SOM PTU QUESTION PAPER SOLUTION CLASS NOTES
SOM PTU QUESTION PAPER
SOLUTION CLASS NOTES
PTU SOM Question answers improves
clarity of the subject. It also increases
the confidence in handling practical
problems. It helps in the selection of
material for a certain application. It is the
ultimate aim of the subject.
2-5 marks question answers
(a). What is slenderness ratio? State the limitations of Euler’s formula.
ANS: slenderness ratio = k = Le/k min
Where Le is the effective length
k min is the minimum radius of gyration
Firstly Le = L for both ends hinged column
Secondly Le =2 L for a cantilever column (One end fixed and the other end is free)
Thirdly Le = L/ for one end hinged and the other end fixed column
Fourthly Le = L/2 for both ends fixed column
k min = I min/A
LIMITATIONS OF THE EULER FORMULA
The column is initially straight.
The load applied is truly axial.
In reality these assumptions cannot be true. Thus these are the limitations of Euler’s formula.
(b). Differentiate between the Centroid and the center of gravity.
ANS; Centroid is the center for a 2-dimensional figure like a rectangle, square, circle, triangle, ellipse etc.
Center of gravity is the center of a 3-dimensional object. At the center of gravity the whole weight of the object acts. For example we talk of center of gravity for a chair, table, television, any object.
(c) What is the practical application of the quantity ‘Moment of inertia’.
ANS: It’s symbol is “I”. It is the second moment of the area about a centroidal axis. Its units are m4.
I = ∫dA y2
It is a resistance to bending. If moment of inertia is high, bending will be less.
(d) What is the radius of gyration of an area and what is the practical use of it?
ANS: Radius of gyration is an imaginary radius about which the whole body is supposed to rotate.
It is property of an area. It’s symbol is “k”. k = (I/A)0.5. Its least value is used to calculate the maximum buckling load for a long and medium column.
(e) Define the moment of inertia of an area about an axis lying in the plane of the area.
ANS: It is the second moment of the area. Mathematically
I = dA y2
I xx = (1/12) bd3 for a rectangle with width b and depth d.(xx is the axis parallel to width b.)
I yy = (1/12) db3 for a rectangle with width b and depth d. (y axis is the axis parallel to depth ‘d’.)
I xx = I yy = d4 for a circular area.
(f) Write the assumptions made in the theory of bending.
Beam is initially straight.
Beam is homogeneous i.e. composition is same throughout.
Material is continuous i.e. no voids or no foreign material is present.
Beam is isotropic material i.e. properties in x, y, z directions are same.
A transverse plane (perpendicular to the axis) remains a transverse plane before and after the bending moment is applied.
A constant bending moment is applied on the beam.
No shear force is acting on the beam
(h) What is Torsional rigidity and Polar modulus?
ANS: Torsion equation is
T/J = τ/r = Gϴ/L
Torsional rigidity is the quantity which produces minimum angle of twist for a given torque.
Hence GJ is torsional rigidity,
greater GJ, lesser will be ϴ.
J is the polar modulus, J solid shaft = d4 m4
J hollow shaft =do4 –di4) m4
(i) How failure of the short and long column takes place?
Failure of short column is by contraction.
Failure of long columns is by buckling.
Buckling is bending due to axial load.
(j) What is crippling load?
ANS: It is a load for a long column. If the value of the external load is greater than the crippling load, the column will buckle.
Pcr = 2E I min/Le2+
Le is the effective length
(a).What is Factor of safety?
For Ductile materials FOS = σyp /σall
For Brittle Materials FOS = σult /σall
σyp = yield point stress
It is possible only for ductile materials. Its value is very near to the value of proportional limit stress.
σall = allowable stress = 1/3 to 2/3 of elastic limit stress
σult = ultimate stress
(b) How stress concentrations affect the life of the material?
Stress concentration reduces the life of a brittle material because the failure will be sudden and unexpected due to crack propagation.
(c)How fatigue strength of a shaft can be improved?
By relieving the residual stresses by annealing.
(d) A load of 400 N has to be raised at the end of a steel wire. If the stress in the wire must not exceed 80 MPa, what will be the diameter of the wire?
Area =( /4)d2 = load/stress=400/80 = 5
(e)What is modulus of rupture?
It is energy required to fracture a material under an impact load.
(h) Give the relationship between modulus of rigidity and Young’s modulus.
E = 2G(1+μ)
Where μ is the Poisson’s ratio, E=Young’s modulus and G= Modulus of rigidity
(i) What is the principal stress and a principal plane?
Principal stress is normal stress with a zero shear stress.
Plane on which there is only normal stress with zero shear stress is called the principal plane.
