s2rdmcasparticuliers.py

Created on June 03, 2024

627 Bytes

cas d'une force ponctuelle au milieu d'une poutre sur deux appuis
N=o
V = - F/2 1[0,L/2] + F/2 1[L/2,L]
M linéaire : F(x)/2 1[0,L/2] + F(L-x)/2 1[L/2,L]
Moment Max FL2/4
Flèche max en L/2 y(L/2) = FL^3/48EI

Cas d'une force ponctuelle à la fin sur un appui encastré
N=0
V(x) = -F 
M(x) = F(x-l) linéaire max en x=0
Flèche max en L Y(0)= -FL

Cas d'une force linéaire sur des appuis 
N=0
V(x)=qx-ql/2
M(x)= - qx2/2 +qLx/2 
Moment max en L/2 ql^2/8
Flèche max en y=L/2  = -5qL^4/384EI

Cas d'une force linéaire dans un appui encastré
N=0
V= qx-qL/2
M(x) = -qx2/2 + qLx/2 -qL2/2
Max en x= 0

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s2rdmcasparticuliers.py