MECH 401 Mechanical Design ApplicationsDr. M. K. OMalley Master NotesSpring 2007Dr. D. M. McStravickRice University

Design ConsiderationsStress Yield Failure or Code ComplianceDeflectionStrainStiffnessStability Important in compressive members

Stress and strain relationships can be studied with Mohrs circleOften the controlling factor for functionality

Deflection [Everythings a Spring]When loads are applied, we have deflectionDepends onType of loadingTensionCompressionBending TorsionCross-section of memberComparable to pushing on a springWe can calculate the amount of beam deflection by various methods

SuperpositionDetermine effects of individual loads separately and add the results [see examples 4-2,3,4]Tables are useful see A-9May be applied ifEach effect is linearly related to the load that produces itA load does not create a condition that affects the result of another loadDeformations resulting from any specific load are not large enough to appreciably alter the geometric relations of the parts of the structural system

Deflection --- Energy MethodThere are situations where the tables are insufficientWe can use energy-methods in these circumstancesDefine strain energy

Define strain energy density**

V volume

Put in terms of s, e

Example beam in bending

Castiglianos Theorem[He was a Grad Student at the Time!!]Deflection at any point along a beam subjected to n loads may be expressed as the partial derivative of the strain energy of the structure WRT the load at that point

We can derive the strain energy equations as we did for bending Then we take the partial derivative to determine the deflection equationPlug in load and solve!AND if we dont have a force at the desired point:If there is no load acting at the point of interest, add a dummy load Q, work out equations, then set Q = 0

Castigliano ExampleBeam AB supports a uniformly distributed load w. Determine the deflection at A.

No load acting specifically at point A!Apply a dummy load Q

Substitute expressions for M, M/ QA, and QA (=0)

We directed QA downward and found A to be positiveDefection is in same direction as QA (downward)Q

StabilityUp until now, 2 primary concernsStrength of a structureIts ability to support a specified load without experiencing excessive stressAbility of a structure to support a specified load without undergoing unacceptable deformationsNow, look at STABILITY of the structureIts ability to support a load without undergoing a sudden change in configurationMaterialfailure

BucklingBuckling is a mode of failure that does not depend on stress or strength, but rather on structural stiffnessExamples:

More buckling examples

BucklingThe most common problem involving buckling is the design of columns Compression membersThe analysis of an element in buckling involves establishing a differential equation(s) for beam deformation and finding the solution to the ODE, then determining which solutions are stableEuler solved this problem for columns

Euler Column Formula

Where C is as follows:

C = ;Le=2LFixed-freeC = 2; Le=0.7071LFixed-pinnedC = 1: Le=LRounded-roundedPinned-pinnedC = 4; Le=L/2Fixed-fixed

BucklingGeometry is crucial to correct analysisEuler long columnsJohnson intermediate length columnsDetermine difference by slenderness ratioThe point is that a designer must be alert to the possibility of bucklingA structure must not only be strong enough, but must also be sufficiently rigid

Buckling Stress vs. Slenderness Ratio

Johnson Equation for Buckling

Solving buckling problemsFind Euler-Johnson tangent point with

For Le/r < tangent point (intermediate), use Johnsons Equation:

For Le/r > tangent point (long), use Eulers equation:

For Le/r < 10 (short), Scr = Sy

If length is unknown, predict whether it is long or intermediate, use the appropriate equation, then check using the Euler-Johnson tangent point once you have a numerical solution for the critical strength

Special Buckling CasesBuckling in very long Pipe

Note Pcrit is inversely related to length squaredA tiny load will cause bucklingL = 10 feet vs. L = 1000 feet:Pcrit1000/Pcrit10 = 0.0001Buckling under hydrostatic Pressure

Pipe in Horizontal Pipe Buckling Diagram

Far End vs. Input Load with Buckling

Buckling Length: Fiberglass vs. Steel

ImpactDynamic loadingImpact Chapter 4Fatigue Chapter 6Shock loading = sudden loadingExamples?3 categoriesRapidly moving loads of constant magnitudeDriving over a bridgeSuddenly applied loadsExplosion, combustionDirect impactPile driver, jack hammer, auto crashIncreasingSeverity

Impact, cont.It is difficult to define the time rates of load applicationLeads to use of empirically determined stress impact factorsIf t is time constant of the system, where

We can define the load type by the time required to apply the load (tAL = time required to apply the load)

Static

Gray area

Dynamic

Stress and deflection due to impactW freely falling massk structure with stiffness (usually large)AssumptionsMass of structure is negligibleDeflections within the mass are negligibleDamping is negligibleEquations are only a GUIDEh is height of freely falling mass before its released is the amount of deflection of the spring/structure

Impact Assumptions

Impact Energy Balance

Energy balanceFe is the equivalent static force necessary to create an amount of deflection equal to d Energy Balance of falling weight, W

Impact, cont.Sometimes we know velocity at impact rather than the height of the fallAn energy balance gives:

Pinger Pulse Setup

Pinger

Pressure Pulse in Small Diameter Tubing

1500 Foot Pulse Test