(j) Write maximum strain energy theory.
(1/2E)(σ12+σ22–2μσ1σ2) σall2/ 2E
Q. What is the energy of distortion or shear strain energy?
ANS: Every object in universe is under a number of forces. Thus, it is under complex loading. From the complex loading, one can determine the three principal stresses σ1, σ2 and σ3 respectively. ASSUME
(i) these principal stresses as tensile and
(ii) σ1> σ2 > σ3
Then energy of distortion OR shear strain energy is
usse=(1/12G)[ (σ1–σ2)2 + (σ2—σ3)2 + (σ3—σ1)2]
NOTE: if any of σ1, σ2 and σ3 is not tensile, then respective sign must be used in the above relation.
Q. Explain the necessity of the theories of failure.
ANS: When identical jobs (Same shape and same size) of different materials are tested in the lab under identical loading, it is found that failure of different materials take place differently. Therefore we need a way to explain these different types of failures under identical loading. Then only these theories of failure help us. Thus these are needed to explain different failures under identical testing.
Q. Define stiffness of a spring or define a spring constant.
ANS: Spring is a highly elastic body. Force required to cause unit deflection is called spring stiffness.
K = W/δ N/mm
Where k is the stiffness
W is the load acting on the spring
δ is the deflection under the load W
Q. For what purpose, cylindrical and spherical vessels are used?
ANS: Cylindrical vessel are used as process vessels as boilers for producing steam.
These vessels are used as storage vessels as LPG cylinders and Oxygen cylinders.
Spherical vessels are used only for storage vessels as these can store more volume in a given space.
Q. State Lame’s equations.
ANS: Lame’s equations are for stresses in a thick vessel. The three stresses are hoop stress
σh, radial stress, σr and longitudinal stress σl
Lame’s equations are
σh =α + β/r2
σr =α — β/r2
Where α and β are Lame’s constant. These are found from the boundary conditions like
use σr =pi at r = ri
and σr =po at r = r0
σl =α and σl is negligible.
Q. List the types of stresses produced in a rotating disc of uniform thickness.
ANS. Variable hoop and radial stresses are produced in a rotating disc of uniform thickness.
Q What is a trapezoidal section? Why is this used in initially curved beams as in a hook of a crane?
ANS: A trapezoidal section is a quadrilateral with two parallel and two non- parallel sides. This is used in a crane hook
because its area is variable which makes stress nearly uniform in the innermost and outermost fibers
Q. Where does the maximum shear stress occurs in an I section?
ANS: It occurs in the center of the web section of the I section.
Q. Explain the importance of the full length leaf in a leaf spring.
ANS: One full length leaf is necessary for mounting of the spring on the vehicle. Any extra full length leaf takes the transverse shear force which is maximum at the ends.
Q. What is the purpose compound cylinders?
ANS: These are used to resist high pressures. It is difficult to make a cylinder from a single thick plate. It is made with number of thin cylinders by hoop shrinking. Such a thick vessel made from thin vessels is called a compound cylinder.
Q. Distinguish between longitudinal and hop stresses in a thick cylinder.
This stress is in the length direction.
Hoop stress is in the tangential direction.
The magnitude of this stress is constant.
Magnitude of hoop stress is variable.
Symbol of this stress is σl.
Its symbol is σh.
Magnitude of this stress is quite small.
Its magnitude is quite large. It is greater than the fluid pressure
This stress is tensile.
It is also tensile.
Q. For the crane hook, locate the plane which is severely stressed.
ANS: The innermost curve of the crane hook is severely stressed.
Q. Define strain energy.
ANS: When an external force acts on a body, strain is produced in the body. Thus force moves through some distance. Hence the work is done. This work is stored in the body as long as it is strained. This work is strain energy. Its symbol is U. Its units are Joules. It is given by the formula
U =(1/2) F δl = σ2x volume/2E=( σ2/2E)x Volume
SECTION- Numerical question answers
Q.2 A hollow steel cylinder of 300 mm length, 150 mm inside diameter and 3mm uniform wall thickness. It is filled with concrete. It is compressed between two rigid parallel plates at the ends by a load of 600 k N. Find the compressive stress in the each material. Also find total shortening of the cylinder.
Young’s modulus of steel =2 x 105 N/mm2
Young’s modulus of concrete is = 2 x 104 N/mm2.
