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Mitchell Matthews
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1 (SO/EC Certified) Model nswer: Summer 7 Code: 17 mportant nstrutions to examiners: 1) The answers should e examined y key words and not as word-to-word as given in the model answer sheme. ) The model answer and the answer written y andidate may vary ut the examiner may try to assess the understanding level of the andidate. ) The language errors suh as grammatial, spelling errors should not e given more importane. (Not appliale for sujet English and Communiation Skills.) ) While assessing figures, examiner may give redit for prinipal omponents indiated in the figure. The figures drawn y the andidate and those in the model answer may vary. The examiner may give redit for any equivalent figure drawn. 5) Credits may e given step wise for numerial prolems. n some ases, the assumed onstant values may vary and there may e some differene in the andidate s answers and the model answer. 6) n ase of some questions redit may e given y judgment on part of examiner of relevant answer ased on andidate s understanding. 7) For programming language papers, redit may e given to any other program ased on equivalent onept Q 1 a) ttempt any SX of the following: 1 i) Define elastiity and modulus of elastiity. Elastiity: - Elastiity is the property of material y virtue of it an regain its original shape and size after removal of deforming fore. 0 Modulus of Elastiity: - t is defined as the ratio of stress to strain within elasti limit. ii) Define angle of oliquity. ngle of Oliquity : - The angle that the line of ation of the resultant stress makes with the normal to the plane is alled the angle of oliquity 0 0 iii) State the parallel axis theorem. t states that the M.. of a plane setion aout any axis parallel to the entroidal axis is equal to the M.. of the setion aout the entroidal axis plus the produt of the area of the setion and the square of the distane etween the two axes. 0 0 age 1 /

2 (SO/EC Certified) Model nswer: Summer 7 Code: 17 1 = G +h iv) What do you mean y eentri load? Show y simple sketh eentrially applied load. Eentri load: - The load whose line of ation does not oinide with the axis of the memer is alled as eentri load. 0 v) State any four assumptions made in the theory of pure torsion. ssumptions in the Theory of ure Torsion. 1. The material of the shaft is homogenous and isotropi and follows Hook s law.. The twist along the shaft is uniform.. The shaft is straight and having uniform irular ross setion throughout.. Cross setions of the shaft whih are plane efore twist remain plane after twist. 5. Stresses do not exeed the proportional limit. ½ mark eah (ny four) 0 age /

3 (SO/EC Certified) Model nswer: Summer 7 Code: 17 1 vi) Define ulk modulus. ulk Modulus: - When a ody is sujeted to three mutually perpendiular like stresses of same intensity then the ratio of diret stress to the orresponding volumetri strain of the ody is onstant and is alled as ulk modulus. 0 0 vii) Define hoop stress. State the formula. Hoop stress the stress whih at in the tangential diretion to the irumferene of the ylinder alled as hoop stress or irumfertial stress Hoop Stress, d t Where, Hoop stress Or Cirumferential stress nternal liquid pressure d = nternal daimeter of thin ylinder t = Thikness 0 viii) State middle third rule. n retangular setion for no tension ondition the load must lie within the middle third shaded area of size and d. This is known as 0 0 middle third rule. age /

4 (SO/EC Certified) Model nswer: Summer 7 Code: 17 Q. 1 ) ttempt any Two of the following: 08 i) metal rod mm diameter and meter long is sujeted to an axial pull of 0 kn. f the elongation of the rod is 0.5mm, find the stress indued and the value of Young s modulus. Given: rea, Stress ( ), d mm, L m 0kN 0 N l 0.5mm? E? d () 5.89mm N mm Young ' s mod ulus( E) E E L l E N / mm ii) simply supported eam of span 9.75m is arrying full span u.d.l. of 10 kn/m. What is the magnitude and position of maximum ending moment developed? To find the support reation, R Fy 0 R ( ) 0 R R 97.5 KN... 1 age /

(SO/EC - 70-005 Certified) Model nswer: Summer 7 Code: 17 1 Taking moment at, 0 1 9.75 R0 109.75 R9.75 0 75.1 9.75R M R R ut R in R R R 75.1 9.75 8.75KN 97.5 eq 8.75 97.5 n R 8.75KN SF alulation, SF.

