UNIT -2 BENDING AND SHEAR STRESSES
This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Bending Stress”.
1. A beam is said to be of uniform strength, if ____________ a) B.M. is same throughout the beam b) Shear stress is the same through the beam c) Deflection is the same throughout the beam d) Bending stress is the same at every section along its longitudinal axis . ANSWER: d Explanation: Beam is said to be uniform strength if at every section along its longitudinal axis, the bending stress is same. 2. Stress in a beam due to simple bending is ____________ a) Directly proportional b) Inversely proportional c) Curvilinearly related d) None of the mentioned . ANSWER: a Explanation: The stress is directly proportional to the load and here the load is in terms of bending. So the stress is directly proportional to bending. 3. Which stress comes when there is an eccentric load applied? a) Shear stress b) Bending stress c) Tensile stress d) Thermal stress . ANSWER: b Explanation: When there is an eccentric load it meANSWER that the load is at some distance from the axis. This causes compression in one side and tension on the other. This causes bending stress. 4. What is the expression of the bending equation? a) M/I = σ/y = E/R b) M/R = σ/y = E/I c) M/y = σ/R = E/I d) M/I = σ/R = E/y . ANSWER: a Explanation: The bending equation is given by M/I = σ/y = E/R where M is the bending moment I is the moment of inertia y is the distance from neutral axis E is the modulus of elasticity R is the radius. 5. On bending of a beam, which is the layer which is neither elongated nor shortened? a) Axis of load b) Neutral axis c) Center of gravity d) None of the mentioned . ANSWER: b Explanation: When a beam is in bending the layer in the direction of bending will be In compression and the other will be in tension. One side of the neutral axis will be shortened and the other will be elongated. 6. The bending stress is ____________ a) Directly proportional to the distance of layer from the neutral layer b) Inversely proportional to the distance of layer from the neutral layer c) Directly proportional to the neutral layer d) Does not depend on the distance of layer from the neutral layer . ANSWER: a Explanation: From the bending equation M/I = σ/y = E/R Here stress is directly proportional to the distance of layer from the neutral layer. 7. Consider a 250mmx15mmx10mm steel bar which is free to expand is heated from 15C to 40C. what will be developed? a) Compressive stress b) Tensile stress c) Shear stress d) No stress . ANSWER: d Explanation: If we resist to expand then only stress will develop. Here the bar is free to expand so there will be no stress. 3. What will be the unit of compressive stress? a) N b) N/mm c) N/mm2 d) Nmm . ANSWER: c Explanation: As the stress is the ratio of force to the area, so it will be N/mm2 . Here mm is normally used in its calculation most of the time. 4. A cast iron T section beam is subjected to pure bending. For maximum compressive stress to be 3 times the maximum tensile stress, centre of gravity of the section from flange side is ____________ a) h/2 b) H/3 c) H/4 d) 2/3h . ANSWER: c Explanation: H/4 when the applied moment is sagging. Otherwise, I.e. if the applied moment is hogging it is H/4. as in the options both are not given meANSWER we have to take hogging. 5. A solid circular shaft of diameter d is subjected to a torque T. the maximum normal stress induced in the shaft is ____________ a) Zero b) 16T/πd3 c) 32T/πd3 d) None of the mentioned . ANSWER: b Explanation: The maximum torque trANSWERmitted by a circular solid shaft is obtained from the maximum shear stress induced at the outer surface of the solid shaft and given by T = πD3 /16 x normal stress, So, normal stress = 16T/πd3 . 