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SOLID MECHANICS ECA
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SOLID MECHANICS
SYLLABUS: Elasticity constants, types of beams- determinate and indeterminate, bending moment and shear force diagrams of simply supported, cantilever and over hanging beams. Moment of area and moment of inertia for rectangular & circular sections, bending moment and shear stress for tee, channel and compound sections, chimneys, dams retaining walls, eccentric loads, slope, deflection of simply supported and cantilever beams, critical load and columns, Torsion of circular shaft.
TOPICS COVERED
STRENGTH OF MATERIALS DETERMINACY AND INDETERMINACY SHEAR FORCE AND BENDING MOMENT BENDING STRESS IN BEAMS SHEAR STRESSES IN BEAMS PRINCIPLE STRESSES IN BEAM DEFLECTION OF BEAMS COLUMNS AND MOMENT OF INERTIA TORSION IN CIRCULAR SHAFTS
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UNIT 1 STRENGTH OF MATERIALS
1. The stress-strain curve for an ideally plastic material is
2. Match List I (Properties) with List II (Stress points labeled 1, 2, 3 and 4) in the stress-strain figure and select the correct answer:
Codes: A B C D (a) 3 4 1 2 (b) 4 3 1 2 (c) 3 4 2 1 (d) 4 3 2 1 3. Consider the following salient points in a stress-strain curve of a mild steel bar. (1) Yield point (2) Breaking point (3) Yield plateau (4) Proportionality limit (5) Ultimate point The correct sequence in which they occur while testing the mild steel bar in tension form initial zero strain to failure is: (a) 4, 1, 2, 3 and 5 (b) 1, 4, 3, 5 and 2 (c) 4, 1, 3, 5 and 2 (d) 1, 4, 2, 3 and 5
4. If a material has identical elastic properties in all directions it is said to be (a) elastic (b) isotropic (c) orthotropic (d) homogeneous 5. Modulus of rigidity is defined as the ratio of (a) Longitudinal stress to longitudinal strain (b) Shear stress to shear strain (c) Stress to strain (d) Stress to volumetric strain 6. For an isotropic, homogenous and elastic material obeying Hooke’s law, number of independent elastic constants is (a) 2 (b) 3 (c) 9 (d) 1 7. Elastic limit is the point (a) up to which stress is proportional to strain (b) at which elongation takes place without application of additional load (c) up to which if the load is removed, original volume and shape are regained (d) at which the toughness is maximum 8. Poisson’s ratio is defined as the ratio of (a) Longitudinal stress and longitudinal strain (b) Lateral strain and longitudinal strain (c) Longitudinal stress and lateral stress (d) Lateral stress and longitudinal stress 9. Which of the following statements are correct? 1. Strain in the direction of applied stress is known as longitudinal strain. 2. Tensile stress results in tensile strain in linear and lateral directions. 3. Strains in all directions perpendicular to the applied stress are known as lateral strain. 4. Ratio of change in volume to original volume is known as volumetric strain. (a) 1, 2 and 3 only (b) 1, 3 and 4 only (c) 3 and 4 only (d) all of the statements 10. As per the elastic theory of design, the factor of safety is the ratio of (a) Working stress to stress at the limit of proportionality (b) Yield stress to working stress
List I List II A. Yield point B. Proportional limit C. Rupture strength D. Ultimate strength
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(c) Ultimate stress to working stress (d) None of the above 11. Limiting value of Poisson’s ratio are (a) – 1 and 0.5 (b) – 1 and – 0.5 (c) 1 and – 0.5 (d) 0 and 0.5 12. What is the correct sequence of the following metals in the decreasing order of their Poisson’s ratio? 1. Aluminium 2. Cast iron 3. Steel Select the correct answer using the code given below: (a) 1, 2, 3 (b) 2, 1, 3 (c) 1, 3, 2 (d) 3, 1, 2 13. A circular rod of diameter 30 mm and length 200 mm is subjected to a tensile force. The extension in rod is 0.09 mm and change in diameter is 0.0045 mm. What is the Poisson’s ratio of the material of the rod? (a) 0.30 (b) 0.32 (c) 0.33 (d) 0.35 14. A bar of 40 mm diameter and 400 mm length is subjected to an axial load of 100 kN. It elongates by 0.150 mm and the diameter decreases by 0.005 mm. What is the Poisson’s ratio of the material of the bar? (a) 0.25 (b) 0.28 (c) 0.33 (d) 0.37 15. Given E as the Young’s modulus of elasticity of a material, what can be the minimum value of its bulk modulus of elasticity? (a) E/2 (b) E/3 (c) E/4 (d) E/5 16. Poisson’s ratio of a material is 0.3. Then the ratio of Young’s modulus to bulk modulus is (a) 0.6 (b) 0.8 (c) 1.2 (d) 1.4 17. The relationship between Young’s Modulus E, Modulus of Rigidity C and Bulk Modulus K in an elastic material is given by the relation
(a) E = 9KC
3K+ C (b) E =
3KC3K+ C
(c) E = 9KC
9K+ C (d) E =
3KC9K+ C
18. If G is the modulus of rigidity, E the modulus of elasticity and µ the Poisson’s ratio for a material, then what is the expression for G?
(a) G = 3E
2(1+ 2μ) (b) G =
5E(1+ μ)
(c) G = E
2(1+ μ) (d) G =
E(1+2μ)
19. If the Poisson’s ratio for a material is 0.5, then the elastic modulus for the material is (a) Three times its shear modulus (b) Four times its shear modulus (c) Equal to its shear modulus (d) Not determinable 20. The value of modulus of elasticity for a material is 200 GN/m2 and Poisson’s ratio is 0.25. What is its modulus of rigidity? (a) 250 GN/m2 (b) 320 GN/m2 (c) 125 GN/m2 (d) 80 GN/m2
21. A metal bar of 10 mm diameter when subjected to a pull of 23.5 kN gave an elongation of 0.3 mm on a gauge length of 200 mm. The Young’s modulus of elasticity of the metal will nearly be (a) 200 kN/mm2 (b) 300 kN/mm2 (c) 360 kN/mm2 (d) 400 kN/mm2
22. The longitudinal strain of cylindrical bar of 25 mm diameter and 1.5 mm length is found to be 3 times its lateral strain in a tensile test. What is the value of Bulk Modulus by assuming E = 1 x 105 N/mm2? (a) 2 x 105 N/mm2 (b) 1.1 x 105 N/mm2 (c) 1 x 105 N/mm2 (d) 2.1 x 105 N/mm2
23. The ratio of intensity of stress in case of a suddenly applied load to that in case of a gradually applied load is (a) 1/2 (b) 1 (c) 2 (d) more than 2
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24. Proof resilience is the maximum energy stored at (a) limit of proportionality (b) elastic limit (c) plastic limit (d) none of the above 25. Strain energy stored in a member is given by (a) 0.5 x stress x volume (b) 0.5 x strain x volume (c) 0.5 x stress x strain x volume (d) 0.5 x strain x stress 26. A rectangular block of size 200 mm x 100 mm x 50 mm is subjected to a shear stress of 100 N/mm2. If modulus of rigidity of material is 1 x 105 N/mm2, strain energy stored will be (a) 10 N.m (b) 25 N.m (c) 50 N.m (d) 100 N.m 27. If the depth of a beam of rectangular section is reduced to half, strain energy stored in the beam becomes (a) 1/4 times (b) 1/8 times (c) 4 times (d) 8 times 28. The material in which large deformation is possible before the absolute failure or rupture is termed as (a) Ductile (b) Plastic (c) Brittle (d) Elastic 29. Creep of a material is property indicated by (a) A time-dependent strain of the material. (b) Elongation of the material due to changes in the material properties. (c) Shortening caused by shrinkage of the member. (d) The decrease in the volume of the material affected by the weather conditions. 30. Resilience is (a) Maximum strain energy (b) Recoverable strain energy (c) Total potential energy (d) Shear strain energy (Beyond Hooke’s Law)
31. What is the nature of stress in a ceiling fan rod? (a) Bending (b) Tensile (c) Compressive (d) Shear 32. Some structural members subjected to long time sustained loads deform progressively with time especially at elevated temperatures. What is such a phenomenon called? (a) Fatigue (b) Creep (c) Creep Relaxation (d) Fracture 33. Which one of the following represents constitutive relationship? (a) Vertical displacements in a structure (b) Rotational displacements in a structure (c) System of forces in equilibrium (d) Stress-strain behavior of a material 34. Impact test enables one to estimate the property of (a) hardness (b) toughness (c) strength (d) creep 35. The phenomenon of decreased resistance of a material to reversal of stress is called (a) creep (b) fatigue (c) resilience (d) plasticity 36. The stress below which a material has a high probability of not failing under reversal of stress is known as (a) tolerance limit (b) elastic limit (c) proportional limit (d) endurance limit 37. The principle strains at a point are: +800 x 10-6, +400 x 10-6 and -1200 x 10-6 The volumetric strain is w equal to (a) +1200 x 10-6 (b) +800 x 10-6 (c) -1200 x 10-6 (d) Zero 38. A mild steel bar of square cross-section 40 mm x 40 mm is 400 mm long. It is subjected to a longitudinal tensile stress of 440 N/mm2 and lateral compressive stress is 200 N/mm2 in perpendicular directions. E = 2 x 10-5 N/mm2, µ = 0.3. What is the approximate elongation of the bar in the longitudinal direction? (a) 0.44 mm (b) 0.88 mm
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(c) 0.22 mm (d) 1 mm 39. A square steel bar of 50 mm side and 5 m long is subjected to a load whereupon it absorbs strain energy of 100 J. What is its modulus of resilience? (a) 1/125 Nmm/mm3 (b) 125 mm3/Nmm (c) 1/100 Nmm/mm3 (d) 100 mm3/Nmm 40. A mild steel rod tapers uniformly from 30 mm diameter to 12 mm diameter in a length of 300 mm. The rod is subjected to an axial load of 12 kN. E = 2 x 105 N/mm2. What is the extension of the rod in mm? (a) 4π/5 (b) 2/5π (c) π/5 (d) 1/5π 41. If all the dimensions of a prismatic bar are doubled, then the maximum stress produced in it under its own weight will (a) decrease (b) remains unchanged (c) increase to 2 times (d) increase to 4 times 42. The elongation of a conical bar under its own weight is equal to (a) that of a prismatic bar of same length (b) half that of a prismatic bar of same length (c) one third that of a prismatic bar of same length (d) none of these 43. Two bars of different materials are of the same size and are subjected to same tensile forces. If the bars have unit elongations in the ratio of 4 : 7, then the ratio of moduli of elasticity of the 2 materials is (a) 7 : 4 (b) 4 : 7 (c) 4 : 17 (d) 16 : 49 44. If a composite bar of steel and copper is heated, then the copper bar will be under (a) tension (b) compression (c) shear (d) torsion
UNIT 2 DETERMINACY AND INDETERMINACY
45. The simply supported beam shown in the figure
(a) Determinate and stable (b) Determinate and unstable (c) Indeterminate and stable (d) Indeterminate and unstable 46. The degree of indeterminacy of the beam given below is
(a) 0 (b) 1 (c) 2 (d) 3 47. Which one of the following is true of a statically determinate beam? (a) One end is fixed and the other end is simply supported. (b) Both the ends are fixed. (c) The beam overhangs over two supports. (d) The beam is supported on three supports. 48. A prismatic beam is shown in the figure given below. Consider the following statements.
