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clear
//Given
shear_v = 3000 //N - Transmitted vetical shear
shear_al = 700 //N - The maximum allowable
//We will divide this into two parts
l_1 = 50.0 //mm
l_2 = 200.0 //mm
b_1 = 200.0 //mm
b_2 = 50.0 //mm
A_1 = l_1* b_1 //sq.mm - area of part_1
y_1 = 25.0 //mm com distance
A_2 =l_2*b_2 //sq.mm - area of part_1
y_2 = 150.0 //in com distance
y_net = (A_1*y_1 +A_2*y_2)/(A_1+A_2) //mm - The com of the whole system
c_max = (4-y_net) //mm - The maximum distace from com to end
c_min = y_net //mm - the minimum distance from com to end
I_1 = b_1*(l_1**3)/12 + A_1*((y_1-y_net)**2) //Parallel axis theorem
I_2 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)
I_net = I_1 + I_2 //mm4 - the total moment of inertia
Q = A_1*(-y_1+y_net) //mm3
q = shear_v*Q/I_net //N/mm - Shear flow
d = shear_al/q // The space between the nails
printf("\n The minimal space between the nails %0.1f mm",d)
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