MECH 401 Mechanical Design Applications

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1 MECH 40 Mechanical Design Applications Dr. M. K. O Malley Master Notes Spring 008 Dr. D. M. McStravick Rice University

2 Design Considerations Stress Deflection Strain Stiffness Stability Often the controlling factor Stress and strain relationships can be studied with Mohr s circle

3 Deflection When loads are applied, we have deflection Depends on Type of loading Tension Compression Bending Torsion Cross-section of member Comparable to pushing on a spring We can calculate the amount of beam deflection by various methods

4 Superposition Determine effects of individual loads separately and add the results Tables are useful see A-9 May be applied if Each effect is linearly related to the load that produces it A load does not create a condition that affects the result of another load Deformations resulting from any specific load are not large enough to appreciably alter the geometric relations of the parts of the structural system

5 Deflection There are situations where the tables are insufficient We can use energy-methods in these circumstances Define strain energy Define strain energy density** V volume Put in terms of σ, ε 0 Fd U dv du μ d ε σ μ ε 0 dv E U du dv dv du E d E d E σ μ μ σ ε σ μ ε ε ε σ ε σ

6 Eample beam in bending ( ) d EI M U d EI da y M dad EI y M dv EI y M U da y I ) ( ) ( f EI M dad dv dv EI y M U dv E U I My σ σ

7 Castigliano s Theorem Deflection at any point along a beam subjected to n loads may be epressed 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 equation Plug in load and solve! If there is no load acting at the point of interest, add a dummy load Q, work out equations, then set Q 0 i U F i

8 Castigliano Eample Beam AB supports a uniformly distributed load w. Determine the deflection at A. No load acting specifically at point A! Apply a dummy load Q Q Substitute epressions for M, M/ Q A, and Q A (0) We directed Q A downward and found A to be positive Defection is in same direction as Q A (downward) 4 wl 8EI M ( ) M Q U Q Q L A ( w )( ) A A A EI A 0 A L 0 M M EI Q w A d d 4 wl 8 EI

9 Stability Up until now, primary concerns Strength of a structure Material failure It s ability to support a specified load without eperiencing ecessive stress Ability of a structure to support a specified load without undergoing unacceptable deformations Now, look at STABILITY of the structure It s ability to support a load without undergoing a sudden change in configuration

10 Buckling Buckling is a mode of failure that does not depend on stress or strength, but rather on structural stiffness Eamples:

11 More buckling eamples

12 Buckling The most common problem involving buckling is the design of columns Compression members The 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 stable Euler solved this problem for columns

13 Euler Column Formula P crit cπ EI L Where C is as follows: P crit π EI L e C ¼ ;LeL Fied-free C Fied-pinned C Rounded-rounded Pinned-pinned C 4; LeL/ Fied-fied

14 Buckling Geometry is crucial to correct analysis Euler long columns Johnson intermediate length columns Determine difference by slenderness ratio The point is that a designer must be alert to the possibility of buckling A structure must not only be strong enough, but must also be sufficiently rigid

15 Buckling Stress vs. Slenderness Ratio

16 Johnson Equation for Buckling

17 Solving buckling problems Find Euler-Johnson tangent point with L e ρ π E S y For L e /ρ < tangent point ( intermediate ), use Johnson s Equation: For L e /ρ > tangent point ( long ), use Euler s equation: For L e /ρ < 0 ( short ), S cr S y S S cr cr π E S y Le ρ S y 4π E Le ρ 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

18 Special Buckling Cases Buckling in very long Pipe P crit cπ EI L Note Pcrit is inversely related to length squared A tiny load will cause buckling L 0 feet vs. L 000 feet: Pcrit000/Pcrit Buckling under hydrostatic Pressure

19 Pipe in Horizontal Pipe Buckling Diagram

20 Far End vs. Input Load with Buckling

21

22 Buckling Length: Fiberglass vs. Steel

23 Impact Dynamic loading Impact Chapter 5 Fatigue Chapter 7 Shock loading sudden loading Eamples? 3 categories Rapidly moving loads of constant magnitude Driving over a bridge Suddenly applied loads Eplosion, combustion Direct impact Pile driver, jack hammer, auto crash Increasing Severity

24 Impact, cont. It is difficult to define the time rates of load application Leads to use of empirically determined stress impact factors If τ is time constant of the system, where τ π m k We can define the load type by the time required to apply the load (t AL time required to apply the load) Static Gray area Dynamic t AL > 3τ τ < t AL < 3τ t AL < τ

25 Stress and deflection due to impact W freely falling mass k structure with stiffness (usually large) Assumptions Mass of structure is negligible Deflections within the mass are negligible Damping is negligible Equations are only a GUIDE h is height of freely falling mass before its release is the amount of deflection of the spring/structure

26 Impact Assumptions

27 Impact Energy Balance

28 Energy balance F e is the equivalent static force necessary to create an amount of deflection equal to ½ because spring takes load gradually ( ) s e static e e static e W F W F k F k W F h W s e s s s s h W F h h W h W ) (

29 Impact, cont. Sometimes we know velocity at impact rather than the height of the fall An energy balance gives: s e s s g v W F g v gh v

30 Pinger Pulse Setup

31 Pinger

32 Pressure Pulse in Small Diameter Tubing

33 500 Foot Pulse Test

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