Do you want to learn how to calculate the ultimate moment resistance of a concrete section using the strain compatibility method? If the answer is yes then this video is for you. I’m going to go through an example and show you how to calculate the moment capacity of a cross-section using the strain compatibility method. I’ve already drawn my cross-section. As you can see the top flange width is a 1200mm and it is 200mm and our web is 1m wide and 1050mm deep and we’re using 16 number 40mm diameter longitudinal reinforcement and I have already calculated our effective depth which is 1168mm and we are using rectangular stress block for this example I have drawn that stress block and the compressive forces. Also the strains diagram here. fck is is 40MPa, gamma c is 1.5, we will be using alfa cc as 0.85. For the reinforcement fyk is 500MPa and gamma s is 1.15. First thing I’m going to do is calculate the area of for longitudinal reinforcement which is As Pi times 40 over 2 squared times 16 number bars. This is equal to and fcd design value of our concrete strength is going to be alfa cc times fck over gamma c which is 0.85 times 40 over 1.5. This comes to 22.67MPa. Now we are going to calculate fyd for our enforcement which is fyk over gamma s. This is equal to 500 over 1.15. It is going to be 434.78MPa. One thing I forgot to mention is reinforcing steel modulus of elasticity is going to be Es 200GPa. Then we can calculate the yield strain of reinforcement which is epsilon s,yield is equal to fyd over Es which is 434.78 over this is in MPa and 200GPa. It is going to be 0.002. Effective height of compression zone factor which is lambda. If our fck is below 50MPa then our lambda is going to be 0.8 and the effective strength factor eta for fck below 50MPa is going to be 1 and I’ve already mentioned it epsilon cu3 is 0.0035. What I’ve done here in this example, I’ve already calculated what the correct neutral axis depth is. Which is x value here and I’m going to use that for this example. Otherwise this is going to be a lengthy repetitive process. If you’d like me to explain this process, iterative process in detail please let me know I will do another video on that topic. Assume effective depth x is 432mm, Hence we can calculate effective height of compression zone. In here we’ve already mentioned it is lambda x. Lambda x is equal 0.8 times 432 which is 345.6mm. Then what we need to do is calculate the centroid of compression zone within the flange which is in this area I’ve noted their as l1. As you can see our neutral axis is lying below the flange. So we, we can easily calculate this l1. l1 is half of the depth of the flange which is 200 over 2. It is going to be 100mm. Then we need to calculate the centroid of compression zone in the web which is this section below this dashed line. You need to calculate the centroid and the distance from the top of the section to the centroid which is l2. We can calculate l2. We can write l2 as lambda x minus the depth of the flange over 2 plus the depth of the flange which is 345.6 minus 200 over 2, plus 200. l2 is equal to 272.8mm. Then what we need to do is calculate Fc1 and Fc2. These are the compressive forces in the compression zone within the flange and the compression zone within the web. We can write Fc1 is equal eta tims fcd times width of the flange times the depth of the flange. If you convert the units then this value is going to be 5440kN. Then we’re going to calculate Fc2 which is this compressive force. Fc2 is equal to eta times fcd times width of the web times, this distance which we can write as lambda x minus height of flange. This is equal to… If you convert the units then this value would be equal to 3300.27kN. Then we can calculate the total compressive force which is Fc is equal to Fc1 plus Fc2 which is equal to 5450 plus 3300. This is equal to 8740kN. Then we can calculate the reinforcement strain. Epsilon s is effective depth minus the depth to the neutral axis over depth to the neutral axis times epsilon cu3. That was derived from these diagrams. This is equal to… And the reinforcement strain 0.006 proves that this reinforcement steel is yielding. Because it’s bigger than the yield strain of reinforcement. Hence we can calculate our tension force in steel. Fs is equal to fyd times the area of steel which is 434.78 times the area. And if you convert the units correctly, this value is going to be 8741.8kN. You can see that this Fs value which is 8742kN is almost equal to total Fc value which is 8740kN So our compressive and tensile forces are in balance. Hence what we can do is we can take moments about any point using these forces and calculate bending moment capacity which is M is equal to… in this example I’m going to take bending moment about this point here. So we can write our bending capacity is equal to Fs times d which is the effective depth and Fc1 times l1 which is the force Fc1 and lever arm is l1. Same goes for Fc2 and lever arm for Fc2 is l2. And if you calculate the value then you would get 8766kNm. If you have any questions please post your questions in the comments section. If you like the video please like the video and if you would like to see more videos like this one please subscribe to our channel and if you have any suggestions for future videos, if you have any questions that you would like us to cover in a video please let us know. Thank you for watching.

Hey. I have some queries regarding stress-strain relation non-linear structural analysis (3.1.5, EC-2). Are there some examples or book you can refer me to?

If this is interesting to you, i can discuss more in detail about my research topic.

Kind Regards

Hi. how did you get the value Fc2? The answer i got was different for Fc2. The correct answer is 330kN.

When the FC and FS value is different, may you show us the further step?

Kindly refer towards an example from where I can understand how to calculate x