5 (SO/EC Certified) Model nswer: Summer 7 Code: 17 1 Taking moment at, R R R M R R ut R in R R R KN 97.5 eq n R 8.75KN SF alulation, SF. 0 L SF R SF L SF L SF. 0 R s S. F. is zero at pt. C To find pt of ontra shear SF. 0 C SF C m MD 1 Mark. M Calulation, M. 0 M M. C M kn m C. M at C is the max. M. as at C S. F. is zero age 5 /

6 (SO/EC Certified) Model nswer: Summer 7 Code: 17 1 iii) irular eam of 10mm diameter is simply supported over a span of 10m and arries u.d.l of 1000 N/m. find the maximum ending stress produed. Given data : d 10 mm, L 10m 10000mm w 1000 N m 1000 w N mm 1N mm 1000? Max. ending Moment, M wl M M N mm Moment of nertia, 6 6 d mm d 10 y 60mm y y y 60mm Max. ending stress, M y t N mm ttempt any FOUR of the following: 16 a) i) What is meant y modular ratio? ii) State any four assumptions made in Euler s theory. i) Modular Ratio: The ratio of modulus of elastiity of two different materials is alled as modular ratio. t is denoted y m. 0 m E E 1 E 1 >E age 6 /

7 (SO/EC Certified) Model nswer: Summer 7 Code: 17 Following are the assumptions in the Euler s theory. a) The material of the olumn is perfetly homogenous and isotropi. ) The olumn is initially perfetly straight and is axially loaded. ) The ross setion of the olumn is uniform. d) The length of olumn is very large ompared to the lateral ½ marks eah (any four) dimensions. e) The self-weight of olumn is negleted. f) The olumn will fail y ukling only. ) irular steel ar of 10 mm diameter and 1.m long is sujeted to a ompressive load in testing mahine. ssuming oth ends hinged, alulate Euler s rippling load. lso alulate safe load y onsidering fator of safety as. Take E=x10 5 N/mm. Given data : d 10 mm, L 1.m 100 mm, fos, E 10 N / mm r? safe load? 5 For olumn with oth ends hinged Le l 100mm M.. for irular se tion, d 6 (10) mm Eulers rippling load, r r E Le (100) r N age 7 /

8 (SO/EC Certified) Model nswer: Summer 7 Code: 17 rippling load Safe load F. O. S Safe load Safe load.9n ) steel ue lok of 50mm side is sujeted to a fore of 6 kn (tensile) along X-diretion; 8 kn. (ompressive) along Y diretion and kn (tensile) along Z diretion. Determine hange in the 10 volume of the lok. Take E=00 Ga and m. Given data : l t 50mm x 6kN 6000 N ( Tensile) y 8kN 8000 N ( omprtessive) z kn 000 N ( Tensile) E 00Gpa 00 N / mm 10 m, 0. Find : v? Stress along x diretion x 6000 x. N / mm 5050 Stress along y diretion y 8000 y 5050 Stress along z diretion z N / mm z z. N / mm ( ompressive) original volume ( V ) V Lt V V 1510 mm age 8 /

9 (SO/EC Certified) Model nswer: Summer 7 Code: 17 For triaxial stress system x y z ev (1 ) E v x y z (1 ) V E v (1 0.) v v v 0.mm Change in volume is 0.mm d) onrete olumn 00mm X 00mm is reinfored with ars of 0mm diameter and arries a ompressive load of 00kN. The modular ratio is 15. Calulate the stresses in steel and onrete. lso alulate the load shared y eah material. Given data : rea of onrete olumn, 0000mm Diameter of steel ar, d 0mm of steel ar, n Load, p 00kN 00 N m 15???? s s rea of steel ar( ) s n d s 0 s s mm age 9 /

