Rewritten max_deflection method#29703
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For few cases like below, P1, P2, M = symbols('P1, P2, M')
E, I = symbols('E, I', positive = True)
b = Beam(50, 20, 30)
b.apply_load(-10, 2, -1)
b.apply_load(15, 26, -1)
b.apply_load(P1, 10, -1)
b.apply_load(-P2, 40, -1)
b.apply_load(90, 5, 0, 23)
b.apply_load(10, 30, 1, 50)
b.apply_load(M, 15, -2)
b.apply_load(-M, 30, -2)
p50 = b.apply_support(50, "pin")
p0, m0 = b.apply_support(0, "fixed")
p20 = b.apply_support(20, "roller")
p=b.max_deflection()
print(p) |
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References to other Issues or PRs
Fixes #29694
Brief description of what is fixed or changed
Rewritten max_deflection method, approach/logic similar to #29650, #29233 and #29441.
Other comments
Illustrations given below used to fail before as shown in #29694. Now they work.
Example: 1
Output: (4.00000000000000, 302.222222222222)
It can be seen maximum deflection is at (4.00000000000000, 302.222222222222)
Example: 2
Output: (2, 25/12)
It can be seen maximum deflection is at (2, 25/12)
Example: 3
Output: (0, 12000)
It can be seen maximum deflection is at (0, 12000)
Example: 4
Output: (5, 625/3)
It can be seen maximum deflection is at (5, 625/3).
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