2 Marks Question with Answers - Vidyarthiplus

2 Marks Question with Answers

UNIT -1

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 α Strain

σ α 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 ration. 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 (µ or 1/m) = Lateral strain /Longitudinal strain

6. State the relationship between Young’s Modulus and Modulus of Rigidity. E = 2G (1+1/m)

Where,

E - Young’s Modulus

K - Bulk Modulus 1/m - Poisson’s ratio

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Mechanics Of Solids

7. Define strain energy Whenever a body is strained, some amount of energy is absorbed in the body. The

energy which 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. 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.

13. 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.

14. 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

15. Define – Young’s modulus The ratio of stress and strain is constant with in the elastic limit.

E = Stress Strain

16. Define Bulk-modulus The ratio of direct stress to volumetric strain.

K = Direct stress

Volumetric strain



17. 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.

18. Define- longitudinal strain The strain right angle to the direction of the applied load is called lateral strain.

19. 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

20. Define- Rigidity modulus The shear stress is directly proportional to shear strain.

N = Shear stress

Shear strain

UNIT –II

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. Derive an expression for the longitudinal stress in a thin cylinder subjected to a uniform internal fluid pressure.

Force due to fluid pressure = p x /4 xd2

Force due to longitudinal stress = f2 x d x t

p x /4 xd2 = f2 x d x t f2 = pd/4t



26. 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.

27. When will bending moment is maximum? BM will be maximum when shear force change its sign.

28. 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 29. 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.

30. 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.

31. 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.

32. What are the types of beams?

1. Cantilever beam

2. Simply supported beam

3. Fixed beam

4. Continuous beam

33. What are the types of loads?

1. Concentrated load or point load

2. Uniform distributed load

3. Uniform varying load 34. In which point the bending moment is maximum?

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



UNIT –III

35. In case of equal like principle stresses, what is the diameter of the Mohr’s

circle?

Answer: Zero

36. 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.

37. 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.

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

Vertical and horizontal shear stress

39. 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.

40. Define: Shear stress distribution

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

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

Qmax is 1.5 times the Qave.

42. 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 Qave.

43. What is the maximum value of shear stress for triangular section? Qmax=Fh2/12I



h- Height F-load

44. 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

45. What is the value of maximum of minimum shear stress in a rectangular

cross section? Qmax=3/2 * F/ (bd)

UNIT- IV

49. 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.

50. What are the assumptions made in Torsion equation o The material of the shaft is homogeneous, perfectly elastic and obeys Hooke’s

law.

o Twist is uniform along the length of the shaft

o The stress does not exceed the limit of proportionality o The shaft circular in section remains circular after loading o Strain and deformations are small.

51. Define polar modulus

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

£ = polar moment of inertia = J

Radius R

52. Write the polar modulus for solid shaft and circular shaft. £ = polar moment of inertia = J

Radius R

J = π D4

32



53. 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.

54. 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

55. Write down the expression for power transmitted by a shaft P=2πNT/60 N-speed in rpm T-torque

56. Write down the expression for torque transmitted by hollow shaft T= (π/16)*Fs*((D4-d4)/d4 T-torque q- Shear stress D-outer diameter D- inner diameter

57. Write the polar modulus for solid shaft and circular shaft

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

58. 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 * Fs*D3 T-torque q Shear stress D diameter

59. Define torsional rigidity

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

60. 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.

61. 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.



62. State any two functions of springs. 1 . To measure forces in spring balance, meters and engine indicators. 2 . To store energy.

63. What are the various types of springs?

i. Helical springs

ii. Spiral springs

iii. Leaf springs

iv. Disc spring or Belleville springs

64. Classify the helical springs.

1. Close – coiled or tension helical spring.

2. Open –coiled or compression helical spring.

65. 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.

66. 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 = nt x d

Where, nt = total number of coils.

67. 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 + Ymax + 0.15 Ymax

68. 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

69. 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

70. 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

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

The spring wires are coiled very The wires are coiled such that there

closely, each turn is nearly at right is a gap between the two consecutive

angles to the axis of helix turns.

