1 static failure theories ductile r1

MACHINE DESIGN: STATIC FAILURE THEORIES

Dr. Himanshu ChaudharyAssociate ProfessorDept. of Mech. Eng.

MNIT Jaipur

Static failure theories• Why do parts fail?• You may say “Parts fail because their stresses exceed their

strength”• Then what kind of stresses cause the failure: Tensile?

Compressive? Shear?• Answer may be: It depends. • It depends on the material and its relative strength in

compressive, tension, and shear. • It also depends on the type loading (Static, Fatigue, Impact) and• presence of the cracks in the material

Static failure theories• The failure may be elastic or fracture• Elastic failure results in excessive

deformation, which makes the machine component unfit to perform its function satisfactorily

• Fracture results in breaking the component into two parts

Static failure theories• Question: How do one compare stresses

induced to the material properties?– Generally machine parts are subjected to combined

loading and to find material properties under real loading condition is practically not economical

– Thus, material properties are obtained from simple tension/torsion test

– These data like Syt, Sut etc are available in form of table (Design Data Book)

Static failure theories• Theories of failure provide a relationship between the

strength of machine component subjected to complex state of stress with the material properties obtained from simple test (Tensile)

Strength of machine component subjected to complex state of stress

Strength of standard component subjected to uniaxial state of stress

Static failure theories• Loads are assumed to not vary over time• Failure theories that apply to:

– Ductile materials– Brittle materials

• Why do we need different theories ??

Stress-strain curve of a ductile material Stress-strain curve of a brittle material

Static failure theories: Tension test

Why nearly 0o ??

Failure along principal shear stress plane

Failure along principal normal stress plane

Cast iron has C between 2.1% to 4% and Si between 1% and 3% C contents less than 2.1% are steels.

Static failure theories:Compression test

Why doesn’t it fail ?? Why does it fail ??Why nearly 45o ??

Does not“fail” Shear failure

Failure along principal shear stress plane Failure along principal normal stress plane

Static failure theories• In general, ductile, isotropic materials are limited by their

shear strengths. • Brittle materials are limited by their tensile strengths. • If cracks are present in a ductile material, it can suddenly

fracture at nominal stress levels well below its yield strength, even under static loads.

• Static loads are slowly applied and remain constant with time.

• Dynamic loads are suddenly applied (impact), or repeatedly varied with time (fatigue), or both.

Static failure theories• In dynamic loading, the distinction between

failure mechanisms of ductile and brittle materials blurs.

• Ductile materials often fail like brittle materials in dynamic loading.

Static failure theories• Total strain energy theory• Distortion energy theory• Pure shear-stress theory• Maximum shear-stress theory• Maximum normal stress theory (limited application)

Accepted failure theories that apply to ductile materials:

Accepted failure theories that apply to brittle materials:• Maximum normal stress theory (even material)• Maximum normal stress theory (uneven material)• Coulomb-Mohr theory• Modified Mohr theory

Maximum Normal Stress Theory

• Credited to the English scientist and educator W. J. M. Rankine (1802–1872)

Maximum Normal Stress Theory • This predicts that failure of machine component, subjected to complex loading, occurs

if the maximum normal principal stress tends to exceeds the uniaxial tensile yield (ductile) or the ultimate tensile strength (brittle) of the material

• Providing a square failure boundary with Sy as the principal stress for ductile materials • Note: not a safe theory for ductile materials

Exercise: Draw failure envelop for two dimension case?

Maximum Normal Stress Theory

• Let 1, 2, 3 are the principal stresses at the critical point in component due to applied loading, and let 1> 2> 3

• Then failure occurs when𝜎 1≥𝑆 𝑦𝑡𝑜𝑟 𝜎1≥𝑆𝑢𝑡

If we want to include uncertainty of data available and loads acting on the component, i.e., factor of safety, then to avoid failure:

𝜎 1=𝑆𝑦𝑡

𝐹𝑆𝑜𝑟 𝜎 1=𝑆𝑢𝑡

𝐹𝑆

Maximum Shear-Stress Theory• The oldest failure theory, originally proposed

by the great French scientist C. A. Coulomb (1736–1806)

• Tresca modified it in 1864, and J. J. Guest• Validated by experiments around 1900 • For these reasons the maximum-shear-stress

theory is sometimes called the Tresca-Guest theory

Maximum Shear-Stress TheoryThis theory states that a material subjected to any combination of loads will fail (by yielding or fracturing) whenever the maximum shear stress exceeds the shear strength (yield or ultimate) of the material. The shear strength, in turn, is usually assumed to be determined from the standard uniaxial tension test.

Exercise: Draw hexagonal failure envelop for two dimesion case?

