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 vary but the examiner may try to assess the understanding level of the candidate. ) The language errors such as grammatical, spelling errors should not be given more importance. (Not applicable for subject English and Communication Skills.) ) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures drawn by the candidate and those in the 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 the 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. ------------------ Q. (a) Attempt any SIX of the following : (i) Define ductility and state names of two ductile metals. Ductility: It is the property of material to undergo a considerable deformation under tension before rapture.. Ductile Metals Steel, Aluminum, Copper. (any two) (ii) Define principal plane and principal stress. Principal Plane: A plane which carries only normal stress and no shear stress is called a principal plane. Principal Stress: The magnitude of normal stress acting on the principal plane is called principal stress. (iii) State theorem of parallel axis for moment of inertia along with a diagram. It states that the M. I. of a plane section about any axis parallel to the centroidal axis is equal to the M. I. of the section about the centroidal axis plus the product of the area of the section and the square of the distance between the two axes. MI about PQ = I PQ = I G +Ah Page
Q. (iv) Define axial load and eccentric load. Axial load: When a load whose line of action coincides with the axis of a member or whose line of action acts at a centroid of a section of member then it is called as axial load. Eccentric load: A load acts away from the centroid of the section or a load whose line of action does not coincide with the axis of member is called as eccentric load. (v) State an expression for power transmitted by a shaft giving meaning of each term used in it. πnt P= Watt 60 Where, P = Power transmitted by shaft (Watt) N = Number of revolutions of shaft per minute (r.p.m) T = Average or mean torque (N-m) OR πnt P= H.P. 500 Where, P = Power transmitted by shaft (H.P) N = Number of revolutions of shaft per minute (r.p.m) T = Average or mean torque (kg-m) (vi) Define Poison s ratio. Also state common value of Poison s ratio for C.I. Poison s ratio ( µ or /m ) : When a homogeneous material is loaded within its elastic limit, the ratio of the lateral strain to the linear strain is constant is known as Poison s ratio. Common value of Poison s ratio for C.I. = 0. to 0.6 (vii) Define hoop stress and longitudinal stress. Hoop Stress (σ c ): The stresses which act in the tangential direction to the perimeter (circumference) of the cylinder are called as hoop stress or circumferential stress. Longitudinal Stress (σ L ): The stresses which act parallel to the longitudinal axis of cylinder are called as longitudinal Stress. Page
Q. (viii) In relation with eccentric load, draw stress distribution diagram for ) Direct stress > bending stress and ) Direct stress = bending stress ) Direct stress > bending stress i.e. σ 0 > σ b ) Direct stress = bending stress i.e. σ 0 = σ b (b) (i) Attempt any TWO of the following : Calculate minimum diameter of steel wire to lift a load of 8. kn, if the permissible stress in wire is 0 MPa. Given : P = 8. kn, σ = 0 MPa = 0 N/mm To find : d min P σ= A P = π (d ) P 8. 0 (d )= = π π (σ) (0) 8. 0 d= = 87.00 π (0) d =9.7 mm min. 8 Page
