages and disadvantages of welded joints over other method joints. Advantages 1. The welded structures are usually lighter than riveted structures. This is due to the reason that in welding, gussets or other connecting components are not used. 2. The welded joints provide maximum efficiency (may be 100%) which is not possible in case of riveted joints.

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3. Alterations and additions can be easily made in the existing structures 4. As the welded structure is smooth in appearance, therefore it looks pleasing. 5. In welded connections, the tension members are not weakened as in the case of riveted joints. 6. A welded joint has a great strength. Often a welded joint has the strength of the parent metal itself. 7. Sometimes, the members are of such a shape (i.e. circular steel pipe) that they afford difficulty for riveting. But they can be easily welded. 8. The welding provides very rigid joints. This is in line with the modern trend of providing rigid frames. 9. It is possible to weld any part of a structure at any point. But riveting requires enough clearance. 10. The process of welding takes less time than the riveting. Disadvantages 1.Since there is an uneven heating and cooling during fabrication, therefore the member may get distorted or additional stresses may develop. 2. It requires a highly skilled labour and supervision. 3. Since no provision is kept for expansion and contraction in the frame, therefore there is a possibility of cracks developing in it. 4. The inspection of welding work is more difficult than riveting work.

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(c) Design of Oval Flanged Pipe Joint Consider an oval flanged pipe joint .A spigot and socket is provided for locating the pipe bore in a straight line. A packing of trapezoidal section is used to make the joint leak proof. The thickness of the pipe is obtained as discussed previously. The force trying to separate the two flanges has to be resisted by the stress produced in the bolts. If a length of pipe, having its ends closed somewhere along its length, be considered, then the force separating the two flanges due to fluid pressure is given by F1 = π/4 x D2 X p where D = Internal diameter of the pipe. The packing has also to be compressed to make the joint leakproof. The intensity of pressure should be greater than the pressure .of the fluid inside the pipe. For the purposes of calculations, it is assumed that the packing material is compressed to the same pressure as that of inside the pipe.Therefore the force tending to separate the flanges due to pressure in the packing is given by F2 = π/4 x[( D1)2 – (D)2 ] p where D1 = Outside diameter of the packing. There fore Total force trying to eparate the two flanges F = F1 + F2 = π/4 x D2 X p + π/4 x*( D1)2 – (D)2 ] p Since an oval flange is fastened by means of two bolts, therefore load taken up by each bolt is Fb = F/2. If dc is the core diameter of the bolts, then Fb = π/4 (dc)2 ϭtb where ϭtb is the allowable tensile stress for the bolt material. The value of

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ϭtb is usually kept low to allow for initial tightening stress in the bolts. After the core diameter is obtained, then the nominal diameter of the bolts i chosen from table( In the absence of table . nominal diameter = -Core diameter/0.84 ). It may be noted that bolts of less than 12 mm diameter should never be used for hydraulic pipes, because very heavy initial tightening stresses may be induced in smaller bolts. The bolt centres hould be as near the centre of the pipe as possible to avoid bending of the flange. But sufficient clearance between the bolt head and pipe surface must be provided for the tightening of the bolts without damaging the pipe material. The thickness of the flange is obtained by considering the flange to be under bending stress due to the forces acting in one bolt. The maximum bending stress will be induced at the section x-x. The bending moment at this section is given by Mxx = Fb x e = F/2 x e And section modulus. Z = 1/6 x b (tf)2 where b = Width of the flange at the section X-X, and tf= Thickness of the flange Using the bending equation, we have Mxx =ϭb.Z Or Fb x e = ϭb x 1/6 x b (tf)2 Where ϭb = Permissible bending stress for the flange material. From the above expression, the value of tf may be obtained, if b is known. The width of the flange is estimated from the layout of the flange. The hydraulic joints with oval flanges are known as Armstrong's pipe joints. The various dimensions for a hydraulic joint may be obtained by using the following empirical relations: Nominal diameter of bolts, d =0.75 t + 10 mm Thickness of the flange, tf = 1.5 t + 3 mm Outer diameter of the flange, Do =D+2t+4.6d Pitch circle diameter, Dp =Do - (3 t + 20 mm)

