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Max bending in tapering rod 
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#1
Feb2013, 01:18 PM

P: 144

A uniformly tapering cantilever of solid circular crosssection has a length L and
carries a concentrated load at the free end. The diameter at the fixed end is D and at the free end d. Show that the position of maximum bending stress occurs at a section (d/(2(Dd))*L distance from the free end. I know that the max bending is when shear = 0 I cant begin to use any formulas because to take moments i need to know where the centre of gravity would be for a tapering rod, which i dont, also how would i use the ∏*d^4/64 to get the second moment of area on a tapering rod? Thanks 


#2
Feb2013, 02:41 PM

P: 696

quote: "I know that the max bending is when shear = 0 ". This is true of bending moment but not necessarily of bending stress. You need to look at the bending stress formula.



#3
Feb2013, 02:43 PM

P: 696

"how would i use the ∏*d^4/64 to get the second moment of area on a tapering rod?"
Express d in the above formula as a function of x. It's not the same d as given in the question. 


#4
Feb2013, 02:44 PM

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Max bending in tapering rod
A concentrated load is applied at the free end. You can construct a shear and bending moment diagram without knowing anything about the beam structure except that it is continuous and it is fixed at one end.
Once you have the bending moment calculated as a function of position along the length of the beam, then you can apply what you know about the cross section to identify the location of maximum bending stress. 


#5
Feb2013, 02:44 PM

P: 696

"i need to know where the centre of gravity would be" Why assume that the cantilever is lying horizontal?



#6
Feb2013, 03:09 PM

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My point is, if a concentrated load is applied at the free end of the cantilever, knowledge of the center of gravity of the beam is not required in order to calculate the bending moment.



#7
Feb2013, 04:12 PM

P: 144




#8
Feb2013, 04:26 PM

P: 696

Consider a section X distance x from the free end. Can you express the moment and shear as functions of x (ignoring selfweight)? Can you express the internal and external diameters at section X as a function of x? If necessary draw graphs.



#9
Feb2013, 05:17 PM

P: 144




#10
Feb2013, 07:09 PM

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You are given the diameter D at the fixed end, and diameter d at the free end.
You are also told that the diameter of the beam tapers uniformly (what does that suggest?) How would the bending moment vary as a function of location from the fixed end? What is the formula for bending stress given the value of a moment M(x)? There are no numbers to work with. The answer is given in algebraic terms. You have received several hints about how to work this problem. Now is the time to put pencil to paper and do some work. 


#11
Feb2113, 03:41 PM

P: 144

I have done some work, thats why i am posting on here because i have had no luck. I dont know the equation for the maximum bending position 


#12
Feb2113, 04:16 PM

P: 144

I have the
shear force at x = concentrated laod at free end(W) Bending moment = W*x diameter equation D((Dd)/L)*x Is this correct? Where do i go next? 


#13
Feb2113, 04:53 PM

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What is the bending stress, given what you know about the bending moment and the diameter of the beam at a location x?



#14
Feb2113, 05:16 PM

P: 696

Your bending moment equation is inconsistent with your diameter equation. From which end do you measure x? It has to be the same in each case. My suggestion to measure it from the free end avoids having to know the reactions at the fixed end (although that isn't a big issue).



#15
Feb2113, 05:18 PM

P: 144

I get stress = ((w*x)*(0.5*(D(Dd*x)/L))) / pi*(D(Dd*x)/L)^4/64



#16
Feb2113, 05:22 PM

P: 144




#17
Feb2213, 05:01 AM

P: 144

Can anybody help me with the equation for the position of max stress. i dont see how i can get to my answer from the M/I = stress/y equation tihout having terms of W and Pi in my answer



#18
Feb2213, 05:49 AM

P: 696

Answer to post #16 is that you can test this proposal for yourself by letting x=0 and x=L to see if you get sensible answers (You don't). But it's nearly right.



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