UNIT -3 TORSION AND STRAIN ENERGY

 UNIT -3 TORSION AND STRAIN ENERGY

This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Definition of Torque”. 

1. Torque is __________ moment. 

a) Twisting b) Shear c) Bending d) Couple . 

ANSWER: a Explanation: A cylindrical shaft is subjected to twisting moment or torque when a force is acting on the member tangentially at some radius in a plane of its cross section. 

2. Twisting moment is a product of __________ and the radius. a) Direction b) Velocity c) Force d) Acceleration . 

ANSWER: c Explanation: Twisting moment will be equal to the product of force and radius. When a shaft is subjected to a twisting moment, every cross section of the shaft will surely experience shear stress. 

3. Torsion is denoted by __________ a) R b) Q c) T d) N . 

ANSWER: c Explanation: If the moment is applied in a plane perpendicular to the longitudinal axis of the beam (or) shaft it will be subjected to torsion. Torsion is represented or denoted by T. 

4. The SI units for torsion is __________ a) N m b) N c) N/m d) m . 

ANSWER: a Explanation: As torsion is a product of perpendicular force and radius, the units will be N m. Torque is also known as torsion or twisting moment or turning moment. 

5. _____________ torsion is produced when twisting couple coincides with the axis of the shaft. a) Exact b) Pure c) Nominal d) Mild . 

ANSWER: b Explanation: When a member is subjected to the equal and opposite twisting moment at its ends, then the member is said to be subjected under pure torsion. Pure Torsion is often produced when the axis of the twisting couple coincides with the axis of the shaft 

This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Definition of Torque”. 

1. Torque is __________ moment. a) Twisting b) Shear c) Bending d) Couple . ANSWER: a Explanation: A cylindrical shaft is subjected to twisting moment or torque when a force is acting on the member tangentially at some radius in a plane of its cross section. 

2. Twisting moment is a product of __________ and the radius. a) Direction b) Velocity c) Force d) Acceleration . ANSWER: c Explanation: Twisting moment will be equal to the product of force and radius. When a shaft is subjected to a twisting moment, every cross section of the shaft will surely experience shear stress. 

3. Torsion is denoted by __________ a) R b) Q c) T d) N . 

ANSWER: c Explanation: If the moment is applied in a plane perpendicular to the longitudinal axis of the beam (or) shaft it will be subjected to torsion. Torsion is represented or denoted by T. 

4. The SI units for torsion is __________ a) N m b) N c) N/m d) m . 

ANSWER: a Explanation: As torsion is a product of perpendicular force and radius, the units will be N m. Torque is also known as torsion or twisting moment or turning moment. 

5. _____________ torsion is produced when twisting couple coincides with the axis of the shaft. a) Exact b) Pure c) Nominal d) Mild . 

ANSWER: b Explanation: When a member is subjected to the equal and opposite twisting moment at its ends, then the member is said to be subjected under pure torsion. Pure Torsion is often produced when the axis of the twisting couple coincides with the axis of the shaft 

This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Torsion Equation”. 

1. Torsional sectional modulus is also known as _________ a) Polar modulus b) Sectional modulus c) Torsion modulus d) Torsional rigidity . 

ANSWER: a Explanation: The ratio of polar moment of inertia to radius of section is called Polar modulus or Torsional section modulus. Its units are mm3 or m3 (in SI).

2. ________ is a measure of the strength of shaft in rotation. a) Torsional modulus b) Sectional modulus c) Polar modulus d) Torsional rigidity . 

ANSWER: c Explanation: The polar modulus is a measure of the strength of shaft in rotation. As the value of Polar modulus increases torsional strength increases. 

3. What are the units of torsional rigidity? a) Nmm2 b) N/mm c) N-mm d) N . 

ANSWER: a Explanation: The product of modulus of rigidity (C) and polar moment of inertia (J) is called torsional rigidity. Torsional rigidity is a torque that produces a twist of one radian in a shaft of unit length. 

4. The angle of twist can be written as ________ a) TL/J b) CJ/TL c) TL/CJ d) T/J . ANSWER: c Explanation: The angle of Twist = TL/CJ Where T = Torque in Nm L = Length of shaft CJ = Torsional rigidity. 