Q. (ii) A cantilever beam, fixed at A has span of.5 m. Beam is loaded with uniformly distributed load of kn/m over entire span and downward point load of kn at free end B. Draw shear force and bending moment diagrams for the beam. Reaction at A = RA = ( x.5) + = 8 kn SF calculations - SF at A = + 8 kn SF at B = +8-6 = + kn L SF at B = + - = 0 ( ok) BM calculations - BM at B = 0 ( free end). 5 BM at A = -(.5)-.5 = -7.5kNm (iii) A simply supported beam of span 5 m is subjected to downward point load of 0 kn at m from left end. Cross section of beam is 00 mm wide and 00 mm deep. Calculate maximum bending stress developed in beam material. Also draw bending stress distribution across the section of beam. Page
Q. Given : l = 5 m, W = 0 kn at m from left, b = 00 mm, d = 00 mm To find : (σ b ) max Wab 0 6 BM max =M max = = = knm= 0 Nmm L 5 bd 00 00 6 I= = =50 0 mm d 00 y= = =50 mm By using flexural equation M σ = I y b 6 M y 0 50 σ= = =8 N/mm 6 I 50 0 σ =8 MPa max Page 5
No. Q. Sub. Attempt any FOUR of the following : 6 (a) (i) Define composite section and modular ratio. Composite Section: If two or more members of different materials are connected together and are subjected to loads, then such section is called composite section. Modular Ratio (m): It is defined as the ratio of modulii of the two different materials. E m= E Where, E = Modulus of elasticity of material E = Modulus of elasticity of material (ii) State equivalent length for column which is fixed at one end and hinged at other. Equivalent length for column which is fixed at one end and hinged at other is given by L l e (b) A column fixed at one end and free at other has effective length of 6 m. Calculate its actual length. Given : Column is fixed at one end and free at other end, l e = 6m To find : L l = L e 6= L 6 L= =m Actual length =m (c) A steel rod mm dia and. m in length is at 0 C. Find expansion of rod if the temperature is raised to 0 C. If this expansion is fully prevented, find the magnitude and nature of the stress induced in the rod. Take E=. x 0 5 N/mm and α = x 0-6 / C. Given : d = mm, L =. m, T = 0 C, T = 0 C, E=. x 0 5 N/mm and α = x 0-6 / C To find : δl, σ, Nature of stress Page 6
No. Q. Sub. Free expansion δl=l α T=L α (T -T ) -6 =. 0 0 (0-0) =.88 mm -6 = 0 If the expansion is prevented, compressive stress is developed in the steel rod. Compressive stress ( ) σ= T E = (T -T ) E (0-0).0 =76.0 N/mm (Compressive) 5 (d) A metal rod of 0 mm diameter and.8 m long when subjected to an axial tensile force of 58 kn showed an elongation of. mm and reduction in diameter was 0.006 mm. Calculate Poisson s ratio and modulus of Elasticity. Given : d =0 mm, L =.8 m, P = 58 kn, δ L =. mm, δ d = 0.006mm To find : µ, E P L 58 0.8 0 E= = A δl π (0). =505.5N/mm E=.5 0 N/mm 5 δd 0.006 LateralStrain d 0 μ= = = Linear Strain δl. L 800 μ=0.5 Page 7
Q. (e) At a point, the normal stress σ is associated with a shearing stress q. If the principal stresses at the point are 80 MPa (tensile) and 0 MPa (compressive), determine values of σ and q. σx σx σ n = + +q σx σx 80= + +q -----(i) σx σx σ n = - +q σx σx -0= - +q -----(ii) Adding equation (i) and (ii) σx σ x σx σ x 80-0= + +q + - +q σx 50= =σx σ x =50 N/mm Substituting the value of σ in equation (i) 50 50 80= + +q 80-5= 5 +q (55) = 5 +q q =(55) - 5 =00 q=8.99 N/mm x Page 8
