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:
(a) T = W/2(1-tanθ) (b) 39.81°
Explanation:
(a) The equation for tension (T) can be derived by considering the summation of moment in the clockwise direction. Thus:
Summation of moment in clockwise direction is equivalent to zero. Therefore,
T*l*(sinθ) + W*(l/2)*cosθ - T*l*cosθ = 0
T*l*(cosθ - sinθ) = W*(l/2)*cosθ
T = W*cosθ/2(cosθ - sinθ)
Dividing both the numerator and denominator by cosθ, we have:
T = [W*cosθ/cosθ]/2[(cosθ - sinθ)/cosθ] = W/2(1-tanθ)
(b) If T = 3W, then:
3W = W/2(1-tanθ),
Further simplification and rearrangement lead to:
1 - tanθ = 1/6
tanθ = 1 - (1/6) = 5/6
θ = tan^(-1) 5/6 = 39.81°
Answer:
... spilling water or getting anything cascading onto the floor
Correcto no se muy bien de que se trata el tema porque está en inglés.
Sorry