Helix angle is less than 10o Helix angle is large (>10

o)

72. 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 73. What is whal’s stress factor? C = 4C-1 + 0.615

4C-4 C 74. 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.

75. 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.

76. 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.

77. 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.

78. 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



UNIT IV

79. Define –column Column or strut is defined as a member of a structure, which is subjected to axial compressive load. If the member the structure is vertical and both of its ends are rigidly fixed while subjected to axial compressive load.

80. What are the causes to fail the column?

1. Direct compressive stress

2. Buckling stresses

3. Combined of direct and compressive stresses

81. What is buckling or cripping load?

The load at which the column just buckle is known is buckling load

82. What are the causes to fail the long column? The column fails due to maximum stresses is more than the crushing stresses

83. What are the assumptions made in the Euler theory? 1. The column is initially straight and the load applied axially

2. The cross section of the column is uniformly throughout the length 3. The column material is perfectly elastic, homogeneous and isotropic and obeys hookes

law.

84. List the end conditions of the column? 1. Both the ends of the column is hinged

2. One end is fixed and other end is free

3. Both the end of the column is fixed

4. One end is fixed and other is pinned

85. What is effective length? The effective length of the given column with given and conditions is the length of an equivalent column of the same material and cross section with hinged ends, and having the value of the cripping load equal to the given column.

86. Define - slenderness ratio The ratio of the actual length of a column to the least radiation of gyration of the column.

UNIT-V

87. 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.

88. 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.



89. Distinguish between cylindrical shell and spherical shell. Cylindrical shell Spherical shell

1. Circumferential stress is twice the longitudinal stress. 2. It withstands low pressure than spherical shell for the same diameter. 1. Only hoop stress presents. 2. It withstands more pressure than cylindrical shell for the same diameter.

90. 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. In thin spherical shell, volumetric strain is -------- times the circumferential strain. Three.

91. 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.

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

1. Circumferential stresses (or) hoop stresses

2. Longitudinal stresses

93. Define – hoop stress? The stress is acting in the circumference of the cylinder wall (or) the stresses induced perpendicular to the axis of cylinder.

94. Define- longitudinal stress? The stress is acting along the length of the cylinder is called longitudinal stress.

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

Hoop stress σh=pd/2t Longitudinal stress

σl=pd/4t p- Pressure (gauge) d- Diameter t- Thickness

97. State principle plane. The planes, which have no shear stress, are known as principal planes. These

planes carry only normal stresses.

98. 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.

99. What is the radius of Mohr’s circle? Radius of Mohr’s circle is equal to the maximum shear stress.

100. What is the use of Mohr’s circle? To find out the normal, resultant stresses and principle stress and their planes.

101. List the methods to find the stresses in oblique plane?

1. Analytical method

2. Graphical method

102. 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 mm2 Load, P = 50KN

θ = 30°

Stress, σ = Load/Area = 50*102/600

= 83.33 N/mm2 Normal stress, σn = σ cos2θ

83.33*cos230°

62.5 N/mm2



PART B – 16 MARKS QUESTIONS

UNIT - I

1. i) Derive a relation for strain energy due to shear force.

ii) Derive the relation for minimum defection of a simply

supported beam with uniformity distributed load over entire span.

Use strain energy method.

2. Determine the deflection of the beam given in Fig. Use principal of

virtual work.

W

L/2

A

B C

3. Find the deflection at one third point from left end of the simply

supported beam of span 6 m subjected to uniformly distributed load of

20 kN/m by strain energy principle.

4. Find the deflection at E in the truss showh in Fig. 1. Areas of members

are given in brackets in square cm. Assume E = 2 x 105 N/mm

2 ,

3 m

D

(A)

(6) (6) (6) (6)

(A) E (A)

A 3m 3m B

20 KN

5. i) Derive a relation for strain energy due to torsion.



ii) A hollow shaft having the external diameter, twice the internal

diameter, subjected to a pure torque, attains a maximum shear stress τ . Show that The strain energy stored per unit volume of the shaft is

5τ2 / 16G. Such a shaft Is required to transmit 4500 kW at 110 r.p.m.

with uniform torque, the maximum stress not exceeding 70

MN/m2. Calculate the shaft diameter and the energy stored per m

3 when transmitting this power G =83 GN/ m

2.

