Structural Design
Introduction
• It is necessary to evaluate the structural
reliability of a proposed design to ensure
that the product will perform adequately
during assembly and service
• The objective of the structural design
process is to generate a part design that will
withstand the loads or imposed deflections
that will be encountered during service
Introduction
• Since it is not always possible to quantify
the loading conditions with certainty, a
series of structural design calculations are
performed
– At anticipated loading during normal service
conditions
– Under conditions that represent the worst case
scenario
Design Methodology
• Design decisions can be made by using structural
engineering relations that allow a designer to
estimate the stresses or strains that occur when a
product is subjected to a mechanical load or
imposed deflection
• Simplification are made regarding material
properties, regularity of form and boundary
conditions to obtain the estimate of how the part
will perform
Design Methodology
• Use classical formulas for stress and strain
– Assume isotropic, homogeneous, elastic behavior
– Compare calculated stresses and strains with the
plastics limits
• Can use finite element analysis program
• Accuracy of results depends on the ability of the
designer to quantify the problem correctly
Quantifying the Design Problem
• The designer needs to assess and specify a number
of factors before beginning the structural
calculations
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–
–
–
–
Part geometry
Type of support
Loading conditions
Environmental conditions
Mechanical properties
Safety factors
Quantifying the Design Problem
• Once these item have been quantified, the
designer
–
–
–
–
–
Performs a series of design calculations
Examines the results
Alters the design if needed
Recalculate
Iterate until the desired results are obtained
Simplifying Part Geometry
• Classical formulas for stress and strain have been
developed for most regular geometries
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–
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Straight, tapered or curved beams
Columns
Plates
Shells
• Common practice to divide parts with a complex
geometries into a series of subparts having regular
geometries and evaluating the stresses and strains
in each subpart
Stress Concentrations
• Stress distribution in parts are influenced by
the presence of features such as corners,
holes or any discontinuity in the design
geometry
• Stresses concentrate at the these
discontinuities, leading to local stress values
that are significantly higher than those in
areas nearby
Stress Concentration Problem
0.5”
””
2”
• The yield stress for a
polymer is 100,000
psi. What is the
allowable design stress
for the polymer in the
following application?
Stress Concentration Problem
 max 
 normal
K
K  3  3.13(d / w)  3.66(d / w) 2  1.53(d / w)3
w  2.0inches
d  0.5inches
K  2.42
 max  41322psi
Stress Concentration Chart
Types of Support
• In order for loaded parts to remain in
equilibrium, the balancing forces are the
reaction forces at the supports
• Most real life products have support
geometries which differ from the idealized
case
• Designer must select the conservative case
Types of Support
• Guided is support at the end of the beams
that prevent rotation, but permits
longitudinal and transverse displacement
• Free or unsupported is when the beam is
totally free to rotate in any direction
• Held is support at the end of the beam that
prevents longitudinal and transverse
displacement but permits rotation
Types of Support
• Simply Supported is support at the end of
the beam that prevents transverse
displacement, but permits rotation and
longitudinal displacement
• Fixed is support at the ends of the beam that
prevents rotation and transverse
displacement, but permits longitudinal
displacement
Idealized Supports
Idealized Supports
Loading Conditions
• Once the part geometry and support conditions
have been established, the loads acting on the
product must be defined and quantified into a form
that is suitable for a stress calculation
• Must quantify the magnitude, direction and type of
load
• Load type can be concentrated at a point, line or
boundary, or distributed over an area
Concentrated and Distributed Loads
Multiple Static Loads
• When a series of loads are acting on a
product, the total stress or deflection can be
found by superposition
• Since stress and deflection are linear in
relation to load, the loads can be added
together to get the total load
Plastic Material Properties
• The stress strain behavior of the plastic
must also be quantified
• The maximum stress or strain estimates that
are calculated from loading must be
compared to the materials stress strain
behavior to determine if these values are
within acceptable limits
Safety Factors
• There are a large number of uncertainties
associated with plastic part design and the
manufacturing process
• The permissible design stress level can be
determined by combining the material
percentage utilization factor, ST, the yield
stress and the stress concentration factor, K
– Design stress < (ST * Yield stress)/K
Safety Factors
• ST=S1*S2*S3*S4
– S1 is risk of injury to people in event of failure
• 1.0 if no risk of injury
• 0.7 is possible injury
• 0.5 if probable injury
– S2 is the processing factor
• 1.0 for neat polymer
• 0.8 for fiber orientated in the direction of maximum stress
• 0.5 for fiber oriented perpendicular to the direction of
maximum stress
Safety Factors
• S3 is the stress calculation accuracy
– 1.0 for finite element analysis
– 0.75 for classical formula
– 0.5 for estimates
• S4 is material degradation
– <1.0 values are material specific
Beams
• A structural member designed to support loads at
various points along the member
• Generally long, straight bars with a cross section
designed to provide the most effective resistance
to shear and bending
• The classical formulas used to calculate stress and
deflection are based on assumption that are not
realistic for plastics
– However the formulas are useful to provide an estimate
Beams
Beams
• The behavior of a beam under load is most easily
described using an example when the beam is
horizontal and the load and reaction are vertical
• Once loaded the beam will deflect
– Material on the concave side will shorten
• It is in compression
– Material on the convex side will lengthen
• It is in tension
Beams
• The neutral surface of the beam
– Is the plane where stress and strain are zero
– Is normal to the plane of the load and is also the
horizontal centroid axis
• Fiber stresses and strains are proportional to
the distance from the neutral surface
Beams
• Fiber stress at any point P, within the boundary of
the beam is
 p  Mp * y / I
• The maximum stress occurs at the points furthest
from the neutral surface on the outer surface
 m  Mm * c / I
• M is the bending moment, I is the moment of inertia, y is the distance
from the neutral surface to p and c is the distance from the neutral
surface to the outer surface
Plane Areas
• Plane areas are cross sections of beams
• In order to design a beam with effective cross
section for a loading application, the area
properties of a beam cross section must be
determined
– The area properties include
•
•
•
•
The centroid axis
The bending moment
The moment of inertia
The section modulus
Plane Areas
Centroid Axis
• Centroid is the point in the plane of an area
about any axis the which the moment of the
area is zero
– Coincides with the center of gravity
• The centroid axis of an area is an axis that
passes through the centroid
Statical moment
• First moment of the area
• It’s the sum of the products obtained by
multiplying each element of area, dA, by
the first power of its distance from the
centroid axis
M x   dA* y
M  A* y
Moment of Inertia
• Second moment of area
• Determines the ability of a beam to support
a load
• Moment of inertia is equal to the sum of the
products obtained by multiplying each
element of area, dA, by the square of the
distance for the centroid axis
I x   dA* y
2
Parallel Axis Theorem
• Used to determine moment of inertia values for
complicated geometries by finding the moment of
inertia about any axis
IT  [ I c,i  Ai * yi ]
2
• Where IT is the total moments of inertia about any axis, Ic is the
moments of inertia with respect to the centroid parallel axis and yi is
the distance between the parallel axis and Ai is the cross sectional area
Section Modulus
• The section modulus(Z) for a symmetrical
cross section is defined as the second
moment of area(I) divided by the distance
from the axis to the most remote point of
that section area(c)
I
Z
c
Area Properties
• Area properties for more complex
geometries are given in handbook and can
be calculated by breaking down the
complex geometry into regular shapes to
produce a composite area
• The moment of inertia of a composite area
is equal to the sum of the individual
moment of inertias
IT   I i
Area Properties
• A void in the cross section can be taken into
account by subtracting the moment of
inertia for the void area
• The statical moment of a composite area is
equal to the sum of the individual statical
moments
M x   Mi  ( Ai * yi )
Area Properties
• The total area of the composite area is the
sum of the individual areas
A   Ai
• The distance between the reference axis and
the centroid axis of the composite area is
yc
M