10 (SO/EC Certified) Model nswer: Summer 7 Code: 17 rea of onrete ( ), mm s as, m s 15 s also, ut, s s s ( ) (887.6 ) 00 ( ) s N / mm s, 15 s s N / mm s Load shared steel s s s s s N s kN ½ Load shared y eonrete N 9.859kN ½ age 10 /

11 (SO/EC Certified) Model nswer: Summer 7 Code: 17 e) antilever eam of length 10m arries two points load of magnitude 0kN and 0 kn at m and free end respetively. Draw the S.F.D and.m.d. Re ation at fixed end R R kN S. F. alulation SF.. 0 L S. F. 50 kn R S. F. 50 kn L S. F kn R S. F kn CL SF CR M alulation, M.. 0 C M M. 180 kn m M 0 0. M. 80 kn m age 11 /

12 (SO/EC Certified) Model nswer: Summer 7 Code: 17 OR S. F. alulation SF.. 0 L S. F. 50 kn R S. F. 50 kn L S. F kn R S. F kn CL SF CR M alulation, M.. 0 C M.. 0. M. 10 kn m M M. 0 kn m f) gas ylinder of internal diameter 1.m and thikness mm is sujeted to a maximum tensile stress of 90 Ma. Find the allowale pressure of gas inside ylinder. age 1 /

13 (SO/EC Certified) Model nswer: Summer 7 Code: 17 nternal daimeter d = 1. m = 100mm Thikness, t = mm Tensile stress, 90Ma 90 N / mm r essure of gas, p =? s tensile stress = Hoop Stress = 90 N / mm d. t N / mm 0 0 ttempt any FOUR of the following: 16 a) Draw S.F. and.m. diagram for simply supported eam of span L arrying a entral point load W. State the value of maximum shear fore and maximum ending moment. W Max. S. F = Max..M= WL age 1 /

14 (SO/EC Certified) Model nswer: Summer 7 Code: 17 ) Define point of ontra flexure. How is the point of ontra flexure loated for a eam? oint of ontra-flexure: - The point at whih ending moment diagram hanges the sign from positive to negative or vie versa or the point at whih M is zero is alled as point of ontra-flexure 0 Loation of point of ontra-flexure i) t the point of ontra-flexure.m is zero. ii) Take.M at the point of ontra-flexure and equate with zero. iii) The distane (loation) of point of ontra-flexure will e find from either end of eam. oint D,.M. = 0 oint D is oint of Contraflexure. oint D is at Z m from ) simply supported eam of m span arries two point loads of 5 kn eah at 1 m and m from the left end. Draw the shear fore and ending moment diagram. age 1 /

15 (SO/EC Certified) Model nswer: Summer 7 Code: 17 Step i) To find support reations, Sine loading is symmetial R R 5 kn Step ii) SF alulations SF.. 5 kn SF.. 5 kn L SF = 0 kn R SF.. = 0 kn C L SF.. = -5 kn C R SF.. = -5 kn D L D SF.. = 0 kn D Step iii) M.. Calulation, Sine the supports are simple.m. 0.M. 0 D M kn-m M kn-m C ½ ½ age 15 /

16 (SO/EC Certified) Model nswer: Summer 7 Code: 17 d) eam 6 m long rests on two supports 5 m apart. The right end is overhang y 1 m. the eam arries a u.d.l. of 5 kn/m over the entire length of the eam. Draw S.F. and. M. diagram. To find support reations, F 0 Y R 56 R 0 R R 0...(i) Taking moment at 6 R R 5 0 R 5 90 R 18 kn ut R in equation (i) R R 0 R 18 0 R 1 kn S. F. Calulation, SF.. = 0 L S. F. 1 kn R L R C L C R SF = -1 kn SF = 5 kn SF.. = = 0 SF.. = 0 Shear fore is zero at pt. D and the pt. D is at 'x' m from SF.. = 0 D SF.. = 1 5 x 0 D 5 x 1 x =. m M.. Calulation,.M. 0..M. D M. 5 kn m D SF al. SFD and MD (1 mark eah) M al. age 16 /