6. When a rectangular beam is loaded trANSWERversely, the maximum compressive stress develops on ____________ a) Bottom fibre b) Top fibre c) Neutral axis d) Every cross-section . ANSWER: b Explanation: Loaded meANSWER loaded downwards. In that case, upper fibres will be compressed while lower will be expanded. Hence maximum compressive stress will be developed in top layer. 7. An axial residual compressive stress due to a manufacturing process is present on the outer surface of a rotating shaft subjected to bending. Under a given bending load, the fatigue of the shaft in the presence of the residual compressive stress is ____________ a) Decreased b) Increases or decreased, depending on the external bending load c) Neither decreased nor increased d) Increases . ANSWER: d Explanation: From the Gerber’s parabola that is the characteristic curve of the fatigue life of the shaft in the presence of the residual compressive stress. The fatigue life of the material is effectively increased by the introduction of compressive mean stress, whether applied or residual. This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Bars of varying sections”. 1. If a bar of two different length are in a line and P load is acting axially on them then what will be the change in length of the bar if the radius of both different lengths is same? a) P/E x (L1 + L2) b) PA/E x (L1 + L2) c) P/EA x (L1 + L2) d) E/PA x (L1 + L2) . ANSWER: c Explanation: Change in length of section 1 = PL1/EA1 Change in length of section 2 = PL2/EA2 Since diameter is same for both the sections, the respective area will be the same Total change in length of bar = PL1/EA1 + PL2/EA2 = P/EA x (L1 + L2). 2. If a bar of two sections of different diameters of same length are in a line and P load is acting axially on them then what will be the change in length of the bar? a) PL/E x (1/A1 + 1/A2) b) P/E x (1/A1 + 1/A2) c) P/EL x (1/A1 + 1/A2) d) PE/L x (1/A1 + 1/A2) . ANSWER: a Explanation: Change in length of section 1 = PL1/EA1 Change in length of section 2 = PL2/EA2 Since length is same for both the sections, Total change length of bar = PL/E x (1/A1 + 1/A2). 3. An axial pull of 35000 N is acting on a bar consisting of two lengths as shown with their respective dimensions. What will be the stresses in the two sections respectively in N/mm2 ? a) 111.408 and 49.5146 b) 111.408 and 17.85 c) 97.465 and 49.5146 d) 97.465 and 34.263 . ANSWER: a Explanation: The stress = P/A Where P = 35000N and A is the respective cross section area of the sections. 4. An axial pull of 1kN is acting on a bar of consisting two equal lengths as shown but of dia 10cm and 20cm respectively. What will be the stresses in the two sections respectively in N/mm2 ? a) 0.127 and 0.0031 b) 0.034 and 0.0045 c) 0.153 and 0.003 d) 0.124 and 0.124 . ANSWER: a Explanation: The stress = P/A Where P = 1000N and A is the respective cross section area of the sections. 5. An axial pull of 35000 N on a bar consisting of two lengths as shown with their respective dimensions. What will be the total extension of the bar if the young’s modulus = 2.1 x 105 ? a) 0.153mm b) 0.183mm c) 0.197mm d) 0.188mm . ANSWER: b Explanation: The total extension in the bar = P/E x ( L1/A1 + L1/A1 ) Where P = 35000 N, E = 2.1 x 105 N/mm2 , L1 and L2 are the 20cm and 25cm respectively and A1 and A2 are the area of both the sections respectively. 6. An axial pull of 20 kN on a bar of two equal lengths of 20cm as shown with their respective dimensions. What will be the total extension of the bar if the young’s modulus = 2×105 ? a) 0.200mm b) 0.345mm c) 0.509mm d) 0.486mm . ANSWER: c Explanation: The total extension in the bar = P/E x (L1/A1 + L1/A1) Where P = 2 kN, E = 2 x 105 N/mm2 , L1 and L2 are same of 20cm and A1and A2 are the area of both the sections respectively. 