Area of steel = As= ( /4)(1562 –1502) = 1836 mm2
Area of concrete =Ac=( /4)(1502) = 17671.4586764 mm2
σs = P Es /(As Es+ Ac Ec)
= 600 x2x105/( 1836x2x105 + 17671.4586764x 2 x 104 )
= 8.71 N/mm2
σc = P Ec /(As Es+ Ac Ec)
=600 x2x104/( 1836x2x105 +17671.4586764x 2 x 104) = 0.871 N/mm2
Q.3 Two principal stresses at a point in a bar is 200 N/mm2 (tensile) and 100 N/mm2 (compressive). Determine the resultant stress the magnitude and direction on a plane. Plane is inclined at 600 to the axis of the major principal stress. Also determine the magnitude of the maximum shear stress.
ANS: σR =180.27 N/mm2, θ=4606’, (σt)max =150 N/mm2.
σn=σ1cos2θ + σ2sin2θ provided σ1and σ2 are tensile.
σn=200cos2300 — σ2sin230=200 x 3/4—100×1/4=125
τθ= (σ1—σ2)sin600/2=(200-(–100))x 1/4=75
σR=( σn2 + τθ2)0.5=177 N/mm2
tan α = τθ/ σn =75/125=0.6
(σt)max= (σ1—σ2)/2=(200—(-100))/2=150 N/mm2
4. Two principal stresses at a point in a bar are 200 N/m m 2 (tensile) and 100 N/m m 2 (compression). Determine the resultant stress in magnitude and direction on a plane inclined at 600 to the axis of the major principal stress. Also determine the maximum intensity of the shear stress.
Firstly σR =180.27 N/m m 2, θ=4606’, (σt)max =150 N/m m 2.
Secondly σn=σ1cos2θ + σ2sin2θ provided σ1and σ2 are tensile.
Thirdly σn=200 cos2300 – σ2 sin230 =200 x 3/4—100 x 1/4=125
τθ= (σ1—σ2) sin600/2=(200-(-100))x 1/4=75
σR=( σn2 + τθ2)0.5=177 N/mm2
tan α = τθ/ σn =75/125=0.6
(σt)max= (σ1—σ2)/2=(200—(-100))/2=150 N/mm2
5. A timber beam 0.3 m depth is symmetrical. It is simply supported at the ends with a span of 8 m. What uniformly distributed load (including its own weight) it can carry if the maximum permissible stress is 8 N/ m m 2. The moment of inertia of the section of the beam is 450 x 106 m m 4. Find the maximum bending stress and the radius of curvature at section 1 m from a support. Modulus of elasticity for timber = 12.6 x 103 N/ m m 2.
M/I = σ/y=σmax/ y max
We get w L2/8I =8/150
w x 8000 x 8000/8 x 450 x 106 = 8/150
ANS: w = 3 k N/m
M 1m from the support = [(wL/2)x1 – wx1x0.5] x 106 N mm
= [(3×8/2)x1—3x1x0.5] x 106 N mm
=10.5×106 N mm
Using M/I = σ/y
10.5×106/450 x 106 = σ/150
σ= 3.17 N/ mm2
© Using M/I = E/R
10.5×106/450 x 106 = 12.6 x 103/R
R =540 m
6. A solid circular shaft transmits 75 kW power at 200 RPM. Calculate the shaft diameter. The twist in the shaft is not to exceed 1 degree in a length of 2 meters. Shear stress is limited to 50 MPa. Take modulus of rigidity = 100 x 103 MPa.
P =75 kW= 75000 W
N = 200 RPM
L = 2 m = 2000 mm
τ= 50 MPa
G = 100 x 1000 MPa
Power = 2 NT/60 = 2 200 T/60
T = 3.5809 x 106 N mm
Diameter from strength
We know T/J = τ/r
T=(π/16) d3 τ
3.5809 x 106 =(π/16) d3 50
d = 72 mm
Diameter from stiffness (angle of twist)
T/J = Gϴ/l
J =(π/32)d4 = Tl/Gϴ = 3.5609 x 106 x 2000/(100 x 103 x 0,01745)
d =80.4 mm
ANS larger diameter = 80.4 mm
Q.7 A short concrete column of 250 mm x 250 mm is under an axial load of 300 k N. The column is reinforced with steel bars. The total area of steel is 5600 mm2. If the modulus of steel is 15 times the modulus of concrete, find the stresses in concrete and steel. If the stress in the concrete is not to exceed 4 N/mm2, find the area of steel required so that the column may support a load of 600 k N.
