Use the principle of stress and deformation analysis in
stress analyses problem.
Representing stress on a stress element. Torsion in members having noncircular cross
Mohr’s circle for plane stress
Direct stresses: tension and compression. Vertical shearing stress.
Deformation under direct axial loading. Stress due to bending.
Direct shear stress. Flexural center for beams.
Torsional shear stress. Combined normal stresses: superposition principle.
Torsional deformation. Stress concentrations factors.
Types of loading and stress ratio
REPRESENTING STRESS ON A
STRESS ELEMENT General case of combined stress
Positive shear stresses tend to rotate the element in a
clockwise direction Consider a small stress
element which combined
Negative shear stresses tend to rotate the element in a
Stresses can be normal
stresses and/or shear
Consider only two
dimensional stresses, i.e. in
the x and y direction.
Combination of applied stresses Max shear stress
(normal and shear) that produces
maximum normal stress. ⎛ σ x −σ y ⎞
Max/min principal stress τ max = ⎜
⎜ ⎟ + τ xy 2
⎝ 2 ⎠
σ x +σ y ⎛ σ x −σ y ⎞
σ1 = + ⎜
⎜ 2 ⎟ + τ xy 2
⎟ Angle for maximum shear stress element
2 ⎝ ⎠
φσ = arctan[− (σ x − σ y ) 2τ xy ]
σ x +σ y ⎛σ x −σ y ⎞ 2
σ2 = ⎜ 2 ⎟ + τ xy
− ⎜ ⎟
2 ⎝ ⎠ Average normal stress
Angle of principal stress element σ avg = (σ x + σ y ) 2
Measured from the positive x-axis
(shown in figure). Positive sign
calls for a clockwise rotation of
element and vise versa.
φσ = arctan[2τ xy (σ x − σ y )]
MOHR’S CIRCLE FOR PLANE
The Mohr’s circle enables the following computation
with relative ease and better accuracy:
Max/min principal stresses and their
Max shear stress and the orientation of the
Value of normal stresses that act on the planes
where the max shear stresses act.
Values of the normal and shear stresses that
act on an element with any orientation.
Drawing the Mohr’s circle
Perform stress analysis and determine the magnitudes and
directions of the normal and shear stresses acting at the point of
interest, using the convention:
Tensile stresses are positive; compressive negative.
Shear stresses that tend to rotate element clockwise are
positive; otherwise negative.
Set up the normal and shear stress (σ, τ) plane and locate the
point on it.
Connect the points. The line that cuts the σ-axis is the center of
the circle, O, and radius of the circle, R:
Locating the points: Mohr’
The Mohr’s circle
σ x −σ y
R= + τ xy
Case when the both principle
stresses have the same sign
The true maximum shear stress on the element will not
be found if the two principal stresses are of the same
Hoop stress, σ x = pD 2t Where;
p = internal pressure in cylinder
D = diameter of the cylinder
Longtitudinal stress, σ y = pD 4t t = thickness of the cylinder wall
Use 3-D stress element
Mohr’s circle special stress
conditions Pure torsional shear
Pure uniaxial tension
Pure uniaxial compression Uniaxial tension combined with torsional shear
DIRECT STRESSES: TENSION
Stress: Internal resistance offered by a unit area normal stress, σ =
[N m2 ]
of material to an external load
Perpendicular to element Conditions:
Compressive stresses: Crushing action. Negative • load-
load-carry member must be straight
by convention • line of action of the load must pass through the
centroid of cross section of the member
Tensile: Pulling action. Positive by convention. • member must be of uniform cross section
• material must be homogeneous and isotropic
• member must be short in the case of compression
DIRECT SHEAR STRESS
DIRECT AXIAL LOADING
Occurs when the applied force tends to cut through the
δ= = [m] member as scissors. Ex: tendency for a key to be sheared
EA E off at the section between the shaft and the hub of a
machine element when transmitting torque(see next
δ = total deformation of the member carrying the axial Apply force is assumed to be uniformly distributed
across the cross section.
