MOMENTOS DE EMPOTRAMIENTO PERFECTO P
A
B
P
A
l/2
l/2
l
3𝑃𝑙 16
𝑀𝐵𝐴 = 0
P
A
𝑀𝐴𝐵 = −
B
𝑃𝑙 8
𝑀𝐵𝐴 =
P
A
b
a
𝑀𝐵𝐴 = 0
ω
𝑀𝐴𝐵 = −
B
𝑃𝑎𝑏 2 𝑙2
𝑀𝐵𝐴 =
ω
A
l
𝑀𝐴𝐵 = −
ω𝑙 2 8
𝑀𝐵𝐴 = 0
ω
𝑀𝐴𝐵 = −
ω𝑙 2 12
𝑀𝐵𝐴 =
ω𝑙 2 12
B
A
l
l
ω𝑙 2 15
𝑀𝐵𝐴 = 0
𝑀𝐴𝐵 = −
ω𝑙 2 20
𝑀𝐵𝐴 =
ω
B
A
l/2
l/2
l
5ω𝑙 2 64
𝑀𝐵𝐴 = 0
ω
B
𝑀𝐴𝐵 = −
A
l/2
l/2
5ω𝑙 2 96
𝑀𝐵𝐴 =
ω
5ω𝑙 2 96
B
l/2
l/2
l
9ω𝑙 2 128
l/2
l/2
l
A
ω𝑙 2 30
ω B
A
𝑀𝐴𝐵 = −
B
ω B
𝑀𝐴𝐵 = −
𝑃𝑎2 𝑏 𝑙2
l
A
𝑀𝐴𝐵 = −
B
l
𝑃 𝑎2 𝑏 (𝑎𝑏 2 + ) 2 𝑙 2
A
𝑃𝑙 8
b
a
l
𝑀𝐴𝐵 = −
l/2
l/2
l
𝑀𝐴𝐵 = −
B
l
𝑀𝐵𝐴 = 0
𝑀𝐴𝐵 = −
11ω𝑙 2 192
𝑀𝐵𝐴 =
5ω𝑙 2 192