# Bending moments on a rotating body

1. Sep 4, 2014

### blargh4fun

1. The problem statement, all variables and given/known data

A body (let's call it a rod for simplicity) is in frictionless space, and is composed of 4 smaller sub-rods fused (cannot break) end to end. Each sub-rod has a unique mass (m1, m2, m3 and m4) and length (l1, l2, l3, and l4), but they all have the same diameter d. A force F is applied laterally to the very end of the rod (the end of the m1 rod). The center of mass happens to fall in the third rod due to the values of m1-m4 (which are not necessary).

Calculate the bending moments in each segment of the rod.

I know four blocks is kind of a headache. 3 blocks would be more than enough to help explain the problem.

2. Relevant equations
Eq1. Torque: $\tau = r F = I \alpha$
Eq2. Mom. Inertia of rod about z: $I_{rod} = \frac{mL^2}{12}$
Eq3. Parallel axis theorem: $I=I_{cm} + mr^2$
Eq4. Center of mass: $R=\frac{1}{M}\sum\limits_{i=1}^n m_i r_i$ where $M=\sum m_i$

3. The attempt at a solution
Step1: Center of mass (com) of entire body.
Use Eq4.

Step2: I of entire body about com
Use Eq. 2 and 3

Step3: Calculate alpha of entire body.
Use Eq. 1 using the results from Step1 and Step2:
$r F = I \alpha$
gives $\alpha =\frac{r F}{I}$

Step4: Draw free body diagram for each sub-rod and solve for all forces

note: a, b, c, and d are the vector distances from an intersection to the com.

Body1: $\tau = Fa - F_{2/1}b=I_1 \alpha$
$F_{2/1} = \frac{1}{b} (Fa -I_1 \alpha)$

Body2: $\tau = F_{1/2}b - F_{3/2}c=I_2 \alpha$
$F_{3/2} = \frac{1}{c} (F_{1/2}b -I_2 \alpha)$

Body3: $\tau = F_{2/3}c - F_{4/3}d=I_3 \alpha$
$F_{4/3} = \frac{1}{d} (F_{2/3}b -I_3 \alpha)$ note d<0

Body4: Same as above but no new information is gained.

4. And now...
And this is where I'm stuck. I feel confident about steps 1-3, and 4 seems reasonable. Where do I go from here to get the bending moments? Thanks.

2. Sep 4, 2014

### paisiello2

In your free body diagrams you are missing the internal bending moments.

Also I1, I2, and I3 would also be subjected to the parallel axis theorem.

I would suggest also doing a vertical equilibrium check on each segment to solve for each F segment.

3. Sep 4, 2014

### blargh4fun

paisiello2,

1. Missing internal bending moments. Would you be willing to write out the equation(s) for one of the bodies (say body 1) so I understand what you mean?

2. Parallel axis theorem. I agree, but would they all be about the center of mass?

3. Vertical equilibrium check. I'm confused.

4. Sep 4, 2014

### paisiello2

You have the right equation you just omitted the internal bending moment. Whenever you cut a section through a 3-dimensional body it will have 6 internal forces acting on it (assuming x direction is out of the page):

1. Nz = Axial force in z direction = in this case 0
2. Vx = Shear force in x direction = in this case 0
3. Vy = Shear force in y direction = in this case F2/1 as you are showing it correctly
4. Mx = Bending Moment about x axis = in this case M2/1 which you are missing
5. My = Bending Moment about y axis = in this case 0
6. Mz = Torque about z axis = in this case 0

And sorry, vertical equilibrium check was a bad choice of words. What I meant to say was:

Sum of forces in y direction (vertical) = m1*a_y

5. Sep 4, 2014

### blargh4fun

Ok, try this.

Body1 - y forces:
F-F21=m1*a
(F+F21)*L1/2 - M21 = I1 *alpha

Body2 - y forces:
F12-F32=m2*a
(F12+F32)*L2/2 + M12 - M32 = I2 * alpha

Body3 - y forces:
F23+F43=m3*a (being careful with signs)
(F23-F43)*L3/2 + M23 + M43 = I3 * alpha (being careful with signs)

Body4 - y forces:
-F34=m4*a
(-F34)*L4/2 - M34 = I4 * alpha

alpha and a are constant across all equations and are solved on the system level (all blocks combined into one).