A
i
i
(A * y )


A
i
i
i
Area Properties
• The moment of inertia for the entire composite
section about the composite sections centroid axis
is
IT  [ I c,i  Ai * yi ]
2
• Where IT is the total moments of inertia, Ic is the moments of inertia
with respect to the centroid axis of each section and yi is the distance
between the sections centroid axis and the composites centroid axis
and Ai is the sections cross sectional area
Deflection
• The formula to use depends on the support
condition and the type of load
– See Formulas for Bending Moments
• Deflection formula for a simply supported beam
subjected to a partially distributed load
 5 * w * L4
ymax 
384* E * I
• y is the deflection, w is the load, L in the unsupported length, E is the
modulus and I is the moment of inertia
Structural Design Samples
• Rectangle
• I-beam
Rectangle
.125”
6.0”
3
b*h
I
12
b  6.0"
h  0.125"
I  0.001in
4
I-Beam
Fig 4.26 p218 malloy
I-Beam
1 inch high
0.25 inch wall thickness
1.25 inches wide
I-Beam
minus
Solid 1.25x1.00 in
2 void
areas
0.5x0.5 in
I-Beam
1.25x0.25 in solid
0.25x0.5 in solid
1.25x0.25 in solid
I-Beam
• Solid minus two voids
– Centroid axis for each of the rectangles are the
same as the centroid axis of the entire
composite beam
I-Beam
I T  [ I c ,i  Ai * yi ]
2
y0
I T  I c , solid  2 * I c ,void
3
3
bs * hs 2 * bv * hv
IT 