5. The power trANSWERmitted by shaft SI system is given by __________ a) 2πNT/60 b) 3πNT/60 c) 2πNT/45 d) NT/60 W . 

ANSWER: a Explanation: In SI system, Power (P) is measured in watts (W) ; P = 2πNT/60 Where T = Average Torque in N.m N = rpm = 2πNT/ 45 1 watt = 1 Joule/sec = 1N.m/s. 

1. The intensity of shear stress at a section is ______ to the distance of the section from the axis of the shaft. a) Inversely proportional b) Directly proportional c) Equal d) Parallel . ANSWER: b Explanation: The intensity of shear stress at a section is directly proportional to the distance of the section from axis of the shaft. The shear stress at a distance from the centre of the shaft is given by fs/R × r. 

2. The shear stress is ____________ at the axis of the shaft. a) Minimum b) Maximum c) Zero d) Uniform . ANSWER: c Explanation: The shear stress is zero at the axis of the shaft and the shear stress is linearly increasing to the maximum value at the surface of the shaft. 

3. The shear stress at the outer surface of hollow circular section is _________ a) Zero b) Maximum c) Minimum d) Can’t determined .

 ANSWER: b Explanation: The shear stress in a hollow circular section varies from maximum at the outer surface to a minimum (but not zero) in the inner face. The minimum value should be greater than zero. 

4. The hollow shaft will trANSWERmit greater _______ then the solid shaft of the same weight. a) Bending moment b) Shear stress c) Torque d) Sectional Modulus . 

ANSWER: c Explanation: For the same maximum shear stress, the average shear stress in a hollow shaft is greater than that in a solid shaft of the same area. Hence the hollow shaft will trANSWERmit greater torque than the solid shaft of the same weight. 

This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Polar Moment of Inertia”. 

1. The moment of inertia of a plane area with respect to an axis ____________ to the plane is called a polar moment of inertia. a) Parallel b) Perpendicular c) Equal d) Opposite . 

ANSWER: b Explanation: The moment of inertia of a plane area with respect to an axis perpendicular to the plane of the figure is called a polar moment of inertia with respect to a point, where the axis intersects a plane. 

2. What are the units of Polar modulus? a) mm3 b) mm2 c) mm d) mm4 . ANSWER: a Explanation: The ratio of polar moment of inertia (J) to the radius of section(R) is known as polar modulus or torsional section modulus. Its units are mm3 . 

3. What is the polar modulus for solid shaft? a) π/16 D2 b) π/12 D3 c) π/ 16 D3 d) π/16 D . ANSWER: c Explanation: For solid shaft Z = J/R = π/32 × D4 / D/2. Z = π/16 D3 . 

4. Calculate the polar moment of inertia for a solid circular shaft of 30 mm diameter. a) 76m4 b) 79.5m4 c) 81m4 d) 84m4 . ANSWER: b Explanation: Diameter of the shaft = 30 mm Polar moment of inertia = J = π/32 × (30)4 mm4 J = 79.52 m4 . 

5. A hollow shaft outside diameter 120 mm and thickness 20 mm. Find polar moment of inertia. a) 16.36 × 106 mm4 b) 18.45 × 106 mm4 c) 21.3 × 106 mm4 d) 22.5 × 106 mm4 . 

ANSWER: a Explanation: For hollow circular shaft, outside diameter = 120 mm; d = 120-2×20 = 80 mm the polar moment of inertia = π/32 × (1204 – 804 ). J = 16.36 × 106 mm4 . 

This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Strain Energy”. 