1. The structure is unstable 2. The bending moment is zero at supports and internal hinge 3. It is mechanism 4. It is statically indeterminate Which of the statements given above is/are correct? (a) 1, 2, 3 and 4 (b) 1, 2 and 3 (c) 1 and 2 (d) 3 and 4
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49. Which one of the following is correct? When a load is applied to a structure with rigid joints (a) There is no rotation or displacement of joint (b) There is no rotation of joint (c) There is no displacement of joint (d) There can be rotation and displacement of joint but the angle between the members connected to the joints remains same even after application of the load 50. Which one of the following is correct? A determinate structure (a) cannot be analyzed without the correct knowledge of modulus of elasticity (b) must necessarily have roller support at one of its ends (c) requires only statical equllibrium equations for its analysis (d) will have zero deflection at its ends 51. Which one of the following is correct? A statically indeterminate structure is the one which (a) cannot be analyzed at all (b) can be analyzed at all (c) can be analyzed using equations of statics and compatibility equations (d) can be analyzed using equation of compatibility only 52. Which one of the following is correct? A suspension bridge with a two-hinged stiffening girder is (a) Statically determinate (b) Indeterminate of one degree (c) Indeterminate of two degrees (d) A mechanism 53. What is the total degree of indeterminacy in the continuous prismatic beam shown in the figure?
(a) 1 (b) 2 (c) 3 (d) 4
54. The number of independent equation to be satisfied for static Equilibrium in a space structure is (a) 3 (b) 6 (c) 4 (d) 2
UNIT 3 SHEAR FORCE & BENDING MOMENT
55. Maximum bending moment in a beam occurs where (a) deflection is 0 (b) shear force is maximum (c) shear force is minimum (d) shear force changes sign 56. If the SFD of a simply supported beam is parabolic, then the load on the beam is (a) uniformly distributed load (b) concentrated load at mid span (c) external moment acting at mid span (d) linearly varying distributed load 57. Shear span is defined as the zone where (a) bending moment is zero (b) shear force is zero (c) shear force is constant (d) bending moment is constant 58. Rate of change of bending moment is equal to (a) shear force (b) deflection (c) slope (d) rate of loading 59. The diagram showing variation of axial load along the span is called (a) shear force diagram (b) bending moment diagram (c) thrust diagram (d) influence line diagram 60. Which of the following loads should be applied on a simply supported beam, so that the shear force is constant throughout its span? (a) UDL over the entire span
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(b) two concentrated loads equally spaced in the span (c) a central concentrated load and a UDL over the entire span (d) a couple anywhere in the span 61. Consider the following statements: A cantilever beam of length ‘l’ is loaded with uniformly distributed load ‘w’ on the span. The beam is propped at the free end having its level same as that of the fixed end (EI is constant). (a) The prop reaction is wl/2 (b) The prop reaction is 3/8 wl (c) The bending moment at the prop = wl2/2 (d) SF is zero at x = 3l/8 from the propped end. Which of these statements are correct? (a) 1 and 3 (b) 2 and 3 (c) 1 and 4 (d) 2 and 4 62. Neglecting self weight, which of the following beams will have points of contraflexure? (a) A simply supported beam with uniformly distributed load over part of the structure. (b) An overhanging beam with loading only over supported span and not on overhangs. (c) Fixed beam subjected to concentrated load. (d) Cantilever beam subjected to uniformly varying load with zero load at free end. 63. A portion of a beam between two sections is said to be in pure bending when there is (a) constant bending moment and zero shear force. (b) constant shear force and zero bending moment. (c) constant bending moment and constant shear force. (d) none of the above 64. The shear force diagram of a beam is shown in the figure. The absolute maximum bending moment in the beam is
(a) 2P x a (b) 5P x a (c) 4P x a (d) 7P x a 65. In which of the following the point of contraflexure will not occur? (a) A two span continuous beam of equal spans, simply supported and loaded by UDL over both spans. (b) A simply supported beam loaded by UDL. (c) A fixed beam loaded by UDL. (d) A propped cantilever loaded by UDL. 66. A prismatic beam fixed at both ends carries a uniformly distributed load. The ratio of bending moment at the supports to the bending moment at mid span is (a) 0.5 (b) 1.0 (c) 1.5 (d) 2.0 67. What are the support reactions at the fixed end of the cantilever beam shown in the figure?
(a) 120 kN , 120 KN-m (b) 120 kN, 240 kN-m (c) 240 kN, 120 kN-m (d) 120 kN, 60 kN-m 68. For S.F to be uniform throughout the span of simply supported beam, which of the following loads should be applied on to the beam? (a) Two equally spaced concentric loads (b) A couple at mid span only (c) A couple anywhere in the span (d) UDL over the entire span 69. Consider the simply supported beam AB subjected to the point loads of equal magnitude as shown in the figure. The portion CD of the beam is
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(a) in pure bending (b) in pure shear (c) having maximum bending moment (d) having maximum shear force 70. The beam shown in the figure given below is subjected to concentrated load and clockwise couple. What is the vertical reaction at A?
(a) 10 kN (b) 40 kN (c) 50 kN (d) 30 kN 71. Couple M is applied at C on a simply supported beam AB. What is the maximum shear force for the beam?
(a) zero (b) M (c) 2 M/3 (d) M/3 72. A cantilever beam AB carries loadings as shown in figure below. Which one of the following is the SFD for the beam?
73. For the simply supported beam shown in the figure given below, at what distance from the support A, is the shear force zero?
(a) L/4 (b) L/3 (c) L/2 (d) L/√3 74. A uniform beam of span l carries a UDL of w per unit length as shown in the figure given below. The supports are at a distance of x from either end. What is the condition for the maximum bending moment in the beam to be as small as possible?
(a) x = 0.107 l (b) x = 0.207 l (c) x = 0.237 l (d) x = 0.25 l 75. At what distance from left support of the given beam, is the shear force zero?
(a) 1 m (b) 1.25 m (c) 1.5 m (d) 2.5 m 76. A beam of uniform flexural rigidity supports a set of loads as shown in figure below. What is the value of W if the magnitude of bending moment at mid span and at support of the beam is numerically equal?
(a) 20 kN (b) 40 kN (c) 60 kN (d) 80 kN 77. A simply supported beam is loaded as shown in figure. The bending moment at C is
(a) 4 kN-m (sagging) (b) 4 kN-m (hogging) (c) 8 kN-m (sagging) (d) zero
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78. A simply supported beam AB is subjected to a concentrated load at C, the centre of the span. The area of the SFD from A to C will give (a) BM at C (b) difference between BM values at A and C (c) SF at C (d) both (a) and (b) 79.
A freely supported beam AB of span 4 m is subjected to a UDL of 1 kN/m over the full span and a moment of 2 kN-m at support A as shown in the figure. The resulting BM at mid span C of the beam will be (a) 1 kN-m (sagging) (b) 1 kN-m (hogging) (c) 2kN-m (sagging) (d) 2kN-m (hogging) 80. The BMD of a beam is shown in the figure
The SFD of the beam is represented by
81. In the analysis of beams subjected to loads, the point with nil bending moment can be a 1. Point of contraflexure 2. Point of maximum shear force 3. Point of inflexion Which of the above statements is/are correct? (a) 1 only (b) 2 only (c) 3 only (d) 1, 2 and 3 82. A simply supported beam of span l carries over its full span a load varying linearly from zero at either end to w/m at mid span .The maximum bending moment occurs at (a) quarter points and is equal to wl2/8 (b) quarter points and is equal to wl2/12 (c) midspan and is equal to wl2/8 (d) midspan and is equal to wl2/12 83. The variation of the bending moment in the portion of a beam carrying linearly varying load is (a) linear (b) parabolic (c) cubic (d) constant 84. If a cantilever beam carries a uniformly distributed load over its entire length, then shapes of shear force diagram and bending moment diagram respectively are (a) quadratic parabola and cubic parabola (b) triangle and quadratic parabola
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(c) rectangle and triangle (d) quadratic parabola and triangle
UNIT 4 BENDING STRESSES IN BEAMS
85. Match List 1 with List 2 and select the correct answer:
List-I List-II
A. Assumption in theory of simple bending. B. The point at which the bending stress is maximum for any cross section. C. The point at which the bending stress is zero for any cross section. D. The point in the cross section through which the NA passes.