7. Does the value of stress in each section of a composite bar is constant or not? a) It changes in a relationship with the other sections as well b) It changes with the total average length c) It is constant for every bar d) It is different in every bar in relation with the load applied and the cross sectional area . ANSWER: d Explanation: The value of stress in every section of a composite bar is given by P/A which is it is dependent on the load applied and the cross sectional area of the section. The value of stress in a section does not depend on the dimensions of other sections in the bar. 8. A composite bar of two sections of equal length and equal diameter is under an axial pull of 10kN. What will be the stresses in the two sections? a) 3.18 N/mm2 b) 2.21 N/mm2 c) 3.45 N/mm2 d) 2.14 N/mm2 . ANSWER: a Explanation: The stress = P/A Where P = 1000N and A is the respective cross section area of the sections. Here the stress will be equal in both the sections as the dimensions are the same. 9. A composite bar of two sections of unequal length and equal diameter is under an axial pull of 10kN. What will be the stresses in the two sections? a) 2.145 N/mm2 b) 3.18 N/mm2 c) 1.245 N/mm2 d) 2.145 N/mm2 . ANSWER: b Explanation: The stress = P/A Where P = 1000N and A is the respective cross section area of the sections. Here the stress will be equal in both the sections as the diameter is the same for both the sections. Even if the length is the variable it will not alter the stress value as the length does not depend on the stress. 10. A composite bar of two sections of equal length and given diameter is under an axial pull of 15kN. What will be the stresses in the two sections in N/mm2 ? a) 190.9 and 84.88 b) 190.9 and 44.35 c) 153.45 and 84088 d) 153045 and 44.35 . ANSWER: a Explanation: The stress = P/A Where P = 15000N and A is the respective cross section area of the sections. This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Bars of Composite Sections – 1”. 1. If a bar of sections of two different length and different diameters are in a line and P load is acting axially on them then what will be the change in length of the bar? a) P/E x (L1 + L2) b) P/E x (A1/L1 + A2/ L2) c) P/E x (L1/A1 + L2/A2) d) E/P x (L1/A1 + L2/A2) . ANSWER: c Explanation: Change in length of section 1 = PL1/EA1 Change in length of section 2 = PL2/EA2 Total change in length of bar = PL1/EA1 + PL2/EA2. 2. How does the elastic constant varys with the elongation of body? a) The elastic constant is directly proportional to the elongation b) The elastic constant is directly proportional to the elongation c) The elongation does not depends on the elastic constant d) None of these . ANSWER: b Explanation: Elongation of a composite bar of two sections = P/E x (L1/A1 + L2/A2) E is inversely proportional to bar elongation. 3. A composite rod is 1000mm long, its two ends are 40 mm2 and 30mm2 in area and length are 400mm and 600mm respectively. If the rod is subjected to an axial tensile load of 1000N, what will be its total elongation(E = 200GPa)? a) 0.130m b) 0.197mm c) 0.160mm d) 0.150mm . ANSWER: a Explanation: As elongation of a composite bar of two sections = P/E x (L1/A1 + L2/A2) Putting L1, L2, A1 and A2 400mm2, 600mm2, 40mm2 and 30mm2 and P = 1000 and E = 200 x 103 . 4. A mild steel wire 5mm in diameter and 1m ling. If the wire is subjected to an axial tensile load 10kN what will be its extension? a) 2.55mm b) 3.15mm c) 2.45mm d) 2.65mm . ANSWER: a Explanation: As change in length = PL/AE P = 10x 1000N, L = 1m, A = πd2 /4 = 1.963 x 10-5 m2 , E = 200 x 109 N/m2 . 5. A composite rod is 1000mm long, its two ends are 40mm2 and 30mm2 in area and length are 300mm and 200mm respectively. The middle portion of the rod is 20mm2 in area. If the rod is subjected to an axial tensile load of 1000N, what will be its total elongation (E = 200GPa)? a) 0.145mm b) 0.127mm c) 0.187mm d) 0.196mm . ANSWER: d Explanation: P = 1000N, Area A1 = 40mm2 , A2 = 20mm2 , A30 = 30mm2 Length, L1 = 300mm, L2 = 500mm, L3 = 200mm E = 200GPa = 200x 1000 N/mm2 Total extension = P/E x (L1/A1 + L2/A2 + L3/A3). 