I1, I2, I3, and I4 are all I=I_cm+mr^2 where r is the distance from the COM of the whole to the middle of each block.

I think I have my signs correct, but would you please check them? I'm particularly concerned about body 3. I believe that the COM being located in body3 causes some signs to flip.

6. Sep 4, 2014

### blargh4fun

solving for alpha and a:
I_net * alpha = F*r where r is distance from COM to F. gives alpha = F*r/I_net
M*a=F so a=F/M

7. Sep 4, 2014

### blargh4fun

Honestly, I'm primarily interested in the bending moments at the center of each block. Maybe I'm going about this the wrong way. Maybe I need to rethink my FBDs. Instead of the Bodies 1-4 like above, I create Entities 1-4 like so:

Entity 1:
Left half of Body 1. Solve for unknown moment at right side of Entity 1.

Entity 2:
Body 1 and left half of Body 2. Solve for unknown moment at right side of Entity 2.

Entity 3:
Body 1, Body 2, and left half of Body 3. Solve for unknown moment at right side of Entity 3.

Entity 4:
Body 1, Body 2, Body3, and left half of Body 4. Solve for unknown moment at right side of Entity 4.

and M1-4 would provide the moments at the centers of each block?

The only thing I'm a little shaky on is how to define the I for each entity. Body 2, for example, I think it would be: I_cm_1+m_1*r_1^2+I_cm_half2+m_2/2*r_2half^2
where r1 is the distance from the COM to center of body 1, and r2half is the distance from COM to the center of the left half of body 2.

Last edited: Sep 4, 2014
8. Sep 4, 2014

### blargh4fun

I've made another observation, and now I'm not sure. I'm concerned on Entities 3 and 4 as to whether I can do what I'm showing or not (due to the center of mass). To double check the numbers I have, I setup Entities 3 and 4 from the right side instead of the left, and the numbers I calculated are very different. Any thoughts on what is correct?

When I do my sum of moments = I*alpha, am I restricted as to where I take my moments about? I've been taking them at the same locations as my M1-4 above. Is that ok? Should I be taking the moments about the COM instead? This has been allowing me to ignore my F1-F4 terms in my moment calculations.

9. Sep 4, 2014

### paisiello2

The signs don't really matter, just as long as you stay consistent, including for body3. I didn't check all your equations but they looked alright to me.

Not sure why you want the moments at the center of the block. I would suggest get the moments at the body intersections and then interpolate.

Last edited: Sep 4, 2014
10. Sep 12, 2014

### blargh4fun

Thanks for the feedback!

Unfortunately, I'm still stuck. I've tried solving this problem several of the ways previously discussed, but I get answers that don't make sense and don't match. I'm going to rehash the startup and show where I'm stuck, and hopefully someone can provide some insight.

Fig 1 below shows the same 4 blocks fused end to end. A force F is applied to the left-most side of block 1, and the system rotates about the center of mass (I'm assuming).

Starting from diagram (a), I setup my equations and solve for $\alpha$ and $a_y$.

1. $\sum F_y=F=m_{net} a_y$ where $m_{net} = \sum m_i$
This provides $a_y=F/m_{net}$.

2. $\sum M$(about the COM) $=F d_{F}=I_{net} \alpha$ where $I_{net}$ is the combined moment of inertia of all blocks.
This provides $\alpha = \frac{F}{d_F}$

Now that I have $\alpha$ and $a_y$, I want to solve for the moment at the 1-2 interface. To do this, I limit my FBD to only block 1 (shown in Fig 2) and write the following questions:

3. $\sum F_y=F-F_{21}=m_1 a_y$, and I can solve this equation for $F_{21}$.
4. $\sum M$(about ???) $=F d_{?} + F_{21} d_{??} + M_{21}=I_{??} \alpha$

a. In statics, I can take the sum of moments about any point because it must equal = 0. Is this still the case for dynamics? This will tell me $d_{?}$ and $d_{??}$.
b. Is the moment of inertia used in this equation $I_1$? And if so, does $I_1$ include the offset I from the total body's center of mass (parallel axis theorem)?
d. Because F is applied to the end, is $|a_y|>0$?