17 (SO/EC Certified) Model nswer: Summer 7 Code: 17 M.. 1. kn-m D 5 M M. = -.5 kn m M.. 0 C M.. is zero at pt. E t. E is at Y m from Y. M. E 1Y -5 Y = 0 Y Y = 0 Y =.8 m from e) point in a strained material stresses are sujeted to two mutually perpendiular tensile stresses of 00 Mpa and 100 Mpa. Determine the intensities of normal, shear a resultant stresses on a plane inlined at 0 0 with the axis of minor tensile stress. Given data: 00 Ma, 00Ma x ) nalytial Method??? n t R x y x y n os sin n os 0 0 x n 15 N/mm x y t sin os t sin 0 0. N/mm t age 17 /

18 (SO/EC Certified) Model nswer: Summer 7 Code: 17 Resultant Stress, R R n t (15) (.) R 1.87 N / mm OR ) Mohr s Cirle method (Graphial) f) Find the M.. of a T setion having top flange 00 mm 0 mm and we 00 mm 0 mm aout the entroidal axis X-X and Y-Y. Given data : 0 mm, d 00mm mm, d 0mm d mm d mm as the T se tion is symmetrial aout y axis 00 X 100 mm To find Y y d mm 1 1 age 18 /

19 (SO/EC Certified) Model nswer: Summer 7 Code: 17 y 1 d mm ( Y 1 1) ( Y ) Y (0000) (000 10) Y ( ) Y 155mm To find M.. aout X X xx x 1 x 1 x d x a 1 1h1 1 h mm (000 ) 1 ( ) x mm 6 x x x d ah 1 h mm (000 ) mm XX X1 X XX YY ( ) 6 = mm Y1 Y To find M.. at Y-Y axis d11 d YY a1h1 ah 1 1 h h 0 as symmetrial at Y axis 1 YY YY mm 6 age 19 /

20 (SO/EC Certified) Model nswer: Summer 7 Code: 17 Q. ttempt any FOUR of the following: 16 a) Find the moment of inertia of a square of side a aout its outer edge. i) For a square of side 'a' a. a a xx 1 1 ii) rea of setion, = a a a iii) The outer edge is paralle to XX axis Distane etween XX axis and Outer edge is a h = iv) Using the parallel axis theorom, M.. aout parallel axis = M.. aout entroidal axis + h h xx a a = a 1 a a 1 a 0 ) hannel setion 100m 100m 0m thik. Find the moment of inertia aout entroidal axis X-X and Y-Y. X a x a x a x a a a (100000) 500 (100000) 500 (0000) 150 (100000) (100000) (0000) X 1.667mm XX D d 1 1 age 0 /

21 (SO/EC Certified) Model nswer: Summer 7 Code: 17 XX XX 1 = mm YY Y1 Y1 YY YY 1 Y 1Y 1 YY d h 1 1 Y = Y 1 YY mm Y1 Y1 YY 10 1 YY d h 1 1 = mm YY YY mm ) n isoseles triangular setion C has ase width 80 mm and height 60 mm. Determine the M.. of the setion aout the C. G. of the setion and the ase C. ase ase xx xx h mm h mm d) hole of 100 mm diameter ut from a retangular plate 600 mm wide and 00 mm deep. The enter of hole is at 160 mm from the edge on an axis iseting shorter side. Find M.. of remaining plate aout X-X and Y-Y axis. X a1x1 ax a1 a (600 00) 100 (600 00) X.76mm age 1 /

22 (SO/EC Certified) Model nswer: Summer 7 Code: 17 XX XX XX D d (100) mm D YY h d h 1 6 YY (100) (100) YY mm State any four assumptions made in the theory of simple ending. e) ssumptions 1. The material of the eam is homogeneous and isotropi and follows the Hooke's Law.. The transverse setion of the eam whih is plane efore ending will remain plane after ending.. Young's modulus for the material is same for tension and ompression 1 mark eah (any four). Eah layer is free to expand or ontrat independently. 5. The eam in initially straight and of onstant ross setion. age /