1. Resilience can also be termed as ___________ a) Stress energy b) Strain energy c) Modulus d) Tenacity . ANSWER: b Explanation: The capability of a material to absorb energy when it is deformed elastically and release that energy upon unloading is known resilience. This resilience is also termed as Strain energy. 2. Mathematically, strain energy = _________ a) Power b) Work done c) Young’s Modulus d) Energy . ANSWER: b Explanation: By the principle of work, the amount of strain energy in a body is found. When a load acts on a body there will be deformation, which causes movement of the applied load. This work is done by the applied load. 3. Calculate the Strain energy stored in a body of stress 0.0366 N/mm2 . The cross sectional area is 60 m2 and length of body is 1 m. Take E = 2×105 N/mm2 . a) 0.2009 N.mm b) 0.0416 N.mm c) 0.0987 N.mm d) 0.1316 N.mm . ANSWER: a Explanation: Given that : l = 1000 mm ; A = 60000 mm2 ; f = 0.0366 N/mm2 . Strain energy stored = f2 /2EI × Volume = (0.0366)2 / 2×2×105× (200×300) × (1000) = 0 2009 N.mm. 4. What are the units of measurement for wooden and steel trusses? a) 1 RM b) 1 N.o c) m2 d) m . ANSWER: b Explanation: The units of measurement for wooden and steel trusses is 1 N.o Description of work Units of measurement Earth work excavation 1 m3 Steel reinforcement 1 kN Wooden and steel trusses 1 No 5. Which of the following methods is also known as individual wall method? a) Centre line method b) Alignment method c) Long wall and short wall method d) Voluminous method . ANSWER: c Explanation: Long wall short wall method is tedious and long lasting. In this method, the length of wall running in one direction are measured first out to out and that of running in the perpendicular direction are measured in to in. 6. Centre line method is accurate method. a) False b) True . ANSWER: b Explanation: The estimates can be prepared quickly by using center line method. This is not only an accurate method but also a very quick method. 7. _______ gives the nature and class of work. a) Estimate b) Specifications c) Tenders d) Survey . ANSWER: b Explanation: Drawings cannot give every information about materials and quality. The specifications give the nature and class of work, quantity of materials and workmANSWERhip. They are very useful during the execution of work. This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Definition of Strain Energy”. 1. What is the strain energy stored in a body due to gradually applied load? a) σE/V b) σE2 /V c) σV2 /E d) σV2 /2E . ANSWER: d Explanation: Strain energy when load is applied gradually = σ2V/2E. 2. Strain energy stored in a body to uniform stress s of volume V and modulus of elasticity E is __________ a) s2V/2E b) sV/E c) sV2 /E d) sV/2E ANSWER: a Explanation: Strain energy = s2V/2E. 3. In a material of pure shear stress τthe strain energy stored per unit volume in the elastic, homogeneous isotropic material having elastic constants E and v will be: a) τ2 /E x (1+ v) b) τ2 /E x (1+ v) c) τ2 /2E x (1+ v) d) τ2 /E x (2+ v) . ANSWER: a Explanation: σ1=τ, σ2= -τσ3=0 U = (τ2+- τ 2 -2μτ(-τ))V = τ2 /E x (1+ v)V. 4. PL3 /3EI is the deflection under the load P of a cantilever beam. What will be the strain energy? a) P 2L 3 /3EI b) P2L 3 /6EI c) P2L 3 /4EI d) P2L 3 /24EI . ANSWER: b Explanation: We may do it taking average Strain energy = Average force x displacement = (P/2) x PL3 /3EI = P2L 3 /6EI. 5. A rectangular block of size 400mm x 50mm x 50mm is subjected to a shear stress of 500kg/cm2 . If the modulus of rigidity of the material is 1×106 kg/cm2 , the strain energy will be __________ a) 125 kg-cm b) 1000 kg-cm c) 500 kg-cm d) 100 kg-cm ANSWER: a Explanation: Strain energy stored = τ2V/2G = 5002 /2×106 x 40x5x5 = 125 kg-cm. 6. A material of youngs modulus and Poissons ratio of unity is subjected to two principal stresses σ1 and σ2 at a point in two dimensional stress system. The strain energy per unit volume of the material is __________ a) (σ1 2 + σ2 2 – 2σ1σ2 ) / 2E b) (σ1 2 + σ2 2 + 2σ1σ2 ) / 2E c) (σ1 2 – σ2 2 – 2σ1σ2 ) / 2E d) (σ1 2 – σ2 2 – 2σ1σ2 ) / 2E . ANSWER: a Explanation: Strain energy = (σ1ε1+ σ1ε1 ) / 2E = (σ1 2 + σ2 2 – 2σ1σ2 ) / 2E. 7. If forces P, P and P of a system are such that the force polygon does not close, then the system will __________ a) Be in equilibrium b) Reduce to a resultant force c) Reduce to a couple d) Not be in equilibrium . ANSWER: d Explanation: The forces are not concurrent so the resultant force and couple both may be present. Thus the best choice is that forces are not in equilibrium. 8. The strain energy in a member is proportional to __________ a) Product of stress and the strain b) Total strain multiplied by the volume of the member c) The maximum strain multiplied by the length of the member d) Product of strain and Young’s modulus of the material . ANSWER: d Explanation: Strain energy per unit volume for solid = q2 / 4G.

