Answer:
I)E= 40.95 GPa
II)E=5.29 GPa
Explanation:
I)
Given that
E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45
E₂ = 72.5 GPa ,V₂=0.55
Longitudinal moduli given as ;
E= E₁V₁+E₂V₂
E= 2.41 x 0.45 + 72.5 x 0.55 GPa
E= 40.95 GPa
II)
E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45
E₂ =230 GPa ,V₂=0.55
Transverse moduli given as:
E=5.29 GPa
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_abσ_allow
σ_cd = F_cd/A_cdσ_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
<u>Sector AB: </u>
σ_ab = F_ab/A_ab σ_allow=300 KPa
= w/10*10^-6 σ_allow=300 KPa
w_ab = 3*10^-3 kN/m
<u>Sector CD: </u>
σ_cd = F_cd/A_cd σ_allow=300 KPa
= 2*w/15*10^-6 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
<u>The solution is: </u>
w=2.25 N/m
note:
find the attached graph
