Thank you for the great video! I believe the last calculation adds the displacement at E and the stretch of EB. It should be .0016858+.03161=.0332858, slightly different from video
To find the Y which is the displacement of E. I think you forgot sir to convert the 8 ft and 2 ft to inches. Because the results of delta's EB and CF are in inches. But you've use 10' and 2' for the similar triangle.
It wouldn't matter because he's only doing a ratio; 10 feet is to .008429 inches (change in position of c) as 2 feet is to y inches (change in position of e.) The equation is y=(1/5)*.008429 whether or not he converts those lengths to inches before evaluating for y.
Why are we adding displacement of e and the stretch of the member to find displacement of b? And what's the difference between the stretching of a member and a point? Someone please respond quick. My exam is tomorrow😭😭
Might be a tad late but pretty sure it’s because the displacement of E is not accounted for in the elongation of tension EB, basically you have a bar which is tilted down prior and on top of that is a stretch of the tension, look at it in a sequence, CF elongates and points C and E go down, after they’re down, the tension in BE elongates from the 2.5 kips on the bottom causing point B to go even further on top of what E was displaced. lmk if this helps I yapped a lil
Thank you for the great video! I believe the last calculation adds the displacement at E and the stretch of EB. It should be .0016858+.03161=.0332858, slightly different from video
0.0332958* (;
So practically the displacement of B its the same of E??
This is presented so beautifully, thank you so much! God bless you.
Thank you for this excellent explanation.
You're better than my college teacher.
dude every online lecturer is better than your college teacher, c'mon you study in an arabic country.
The displacement for B is 0.03161+0.0016858=0.0332958, not 0.033468.
Now im wondering it the displacement of B and E are equal
the nek is professionally perfect
Would the weight of the members be relevant in a real-world application or are they small enough that they are negligible?
Wonderful Lectures ! Thanks.
To find the Y which is the displacement of E. I think you forgot sir to convert the 8 ft and 2 ft to inches. Because the results of delta's EB and CF are in inches. But you've use 10' and 2' for the similar triangle.
It wouldn't matter because he's only doing a ratio; 10 feet is to .008429 inches (change in position of c) as 2 feet is to y inches (change in position of e.) The equation is y=(1/5)*.008429 whether or not he converts those lengths to inches before evaluating for y.
Love you, 😍💜
Isn't y the change in height of eb? so isnt it the same as δeb? why are the values for δeb and y different?
for calculation 12.5 mm radius should converted into inches 0.4921 inch= 12.5mm
Why are we adding displacement of e and the stretch of the member to find displacement of b? And what's the difference between the stretching of a member and a point?
Someone please respond quick. My exam is tomorrow😭😭
did you find out? i have that same question
Might be a tad late but pretty sure it’s because the displacement of E is not accounted for in the elongation of tension EB, basically you have a bar which is tilted down prior and on top of that is a stretch of the tension, look at it in a sequence, CF elongates and points C and E go down, after they’re down, the tension in BE elongates from the 2.5 kips on the bottom causing point B to go even further on top of what E was displaced. lmk if this helps I yapped a lil
Shouldn't the members FC & BE be in compression? Just a little confused on why they are stretching.
If you remove link FC beam DC would move down so FC must be holding it up, i.e. tension.
Same with link BE.
why did you add together the A and B at the end for the last displacement