Answer:
Sorry it doesnt tall me anythikng
Explanation:
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
Hello there,
In the problems given in the question, the driver's license is confiscated and suspended.
So our answer is: A)
Achievements.
Answer:
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Explanation:
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Answer:
Explanation:
Given that : -
The desirable limit is 500 mg / l , but
allowable upto 2000 mg / l.
The take volume is V = 160.000 m3
V = 160 , 000 x 103 l
The crainage gives 150 mg / l and lake has initialy 100 mg / l
Code of tpr frpm drawn = 150 x 60, 000 x 1000
Ci = 9000 kg / gr
Cl = 100 x 160,000 x 1000
Cl = 16, 000 kg
Since allowable limit = 2000 mg / l
Cn = ( 2000 x 160, 00 x 1000 )
= 320, 000 kg
so, each year the rate increases, by 9000 kg / yr
Read level = ( 320, 000 - 16,000 )
Li = 304, 000 kg
Tr=<u>304,000</u>
900
=33.77