1. Neutral axis 2. Centroid 3. The plane sections remain plane. 4. Extreme fibre 5. The cross section is circular.
Codes: A B C D (a) 5 4 1 2 (b) 3 1 2 4 (c) 5 1 2 4 (d) 3 4 1 2 86. The relationship between the radius of curvature R, bending moment M and flexural rigidity EI is given by (a) R = M/EI (b) M = EI/R (c) EI = R/M (d) E = MI/R 87. Of the several prismatic beams of equal lengths, the strongest in flexure is the one having maximum (a) moment of inertia (b) section modulus (c) tensile strength (d) area of cross section 88. Two beams, one of circular cross section and other of square cross section, have equal areas of cross section. If subjected to bending (a) circular section is more economical (b) square section is more economical
(c) both sections are equally strong (d) both sections are equally stiff 89. A prismatic bar when subjected to pure bending assumes the shape of (a) catenary (b) cubic parabola (c) quadratic parabola (d) arc of a circle 90. Two beams carrying identical loads, simply supported are having same depth but beam A has double the width as compared to that of beam B. The ratio of the strength of beam A to that of beam B is (a) 1/2 (b) 1/4 (c) 2 (d) 4 91. A steel wire of 20 mm diameter is bent into a circular shape of 10 m radius. If E, the modulus of elasticity is 2 x 106 kg/cm2, then the maximum stress induced in the wire is (a) 103 kg/cm2 (b) 2 x 103 kg/cm2 (c) 4 x 103 kg/cm2 (d) 6 x 103 kg/cm2 92. A beam has the same section throughout its length with I = 1 x 108 mm4.It is subjected to uniform B.M = 40 kN-m. E= 2 x 105 N/mm2. What is the radius of curvature of the circle into which the beam will bend in the form of an arc of a circle? (a) 1000 m (b) 500 m (c) 400 m (d) 350 m 93. A beam of symmetrical I section, made of structural steel has an overall depth of 300 mm. If the flange stresses developed at the top and bottom of the beam are 1200 kg/cm2 and 300 kg/cm2 respectively, then the depth of N.A from the top of beam would be (a) 250 mm (b) 240 mm (c) 200 mm (d) 180 mm 94. A structural steel beam has an unsymmetrical I- cross section. The overall depth of the beam is 200 mm. The flange stresses at the top and bottom are 120 N/mm2 and 80 N/mm2 respectively. The depth of the neutral axis from the top of the beam will be (a) 120 mm (b) 100 mm (c) 80 mm (d) 60 mm
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95. A thin steel ruler having its cross-section of 0.0625 cm x 2.5 cm is bent by couples applied at its ends so that its length l equals to 25 cm, when bent, as a circular arc, subtends a central angle θ = 600. Take E = 2 x 106 kg/cm2. The maximum stress induced in the ruler is (a) 2618 kg/cm2 (b) 2512 kg/cm2 (c) 2406 kg/cm2 (d) 2301 kg/cm2 96. A steel plate is bent into a circular arc of radius 10 m. if the plate section be 120 mm wide and 20 mm thick , with E = 2 x 105 N/mm2 , then the maximum bending stress induced is (a) 210 N/mm2 (b) 205 N/mm2 (c) 200 N/mm2 (d) 195 N/mm2 97. A simply supported beam of T-section is subjected to a uniformly distributed load acting vertical downward. Its N.A is located at 25 mm from the top of the flange and the total depth of the section is 100 mm. The ratio of maximum tensile stress to maximum compressive stress in the beam is (a) 2 (b) 2.5 (c) 3 (d) 4 98. A cantilever beam of T cross section carries uniformly distributed load, where does the maximum magnitude of the bending stress occur? (a) at the top of cross section (b) at the junction of flange and web (c) at the mid depth point (d) at the bottom of the section 99. A structural beam subjected to sagging bending has a cross section which is an unsymmetrical I-section. The overall depth of the beam is 300 mm. The flange stresses in the beam are: σtop =200 N/mm2 ;σbottom = 50 N/mm2 What is the height in mm of the neutral axis above the bottom flange? (a) 240 mm (b) 60 mm (c) 180 mm (d) 120 mm 100. In a simply supported wooden beam under uniformly distributed load, a hole has to
be made in the direction of width at midspan to provide a pipeline. From structural strength point of view, it would be advisable to have the hole made at (a) the bottom (b) the top (c) mid-depth (d) 1/4 depth either from the top or the bottom 101. A square section as shown in the figure above is subjected to bending moment M. What is the maximum bending stress?
(a) σbc = σbc= 12 M/h3 (b) σbc = σbc = 6 M/h3 (c) σbc = σbc = 9 M/2h3 (d) σbc = σbc = 9 M/h3 102. A square beam laid flat is then rotated in such a way that one of its diagonal becomes horizontal. How is its moment capacity affected? (a) increases by 41.4% (b) increases by 29.27% (c) decreases by 29.27% (d) decreases by 41.4% 103. A steel beam is replaced by a corresponding aluminium beam of same cross-sectional shape and dimensions, and is subjected to same loading. The maximum bending stress will (a) be unaltered (b) increase (c) decrease (d) vary in proportion to their modulus of elasticity 104. A rectangular beam of dimensions b x d is to be cut from a circular log of wood of diameter D. For the beam to be strongest in bending, the dimensions will be
(a) D√2
and D�23 (b) D
√3 and D�2
3
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(c) D√2
and D�2D3
(d) D√3
and �2D3
105. A beam of rectangular cross-section is 100 mm wide and 200 mm deep. If the section is subjected to a shear force of 20 kN, then the maximum shear stress in the section is (a) 1 N/mm2 (b) 1.125 N/mm2 (c) 1.33 N/mm2 (d) 1.5 N/mm2 106. The portion, which should be removed from top and bottom of a circular cross section of diameter d in order to obtain maximum section modulus is (a) 0.01 d (b) 0.1 d (c) 0.011 d (d) 0.11 d
UNIT 5 SHEAR STRESSES IN BEAMS
107. The state of pure shear is produced by (a) tension in one direction and equal compression in perpendicular direction (b) equal tension in two directions at right angles (c) equal compression in two directions at right angles (d) none of the above 108. A symmetrical I section is subjected to shear force. The shear force induced across the section is maximum at which location? (a) extreme fibre (b) at the bottom of flanges in flange (c) at the bottom of flange in web portion (d) at the neutral axis 109. What is the ratio of maximum shear stress to average shear stress for a circular section? (a) 2 (b) 3/2 (c) 4/3 (d) 3/4
110. The shear stress distribution shown in the figure represents a beam with cross-section
111. A beam of triangular cross section is placed with its base horizontal .The maximum shear stress intensity in the section will be (a) at the neutral axis (b) at the base (c) above the neutral axis (d) below the neutral axis 112. Consider the following statements: When a beam of square cross section is used with a diagonal in a vertical position 1. The shear stress distribution across the section of the beam will be maximum at neutral axis. 2. The shear stress distribution across the section of the beam will be zero at both top and bottom. 3. The maximum stress does not occur at neutral axis. Which of these statements are correct? (a) 1, 2 and 3 (b) 1 and 3 (c) 2 and 3 (d) 1 and 2 113. Which one of the following shear stress distribution diagrams is correct for cross bar section shown in the given figure?
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114. A rectangular beam of width 100 mm is subjected to a maximum shear force of 60 kN. The corresponding maximum shear stress in the cross section is 4 N/mm2. The depth of the beam should be (a) 150 mm (b) 225 mm (c) 200 mm (d) 100 mm 115. A beam has a triangular cross section having base 40 mm and altitude 60 mm. If this section is subjected to a shear force of 36000 N, the maximum shear stress in the cross section would be (a) 60 N/mm2 (b) 36 n/mm2 (c) 45 N/mm2 (d) 30 N/mm2 116. A rectangular beam of width 200 mm and depth 300 mm is subjected to a shear force of 200 kN. The maximum shear stress produced in the beam is (a) 10 MPa (b) 7.5 MPa (c) 5 MPa (d) 3.33 MPa 117. The shear stress distribution for a section under the action of shear force S is shown below. The rectangular section is b x d. Select the correct shear stress distribution from the following:
118. A timber beam is 100 mm wide and 150 mm deep. The beam is simply supported and carries a central concentrated load W. If the maximum stress in shear is 2 N/mm2, what would be the corresponding load W on the beam? (a) 20 kN (b) 30 kN (c) 40 kN (d) 25 kN 119. A rectangular beam of width 100 mm is subjected to a maximum shear force of 60 kN. The corresponding maximum shear stress in the cross section is 4 N/mm2. The depth of the beam should be (a) 200 mm (b) 150 mm (c) 100 mm (d) 225 mm 120. A simply supported beam of length 4 m is subjected to a uniformly distributed load of 2 kN/m. What is the maximum shear stress if the cross section is rectangular, 100 mm wide and 200 mm deep? (a) 0.2 N/mm2 (b) 0.1 N/mm2 (c) 0.4 N/mm2 (d) 0.3 N/mm2 121. A beam of square cross section with side 100 mm is placed with one diagonal vertical. If the shear force acting on the section is 10 kN, the maximum shear stress is (a) 1 N/mm2 (b) 1.125 N/mm2 (c) 2 N/mm2 (d) 2.25 mm2
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122. Two planks each of 50 mm x 50 mm section are glued together along the length to form a section 50 mm x 100 mm, and used as a beam. If the shear force at a section is 1000 N, what is the maximum shear stress on the glue? (a) 0.15 MPa (b) 0.3 MPa (c) 0.6 MPa (d) 2.4 MPa
UNIT 6 PRINCIPLE STRESSES
123. The sum of normal stresses is (a) constant (b) variable (c) depends on the planes (d) none of the above 124. Which of the following statements are correct for stresses acting on mutually perpendicular faces of a plane element? 1. The sum of the normal stresses in mutually perpendicular planes is equal to the sum of principle stresses. 2. The shearing stresses in two mutually perpendicular planes are equal in magnitude and direction. 3. Maximum shear stress is half of the difference between principle stresses. (a) 1, 2 and 3 only (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only 125. If a body carries two unlike principle stress, what is the maximum shear stress? (a) half the difference of magnitude of the principle stresses (b) half the sum of magnitude of the principle stresses (c) difference of the magnitude of principle stresses (d) sum of the magnitude of principle stresses 126. The figure shows the stress condition of an element. The principle stresses are
(a) ± 2τ (b) ± τ/2 (c) ± τ (d) ± 2τ/3 127. If the principle stresses at a point in a stressed body are 150 kN/m2 tensile and 50 kN/m2 compressive, then maximum shear stress at the point will be (a) 100 kN/m2 (b) 150 kN/m2 (c) 200 kN/m2 (d) 250 kN/m2 128. For the plane stress situation shown in the diagram, what is the maximum shear stress?
(a) Zero, when X and Y axis are rotated 45° clockwise (b) Zero, at all positions of orientation of X and Y axes (c) 20 MPa, at all positions of orientation of X and Y axes (d) – 20 MPa, when X and Y axes are rotated 45° anti-clockwise. 129. A thin rod of 10 mm diameter is subjected to a tensile force of 7580 N. What are the principle stresses and maximum shear stress? (a) 70 MPa, 50 MPa, 10 MPa (b) 100 MPa, Zero, 50 MPa (c) 100 MPa, 50 MPa, 25 MPa (d) 100 MPa, Zero, Zero
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130.
The principle stresses in N/mm2 on a rectangular element as shown in the above figure. The intensity of normal stress 𝜎𝜎Rn on the oblique plane BE is (a) 125 N/mm2 (b) 275 N/mm2 (c) 375 N/mm2 (d) 250 N/mm2 131. In a strained material, the principle stresses in the x and y directions are 100 N/mm2 (tensile) and 60 N/mm2 (compressive). On an inclined plane, the normal to which makes an angle 30° to the x-axis, what is the tangential stress in N/mm2? (a) 30√3 N/mm2 (b) 40√3 N/mm2 (c) 60 N/mm2 (d) 40 N/mm2 132. At a point in the web of a girder, the bending and the shearing stresses are 90 N/mm2 (tensile) and 45 N/mm2 respectively. The principle stresses are (a) 108.64 N/mm2 (tensile) and 18.64 N/mm2 (compressive) (b) 107.60 N/mm2 (compressive) and 18.64 N/mm2 (tensile) (c) 108.64 N/mm2 (compressive) and 18.64 N/mm2 (tensile) (d) 0.64 N/mm2 (tensile) and 0.78 N/mm2 (compressive) 133. A mild steel bar is subjected to an axial force P, resulting in an axial stress σx= 100 N/mm2. What would be normal stress σn on a plane n-n making an angle θ = 45o with its axis? (a) 25 N/mm2 (b) 40 N/mm2 (c) 50 N/mm2 (d) 100 N/mm2 134. The state of stress at a point in a 2-D stress system is characterized by direct stresses
of 40 MPa compressive and 80 MPa tensile, on mutually perpendicular planes. Shear stress is absent on these planes. The maximum shear stress at this point (along a duly identified plane) is (a) 20 MPa (b) 40 MPa (c) 60 MPa (d) 80 MPa 135. The radius of Mohr’s circle for two equal unlike principle stresses of magnitude p is (a) p (b) p/2 (c) zero (d) none of the above 136. Consider the following statements: Mohr’s Circle is used to determine the stress on an oblique section of a body subjected to 1. Direct tensile stress on one plane and accompanied by a shear stress 2. Direct tensile stresses in two mutually perpendicular directions accompanied by a simple shear stress 3. Direct tensile stress in two mutually perpendicular directions 4. A simple shear stress Select the correct answer using the codes given below: (a) 1 and 4 (b) 2 and 3 (c) 1, 2, 3 and 4 (d) 1, 2 and 3 137. The radius of Mohr’s circle is zero when the state of stress is such that (a) shear stress is zero (b) there is pure shear (c) there is no shear stress but identical direct stresses (d) there is no shear stress but equal direct stresses, opposite in nature, in two mutually perpendicular directions 138. Mohr’s stress circle helps in determining which of the following? 1. Normal stresses on one plane 2. Normal and tangential stresses on two planes 3. Principle stresses in all three directions 4. Inclination of principle planes Select the correct answer using the code given below:
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(a) 1 and 2 only (b) 2 and 3 only (c) 3 and 4 only (d) 2 and 4 only 139. What is the diameter of Mohr’s circle of stress for the state of equal stress shown below?
(a) 20 (b) 10√2 (c) 10 (d) zero
UNIT 7 DEFLECTION OF BEAMS
140. Which of the following is/are determined at a point of a given beam by moment area method? 1. Shear force 2. Bending moment 3. Slope 4. Deflection Select the correct answer using the codes given below: (a) 1 and 2 (b) 3 alone (c) 4 alone (d) 3 and 4 141. A mild steel bar of uniform cross section ‘A’ and length L is subjected to an axial load ‘W’. The strain energy stored in the bar would be (a) WL/2AE (b) W2L/4AE (c) WL/4AE (d) W2L/2AE 142. A simply supported beam of span ‘L’ is subjected to a concentrated load W at midspan. The strain energy due to bending in the beam would be (a) W2L3/48EI (b) W2L3/96EI (c) W2L3/24EI (d) WL3/96EI 143. Slope at the end of a simply supported beam of span 1 with uniformly distributed load w/m over the entire span is given by (a) WL2/16EI (b) WL3/16EI (c) WL3/24EI (d) WL2/24EI
144. A simply supported beam ‘A’ carries a point load at its midspan. Another identical beam ‘B’ carries the same magnitude of load but is uniformly distributed over the entire span. The ratio of the maximum deflections of beams ‘A’ and ‘B’ will be (a) 8/3 (b) 2/3 (c) 3/5 (d) 8/5 145. The maximum deflection of simply supported beam occurs at zero (a) bending moment location (b) shear force location (c) slope location (d) shear force location and also zero bending moment location 146. Moments of the same sense are applied to both the ends of a simply supported beam. The ratio of the rotation of the two ends is 2. What is the ratio of the applied moments? (a) 3/2 (b) 4/3 (c) 5/4 (d) 6/5 147. Simply supported beam AB of span 4 m is subjected to terminal couples as shown in the figure. If EI is in kN/m2, what is the magnitude of the central deflection of the beam in metres?
(a) 4/EI (b) 8/EI (c) 2/EI (d) 16/EI 148. If in a fixed beam of length L, concentrated load of P acts at the centre of the beam, what is the slope below the load (a) PL2/48EI (b) PL3/48EI (c) PL/24EI (d) zero 149. If a fixed beam of length L is loaded by a UDL of w kN/m over its full span, find the maximum deflection (a) wL3/96EI (b) wL4/384EI (c) wL3/384EI (d) 0
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150. If the length of a simply supported beam carrying a concentrated load at the centre is doubled, the deflection at the centre will become (a) two times (b) four times (c) eight times (d) sixteen times 151. A simply supported beam with rectangular cross section is subjected to central concentrated load. If the width and depth of the beam are doubled, then the deflection at the centre of the beam will be reduced by (a) 50% (b) 25% (c) 12.5% (d) 6.25% 152. A straight cantilever of uniform cross section carries a load ‘W’ distributed evenly over its entire length. If the free end of the cantilever is now propped upto the level of the fixed end, the vertical force required at the prop is (a) 3/8 W (b) 5/8 W (c) 3/4 W (d) W 153. A propped cantilever of span 4 m is fixed at A and propped at B. The beam carries a u.d.l. of 1 t/m over the entire span. The reaction at B is (a) 2.5 t (b) 2 t (c) t (d) 1.5 t 154. If the hinged end of a propped cantilever of span L settles by an amount 𝛿𝛿, then the rotation of the hinged end will be (a) δ/L (b) 2δ/L (c) 3δ/2L (d) 4δ/3L 155. If a cantilever beam of span L and flexural rigidity EI carries a moment M at the free end, the deflection at that end is (a) ML/24EI (b) ML2/12EI (c) ML/6EI (d) ML2/2EI 156. If the deflection at the free end of a uniformly loaded cantilever beam is 15 mm and the slope of the deflection curve at the free end is 0.02 radian, then the length of the beam is (a) 0.8 m (b) 1.0 m (c) 1.2 m (d) 1.5 m
157. If the deflection at the free end of a uniformly loaded cantilever beam of length 1 m is equal to 7.5 mm, then the slope at the free end is (a) 0.01 radian (b) 0.015 radian (c) 0.02 radian (d) none of the above 158. A cantilever beam carries a uniformly distributed load from fixed end to the centre of the beam in the first case and a uniformly distributed load of same intensity from center of the beam to the free end in the second case. The ratio of deflections at the free end in the two cases is (a) 1/2 (b) 3/11 (c) 5/24 (d) 7/41
UNIT 8 COLUMNS & MOMENT OF INERTIA
159. When both ends of a column are fixed, the crippling load is P. If one end of the column is made free, the value of crippling load will be changed to (a) P/16 (b) P/4 (c) P/2 (d) 4P 160. Buckling load for a given column depends upon (a) length of column only (b) least lateral dimension only (c) both length and least lateral dimension (d) none of the above 161. Euler’s formula for a mild steel long column hinged at both ends is not valid for slenderness ratio (a) greater than 80 (b) less than 80 (c) greater than 180 (d) greater than 120 162. Match List 1 (end conditions of columns) with List 2(effective length, le) and select the correct answer
List-I List-II A. Both ends fixed B. Both ends hinged
1. Le = 2l 2. Le = l/2
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C. One end fixed, other free D. One end fixed, other hinged
3.Le = l/√2 4. Le = l
Codes: A B C D (a) 2 4 1 3 (b) 3 1 2 4 (c) 2 4 3 1 (d) 3 1 4 2 163. For a solid circular section of diameter d, the stress in a column will be compressive only if the eccentricity of the line of action of the compression force is within (a) d/4 (b) d/8 (c) d/6 (d) d/16 164. If the Euler load for a steel column is 1000 kN and crushing load is 1500 kN, the Rankine load is equal to (a) 2500 kN (b) 1500 kN (c) 1000 kN (d) 600 kN 165. A long column has maximum crippling load when its (a) both ends are hinged (b) both ends are fixed (c) one end is fixed and other is hinged (d) one end is fixed and other end is free 166. Effective length of a chimney of 20 m height is taken as (a) 10 m (b) 20 m (c) 28.28 m (d) 40 m 167. Slenderness ratio of a 5 m long column hinged at both ends and having a circular cross-section with a diameter 160 mm is (a) 31.25 (b) 62.5 (c) 100 (d) 125
MOMENT OF INERTIA
168. The polar moment of inertia of the cross section of the member is required to assess the strength of the member in (a) bending (b) torsion
(c) axial force (d) shear 169. The polar modulus of a circular shaft of diameter d is (a) πd3/16 (b) πd3/32 (c) πd3/64 (d) πd2/32 170. What is the polar modulus of a solid circular metal shaft of diameter 8 cm? (a) 64 π cm3 (b) 32 π cm3 (c) 16 π cm3 (d) 8 π cm3
UNIT 9 TORSION IN CIRCULAR SHAFTS
171. In a circular shaft of diameter d, subjected to a torque T, the maximum shear stress induced is (a) proportional to d3 (b) proportional to d4 (c) inversely proportional to d3 (d) inversely proportional to d4 172. If a shaft of diameter d is subjected to a torque T, the maximum shear stress is (a) 32T/πd3 (b) 16T/πd2 (c) 16T/πd3 (d) 64T/πd4
