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The ACI 318-19 code has a few things to say about multiple spans. ACI 435 committee design guides will have more information on this topic. Code Section 24.2.3.6 permits us to average the effective moment of inertia (I_e) taken from the critical positive and negative moment regions. Section 24.2.3.7 also permits us to use I_e computed only at midspan for continuous spans. Take your pick. Either way, these seem to imply that you could compute I_e for each span and then compute a deflection. Keep in mind that the purpose of these code sections is largely for control of maximum deflections, not always exact calculation of a deflected shape. Personally, I'd just use the midspan value if it is less than the value at the supports rather than taking an average. The other approach that I like is suggested by Bischoff (2007): take a weighted average of the flexibilities. For example, using something like I_span = 1 / (0.70*(1/I_mid) + 0.15*(1/I_end1) + 0.15*(1/I_end2)), where I_span would be the overall effective moment of inertia of the span, and I_mid, I_end1, and I_end2 are the effective moments of inertia calculated at midspan and the two ends, respectively. Of course, you are always permitted to do a more rigorous analysis, and that might be preferable if you want a really accurate profile of the deflections along the entire length of the beam. You may compute the effective moment of inertia at each section, and then find the curvature = M/(E*I) along the entire length. Finding the displacement would then require you to integrate the curvature twice (or have some software do it for you).

@@StructuresProfH thank you so much for your answer. This was very helpful. Keep doing those great videos

Deflection limits like L/360 work for any kind of unit, no need to convert. If length L is in inches, then the resulting limit will be inches. If the length L is in millimeters, than the resulting limit will be in millimeters.

@@StructuresProfH thank you for the clarification sir, your help is much appreciated :)

Sorry I wasn't clear - the specified limit would also apply to beams supporting the floor. Beams supporting roofs will often times have different (typically more lenient) limits.

There is no such provision in ACI. The (n-1)As' comes from mechanics. Normally a transformed area would be n*As, where n = Es/Ec, and this is valid for the tension steel. However, the compression steel is displacing an area of uncracked concrete which is also carrying compression, so we use (n-1) to reflect this.

What are the stress and strain in the link180 elements? If it is less than yielding (415 MPa) then the bilinear yielding behavior will obviously do nothing - the material is still elastic. Otherwise, it’s impossible for me to say what is going on here.