9. A bar of cross-section A and length L is subjected to an axial load W. the strain energy stored in the bar would be __________ a) WL / AE b) W2L / 4AE c) W2L / 2AE d) WL / 4AE . ANSWER: c Explanation: Deformation in the bar = WL / AE Strain energy = W/2 x WL / AE = W2L / 2AE. 10. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5m long. What is the stretch in the rod if E = 2×105 N/mm2 ? a) 1.1mm b) 1.24mm c) 2mm d) 1.19mm . ANSWER: d Explanation: Stress = Load/ area = 60,000 / (π/4 D2 ) = 470746 N/mm2 So stretch = stress x length / E = 1.19mm. advertisement 11. A tensile load of 50kN is gradually applied to a circular bar of 5cm diameter and 5m long. What is the strain energy absorbed by the rod (E = 200GPa)? a) 14 N-m b) 15.9 N-mm c) 15.9 N-m d) 14 N-mm . ANSWER: c Explanation: Stress = 50,000 / 625π = 25.46 Strain energy = σ2V/2E = 25.46×25.46×9817477 / (2×200000) = 15909.5 N-mm = 15.9 N-m. 12. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5m long. What is the strain energy in the rod if the load is applied suddenly (E = 2×105 N/mm2 )? a) d143.23 N-m b) 140.51 N-m c) 135.145 N-m d) 197.214 N-m . ANSWER: a Explanation: Maximum instantaneous stress = 2P / A = 95.493 Strain energy = σ2V/2E = 143288N-mm = 143.238 N-m. This set of Strength of Materials Multiple Choice Questions & ANSWERs (MCQs) focuses on “Resilience”. 1. The ability of a material to absorb energy when elastically deformed and to return it when unloaded is called __________ a) Elasticity b) Resilience c) Plasticity d) Strain resistance . ANSWER: b Explanation: Resilience is the ability of a material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. 2. The strain energy stored in a specimen when stained within the elastic limit is known as __________ a) Resilience b) Plasticity c) Malleability d) Stain energy . ANSWER: a Explanation: Resilience is the ability of a material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. Malleability is the property by which any material can be beaten into thin sheets. 3. The maximum strain energy stored at elastic limit is __________ a) Resilience b) Proof resilience c) Elasticity d) Malleability . ANSWER: b Explanation: Proof resilience is the maximum stored energy at the elastic limit. Resilience is the ability of material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. Malleability is the property by which any material can be beaten into thin sheets. 4. The mathematical expression for resilience ‘U’ is __________ a) U = σ2 /E x volume b) U = σ2 /3E x volume c) U = σ2 /2E x volume d) U = σ/2E x volume . ANSWER: c Explanation: The resilience is the strain energy stored in a specimen so it will be U = σ2 /2E x volume. 5. What is the modulus of resilience? a) The ratio of resilience to volume b) The ratio of proof resilience to the modulus of elasticity c) The ratio of proof resilience to the strain energy d) The ratio of proof resilience to volume . ANSWER: d Explanation: The modulus of resilience is the proof resilience per unit volume. It is denoted by σ. 6. The property by which an amount of energy is absorbed by material without plastic deformation is called __________ a) Toughness b) Impact strength c) Ductility d) Resilience . ANSWER: d Explanation: Resilience is the ability of a material to absorb energy when elastically deformed and to return it when unloaded. 