S.Y. Diploma : Sem. III [CE/CS/CR/CV] Mechanics of Structure Time: Hrs.] Prelim Question Paper Solution [Marks : 70 Q.1(a) Attempt any SIX of the following. [1] Q.1(a) Define moment of Inertia. State MI of triangular section about its [] base. Ans.: It is the second moment of area which is equal to product of area of the body and square of the distance of its centroid from that axis, is called as moment of Inertia. OR Moment of inertia of a body about an axis is defined as the sum of second moment of all elementary areas about that axis. bh MI of triangular section about base I base = 1 Where, b = Base of triangle and h = Height of triangle Q.1(b) If polar moment of inertia of circular section is 000 mm 4 calculate diameter of the section. Ans.: Given: I P = 000mm for circular section I P = I xx + I yy I P = 4 4 D D 64 64 000 = D 4 64 D = 11.946 mm then [] Q.1(c) Define elastic body, giving two examples. [] Ans.: A body is said to be elastic if it regain s its original size and shape when an externally applied force causing deformation is entirely removed. Examples (i) Rubber band (ii) Golf Ball (iii) Soccer Ball Q.1(d) State Hooke's law. [] Ans.: Generalized Hooke s Law : Within elastic limit, the stress and strain vary linearly. i.e. within elastic limit : const. E where EYoung s modulus (1) - 1 -
Vidyalankar : S.Y. Diploma MOS Q.1(e) Define slenderness ratio. [] Ans.: It is the ratio of effective length of column to minimum radius of gyration, is called as slenderness ratio. Q.1(f) Define Resilience and modulus of resilience. [] Ans.: Resilience: It is the energy stored in the body or material, when loaded within elastic limit is called as strain energy or resilience. Modulus of Resilience: It is the proof resilience per unit volume, called as modulus of resilience is called as modulus of resilience. OR It is the maximum strain energy stored in body per unit volume is called modulus of resilience. Q.1(g) State meaning of effective length of column. [] Ans.: The material of the beam is homogeneous and isotropic i.e. the beam made of the same material throughout and it has the same elastic properties in all the directions. The beam is subjected to pure bending that is shear stress is totally neglected. The beam material is stressed within its elastic limit and thus obeys Hooke s law. The transverse sections which where plane before bending remains plane after bending. Each layer of the beam is free to expand or contact independently of the layer above or below it. Young s modulus (E) for the material has the same value in tension and compression. The radius of curvature is large as compared to the dimensions of the cross section. Q. Attempt any THREE of the following : [1] Q.(a) Determine moment of Inertia about the centroidal axes X-X and Y- [4] Y of an Unsymmetrical I section with following details. Top flange - 100 mm 0 mm Bottom flange - 160 mm 0 mm Web - 80 mm 0 mm Ans.: (i) Calculation of centroid: As given section is unsymmetrical about y-y axis, - -
Prelim Question Paper Solution Largeflangewidth 160 X = = 80mm A 1 = 160 0 = 00mm, A 1 = 80 0 = 1600mm, A = 100 0 = 000mm Y 1 = 0 = 10mm, Y = 0 + 80 = 60mm, Y = 0 + 0 + 0 = 110mm, 00 10 1600 60 000 110 Y = = 51.17mm, 6800 (ii) Calculation of I xx : I xx = I xx1 + I xx + I xx I xx = I A h I A h I A h G1 1 1 G G I xx = bd bd bd Ah Ah Ah 1 1 1 1 1 Here, h 1 = YY = 1.17 10 = 41.17mm h = Y 1 Y = 60 51.17=8.8mm h = Y Y = 110 51.17 = 58.8mm I xx = 160 0 0 80 00 41.17 1600 8.8 1 1 + 100x0 000 58.8 1 I xx = 1.496 10 6 mm 4 - -
