structural design. introduction it is necessary to evaluate the structural reliability of a proposed...

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– 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– 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

• The yield stress for a polymer is 100,000 psi. What is the allowable design stress for the polymer in the following application?

2” 0.5”””

Stress Concentration Problem

psi

K

inchesd

inchesw

wdwdwdK

Knormal

41322

42.2

5.0

0.2

)/(53.1)/(66.3)/(13.33

max

32

max

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

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

• 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

• The maximum stress occurs at the points furthest from the neutral surface on the outer surface

• 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

IcMmm /*

IyM pp /*

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

yAM

ydAM x

*

*

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

2* ydAI x

Parallel Axis Theorem

• Used to determine moment of inertia values for complicated geometries by finding the moment of inertia about any axis

• 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

]*[ 2, iiicT yAII

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)

c

IZ

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 II

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

)*( iiix yAMM

Area Properties

• The total area of the composite area is the sum of the individual areas

• The distance between the reference axis and the centroid axis of the composite area is

iAA

i

ii

i

ic A

yA

A

My

)*(

Area Properties

• The moment of inertia for the entire composite section about the composite sections centroid axis is

• 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

]*[ 2, iiicT yAII

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

• y is the deflection, w is the load, L in the unsupported length, E is the modulus and I is the moment of inertia

IE

Lwy

**384

**5 4

max

Structural Design Samples

• Rectangle

• I-beam

Rectangle

.125”

6.0”

4

3

001.0

"125.0

"0.612

*

inI

h

b

hbI

I-Beam

Fig 4.26 p218 malloy

I-Beam

1.25 inches wide

1 inch high 0.25 inch wall thickness

I-Beam

Solid 1.25x1.00 in 2 void areas 0.5x0.5 in

minus

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

4

33

33

,,

2,

0938.0

12

)5.0(*5.0*2

12

)1(*25.1

12

**2

12

*

*2

0

]*[

inI

ininininI

hbhbI

III

y

yAII

T

T

vvssT

voidcsolidcT

iiicT

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

Beam centroid axis Middle centroid axis

Bottom centroid axis

Top centroid axis

0.25

0.125

0.375

1

3

2

I-Beam

431

23

1

311

1,

1

1

1

2,

0456.0

)375.0(*)25(.*)25.1(12

)25(.*)25.1(

12

*

25.0

25.1

375.0

]*[

inII

ininininin

I

hbI

inh

inb

iny

yAII

c

iiicT

I-Beam

4

321

42

3

2

322

2,

2

2

2

2,

0938.0

0026.0

12

)50.0(*)25.0(

12

*

50.0

25.0

0

]*[

inI

IIII

inI

ininI

hbI

inh

inb

y

yAII

T

T

c

iiicT

Use of Ribs

• Ribs are used to improve stiffness by increasing the moment of inertia of a part

• Ribs are usually placed along the non-appearance surface in the direction of maximum stress

Rib Calculation

0.15” at base with 1draft angle for ribs

0.5”

4.0”

0.125”

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

inb

ininb

hBb

aBb

ha

1325.0

deg)1tan(*)5.0(*215.0

tan**2

*2

tan*

b

h

B

a

Rib Calculation – Trapezoid Formulas

Centroid

d

c

y1

b

)(

)**4(*

36

)(

)*2(*

3

)(*2

223

1

cb

ccbbdI

cb

cbdy

cbd

A

c

4

1

2

0015.0

2552.0

0706.0

1325.0

15.0

5.0

inI

iny

inA

inc

inb

ind

c

Rib Calculation – Rectangle Formulas

y=0.5625in 0.5in

0.0625in

4

3

2

000651.0

12

12504

5625.0

5.0

)125.0(*)0.4(

inI

in).in)*((I

y

inA

ininA

c

c

Rib Calculation – Centroid Axis Location

Rectangle centroid axis

Trapezoids centroid axis

Composite centroid axis

Reference Axis

iny

inin

ininininy

A

yAy

yAM

A

My

c

c

i

iic

iii

i

ic

4711.0

)0706.0*35.0(

)2552.0*0706.0(3)5625.0*5.0(

)*(

)*(

Reference to Relative LocationAxis Centroid Composite

22

22

Rib Calculation – Total Moment of Inertia

4

224

224

2,

0192.0

)4711.02552.0(*)0706.0(0015.0[*3

])4711.05625.0(*)5.0(0065.0[

]*[

section ribbedfor Inertia of Moments

inI

inininin

ininininI

yAII

T

T

iiicT

Deflection Calculation

0.15” at base with 1draft angle for ribs

0.5”

4.0”

0.125”

Force=100lbs distributed

inchesy

ininlb

ininlb

y

IE

Lwy

inL

inlbsw

inI

psiEpe

0174.0

0192.0*000,250*384

)4(*25*5

**384

**5

4

4/100

0192.0

000,250

max

42

4

max

4

max

4

Formulas Bending Moments

Properties of Sections, Moment of Inertia

structural design. introduction it is necessary to evaluate the structural reliability of a proposed...

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