Mechanics of solid

Stress and StrainBy Kaushal Patel

• 𝜎 = 𝐹/𝐴

Stress

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• 휀 = 𝛿𝑙/𝐿

Strain

h

l

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• When a body is subjected to two equal and opposite axial pulls P (also calledtensile load) , then the stress induced at any section of the body is known astensile stress.

• Tensile load, there will be a decrease in cross-sectional area and an increasein length of the body. The ratio of the increase in length to the original lengthis known as tensile strain.

Tensile Stress and Strain

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• When a body is subjected to two equal and opposite axial pushes P (alsocalled compressive load) , then the stress induced at any section of the bodyis known as compressive stress

• Compressive load, there will be an increase in cross-sectional area and adecrease in length of the body. The ratio of the decrease in length to theoriginal length Is known as compressive strain

Compressive Stress and 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 isdirectly proportional to strain,

σ ∝ ε or σ = E × ε

• 𝐸 =𝜎=

𝑃.𝑙

𝐴.𝛿𝑙

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• When a body is subjected to two equal and opposite forces actingtangentially across the resisting section, as a result of which the body tends toshear off the section, then the stress induced is called shear stress (τ), Thecorresponding strain is known as shear strain (φ)

• Shear stress, 𝜏 =𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑓𝑜𝑟𝑐𝑒

𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑎𝑟𝑒𝑎

Shear Stress and Strain

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

Mathematically τ ∝ φ or τ = G . φ or τ / φ = G

• where, τ = Shear stress,

• φ = Shear strain,

• G = Constant of proportionality, known as shear modulus or modulus ofrigidity.

It is also denoted by N or C.

Shear Modulus or Modulus of Rigidity

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Stress-Stress Diagram

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• A composite bar may be defined as a bar made up of two or more different materials, joinedtogether, in such a manner that the system extends or contracts as one unit, equally, whensubjected to tension or compression.

Stress in Composite Bar

1. The extension or contraction of the bar is being equal

2. The total external load on the bar is equal to the sum of the loads carried by different materials.

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• P1 = Load carried by bar 1,• A1 = Cross-sectional area of bar 1,• σ1 = Stress produced in bar 1,• E1 = Young's modulus of bar 1,

• P2, A2, σ2, E2 = Corresponding values of bar 2,

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

• Stress in bar 1, 𝜎1 =𝑃1

𝐴1

• strain in bar 1, 휀 =𝜎1

𝐸1=

𝑃1

𝐴1𝐸1

Continue…

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• Elongation in bar -1: 𝛿𝑙1 =𝑃1𝑙

𝐴1𝐸1

• Elongation in bar -2: 𝛿𝑙2 =𝑃2𝑙

𝐴2𝐸2

There fore,

δl1 = δl2 𝑃1𝑙

𝐴1𝐸1=

𝑃2𝑙

𝐴2𝐸2

𝜎1

𝐸1=

𝜎2

𝐸2

𝑃 = 𝑃1 + 𝑃2 = 𝜎1𝐴1 + 𝜎2𝐴2

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

Continue…

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• 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 areasare A1, A2, A3

• E = Young’s modulus of the material

• P = applied axial load.

Bars with Cross Section Varying in Steps

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• Forces acting on the cross-sections of the three portions. It is obvious that tomaintain equilibrium the load acting on each portion is P only.

Continue…

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Stress Strain and Extension of each BarPortion Stress Strain Extension

1 σ1 = P/ A1 e1 = σ1 / E δ1 = P L1 / A1 E

2 σ2 = P/ A2 e2 = σ2 / E δ2 = P L2 / A2 E

3 σ3 = P/ A3 e3 = σ3 / E δ3 = P L3 / A3 E

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• Total elongation,

δ = δ1 + δ2 + δ3 = [P L1 / A1 E] + [P L2 / A2 E] + [P L3 / A3 E]

Continue…

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• Stresses due to Change in Temperature

• Whenever there is some increase or decrease in the temperature of a body, itcauses the body to expand or contract.

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

• But, if the deformation of the body is prevented, some stresses are induced inthe body. Such stresses are known as thermal stresses.

Thermal Stresses

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• l = Original length of the body,

• t = Rise or fall of temperature,

• α = Coefficient of thermal expansion,

∴ Increase or decrease in length,

δl = l × α × t

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

휀𝑐 =𝛿𝑙

𝑙=𝑙 𝛼 𝑡

𝑙= 𝛼 𝑡

∴ Thermal stress,𝜎𝑡ℎ = 휀𝑐𝐸 = 𝛼𝑡𝐸

Continue…

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• 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 δl

and the diameter decreases by

an amount δd

• Similarly, if the bar is subjected

to a compressive force,

• Every direct stress is accompanied by a strain in its own direction is known as linear strainand an opposite kind of strain in every direction, at right angles to it, is known as lateralstrain.

Linear and Lateral Strain

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

𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑆𝑡𝑟𝑎𝑖𝑛

𝐿𝑖𝑛𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛= 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

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

Poisson’s Ratio

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

• The ratio of the change in volume to the original volume is known as volumetric strain.

• Volumetric strain, εv = δV / V ; δV = Change in volume ; V = Original volume.

• Volumetric strain of a rectangular body subjected to an axial force is given as

휀𝑣 =𝛿𝑣

𝑣= 휀 1 −

2

𝑚

• Volumetric strain of a rectangular body subjected to three mutually perpendicular forces is given by

εv = εx + εy + εz

Volumetric Strain

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

• It is usually denoted by K.

• Bulk modulus,

𝐾 =𝑆𝑡𝑟𝑒𝑠𝑠

𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛=

𝜎

𝛿𝑉𝑉

Bulk Modulus

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Thank You

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