6. A simply supported beam of span 8 m carries a udl of 4 kN/m over

the entire span and two point loads of 2 kN at 2 m from each support.

Find the mid-span deflection using strain energy method. E = 200

kN/mm2

, I = 16 x 108 mm

4.

7. (i) Derive the expression for strain energy due to bending.

(ii) Derive the expression for strain energy due to torsion.

8. Calculate the strain energy stored in a cantilever beam of 4 m span,

carrying a point load 10 kN at free end. Take EI = 25000 KNm2.

9. Find the deflection at mild span of a simply supported beam carrying

an uniformly distributed load of 2 KN/m over the entire span using

principal of virtual work. Take span = 5 m; EI = 20000 KN/m2.

10. A simply supported beam of length 10 m is subjected to udl. 10 kN/m

over the left half of the span and a concentrated load 4 kN, 2.5 m

from the right support. Find bending strain energy. Flexural rigidity is

uniform and equal to EI.

11. State and explain the Engesser’s theorem and Castigliano’s theorem.



UNIT - II

1. A simply supported beam of span 10 m carries a uniformly distributed

load 1153 N per unit length. The beam is propped at the middle of the

span. Find the amount by which the prop should yield, in order to make

all the three reactions equal. Take E =2 x 105 N/mm

2 and I for beam

= 105 mm

4.

2. A fixed beam AB of length 6 m carries point loads of 160 KN and 120

KN at a distance of 2 m and 4 m from the left end A. Find the fixed end

moments and the reactions at the supports. Draw BM and SF diagrams.

3. Find the support moments and reactions in a fixed beam of span of 9

m. it is subjected to concentrated loads of 36 KN and 54 KN at 3 m

and 6 m from left Support respectively. Also draw shear force and

bending moment diagrams.

4. A continuous beam ABCD is simply supported at A, B, C and D. AB =

BC =CD = 4 m. Span AB carries a load of 24 KN at 3 m from A.

Span BC carries an uniformly distributed load of 24 KN/m. Span

carries a central concentrated load of 48 KN. Draw shear force and

bending moment diagrams.

5. Draw shear force and bending moment diagram for a simply supported

beam with a uniformly distributed

load over entire span and

propped at the centre. Also derive relations for slope at the ends and

maximum and support reactions.

6. A fixed beam of ACB of span 6 m is carrying a uniformly distributed

load of 4 KN/m over the left half of the span AC. Find the fixing

moments and support reactions.

7. A fixed beam AB of length 6 m carries point loads of 200 kN and 150

kN at distance of 2mm and 4m, respectively from the left support A.

Find the fixed end moments and the reactions as the support. Draw the

BM and SF diagrams.

8. A continuous beam ABC has two spans AB and BC of length 4 m and

6 m simply supported at A, B and C. the beam carries a udl of 60 kN/m



on the span AB and 100 kN/m on the span BC. Determine the support

moments at A, B and C. Draw the BM and SF diagrams. Use theorems

of three moments.

9. Analyse the beam shown in Fig.

60 kN

10 kN/m

A C

B

3m 3m 6m

10. Analyze beam shown in Fig. EI = constant. Draw the bending

moment diagram.

300 kN 300 kN

A B

2m 2m 2m

11. A beam AB of span 8 m fixed at ends carries point loads of 10 KN, 30

kN and 10 kN at 2 m, 5 m and 6 m respectively from the left end. EI =

1.72 x 1010

KN/mm2. Find the fixed end moments at A and B.

Also find the deflection under the loads and maximum deflection.

12. A continuous beam ABC consists of two consecutive spans AB and BC

4 m each and carrying an UDL of 60 KN/m. The end A is fixed and C

is simply supported. Find the support moments by using three moment

equation.

13. Using the theorem of three moments draw the shear force and bending

moment diagrams for the following continuous beam.



4 kN/m 6 kN 8 kN

A B C D

4m 2m 1m 1m 3m

14. Using unit load method, find the vertical deflection of joint F and

horizontal deflection of joint D of the following truss. Axial rigidity

AE is constant for all members.

