Solution of Som Paper

8/13/2019 Solution of Som Paper

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SOLUTION

PUT PAPER STRENGTH OF MATERIALS

Q1 a)

i) Ductility:-

It is the ability of a material to under go deformation under tension without rupture. It is the

property of material by virtue of which it can be drawn into wire. Ductility of the material

depends on the grain size of the matal crystal.

In quantitative terms it is normally measured as the ratio of elongation of the material at fracture

during the tensile test to the original length, expressed as a percentage. The final value of

elongation obtained during the tensile test immediately after fracture could be taken as the

ductility.

Gold is most ductile material

The ductility of a material in tension can be characterized by its elongation and by the reductionin area at the cross section where fracture occurs.

It is the ratio of the extension in length of the specimen after fracture to its initial gauge length,expressed in percent.

ii) HARDNESS:-

Hardness is the property of the material by virtue of which it resist the penetration and skrech of

another harder body. It also means the ability to cut another material. The hardness of material

depends upon the type of bondind forces between atoms, ions or molecules.

Diamond is the hardest known material

Hardness of a material can be measured by following test:-

1) Brinell hardnes test

2) Rockwell hardness test

3) Vickers hardness test etc

iii) TOUGHNESS:-

toughness is the ability of a material to absorb energy just before fracture when load is applied.It is strength of a material which opposes rupture. It can be considered as the total area under the

stress strain. Since area under curve shows the amount of work done on the material without

causing fracture. Thus toughness can be considered as a parameter consisting of both strength

and ductility.

Toughness of the material is measured by the impact test. 1) Izod test 2) charpy test


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iv) YIELD STRENGTH:-

A yield strength oryield point of a material is defined in engineering and materials science as

the stress at which a material begins to deform plastically. Prior to the yield point the material

will deform elastically and will return to its original shape when the applied stress is removed.Once the yield point is passed, some fraction of the deformation will be permanent.

v) POISONS RATIO:-


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Q1 b)


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OR

Q2 a)


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Q2 b)


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Q3 a)

Q3 b)


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OR

Q4 a)

When the pressure vessels subjected to internal fluid pressure the following Stresses generated in

the pressure vessels

1) Hoop or circumferential stress

2) Longitudinal stress

3) Radial stress


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Hoop stress:-

Longitudinal stress:-

Radial stress:-

Radial stresses act along the radius. In thin cylinders the thickness is less so radial stresses are

neglected in thin pressure vessels.


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Expression for volumetric strain:-

Q4 b)


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Q5 a) Assumptions:-


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Now

This the bending equation.


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Q5 b)

OR


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Q6 b)

Derivation of equation for shearing stress :

Assumptions :

1. Stress is uniform across the width (i.e. parallel to the neutral axis)

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

It may be noted that the assumption no.2 cannot be rigidly true as the existence of shear stress

will cause a distortion of transverse planes, which will no longer remain plane.

In the above figure let us consider the two transverse sections which are at a distance dx' apart.

The shearing forces and bending moments being F, F + dF and M, M + dM respectively. Nowdue to the shear stress on transverse planes there will be a complementary shear stress onlongitudinal planes parallel to the neutral axis.

Let tbe the value of the complementary shear stress (and hence the transverse shear stress) at a

distance Y'0 from the neutral axis.

z is the width of the x-section at this position

A is area of cross-section cut-off by a line parallel to the neutral axis.

= distance of the centroid of Area from the neutral axis.


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Let s , s + ds are the normal stresses on an element of area dA at the two transverse sections,

then there is a difference of longitudinal forces equal to ( ds . dA) , and this quantity summed

over the area A is in equilibrium with the transverse shear stress t on the longitudinal plane ofarea z dx .

The figure shown below indicates the pictorial representation of the part.

So substituting


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Where z' is the actual width of the section at the position where t ' is being calculated

I is the total moment of inertia about the neutral axis.

Q7 a)


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Q7 b)

OR


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Q8 a)


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Q8 b)

i) TORSIONAL RIGIDITY

Torsional rigidity or stiffness of the shaft is defined as the product of modulus of rigidity (C) and

polar moment of inertia of the shaft (J). Hence mathematically, the Torsional rigidity is given as,

Torsional rigidity=C x J.

Torsional rigidity is also defined as the torque required to produce a twist of one radian per

unit length of the shaft.

Let a twisting moment T produces a twist of radians in a shaft of length L.

Using equation, we have,

T / J = C / L

C x J = T x L /

But C x J = Torsional Rigidity

Torsional Rigidity = T x L / ii)

Q9 a) EQUIVALENT LENGTH OF COLUMN:-


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ii) Columns and struts: Structural members subjected to compression load and which arerelatively long compared to their lateral dimensions are called columns or Struts.

columns are vertical members subjected to compression load and fixed rigidly at both ends

struts are inclined members subjected to compressive load and ends may be fixed, pin jointed or

hinged.

Examples: strut in a truss, Piston rods, side links in forging machines, connecting rods etc.

iii) Slenderness ratio: It is the Ratio of the effective length of the column to the least radius of

gyration of the cross sectional ends of the column.

Slenderness ratio, S =Le /k

Least radius of gyration, k= Imin/A

Imin is the least of I xx and I yy

L =actual length of the column

Le=effective length of the column

A= area of cross section of the column


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Q9 b)

Q10 a) Eulers Theory (For long columns)

Assumptions:

1. The column is initially straight and of uniform lateral dimension

2. The material of the column is homogeneous, isotropic, obeys Hookes law

3. The stresses are within elastic limit

4. The compressive load is axial and passes through the centroid of the section

5. The self weight of the column itself is neglected.

6. The column fails by buckling alone

Eulers Formula for Pin-Ended Beams (both ends hinged)

Consider an axially loaded long column AB of length L. Its both ends A and B are hinged. Due

to axial compressive load P, let the deflection at distance x from A be y. The bending moment at

the section is given by


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Eulers Formula for Pin-Ended Beams

(both ends hinged)

Consider an axially loaded long column AB of length L. Its both

ends A and B are hinged. Due to axial compressive load P, let the

deflection at distance x from A be y.

d2y

dx2

d2y

dx2+

Py

EI= 0

L

y x

P

PA

BThe bending moment at the section is given

byEI = - P y

-ve sign on right hand side, since as x

increases curvature decreases

At x =0, y =0,we get c1=0 (from eq.1)

Also at x=L, y =0 we get

c2 .sin [LP/(EI)] =0

If c2 = 0, then y at any section is zero, which means there is no

lateral deflection which is not true

Therefore sin [LP/(EI)] =0

This is the linear differential equation, whose solution is

Y = c1.cos [xP/(EI)] + c2.sin[x P/(EI)] (1)

Where c1 and c2 are the constants of integration. They can befound using the boundary conditions.


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sin [LP/(EI)] =0

=> [LP/(EI)] = 0, , 2 ,n

Taking least non zero value we get

[LP/(EI)] =

Squaring both sides and simplifying

PE

=2E I

L2

This load is called critical or buckling load or crippling load

Q10 b)

i) Maximum principal stress theory


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Maximum shear stress theory

iii) Distortion energy theory

This theory states that the failure occurs when the maximum shear strain energy component forthe complex state of stress system is equal to that at the yield point in the tensile test.

Hence the criterion for the failure becomes

As we know that a general state of stress can be broken into two components i.e,

(i) Hydrostatic state of stress ( the strain energy associated with the hydrostatic state of stress is

known as the volumetric strain energy )

(ii) Distortional or Deviatoric state of stress ( The strain energy due to this is known as the shear

strain energy ) As we know that the strain energy due to distortion is given as

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