1. Ken Youssefi Mechanical Engineering Department 1 Normal & Shear components of stress Normal stress is perpendicular to the cross section,  (sigma). Shear stress is parallel to the cross section,  (tau). 3D case x y z  xy xx  xz yy  yz  yx zz  zx  zy Second subscript indicates the positive direction of the shear stress  xy First subscript indicates the axis that is perpendicular to the face Due to equilibrium condition;  xy =  yx  zy =  yz  zx =  xz

2. Ken Youssefi Mechanical Engineering Department 2 Normal & Shear components of stress Two Dimensional Case  xy yy xx xx yy

3. Ken Youssefi Mechanical Engineering Department 3 Normal Stress Due to Axial Load A positive sign is used to indicate a tensile stress (tension), a negative sign to indicate a compressive stress (compression) Uniform stress distribution across the cross sectional area

4. Direct Shear Ken Youssefi Mechanical Engineering Department 4 Direct shear is produced where there is no bending (or stress caused by bending is negligible

5. Ken Youssefi Mechanical Engineering Department 5 Normal Stress Due to Bending Load Typical loads on a barbell Stress distribution Maximum stress at the surface Where I is area moment of inertia I = π (d) 4 / 64 (round cross section)

6. Transverse Shear Ken Youssefi Mechanical Engineering Department 6 In beam loading, both bending stress and shear stress due to transverse loading are applied to particular section. Maximum shear stress due to bending

7. Ken Youssefi Mechanical Engineering Department 7 Shear Stress Due to Torque (twisting) Torsional stress is caused by twisting a member Stress distribution Maximum shear stress at the surface Where J is polar area moment of inertia J = π (d) 4 / 32 (round cross section)

8. Ken Youssefi Mechanical Engineering Department 8 Torsional Stress - examples Structural member Power transmission Mixer

9. Ken Youssefi Mechanical Engineering Department 9 Combined Stresses - examples Bicycle pedal arm and lug wrench, bending and torsion stresses Trailer hitch, bending and axial stresses Power transmission, bending and torsion stresses Billboards and traffic signs, bending, axial and torsion stresses

10. Ken Youssefi Mechanical Engineering Department 10 Principal Stresses – Mohr’s Circle 3D Case  1 >  2 >  3 2 2 2  3 - (  x +  y +  z )  2 + (  x  y +  x  z +  y  z -  xy -  xz -  yz )  - (  x  y  z - 2  xy  xz  yz -  x  yz -  y  xz -  z  xy ) = 0 2 2 2 The three non-imaginary roots are the principal stresses 2D Case  1,  2 = (  x +  y ) / 2 ± [ (  x -  y ) / 2 ] 2 + (  xy ) 2  1 >  2

11. Ken Youssefi Mechanical Engineering Department 11 Equivalent Stress - von Mises Stress Using the distortion energy theory, a single equivalent or effective stress can be obtained for the entire general state of stress given by  1,  2 and  3. This equivalent (effective) stress can be used in design and is called von Mises stress (  ′).  ′ = (  1 +  2 +  3 -  1  2 -  1  3 -  2  3 ) 1/2 2 2 2 3D Case  ′ = (  x +  y -  x  y + 3  xy ) 1/2 2 2 2 Substituting for  1,  2 from Mohr circle, we have the von Mises stress in terms of component stresses.  ′ = (  x + 3  xy ) 1/2 2 2 In most cases  y = 0 2D Case,  ′ = (  1 +  2 -  1  2 ) 1/2 2 2  3 = 0

12. Ken Youssefi Mechanical Engineering Department 12 Maximum Shear Stress – Mohr’s Circle 33 11 22 Mohr’s circles for a 3D case  12  13  23  max = largest of the three shear stresses, in this case  13  

13. Ken Youssefi Mechanical Engineering Department 13 Maximum Shear Stress – Mohr’s Circle Mohr’s circles for a 2D case  3 =0 11 22  12  13  23  13  12  23  1 and  2 have the same sign, both positive or negative.  3 =0 11 22  1 and  2 have the opposite sign.  max =  13 = 11 2  max =  12 =  1 -  2 2

14. Ken Youssefi Mechanical Engineering Department 14 Stress – Strain Relationship Poisson’s Ratio, v Load v = Strain in the y direction Strain in the x direction = εyεy εxεx x y z Uniaxial state of stress xx xx εxεx = xx E εyεy = - v ε x εzεz

15. Ken Youssefi Mechanical Engineering Department 15 Stress – Strain Relationship Biaxial state of stress xx xx yy yy Triaxial state of stress xx xx yy yy zz zz εxεx = xx E = xx E yy E v εyεy = yy E v ε x yy E xx E v = εzεz = v ε y εxεx = xx E yy E zz E v v εyεy = yy E xx E zz E v v εzεz = zz E xx E yy E v v Strain in the x direction due to the force in the y direction