23 (SO/EC Certified) Model nswer: Summer 7 Code: 17 f) eam 100mm wide and 50mm deep is sujeted to a shear fore of 0KN at a ertain setion find the maximum shear stress and draw the shear stress variation diagram. q q av max S S 0 d 500 q av max 1.6 N / mm 1.5 q q av. N / mm age /

24 (SO/EC Certified) Model nswer: Summer 7 Code: 17 5 ttempt any FOUR of the following: 16 a) timer eam has a ross setion 10 mm X 00 mm. t is simply supported over a span of m and arries a u.d.l. of 1 kn/m over the entire span. Calulate the maximum ending stress indued in eam and the radius of urvature to whih the eam will end at the setion. M E y R wl M 10 N mm ½ d mm 1 1 d 00 Y 100 mm 6 M 10 Y N / mm 80 M E OR y y R E E R OR R Y M E E R 6 OR R ½ R 0E OR R 0E Note: - f suitale value of E is assume should e onsider ) irular setion of diameter d is sujeted to load eentri to the axis the eentriity of load is e otain the limit of eentriity suh that no tension is indued at the setion. To find : e=? Diret stress, σ= 0 ending stress, M σ= Z age /

25 (SO/EC Certified) Model nswer: Summer 7 Code: 17 5 for irular se tion, d 6 d y d Z 6 d y d for no tension ondition, 0 M Z e Z 1 e Z d Z e d d e 8 The ore of a setion is d d irle of radius e = or diameter 8 ) retangular olumn 150mm wide and 100mm thik arries a load of 150 kn at an eentriity of 50mm in the plane iseting the thikness. Find the maximum and minimum intensities of stress. lso draw stress distriution diagram. Given data: = 150 mm, d = 100mm = 150 kn, e = 50mm σ =? σ =? max min rea of setion, = d= =15 10 mm age 5 /

26 (SO/EC Certified) Model nswer: Summer 7 Code: 17 5 Diret stress, σ= σ= σ 0 =10 N mm (C) ½ ending Stress, M σ= Z yy e σ= d 6 6e σ= d Diag. ½ σ= 1050 σ = 0 N mm σ = σ + σ = σ max 0 max = 0 N mm (C) 0 ½ σ = σ - σ = 10-0 σ min 0 min = 10 N mm (T) ½ d) square olumn 00mm X 00 mm arries an axial load of 00 kn. Find the position of 0 kn load ating along the axis iseting the width of the ross setion so that the stress developed at the other extreme of the olumn will e zero N / mm M Y 150 mm mm 1 1 e age 6 /

27 (SO/EC Certified) Model nswer: Summer 7 Code: 17 5 M Y 0 0 e For no tension ondition, e 8.5mm e e 0 e) square pillar is 600mm X 600mm in setion. t what eentriity a point load of 6000 kn is plaed on one of the entroidal axis of the setion so as to produe no tension in the setion. For no tension ondition, 0 M Z.. ey e Y e 100mm f) mild steel flat 50mm wide and 5mm thik is sujeted to load ating in the plane iseting the thikness at a point 10mm away of the entroid of the setion. f the tensile stress is not to exeed 150 Ma, alulate the magnitude of. Given data : 50 mm, d 5 mm, e 10 mm, max? 150Ma 150 N mm ( tensile) d mm age 7 /

28 (SO/EC Certified) Model nswer: Summer 7 Code: M e 6 e Z d 6 d yy max N kN 6 ttempt any FOUR of the following: 16 a) hollow shaft is of the same external diameter as that of the solid shaft. The inside diameter of the hollow shaft eing half the external diameter. oth the shafts have the same material and length. Then show that the ratio of torque transmitted y hollow shaft to the torque transmitted y solid shaft is T Gθ = J L G.θ T Hollow = J L G.θ T Solid = JSolid L G.θ J T Hollow L = T Solid G.θ J L THollow JHollow = T J Solid Solid Hollow Hollow Solid ut, π J Hollow = D -d π D J Hollow = D - age 8 /

