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clear
//Given
P_Max = 10 //N - the maximum distribution in a triangular distribution
L = 3 //mt the total length of force distribution
L_X = 5 //mt - the horizantal length of the rod
//caliculations
F_y = P_Max*L*0.5 //N - The force due to triangular distribition
L_com = 2*L /3 //mt - the resultant force acting as a result of distribution acting position
//F_X = 0 forces in x directions
R_A_X = 0 // since there are no forces in X-direction
R_B_X = 0
//M_A = 0 momentum at point a is zero
//F_y*L_com - R_B_Y*L_X = 0
R_B_Y = F_y*L_com/L_X
//M_B= 0 momentum at point b is zero
//- R_A_Y*L_X = F_y*(L_X-L )
R_A_Y = - F_y*L/L_X
//For a---a section
l_a = 2 //mt - a---a section from a
l_com_a = 2*l_a/3
v_a = R_A_Y + 0.5*l_a*(10.0*2/3) //*(10*2/3) because the maximum moves
M_a = (10.0*0.66)*l_a*(0.33) + R_A_Y*l_a
//For b---b section
v_b = F_y + R_A_Y //equilabrium conditions
M_b = (F_y + R_A_Y)*(-1)
printf("\n The force and moment in section a--a are %0.2f kN %0.3f kN-m",v_a,M_a)
printf("\n The force and moment in section b--b are %0.3f kN %0.3f kN-m",v_b,M_b)
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