Normal & Shear components of stress

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

xy x Normal & Shear components of stressTwo Dimensional Case y xy x xy xy y Mechanical Engineering Department

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) Mechanical Engineering Department

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

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

Power transmission Mixer Torsional Stress- examples Structural member Mechanical Engineering Department

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

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

3D Case ′ = (1+ 2 + 3 - 12 - 13 - 23)1/2 2 2 2 3= 0 2D Case, Substituting for 1, 2 from Mohr circle, we have the von Mises stress in terms of component stresses. ′ = (x+ y - xy + 3xy)1/2 ′ = (x+ 3xy)1/2 ′ = (1+ 2 - 12)1/2 2 2 2 2 2 2 2 In most cases y= 0 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 (′). Mechanical Engineering Department

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

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

Normal & Shear components of stress