**Q 1. Define stress.**

When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The magnitude of the resisting force is numerically equal to the applied force. This internal resisting force per unit area is called stress.

Stress = Force/Area

**2. Define strain**

When a body is subjected to an external force, there is some change of dimension in the body. Numerically the strain is equal to the ratio of change in length to the original length of the body.= P/A unit is N/mm^{2}

Strain = Change in length/Original length

e = δL/L

**3. State Hooke’s law.**

It states that when a material is loaded, within its elastic limit, the stress is directly proportional to the strain.

Stress ∝ Strain

σ ∝ e

σ = Ee

E = σ/e unit is N/mm^{2}

Where,

E - Young’s modulus

σ - Stress

e - Strain

**4. Define shear stress and shear strain.**

The two equal and opposite force act tangentially on any cross sectional plane of the body tending to slide one part of the body over the other part. The stress induced is called shear stress and the corresponding strain is known as shear strain.

**5. Define Poisson’s ratio.**

When a body is stressed, within its elastic limit, the ratio of lateral strain to the longitudinal strain is constant for a given material.

Poisson’ ratio (ì or 1/m) = Lateral strain /Longitudinal strain

**6. State the relationship between Young’s Modulus and Modulus of Rigidity.**

E = 2K (1+1/m)

Where,

E - Young’s Modulus

K - Bulk Modulus

1/m - Poisson’s ratio

**7. Define strain energy**

Whenever a body is strained, some amount of energy is absorbed in the body. The energy that is absorbed in the body due to straining effect is known as strain energy.

**8. What is resilience?**

The total strain energy stored in the body is generally known as resilience.

**9. State proof resilience**

The maximum strain energy that can be stored in a material within elastic limit is known as proof resilience.

**10. Define modulus of resilience**

It is the proof resilience of the material per unit volume Modulus of resilience = Proof resilience / Volume of the body

**11. Give the relationship between Bulk Modulus and Young’s Modulus.**

E = 3K (1-2/m) Where, E - Young’s Modulus K - Bulk Modulus 1/m - Poisson’s ratio

**12. Define- Rigidity modulus**

The shear stress is directly proportional to shear strain. N = Shear stress Shear strain

**13. What is compound bar?**

A composite bar composed of two or more different materials joined together such that system is elongated or compressed in a single unit.

**14. What you mean by thermal stresses?**

If the body is allowed to expand or contract freely, with the rise or fall of temperature no stress is developed but if free expansion is prevented the stress developed is called temperature stress or strain.

**15. Define- elastic limit**

Some external force is acting on the body, the body tends to deformation. If the force is released from the body its regain to the original position. This is called elastic limit

**16. Define – Young’s modulus**

The ratio of stress and strain is constant with in the elastic limit.

E = Stress / Strain

**17. Define Bulk-modulus**

The ratio of direct stress to volumetric strain.

K = Direct stress / Volumetric strain

**18. Define- lateral strain**

When a body is subjected to axial load P. The length of the body is increased. The axial deformation of the length of the body is called lateral strain.

**19. Define- longitudinal strain**

The strain right angle to the direction of the applied load is called lateral strain.

**20. What is principle of super position?**

The resultant deformation of the body is equal to the algebric sum of the deformation of the individual section. Such principle is called as principle of super position

**21. Define point of contra flexure? In which beam it occurs?**

Point at which BM changes to zero is point of contra flexure. It occurs in overhanging beam.

**22. What is mean by positive or sagging BM?**

BM is said to positive if moment on left side of beam is clockwise or right side of the beam is counter clockwise.

**23. What is mean by negative or hogging BM?**

BM is said to negative if moment on left side of beam is counterclockwise or right side of the beam is clockwise.

**24. Define shear force and bending moment?**

SF at any cross section is defined as algebraic sum of all the forces acting either side of beam. BM at any cross section is defined as algebraic sum of the moments of all the forces which are placed either side from that point.

**25. What is meant by transverse loading of beam?**

If load is acting on the beam which is perpendicular to center line of it is called transverse loading of beam.

**26. When will bending moment is maximum?**

BM will be maximum when shear force change its sign.

**27. What is maximum bending moment in a simply supported beam of span ‘L’ subjected to UDL of ‘w’ over entire span**

Max BM =wL2/8

**28. In a simply supported beam how will you locate point of maximum bending moment?**

The bending moment is max. when SF is zero. Write SF equation at that point and equating to zero we can find out the distances ‘x’ from one end .then find maximum bending moment at that point by taking all moment on right or left hand side of beam.

**29. What is shear force?**

The algebric sum of the vertical forces at any section of the beam to the left or right of the section is called shear force.

**30. What is shear force and bending moment diagram?**

It shows the variation of the shear force and bending moment along the length of the beam.