(d) The Method of SectionsI; In the method of sections, a frame is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the frame. Since frame members are subjected to only tensile or compressive forces along their length, the internal forces at the cut member will also be either tensile or compressive with the same magnitude. This result is based on the equilibrium principle and Newton’s third law. Steps for Analysis 1. Decide how you need to “cut” the frame. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general). 2. Decide which side of the cut frame will be easier to work with(minimize the number of forces you have to find). 3. If required, determine the necessary support reactions by drawing the

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FBD of the entire frame and applying the E-of-E. 4.Draw the FBD of the selected part of the cut truss. You need to indicate the unknown forces at the cut members. Initially we assume all the members are in tension, as we did when using the method of joints. Upon solving, if the answer is positive, the member is in tension as per your assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either in tension or compression by inspection as was done in the figures above.) 5. Apply the E-of-E to the selected cut section of the truss to solve for the unknown member forces. Note that in most cases it is possible to write one equation to solve for one unknown directly.

(e)

A. Proportional limit: Hooke's law holds good up to point A and it is known as proportional limit. It is defined as that stress at which the stress-strain curve begins to deviate from the straight B. Elastic limit: The material has elastic properties up to the point B. This point is known as elastic limit. It is defined as the stress developed in the material without any permanent set C & D. Yeild Point: There are two yield points C and D. The points C and D are called the upper and lower yield points respectively. E. Ultimate stress: At E, the stress, which attains its maximum value is known as ultimate stress. F. Breaking strength: Failure is complete

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4. (A) Attempt any THREE of the following: 3Χ4=12

(a) Stdardization is the process of developing and implanting technical. Stdardization can help to maximize compatibility, interoperability, safety, repeatability or quality. It can also facilitate commoditization of formerly custom process. Stdardization has following advantages

1) Enhanced network effectd.Stadards increase compatibility

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Ductile material

Brittle material



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and interoperability between products, allowing information to be share within larger network and attracting more consumers to use the new technology, further enhancing network effects.

2) Other benefits of standardation to consumers are reduced uncertainly, because consumers can be more certain that they are not choosing the wrong product, and reduced lock in because the standard makes it more likely that there will be competing product in the space.

3) Consumers may also get the benefit of being able to mix and match components of a system to align with their preferences

4) It helps interchangeably of the basic components of the products so as the benefits of mass product and the largely required basic components can be made at a cheaper price.

Types of standards: (Any two) ISO 8402/IS:13999, ISO 9000/IS:14000, ISO 9001/IS:14001, ISO 9002/IS:14002, ISO9003/IS:14003, ISO 9004/IS:14004

In industrial design, preferred numbers (also called preferred values)

are standard guidelines for choosing exact product dimensions within

a given set of constraints. Product developers must choose

numerous lengths, distances, diameters, volumes, and other

characteristic quantities. While all of these choices are constrained by

considerations of functionality, usability, compatibility, safety or cost,

there usually remains considerable leeway in the exact choice for

many dimensions.

Preferred numbers are usedv for following two purposes:

1. Using them increases the probability of compatibility between

objects designed at different times by different people. In

other words, it is one tactic among many instandardization,

whether within a company or within an industry, and it is

usually desirable in industrial contexts (unless the goal

is vendor lock-in or planned obsolescence)

2. They are chosen such that when a product is manufactured in

many different sizes, these will end up roughly equally

spaced on a logarithmic scale. They therefore help to

minimize the number of different sizes that need to be

manufactured or kept in stock.

02

marks.

