MECH 210 080827 - Stress Strain

# MECH 210 080827 - Stress Strain - MECH 210 080827 Stress...

This preview shows pages 1–2. Sign up to view the full content.

Page 1 of 4 MECH 210 080827 Stress and Strain Stress: applied force per unit area in MPa (= 10 6 N/m 2 ) or psi (= lb f /in 2 ) 1 MPa = 145 psi Engineering Stress = area ional sect cross Original load Applied Normal stress (tensile or compressive), σ = area ional sect cross Originalc surface to lar perpendicu Force = o A F Shear stress, τ = o A F area surface Original force Shear = True Stress = area sectional cross ous Instantane time the at load Applied True normal stress, σ T i A F area sectional cross ous Instantane surface to lar perpendicu Force = = Strain: rate of dimensional change in m / m , in / in , or simply unitless Engineering Strain = length Original length in Change Tensile or compressive strain, ε = o o i o l l l l l length Original force of direction along length in Change - = Δ = where l i is the instantaneous length. Shear strain, ν θ Δ tan h l = = True Strain , ε T = ln (1 + ε ) Relationship between engineering stress and true stress: σ T = σ (1 + ε ) Relationship between engineering strain and true strain: ε T = ln (1 + ε ) The normal stress σ ’ and shear stress ε ’ on a plane at an angle with the applied force: + = = 2 2 cos 1 cos ' 2 θ σ θ σ σ and = = 2 2 sin cos sin ' θ σ θ θ σ ε where θ is the angle between the plane and the surface on which the normal force is applied.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern