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INDUS INSTITUTE OFTECHNOLOGY & ENGINEERING

DEPARTMENT OF

MECHANICAL ENGINEERING

Name : Keval Patel

Subject Code :Subject Name : Mechanics of solid

B.E Mechanical Engineering , 1 st Semester 1


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LOAD LOAD:- any external force acting upon a machine part

2

Force

Dead

Live

Suddenlyapplied

Impact


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Stress () = F / A

F =R =

3

1 MPa = 1 10 6 N/m 2 = 1 N/ mm 2


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Strain ( ) = l / L

h

l= L - l l 4


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Tensile Stress and Strain When a body is subjected to two equal and opposite axial pulls

P (also called tensile load) , then the stress induced at anysection of the body is known as tensile stress

Tensile load, there will be a decrease in cross-sectional area and

an increase in length of the body. The ratio of the increase inlength to the original length is known as tensile strain.

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Compressive Stress and Strain

When a body is subjected to two equal and opposite axial

pushes P (also called compressive load) , then the stress inducedat any section of the body is known as compressive stress

Compressive load, there will be an increase in cross-sectionalarea and a decrease in length of the body. The ratio of the

decrease in length to the original length Is known ascompressive strain

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Young's Modulus or Modulus of Elasticity

Hooke's law :- states that when a material is loaded within

elastic limit, the stress is directly proportional tostrain,

or = E

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Example

1. A circular rod of diameter 16 mm and 500 mm longis subjected to a tensile force 40 kN. The modulusof elasticity for steel may be taken as 200 kN/mm2.Find stress, strain and elongation of the bar due to

applied load.

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Shear Stress and Strain

When a body is subjected to two equal and opposite forces

acting tangentially across the resisting section, as a result ofwhich the body tends to shear off the section, then the stressinduced is called shear stress () , The corresponding strain isknown as shear strain ()

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Shear Modulus or Modulus of Rigidity

It has been found experimentally that within the elastic limit,the shear stress is directly proportional to shear strain.

Mathematically or = G . or / = G

where, = Shear stress,

= Shear strain,

G = Constant of proportionality, known as shear modulus ormodulus of rigidity.

It is also denoted by N or C. 12


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Stress-strain Diagram

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Working Stress

When designing machine parts, it is desirable to keep the stresslower than the maximum or ultimate stress at which failure ofthe material takes place. This stress is known as the workingstress or design stress .

It is also known as safe or allowable stress .

working stress = design stress = safe stress = allowable stress

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Factor of Safety

The ratio of the maximum stress to the working stress .

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1: A hollow steel tube is to be used to carry an axial compressive load of160 kN. The yield stress for steel is 250 N/mm2. A factor of safety of1.75 is to be used in the design. The following three class of tubes ofexternal diameter 101.6 mm are available .

Class Thickness Light 3.65 mm Medium 4.05 mm Heavy 4.85 mm

Which section do you recommend?

2: A specimen of steel 20 mm diameter with a gauge length of 200 mmis tested to destruction. It has an extension of 0.25 mm under a loadof 80 KN and the load at elastic limit is 102 KN. The maximum load is130 KN. The total extension at fracture is 56 mm and diameter atneck is 15 mm. Find

(i) The stress at elastic limit. (ii) Youngs modulus. (iii) Percentage elongation. (iv) Percentage reduction in area.

(v) Ultimate tensile stress. 19


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Stresses in Composite Bars

A composite bar may be defined as a bar made up of two or more

different materials, joined together, in such a manner that the systemextends or contracts as one unit, equally, when subjected to tensionor compression.

1. The extension or contraction ofthe bar is being equal

2. The total external load on the bar is

equal to the sum of the loads carriedby different materials.

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d d b b


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P1 = Load carried by bar 1, A1 = Cross-sectional area of bar 1, 1 = Stress produced in bar 1,

E1 = Young's modulus of bar 1,

P2, A2, 2, E2 = Corresponding values of bar 2 ,

P = Total load on the composite bar, l = Length of the composite bar, and l = Elongation of the composite bar. We know that P = P1 + P2

Stress in bar 1,

strain in bar 1, 21


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Elongation in bar -1:

Elongation in bar -2:

There fore,

l 1 = l 2

The ratio E1 / E2 is known as modular ratio of the two materials

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P = P1 + P2=

=

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l 1 = l 2

P = P1 + P2 = 1.A1 + 2.A2

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EX 3


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EX:3A bar 3 m long is made of two bars, one of copper havingE = 105 GN/m2 and the other of steel having E = 210 GN/m2.

Each bar is 25 mm broad and 12.5 mm thick. This compoundbar is stretched by a load of 50 kN. Find the increase in lengthof the compound bar and the stress produced in the steel andcopper. The length of copper as well as of steel bar is 3 meach.

Ans:Pc=16.67NPs=33.33N

l=1.52 mm

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EX 4


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EX:4 A central steel rod 18 mm diameter passes through a

copper tube 24 mm inside and 40 mm outside diameter, as

shown in Fig. It is provided with nuts and washers at eachend. The nuts are tightened until a stress of 10 MPa is setup in the steel. Find out stress generated in copper tube.

Hint: Ps=Pc

s As = c Ac


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BARS WITH CROSS-SECTIONS VARYING IN STEPS A typical bar with cross-sections varying in steps and subjected

to axial load length of three portions L1, L2 and L3 and the respective cross-

sectional areas are A1, A2, A3 E = Youngs modulus of the material P = applied axial load.