173. A solid shaft has diameter 80 mm. It is subjected to a torque of 4 kNm. The maximum shear stress induced in the shaft would be (a) 75/π N/mm2 (b) 250/π N/mm2 (c) 125/π N/mm2 (d) 150/π N/mm2 174. A hollow shaft of 16 mm outside diameter and 12 mm inside diameter is subjected to a torque of 40 N-m. The shear stresses at the outside and inside of the material of the shaft are respectively (a) 62.75 N/mm2 and 50 N/mm2 (b) 72.75 N/mm2 and 54.54 N/mm2 (b) 79.75 N/mm2 and 59.54 N/mm2 (b) 80 N/mm2 and 40 N/mm2 175. A 40 mm diameter shaft is subjected to a twisting moment Mt. If max. shear stress
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developed in shaft is 5 N/mm2, what is the value of the twisting moment ? (a) 628.8 Nm (b) 328.4 Nm (c) 62.8 Nm (d) 30.4 Nm 176. Which of the following terms represents the torque corresponding to a twist of one radian in a shaft over its unit length? (a) torsional stress (b) torsional rigidity (c) flexural rigidity (d) moment of resistance 177. Strain energy in torsion of a shaft per unit volume is given by (q is shear stress, E- modulus of Elasticity and G is modulus of rigidity) (a) q2/2G (b) q2/2E (c) q2/4G (d) q2/4E 178. The ratio of the torsional moments of resistance of a solid circular shaft of diameter D, and a hollow circular shaft having external diameter D and internal diameter d is given by
(a) D4
D4− d4 (b) D4 − d4
D4
(c) D3− d3
D3 (d) D3
D3− d3
179. The failure surface of a standard cast iron torsion specimen, subjected to a torque is along (a) the surface helicoidal at 450 to the axis of the specimen (b) the curved surface at the grips (c) the plane surface perpendicular to the axis of the specimen (d) the curved surface perpendicular to the axis of the specimen. 180. A hollow shaft will transmit ………………. Power than a solid shaft of same weight and material (a) less (b) same (c) more (d) none of the above 181. Two shafts one of solid section and the other of hollow section, of same material and weight having same length are subjected to equal torsional force. What is the torsional stiffness of hollow shaft? (a) equal to that of the solid shaft
(b) less than that of the solid shaft (c) more than that of the solid shaft (d) exactly half of that of the solid shaft 182. Torsion applied to a circular shaft results in a twist of 10 over a length of 1 m. The maximum shear stress induced is 120 N/mm2 and the modulus of rigidity of the shaft material is 0.8 x 105 N/mm2. What is the radius of the shaft? (a) 300/π (b) 180/π (c) 90/π (d) 270/π 183. A solid shaft rotating at 180 rpm is subjected to a mean torque of 5000 Nm. What is the power transmitted by the shaft in kW? (a) 25 π (b) 20 π (c) 60 π (d) 30 π 184. If a shaft is turning at N r.p.m and the mean torque to which the shaft is subjected is T N-m, the power transmitted by the shaft in kW would be
(a) 2πNT45000
(b) 2πNT60000
(c) 2πNT30000
(d) 2πNT33000
185. What is the power transmitted by a 100 mm diameter solid shaft at 150 rpm without exceeding a maximum stress of 60 N/mm2? Take π2 =10 (a) 187.5 kW (b) 18.75 kW (c) 1.875 kW (d) 1875 kW
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ANSWER KEY 1. (c) 21. (a) 41. (c) 61. (d) 81. (a)
2. (a) 22. (c) 42. (c) 62. (c) 82. (d)
3. (c) 23. (c) 43. (a) 63. (a) 83. (c)
4. (b) 24. (b) 44. (b) 64. (b) 84. (b)
5. (b) 25. (c) 45. (b) 65. (b) 85. (d)
6. (a) 26. (c) 46. (b) 66. (d) 86. (b)
7. (c) 27. (d) 47. (c) 67. (a) 87. (b)
8. (b) 28. (a) 48. (b) 68. (c) 88. (b)
9. (b) 29. (a) 49. (d) 69. (a) 89. (c)
10. (b) 30. (b) 50. (c) 70. (a) 90. (c)
11. (a) 31. (b) 51. (c) 71. (d) 91. (b)
12. (c) 32. (b) 52. (b) 72. (a) 92. (b)
13. (c) 33. (d) 53. (d) 73. (d) 93. (b)
14. (c) 34. (b) 54. (b) 74. (b) 94. (a)
15. (b) 35. (b) 55. (d) 75. (c) 95. (a)
16. (c) 36. (d) 56. (d) 76. (b) 96. (c)
17. (a) 37. (d) 57. (c) 77. (d) 97. (c)
18. (c) 38. (d) 58. (a) 78. (d) 98. (d)
19. (a) 39. (a) 59. (c) 79. (a) 99. (b)
20. (d) 40. (d) 60. (d) 80. (a) 100. (c)
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101. (b) 121. (b) 141. (d) 161. (b) 181. (c)
102. (c) 122. (b) 142. (b) 162. (a) 182. (d)
103. (a) 123. (a) 143. (c) 163. (b) 183. (d)
104. (b) 124. (d) 144. (d) 164. (d) 184. (b)
105. (d) 125. (b) 145. (c) 165. (b) 185. (a)
106. (c) 126. (c) 146. (c) 166. (d)
107. (a) 127. (a) 147. (b) 167. (d)
108. (d) 128. (b) 148. (d) 168. (b)
109. (c) 129. (b) 149. (b) 169. (a)
110. (a) 130. (b) 150. (c) 170. (b)
111. (c) 131. (b) 151. (d) 171. (c)
112. (c) 132. (a) 152. (a) 172. (c)
113. (a) 133. (c) 153. (d) 173. (c)
114. (b) 134. (c) 154. (c) 174. (b)
115. (c) 135. (a) 155. (d) 175. (c)
116. (c) 136. (c) 156. (b) 176. (b)
117. (b) 137. (c) 157. (a) 177. (c)
118. (c) 138. (d) 158. (d) 178. (a)
119. (d) 139. (d) 159. (a) 179. (a)
120. (d) 140. (d) 160. (c) 180. (c)
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......SOLUTION...... UNIT 1 PROPERTIES OF METALS & APPLICATION OF HOOKE’S LAW
1. (c) An ideal plastic material experiences no work (non strain) hardening during plastic deformation. 2. (a)
OA = Linear curve A = Proportional limit B = Elastic limit C = Upper yield point D = Lower yield point DE = Plastic region EF = Strain hardening region FG = necking region F = Ultimate stress point G = Fracture point Thus, Point 1 corresponding to rupture strength Point 2 corresponding to ultimate strength Point 3 corresponding to yield point Point 4 corresponding to proportional limit 5. (b) Modulus of rigidity or Shear modulus (G) is the ratio of shear stress (𝜏𝜏) to shear strain (ф) G = τ
ф
6. (a) For an isotropic homogenous material the total number of elastic constants are 4 Young’s Modulus (E), Shear modulus (G), Bulk modulus (K) and Poisson’s ratio (𝜇𝜇). Out of these 2 are independent (E and 𝝁𝝁)
7. (c) When material is unloaded before elastic limit original dimension of the member is regained instantly.
8. (b) μ = − Lateral strain
Longitudinal strain
9. (b) Tensile stress results in tensile strain in longitudinal and compressive strain in lateral directions. Statement (i), (iii) and (iv) are true. 10. (b) In elastic theory of design
FOS = Yield stress
Working stress
11. (a) The value of Poisson’s ratio lies between − 1 and 0.5. Generally for all engineering materials it is more than 0. μ < 0 , for rheopetic materials (eg. Human tissues) μ = 0.286 for Mild steel μ = 0.33 for Aluminium 𝜇𝜇 = 0.5 for Perfectly incompressible materials. 12. (c)
Metal Poisson’s ratio (𝝁𝝁) Aluminium Cast iron Steel
0.334 0.21 -0.6 0.303
13. (c) Poisson’s ratio, μ = − Lateral strainLongitudinal strain
= −�−0.0045
30 �
�0.09200 �
= 13 = 0.33
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14. (c) Poisson’s ratio = − Lateral strainLongitudinal strain
=
−0.00540
0.15400
= 0.33
15. (b) E = 3k (1 − 2𝜇𝜇)
K = E
3(1−2μ)
For minimum value of K, (1-2𝜇𝜇) should be maximum for (1- 2𝜇𝜇) to b maximum, 𝜇𝜇 should be minimum, for all practical purposes, minimum value of 𝜇𝜇 = 0
Kmin = E3
16. (c) E = 3K (1-2𝜇𝜇) EK
= 3 (1 − 2 x 0.3) = 65 = 1.2
17. (a) We know E = 3K (1-2𝜇𝜇) and also E = 2C (1+ 𝜇𝜇)
⇒ 𝜇𝜇 = � 𝐸𝐸
2𝐶𝐶 � -1 putting value of 𝜇𝜇 𝑖𝑖𝑖𝑖 𝑒𝑒𝑒𝑒. (𝑖𝑖)
E =3K �1 − 2E2C
+ 2�
E = 3K �3− EC�
E�1 + 3KC� = 9K
E = 9KC
3K+C
18. (c) E = 2G (1 + 𝜇𝜇) 19. (a) Elastic modulus = E Shear modulus = G E = 2G (1 + 𝜇𝜇) Given, 𝜇𝜇 = 0.5 E = 2 x 1.5 x G = 3G 20. (d) E = 2G (1+ 𝜇𝜇) 200 = 2G (1 + 0.25) G = 80 GN/m2
21. (a)
∆𝑙𝑙 = 0.3 mm
⇒ 𝜀𝜀 = ∆𝑙𝑙
𝑙𝑙 =
0.3200
⇒ σ =
PA
= εE
⇒ 23.5 x103
π4 x102 =
0.3200
x E
⇒ E =
2350.785
x 2000.3
= 200 x 103 N/mm2 22. (c) |longitudinal strain| = 3 x |lateral strain|
Poisson’s ratio, 𝜇𝜇 = � lateral strainlongitudinal strain
�
𝜇𝜇 =13
E = 3K (1-2𝜇𝜇)
Bulk modulus, K = E
3�1−2 x 13�
K = E = 1 x 105 N/mm2 23. (c) Gradually applied load: When a load is applied in installments i.e if a load of 100 N is to be applied. First a load of 5 N then 10 N, 15 N, 20 N................. 100 N is applied. Overall it causes less stress and less strain as compared to suddenly applied load.
σGradually applied load = FA
Stress strain diagram will be linear (triangular area of stress strain will be covered). Suddenly applied load: When the total force is applied in one instalment i.e. force of 100 N is applied in one instalment, it causes 2 times the stress as compared to when the same load is applied gradually.
σsuddenly ,applied load = 2FA
(In this case stress strain curve will be a rectangle) 24. (b) Proof stress is defined for those ductile metals which do not clearly represent Yield Point like Aluminium, copper , gold, etc. In such metals Proof stress is used as design stress which is similar to yield stress. Resilience means elastic strain energy. So proof resilience means elastic strain energy.
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Note: For most of the materials Yield stress and elastic limit are very close. So for all practical purposes Yield stress is taken as elastic stress (𝝈𝝈𝒚𝒚 = 𝝈𝝈𝒆𝒆) 25. (c) Elastic Strain energy (Resilience) stored in a member is the total energy which can be stored in a member when loading is done within elastic limit and the same energy is released on loading. For linear elastic metals resilience is given by
0.5 x stress x strain x volume �12
x σ x ε x V�
or 0.5 x Load applied x elastic
deflection�12
x P x ∆�
26. (c) Shear stress (τ)= 100 N/mm2
Modulus of rigidity (G)= 1 x 105 N/mm2 Strain energy (U)= area under stress strain curve x volume of the block
Shear strain (ф) = 𝜏𝜏𝐺𝐺
= 100
1 x 105 = .001
U = 12 x τ x ф x (200 x 100 x 50)
U = 0.5 x 100 x .001 x (200 x 100 x 50) U = 50 x 103 N-mm = 50 N-m 27. (d) Strain energy stored in a beam of span L is obtained by integrated the equation:
U = ∫ M2
2EIL
0 ds
For a rectangular section, I = bd3
12
So if depth is reduced to half, U will decrease by 1/8. 29. (a) Creep is the property of material by virtue of which material continues to deform over a period of time under sustained loading. 30. (b) Resilience is the property of material to absorb strain energy when it is deformed upon unloading to have this energy recovered. Area under load-deformation curve within elastic limit is called resilience.
