in y(x)=mx+b, if X=0 then y(x)=b and looks like a horizontal line on a graph that can move up or down (b values) along the Y axis but remains horizontal this is 0 degree. in (y)x=mx+b, y(x) is a line, that can have a slope; this is 1 degree in y(x)=mx^2+b y(x) is a parabole, this is 2 degree y(x)=mx^3+b is 3 degree y(x)=mx^4 is 4 degree can also be in the form of y(x)=x^0 [0 degree] y(x)=x^1 [1 degree] y(x)=x^2 [2 degree] y(x)=x^3 [3 degree] and so on.
you saved my life sir
Thanks
please compute both varying load and uniform distributed load
You made a great help for me sir .. thankyou for the video
Glad it helped
Can u share some video math of beam deflection ,sir? It'll be a bunch of help for me .❤️
Thank You so much sir for your given useful knowledge ❣️❣️❣️👍👍😊
Thanks for sharing! Can you find the deflection?
At 8:28 why you use concave down not up ?
What if the support is on the right side? How do I do with that?
same principle
Mind explaining what is 1 degree 2 degrees and 3 degrees?
in y(x)=mx+b, if X=0 then y(x)=b and looks like a horizontal line on a graph that can move up or down (b values) along the Y axis but remains horizontal this is 0 degree.
in (y)x=mx+b, y(x) is a line, that can have a slope; this is 1 degree
in y(x)=mx^2+b y(x) is a parabole, this is 2 degree
y(x)=mx^3+b is 3 degree
y(x)=mx^4 is 4 degree
can also be in the form of
y(x)=x^0 [0 degree]
y(x)=x^1 [1 degree]
y(x)=x^2 [2 degree]
y(x)=x^3 [3 degree] and so on.