**31. What are the types of beams?**

1. Cantilever beam

2. Simply supported beam

3. Fixed beam

4. Continuous beam

**32. What are the types of loads?**

1. Concentrated load or point load

2. Uniform distributed load

3. Uniform varying load

**33. In which point the bending moment is maximum?**

When the shear force change of sign or the shear force is zero

**34. Write the assumption in the theory of simple bending?**

1. The material of the beam is homogeneous and isotropic.

2. The beam material is stressed within the elastic limit and thus obey hooke’s law.

3. The transverse section which was plane before bending remains plains after bending also.

4. Each layer of the beam is free to expand or contract independently about the layer, above or below.

5. The value of E is the same in both compression and tension.

**35. Write the theory of simple bending equation?**

M/ I = F/Y = E/R

M - Maximum bending moment

I - Moment of inertia

F - Maximum stress induced

Y - Distance from the neutral axis

E - Young’s modulus

R - Constant.

**36. What types of stresses are caused in a beam subjected to a constant shear force ?**

Vertical and horizontal shear stress

**37. State the main assumptions while deriving the general formula for shear stresses**

The material is homogeneous, isotropic and elastic

The modulus of elasticity in tension and compression are same.

The shear stress is constant along the beam width

The presence of shear stress does not affect the distribution of bending stress.

**38. Define: Shear stress distribution**

The variation of shear stress along the depth of the beam is called shear stress distribution

**39. What is the ratio of maximum shear stress to the average shear stress for the rectangular section?**

Qmax is 1.5 times the Qavg.

**40. What is the ratio of maximum shear stress to the average shear stress in the case of solid circular section?**

Qmax is 4/3 times the Qavg.

**41. What is the maximum value of shear stress for triangular section?**

Qmax=Fh2/12I

h- Height

F-load

**42. What is the shear stress distribution value of Flange portion of the I-section?**

q= f/2I * (D2/4 - y) D-depth y- Distance from neutral axis

**43. What is the value of maximum of minimum shear stress in a rectangular cross section?**

Qmax=3/2 * F/ (bd)

**44. Define Torsion**

When a pair of forces of equal magnitude but opposite directions acting on body, it tends to twist the body. It is known as twisting moment or torsional moment or simply as torque.

Torque is equal to the product of the force applied and the distance between the point of application of the force and the axis of the shaft.

**45. What are the assumptions made in Torsion equation**

• The material of the shaft is homogeneous, perfectly elastic and obeys Hooke’s law.

• Twist is uniform along the length of the shaft

• The stress does not exceed the limit of proportionality

• The shaft circular in section remains circular after loading

• Strain and deformations are small.

**46. Define polar modulus**

It is the ratio between polar moment of inertia and radius of the shaft.

£ = polar moment of inertia/Radius = J/ R

**47. Write the polar modulus for solid shaft and circular shaft.**

£ = polar moment of inertia/Radius = J/ R J = π D^{4}/32

**48. Why hollow circular shafts are preferred when compared to solid circular shafts?**

• The torque transmitted by the hollow shaft is greater than the solid shaft.

• For same material, length and given torque, the weight of the hollow shaft will be less compared to solid shaft.

**49. Write torsional equation**

T/J=Cθ/L=q/R

T-Torque

J- Polar moment of inertia

C-Modulus of rigidity

L- Length

q- Shear stress

R- Radius

**50. Write down the expression for power transmitted by a shaft**

P=2πNT/60

N-speed in rpm

T-torque

**51. Write down the expression for torque transmitted by hollow shaft**

T= (π/16)*F_{s}*((D^{4}-d^{4})/d^{4}

T-torque

q- Shear stress

D-outer diameter

D- inner diameter

**52. Write the polar modulus for solid shaft and circular shaft**

It is ratio between polar moment of inertia and radius of shaft

**53. Write down the equation for maximum shear stress of a solid circular section in diameter ‘D’ when subjected to torque ‘T’ in a solid shaft shaft.**

T=π/16 * F_{s}*D^{3}

T-torque

q Shear stress

D diameter

**54. Define torsional rigidity**

Product of rigidity modulus and polar moment of inertia is called torsional rigidity

**55. What is composite shaft?**

Some times a shaft is made up of composite section i.e. one type of shaft is sleeved over other types of shaft. At the time of sleeving, the two shaft are joined together, that the composite shaft behaves like a single shaft.

**56. What is a spring?**

A spring is an elastic member, which deflects, or distorts under the action of load and regains its original shape after the load is removed.

**57. State any two functions of springs.**

1 . To measure forces in spring balance, meters and engine indicators.

2 . To store energy.

**58. What are the various types of springs?**

i. Helical springs

ii. Spiral springs

iii. Leaf springs

iv. Disc spring or Belleville springs

**59. Classify the helical springs.**

1. Close – coiled or tension helical spring.

2. Open –coiled or compression helical spring.

**60. What is spring index (C)?**

The ratio of mean or pitch diameter to the diameter of wire for the spring is called the spring index.

**61. What is solid length?**

The length of a spring under the maximum compression is called its solid length. It is the product of total number of coils and the diameter of wire.

Ls = n_{t} x d

Where, n_{t} = total number of coils.