6. A rod of two sections of area 625mm2 and 2500mm2 of length 120cm and 60cm respectively. If the load applied is 45kN then what will be the elongation (E = 2.1x 105 N/mm2 )? a) 0.462mm b) 0.521mm c) 0.365mm d) 0.514mm . ANSWER: a Explanation: P = 45,000N, E =2.1x 105 N/mm2 , Area, A1 = 625mm2 , A2 = 2500mm2, Length, L1 = 1200mm, L2 = 600mm Elongation = P/E x (L1/A1 + L2/A2). 7. What will be the elongation of a bar of 1250mm2 area and 90cm length when applied a force of 130kN if E = 1.05x 105 N/mm2 ? a) 0.947mm b) 0.891mm c) 0.845mm d) 0.745mm . ANSWER: b Explanation: As change in length = PL/AE P = 130x 1000N, L = 900mm, A = 1250 mm2 , E = 1.05 x 105 N/m2 . 8. A bar shown in diagram is subjected to load 160kN. If the stress in the middle portion Is limited to 150N/mm2, what will be the diameter of the middle portion? a) 3.456 cm b) 3.685 cm c) 4.524 cm d) 4.124 cm . ANSWER: b Explanation: Let L2 and D2 be the dimensions of the middle portion and L1 and D2 be the end portion dimensions. For middle portion area = load / stress This gives area by which diameter can be calculated. 9. A steel bar of 20mm x 20mm square cross-section is subjected to an axial compressive load of 100kN. If the length of the bar is 1m and E=200GPa, then what will be the elongation of the bar? a) 1.25mm b) 2.70mm c) 5.40mm d) 4.05mm . ANSWER: a Explanation: Elongation in bar = PL/ AE = (100x1000x1) / (0.2×0.2x200x106 ) = 1.25mm. 10. A solid uniform metal bar is hanging vertically from its upper end. Its elongation will be _________ a) Proportional to L and inversely proportional to D2 b) Proportional to L2 and inversely proportional to D c) Proportional of U but independent of D d) Proportional of L but independent of D . ANSWER: a Explanation: Elongation = WL / 2AE = 4WL / 2πD2E α L/D2 . This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Moment of Inertia”. 1. The axis about which moment of area is taken is known as ____________ a) Axis of area b) Axis of moment c) Axis of reference d) Axis of rotation . ANSWER: c Explanation: The axis of reference is the axis about which moment of area is taken. Most of the times it is either the standard x or y axis or the centeroidal axis. 2. Point, where the total volume of the body is assumed to be concentrated is ____________ a) Center of area b) Centroid of volume c) Centroid of mass d) All of the mentioned . ANSWER: b Explanation: The centroid of the volume is the point where total volume is assumed to be concentrated. It is the geometric centre of a body. If the density is uniform throughout the body, then the center of mass and center of gravity correspond to the centroid of volume. The definition of the centroid of volume is written in terms of ratios of integrals over the volume of the body. 3. What is MOI? a) ml2 b) mal c) ar2 d) None of the mentioned . ANSWER: c Explanation: The formula of the moment of inertia is, MOI = ar2 where M = mass, a = area, l = length, r = distance. 4. What is the formula of radius of gyration? a) k2 = I/A b) k2 = I2 /A c) k2 = I2 /A2 d) k2 = (I/A)1/2 . ANSWER: a Explanation: The radius of gyration of a body about an axis is a distance such that its square multiplied by the area gives moment of inertia of the area about the given axis. The formula of radius of gyration is given as k2 = I/A. 5. What is the formula of theorem of perpendicular axis? a) Izz = Ixx – Iyy b) Izz = Ixx + Ah2 c) Izz – Ixx = Iyy d) None of the mentioned . ANSWER: c Explanation: Theorem of perpendicular axis stares that if IXX and IYY be the moment of inertia of a plane section about two mutually perpendicular axis X-X and Y-Y in the plane of the section then the moment of inertia of the section IZZ about the axis Z-Z, perpendicular to the plane and passing through the intersection of X-X and Y-Y is given by the formula Izz – Ixx = Iyy. 6. What is the formula of theorem of parallel axis? a) IAD = IG + Ah b) IAB = Ah2 + IG c) IAB = IG – Ah2 