A column=250 x 250, A steel=5600 mm2, A con = 56900 mm2,
E steel=15 Econ
σcon=P Econ/(Es As + Ec Ac) =300 x 1000 x Econ/(15 Econ5600 +56900 Econ) =2.1 N/mm2
σsteel=PE steel/(EsAs + EcAc) =300 x 1000 x 15 Econ/(15 Econ5600 +56900 Econ) =31.5 N/mm2
(B) given σcon=4 N/m m 2,
Using σcon=PE con/(E s As + E c Ac)=600 x 1000 x Econ/(15E con As +(62500–As) Econ) =4
600 x 1000/(15As + 62500 –As)=4
Q.8 Compare the strength of solid circular column and hollow circular column. Solid is of diameter 200 mm. Hollow circular column of the same cross sectional area and thickness 30 mm. The other parameters are same for both the sections.
Assume these columns as long columns.
A solid = (π /4) d2 =(π /4) 2002
A hollow = ( π /4)( do2 –(do –60)2)
Given A solid= A solid
(π /4) 2002 = (π /4)( do2 –(do –60)2)
do = 363.33 mm
I solid = ( π /64) d4
I hollow = (π /64)( do4 –di4) =( π /64)( do4 –(do –60)4)
P Crippling solid = 2 EI solid/Le2
P Crippling hollow = 2 EI hollow/Le 2
Ratio of strengths = P Crippling solid/ P Crippling hollow
=I solid/I hollow
Ratio of strengths =( π /64) d4/( π /64)( do4 –(do –60)4)
= 2004/[(363.334 –(363.33 -60)4]
Q.9 A 4 m long simply supported wooden beam at its ends. It is carrying a point load of 7.25 k N at its center. The cross sectional area of the beam is 140 mm wide and 240 mm deep. If E = 6 x 10 3 M Pa, find the deflection at the center using the double integration method.
Prove the article for yc = WL3/48EI
Deflection at the center = WL3/48EI
= 7250 x 40003/48 x6x103x1/12×140 x 2403
Q. Compare the strains produced in a body subjected to same amount of load when applied gradually and suddenly.
gradually applied load =P
area of cross section of the body=A
Then stress = P/A
Gradually applied load is twice of suddenly applied load.
Let P is suddenly applied load.
Stress = 2P/A
Now stress is two times. Strain is also two times.
Gradually applied load is two times the suddenly applied load
W.D. in suddenly applied load = P sudden δl
W.D. under gradually applied load= 0.5 PGAL δl
where GAL is gradually applied load
In order to find the equivalent gradually applied load, the W.D. and extension are same. Therefore equate the W.D. in the two cases.
P sudden δl = 0.5 PGAL δl
PGAL = 2Psudden
2. For a thin cylindrical shell, the length diameter ratio is 3 and its volume is 20 m3. The safe tensile stress for the shell material is 100 M Pa. Determine the cylinder diameter and wall thickness. It contains water at an absolute pressure of 2 M Pa.
ANS: Given L/D =3 and volume V =D2L= D2 3D= 20
D3 = 20
D = 2.04 m
Now hoop stress is the highest stress which taken equal to the safe stress.
σh=pi D/2t = 100
t = 2x 2.04 x 1000/2×100 = 20 mm
3. Distinguish between energy of dilation and energy of distortion.
Every object in universe is under a number of forces. Thus it is under complex loading. From the complex loading, one can determine the three principal stresses σ1, σ2 and σ3 respectively. ASSUME
these all principal stresses as tensile and
σ1> σ2 > σ3
Energy of dilation deals with stresses that cause a change in size (length, width, thickness, depth, area, volume). Definitely these are tensile and compressive stresses. Principal stresses are also tensile and compressive stresses. Then the energy of dilation is given below
u 3d=(1/2E)[ σ12 + σ22 + σ32 -2μ σ1σ2-2μ σ2σ3-2μ σ3σ1]
u 2d=(1/2E)[ σ12 + σ22 + -2μ σ1σ2]
where μ is the Poisson’s ratio
It will be used in Maximum Strain Energy Theory of Elastic Failure.
Then energy of distortion OR shear strain energy theory (sse)
applies when there is no change of size but there is twist or distortion in the body. It is given by the relation given below.
u sse=(1/12G)[ (σ1–σ2)2 + (σ2—σ3)2 + (σ3—σ1)2]
u sse=(1/12G)[ (σ1–σ2)2]
It is used in Maximum Strain Energy of Distortion Theory of Elastic Failure.
NOTE: if any of the three principal stresses σ1, σ2 and σ3 is not tensile, then respective sign must be used in the above relation.
4. State maximum principal stress theory.
ANS: Every object in universe is under a number of forces. Thus it is under complex loading. From the complex loading, one can determine the three principal stresses σ1, σ2 and σ3 respectively. ASSUME
these all principal stresses as tensile and
σ1> σ2 > σ3
Maximum principal stress theory
states that a material will fail if largest principal stress in the material exceeds the allowable stress.
Therefore when σ1 = σallowable =σelastic/Factor of safety
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