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f ti th ti f th th ti It


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forces acting on the cross-sections of the three portions. Itis obvious that to maintain equilibrium the load acting oneach portion is P only.

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St t i d t i f h f th ti


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Stress, strain and extension of each of these portions are :

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Portion Stress Strain Extension

1 1

= P/ A1 e

1 =

1/ E

1= P L

1 / A

1 E

2 2 = P/ A2 e 2 = 2 / E 2 = P L2 / A2 E

3 3 = P/ A3 e 3 = 3 / E 3 = P L3 / A3 E


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Total Elongation:

1 + 2 + 3 = [P L1 / A1 E] + [P L2 / A2 E] + [P L3 / A3 E]

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EX:5


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EX:5

The bar shown in Fig. 8.16 is tested in universal testingmachine. It is observed that at a load of 40 kN the total

extension of the bar is 0.280 mm. Determine the Youngs modulus of the material.

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Thermal StressesStresses due to Change in Temperature

Whenever there is some increase or decrease in thetemperature of a body, it causes the body to expand orcontract.

If the body is allowed to expand or contract freely, with the riseor fall of the temperature, no stresses are induced in the body.

But, if the deformation of the body is prevented, some stressesare induced in the body. Such stresses are known as thermalstresses.

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l = Original length of the body, t = Rise or fall of temperature,

= Coefficient of thermal expansion,

Increase or decrease in length, l = l t

If the ends of the body are fixed to rigid supports, so that itsexpansion is prevented, then compressive strain induced in thebody,

Thermal stress,33

If the free expansion is prevented fully


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If the free expansion is prevented fully Since support is not permitting it, the support force P develops

to keep it at the original position.

Magnitude of this force is such that contraction is equal to freeexpansion

l = l t

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If free expansion is prevented partially


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If free expansion is prevented partially

Expansion prevented = tL

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EX:6: A steel rail is 12 m long and is laid at a temperature of 18C


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EX:6: A steel rail is 12 m long and is laid at a temperature of 18 C.The maximum temperature expected is 40C.

(i) Estimate the minimum gap between two rails to be left so that

the temperature stresses do not develop.

(ii) Calculate the temperature stresses developed in the rails, if:(a) No expansion joint is provided.(b) If a 1.5 mm gap is provided for expansion.

(iii) If the stress developed is 20 N/mm2, what is the gap provided

between the rails?

Take E = 2 105 N/mm2 and = 12 10 6/C.

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Li d L t l St i


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Linear and Lateral Strain Consider a circular bar of diameter d and length l, subjected to a

tensile force P

Due to tensile force, the length

of the bar increases by an amount land the diameter decreases byan amount d

Similarly, if the bar is subjectedto a compressive force,

Every direct stress is accompanied by a strain in its own direction isknown as linear strain and an opposite kind of strain in everydirection, at right angles to it, is known as lateral strain .

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Poisson's Ratio

When a body is stressed within elastic limit, the lateral strain bears aconstant ratio to the linear strain.

This constant is known as Poisson's ratio and is denoted by(1/ m ) or .

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l


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Volumetric Strain

When a body is subjected to a system of forces, it undergoes somechanges in its dimensions. The volume of the body is changed.

The ratio of the change in volume to the original volume is known asvolumetric strain .

Volumetric strain , v = V / V ; V = Change in volume,; V = Original volume.

Volumetric strain of a rectangular body subjected to an axial forceis given as

Volumetric strain of a rectangular body subjected to three mutuallyperpendicular forces is given by

v = x + y + z 39

lk d l


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Bulk Modulus

When a body is subjected to three mutually perpendicular stresses,of equal intensity, then the ratio of the direct stress to thecorresponding volumetric strain is known as BULK MODULUS.

It is usually denoted by K.

Bulk modulus,

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Relation Between Bulk Modulus and Youngs Modulus

Relation Between Youngs Modulus and Modulus of Rigidity

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EX:7


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EX:7A bar of 25 mm diameter is tested in tension. It is observed thatwhen a load of 60kN is applied, the extension measured over agauge length of 200 mm is 0.12 mm and contraction in diameter is0.0045 mm. Find Poissons ratio and elastic constants E, G, K.

EX:8

A circular rod of 25 mm diameter and 500 mm long is

subjected to a tensile force of 60 KN. Determine modulus ofrigidity, bulk modulus and change in volume if Poissons ratio= 0.3 and Youngs modulus E = 2 105 N/mm2.

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EX:9 A 400 mm long bar has rectangular cross-section 10 mm 30


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EX:9 A 400 mm long bar has rectangular cross section 10 mm 30mm. This bar is subjected to

(i) 15 kN tensile force on 10 mm 30 mm faces,

(ii) 80 kN compressive force on 10 mm 400 mm faces, and(iii) 180 kN tensile force on 30 mm 400 mm faces.Find the change in volume if E = 2 105 N/mm2 and = 0.3.

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S i


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Stress concentration

Whenever a machine component changes the shape of its cross-section, the simple stress distribution no longer holds good.

This irregularity in the stress distribution caused by abrupt changes ofform is called stress concentration.

It occurs for all kinds of stresses in the presence of fillets, notches,holes, keyways, splines, surface roughness or scratches etc.

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Thank You

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