me 270 class notes - part 3
DESCRIPTION
Purdue ME 270 class notesTRANSCRIPT
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Shear-Force & Bending-Moment Diagrams
Distributed Loads
Learning Objectives
1) To evaluate the shear-force and bending-moment diagrams for systems
with discrete loads.
2) To do an engineering estimate of these quantities.
Beam Sign Convention
Distributed load - An upward load is positive
Shear Force - A positive internal shear force causes a clockwise
rotation of beam segment. (i.e., it pushes a left-
facing cross-section upward or a right-facing cross-
section downward).
Bending Moment - A positive internal moment causes compression in
the top fibers of the segment (i.e., clockwise on a
left-facing cross-section or counter-clockwise on a
right-facing cross-section).
Procedure
1. Determine support reactions
2. Specify beam sections origin (left end) to between each discrete load
(force or moment). Be sure V and M are shown acting in the positive
sense.
3. Sum forces vertically to determine V
4. Sum moments at sectioned end to determine M. (This eliminates V
from the moment equation).
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Shear-Force & Bending-Moment Diagrams
Graphical Methods
Learning Objectives
1) To evaluate the shear-force and bending-moment diagrams for systems
with discrete loads.
2) To do an engineering estimate of these quantities.
Beam Sign Convention
Distributed load - An upward load is positive
Shear Force - A positive internal shear force causes a clockwise
rotation of beam segment. (i.e., it pushes a left-
facing cross-section upward or a right-facing cross-
section downward).
Bending Moment - A positive internal moment causes compression in
the top fibers of the segment (i.e., clockwise on a
left-facing cross-section or counter-clockwise on a
right-facing cross-section).
Procedure
1. Determine support reactions
2. Specify beam sections origin (left end) to between each discrete load
(force or moment). Be sure V and M are shown acting in the positive
sense.
3. Sum forces vertically to determine V
4. Sum moments at sectioned end to determine M. (This eliminates V
from the moment equation).
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Normal Stress and Axial Strains
Normal Stress - The stress acting normal to a
surface along an axis of a member.
= Normal Stress = 0
lim = A
F dF
A dA
AVGF= , dA = AA
y z
AVG
F Axial Load = =
A Cross-Sectional Area
where the units are given by:
AVG 2 2
lbs N = = psi or = Pa
in m
Note:
AVG AVG > 0 (Tension) ; < 0 (Compression)
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Axial Strain ( ) – The elongation (or shortening) of a part
along an axis.
0
- xAxial Strain = lim
x
x
x
L* - L L*Avg. Axial Strain = = - 1
L L
L* - 1 ,
L
εx =
(εx )AVG =
εave =
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Hooke’s Law
Hooke’s Law – The nearly linear relationship between
normal stress and axial strain in the proportional region
(typically under low strains)
where
E = Modulus of Elasticity or Young’s Modulus, which is a
measure of the stiffness of a material (E has units of psi or
Pa).
Elongation is the axial direction causes contraction in
transverse directions.
where
= Poisson's ratio (dimensionless)
= Eε
εy = εx = - εx
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Stress-Strain Regions:
Proportional Region (A-B): Nearly linear relationship between
σ and ε, whose slope is E. Hooke’s
Law is valid up to the proportional
limit (σPL).
Elastic Region (A-C): Similar to the proportional region
except between B and C the
material is still behaving elastically
up to the upper yield point (σγP)u,
but is not linear (Hooke’s Law is
not valid).
Yielding Regions (C-E): Plastic deformation begins. A
slight material “softening” occurs
from C-D, down to the lower yield
point (σγP)l. Perfectly plastic
behavior occurs from D-E as the
material strain increases without an
increase in stress.
Strain Hardening Region (E-F): Strain hardening occurs from E-
F up to the “ultimate stress”
(σU).
Necking Region (F-G): Above σU, the material “necks
down” resulting in a significant
reduction in cross-sectional area
and ultimately in fracture (G) at the
fracture stress (σF).
