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Hi,
The following is the equation in which , the value "f" should be zero, but the integration limit has to change with the y1 value,
i.e if y1<z1 the equation will be same as shown,
if z1<y1<z2 , then y1 must be in the P2(z) integration limit ,
also if z2<y1,then y1 must be in the P3(z) integration limit
Keeping the fact that using trial and error "f "must be zero and the equation must change the integration limit as per the "y1" value.
y1 = | 3.26 | Solve by trial & error until f = 0 | Q = | 1192 KN | ||||||
z1 | z2 | y1 | ||||||||
f = | Q - ∫(P1(z,ϕ1)dz - ∫(P2(z,ϕ2)dz - ∫(P3(z,ϕ3)dz | |||||||||
0 | z1 | z2 | ||||||||
f = | 0.00 |
There are two "If" statements.
One is programmed if xxx if Boolean test
else xxxx
The second is a function "if" very similar to EXCEL: if([Boolean test], if true, if false)
Either could be used to develop your statements f(y):=if(y<z1first expression, if(y<z2,second expression, third expression))
Thank you Fred for the quick reply,
In addition to this the value of "f" should also "zero" using trial and error assuming a specific value of "y1".
i.e if y1<z1, the equation must change its integration value and solve for f=0 , until "f=0" , "y1" should increase from "0 to z2".and the equation changes accordingly.
Example no. 1. (when y1<z1, the value of y is increased till 2.3680 from 0 , till "f=0"
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Similarly,
Example no. 2. (when y1<z3, the value of y is increased till 3.8482 from 0 , till "f=0",and the equation changes accordingly.
Thanks you.