10 kN D E 20 kN F

4m

A B C

UNIT – III

1. Derive an expression for crippling load when one end of the column

is fixed and the other end is free.

2. Calculated the Euler’s critical load for a strut of T-section. the flange

width being 10 cm, overall depth 8 cm and both flange and stem 1 cm

thick. the strut is 3 m long and is built in at both ends. Take E = 2 x

10 N/mm3.

3. Find the Euler’s critical load for a cast iron hollow column of external

diameter 200 mm diameter, 25 mm thick and of length 6 m hinged at

both ends. E = 0.8 x 104 N/mm

2 . Compare Euler’s load with

Rankine’s critical load. Assume fc =550 N/mm2 and α =1/1600. Find

the length of column at which both critical loads are equal.



4. Derive the Euler’s buckling load for a column with both ends hinged.

5. Find the ratio of buckling strength of a solid column to that of a

hollow column of the same material and having the same cross-

sectional area. the internal diameter of the hollow column is half of

its external diameter. Both the columns are hinged and the same

length.

6. i) A pipe of 400 mm internal diameter and 100 mm thick contains a

fluid at a pressure of 10 N/mm2 Find maximum and minimum hoop

stress across the section. Also sketch the stress distribution.

ii) Find the thickness of steel cylindrical shell of internal diameter

200 mm to withstand an internal pressure of 35 N/mm2. Maximum

hoop stress in the section not to exceed 120 N/mm2.

7. A ‘T’ section 150 mm x 120 mm x 20 mm is used as a strut of 4 m

long with hinged at its both ends. Calculate the Crippling load, if

Young’s modulus for the materials is 200 GPa.

8. A steel cylinder is 1 m inside diameter and is to be designed for an

internal pressure of 8 MN/m2. Calculate the thickness if the maximum

shearing stress is not to exceed 35 MN/m2. Calculate the increase in

volume, due to working pressure, if the cylinder is 6 m long with

closed ends. E = 200 GN/ m2, Poisson’s ratio = 1/3.

9. The internal and external diameters of a thick cylinder are 80 mm and

120 mm, respectively. It is subjected to an external pressure of 40

N/mm2 when the internal pressure is 120 N/mm

2. Find the

circumferential stress at the external and internal surface and

determine the radial and circumferential stresses at the mean radius.

10. Derive the expression for buckling load of a long column fixed at

once end and Hinged at the other end.

11. Find the greatest length of mild steel bar 25 mm x 25 mm in cross-

section which can be used as compression member with one end

fixed and the other end free to carry a working load of 35 kN. Allow

a factor of safety of 4. Take α = 1__ and f c = 320 N/mm2.

7500



12. Find the thickness of metal necessary for a steel cylinder of internal

diameter 200 mm to withstand an internal pressure of 50 N/mm2. The

maximum hoop stress in the section is not to exceed 150 N/mm2.

Assume thick cylinder.

13. Derive the expression for buckling load of a column fixed at one end

and free at the other end.

14. A hollow cylinder cast iron column is 4 m long and fixed at the ends.

Design the column to carry an axial load of 250 KN. Use Rankineg’s

formula and adopt a factor of safety of 5. The internal diameter

may be taken as 0.8 times the external diameter. Taken Fc = 550

N/mm2 and Rankine’s constant 1__

1600

15. A compound tube is composed of 250 mm internal diameter and 25

mm thick Shrunk on tube of 250 mm external diameter and 25 mm

thick. The radial pressure at the function is 8 N/mm2. Find the

variation of hoop stress over the wall of the compound tube.

16. Using Euler’s theory, find the buckling load for the column with

following Boundary conditions: i) Fixed-free ii) Fixed-hinged

17. A column with one end hinged and the other end fixed has a length of

5 m and a hollow cylinder cross section of outer diameter 100 mm

and wall thickness 10 mm. If E = 1.60 x 105 N/mm

2 and crushing

strength σc = 350 N/mm2, find the load that the column may carry

with a factor of Safety of 2.5 according to Euler theory and Rankine-

Gordon theory. If the Column is hinged on both ends, find the safe

load according to the two Theories.

