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In the case of a simple cantilever beam having a uniform shape and being loaded at its end - the calculation of the deflection at any point along the beam is well known. What happens if we split it along its longitudinal axis?

cantilever beam composed of separate planks - taken from http://www.bu.edu/moss/mechanics-of-materials-bending-shear-stress/

I realize we are eliminating the shear forces along that axis between the beam segments, and this allows each segment to slide relatively to its neighbors. However, I'm not sure how this phenomena affects the beam deflection. How can I calculate it?

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  • $\begingroup$ How do you define $I_{yy}$ the second area moment for each state? $\endgroup$
    – Sam Farjamirad
    Sep 5 '18 at 5:54
  • $\begingroup$ Have you considered the average stiffness calculated by the relative thicknesses? $\endgroup$
    – Solar Mike
    Sep 5 '18 at 6:06
  • $\begingroup$ If I assume all the segments (planks in this case) are acting together, I will end up with a total second moment of area identical to the one of the original beam. The same is true for the stiffness, I guess. But, I realize that this assumption is probably faulty. $\endgroup$
    – Yaniv Ben David
    Sep 5 '18 at 6:12
  • $\begingroup$ I you follow the Bernoulli beam theory, then we you'll see the shear force is equal to the constant times the forth derivative of deflection. The only parameter here which is btw geometric dependent is the second area of moment. If they bonded together then the Bernoulli model doesn't work here anymore. As they are not bonded to gather then the load in each beam is dependent on each second area of moment and superposition gives the correct results. $\endgroup$
    – Sam Farjamirad
    Sep 5 '18 at 6:38
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    $\begingroup$ Your pictures may be misleading you, because of a rather important mistake. In "boards bonded together" the free end of the beam should be perpendicular to the beam, like the three separate beams in "boards not bonded together", not as you have drawn it. Think about how that makes the change in length (i.e. the tension or compression) different in the top and bottom boards, in the two pictures. $\endgroup$
    – alephzero
    Sep 5 '18 at 11:04
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in a cantilever beam the deflection is $$\delta_{max} = \frac {PL^3}{3EI} $$

In this case assuming free sliding between the planks the load P is going to be supported equally between the 3 planks.

So the deflection will be $$ \delta_{max} = \frac {(P/3)L^3}{3EI_{\text{single board}}} $$

Because $I$ is proportional to the cube of the board's height (in this case, its thickness), the single board's inertia will be $(1/3)^3=1/27$ that of the bonded boards. Therefore the unbonded deflection will be greater than the bonded boards by a factor of

$$\begin{gather} \dfrac{\left(\frac{1}{3}\right)}{\left(\frac{1}{27}\right)} = \frac{27}{3} = 9 \\ \therefore \delta_{unbonded}= 9\delta_{bonded} \end{gather}$$

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    $\begingroup$ @Wasabi, thanks for the edit. I am liable to make spelling and arithmetic errors because I use my cell phone to write my answers. $\endgroup$
    – kamran
    Sep 5 '18 at 19:06
  • $\begingroup$ The three beams are subject to a point load at the end, but also act on each other unless the point load is perfectly distributed among perfectly equal beams (even if we ignore friction there will be a vertical component) $\endgroup$
    – mart
    Aug 7 '20 at 5:23
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Thanks to @kamran for his answer.

I simulated the problem in ANSYS student v19 to verify his approach. In the pictures below, the upper beam is solid, the middle one is split into two segments and the lower one is split into 3 segments. Each segment is allowed to slide in respect to its neighbors. It is clear that the deflection of the 3 segments beam is 9 times the full one.

beam deflection ANSYS simulation results - for the case of frictionless contacts

In the case the segments are bonded together (i.e cannot slip in respect to each other) - We get the same results for all the case. All the beam act like a full solid body in bending:

beam deflection ANSYS simulation results - for the case of bonded contacts

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Although I agree with @kamran, I have another way of thinking about it

The deflection of the structure is

$$\delta=\frac{P\cdot L}{3\cdot E\cdot I}$$

The only difference in this problem between then bonded and the decoupled is the I. Assuming b: breadth, and h: height

For the coupled: $I_{coupled}= \frac{b\cdot h^3}{12}$

For the decoupled, the things slightly more complex. The resulting $I_{decoupled}=3 I_{(single\ board)}$. Each board has a $h_{single}=\frac{h}{3}$. Therefore the total moment of inertia is (Exactly as @kamran):

$$I_{decoupled}=3 \frac{b\cdot (\frac{h}{3})^3}{12} = \frac{1}{9} \frac{b\cdot h^3}{12} = \frac{1}{9} I_{coupled}$$.

Therefore, the deflection of the coupled (bonded) compared to the decoupled (for the same load) is:

$$\delta_{decoupled} = 9 \delta_{coupled}$$

An interesting historical fact about your problem is that the Viking shipbuilders new about that and used it in dragon ships for the bow. I.e. it was difficult to find a properly shaped tree to shape the bow of the ship. So what the "decouple" the different layers and it was possible (and easier) to bend without breaking.

Makes me wonder what other craftsmen have intuitively found out, without the need of maths.

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