1. Oct 22, 2009

### Xaspire88

Hi

Im trying to help a friend of mine out with this problem he has in a solid mechanics class he is taking, but it has been awhile since I took that course so i was hoping you guys could help me help him :)

Here is the problem he gave me.

in that image p1 p2 and E1 E2 are the density and Youngs modulus respectively

So what we are needing to find is the deformation of the 2 material beam as a function of postion (x). He said his prof. instructed him to first find the 'supportive' forces P1 P2 and then solve for the deformation in terms of those. Since this beam is fully constrained we know that the total deformation will be zero (0) and that P1+P2= W where W is equal to the weight force downward acting on the entire beam. Where I have a hard time is knowing how to solve for these P1 & P2 forces, with the equations:

1: P1 + P2 = W
2: 0=(deformation part 1)+(deformation part 2)

first since when I make my 'cuts' in region 1 and 2 to solve for the deformation in that portion of the beam I have another force say p(x) which is dependent on those other forces.

Any insight, helpful links would be appreciated. I know you guys used to help me out with my HW so hopefully you can help me on this. Thanks in advance.

2. Oct 22, 2009

### Xaspire88

Some thoughts that I need clarifying.

1. In the sketch in my last post are P1 and P2 constant through out the beam? Like I said it's been a while since I have studied this material, but I want to say yes.

2. If they are constant, do P1 and P2 share the load W(weight of the beam) equally? Leading me to: Are P1 and P2 = 1/2 W?

Edit:
I fear I may of made a huge mistake on those pages I posted so I took them down.

Equations I used based off of above assumptions.

1. P1=P2=1/2W(total weight)=1/2(p1gAL1+p2gaL2)

2. In region 1 p(x)= P2-w(x)=1/2Wt-w(x)=gA[1/2(p1L1+p2L2)-p1x]

3. In region 2 p(x)= P2-w(x)=1/2Wt-w(x)=gA[1/2(p1L1+p2L2)-p2x]

then using these p(x) solving for$$\delta$$ in each region 1 & 2

Region 1: limits of integration from 0 to x

$$\delta$$(x)=(gx)/(2E1)[(p1L1+p2L2)-p1x]

Region 2: limits of integration from (0 to L1) + (L1 to x)

$$\delta$$(x)=(gL1)/(2E1)[(p1L1+p2L2)-p1L1]+g/(2E2)[[(p1L1+p2L2)x-p2x^2]-[(p1L1+p2L2)L1-p2L1^2]]

Last edited: Oct 22, 2009
3. Oct 25, 2009

### nvn

Xaspire88: I will herein change your reaction forces P1 and P2 at the fixed supports to R1 and R2, to avoid confusion. And like you, I will use p1 and p2 for the densities.

No, the axial force throughout the beam is not constant. You made a mistake by assuming R1 = R2.

When you make your section cut in region 1 (with the free body below the cut), you can see the upward force acting on the section cut is F1(x) = p1*g*A*x - R1, for 0 ≤ x ≤ L1. And when you make your section cut in region 2 (with the free body below the cut), you can see the upward force acting on the section cut is F2(x) = p1*g*A*L1 + p2*g*A*(x - L1) - R1, for L1 < x ≤ (L1 + L2).

The deformation in region 1 is delta1(x) = integral{[F1(x)/(E1*A)]*dx}, integrated from x = 0 to x. The deformation in region 2 is delta2(x) = delta1(L1) + integral{[F2(x)/(E2*A)]*dx}, integrated from x = L1 to x.

Now, to find R1, we know delta2(L1 + L2) = 0, right? Therefore, solve delta2(L1 + L2) = 0 for R1. Afterwards, you have delta(x) = {delta1(x) for 0 ≤ x ≤ L1, delta2(x) for L1 < x ≤ (L1 + L2)}.

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4. Oct 25, 2009

### Xaspire88

Here is a simplified version of this problem.

In this version we just have one material.

Can I do what I have done in that image?

Since total deformation of the beam is 0 can we think of it as the beam first being compressed by its own weight, and then be extended by the upward force F1?

Also the weight force varies over the object but does F1 in this case? Should my limits of integration for F1 be from 0 -> x?

Last edited: Oct 25, 2009
5. Oct 25, 2009

### nvn

To answer your second question in post 4, yes, you can use superposition to solve that problem. To answer your third question, if you apply only an end force to the rod, but you apply no other applied load (not even gravity), then the force in the rod is constant along the length of the rod. Yes, the limits of integration should be 0 to x.

In your post 4 image, delta_w(x) + delta_F1 = 0 is incorrect, and should instead be delta_w(L) + delta_F1 = 0. Otherwise, the image looks OK.

6. Oct 25, 2009

### Xaspire88

*Edit: "To answer your third question, if you apply only an end force to the rod, but you apply no other applied load (not even gravity), then the force in the rod is constant along the length of the rod. Yes, the limits of integration should be 0 to x." You say the force in the rod is constant along the length of the rod but if I integrate from 0->x then i get a F1 as a function of x... so then it would be dependent upon the position within the rod?*

*First Scenario delta_w integrated from 0->L and delta_F1 integrated from 0->x:*

So my limits of integration for my delta_w were wrong and should instead be 0->L in which case:

delta_F1= delta_w(L)

(F1*x)/(A*E)=(p*g*L^2)/(2*E)

and

F1=(p*g*A*L^2)/2x

then from this make a cut in the beam and there will be a p(x) upward on the bottom portion of the beam below the cut p(x)= w(of the top portion) - F1 = p*g*A(L-x) - (p*g*A*L^2)/x

p(x)= p*g*A[(L-x)-(L^2)/2x]

and my delta_x(deformation as a function of x for the whole beam) =

$$\int$$(p*g*A[(L-x)-(L^2)/2x])/(A*E)dx from 0 -> x

works out to

delta_x = (p*g[L*x-(x^2)/2-(L^2)/2*ln(x)])/E

*Second Scenario delta_w integrated from 0->L and delta_F1 integrated from 0->L*

Then F1 = (p*g*A*L)/2

and total deformation:

delta_x= (p*g(L*x-x^2))/2E

I get the feeling I am making this way harder than it actually is and it is driving me crazy

Last edited: Oct 25, 2009
7. Oct 25, 2009

### nvn

In the meantime, post 3 gives you a clear solution to your question in post 1.

8. Oct 26, 2009

### Xaspire88

The more I work on this the more confused I get. For now I just want to be able to clearly understand the simple one material model, and have an answer in front of me that I don't have to second guess. ugh.

9. Oct 26, 2009

### nvn

OK, for a simple one-material model, use post 3, letting L2 = 0. This transforms the solution to a one-material model. Therefore, you already have a clear solution, given in post 3, even for the one-material model.