Vidyalankar : S.Y. Diploma MOS (iii) Calculation of I yy : I YY = I YY1 + I YY + I YY I A h I A h I Ah G1 1 1 G G I YY = db db db 1 1 1 I YY = 0 160 80x0 0 100 1 1 1 I YY = 8.546 10 6 mm 4 I YY = Q.(b) Find the least moment of Inertia about the centroidal axes XX and YY of an unequal angle section 15 mm 75 mm 10 mm as shown in figure. 15 mm 10 mm [4] 10 mm Ans.: (i) Calculation centroid: A 1 = 75 10 = 750mm, A = 115 x 10 = 1150mm X 1 = 75 = 7.5m, X = 10 = 5mm Y 1 = 10 = 5mm, Y = 10 + 115 = 67.5mm AX AX AY AY 1 1 1 1 X = = 17.8mm, Y = = 4.8mm A A A A 1 1 75 mm - 4 -
(ii) Calculation of I xx : I xx = I xx1 + I xx I A h I A h I xx = G1 1 1 G I xx = I A h I A h G1 1 1 G Here, h 1 = Y - Y = 4.8 5 = 7.8mm h = Y Y 67.5 4.84.67mm I xx = 75 10 10 115 750 7.8 1150 4.67 1 1 I xx =.046 10 6 mm 4 (iii) Calculation of I YY : I yy = I yy1 + I yy I Ah I Ah I yy = G1 1 G 4 Here, h = X X 7.5 17.8 19.67mm h 4 = X X 17.8 5 1.8mm I yy = 10 75 115 10 750 19.67 1150 1.8 1 1 I yy = 840.67 10 mm 4 Prelim Question Paper Solution Q.(c) Find the least M.I. of a symmetrical I-section having following details: Flanges : 100 mm 0mm Overall depth : 80 mm Thickness of web : 10 mm Ans.: [4] I xx and I yy =? - 5 -
Vidyalankar : S.Y. Diploma MOS Above figure symmetrical @ XX and YY axis So, M.I. of I section @ xx-axis M.I xx = M.I. AABCD M.I. PQRS M.I. LMNO 100 80 M.I. ABCD = = 18.9 10 6 mm 4 1 45 40 M.I. PQRS = = 51.84 10 6 mm 4 1 45 40 M.I. LMNO = = 51.84 10 6 mm 4 1 From (i) M.I xx = 18.9 10 6 - [ 51.84 16 6 ] M.I xx = 79.5 10 6 mm 4 M.I. of I- section @ Y Y axis I yy = I yy1 + Iyy + I yy I yy1 = b 0 100 = 1.67 10 6 mm 4 1 1 I yy = b 40 10 = 0 10 mm 4 1 1 I yy = I yy1 = 1.67 10 6 mm 4 From equation (ii) I yy =1.67 10 6 + 0 10 + 1.67 10 6 I yy =.6 10 6 mm 4 least M.I. (i) (ii) Q.(d) A column having diameter 00 mm is of length meters. Both ends of a column are hinged. Find Euler's crippling load. Take E = 10 5 MPa. Ans.: Given: d = 00 mm L = m = 000mm E = 10 5 mpa P E = Euler s crippling load =? Both end of column are hinged Le = L 000 mm We have Euler s crippling load EI P E = (i) Le For circular column, M.I. is [4] I = 4 64-6 -
Prelim Question Paper Solution 4 I = 00 64 I = 78.5 10 6 mm 4 From equation (i) 10 78.510 P E = 000 P E = 17. 10 6 N 5 6 Q. Attempt any THREE of the following : [1] Q.(a) A bar of uniform cross sectional area 100 mm is subjected to axial [4] forces as shown in Fig. Calculate the net change in length of the bar. Take E = 10 5 N/mm. A B C D 1 kn kn P kn 4 kn 00 mm Ans.: A = 100 mm E = 10 5 N/mm L =? To calculate, P 1 + 4 = P P = 5 P = kn L = - L AB + L BC - L CD L = - PL PL PL AE AB AE BC AE CD L =- 1 10 00 10 10 600 100 00 10 100 00 10 100 00 10 L = 0.015 + 0.04 0.06 L = - 0.05mm 400 mm 600 mm AB BC CD Q.(b) A steel tube with 40 mm inside diameter and 4 mm thickness is filled with concrete. Determine load shared by each material due to axial thrust of 60 kn. Take E steel = 10 10 N/mm E concrete = 14 10 N/mm Ans.: Given: d = 400mm t = 4mm P = 60KN E S = 10 10 N/mm E C = 14 10 N/mm D = d + t = 40 + ( 4) D = 48 mm [4] - 7 -
Vidyalankar : S.Y. Diploma MOS 4 48 40 4 A s = 55.9mm A s = D d A s = A C = d 4 A c = 40 4 Ac = 156.67mm m = ES 10 10 E 14 10 = 15 C es = ec s c E E S C s = E S EC s = 15 c P = P S + P C c... (i) (ii) P = s A S + c A C 60 10 = 15 c 55.9 + c 156.67 60 10 c = 9550.47 c = 6.84 N/mm P c = c A C P c = 6.84 156.67 P c = 7894.74N P c = 7.89kN s = 15c s = 15 6.84 s = 94.6 N/mm P s = s A S P s = 94.6 55.9 P s = 5104.96N P s = 5.104kN - 8 -
Prelim Question Paper Solution Q.(c) Draw stress-strain curve for mild steel under tensile loading showing important points on it. Ans.: [4] Q.(d) State any four assumptions made in theory of pure bending. [4] Ans.: Assumption in the theory of pure bending : (i) The material of the beam is homogenous & isotrophic. (ii) The beam material is stressed to obey Hooke s law. (iii) The transverse sections, which were plane before bending, remain plane after bending. (iv) Each layer of beam is free to expand or contract independently. (v) Value of E is the same in tension & compression. Q.4 Attempt any TWO of the following : [16] Q.4(a) A cube of 00 mm side is subjected to a compressive force of [8] 600 kn on all its faces. The change in the volume of cube is found to be 5000 mm. Calculate the Bulk modulus. If = 0.8, find the Young s modulus. P 00 10 Ans.: = A 00 00 = 90N/mm V = L = (00) V = 8 10 6 mm V xyz 1 V E E = 1 V V - 9 -
