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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.

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3: Linear elasticity – axially loaded structures

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  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

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