No. Q. Sub. (f) A cylindrical shell is m long with internal diameter 900 mm and thickness 0 mm. If the tensile stress in the material is not to exceed 5 MPa, determine the maximum fluid pressure which can be allowed in shell. Given : L = m, d = 900 mm, t = 0 mm, σ = 5 MPa To find : P max P d σ= c t P 900 5= 0 P=.N/mm Allowable maximum fluid pressure =.N/mm Q. Attempt any FOUR of the following : 6 (a) A simply supported beam of span L is subjected to downward point load of w at a distance of a from left support and b from right support. Draw S.F. and B.M. diagrams. Take a > b. i) Support reactions Wb Wa R A = R B = L L ii) Shear force reactions Wb S.F. at A = + L Wb C L = + L Wb Wa C R = + - W = - L L Wa B L = - L Wa Wa B = - + = 0 ( OK) L L iii) Bending Moment calculations BM at A and B is equal to zero (Supports are simple) Wb BM at C = + x a = L Wab L Page 9
No. Q. Sub. (b) State relation between rate of loading, shear force and bending moment. a) Relation between rate of loading and shear force df = W dx The rate of change of shear force with respect to the distance is equal to the intensity of loading. b) Relation between shear force and bending moment. dm = F dx The rate of change of bending moment at any section is equal to the shear force at that section with respect to the distance. (c) A cantilever AD,.5 m long, carries point loads of 500 N, 700 N and 900 N at 0.5 m,.0 m and.5 m from fixed end A respectively. Draw S.F. and B. M. diagrams for cantilever. Neglect self weight of the beam. Page 0
Q. i) Support Reaction calculations Fy = 0 RA 500 700 900 = 0 RA = 00 N ii) Shear Force calculations SF at A = +00 N B L = +00 N B R = +00 500 = +600 N C L = +600 N C R = +600 700 = +900 N D L = +900 N D = +900 900 = 0 ( OK) iii) Bending Moment calculations BM at free end (i.e. at D ) = 0 BM at C = - (900 x 0.5) = - 50 Nm B = - (900 x ) (700 x 0.5) = - 50 Nm A = - (900 x.5) (700 x ) (500 x 0.5) = -00 Nm Page
Q. (d) Figure No. shows a shear force diagram for a simply supported beam ABCD of span 5m. Draw loading diagram and locate the position from support A where bending moment will be maximum. (There is no couple acting on the beam) (i) At A : The SF increases from 0 to 50 kn therefore R A = 50 kn. (ii) Between A to B : The SF diagram is an inclined straight line in m length, indicating that the beam carries udl between A to B. AB having intensity = 50-(-0) =0kN/m (iii) Between BC, the SF remains constant therefore there is no load. (iv) At C : There is sudden drop from (-0 kn) to (-50kN). Therefore there is a point load of (-0)-(-50) =0kN (v) Position of maximum bending moment from support A Let x be the distance of point of maximum bending moment from support A. Maximum bending moment occurs at point of contrashear. SF equation at point of contrashear is 50 = 0 x x =.5 m Maximum Bending Moment at.5m from support A. Page
No. Q. Sub. (e) A simply supported beam ABC is supported at A and B 5 m apart with an overhang BC.5 m long, carries uniformly distributed load of 0 kn/m over AB and downward point load of 0 kn at C. Draw S.F. and B.M. diagrams. State the value of maximum positive B.M. Calculations for point of contraflexure not expected. i) Support Reaction calculations MA = 0 (RA x 0) + (50x.5) (RBx5) + (0x6.5) = 0 0 + 5-5 RB + 0 = 0 55-5 RB = 0 RB = 5 kn Fy = 0 RA 50 + RB 0 = 0 RA 50 + 5 0 = 0 RA = 9 kn ii) Shear Force calculations SF at A = +9 kn B L = +9 50 = - kn B R = - + 5 = + 0 kn C L = +0 kn C = +0-0 = 0 ( OK) Page
iii) Bending Moment calculations BM at A and C = 0 ( A is simple support and C is free end) BM at B = - (0x.5) = - 0 knm iv) Maximum Bending Moment Maximum bending moment occurs at point of contrashear. Let AD = x, where D is point of contrashear at which maximum positive bending moment occurs. SF equation at D is 9 = 0x x =.9m from support A M D = M max = +(9x.9) (0x.9x.9/) = 6. 8.05 = 8.05 knm Page