UNIT – IV

1. The normal stress in two mutually perpendicular directions are 600

N/mm2

and 300 N/mm2

both tensile. The complimentary shear

stresses in these directions are of intensity 450 N/mm2 . Find the

normal and tangential stresses in the two planes which are equally

inclined to the planes carrying the normal stresses mentioned above.



2. A solid circular shaft is subjected to a bending moment of 40 KN.m

and a Torque of 10 KN.m Design the diameter of the shaft according

to

(i) Maximum principal stress theory

(ii) Maximum shear stress theory

(iii) Maximum strain energy theory.

3. A rectangular block of size 250 mm x 100 mm x 80 mm is subjected

to axial loads as follows. A tensile force of 480 kN in the direction of

it’s length. A tensile force of 900 KN in 250 mm x 80 mm faces. A

compressive force of 1000 kN in 250 mm x 100 mm faces. Assuming

Poisson’s ratio of 0.25 and modulus of elasticity of 2 x 105 N/mm

2.

Find strain and change in length in each direction. Also find

volumetric strain and change in volume.

4. A rectangular block is subjected to a tensile stress of 100 N/mm2 on

one plane and 50 N/mm2 on a plane at right angles together with a

shear stress of 60 N/mm2 on the same plane. Find the direction of

principal plan, magnitude and Nature of principal stresses. Also find

the maximum shear stress.

5. In a triaxial stress system, the six components of the stress at a point

are given below: σx = 6 MN/m2

, σy = 5 MN/m2

, σz = 4MN/m2

,

τxy = τxy = 1 MN/ m2

, τyz = τyz = 3 MN/m2 , τzx = τxz = 2 MN/m

2

Find the magnitudes of three principal stresses.

6. In a two dimensional stress system, the direct stresses on two

mutually perpendicular planes are σ and 120 MN/m2. In addition

these planes carry a shear stress of 40 MN/m2. Find the value of σ at

which the shear strain energy is least. If failure occurs at this value of

the shear strain energy, estimate the elastic limit of the material in

simple tension. Take the factor of safety on elastic limit as 3.

7. A steel drum 600 mm diameter is required to hold gas under a

pressure of 3.5N/ mm2. Calculate the thickness of the drum required

according to (i) maximum principal stress theory and (ii) maximum

shear strain energy. The allowable tensile strength of steel is 120

N/mm2. E = 200 kN/mm

2, µ = 0.3.



8. A simply supported wooden beam 3 m long supports a total udl of 8

kN. The Cross-section of the beam is 100 mm x 150 mm. The applied

load acts in a plane making an angle of 300 with the vertical plane.

Find the maximum bending stress developed in the beam. Self weight

of the beam may be neglected.

UNIT - V

1. A curved bar is formed of a tube of 120 mm outside diameter and 7.5

mm Thickness. The centre line of this beam is a circular arc of radius

225 mm. A bending moment of 3 KNm tending to increase curvature

of the bar is applied. Calculate the maximum tensile and compressive

stresses set up in the bar.

2. A 80 mm x 80 mm x 10 mm angle section shown in fig is used as a

simply supported beam over a span 2.4 m. It carries a load of 400 N

along the line YG, where G is the centroid of the section. Calculate

the stresses at the points A, B and C of the mild section of the beam

Stresses at the points, A, B and C of the mild section of the beam.

Deflection of the beam at the mild section and its direction with the

Load line. Position of the neutral axis. Take E = 200 GN/m2

10 mm Y

A

X G X

80 mm

10 mm

B C

K = 80 mm

3. Find the principal moment of inertia of angle section 60 mm x 40 mm

x 6 mm.



4. A rectangular section of 80 mm wide and 120 mm deep is subjected

to a bending moment of 12 KNm. The trace of plane of loading is

inclined at 450 to YY axis of the section. Locate neutral axis and

find the maximum stress induced in the section.

5. Find the centroidal principal moments of inertia of a equal angle

section 30 x30x10 mm.