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General Stress-Strain Curves
for Various Materials
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Design for Allowable Stress
In Axially-Loaded Members
Structural Design Strategies
Design for strength
Design for stiffness (avoidance of large deformations)
Design for ductility (avoidance of fracture)
Design for strength (two possibilities)
Yield strength (σYP)
Ultimate Strength (σU)
Factor of Safety (FS)
To avoid failure FS > 1
Typical FS values 1.3 < FS < 3
Failure LoadFS =
Allowable Load
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For yield failure,
For ultimate failure,
allow
FS = YP
YPallow =
FS
allow
FS = U
Uallow =
FS
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Shear Stress Due to Pure Torsion
in Circular Members
Local Shear Stress at radius ( )
where T = applied Torque
ρ = radius at some location in the shaft
J = polar area moment of inertia of the shaft
cross-section
The maximum shear stress occurs at the outer perimeter
of the shaft when (ρ = r)
Two Common Circular Configurations:
Solid Shaft
Tubular Shaft
T( )
J( )
MAX
Tr
J
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Stress Distribution in a Solid Shaft
where r = radius of shaft
and J = 4π
r2
4
MAX 3
T(ρ) 2(ρ) = = (ρ)
Tr 2T = =
J πr
T
J r
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Stress Distribution in a Tubular Shaft
where ro = outer radius of tubular shaft
ri = inner radius of tubular shaft
J = 4 4π
(r - r )2
o i
4 4
MAX 4 4
T(ρ) 2(ρ) = = (ρ)
- r
Tr 2Tr = =
J π - r
o i
o o
o i
T
J r
r
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Power Transmission
where P = Power (work performed per unit of time)
T = Torque
W = Angular speed of shaft
Power often measured in horsepower (hp)
Power can also be expressed as a function of frequency
(f)
where f = frequency (in cycles/sec or Hz)
Note: ω = 2πf
P = T
ft-lb1 hp = 550
s
P = 2 f T
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Shaft Design
When the Power (p) and frequency (f) of a shaft are
known, the torque developed can be determined
Knowing the torque (T) and the allowable shear stress
(τAllow), the size of the shaft’s cross-section can be
determined (assuming small strains in the linear elastic
range).
A common geometric (design) parameter is
Remember, J = 4π
r 2
Solid Shaft
J = 4 4
o i
π r - r
2 Tubular Shaft
T = P / 2 f
Allow
T =
τ
J
r
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Stresses Due to Pure Bending
Pure Bending
Sign Convention
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Pure Bending in a Beam
When V = 0 and M ≠0 => Pure Bending
Below Pure Bending occurs between B and C.
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Stress and Strain Distributions Due to Pure Bending
x x
1 = ε (y) = (y)ε
x x x
E = (y) = E(ε ) = (y)
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Stress Distribution
where M = resultant internal moment about the neutral
axis of the cross-section
y = perpendicular distance from the neutral axis
I = second area moment of inertia of the cross-
sectional area about the neutral axis
E = Young’s modulus of elasticity
Second Area Moment of Inertia
x x
- M(y) = ( ) =
Iy
A
2I = y dA
xx x
- M(y) = (y) = =
E E Iε
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Maximum Normal Stress
1. Only normal stress σx exist at the cut.
2. Neutral axis passes through centroid.
3. Normal stresses vary linearly in the y-direction.
4. Normal stresses are constant in the z-direction.
5. Normal stress is zero at neutral axis.
6. Max normal stress exists at most outer surface of
beam.
MAXx MAX
M y =
I
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Second Area Moment of Inertia By Integration
Recall
Similar to centroids by integration except there is a y2
(rather than a y) term in the integrand.
2
xA
2
yA
2
o x yA
I = y dA
I = x dA
J = r dA = I + I
A
2I = y dA
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Stresses Due to Combined
Transverse Shear and Bending
Previously, we considered pure shear and pure bending
separately.
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For combined transverse shear and bending
where: V = shear force at cross-section
A* = cross-sectional area above the element
y* = centroid of the area above the element
I = centroidol second area moment for the entire
cross-section
t = depth of the beam at the location of the
stress element of interest
For a rectangular cross-section beam of height (h) and
width (t), (remember 31
I = th12
),
V A y =
I t
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Combining equations gives,
hA* = ( - y) t ,
2
1 hy* = ( +y) .
2 2
3
h 1 hV - y t ( + y)
2 2 2
th t
12
=
22
2
6V h = - y
Ah 4
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Note:
1. τ distribution is quadratic
2. τ = 0 at y = ± h/2
3. τ = max at the neutral axis (y = 0)
MAX
3V =
2A
where:
V = shear force at cross-section
A = cross-sectional area