12
12
1.25in * (1in) 3 2 * 0.5in * (0.5in) 3
IT 

12
12
I T  0.0938in 4
I-Beam
• Three solid sections
– The centroid axes for the outer sections not the
same as the inner section or as the composite
beam
– Need to use the parallel axis theorm
I-Beam
1
Top centroid axis
0.375
0.25
Beam centroid axis
Middle centroid axis
0.125
Bottom centroid axis
2
3
I-Beam
I T  [ I c ,i  Ai * yi ]
2
y1  0.375in
b1  1.25in
h1  0.25in
3
b1 * h1
I c ,1 
12
(1.25in ) * (.25in )3
I1 
 (1.25in ) * (.25in ) * (0.375in ) 2
12
I1  I 3  0.0456in 4
I-Beam
I T  [ I c ,i  Ai * yi ]
2
y2  0
b2  0.25in
h2  0.50in
3
b2 * h2
I c,2 
12
(0.25in) * (0.50in) 3
I2 
12
I 2  0.0026in 4
I T  I1  I 2  I 3
I T  0.0938in 4
Use of Ribs
• Ribs are used to improve stiffness by
increasing the moment of inertia of a part
• Ribs are usually placed along the nonappearance surface in the direction of
maximum stress
Rib Calculation
4.0”
0.125”
0.5”
0.15” at
base with
1
draft
angle for
ribs
Rib Calculation
• Break into four parts
– Rectangle
– 3 trapezoids
• Find dimensions of trapezoid
• Find areas, centroid axis location and
moment of inertia for each part
• Find overall centroid axis location
• Find overall moment of inertia
Rib Calculation – Trapezoid
Geometry
B
a  h * tan

h
b  B  2*a
b  B  2 * h * tan
b  0.15in  2 * (0.5in) * tan(1deg)
b  0.1325in
b
a
Rib Calculation – Trapezoid
Formulas
b
Centroid
d
y1
c
d
A  * (b  c)
2
d (2 * b  c)
y1  *
3
(b  c)
b  0.15in
d 3 (b 2  4 * b * c  c 2 )
Ic 
*
36
(b  c)
y1  0.2552in
d  0.5in
c  0.1325in
A  0.0706in 2
I c  0.0015in 4
Rib Calculation – Rectangle
Formulas
0.0625in
y=0.5625in
A  ( 4.0in ) * (0.125in )
A  0.5in 2
y  0.5625
( 4in)*(0.125in)3
Ic 
12
I c  0.000651in 4
0.5in
Rib Calculation – Centroid Axis
Location
Rectangle centroid axis
Composite centroid axis
Trapezoids centroid axis
CompositeCentroidAxis LocationRelativeto Reference
M
A
 M  ( A * y )
( A * y )
y 
A
yc 
i
i
i
i
i
i
i
c
i
(0.5in 2 * 0.5625in )  3(0.0706in 2 * 0.2552in )
yc 
(0.5in 2  3 * 0.0706in 2 )
yc  0.4711in
Reference Axis
Rib Calculation – Total Moment
of Inertia
Momentsof Inertiafor ribbed section
I T  [ I c ,i  Ai * yi ]
2
I T  [0.0065in 4  (0.5in 2 ) * (0.5625in  0.4711in)2 ]
 3 * [0.0015in 4  (0.0706in 2 ) * (0.2552in  0.4711in)2
I T  0.0192in 4
Deflection Calculation
Force=100lbs distributed
E pe  250,000psi
I  0.0192in 4
4.0”
w  100lbs / 4in
0.125”
L  4in
0.5”
ymax
0.15” at
base with
1

draft
angle for
ribs
 5 * w * L4

384* E * I
lb
* ( 4in ) 4
in

lb
384* 250,000 2 * 0.0192in 4
in
 0.0174inches
 5 * 25
ymax
ymax
Formulas Bending Moments
Formulas Bending Moments
Formulas Bending Moments
Formulas Bending Moments
Formulas Bending Moments
Properties of Sections, Moment of Inertia
Properties of Sections, Moment of Inertia
Properties of Sections, Moment of Inertia
Properties of Sections, Moment of Inertia
Properties of Sections, Moment of Inertia
Properties of Sections, Moment of Inertia
Properties of Sections, Moment of Inertia
Properties of Sections, Moment of Inertia