29 (SO/EC Certified) Model nswer: Summer 7 Code: 17 6 π J Solid = D π D D- T Hollow = T π Solid D 16D -D = 16D T T Hollow Solid =0.975 ) shaft is transmitting 150 kw at 00 RM. f allowale shear stress is 80N/mm and allowale twist is per m, find the diameter of shaft. Take C = 0.8 X 10 5 N/mm ower =150 kw= W Speed N=00rpm Shear stress f =80N/mm 1.5 π 0 θ=1.5 = rad 180 Length L= m=000mm S C= N/mm 5 Find Case i) mean D=? πnt = watts 60 π 00 T = 60 T = N.m mean T = N.mm mean Case ii) Diameter ased on shear stress: T mean = T max Using relation, π T max= f S D 16 π = 80 D 16 D = mm age 9 /

30 (SO/EC Certified) Model nswer: Summer 7 Code: 17 6 max = 180 π 000 Case iii) Diameter ased on angle of twist Using relation, T Cθ = L D D=108.6 mm π Note: - dopt higher value of Diameter i.e mm eause it will satisfy oth shear stress and angle of twist. ) Calulate the suitale diameter of the solid shaft to transmit 0 kw at 150 rpm if the permissile shear stress is 68 Ma. Given data: ower = 0 kw = 0 10 W Speed N = 150 rpm Shear stress, f = 68 Ma = 68 N/mm S find : D i) Using the relation, = 0 10 = πnt watts 60 π 150 T 60 T = N.m T = ii)using the relation, 10 N -mm π T= f S D 16 π = 68 D 16 D = 1.6 mm age 0 /

31 (SO/EC Certified) Model nswer: Summer 7 Code: 17 6 d) Selet a suitale diameter for a solid shaft to transmit 00 H at 180 rpm. The allowale shear stress is 80 N/mm. Given data: ower =00H Speed N=180rpm Shear stress f =80N/mm π 0 θ=1 = rad 180 Length L=m=000mm S C= N/mm 5 Find D=? f the power is given in terms of H then πnt = H 500 π 180 T 00= 500 T=795.77kg.m T= = N.m T= N.mm Diameter ased on shear stress Using relation, π T= f S D 16 π = 80 D 16 D=79.1 mm Note: - f suitale value modulus of rigidity assumed to alulate diameter of solid shaft should e onsider. age 1 /

32 (SO/EC Certified) Model nswer: Summer 7 Code: 17 6 e) hollow shaft is of external diameter and internal diameter 00mm and 00mm respetively. Find the maximum torque is an transmit, if the angle of twist is not to exeed in a length of 10m. take C = 0.8X10 5 N/mm Given data: External diameter, D = 00mm nternal diameter, d=00mm 0 θ =1.5 =1.5 = 0.06 rad Length L=10m=10 10 mm π 180 C= N/mm Find T=? 5 y using torsional relation T Cθ = L Cθ π Cθ T = = (D - d ) L L 5 π T = (00-00 ) π 10 T = T = N.mm T = 9.8 kn.m f) (i) Differene etween pure ending and ordinary ending (ii) Write the equation of torque transmitted y the O.C. shaft giving meaning and unit of eah term. age /

33 (SO/EC Certified) Model nswer: Summer 7 Code: 17 6 (i) Differene pure ending and ordinary ending ure ending a) n pure ending the eam deflets into an ar of irle. ) eam is sujeted to normal (ending )stresses of tensile or ompressive in nature Ordinary ending a) n ordinary ending eam dose not deflets into an ar of irle. ) eam is sujeted to normal and shear stresses in it 0 (ii) The equation of torque transmitted y the O.C. shaft a) ased on angle Twist: T G.θ = L G.θ T= L ) ased on shear stress: T q = R q T= R Where, T = Torque ating on shaft (N-mm) R = Radius of urvature shaft (mm) G = Modulus of rigidity (N/mm ) L = Length of shaft (mm) = olar M.. of shaft setion (mm ) = ngle of twist in radians q = Max. shear stress at outer most fire of shaft (N/mm ) age /

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Sub. Code: Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may