7. Resilience of a material plays important role in which of the following? a) Thermal stress b) Shock loading c) Fatigue d) Pure static loading . ANSWER: b Explanation: The total strain energy stored in a body is commonly known as resilience. Whenever the straining force is removed from the strained body, the body is capable of doing work. Hence the resilience is also define as the capacity of a strained body for doing work on the removal of the straining force. 8. A steel has its yield strength of 200N/mm2 and modulus of elasticity of 1x105MPa. Assuming the material to obey hookes law up to yielding, what will be its proof resilience? a) 0.8 N/mm2 b) 0.4 N/mm2 c) 0.2 N/mm2 d) 0.6 N/mm2 . ANSWER: c Explanation: Proof resilience = σ2 /2E = (200)2 / (2 x 105 ) = 0.2 N/mm2 . 9. A 1m long bar of uniform section extends 1mm under limiting axial stress of 200N/mm2 . What is the modulus of resilience for the bar? a) 0.1 units b) 1 units c) 10units d) 100units . ANSWER: a Explanation: Modulus of resilience, u = f2 /2E, where E = fL/δL Therefore, u = 200×1 / 2×1000 = 0.1units. 10. A square steel bar of 10mm side and 5m length is subjected to a load whereupon it absorbs a strain energy of 100J. what is its modulus of resilience? a) 1/5 N-mm/mm3 b) 25 N-mm/mm3 c) 1/25 N-mm/mm3 d) 5 N-mm/mm3 . ANSWER: a Explanation: Modulus of resilience is the strain energy stored in the material per unit volume. u = U/v = ( 100 x 1000 ) / ( 10 x 10x 5x 1000) = 1/5 N-mm/mm3 . UNIT -4 PRINCIPAL STRESSES AND STRAINS 1.A principal plane is a plane of (a) Zero tensile stress (b) Zero compressive stress (c) Zero shear stress (d) None (ANSWER: c) 2.A principal plane is a plane of (a) Only normal stress (b) Only shear stress (c) Only bending stress (d) None (ANSWER: a) 3.There are in all (a) Two principal planes (b) Three principal planes (c) Four principal planes (d) None (ANSWER: b) 4.There are in all (a) Two principal stresses (b) Three principal stresses (c) Four principal stresses (d) None (ANSWER: b) 5.There are in all (a) Two principal strains (b) Three principal strains (c) Four principal strains (d) None (ANSWER: b) 6.Identify the principal stress (a) Shear stress (b) Bending stress (c) Compressive stress (d) None (ANSWER:c) 7.On the planes of maximum shear, there are (a) Normal stresses (b) Bending stresses (c) Bucking stresses (d) None (ANSWER: a) 8.Maximum shear stress is (a) Average sum of principal stresses (b) Average difference of principal stresses (c) Average sum as well as difference of principal stresses (d) None (ANSWER: b) 9.The magnitude of principal stresses due to complex stresses is (a) (1/2)[ (σx + σy) ± ((σx –σy)2 + 4 τ2 ))0.5] (b) (1/2)[ (σx + σy) ± (1/2)((σx –σy)2 + 4 τ2 ))0.5] (c) (1/2)[ (σx + σy) ± ((1/2)(σx –σy)2 + 4 τ2 ))0.5] (d) None (ANSWER: a) 10.The equations in the various questions are valid