(b) Perfect frame : A pin-jointed frame which has got just sufficient number of members to

http://en.wikipedia.org/wiki/Industrial_design

http://en.wikipedia.org/wiki/Quantities

http://en.wikipedia.org/wiki/Standardization

http://en.wikipedia.org/wiki/Vendor_lock-in

http://en.wikipedia.org/wiki/Planned_obsolescence

http://en.wikipedia.org/wiki/Logarithmic_scale



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resist the loads without undergoing appreciable deformation in shape is called rigid or perfect frame. The perfect frame obeys the following condition viz. n = 2 j – 3 where, n= no. of links and j= no. of joints Deficient frame: A frame is said to be deficient if the number of members in it is less than that required for a perfect frame. Such frames can’t retain their shape when loaded. Redundant frame: A frame is said to be redundant if the number of members it is more than that required for a perfect frame. Such frame can be analyzed by making use of equations of equilibrium alone.

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(c)

Axially loaded unsymmetrical welded section Sometimes unsymmetrical sections such as angles, channels-sections etc., welded on the flange edges are loaded axially as shown in Fig. In such cases, the lengths of weld should be proportioned in such a way that the sum of resisting moments of the welds about the gravity axis is zero. Consider an angle section as shown in Fig. Let la = Length of weld at the top, lb = Length of weld at the bottom, l = Total length of weld = la + lb

P = Axial load, a = Distance of top weld from gravity axis, b = Distance of bottom weld from gravity axis, and f = Resistance offered by the weld per unit length Moment of the top weld about gravity axis = la x f x a and moment of the bottom weld about gravity axis = lb x f x b Since the sum of the moments of the weld about the gravity axis must be zero, therefore, la x f x a - lb x f x b or la X a = Ib x b ……..(i) We know that l = la + lb …….(ii) From equations (i) and (ii), we have la= l x b /a + b and lb = l x a /a + b

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(d) Given ϭb= 380N/mm2,M= 250 N-m T= 500 N-m Me=1/2{M + √ (M2 + T2)}= 1/2{250 + √ (2502 + 5002)}=404.50N-m Me= 404.50 x 103 N-mm There for Me= π/32 x ϭb x d3

404.50 x 103 = π/32 x 380 x d3 There fore d=22.13mm Selet dia of 25 mm

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(B) Attempt any ONE of the following: 1Χ6=06

(a) Given ϭt=50N/mm2,W=15KN,n=4,L1=60mm,L2=350mm,L=450mm Shear load Ws= W/n=15/4=3.75KN Maximum tensile load Wt=W.L.L2 / 2[L12 + L22] Wt=15 x 45 x 350 / 2[602 + 3502]=9.36 KN Equivalent tensile load Weq=1/2[Wt + √ (Wt2 + 4 Ws2)]= 1/2[9.36 + √ (9.362 + 4 3.752)] Weq=10.67KN

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(b) Design of Cylinder Covers 1. Design of bolts or studs

In order to find the size and number of bolts or studs, the following procedure may be adopted. Let D = Diameter of the cylinder, p = Pressure in the cylinder, dc = Core diameter of the bolts or studs, n = Number of bolts or studs, and ϭtb= Permissible tensile stress for the bolt or stud material. We know that upward force acting on the cylinder cover, P =π/4 (D)2 -------------(i) This force is resisted by n number of bolts or studs provided on the cover. Therefore Resisting force offered by tt number of bolts or studs P = π/4 (dc)2ϭtb X n -------------(ii) From equations i) and (ii), we have π/4 (D)2= π/4 (dc)2ϭtb X n

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The tightness of the joint also depends upon the circumferential pitch of the bolts or tuds. The circumferential pitch should be between 20 √d1 and 30 √d2 ,where d) is the diameter of the hole in mm for bolt or stud. The pitch circle diameter (Dp) is usually taken as D + 2t + 3d1 and outside diameter of the cover is kept as Do = Dp + 3d1 = D + 2t + 6d1 where t = Thickness of the cylinder wall 2. Design of cylinder cover plate

Semi cover plate of cylinder The thickness of the cylinder cover plate (t)) and the thickness of the cylinder flange (t2) may be determined as discussed' below: We know that the bending moment at A-A, M = (Total bolt load/2)(OX- OY)= P/2(0.318Dp – 0.212Dp) = (P/2) X 0.106Dp = 0.053P X Dp Section modulus, Z =(1/6)w(t1)2 where w = Width of plate = Outside dia. of cover plate - 2 x dia. of bolt hole =Do-2d1 Knowing the tensile stress for the cover plate material, the value of t1, may be determined by using the bending equation, i.e. ϭt =M/Z. 3.Design of cylinder flange

