150
A
Explanation:
V
s
V
p
=
N
s
N
p
(
1
)
N
refers to the number of turns
V
is voltage
s
and
p
refer to the secondary and primary coil.
From the conservation of energy we get:
V
p
I
p
=
V
s
I
s
(
2
)
From
(
1
)
:
V
s
V
p
=
900
00
3
00
=
300
∴
V
s
=
300
V
p
Substituting for
V
s
into
(
2
)
⇒
V
p
I
p
=
300
V
p
×
0.5
∴
I
p
=
150
A
Seems a big current.
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
C. seems like the best answer. i may be wrong so don’t quote me on that