only when (a)σx and σy are both tensile (b) σx is compressive and σy is tensile (c) σx is tensile and σy is compressive (d) None (ANSWER: a) 11.The magnitude of maximum shear stress will be (a) ± (1/2)[ ((σx –σy)2 + 4 τ2 ))0.5] (b) ± (1/2)[ (1/2)((σx –σy)2 + 4 τ2 ))0.5] (c) ± (1/2)[ ((1/2)(σx –σy)2 + 4 τ2 ))0.5] (d) None (ANSWER: a) 12.A complementary shear stress is equal in magnitude and opposite in rotational tendency of an applied (a) Tensile stress (b) Compressive stress (c) Shear stress (d) None (ANSWER:c) 13.All the principal stresses are at an angle of (a) 450 (b) 600 (c) 750 (d) None (ANSWER:d) 14.All the principal stresses are at an angle of (a)900 (b) 450 (g) 1350 (h) None ANSWER(a) 15.All the principal strains are at an angle of (a) 450 (b) 600 (c) 750 (d) None (ANSWER:d) 16.All the principal strains are at an angle oft (a) 450 (b) 900 (c) 1350 (d) None (ANSWER:b) 17.Total number of maximum shear stresses is (a) One (b) Three (c) Five (d) None (ANSWER: b) 18.All the maximum shear stresses are at an angle of (a)450 (b) 900 (c) 1350 (d) None (ANSWER:b) 19.Does a plane of maximum shear stress contain a (a) Normal stress (b) Bending stress (c) Torsional shear stress (d) None (ANSWER: a) 20.Mohr’ s circle is a graphical method to find (a) Bending stresses (b) Bucking stresses (c) Maximum shear stresses (d) None (ANSWER: c) 21.Mohr’ s circle is a graphical method to find (a) Bending stresses (b) Bucking stresses (c) Torsional shear stresses (d) None (ANSWER: d) 22.Mohr’ s circle is a graphical method to find (a) Bending stresses (b) Principal stresses (c) Torsional shear stresses (d) None (ANSWER: b) 23.Mohr’s stress circle method is used to analyze a body under (a) Complex stresses (b) Tensile and compressive stresses (c) Axial and longitudinal stresses (d) None (ANSWER:a) 24.When does Mohr’s stress circle method fail (a) the given two normal stresses are of the same magnitude and same nature (b)the given two normal stresses are of the same magnitude and are of opposite nature (c) the given two normal stresses are of the unequal magnitude and same nature (d) None (ANSWER: a) 25.The abscissa of the Mohr’s circle is a (a) Shear stress (b) Normal stress (c) Normal as well as shear stress (d) None (ANSWER: b) 26.The ordinate of the Mohr’s circle is a (a) Shear stress (b) Normal stress (c) Normal as well as shear stress (d) None (ANSWER: a) 27.The principal strain due to σ1(tensile) and σ2(Compressive ) stress will be (a) (1/E)( σ1 + σ2) (b) (1/E)( σ1 +µ σ2) (c) (1/E)( σ1 -µ σ2) (d) None (ANSWER: b) 28.The principal strain due to σ1 (compressive) and σ2 (tensile) stress will be (a) (1/E)( -σ1 + σ2) (b) (1/E)( -σ1 +µ σ2) (c) (1/E)(- σ1 -µ σ2) (d) None (ANSWER: c) 29.The relation between the elastic constant is (a) E = 2G (1–2µ) (b) E = 2G (1+2µ) (c) E = 2G (1+µ) (d) None (ANSWER:c) 30.The relation between the elastic constant is (a) E = 3 KG /(3K + G) (b) E = 6 KG / (3K + G) (c) E = 9GK /(3K + G) (d) None (ANSWER:c) 31.The relation between the elastic constant is (a) E = 3 K(1 – 2µ) (b) E = 6 K (1 – 2µ) (c) E = 9K(1 – 3µ) (d) None (ANSWER:a)