Static failure theories

This theory states that failure occurs when:

Ductile materials: maximum shear-stress theory

𝑆 𝑦𝑠=0.5𝑆𝑦

(Failure occurs when maximum shear stressexceeds the shear stress at yield in pure tension)

Mohr’s circle:pure tension

𝑆𝐹=𝑆 𝑦 𝑠

𝜏𝑚𝑎𝑥𝜏𝑚𝑎𝑥≤𝑆𝑦𝑠

Static failure theoriesDuctile materials: maximum shear-stress theoryProviding a hexagonal failure envelope that is more conservative than

the distortion energy theory

Maximum Shear Stress Theory

Static failure theoriesDuctile materials

Total strain energy U:

Elastic range assuming stress strain curve is linear upto yield point

𝑈=12 𝜎𝜀

𝑈=12 𝜎𝜀=

12 (𝜎 1𝜀1+𝜎2𝜀2+𝜎3𝜀3)

Principal stresses and strains

where

Static failure theoriesDuctile materials: total strain energy

Using previous expressions, total energy is:

𝑈=12 𝜎𝜀=

12𝐸 ¿

which can be expressed as 𝑈=𝑈 h+𝑈𝑑

Hydrostatic energy Deformation energy

𝑈 h=32

(1−2𝜐)𝐸 𝜎 h

❑2

𝜎 h=𝜎 1+𝜎2+𝜎3

3Obtained by setting:𝑈 h=𝑈 (𝜎1=𝜎 2=𝜎3=𝜎h)

Obtained by setting:𝑈𝑑=𝑈 −𝑈 h

Static failure theoriesDuctile materials: distortion energy theory

𝑈𝑑=1+𝜐3𝐸 ¿

If uniaxial yield stress state (failure state):

Therefore: 𝑈𝑑=1+𝜐3𝐸 𝑆 𝑦❑

2 Using uniaxial yieldstress state (failurestate)


𝑈𝑑=1+𝜐3𝐸 [𝜎 1❑

2 +𝜎 2❑2 +𝜎 3❑

2 − (𝜎1𝜎2+𝜎2𝜎3+𝜎3𝜎1 ) ]

For any other state of stresses:

Failure criterion is obtained by setting:1+𝜐3𝐸 ¿

Distortion energy:uniaxial stress atyield

Distortion energy:any other state ofstresses

𝜎 1❑2 +𝜎 2❑

2 +𝜎 3❑2 −𝜎1𝜎 2−𝜎2𝜎3−𝜎3𝜎1≤𝑆 𝑦❑

2


𝑆 𝑦❑2 =𝜎 1❑

2 +𝜎 2❑2 +𝜎 3❑

2 −𝜎1𝜎2−𝜎 2𝜎 3−𝜎 3𝜎1

For a 2D stress where 𝜎2= 0, 𝑆 𝑦❑2 =𝜎 1❑

2 +𝜎 3❑2 −𝜎1𝜎3

• The 2D distortion energy equation is described in an ellipse

• The interior of the ellipse show the biaxial safe stress sage against yielding under static loads


𝑆 𝑦❑2 =𝜎 1❑

2 +𝜎 2❑2 +𝜎 3❑

2 −𝜎1𝜎2−𝜎 2𝜎 3−𝜎 3𝜎1

•The 3D distortion energy equation is described in cylinder inclined to principal stress axes •The interior of the cylinder show the region safe against yielding for combined stresses 1, 2, & 3 stress sage under static loads


𝑆 𝑦❑2 =𝜎 1❑

2 +𝜎 2❑2 +𝜎 3❑

2 −𝜎1𝜎2−𝜎 2𝜎 3−𝜎 3𝜎1

Intersection of the cylinder with each of the principal planes are ellipses similar to the 2D stresses

Static failure theoriesDuctile materials: distortion energy theoryVon Mises effective stress

𝑆 𝑦❑2 =[𝜎 1❑

2 +𝜎 2❑2 +𝜎 3❑

2 −𝜎1𝜎2−𝜎 2𝜎3−𝜎 3𝜎1]≡ (𝜎 ′ ) 2

Definition:𝜎 ′=√ [𝜎 1❑

2 +𝜎 2❑2 +𝜎 3❑

2 −𝜎 1𝜎 2−𝜎2𝜎 3−𝜎3𝜎 1 ] (Yield surface)

von Mises effective stressvon Mises effective stress: uniaxial stress that would create the same distortion energy as is created by actual combination of applied stresses

Distortion Energy Theory (DET)• The failure theory based on distortion

energy is also known as von Mises-Hencky𝜎 ′≤𝑆 𝑦

𝑆𝐹=𝑆𝑦

𝜎 ′

Yield strength of the material

von Mises effective stress


Example: pure shear loadas in case of torsional failure Mohr’s circle:

pure shear𝜎1= max : 𝜎3 =- max and 𝜎2 = 0

𝑆 𝑦❑2 =𝜎 1❑

2 +𝜎 2❑2 +𝜎 3❑

2 −𝜎1𝜎2−𝜎 2𝜎 3−𝜎 3𝜎1

Using:

𝑆 𝑦❑2 =3𝜏2𝑚𝑎𝑥

𝑆 𝑦𝑠≡𝜏𝑚𝑎𝑥=1√3

𝑆𝑦=0.577𝑆𝑦

Maximum stress before failure, in this case, is:

Static failure theories: experimental verificationsDuctile & brittle materials

Applicable examples• A circular rod is subjected to combined

loading consisting of a tensile load P = 10 kN and a torque T = 5 kN-m. Rod is 50 mm in diameter.– 1) Draw stress element (cube) at the most highly stressed

location on the rod, and– 2) draw corresponding Mohr’s circle(s).

Example: Failure of ductile material under static loadingProblem: Determine the safety factors for the bracket rod based on the both the distortion-energy theory and the maximum shear theory and compareGiven: Yield strength Sy=324 MPa Rod length L=150 mm

Arm length a=200 mm Rod diameter d=38 mmLoad F=4450 N

• Limiting to bending of cantilever and in torsion

• The shear and moment diagrams will be similar to a cantilever beam loaded at its end

1 static failure theories ductile r1

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