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

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

Ken Youssefi Mechanical Engineering Department 2

Normal & Shear components of stress

Two Dimensional Case

xy

y

xy

x

xy

xy

x

y

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

Direct Shear

Ken Youssefi Mechanical Engineering Department 4

Direct shear is produced where there is

no bending (or stress caused by bending

is negligible

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)

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

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)

Ken Youssefi Mechanical Engineering Department 8

Torsional Stress - examples

Structural member Power transmission

Mixer

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

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

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 - 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

Ken Youssefi Mechanical Engineering Department 12

Maximum Shear Stress – Mohr’s Circle

3 1 2

Mohr’s circles for a 3D case

12

13

23

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

Ken Youssefi Mechanical Engineering Department 13

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

3=0 1 2

12

13

23

13

12

23

1 and 2 have the same sign, both

positive or negative.

3=0 1 2

1 and 2 have the opposite sign.

max = 13 = 1

2 max = 12 =

1 - 2

2

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 =

εy

εx x

y

z

Uniaxial state of stress

x x εx = x

E

εy = - v εx

εz = - v εx

Ken Youssefi Mechanical Engineering Department 15

Stress – Strain Relationship Biaxial state of stress

x x

y

y

Triaxial state of stress

x x

y

y

z

z

εx = x

E = x

E

y

E v

εy = y

E v εx

y

E

x

E v =

v εx εz = v εy

εx = x

E

y

E

z

E v v

εy = y

E

x

E

z

E v v

εz = z

E

x

E

y

E v v

v εy

Strain in the x direction due

to the force in the y direction