6. An equal angle section 150 mm x 150 mm x 10 mm is used as a

simply supported beam of 4 m length is subjected to a vertical load

passing through the centroid. Determine bending stress at point A as

shown in Fig.

10mm

A

150 mm

10 mm

B C

150 mm

7. An I section of a beam consists of top flange 140 mm x 40 mm and

bottom flange 140 mm x 40 mm. The web is 20 mm x 220 mm. The

centre line of the web is 80 mm from the left edge of the flanges and

60 mm from the right edges of the flanges. Determine the position of

shear centre for the beam.

8. A three span beam ABCD has spans AL = 6 m, BC = 5 m and CD = 4

m. All the supports are at same level and also simple supports. The

spans AB and BC are loaded with 9 kN and 8 kN respectively at 2 m

from A in span AB and B in span BC. The span CD’s carrying a UDL

of 3 kN/m. If EI is constant throughout analyse the continuous beam

using theorem of three moment and draw the BMD.



9. Find the product moment of inertia of a quadrant of circle about the

Perpendicular axes OX and OY as shown in Fig.

Y

O

X

R

10. Find the centroidal principal moments of inertia of an equal angle

section 30 mm x 30 mm x 10 mm.

11. Determine the principal stresses and principal directions for the

following 3D-stress field.

( 30 15 20 )

[σ] =

( 15 20 25 )

Mpa.

( 20 25 40 )

12. A beam of constant M.I. spans over 8 m. It is subjected to point loads

5 kN and 10 kN respectively at left and right quarter points. If E for

the beam material is 210 GPa and the permissible deflection at 10 kN

load is 25 mm, find the moment of inertia of the beam section by

using energy methods.

13. A Overhanging beam ABC with simple supports at A and B 5 m

apart over hangs by 2 m. It is subjected to a point load of 10 kN

acting at the free end C calculated the slopes at A and B.

14. RSJ 400 x 200 mm is used as a strut with fixed ends for a length of 6

m. Find the crippling load using Euler’s approach. Assume the

thickness of the web and flanges to be 20 mm and E = 210 GPa.



15. A built column made up of two channels ISJC 200 x 75 mm Back to

Back at 100 m was two plates 250 x 10 mm attached on either side.

For the channel section Ixx =11.6 x 106 mm

4; Iyy = 0.84 x 10

6mm

4;

A = 1777 mm2, Cxx= 19.7 mm and the crushing stress is 300 MPa. If

the f the column is 6 m and the ends fixed, find the safe load.

Rankines constant is 1/7500.

16. A steel bar of section 60 x 40 mm is arranged as a cantilever

projecting Horizontally 0.6 m beyond a support. The wider face of

the section makes 300 with the horizontal. A load of 500 N is hung

from the free end. Locate the Neutral axis and find the orizontal and

vertical deflections of the free end. Also find the maximum tensile

stress. Take E = 2 x 105 MPa.

17. A closed ring made up of 40 mm Ǿ steel bar carries a pull of 20 kN,

the line of action of which passes through its centre. The mean radius

of the ring is 10 cm. Find the extreme fiber stresses in the ring.

18. Locate the shear centre for a channel section used with its web

vertical. The size of the channel is 200 x 100 mm with 10 mm

uniform thickness. Also draw the shear flow diagram.

19. A shrunk cylinder consists of an inner cylinder of 180 mm and outer

diameter 200 mm and outer cylinder of external diameter of 240 mm

and thickness 20 mm. The pressure due to shrinking 8 MPa. If an

external of pressure of 60 MPa acts find the resultant stresses across

the wall.

20. A semicircular bar of circular cross section with radius 20 mm is

fixed at one loaded at the other end as shown in the figure. Find the

stresses at points A and B.

21. A cylinder of outer diameter 280 mm and inner diameter 240 mm

shrunk over Another cylinder of outer diameter slightly more than

240 mm and inner diameter 200 mm to form a compound cylinder.

The shrink fit pressure is 10 N/ mm2. If an internal pressure of 50 N/

mm2 is applied to the compound Cylinder, find the final stresses

across the thickness. Draw sketches showing their variations.



2 Marks Question with Answers - Vidyarthiplus

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