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# Beam deflections using singularity functions

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### Beam deflections using singularity functions

1. 1. MEEG 202 Strength of Materials 26-06-201226-06-2012 Beam Deflections Using Singularity Functions
2. 2. Admin • Preview Example Problems BEFORE Class • Homework #4 Due Thursday • Next class Thursday 1-4  Please be on Time
3. 3. Lesson Objectives 1. Understand the meaning of a singularity functions. 2. Be able to integrate singularity functions. 3. Be able to calculate the deflection of a beam by using singularity functions.beam by using singularity functions.
4. 4. Surprise Quiz Name:_____________________________ Roll #____________________ 1) The drawing below shows a simply supported beam with a center load. The cross section is triangular. a) At what point in the cross section do you expect the transverse shear stress (due to the shear load) to be the greatest? b) What is its value? Sketch: x A m2 C m1 kN50 B mm90 mm90 My bhIxx  3 36 1 2) What was the subject of the pre-class example problem for today? Ib SQ I My    
5. 5. Beam Deflection by Integration Summary F x Deflection at any point is y dx yd EI   4 4  xfy dx dy dx yd EI M dx yd EI S dxEI       2 2 3 3 4
6. 6. Review Beam Deflection by Integration ProcessF x Deflection at any point is y dx yd EI   4 4 xF S M A Fl FlFxM  2) Put M(x) into the integral:  Mdx dx dy EI    dxFlFx dx dy EI 3) Integrate to get slope function 21 CFlxFx dy EI  C A 1) Write Moment Function as a function of x  xfy dx dy dx yd EI M dx yd EI S dxEI       2 2 3 3 4 3) Integrate to get slope function 1 2 2 1 CFlxFx dx dy EI  4) Integrate again to get deflection 21 23 1 2 2 1 6 1 2 1 CxCFlxFxEIy dxCFlxFxEIy          5) Apply slope and deflection boundary conditions to get C1 and C2.     00 00   y  At y=0 0 0 2 1   C C Therefore:
7. 7. Superposition Example  mkN /10 x A m2 B kN40 mm150 mm100 F x A l EI Fl y 3 3 max  lx EI Fx 2 2   mN / x A l B EI l y 8 4  max  22 33 6 xlxl EI x     lx EI Fx y 3 6 2   22 2 46 24 xlxl EI x y  
8. 8. Singularity Functions • Know as Macaulay’s Method for finding beam deflections Unit Step FunctionUnit Step Function x 1 a   0 axxf  1 0 0 0   ax ax What does the Integral of ax ax   0 ax  Look like? dxax  0
9. 9. Singularity Functions • Unit Ramp Function 1   1 axxf  ax  1 0 ax  10 axdxax    1 axxf  x 1 a axax ax   1 0 What is the Integral of ax ax   1 ax  ? 1 2 2 1 ax dxax     22 0 2 22 2 axax ax      ax ax  
10. 10. Singularity Functions What is the Derivative of the Unit Step Function? x   0 axxf  1 a This is the Unit Impulse (Concentrated Force Function)   1  axxf 1 0 1 1     ax ax ax ax   01 axdxax   x   1  axxf  a This is the Unit Impulse (Concentrated Force Function)
11. 11. Singularity Functions x   1  axxf  a What is the Derivative of the Unit Impulse Function? The derivative of the Unit impulse is the unit doublet (or concentrated moment function)?   2  axxf     2 2 0 ax ax ax ax   12   axdxax x   2  axxf a
12. 12. Singularity Function Summary     2 2 0 ax ax ax ax   12   axdxax x   2  axxf a 1 0 1 1     ax ax ax ax   01 axdxax     1  axxf  x 1 a   0 axxf  1 0 0 0   ax ax ax ax   10 axdxax  axdxax  x a x 1 a   1 axxf  axax ax   1 1 0 ax ax   1 2 2 1 ax dxax  
13. 13. Singularity Function Process 1) Write the load function w(x) in terms of singularity functions. 2) Integrate again to get S(x) 3) Integrate twice to get M(x) 4) Integrate again to get EIθ(x) plus an dx yd EI S dx yd EI    3 3 4 4 4) Integrate again to get EIθ(x) plus an integration constant 5) One more integration gets you EIy(x) with another integration constant 6) Use boundary conditions to find integration constants.  xfy dx dy dx yd EI M dxEI     2 2 3
14. 14. Find a function that describes the deflection of the beam shown at right as a function of x. Solution: Problem Type: Find: Given: The figure of the simply supported beam at right. y(x) Beam Deflections Example 10 – Beam Deflection Using Singularity Functions First find the reactions. l Fa R FalR M C C A    0 0 Now write an equation for the loading in terms of singularity functions. Sketch: x A l C a F x A l C a F B B AR CR l Fb R FblR M A A C    0 0 b Now write an equation for the loading in terms of singularity functions. Next, Integrate w(x) to get S(x), the shear loading function No constant of integration is needed since this completely describes the shear.   111   lx l Fa axFx l Fb xw   000 lx l Fb axFx l Fa xS  Next, Integrate S(x) to get M(x), the bending moment function No constant of integration is needed again since this completely describes the moment function.   111 lx l Fa axFx l Fb xM 
15. 15. Note that the first term in the singularity function will always be evaluated as x2 since x is always greater than 0 and the last term will always be zero since x≤l. We can then simplify the last two equations. Now integrate to get EIθ(x)   1 222 222 Clx l Fa ax F x l Fb xEI  Now we need to have a constant of integration to make sure any physical boundary conditions are met. Lastly, integrate one more time to get EIy(x)   21 333 666 CxClx l Fa ax F x l Fb xEIy  We need 1 more constant of integration.   1 22 22 Cax F x l Fb xEI    21 33 66 CxCax F x l Fb xEIy  66l     0 00   ly y Now apply Boundary Conditions And      0 0000 00 6 0 6 0 66 2 2 21 33 21 33     C C CCa F l Fb EI CxCax F x l Fb xEIy          22 1 1 33 1 33 1 33 21 33 6 66 0 66 0 0 66 0 66 bl l Fb C lCb F l l Fb lCal F l l Fb ClCal F l l Fb EI CxCax F x l Fb xEIy      alb sub 
16. 16. Now substitute C1 and C2 into the deflection function and solve for y(x) This equation completely describes the displacement, if we want we can split it into two continuous functions. One for the interval 0<x≤a and one for a≤x<l              3222 2233 2233 6 6 666 1 axlblxbx EIl F xy xblbaxlbx EIl F xy xbl l Fb ax F x l Fb EI xy             222 6 blx EIl Fbx xy  For 0<x≤a For a≤x<l       axlblxbx EIl F xy 6 3222  Reflection: This result can now be used for any simply supported beam with a single load in between the supports. It can also be used in conjunction with superposition.                   lxax EIl xlFa xy laxalxaalxax EIl F xy axlallxxal EIl F xy alb EIl 2 6 23 6 6 6 22 32323 3222      
17. 17. To do for next time • Next Class on Thursday 1-4  We will practice beam deflection problems
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