d) IAB = IG + Ixx . ANSWER: b Explanation: The theorem of parallel axis states that if the moment of inertia of a plane area about an axis in the plane of area theough the C.G. of the plane area be represented by IG, then the moment of the inertia of the given plane area about a parallel axis AB in the plane of area at a distance h from the C.G. is given by the formula IAB = Ah2 + IG. 7. What is the unit of radius of gyration? a) m4 b) m c) N d) m2 . ANSWER: b Explanation: The radius of gyration = (length4 /length2 )1/2 = length So its unit will be m. This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Bending Equation”. 1. In simple bending, ______ is constant. a) Shear force b) Loading c) Deformation d) Bending moment . ANSWER: d Explanation: If a beam is undergone with simple bending, the beam deforms under the action of bending moment. If this bending moment is constant and does not affect by any shear force, then the beam is in state of simple bending. 2. If a beam is subjected to pure bending, then the deformation of the beam is_____ a) Arc of circle b) Triangular c) Trapezoidal d) Rectangular . ANSWER: a Explanation: The beam being subjected to pure bending, there will be only bending moment and no shear force it results in the formation of an arc of circle with some radius known as radius of curvature. 3. When a beam is subjected to simple bending, ____________ is the same in both tension and compression for the material. a) Modulus of rigidity b) Modulus of elasticity c) Poisson’s ratio d) Modulus of section . ANSWER: b Explanation: It is one of the most important assumptions made in the theory of simple bending that is the modulus of elasticity that is Young’s modulus [E] is same in both tension and compression for the material and the stress in a beam do not exceed the elastic limit. 4. E/R = M/I = f/y is a bending equation. a) True b) False . ANSWER: a Explanation: The above-mentioned equation is absolutely correct. E/R = M/I = f/y is a bending equation. It is also known as flexure equation (or) equation for theory of simple bending. Where, E stands for Young’s modulus or modulus of elasticity. R stands for radius of curvature. M stands for bending moment I stand for moment of inertia f stands for bending stress y stands for neutral axis. 5. Maximum Shearing stress in a beam is at _____ a) Neutral axis b) Extreme fibres c) Mid span d) Action of loading . ANSWER: a Explanation: Shearing stress is defined as the resistance offered by the internal stress to the shear force. Shearing stress in a beam is maximum at a neutral axis. 6. At the neutral axis, bending stress is _____ a) Minimum b) Maximum c) Zero d) Constant . ANSWER: c Explanation: Neutral axis is defined as a line of intersection of neutral plane or neutral layer on a cross section at the neutral axis of that section. At the NA, bending stress or bending strain is zero. The first moment of area of a beam section about neutral axis is also zero. The layer of neutral axis neither contracts nor extends. 7. Curvature of the beam is __________ to bending moment. a) Equal b) Directly proportion c) Inversely proportion d) Coincides . ANSWER: b Explanation: From the flexural equation, we have 1/R is called as the “curvature of the beam”. 1 / R = M / EI Hence the curvature of the beam is directly proportional to bending moment and inversely proportional to flexural rigidity (EI). 8. What are the units of flexural rigidity? a) Nm2 b) Nm c) N/m d) m/N3 . ANSWER: a Explanation: The product of young’s modulus (E) of the material and moment of inertia (I) of the beam section about its neutral axis is called flexural rigidity. Units for E are N/m2 Units for I are m4 Their product is Nm2 . 9. What are the units for section modulus? a) m2 b) m4 c) m3 d) m . ANSWER: c Explanation: The ratio of moment of inertia to the distance to the extreme fibre is called modulus of section or section modulus. It is generally denoted by the letter Z. Section modulus is expressed in m3 Z = I/y = m4 / m = m3 . 