31. (b)
Nature of stress in a ceiling fan rod is tensile. Due to friction, bearing shear stress will also develop but its magnitude will be very less and hence neglected. 32. (b) Creep is defined as strain under sustained loading. The property by virtue of which a material undergoes additional deformation with passage of time under sustained loading within elastic limit is called creep. 33. (d) Constitutive relation is relation between two physical quantities that is specific to the material. Hence stress strain behaviour of solid material is constitutive relationship. 34. (b) Toughness is the maximum strain energy which can be stored in the metal before fracture when all the load is applied instantly . Note: 1. Toughness is desirable to prevent fracture against impact loading. 2. Greater is the fracture strain greater is the toughness. 35. (b) Due to cyclic or reverse cyclic loading fracture failure may occur if total stored stimulating strain energy exceeds the toughness. Toughness of a material starts reducing with time. Some of the examples of fatigue failure are:
1. Failure of fly wheel 2. Cracks in turbine blades 3. Breaking of metal wire under cyclic
bending
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36. (d)
Endurance limit is the stress level below which there is high probability of no failure even at infinite number of load cycles. In design factor of safety should be adopted in such a way that maximum stress level remains below endurance limit
37. (d) Volumetric strain, ∆VV
= εx + εy+ εy
= 800 x 10-6 + 400 x 10-6 + (-1200) x 10-6
= 0
38. (d) εx = σxE
- μ σy
E
∆𝑙𝑙𝑙𝑙
= 440
2 x 105 -0.3(−200)
2 x 105
∆𝑙𝑙
400 =
1400
∆𝑙𝑙 = 1 mm 39. (a) Modulus of resilience = Strain energy absorbed per unit volume upto elastic limit
= 100 J
50 x 50 x 5000
= 100 x 103
50 x 50 x 5000N−mm
mm 3
= 1
125 N-mm/mm3
Note: 1 J = 1 N- m = 1 x 103 N-mm 40. (d) Extension of tapered rod due to axial force P
∆ = PL
πd1d24
x E
= 12 x 103 x 300 x 4
π x 30 x 12 x 2 x 105
= 1
5π mm
41. (c) The deflection for prismatic bar under
its own weight (∆) = γL2
2E
Dimension in prismatic bar can be its length (L) and diameter (D) only Then ∆new = 4∆
Longitudinal strain (εl = ∆new2L
) = 2∆L
= 2εL
Stress (σ)= εL x E New stress = 2εL x E = 2σ 42. (c) The deflection of a conical bar is given by
∆conical = γL2
6E
= 13�γL2
2E� and �∆prismatic =
γL22E
43. (a) ∆1= PLAE1
& ∆2= PLAE2
∆1∆2
=� PL
A E 1�
� PLA E 2
�
∆1∆2
= 47 =� PL
A E 1�
� PLA E 2 �
∴ E1E2
= 74
44. (b) αsteel = 12 x 10−6 /°C αcopper = 16 x 10−6 /°C ∴ αsteel < αcopper The composite bar will behave as a single material, so its steel portion will be in tension and copper portion will be in compression.
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UNIT 2 DETERMINACY AND INDETERMINACY OF BEAMS 45. (b)
Ds = No. of unknowns – no. of equilibrium equation available
= 2 – 3 = - 1 So the structure is determinate and unstable. 46. (b) For static indeterminacy of beam, beam is made cantilever by adding constraint and removing all other support reactions. Ds = Support removed – constraint added
Ds = 2 – 1 = 1 47. (c)
48. (b)
Degree of indeterminacy = Number of unknown reactions – Number of equilibrium capacity = 3 – (3 + 1) Ds = – 1 Ds < 0, so unstable structure. Supports are hinge, so BM is zero at supports and internal hinge. Beam is unstable and will form a mechanism. 49. (d)
The joints can rotate as well as sway but angles between AB and AC will remain as it is. 50. (c)
This determinate structure may not have zero deflection at its ends. 51. (c) In statically indeterminate structures, in addition to equations of statics, additional compatibility equations, equal to degree of static indeterminacy are required to analyze the structure. 52. (b) A suspension bridge with a two hinged stiffening girder is statically indeterminate of one degree. A suspension bridge with a three hinged stiffening girder is statically determinate 53. (d) As the loading is shown hence, we consider determinacy from the point of view of the vertical loading only
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⇒ Ds = 4
Two vertical reactions at the roller and two reactions at end B. 54. (b) In a space structure or 3-D structure (in which members and forces are in 3 co-ordinates direction). The equations of equilibrium are ∑ Fx = 0, ∑Mx = 0 ∑ Fy = 0, ∑My = 0 ∑ Fz = 0, ∑Mz = 0 Thus there are total 6 number of equations
UNIT 3 SFD AND BMD UNIT
56. (d) dVdx
= w
If SFD is parabolic (2nd degree), then the load of the beam is linearly varying distributed load (2 -1 = 1 degree) 57. (c) Shear span is the zone in which SF is constant
58. (a) Relation between Shear force (S.Fx)
and bending moment (M) is dMdx
= S.Fx
Relation between Shear force (S.Fx) and
loading rate is – d SF xdx
= Wx
59. (c) The variation of axial load (Horizontal load) is called Horizontal thrust diagram or thrust diagram. 60. (d) For a constant shear force dVdx
= w = 0
Only couple is acting in the span.
61. (d)
• At end B
Deflection due to UDL = Deflection due to RB
wl4
8EI =
RB l3
3EI
RB= 38 wl
• End B is hinged so BM is zero • SF is zero at a distance x from end B
RB = wx
x = 38 l
62. (c) Point of contraflexure is the point where bending moment changes its sign
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63. (a) If over a span Bending Moment is constant and shear force is zero then such a span is in pure bending . Under pure bending deflected shape will be an arc of a circle. 64. (b)
dMdx
= V
∫dM =∫V dx M2 – M1 = Area of shear force diagram
∫ V dxEA = - 5P x a + 7P x a -2P x 2a + 2P x a
= 0 Assuming bending moment at A = 0 BM at point B = Area of SFD upto B = - 5 Pa BM at point C = -5 Pa + 7 Pa = 2 Pa BM at point D = 2 Pa + (-2P) x 2a = -2 Pa BM at point E = - 2 Pa + 2 Pa = 0 Therefore, Absolute maximum Bending moment in beam = 5 Pa 65. (b) Point of contreflexure is the point where bending moment changes it’s sign
67. (a)
Total load will act through the C.G of loading diagram so equivalent structure for calculating support reaction is
68. (c) dVdx
= w
Shear force is uniform throughout the span so dVdx
= 0
w = 0 Hence, no vertical loading should be there in the span. A couple anywhere in the beam will cause equal and opposite support reactions in the beam, so SFD will be rectangular or uniform, throughout the beam.
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69. (a)
In portion CD shear force is zero so it is in pure bending. 70. (a)
Taking moment about B ∑MB =0 RA x 8 + 120 – 50 x 4 = 0 RA = 10 kN upward 71. (d)
72. (a) In part CB only moment is in action so SF in part CB is zero so, only load is 2 KN will cause reaction support so SF = 2 kN 73. (d) Resultant of loading will act at the centroid of loading diagram Resultant R = Area of loading diagram 12 x L x W =
WL2
RA =
WL2
x (L 3⁄ )
L =
WL6
RB = WL
2�2L 3⁄
L� =
WL3
Let us assume shear force is zero at a distance x from end A. Therefore SFA = Area of loading diagram upto x
�V = �wdx�
SFA = 12 x × x
WxL
WL
6 =
Wx 2
2L
x = L√3
74. (b)
At support A and B maximum hogging moment will be generated and at mid span of AB, maximum sagging moment will be generated (if A and B are not very close) For maximum moment to be as small as possible hogging bending moment at A or B = Sagging bending moment at mid span of AB wx2
2 =
w(l−2x)2
8 -
wx 2
2
x2 = (l−2x)2
8
√8x = l - 2x 8x2 = l2 + 4x2 - 4xL On solving
x = l
�2+ √8�
= 0.207 l
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75. (c)
Taking moment about B ∑MB = 0 RA x 4 -15 x 2 x 3 = 0 RA = 22.5 kN Shear force is zero at a distance x from A V(x) = RA - 15x ; {0 < x < 2} 0 = 22.5 - 15x x = 1.5 m 76. (b)
At support, MA = 10 x 2 = 20 kN-m Given: MA = MC
B.M at mid span
MC = WL
4 - 20 =
W x 44
- 20
= W – 20
⇒ 20 = W – 20
W = 40 kN 77. (d)
Taking moment about B ∑MB = 0 RA + 4 + 4 – 4 – 4 x 2 = 0
RA = 4 x 2
4 = 2 kN
B.M at C, MC = RA x 2 – 4 = 2 x 2 – 4 = 0
78. (d) dMdx
= V
∫ dMCA =∫ VdxC
A
MC – MA = Area of SFD from A to C 79. (a)
Taking moment about B ∑MB = 0
RA x 4 - 2 – 1 x 42
2 = 0
RA = 2.5 kN
B.M at C, MC = RA x 2 – 2 -1 x 22
2
= 2.5 x 2 - 2 -2 = 1 kN-m (Sagging) 80. (a) Slope of B.M.D gives shear force As given B.M.D is linear therefore shear force must be constant Slope of BMD change its sign at mid-span therefore shear force changes its sign at mid-span Hence option (a) is correct. 81. (a)
82. (d)
RA + RB = 𝑤𝑤𝑙𝑙
2
Since load is equally distributed so
RA = RB = 𝑤𝑤𝑙𝑙4
B.M is maximum where S.F is zero RA −
wx2
= 0
⇒ x = L
2
B. Mcentre = RA x 𝑙𝑙2− 𝑤𝑤𝑙𝑙
4x �1
3 x 𝑙𝑙
2�
= w 𝑙𝑙2
8− w 𝑙𝑙2
24
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= w 𝑙𝑙2
12 and occurs at
𝑙𝑙2 (midspan)
83. (c)
Type of loading Variation of BMD
1.For point loads 2. For udl 3. For linearly varying load
Linear Parabolic Cubic
UNIT 4 BENDING STRESSES IN BEAMS
85. (d)
Bending stress, σ = My
I
At the extreme fibre y is maximum so bending stress is also maximum. At neutral axis, y = 0 σ = 0 Hence bending stress is zero at NA Neutral axis passes through the centroid of cross section (If Hooke’s law is valid). 86. (b) The bending equation is given by MI
= σy
= ER
∴ M = EIR
87. (b) M = σmax x z If 𝜎𝜎max for different section remains constant Then M ∝ Z, it means a strongest section in bending will have maximum value of Z ∴ Z is called flexural strength. 88. (b) Z (Isection ) > Z (rectangle) > Z (square) > Z (circle) 90. (c) Maximum bending moment resisted by a section without undergoing failure is M.R (Moment of Resistance) M.R = σper x z σper =
σy
F.O.S
The larger the value of section modulus stronger is beam
So, ratio of strength of beam A to that of beam B
= (Section modulus )A(Section modulus )B
Let b and D be width and depth respectively
= ZAZB
= 2bD 2 6⁄bD 2 6⁄
= 2
91. (b) From flexure formula MI
= fy =
ER
fy =
ER
fmax(2 2)⁄
= 2 x 106
1000�for y → ymax
f → fmax�
fmax = 2 x 103 kg /cm2 92. (b) From flexure formula
MI
= fy
= ER
MI
= ER
R = EIM
= 2 x 105 x 108
40 x 106
= 500000 mm = 500 m 93. (b)
From stress diagram, using similar triangle property
1200y
= 300
300− y
y = 240 mm 94. (a) Overall depth = 200 mm
σtop =120 N mm 2⁄ σbottom =80 N mm 2⁄
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Let the depth of neutral axis from top = y
∴f1
y1=
f2
y2
120y
= 80
200 − y
2(200 − y) = 2y
600 - 3y = 2y
Y = 600
5
= 120 mm from top 95. (a) Given Cross section of ruler = 0.0625 cm x 2.5 cm
Angle subtended = 600 Therefore l = R 25 = R x
𝜋𝜋180
x 60
⇒ R = 23.87 cm
Max. Stress in the ruler will be on the extreme end
∴ fy
= ER
f0.0625
2
= 2 x 106
23.87
⇒ f = 2618 kg/cm2
96. (c)
σy
=MI
=ER
σ = My
I= Ey
R
= 2 x 105x 20
210 x 103
= 200 N/mm2 97. (c) In simply supported beam loading is downward therefore sagging bending moment will be generated in beam. Top fibre will be in compression and bottom fibre will be in tension.