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A portion of cylinder flange. The thickness of the cylinder flange (t2) may be determined from bending consideration. A portion of the cylinder flange under the influence of one bolt is shown in Fig. The load in the bolt produces bending stress in the section X-X. From the geometry of the figure, we find that eccentricity of the load from section X-X is e = Pitch circle radius - (Radius of bolt hole +Thickness of cylinder wall) = (Dp / 2) – {(d1/2)+ t} Bending moment, M = Load on each bolt x e = P/n x e Radius of the section X-x, R = Cylinder radius + Thickness of cylinder wall = D/2 + t Width of the section X-X, w = 2π R /n where n is the number of bolts.



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Section modulus, Z = 1/ 6 x w(t2)2 Knowing the tensile stress for the cylinder flange material, the value of t2 may be obtained by using the bending equation i.e. ϭt =M / Z

5. Attempt any TWO of the following: 2Χ8=16

(a) Given k= 0.8= di/do,P= 400KN,N=225rpm,M= 5000N.m, דmm= 50MPa T = 60P/2 π N =60 x 400 x 103 / 2 x π x 225 T=16976 N.m Equivantent twisting moment Te = √ (M2 + T2) = √ (50002 + 169762)= Te= 17697N.m= 17697 x 103N.mm Take service facor as 1.5 max /1.5= 50/1.5 ד = ד MPa 33.33 = דTe= π/16 x do3 (1-k4)ד

Te= π/16 x do3 (1-k4)ד 17697 x 103= π/16 x do3 (1-0.84) x 33.33 Therfore do=166.11 say 170mm And di= 0.8do= 0.8 x 170=132mm Select di=135mm

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(b) Question is vague and hence the standard procedure for the design of weld joint is considered.

(c)

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Design of Circumferential Lap Joint for a Boiler The following procedure is adopted for the design of circumferential lap joint for a boiler. 1. Thickness of the shell and diameter of rivets. The thickness of the boiler shell and the diameter of the rivet will be same as for longitudinal joint. 2. Number of rivets. Since it is a lap joint, therefore the rivets will be in single shear. Shearing resistance of the rivets, Ps =n x π /4 x d2x ד ... (i) where n = Total number of rivets. Knowing the inner diameter of the boiler shell (D), and the pressure of steam (P), the total shearing load acting on the circumferential joint, Ws = π /4 x D2x P -----(ii) From equations (i) and (ii), we get n x π /4 x d2x ד = π /4 x D2x P n =(D/d)2 x (P/ד) 3. Pitch of rivets. If the efficiency of the longitudinal joint is known, then the efficiency of the circumferential joint may be obtained. It is generally taken as 50% of tearing efficiency in longitudinal joint, but if more than one circumferential joints is used, then it is 62% for the intermediate joints. Knowing the efficiency of the circumferential lap joint (TIc)' the pitch of the rivets for the lap joint (P1) may be obtained by using the relation: ɳc =(P1 - d)/P1 4. Number of rows. The number of rows of rivets for the circumferential joint may be obtained from the following relation: Number of rows = Total number of rivets / Number of rivets in one the number of rivets in one row π (D + t) /p1 where D = Inner diameter of shell. 5. After finding out the number of rows, the type of the joint (i.e. single riveted or double riveted etc.) may be decided. Then the number of rivets in a row and pitch may be re-adjusted. In order to have a leak-proof joint, the pitch for the joint should be checked from Indian Boiler regulations. 6. The distance between the rows of rivets (i.e. back pitch) is calculated by using the relations discussed in the previous article. 7.After knowing the distance between the rows of rivets (Pb)' the overlap of the plate may be fixed by using the relation, Overlap = (No. of rows of rivets - 1) Pb + m where m = Margin

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6. Attempt any FOUR of the following: 4Χ4=16

(a)



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The transverse fillet welds are designed for tensile strength. Let us consider a single and double transverse fillet welds as shown in Fig.(a)and (b)respectively.