10. What are the units of axial stiffness? a) m3 b) m2 c) N/ m d) -m . ANSWER: c Explanation: Axial rigidity is a product of young’s modulus (E) and the cross-sectional area (A) of that section. Axial rigidity per unit length is known as axial stiffness the si units of axial stiffness are Newton per metre (N/m). This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Section Modulus”. 1. What is the section modulus (Z) for a rectangular section? a) bd2 /6 b) a3 /6 c) BD3 -bd3 d) D4 -d 4 . ANSWER: a Explanation: The modulus of section may be defined as the ratio of moment of inertia to the distance to the extreme fibre. It is denoted by Z. Z= I/y ; For rectangular section, I = bd3 /12 & y = d/2. Z= bd2 /6. 2. Find the modulus of section of square beam of size 300×300 mm. a) 4.8 × 106 mm3 b) 4.5 × 106 mm3 c) 5.6 × 106 mm3 d) 4.2 × 106 mm3 . ANSWER: b Explanation: Here, a = side of square section = 300 mm. I = a4 /12. y= a/2. Z = I/y = a3 /6 = 3003 /6 = 4.5 × 106 mm3 . 3. _________ of a beam is a measure of its resistance against deflection. a) Strength b) Stiffness c) Deflection d) Slope . ANSWER: b Explanation: A beam is said to be a strength when the maximum induced bending and shear stresses are within the safe permissible stresses stiffness of a beam is a measure of its resistance against deflection. 4. To what radius an Aluminium strip 300 mm wide and 40mm thick can be bent, if the maximum stress in a strip is not to exceed 40 N/mm2 . Take young’s modulus for Aluminium is 7×105 N/mm2 . a) 45m b) 52m c) 35m d) 65m . ANSWER: c Explanation: Here, b = 300mm d= 40mm. y= 20mm. From the relation; E/R = f/y R= E×y/f =70×103 × 20 / 40 = 35m. 5. The bending stress in a beam is ______ to bending moment. a) Less than b) Directly proportionate c) More than d) Equal . ANSWER: b Explanation: As we know, the bending stress is equal to bending moment per area. Hence, as the bending (flexure) moment increases/decreases the same is noticed in the bending stress too. 6. The Poisson’s ratio for concrete is __________ a) 0.4 b) 0.35 c) 0.12 d) 0.2 . ANSWER: d Explanation: The ratio of lateral strain to the corresponding longitudinal strain is called Poisson’s ratio. The value of poisons ratio for elastic materials usually lies between 0.25 and 0.33 and in no case exceeds 0.5. The Poisson’s ratio for concrete is 0.20. 7. The term “Tenacity” meANSWER __________ a) Working stress b) Ultimate stress c) Bulk modulus d) Shear modulus . ANSWER: b Explanation: The ultimate stress of a material is the greatest load required to fracture the material divided by the area of the original cross section in the point of fracture The ultimate stress is also known as tenacity. 8. A steel rod of 25 mm diameter and 600 mm long is subjected to an axial pull of 40000. The intensity of stress is? a) 34.64 N/mm2 b) 46.22 N/mm2 c) 76.54 N/mm2 d) 81.49 N/mm2 . ANSWER: d Explanation: Cross sectional area of steel rod [Circular]be 490.87 mm2 . The intensity of stress = P/A = 40000/490.87 = 81.49 N/mm2 . 9. The bending strain is zero at _______ a) Point of contraflexure b) Neutral axis c) Curvature d) Line of action of loading . ANSWER: b Explanation: The neutral axis is a line of intersection of neutral plane or neutral layer on a cross section. The neutral axis of a beam passes through the centroid of the section. At the neutral axis bending stress and bending strain is zero. 10. Strength of the beam depends only on the cross section. a) True b) False . ANSWER: b Explanation: The strength of two beams of the same material can be compared by the section modulus values. The strength of beam depends on the material, size and shape of cross section. The beam is stronger when section modulus is more, strength of the beam depends on Z.
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