Ken Youssefi Mechanical Engineering Department 1 Normal & Shear components of stress Normal stress...
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Transcript of Ken Youssefi Mechanical Engineering Department 1 Normal & Shear components of stress Normal stress...
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
Ken YoussefiMechanical Engineering Department 2
Normal & Shear components of stressTwo Dimensional Case
xy
y
xy
x
xy
xy
x
y
Ken YoussefiMechanical 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 YoussefiMechanical Engineering Department 4
Direct shear is produced where there is no bending (or stress caused by bending is negligible
Ken YoussefiMechanical 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 YoussefiMechanical 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 YoussefiMechanical 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 YoussefiMechanical Engineering Department 8
Torsional Stress - examples
Structural member Power transmission
Mixer
Ken YoussefiMechanical 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 YoussefiMechanical Engineering Department 10
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
Ken YoussefiMechanical 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 YoussefiMechanical Engineering Department 12
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
Ken YoussefiMechanical Engineering Department 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
Ken YoussefiMechanical 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
xxεx = x
E
εy = - v εx
εz = - v εx
Ken YoussefiMechanical Engineering Department 15
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