Vidyalankar : S.Y. Diploma MOS 90 E = 5000 10.8 6 810 E = 1.9 10 5 N/mm Q.4(b) An over changing beam ABC is such that AB = 4m, BC = 1 m, is supported at A and B. The beam ABC is subjected to vdl of 0 kn/m over the entire length ABC. It is subjected to point load 50 kn at the free end C. Draw SFD and BMD with calculations and locate the point of contra flexure. Ans.: (i) To calculate the reactions at supports: R B 4 = (0 5.5) + 50 5 R B = 1565kN R A = (0 5 50) 156.5 R A = 4.75kN [8] (ii) Shear force calculations SF at A = 4.75 kn SF at B L = 4.75 0 40 = -76.5 kn SF at B R = - 76.5 + 156.5 = 80 kn SF at C L = 80 0 1 = 50 kn SF at C = 50 50 = 0 (OK) (iii) Bending moment calculations BM at A and C = 0 BM at B = - 50 1 0 1 1 = - 65 kn-m (i) To calculate Maximum Bending Moment SF at x = 0 4.75 0 = 0 x = 1.458m from A 1.458 BM max = 4.75 1.458 0 BM max = 1.90kN m (ii) To locate point of contraflexure BM at x = 0 4.75 x 0 x' = 0 X =.916 m from A - 10 -
Prelim Question Paper Solution Q.4(c) Draw SFD and BMD of a beam as shown in figure. Also find the point of contra flexure. 0 kn/m B A C [8] 8 m m Ans.: (i) To find reactions R A and R B R A + R B = 0 10 R A + R B = 00 (i) M A = 0 10 10 R B 8 M A = 1000 8 R B 8R B = 1000 R B = 15 kn R A = 00 R B R A = 00 15 R A = 75 kn (ii) Shear force calculation F C = 0 F BR = 0 = 40 kn F BL = 40 15 F BL = 85 kn - 11 -
Vidyalankar : S.Y. Diploma MOS F A = 85 + 0 8 F A = 85 + 160 F A = 75 kn (iii) B.M. calculation M C = 0 kn.m M B = (0 ) = 0 1 = 40 kn.m M A = 0 kn.m (iv) To locate the position of point of contrashear 85 8 x = 75 x 85x = 75 (8 x) 85x = 600 75x 160x = 600 x =.75 m from left M XX = M PCS = 75 x 0 x x M XX = 75.75 0.75.75 M XX = 81.5 140.65 M XX = 140.65 kn.m (sagging) To locate the point of contraflexure (P CF ) M x 1 = 75 x 1 0 x 1 0 = 75 x 1 10x 1 0 = x 1 (75 10x 1 ) 75 = 10x 1 = 0 75 = 10x 1 x 1 = 7.5 m from A x 1 Point of contraflexrure (P CF ) = 7.5 cm from A - 1 -
Prelim Question Paper Solution Q.4(d) State the flexural formula, giving meaning of the symbols used in it. [4] Ans.: Flexural formula M E I y R Where m = max bending moment which is equal to moment of resistance of beam. I = M.I. of beam section about the neutral axis since neutral axis always lies at the centroid of the section. I = I NA = I XX = Bonding stress in a layer at a distance y from N.A. Y = distance of the layer from the N.A. of the beam material. R = Radius of curvature of the bentup beam. Q.5 Attempt any TWO of the following : [1] Q.5(a) A cantilever beam is loaded as shown in figure. Draw the S.F.D. kn 1kN/m [6] and B.M.D. A C B m 4 m Ans.: (i) Support reaution Fy = 0 RA = + (1 4) = 6kN (ii) Shear force calculation [+ve -ve] S.F at just left of A = 0 kn - 1 -
Vidyalankar : S.Y. Diploma MOS S.F at just right of A = R A = 6 kn S.F at just left of C = 6kN S.F. at just right of C = 6 = 4 kn S.F at B = 4 (1 4) = 0 kn Q.5(b) A T section of flange 160 mm x 0 mm and web 180 x 0 mm is simply supported at the both ends. It carries two concentrated loads of 100 kn each acting m distance from each support. Span of the beam is 8m. Determine the maximum bending stress induced in the beam and draw bending stress distribution diagram and also find bending stress at the layer 100 mm from the bottom. ay ay 160 0190 180 090 1 1 Ans.: Y = a a 00 600 1 Y = 17.05 mm from base I NA = I xx = I xx1 + I xx I NA = I xx = I A h I A h G1 1 1 G I NA = I xx = bd bd Ah Ah 1 1 1 1 I NA = I xx = 160 0 0 180 00 5.94 600 47.06 1 1 I NA = I xx = 676784.15 mm 4 Maximum bending moment M = 100 = 00 kn-m = 00 10 6 N mm c = t = t = M y c I M y t I M yt100 I = 6 00 10 676784.15 6.94 = 470.65 N/mm (C) = 6 00 10 676784.15 17.06 = 104.06 N/mm (T) = 6 00 10 676784.15 7.06 = 76.899 N/mm (T) [6] - 14 -
