1 / 23

3: Linear elasticity – axially loaded structures

3: Linear elasticity – axially loaded structures. . Consider the following prismatic bar of x-sectional area A i)unloaded and ii) in tension:. P.

aurora
Télécharger la présentation

3: Linear elasticity – axially loaded structures

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 3: Linear elasticity – axially loaded structures  Consider the following prismatic bar of x-sectional area A i)unloaded and ii)in tension: P If the load acts through the centroid of the x-section the uniform stress will be =P/A. If it is a homogeneous material then strain will be = /L. If it is linear elastic then E= Stress/Strain. We can use these relations to get: EA=axial rigidity

  2. Modelling a prismatic bar as a spring:  stiffness P=k P= (EA/L)   = P(L/EA) flexibility

  3. Example: Precision measuring device; based on example in Gere and Timoshenko C B A 120mm 180mm 180mm 100mm X-section area CE=150mm2 X-section area BD=200mm2 D A? 10kg E Both are made of steel E=205GPa

  4. Hor. Force=0 A C B 180mm 100mm T find A we must determine displacements of B and C.1) Produce FBD of beam 2) Take moments about B 3) Sum forces in the vertical direction 4) Calculate changes in length. Use:

  5. Loads on beam and columns: C B A Actual load direction for A: 120mm 180mm 180mm 100mm X-section area CE=150mm2 X-section area BD=200mm2 D A? 10kg E Both are made of steel E=205GPa

  6. Forces Moments about B For column FC=+177N Sum vertical forces=0 For column FB=-275N

  7. Deflections:

  8. F A B C B=-0.81μm A ? 180mm 100mm Produce displacement diagram: C=1.04μm and

  9. Produce displacement diagram: A B C B-C A-C 180mm 100mm

  10. Non-uniform bars- 1) Varying load2) Varying cross-section or modulus When a prismatic bar of linear elastic material is loaded only at its ends we use the following equation for calculating the total deflection: Suppose load or cross-section changes along the length of the bar. What do we do?

  11. Abrupt changes in load: L1 PA A Consider the prismatic bar loaded at points A and B L2 B PB

  12. There are two sets of loads, at A and B. If we cut the bar at A to evaluate the length spanned by L1 we must add all loads in front of and hence supported by this length of bar. The load in segment i is Ni. For instance, the length spanned by L2 only supports one load, PB, so N2= PB A L1 L2 A N1= PA+ PB B N2= PB Total deflection at end:

  13. Varying cross-section or modulus P A E1 L1 N1= P B E2 N2= P L2 C

  14. P Example: Surgical probing tool of UHMWPe (section AB) of area A=3mm2 and Ti6Al4V (section BC) of area A=4mm2. Likely loads are P=-10 N A E1 L1=10mm B L2=10mm E2 C

  15. P Example: Surgical probing tool of UHMWPe (section AB) of area A=3mm2 and Ti6Al4V (section BC) of area A=4mm2. Likely loads are P=-10 N A E1 L1=10mm B L2=10mm E2 C

  16. When the bar consists of several prismatic segments joined together, and there are a number of point loads we can use the following expression to evaluate the bar:

  17. Example: Steel bar ABC and beam BDE; find vertical displacement  at C. A 0.225m 0.25m 1) Measured deflection of spring at E is 10mm 2) Segment AB has L=0.5m and area 200mm2. 3) Segment BC has L=0.25m And area 100mm2. E=205 GPa B E D C k=50N/mm 1000N

  18. Solution: • Produce a FBD of beam; • Take moments about D • Calculate FB 0.225m 0.25m Sum of moments about D=0 E (-500N*0.225m)+(FB*-0.25m)=0 B D 500 N E FB=-450N on the beam! It is therefore +450N ON THE COLUMN! B D 500 N -450 N

  19. Solution continued: 550 N 4) Now draw FBD of column 5) Determine FA A Sum of vertical forces=0 450 N FA+450N-1000N=0 FA=550N C 1000 N FB=-450N on the beam! It is therefore +450N ON THE COLUMN!

  20. 550 N Solution continued: A 6) Dissect the column: produce a FBD of both parts: Section AB is in tension under 550 N force and section BC is in tension under a 1000 N force. 7) Calculate deflections: B 1000 N 550 N B C 1000 N

  21. Continuous variation in load or section A prismatic bar of length L, area A, modulus E and density loaded by self weight. Consider a small segment of length dx at distance x from the free end. dx x The flexibility (deflection/unit load)of the segment dx is: dx/EA The load on it =  gxA Deflection of segment of length dx at location x is d=load*flexibility d=PL/EA=(gxA)(dx)/EA= gxdx/E

  22. Integrate this from x=0 to x=L to get the total deflection from 0 to L: In general, we use: Example: What is the deflection due to self weight of a steel flagpole of constant cross-sectional area30m high?

  23. Flagpole: dx x

More Related