Answer:
w=2.25
Explanation:
It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.
The surface of the cross-section of the stapes was determined:
A_ab= 10 mm^2
A-cd= 15 mm^2
The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.
σ_ab = F_ab/A_ab[tex]\leq[/tex]σ_allow
σ_cd = F_cd/A_cd[tex]\leq[/tex]σ_allow
In the next step we will determine the static size: Picture b).
We apply the conditions of equilibrium:
∑F_x=0
∑F_y=0
∑M=0
∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0
==> F_cd = 2*w*k*N
∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0
==> F_ab = w*k*N
Now we determine the load w
Sector AB:
σ_ab = F_ab/A_ab[tex]\leq[/tex] σ_allow=300 KPa
= w/10*10^-6[tex]\leq[/tex] σ_allow=300 KPa
w_ab = 3*10^-3 kN/m
Sector CD:
σ_cd = F_cd/A_cd[tex]\leq[/tex] σ_allow=300 KPa
= 2*w/15*10^-6[tex]\leq[/tex] σ_allow=300 KPa
w_cd = 2.25*10^-3 kN/m
w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}
==> w=2.25 * 10^-3 kN/m
The solution is:
w=2.25 N/m
note:
find the attached graph