Prelim Question Paper Solution Q.5(c) A metal rod 0 mm diameter and m long when subjected to tensile force of 60 kn shows an elongation of mm and reduction in diameter 0.006 mm. Calculate the modulus of elasticity and modulus of rigidity. Ans.: E = PL AL 5 60 10 10 E = 0 4 E = 1.91 105 N/mm = Lateral Strain LinearStrain d 0.006 = d 0 L L 000 = 0. E = G(1 + ) = G(1 + 0.) 1.91 10 5 = G(1 + 0.) 5 1.9110 G = 1. G = 7.45 10 4 N/mm [6] Q.6 Attempt any TWO of the following : [1] Q.6(a) (i) A simply supported beam of span `L carries central point load [6] `W. Draw SED and BMD (ii) Define shear force and bending moment. Write unit of each. Also state relation between them. Ans.: (i) Step 1 Calculation of Reaction, As, the load is at centre so, Support reaction are equal R A = R B = W Step Share force calculation (a) S.F. at any section between A and C is, F x = + R A = W - 15 -
Vidyalankar : S.Y. Diploma MOS (b) S.F. at any section between B and C is, F x = -R B = - W Step : Bending Moment Calculation Beam is simply supported at the end A and B, M A = M B = 0 M max = MC = + M C = WL 4 W L (ii) Shear force: Shear force at any cross section of the beam is the algebraic sum of vertical forces on the beam acting on right side or left side of the section is called as shear force. OR A shear force is the resultant vertical force acting on the either side of a section of a beam. Unit :- kn or N Bending Moment: Bending moment at any section at any cross section is the algebraic sum of the moment of all forces acting on the right or left side of section is called as bending moment. Unit: kn-m or N-m Relation between shear force and bending movement dm F dx The rate of change of bending moment at any section is equal to the shear force at that section. - 16 -
Prelim Question Paper Solution Q.6(b) A cast iron column 100 mm external diameter and 80 mm internal diameter is m long. It is fixed at one end and hinged at other end. Calculate the safe axial load by Rankine s formula taking factor of safety. Assume c = 550 N/mm and Rankine s constant = 1 1600. Ans.: Given D = 100mm, d = 80mm, L = m = 000mm, FOS =, c = 550N/mm 1, α = 1600 As, the column is fixed at one end and hinged at other end. L 000 Efeective length, (Le) = Le = 1414.mm For hollow circular column. I min = I xx = I yy = 100 4 80 4 64 I min = I xx = I yy = 898119.mm 4 Area, A = 100 80 4 A = 87.4mm [6] K = I A K = 898119. 87.4 K =.056mm Le 1 + a = 1 + K Le 1 + a =.195 K By using Rankine s formula, A c P R = Le 1 a K 550 87.4 P R =.195 P R = 707644.077N 1 1414. 1600.0156-17 -
Vidyalankar : S.Y. Diploma MOS Safe Load = Rankine'scripplingload Factorofsafety Safe Load = 707644.077 Safe Load = 548.069N Safe Load =.548kN Q.6(c) A beam has hollow rectangular section with external dimensions 80 mm 160 mm and uniform thickness of section is 10 mm. Draw shear stress variation diagram. It section is subjected to the shear force 70 kn. Also determine ratio of maximum shear stress and average shear stress. Ans.: [6] A = (BD b d) = (80 160 60 140) A = 4400 mm 1 BD bd = 1 I NA = 1586666.67 mm 4 I NA = q avg = S A = 70 10 4400 q avg = 15.91 N/mm q 1 = SAY bi = 70 10 80 10 75 80 1586666.67 q 1 =.864 N/mm 1 1 (80 160 60 140 ) - 18 -
Prelim Question Paper Solution q = q 1 80 =.864 40 0 q = 15.456 N/mm q add = SAY bi = 70 10 70 10 5 0 1586666.67 q add = 1.6 N/mm q NA = q max = q + q add = 15.456 + 1.6 q NA = 8.078 N/mm Ratio, qmax = 8.078 q 15.91 avg qmax = 1.765 q avg - 19 -