-
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 = L/L
3. State Hookes law.
It states that when a material is loaded, within its elastic
limit, the stress is directly proportional to the strain.
Stress Strain e
= Ee E = /e unit is N/mm^2
Where,
E - Youngs modulus - 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 Poissons 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 ( or 1/m) = Lateral strain /Longitudinal
strain
6. State the relationship between Youngs Modulus and Modulus of
Rigidity. E = 2G (1+1/m)
Where, E - Youngs Modulus
K - Bulk Modulus 1/m - Poissons ratio
www.Vidyarthiplus.com
www.Vidyarthiplus.com
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 Youngs
Modulus. E = 3K (1-2/m)
Where,
E - Youngs Modulus K - Bulk Modulus 1/m - Poissons 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 Youngs 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
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
UNIT III
35. In case of equal like principle stresses, what is the
diameter of the Mohrs 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 hookes
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 - Youngs
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
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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 Hookes 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 = D4
32
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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=C/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=2NT/60 N-speed in rpm T-torque
56. Write down the expression for torque transmitted by hollow
shaft
T= (/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=/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.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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 (>10o)
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 whals 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
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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 h=pd/2t Longitudinal stress l=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.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
Principle plane: The planes which have no shear stress are known
as principal planes.
99. What is the radius of Mohrs circle? Radius of Mohrs circle
is equal to the maximum shear stress.
100. What is the use of Mohrs 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
= 30 Stress, = Load/Area
= 50*102/600 = 83.33 N/mm2
Normal stress, n = cos2
83.33*cos230 62.5 N/mm2
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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/mm2 ,
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.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
ii) A hollow shaft having the external diameter, twice the
internal diameter, subjected to a pure torque, attains a maximum
shear stress . Show that The strain energy stored per unit volume
of the shaft is 52 / 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/ m2.
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 mm4.
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 Engessers theorem and Castiglianos
theorem.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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/mm2 and I for beam = 105 mm4.
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
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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 Eulers 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 Eulers 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/mm2 . Compare Eulers load with
Rankines critical load. Assume fc =550 N/mm2 and =1/1600. Find the
length of column at which both critical loads are equal.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
4. Derive the Eulers 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
Youngs 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, Poissons 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/mm2. 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 = 1__ and f
c = 320 N/mm2. 7500
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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
Rankinegs 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 Rankines 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 Eulers 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/mm2 and
crushing strength c = 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.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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 its 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 Poissons ratio of 0.25 and modulus of elasticity of
2 x 105 N/mm2. 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: x = 6 MN/m2 , y = 5 MN/m2 , z = 4MN/m2
, xy = xy = 1 MN/ m2 , yz = yz = 3 MN/m2 , zx = xz = 2 MN/m2 Find
the magnitudes of three principal stresses.
6. In a two dimensional stress system, the direct stresses on
two mutually perpendicular planes are and 120 MN/m2. In addition
these planes carry a shear stress of 40 MN/m2. Find the value of 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/mm2, = 0.3.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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 CDs 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.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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 ) [] = ( 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 Eulers approach.
Assume the thickness of the web and flanges to be 20 mm and E = 210
GPa.
www.Vidyarthiplus.com
www.Vidyarthiplus.com
-
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 mm4; Iyy = 0.84 x
106mm4;
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 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/ mm
2 is applied to the compound Cylinder, find the final
stresses
across the thickness. Draw sketches showing their
variations.
www.Vidyarthiplus.com
www.Vidyarthiplus.com