Q. (f) A circular disc has diameter of 80 mm. Calculate moment of inertia about its any one tangent. Let tangent PQ parallel to centroidal x-x axis According to parallel axis theorem I PQ = I G + Ah I PQ = I xx + Ah = d 6 d d = 80 80 80 6 = 0.09 x 0 6 mm Page 5
No. Q. Sub. Attempt any FOUR of the following : 6 (a) An angle section 0 mm x 00 mm x 0 mm is placed such as its longer leg is vertical. Calculate M.I. about centroidal horizontal axis only (i.e. I xx only). i) To find the position of Centroid A = 80 x 0 = 600 mm y = 0 mm A = 0 x 0 =00 mm y = 0/ = 60 mm Y = A y A A A y = 6000 00 60 600 00 = 0 mm ii) Moment of Inertia I =(I ) +(I ) XX XX XX bd bd I XX = +Ah + +Ah 80 0 0 0 I XX = +600 (0-0) + +00 (60-0) I = 9. + 80000 I XX XX =5. 0 mm 6 Page 6
Q. (b) Define polar moment of inertia. Also state perpendicular axis theorem of M.I. Polar Moment of Inertia: The moment of inertia of a plane area about an axis perpendicular to the plane of the figure is called polar moment of inertia with respect to the point, where the axis intersects the plane. I P = I zz I P = I xx + I yy Perpendicular axis theorem: It states that if I xx and I yy are the moments of inertia of a plane section about the two mutually perpendicular axes, then moment of inertia of I zz about the third axis z-z perpendicular to the plane and passing through the intersection of x-x and y-y axes. I zz = I xx + I yy (c) An equilateral triangle has base of 00 mm. Using parallel axis theorem, calculate its M.I. about base. Height of triangle (h) h = 00 sin60 = 86.60mm y = h/ = 86.60 / = 8.87mm According to the parallel axis theorem I base = I G + Ay = bh bhy 6 0086.60 = 0086.60 8.87 6 = 5. x 0 6 mm Page 7
No. Q. Sub. (d) A T-section has flange 0 mm x 0 mm and web 5 mm x 0 mm, overall depth 0 mm. Calculate M.I. about its vertical centroidal yy axis. (i.e. I yy only). As the given composite section is symmetrical about y-y axis Moment of Inertia about yy axis (e) I YY = (I yy ) + (I yy ) db db = 05 00 = (880000) (750) I YY =.9 x 0 6 mm State four assumptions made in theory of simple bending.. The martial of the beam homogeneous and isotropic i.e. the beam made of the same material throughout and it has the elastic properties in all the directions.. The beam is straight before loading and is of uniform cross section throughout.. The beam material is stressed within its elastic limit and this obeys Hooke s law.. The transverse sections which where plane before bending remain plane after bending. 5. The beam is subjected to pure bending i.e. the effect of shear stress is totally neglected. 6. Each layer of the beam is free to expand or contact independently of the layer above or below it. 7. Young s modulus E for the material has the same value in tension and compression. each (any four) Page 8
No. Q. Sub. (f) Draw shear stress distribution diagram for circular beam section. State the formula to calculate average shear stress for circular section having diameter d. Average shear stress for circular section S S avg = = A d max = avg OR avg = max Q. 5 Attempt any FOUR of the following : 6 (a) With reference to theory of simple bending, explain neutral axis and moment of resistance. Neutral Axis: The fibers in the lower part of the beam undergo elongation while those in the upper part are shortened. These changes in the lengths of the fibers set up tensile and compressive stresses in the fibers. The fibers in the centroidal layer are neither shortened nor elongated. These centroidal layers which do not undergo any extension or compression is called neutral layer or neutral surface. When the beam is subjected to pure bending there will always be one layer which will not be subjected to either compression or tension. This layer is called as neutral layer and axis of this layer is called Neutral Axis. Page 9
