Answer: The net force in every bolt is 44.9 kip
Explanation:
Given that;
External load applied = 245 kip
number of bolts n = 10
External Load shared by each bolt (P_E) = 245/10 = 24.5 kip
spring constant of the bolt Kb = 0.4 Mlb/in
spring constant of members Kc = 1.6 Mlb/in
combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6) = 0.4 / 2 = 0.2 Mlb/in
Initial pre load Pi = 40 kip
now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them
External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip
So Total net Force on each bolt Fb = P_Eb + Pi
Fb = 4.9 kip + 40 kip
Fb = 44.9 kip
Therefore the net force in every bolt is 44.9 kip
Answer:what are you trying to say
Explanation:
So what happens is the host will not kill the y no se que hacer para no one can see it in
Answer:
Explanation:
The detailed steps is as shown in the attachment.
Answer:
Machine 2 has a higher process capability index, it would be best considered for purchase.
Explanation:
Process capability index: Cpk= Min [(mean-L spec)/3sd; (U spec-mean)/3sd]
For machine 1, mean= 48mm and L spec= 46 and U spec= 50, Standard deviation sd= 0.7
Cpk= [0.952;0.952]= 0.952
For machine 2, mean= 47 and L spec= 46 and U spec= 50, Standard deviation sd= 0.3
Cpk= [1.111;3.333]= 1.111
It is clearly observed from the calculations above that the Cpk value of machine 2 is higher than that of machine 1.
Since machine 2 has a higher process capability index, it would be best considered for purchase.