32.Resilience under principal tensile stresses σ1 and σ2 is (a) (1/2E)( σ12 + σ22 –3µ σ1 σ2) (b) (1/2E)( σ12 + σ22 –4µ σ1 σ2) (c) (1/2E)( σ12 + σ22 –5µ σ1 σ2) (d) None (ANSWER:d) 33.Resilience under principal tensile stresses σ1 and σ2 is (a) (1/2E)( σ12 + σ22 –3µ σ1 σ2) (b) (1/2E)( σ12 + σ22 –4µ σ1 σ2) (c) (1/2E)( σ12 + σ22 –5µ σ1 σ2) (d) None (ANSWER:d) 34.Resilience under principal tensile stresses σ1 and σ2 is (a) (1/2E)( σ12 + σ22 –µ σ1 σ2) (b) (1/2E)( σ12 + σ22 –4µ σ1 σ2) (c) (1/2E)( σ12 + σ22 –2µ σ1 σ2) (d) None (ANSWER:c) 35.Shear strain energy under principal tensile stresses σ1 and σ2 is (a) (1/12E) (σ1 — σ2)2 + σ22— σ12 ) (b) (1/12G) (σ1 — σ2)2 + σ22+ σ12 ) (c) (1/12K) (σ1 — σ2)2 + σ22+ σ12 ) (d) None (ANSWER:b) 36. The angle between normal stress and tangential stress is known as angle of ______ a. declination b. orientation c. obliquity d. rotation ANSWER: obliquity Explanation: No explanation is available for this question! 37. Principal stress is the magnitude of ________ stress acting on the principal plane. a. Normal stress b. Shear stress c. Both a. and b. d. None of the above ANSWER: Normal stress 38. When a component is subjected to axial stress the normal stress σn is maximum, if cos θ is _______ . (σn=σxCos2θ) 1. maximum 2. minimum 3. always one 4. always zero a. 1 and 4 b. 1 and 3 c. 2 and 3 d. 2 and 4 ANSWER: 1 and 3 39. Which of the following stresses can be determined using Mohr's circle method? a. Torsional stress b. Bending stress c. Principal stress d. All of the above ANSWER: Principal stress 40. The graphical method of Mohr's circle represents shear stress (τ) on ______ a. X-axis b. Y-axis c. Z-axis d. None of the above ANSWER: Y-axis 41. In Mohr's circle method, compressive direct stress is represented on ____ a. positive x-axis b. positive y-axis c. negative x-axis d. negative y-axis ANSWER: negative x-axis 42. What is the value of shear stress acting on a plane of circular bar which is subjected to axial tensile load of 100 kN? (Diameter of bar = 40 mm , θ = 42.3o ) a. 58.73 Mpa b. 40.23 Mpa c. 39.60 Mpa d. Insufficient data ANSWER: 39.60 Mpa 43. The maximum tangential stress σt = (σx sin 2θ)/2 is maximum if, θ is equal to ________ a. 45o b. 90o c. 270o d. all of the above ANSWER: 45o 44. Minor principal stress has minimum ________ a. value of shear stress acting on the plane b. intensity of direct stress c. both a. and b. d. none of the above ANSWER: intensity of direct stress 45. Which of the following formulae is used to calculate tangential stress, when a member is subjected to stress in mutually perpendicular axis and accompanied by a shear stress? a. [(σx – σy)/2 ]sin θ – τ cos 2θ b. [(σx – σy)/2 ]– τ cos 2θ c. [(σx – σy)/2 ]sin θ – τ 2 cos θ d. None of the above ANSWER: [(σx – σy)/2 ]sin θ – τ cos 2θ 46-Principal planes are those planes on which a. Normal stress is maximum b. Normal stress is minimum c. Normal stress is either maximum or minimum d. Shear stress is maximum (ANSWER: c) 47-In a general two dimensional stress system, there are a. Two principal planes b. Only one plane c. Three principal planes d. No principal plane (ANSWER: a) 48-Principal planes are mutually inclined at a. 45 degree b. 60 degree c. 90 degree d. 180 degree (ANSWER: c) 49-In a general two dimensional stress system, planes of maximum shear stress are inclined at ___ with principal planes. a. 90 degree b. 180 degree c. 45 degree d. 60 degree (ANSWER: c) 50-In a general two dimensional stress system, the planes on which shear stress is zero a. The normal stress is minimum b. Normal stress is zero c. Normal stress is either maximum or minimum d. None of these (ANSWER: c) 51-Ellipse of stress is used to find a. Resultant stress on any plane in a bi-axial stress system b. Resultant stress on any plane in a general two dimensional system c. Maximum shear stress d. Location of planes of maximum shear stress (ANSWER: a) 52-In Mohr’s circle of strain, y-axis represents a. Shear strain b. Half of shear strain c. Normal strain d. Half of normal strain (ANSWER: b) 53-In a general two dimensional stress system, Ellipse of stress can find a. Principal planes b. Principal stresses c. Planes of maximum shear stress d. Resultant stress on any plane (ANSWER: d)