No. Q. 5 Sub. Moment of resistance: Moment of resistance of the beam is the moment of couple formed by the total compressive force acting at the Centre of gravity of the compressive stress diagram and the total tensile force acting at the Centre of gravity of the tensile stress diagram. Moment of couple = C x Z or T x Z. This moment is called the moment of resistance of the beam and is denoted by M r. (b) Calculate diameter of core of section for circular column section having diameter of 0 mm. Using basic principles draw neat sketch for the same with dimensions. (Do not use direct formula) Given : Diameter of circular column = 0 mm To find: Diameter of core of section. Page 0
No. Q. 5 Sub. π π A= d = (0) =58.9 mm π π I= d = (0) =68606.0 mm 6 6 d 0 y= = =0 mm I 68606.0 Z= = =5768.06 mm y 0 For no tension condition Direct stress = Bending stress σ = σ 0 b P M = A Z P P e = A Z Z e= A 5768.06 e= 58.9 e = 0 mm Diameter of core of section is d'= e= 0 d'= 60 mm (c) A rectangular column 50 mm wide and 00 mm thick carries a load of 0 kn at an eccentricity of 0 mm in the plane bisecting the thickness. Calculate maximum and minimum stress intensities at the base along with their nature. Page
Q. 5 Given: P = 0 kn, b = 50 mm, d = 00 mm, e = 0 mm To find : σ max and σ min at base along with their nature P P Direct stress (σ 0)= = A b d = 0 0 50 00 =.N/mm 6 M=P e=0 0 0=6.0 0 Nmm db 0050 I yy = 785000 mm b 50 y= = =5 mm I 785000 Z= = = 05000mm y 5 6 M 6.0 0 Bending stress (σ b) = ± = ± = ±.56N/mm Z 05000 σ =σ +σ =.+.56 σ max 0 b max =7.67 N/mm (C) σ =σ -σ =.-.56 σ min 0 b min =.59 N/mm (T) (d) A hollow circular column having external and internal diameters 80 mm and 0 mm respectively is subjected to an eccentric vertical load of 0 kn at an eccentricity of 00 mm. Calculate maximum and minimum intensities of stress across the section. Page
Q. 5 Given : D = 80 mm, d = 0 mm, e = 00 mm, P = 0 kn To find : σ max and σ min across the section P P Direct stress (σ 0)= = A D d 0 0 = 80 0 =6.7 N/mm M Bending stress (σ b) = ± Z 6 M=P e=0 0 00= 0 Nmm I = D d = 80 0 885895.mm 6 6 D 80 y= = =0 mm I 885895. Z= = = 9985.6806 mm y 0 max 0 b 6 M 0 (σ b) = ± = ± = ±.09 N/mm Z 9985.6806 σ =σ +σ =6.7+.09 σ max =7.8 N/mm (C) σ =σ -σ =6.7-.09 σ min 0 b min =.57 N/mm (T) (e) A 6 mm diameter rod is bentup to form as offset link as shown in Figure No.. If permissible tensile stress is 90 N / mm, calculate maximum value of P. Page
Q. 5 Given : d = 6 mm, σ max = 90 N/mm To find : P max π π C/S area of section (A) = d = 6 =50.99 mm π π M.I.of section (I)= d = 6 =.756 mm 6 6 d 6 Eccentricity(e) =+ =+ =55 mm d 6 Maximum distance from neutral axis (y) = = =mm Section Modulus (Z) = = y I.756 =75.5mm - Direct stress (σ 0 )= = =.885 0 P max 0 b - 90 =.885 0 P+(0.087)P max b P P A 50.99 M P e P 55 Bending stress (σ ) = ± =± =± = (0.087)P σ =σ +σ 90 =(0.075)P P P =666.9 N =.66 kn Z Z 75.5 (f) Calculate maximum eccentricity for a hollow circular section having external diameter and internal diameter equal to 50 mm and 0 mm respectively, so that stress distribution is of same nature. Given : D = 50 mm, d = 0 mm To find : e max π π A= D -d = 50-0 =7777.65 mm π π I= D -d = 50-0 =856888.mm 6 6 D 50 y= = =5 mm I 856888. Z= = = 5550.706mm y 5 For stress distribution of same nature i.e.for no tension condition Page