In stress diagram from similar triangle
𝜎𝜎𝑡𝑡75
= 𝜎𝜎𝑐𝑐25
⇒σtσc
= 3
98. (d)
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Maximum bending stress 𝜎𝜎 = MyI
Due to cross sectional shape neutral axis will shift towards the flange of T beam ∴ y2 > y1 For a given loading and cross section M and I are constant
⇒ 𝜎𝜎 ∝ y
Hence maximum bending stress will occur at a distance y2 from N.A i.e. at the bottom of the section. 99. (b) As strain variation remains linear so stress variation is also linear (within elastic limit) across the cross section of beam
Height of NA above bottom flange = x
∴ 200
(300− x) = 50x
20050
= 300
x− 1
x = 60 mm 100. (c) Hole should be made at the centre of beam (at neutral axis) otherwise strength of section will decrease more. 101. (b) Maximum bending stress,
σ = MyI
= M.h2�h 4
12� =
6 Mh3
102. (c) Case 1:
Ixx = Iyy = a4
12
ymax = a2
Case 2:
Ixx = Iyy =
a4
12
Ymax = a√2
So, Moment resisting capacity,
M = fy.z = fy.I
ymax
% change in moment capacity
= �M2 −M1M1
� x 100
=
fy .Ia√2
−fy .I
a2
fy .Ia2
x 100
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√2−22
x 100 = -29.28 %
(-) ve sign shows decrease in moment capacity. 103. (a) Max. bending stress σmax = Mmax ymax
1
For same cross sectional shape dimensions and loading, Mmax, ymax and I will remain same. Therefore bending stress will remain unaltered. 104. (b)
From figure, b2 + d2 =D2 M = σz ∴ For the beam to be strongest in bending section modulus should be maximum Section modulus for rectangular beam
z = bd 2
6
z= b�D2−b2�
6 =
16
(D2b − b3) dzdb
= 16
(D2 − 3b2) For z to be maximum, dzdb
= 0
0 = 16
(𝐷𝐷2 − 3𝑏𝑏2) ⇒ b =
𝐷𝐷√3
From equation (i) b2 + d2 = D2 D2
3 + d2 = D2
d2 = 23
D2
d = D�23
105. (d) τavg = VB.D
= 20 x 103
100 x 200 = 1 N/mm2
∴ τmax = 1.5 τavg (For rectangular section) τmax = 1.5 N/mm2 106. (c) A circular section can be made stronger by cutting off top and bottom corner. The maximum percentage increase in z will be 0.7 % when cut off portion (𝛿𝛿) = .011 d
UNIT 5 SHEAR STRESSES IN BEAMS
108. (d) Shear stress is maximum at neutral axis.
Variation of shear stress in I-section 109. (c) Shear stress in circular sections
τ = 43
VA�1 − y2
R2�
τ = 43τav �1 − y2
R2�
Maximum shear stress will occur at y = 0
τmax = 43τav
τmaxτavg
= 43
110. (a) Shear stress at point A is suddenly reduced. It shows sudden increase in width at point A, which is the case with option (a)
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111. (c) For triangular cross-section, the intensity of shear stress at the N.A is 4/3 of the mean intensity. The maximum shear stress occurs at the mid height. 112. (c) Shear stress distribution diagram
113. (a)
Variation of shear stress is parabolic in shape across a rectangular cross section with zero value at the top and bottom and maximum value at neutral axis. Shear stress will be zero at M and will increase parabolically upto N, at N as the width has been suddenly increased so shear stress will reduce suddenly and it will again increase upto neutral axis in parabolic shape. Due to symmetry of structure about NA, shear stress diagram will also be symmetrical about NA. 114. (b) In rectangular beam Max. Shear stress = 1.5 Avg. Shear stress
4 = 1.5 V
b x d
4 = 1.5 x 60 x 103
100 x d
d = 225 mm
115. (c) Average shear stress, τavg = VA
= 36000
12 x 40 x60
= 30 N/mm2
In triangular cross section maximum shear
stress, τmax = 32τavg
= 32
x 30 = 45 N/mm2
116. (c) In rectangular beam Maximum shear stress = 1.5 Average shear stress
= 1.5 x 200 x 103
200 x 300 = 5 MPa
117. (b) In rectangular sections shear stress variation is parabolic with zero value at top and bottom and maximum value equal to 1.5 times the average value, at neutral axis. Shear stress at a distance y from NA
τ = 6Vbd 3 �
d2
4− y2� =
V2I�d2
4− y2�
τmax = 6V
bd 3 �d2
4− 0� =
32
Vbd
= 32τavg �τavg = V
bd�
118. (c)
Maximum SF in beam,
V = W2
Max. Shear stress = 32 Avg. Shear stress
2 = 32 x �V
A�
2 = 32� W
2 x 100 x 150�
W = 40000 N W = 40 kN 119. (d) In rectangular beam
Max. Shear stress = 32 Average shear stress
4 = 32� V
bd�
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4 = 32�60 x 103
100 x d�
d = 225 mm 120. (d) Maximum shear force is at support and is equal to
Vmax = w.L
2 =
2 x 42
= 4 kN
τmax = 1.5 τavg
= 1.5 x 4 x 103
100 x 200
= 0.3 N/mm2
121. (b) The average shear stress (τavg )= Forcearea
= 10 x 103
100 x 100 = 1 N/mm2
τmax = 98
x τavg = 1.125 N/mm2
122. (b)
Shear stress at glue
= 32 Avg. Shear stress
= 1.5 x 1000
50 x 100
= 0.3 MPa
UNIT 6 PRINCIPLE STRESSES
124. (d) Shearing stresses in two mutually perpendicular planes are equal in magnitude but not same in direction. Statement 1 and 3 are correct.
125. (b) Maximum shear stress
= max �σ1−(σ2)2
, σ12
, −σ22�
= �σ1+σ22
�
where 𝜎𝜎1 and 𝜎𝜎2 are principle stresses 126. (c) Principle stress,
σmax = σ1+σ2
2 +��σ1−σ2
2�
2+ τ2
= 0 + √0 + τ2 = τ
σmin = σ1+σ2
2−��σ1−σ2
2�
2+ τ2
= 0−√0 + 𝜏𝜏2 = − 𝜏𝜏 ∴ Principle stresses are ± 𝜏𝜏 127. (a) Maximum shear stress = Radius of Mohr’s circle
=��𝜎𝜎1−𝜎𝜎22
�2
+ 𝜏𝜏2
=���150−(−50)2
�2
+ 02�
= √1002 = 100 kN/m2
128. (b) Radius of Mohr’s circle, `
R = ��𝜎𝜎1−𝜎𝜎22
�2
+ 𝜏𝜏2
= ��20−202
�2
+ 02 = 0
Radius of Mohr’s circle is zero, Mohr’s circle has been reduced to a point so at all positions of orientation of x and y axes shear stresses will be zero.
