# Deflection of a cantilever beam composed of separate (not bonded) planks

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? 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?

• How do you define $I_{yy}$ the second area moment for each state? – Sam Farjamirad Sep 5 '18 at 5:54
• Have you considered the average stiffness calculated by the relative thicknesses? – Solar Mike Sep 5 '18 at 6:06
• 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. – Yaniv Ben David Sep 5 '18 at 6:12
• 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. – Sam Farjamirad Sep 5 '18 at 6:38
• 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. – alephzero Sep 5 '18 at 11:04

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

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

• @Wasabi, thanks for the edit. I am liable to make spelling and arithmetic errors because I use my cell phone to write my answers. – kamran Sep 5 '18 at 19:06

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. 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: 