No. Q. 5 Sub. Direct stress = Bending stress i.e. σ = σ 0 b Z e A 5550.706 e= 7777.65 e = 8.5mm e max = 8.5mm Q. 6 Attempt any FOUR of the following : 6 (a) State torsional equation with meaning of each term. Torsion Equation: - T C θ fs = = I L R P Where, T = Torque Or Turning moment (N.mm) I = Polar momet of inertia of the shaft section = I P xx yy C= Modulus of rigidity of the shaft material (N/mm ) θ = Angle through which the shaft is twisted due to torque s i.e. angle of twist (radians) L= Lenght of the shaft (mm) f = Maximum shear stress induced at the outermost layer of the shaft (N/mm ) R= Radius of the shaft (mm) + I (b) A shaft required to transmit 5 kw power at 80 r.p.m. The maximum torque may exceed the mean torque by 0%. If shear stress is not to exceed 60 N/mm, determine the minimum diameter of the shaft. Given : P = 5 kw = 5 x 0 Watt, N = 80 rpm, T max =. T mean, q max = 60 N/mm To find : Diameter of shaft π N Tmean P= 60 π 80 Tmean 5 0 = 60 T =6.9Nm mean Page 5
No. Q. 6 Sub. T max =. T mean =. 6.9 =7.78Nm =778.55Nmm π T max = q max d 6 π 778.55= 60 d 6 d =5.7mm (c) A solid circular shaft of 0 mm diameter is subjected to torque of 0.8 knm, causing angle of twist of.50 in a length of m. Calculate modulus of rigidity for the material of shaft. Given : d = 0 mm, T = 0.8 knm= 0.8 x 0 6 Nmm, L = m = 000 mm, π θ=.50 =.5 rad = 0.0608 80 To find : G I P d 0 795.56 mm Using torsional formula, T I P G L T L G I P 6 0.80 0 795.56.5 80 58 N / mm G.580 N / mm 5 (d) Compare the torsional strengths of two shafts A and B, made up of same material having equal weight and length. Shaft A is solid and B is hollow circular with D =.6 d. Page 6
Q. 6 Given: D =.6 d To compare the torsional strength of two shafts A and B As the material of shaft A and B is same their shear strengths are same. As their lengths and weights are same their cross sectional areas will be same. Let d be the diameter of solid shaft. D and d be the external and internal diameters of hollow such that D=.6d. Area of (solid) shaft A = Area of (hollow) shaft B d ' D d d '.6d d d'.56 d d'.56d d'.56d d'.5d Torsional strength for solid shaft A, π ' T A = q d 6 π = q.5d 6 π 6 Torsional strength for hollow shaft B, T A =.955 q d () π D - d T B= q 6 D π (.6 d) - d = q 6 (.6 d) π (.6) - () d = q 6 (.6) d π = q.7 d 6 π 6 =.7 q d () Page 7
Q.6 (d) From equation and π.7 q d T B 6 T π A.955 q d 6 TB.7 =.78 T.955 A The torsional strength of shaft B (hollow shaft) is 78% greater than that of shaft B (solid shaft) of same material and same cross sectional area. (e) A hollow shaft, having external diameter.5 times the internal diameter, is to transmit 50 kw at 00 r.p.m. If allowable angle of twist is in a length of m. Calculate diameters of the shaft. Take T max =. T mean. G = 80 GPa. Given : D =.5 d, P = 50 kw = 50 x 0 Watt, N = 00 rpm, L = m = 000 mm, T max =. x T mean, G = 80 GPa = 80 x 0 N/mm, π θ = = =0.09rad 80 To find : d and D π N Tmean P = 60 π 00 Tmean 50 0 = 60 T mean =76.979 Nm T =. T max mean =. 76.979 =859.66 Nm =85966.97 Nmm Using relation based on angle of twist T G θ = I L p Page 8
No. Q. 6 Sub. (f) π I P = (D -d ) π = (.5d) -d ) π = (.5) -( ) d I P=(0.9885)d T G θ = I L p 85966.97 80 0 0.09 = (0.9885)d 000 d= (0.9885) 80 0 0.09 d = 69.67 mm D =.5 d =.5 69.67 =0.05 mm 85966.97 000 (i) Draw bending stress distribution for rectangular beam section which is used for cantilever beam, subjected to downward load. (ii) Define torque and state its S.I. unit. Torque: When a tangential force is applied to a shaft at the circumference, in the plane of its transverse cross-section, the shaft is said to be subjected to a twisting moment called torque. Torque = Force x Radius S. I. Unit - Nm Page 9