129. (b) Nominal stress, σn = PA
= 7850π x 102
4
= 99.94 Major principle stress, 𝜎𝜎1= 100 MPa Nominal stress in other two perpendicular plane = 0 ∴ Minor principle stress = 0 Maximum shear stress under uniaxial tension τmax =
σ2 = 50 MPa
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130. (b)
R = 400−(−100)
2 = 250
σn = 150 + 250 cos 600 = 150 + 125 = 275 N/mm2
131. (b)
R =
100−(−60)2
= 80
𝜏𝜏 = 80 sin2θ τ = 80 sin 60° 𝜏𝜏 = 40√3 N/mm2 132. (a) 𝜎𝜎𝑥𝑥 = 90 N/mm2 σy = 0 τxy = 45 N/mm2
⇒ σ1 2⁄ =
σx +σy
2
= ±��σx−σy
2�
2+ τxy
2
= 90+0
2±��90−0
2�
2+ 452
𝝈𝝈𝟏𝟏 = 108.63 N/mm2
𝝈𝝈𝟐𝟐 = -18.63 N/mm2
133. (c) Mohr circle will be
Normal stress on a plane inclined at 45° will be 50 N/mm2 134. (c) Mohr circle will be
Max. Shear stress = radius of Mohr circle
= 80+40
2 = 60 MPa
135. (a) The radius of Mohr’s circle is given
by (P1−P2)
2 but given (P1 = P and P2 = - P)
∴ r = �P−(−P)2
�
⇒ r = P
136. (c) Whenever we know state of stress on two planes and known angle between those planes, then with the help of Mohr’s circle we can determine the stresses on a oblique section. 137. (c) Radius of Mohr circle,
R =��𝜎𝜎1−𝜎𝜎22
�2
+ 𝜏𝜏2
For R to be zero 𝜏𝜏2 = 0
⇒ 𝜏𝜏 = 0
and �𝜎𝜎1−𝜎𝜎22
�2= 0
⇒𝜎𝜎1 = 𝜎𝜎2
138. (d)
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139. (d) As both stress are equal if we draw Mohr’s circle so it will be a single point so radius = 0 Radius of Mohr circle,
R = ��𝜎𝜎1−𝜎𝜎22
�2
+ 𝜏𝜏2
= ��10−102
�2
+ 02
R = 0 Diameter = 2R = 0
UNIT 7 DEFLECTION OF BEAMS
140. (d) Generally in moment area method
we plot MEI
Diagram and find slope and
deflection at a point. Note: However in indeterminate structures unknown bending moment can also be found out by knowing known slope and deflection. 141. (d) Strain energy stored in the bar
= 12
(P x δ)
= 12 x W x
WLAE
= W 2L2AE
142. (b)
Strain energy due to bending,
U = 2∫ M2dx2EI
L2
0 = ∫�Wx
2 �2
dx
EI
L2
0
= W 2
4EI�x3
3�
0
L 2⁄=
W 2
4L3
3 x 8 x EI
= W 2L3
96EI
144. (d)
∆Rmax = PL 3
48EI
Given: P = WL
W = �PL�
∆Rmax = 5
384WL 4
EI=
5384
PL 3
EI
(∆max )A(∆max )B
= � PL 3
48EI �
� 5384
PL 3EI �
= 85
145. (c) Let y be deflection
for max. y, dydx
= 0
⇒ θ = 0
⇒ Slope = 0
146. (c)
From principle of superposition taking effect of each moment separately At A, MA will cause rotation in clockwise direction and MB will cause rotation in anti-clockwise direction
θA = MA L3EI
− MB L6EI
Similarly, θB= MB L3EI
− MA L6EI
Given θAθB
= 2
M A3 −M B
6M B
3 −M A6
= 2
2MA−MB2MB−MA
= 2
2MA − MB = 4MB − 2MA 4MA = 5MA MA
MB=
54
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147. (b)
RA = RB =4 − 4
4= 0
Using moment area method:
At central point C, the deflection is ∆C and due to symmetry, deflection at C is maximum and slope of elastic curve is zero i.e. θC= 0 Applying moment area equation between point C & B
∆B = ∆C + θC x L + Ax�
0 = ∆c + 0 + �2 x 4EI� x 1
∆c= − 8EI
(-)ve sign shows downward deflection. 148. (d)
For the fixed beam acted upon by a concentrated load at its centre, slope is zero.
150. (c) δmax . = PL3
48EI (For SSB)
If L becomes 2L then new 𝛿𝛿 will become 8 times as δ ∝ L3
151. (d) δmax . = PL3
48EI (For SSB)
If the width and depth of the beam are doubled, then moment of inertia will change (I)
Irectangular = BD 3
12
If B becomes 2B and D becomes 2D then
Irectangular = 16 �BD 3
12 �
∴ Deflection reduces by 16 times
Change in deflection in % = � 𝛿𝛿16� x 100
= 6.25 % of 𝛿𝛿 152. (a)
W is distributed evenly over entire length
∴ udl w = WL
At propped end B, Vertical downward deflection due to udl = Upward deflection due to RB wL4
8EI=
RB L3
3EI
RB = 38
wl = 38 W
153. (d)
At end B Downward deflection due to loading = Upward deflection due to reaction RB
wl 4
8EI =
RB L3
3EI
⇒ RB =
38
wl
= 3 8
x 1 t/m x 4 = 1.5 t 154. (c)
When settlement of hinged end B is 𝛿𝛿, reaction at B is RB.
δ =RB L3
3EI
RB =3EIδ
L3
SOLID MECHANICS ECA
Page 173
Rotation at end B due to this reaction RB
θB =RBL2
2EI=
3EIδL3 x
L2
2EI=
3δ2L
155. (d)
∆B = ∆A + θA x L + Ax�
= 0 + 0 + �−MLEI� x
L2
= −ML 2
2EI
∆B = ML 2
2EI (downward)
156. (b) Deflection of free end = 15 mm
∴ wL 4
8EI= 15 x 10−3 .....(i)
Slope of free end = 0.02 radian
∴ wL 3
6EI= 0.02 .....(ii)
Dividing equation (i) by (ii), we get 6. L
8=
15 x 10−3
0.02
⇒ L = 1 m
157. (a)
δmax = wL4
8EI (For cantilver beam)
δmax = 7.5 mm and 𝑙𝑙 = 1m = 1000 mm
7.5 = wL4
8EI
Also θmax = wL3
6EI
⇒ δ x 8
6l = wL3
6EI
∴ θmax = 7.5 x 86 x 1000
θmax = 0.01 radian 158. (d) We can use area moment method . Case 1: Load is located from centre to the fixed end.
δmax = δcentre + θB x l2
= �w(L 2⁄ )4
8EI�+ w(L 2⁄ )3
6EI x l
2
= 7wl4
384EI
Case 2: Load is located from fixed end to the end of the span
δmax = wl4
8EI
(δmax )case 1
(δmax )case 2=
748
UNIT 8 COLUMNS & MOMENT OF INERTIA
159. (a)
Pc = n2π2EI
L2 or Pc = π2EILeff
2
Case 1: If both ends of the column are fixed
Leff = L2
∴ Pc=4π2EI
L2 Case 2: If one end of the column is made free Leff = 2L
Pc = π2EI4L2
∴ (Pc )case 1(Pc )case 2
=4π2EI
L2π2EI4L2
= 116
160. (c) Buckling Load (Critical load or Euler’s load or crippling load) is given by
Pc = n2π2EI
L2 Moment of Inertia (I) depends upon dimensions of the column
∴ Pc ∝ 1
L2 and also it depends on the
dimension of the column
161. (b) Euler’s formula is valid for 𝐿𝐿𝑟𝑟 ratio
higher than 89 for Mild steel.It does not take into account the direct compressive stress and hence gives the correct results only for long columns.
SOLID MECHANICS ECA
Page 174
162. (a)
163. (b) For compressive stress
− PA− My
I < 0
−PπD2
4
−(−Pe) D
2πD4
64
< 0
e < 𝐷𝐷8
within limit of −D8
to D8
164. (d) 1P
= 1
PC +
1Pe
Given, Pc = 1500 kN Pe = 1000 kN 1P
= 1
1500 +
11000
P = 1500 x 1000
2500 = 600 kN
166. (d) For chimney one end is fixed and other end is free. Unsupported length of chimney is 20 m ∴ Leffective = 2L = 2 x 20 m = 40 m 167. (d) Leffective = L (both ends hinged)
Slenderness ratio(𝜆𝜆) = Leffective
rmin
Minimum radius of gyration (rmin ) = �IminA
= �πd 464πd 2
4
= d4
∴ λ = 5000160
4= 125
168. (b) τ = Trj
j = Polar moment of inertia 𝜏𝜏 = Torsional moment 169. (a) Polar modulus
= Polar moment of inertia
Distance of extreme fibre from neutral axis
= πd 4
32d2
= πd3
16
170. (b) T = πd3
16
= π
16 x 8 x 8 x 8 = 32π cm3
UNIT 9 TORSION
171. (c) In circular shaft
Maximum shear stress, τmax = 16Tπd3
∴ τmax ∝ 1
d3
172. (c) From Torsional equation:
TIP
= GфL
= τr
The maximum shear stress occurs at the
surface of the shaft, i.e at r = d2
∴ Tπd 432
= τmaxd2
⇒ τmax =
16Tπd3
173. (c) Maximum shear stress in the shaft due to torque T
τmax = 16Tπd3 =
16 x 4 x 106
π x 803
= 125 π
N/mm2
174. (b) D0 = 16 mm, τr
= TJ
Di = 12 mm
SOLID MECHANICS ECA
Page 175
𝜏𝜏 = 40 N-m
τ0 = T x r
J
τi = π �164−124
32�
τ0 = 40 x 103x 16
2
π x �164−124�32
= 72.75 N/mm2
τi =72.7
162
x �122�
= 54.56 N/mm2 175. (c) In circular shaft
Max. Shear stress, 𝜏𝜏 = 16TπD3
5 = 16T
π x 403
T = 62831.8 N-mm = 62.8 N-mm 176. (b) From torsion formula
τr
=TJ
=GθL
Tθ
= GJL
For twist of 1 radian over unit length of shaft θ = 1, L = 1, T = GJ T = Torsional Rigidity 177. (c) Strain energy in torsion
U = 12
T. θ
From Torsional formula τr
=TJ
=GθL
∴ U = 12
x τ. Jr
x τLr. G
= 12
. q. πd4
32x d2 x q.L
d2 .G
= q2
4G�πd2
4x L�
= U
volume= q2
4G
178. (a) Torsional moment resistance T = τJr
Tsolid
Thollow=
(τ. J r⁄ )solid(τ. J/r)hollow
= σs�
πD 4
32 � D2�
σsπ�D 4−d 4�
32 �D2��
= D4
D4−d4
179. (a)
Brittle material fails due to normal stress which is maximum at 450 to the axis in case of torsion. 181. (c)
They have same weight
⇒ Area of cross section will be same for the
two shaft
⇒πds
2
4= π(d1
2 − d22)
4
ds2 = d1
2 − d22 ..... (i)
From (i) we can say that d1 > ds For stiffness
Stiffness = GJL
(Stiffness )h(stiffness )s
= JhJs
For same material and length (Stiffness )h(stiffness )s
=d1
4−d24
ds4
�d1ds�
4− �d2
ds�
4 .....(ii)
From (i) and (ii) So we can say stiffness of hollow shaft is greater than solid shaft.
SOLID MECHANICS ECA
Page 176
182. (d) From torsion formula τr
=TJ
=GθL
Given θ = 1° = 1 x 𝜋𝜋
180 radian
τr
= GθL
120r
= 0.8 x 105 x � π
180�
1000
r =270𝜋𝜋
mm
183. (d) Power = Torque x Angular velocity = 5000 x 2πn
= 5000 x 2π x 18060
= 30000 π W
= 30 π kW 184. (b) Power transmitted by the shaft P = Torque x Angular velocity
= T x 2πN
60 Watt
= 2πN
60 x 10−3 kW
= 2πNT
60000 kW
185. (a) T = τ x Ip
r
= τ x π32d4
d2
= τ x πd3
16
Power = T𝜔𝜔
= τ xπd3
16 x 150 x
2π60
= 60 x d3 x 150 x 2π2
16 x 60
= 1003x 150 x 2 x 10 x10−6
16 kW
= 187.5 kW