You could use these design aids as a starting point and modify the required steel by a ratio of actual yield to yield for the design aid. You need to go back at the end and analyze your actual section with actual steel yield at the end.
@@nazirghnim9340 In the picture, I only see : 22" overall length ( MINUS ) 2 x 1.5” - which is concrete cover that is encountered twice as you work towards the center of the column vertically ( MINUS ) 2 x 0.5” - which is the singular #4 stirrups that is encountered twice as you work towards the center of the column vertically ( MINUS ) ** 5 x 1.27” - 5 actual #10 rebars with 1.27" diameter in the column ** So, above description should add up to : 11.65”. Dividing the left over space by equal 4 spaces : 11.65” / 4 spaces = 2.91” So, again, in the original calculation, where did the : 4 x 0.5” come from? So, I am not getting the 4" like What (Dr. David Garber) @David Garber got.
This example uses design aids with the varying phi factor built in. You should go back and analyze your actual design at the end. You can develop an interaction diagram (similar to what you would find in the videos in this playlist ua-cam.com/play/PLLhnAwUfiIPvGb4tM69GD-ammbq0BPS7R.html). The 1.2D + 1.6L is the load combination that will typical control when finding the required strength U (Mu and Pu in this example). You can look to ACI 318-19 section 5.3 for more information.
Wow! Great explanation. Easy to follow. Thank you.
thank you .....so good video
Thanks sir,one question :normally we calculate axial load manually so major moment multipied by eccentricity value=L/500+D/30 or min 20 mm,right?
Great job! you saved me
this helped me alot . thank you
Hi! I hope this gets a reply. What's the reference for this video? I'd appreciate the response. Thank you.
Hi Patrick, The diagrams that I show in this video are from Wight's textbook "Reinforced Concrete: Mechanics and Design" Appendix A.
Can you use this for rectangular columns as well or are there equations unique to square column behavior?
You can use the same procedure for rectangular columns. Make sure you have the column orientation correct for your direction of bending.
Hello, I want to ask that what if fy is not equal to 60 ksi (e.g 40, 50ksi), it seems that ACI has not provided these design charts
You could use these design aids as a starting point and modify the required steel by a ratio of actual yield to yield for the design aid. You need to go back at the end and analyze your actual section with actual steel yield at the end.
spacing between all of the bars : ( 22" - 2(1.5") - 2(0.5") - * 4(0.5") * ) / 4 = 4" - - - Where did you get the 4(0.5)? Video time : 11:42 / 16.09
b bent = 4 d stirrups
@@nazirghnim9340 In the picture, I only see : 22" overall length ( MINUS ) 2 x 1.5” - which is concrete cover that is encountered twice as you work towards the center of the column vertically ( MINUS ) 2 x 0.5” - which is the singular #4 stirrups that is encountered twice as you work towards the center of the column vertically ( MINUS ) ** 5 x 1.27” - 5 actual #10 rebars with 1.27" diameter in the column **
So, above description should add up to : 11.65”.
Dividing the left over space by equal 4 spaces : 11.65” / 4 spaces = 2.91”
So, again, in the original calculation, where did the : 4 x 0.5” come from?
So, I am not getting the 4" like What (Dr. David Garber) @David Garber got.
Where can I download this graphs?
Dude, its a cool video but, the phi factor is 0.65 or 0.75 depending of transversal reinforcement. The 1.2 wd + 1.6 wl is to get the Pn and Mn.
This example uses design aids with the varying phi factor built in. You should go back and analyze your actual design at the end. You can develop an interaction diagram (similar to what you would find in the videos in this playlist ua-cam.com/play/PLLhnAwUfiIPvGb4tM69GD-ammbq0BPS7R.html). The 1.2D + 1.6L is the load combination that will typical control when finding the required strength U (Mu and Pu in this example). You can look to ACI 318-19 section 5.3 for more information.
thank u m.r