F = direct axial load
L = original load length of the member τ=
shearing force F
area in shear
[N m2 ]
E = modulus of elasticity of the material
A = cross-sectional area of the member
σ = direct/normal stress
TORSIONAL SHEAR STRESS
A torque will twist a member, causing a shear stress in
τ max =
[N m2 ]
A general shear stress formula:
Where: The distribution of stress is not uniform across the cross
T = torque
c = radius of shaft to its outside surface
= polar moment of inertia (Appendix)
r = radial distance from the center of the shaft to the
point of interest
Zp = polar section modulus (Appendix)
TORSIONAL DEFORMATION TORSION IN MEMBERS
Where: SECTION SECTION
T = torque τ max =
τ max =
[N m2 ]
L = length of the shaft over which the angle of twist is
being computed θ=
G = modulus of elasticity of the shaft material in shear
J = polar moment of inertia (Appendix)
VERTICAL SHEAR STRESS
Beam carrying transverse loads experience shearing
TORSION IN (V
forces (V) which cause shearing stress:
m 2 ; Q = Ap y [m ]
V = shearing force In the analysis of beams, it is usual to compute the
variation in shearing force across the entire length of
Q = first moment the beam and to draw the shearing force diagram.
I = moment of inertia Vertical shear stress = Horizontal shear stress, because
t = thickness of the section any element of material subjected to a shear stress on
Ap = area of the section above the place where the one face must have a shear stress of the same
shearing force is to be computed magnitude on the adjacent face for the element to be in
y = distance from the neutral axis of the section to the
centroid of the area Ap
STRESS DUE TO BENDING
A beam is a member that carries load transverse to its axis. Where:
Such loads produce bending moments in the beam, which
result in the development of bending stress. M = magnitude of bending moment at the section
Bending stress are normal stresses, that is, either tensile or c = distance from the neutral axis to the outermost
fiber of the beam cross section
The maximum bending stress in a beam cross section will
occur in the part farthers from the neutral axis of the section. I = moment of inertia
At that point, the flexure formula gives the stress:
flexure formula; σ =
[N m2 ]
The flexure formula was developed subject to the following
- Beam must pure bending. No shearing stress and axial loads.
- Beam must not twist or be subjected to torsional load.
- Material of beam must obey Hooke’s law
- Modulus of elasticity of the material must be the same in
both tension and compression.
- Beam is initially straight and has constant cross section.
- No part of the beam shape fails because of buckling or
FLEXURAL CENTER FOR
For design, it is convenient to define the term section BEAMS
modulus, S: To ensure symmetrical bending i.e. no tendency to twist
under loading, action of load pass through the line of
S=I c [m3 ]
The flexure formula then becomes:
σ =M S [ N m2 ]
Then, in design, it is usual to define a design stress, σd ,and
with the bending moment known, then:
S = M σd [m3 ]
If there is no vertical axis symmetry:
When the same cross section of a load-carrying member
is subjected to both a direct tensile and compressive
stress and a stress due to bending, the resulting normal
stress can be computed by the method of superposition:
[N m2 ]
NORMAL EXAMPLE 3.8
Shigley’s text book
Any geometric discontinuities will cause the actual
maximum stress to be higher than the calculated value Diagram shows a round bar subjected to an axial force, F.
Stress concentration factor, Kt= factor by which the Compute maximum stress.
actual maximum stress exceeds the nominal stress Given Kt = 1.60 (Appendix), F = 9800N
(σnom,τnom) predicted by calculations. That is:
σ max = K tσ nom ; τ max = K tτ nom [ N m2 ]
σnom = F/A = ( 9800N)/ [π( 10mm)2/4] = 124.8 MPa
σmax = Kt σnom = (1.60)(124.8) = 199.6 MPa
TYPES OF LOADING AND
From figure below, the highest stress occurs in the fillet
Mean stress, σ m = (σ max + σ min ) 2
Alternating stress, σ a = (σ max − σ min ) 2
Stress ratio, R = σ min σ max
Stress ratio, A = σ a σ m
Static Repeated and Reversed
Load is applied slowly without shock and is held at constant load-
Reversed: when a load-carrying component is subjected to
value. certain level of tensile load followed by a same level of
R = 1.0 because σmax = σmin compressive load.
Repeated: when loading is repeated many thousand times.
Also known as fatigue loading.
Alternating loading with non-zero mean.
R.R Moore Fatigue test
Shock or Impact loading
A special case of fluctuating stress is repeated, one-direction
stress. Loads applied suddenly and rapidly.
E.g. Hammer blow, Rock crushing.
Varying loads that are not regular in their amplitude.
COMPRESSION SHEAR STRESS
SUPERPOSITION DIRECT VERTICAL
PRINCIPLE SHEAR STRESS SHEARING STRESS