**62. Define free length.**

Free length of the spring is the length of the spring when it is free or unloaded condition. It is equal to the solid length plus the maximum deflection or compression plus clash allowance.

Lf = solid length + Y_{max} + 0.15 Y_{max}

**63. Define spring rate (stiffness).**

The spring stiffness or spring constant is defined as the load required per unit deflection of the spring.

K= W/y

Where W -load

Y – deflection

**64. Define pitch.**

Pitch of the spring is defined as the axial distance between the adjacent coils in uncompressed state. Mathematically

Pitch=free length / n-1

**65. Define helical springs.**

The helical springs are made up of a wire coiled in the form of a helix and is primarily intended for compressive or tensile load

**66. What are the differences between closed coil & open coil helical springs?**

The spring wires are coiled very closely, each turn is nearly at right angles to the axis of helix | The wires are coiled such that there is a gap between the two consecutive turns. |

Helix angle is less than 10^{o} |
Helix angle is large (>10^{o}) |

**67. What are the stresses induced in the helical compression spring due to axial load?**

1. Direct shear stress

2. Torsional shear stress

3. Effect of curvature

**68. What is buckling of springs?**

The helical compression spring behaves like a column and buckles at a comparative small load when the length of the spring is more than 4 times the mean coil diameter.

**69. What is surge in springs?**

The material is subjected to higher stresses, which may cause early fatigue failure. This effect is called as spring surge.

**70. Define active turns.**

Active turns of the spring are defined as the number of turns, which impart spring action while loaded. As load increases the no of active coils decreases.

**71. Define inactive turns.**

An inactive turn of the spring is defined as the number of turns which does not contribute to the spring action while loaded. As load increases number of inactive coils increases from 0.5 to 1 turn.

**72. What are the different kinds of end connections for compression helical springs?**

The different kinds of end connection for compression helical springs are

a. Plain ends

b. Ground ends

c. Squared ends

d. Ground & square ends

**73. When will you call a cylinder as thin cylinder?**

A cylinder is called as a thin cylinder when the ratio of wall thickness to the diameter of cylinder is less 1/20.

**74. In a thin cylinder will the radial stress vary over the thickness of wall?**

No, in thin cylinders radial stress developed in its wall is assumed to be constant.

since the wall thickness is very small as compared to the diameter of cylinder.

**75. Distinguish between cylindrical shell and spherical shell.**

Cylindrical shell | Spherical shell |

Circumferential stress is twice the longitudinal stress. | Only hoop stress presents. |

It withstands low pressure than spherical shell for the same diameter. | It withstands more pressure than cylindrical shell for the same diameter. |

**76. What is the effect of riveting a thin cylindrical shell?**

Riveting reduces the area offering the resistance. Due to this, the circumferential and longitudinal stresses are more. It reduces the pressure carrying capacity of the shell.

**77. What do you understand by the term wire winding of thin cylinder?**

In order to increase the tensile strength of a thin cylinder to withstand high internal pressure without excessive increase in wall thickness, they are sometimes pre stressed by winding with a steel wire under tension.

**78. What are the types of stresses setup in the thin cylinders?**

1. Circumferential stresses (or) hoop stresses

2. Longitudinal stresses

**79. Define – hoop stress?**

The stress is acting in the circumference of the cylinder wall (or) the stresses induced perpendicular to the axis of cylinder.

**80. Define- longitudinal stress?**

The stress is acting along the length of the cylinder is called longitudinal stress.

**81. A thin cylinder of diameter d is subjected to internal pressure p . Write down the expression for hoop stress and longitudinal stress.**

Hoop stress

σ_{h}=pd/2t

Longitudinal stress

σ_{l}=pd/4t

p- Pressure (gauge)

d- Diameter

t- Thickness

**82. State principle plane.**

The planes, which have no shear stress, are known as principal planes. These planes carry only normal stresses.

**83. Define principle stresses and principle plane.**

Principle stress: The magnitude of normal stress, acting on a principal plane is known as principal stresses.

Principle plane: The planes, which have no shear stress, are known as principal planes.

**84. What is the radius of Mohr’s circle?**

Radius of Mohr’s circle is equal to the maximum shear stress.

**85. What is the use of Mohr’s circle?**

To find out the normal, resultant stresses and principle stress and their planes.

**86. List the methods to find the stresses in oblique plane?**

1. Analytical method

2. Graphical method

**87. A bar of cross sectional area 600 mm ^{2} is subjected to a tensile load of 50 KN applied at each end. Determine the normal stress on a plane inclined at 30° to the direction of loading.**

A = 600 mm

Load, P = 50KN

θ = 30°

Stress, σ = Load/Area

= 50*102/600

= 83.33 N/mm

Normal stress, σ

= 83.33*cos2*30°

= 62.5 N/mm

Answer: Zero

Force due to fluid pressure = p x π/4 xd

Force due to longitudinal stress = f

p x π/4 xd

f

Three.