ken youssefi mechanical engineering department 1 normal & shear components of stress normal...

Ken YoussefiMechanical Engineering Department 1

Normal & Shear components of stressNormal stress is perpendicular to the cross section, (sigma).

Shear stress is parallel to the cross section, (tau).

3D case

x

y

z

xy

x

xz

y

yz

yx

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


Normal & Shear components of stressTwo Dimensional Case

xy

y

xy

x

xy

xy

x

y


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

Direct Shear


Direct shear is produced where there is no bending (or stress caused by bending is negligible


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)

Transverse Shear


In beam loading, both bending stress and shear stress due to transverse loading are applied to particular section.

Maximum shear stress due to bending


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)


Torsional Stress - examples

Structural member Power transmission

Mixer


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


Principal Stresses – Mohr’s Circle

3D Case1 > 2 > 3

2

223 - (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 22

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


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 - 12 - 13 - 23)1/2

2 2 2

3D Case

′ = (x + y - xy + 3xy)1/2

2 2 2

Substituting for 1, 2 from Mohr circle, we have the von Mises stress in terms of component stresses.

′ = (x + 3xy)1/2

2 2 In most cases y = 0

2D Case,

′ = (1 + 2 - 12)1/2

2 2

3 = 0


Maximum Shear Stress – Mohr’s Circle

3 12

Mohr’s circles for a 3D case

12

13

23

max = largest of the three shear stresses, in this case 13


Maximum Shear Stress – Mohr’s CircleMohr’s circles for a 2D case

3=0 12

12

13

23

13

12

23

1 and 2 have the same sign, both positive or negative.

3=0 12

1 and 2 have the opposite sign.

max = 13 =1

2max = 12 =

1 - 2

2


Stress – Strain Relationship

Poisson’s Ratio, v

Load

v = Strain in the y direction

Strain in the x direction=

εy

εx x

y

z

Uniaxial state of stress

xxεx = x

E

εy = - v εx

εz = - v εx


Stress – Strain RelationshipBiaxial state of stress

xx

y

y

Triaxial state of stress

xx

y

y

z

z

εx = x

E=

x

E

y

Ev

εy = y

Ev εx

y

E

x

Ev=

v εxεz = v εy

εx = x

E

y

E

z

Evv

εy = y

E

x

E

z

Evv

εz = z

E

x

E

y

Evv

v εy

Strain in the x direction due to the force in the y direction

ken youssefi mechanical engineering department 1 normal & shear components of stress normal...

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