Equation for Modified Mohr Theory (4th Quadrant) · PDF fileEquation for Coulomb-Mohr Theory...

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ME 3221 - Spring 2012 Name: Brittle Static Failure February 22, 2012 Equation for Coulomb-Mohr Theory (4th Quadrant) 1 n = σ A S ut + σ B S uc σ A 0 σ B 0 Equation for Modified Mohr Theory (4th Quadrant) n = Sut σ A σ A 0 σ A ≥−σ B 1 n = σ A 1 Sut + 1 Suc + σ B Suc σ A 0 σ A ≤−σ B Note: σ B and S uc are taken to be negative numbers in all of the above equations. Example: Brittle Failure The following shows the state of stress at a critical point in a part fabricated from Class 20 gray cast iron. τxy = 30 MPa σx = 50 MPa x y Determine the static factor of safety based on the state of stress at this point. (Turn over for solution) 1

Transcript of Equation for Modified Mohr Theory (4th Quadrant) · PDF fileEquation for Coulomb-Mohr Theory...

ME 3221 - Spring 2012 Name:Brittle Static Failure February 22, 2012

Equation for Coulomb-Mohr Theory (4th Quadrant)

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n=

σA

Sut+

σB

SucσA ≥ 0 σB ≤ 0

Equation for Modified Mohr Theory (4th Quadrant)

n = Sut

σAσA ≥ 0 σA ≥ −σB

1n

= σA

(1

Sut+ 1

Suc

)+ σB

SucσA ≥ 0 σA ≤ −σB

Note: σB and Suc are taken to be negative numbers in all of the above equations.

Example: Brittle Failure

The following shows the state of stress at a critical point in a part fabricated from Class 20gray cast iron.

τxy = 30 MPa

σx = 50 MPa

x

y

Determine the static factor of safety based on the state of stress at this point.

(Turn over for solution)

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Solution:

From Appendix C-3a:Sut = 152 MPa

Suc = −572 MPa

The principle stresses are first obtained using Mohr’s circle:

τxy (CCW)

30

50

r = 39.05

25σ3 = −14.05 MPa σ1 = 64.05 MPa

σ

Then, using the Modified Mohr theory:

σA = σ1

σB = σ3

σA ≥ −σB

Therefore:

n =Sut

σA

=152

64.05= 2.4

Representing the solution graphically:

σB 64 152 σA

−14

−572

−152

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