Enlarge view of fillet weld. The minimum area of the weld is obtained at the throat BD, which is given by the product of the throat thickness and length of weld. Let t = Throat thickness (BD), s= Leg or size of weld, = Thickness of plate, and l= Length of weld, From Fig. we find that the throat thickness, t = s x sin 45° = 0.707 s

01 Mark for fig.

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Minimum area of the weld or throat area, A = Throat thickness x Length of weld = t x I = 0.707 s x l If ϭt is the allowable tensile stress for the weld metal, then the tensile strength of the joint for single fillet weld, P = Throat area x Allowable tensile stress = 0.707 s x I X ϭt and tensile strength of the joint for double fillet weld, P = 2 x 0.707 s x I X ϭt = 1.414 s x I x ϭt Strength of Parallel Fillet Welded Joints: The parallel fillet welded joints are designed for shear strength. Consider a double parallel fillet welded joint as shown in Fig. We have already discussed in the previous article, that the minimum area of weld or the throat area, A = 0.707 s x I If T is the allowable shear stress for the weld metal, then the shear strength of the joint for single parallel fillet weld, P = Throat area x Allowable shear stress = 0.707 s x I x T and shear strength of the joint for double parallel fillet weld, P= 2 x 0.707 x s x l x T = 1.414 s x l x T

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(b) Question is vague and hence a specific answer is not possible but standard procedure for analysis of support reactions are to be followed.

4m 4m

(c) Consider a standard mirror polished specimen rotating in a fatigue-testing machine and loaded in bending • As the specimen rotates, the bending stresses at the upper fibers. vary from maximum compression to maximum tensile, whereas, at the bottom fibers vary from maximumtensile to maximum compressive. •In other words, the specimen is subjected to completely reversed stresses. This is represented by time stress diagram .

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•A record is kept of number of cycles required to produce a failure at given stress and results are plotted in stress-cycle graph. •A little consideration will show that, if the stress is kept lower than a certain value as shown by dotted line, the material will not fail, whatever may be the number of cycles.This stress shown by dotted line is known as endurance or fatigue limit Endurance or fatigue limit is defined as 'maximum value of completely reversed bending stress, which a standard specimen can withstand without failure, for infinite number of cycles of loads.

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(diagm.)

d Given w= 24mm,t=14mm,P=25KW,N=300rpm,FOS=3,ϭty=300N/mm2,d=40mm ϭt=ϭty/FOS =300/3=100MPa Consideing maximum shear stresstheory max= ϭt /2 = 100/2 = 50MPaדConsidering Key under shear failure T = l x w x דmax x d/2 -------------1. T= 60P/ 2πN = 60 x 25 x 103 /2 x π x 300=795.77 N.m T=795.77 x 103N.mm Therefore equation 1.becomes 795.77 x 103 = l x 24 x 50 x 40/2 Therefore l=33.15mm Consider failure of key in crushing T= l x t/2 x ϭc x d/2 Assumeϭc=ϭt=100Mpa 795.77 x 103 = l x 14/2 x 100 x 40/2 l= 56.84 say 60mm

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Hence length of the key should be taken as say 60mm 01 mark

e following are the general considerations in designing a machine component:

1. Type of load and stresses by the load 2. Motion of the parts or kinematics of the machine 3. Selection of materials 4. Form and size of the parts 5. Frictional resistance and lubrication 6. Convenient and economical features 7. Use of standard parts 8. Safety of operation 9. Workshop facilities 10. Number of machines to be manufactured. 11. Cost of construction 12. Assembling.

½ mark for each consider

ation (any

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WINTER 14 EXAMINATIONS Subject Code: 12310 Model …msbte.engg-info.website/sites/default/files/12310_winter_2014... · WINTER – 14 EXAMINATIONS Subject Code: 12310 Model Answer

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