A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in.2 The
allowable stress for the hoops is σallos=11.4 ksi. Determine the maximum spacing s along the pipe so that the pipe can resist an internal gauge pressure of 4 psi.
1 answer:
Answer:
31.67 in
Explanation:
Given:
Diameter of the pipe, D = 3ft = 36 in
cross-sectional area of the steel = 0.2 in²
Note: Refer to the figure attached
From the free body diagram represented in the figure, we have
ΣFx = 0
or
pressure × projected area = 2 × Force in steel
Now, the projected area = spacing (s) × diameter of the wood pipe
force in steel = stress in steel (σ) × cross-sectional area of the steel
on substituting the values we get
4 ksi × (s × 36 in) = 2 × σ × 0.2 in²
also, allowable hoop stress = 11.4 ksi
thus,
σ = 11.4 ksi = 11.4 × 10³ psi
therefore, we have
4 psi × (s × 36 in) = 2 × 11.4 × 10³ psi × 0.2 in²
thus,
s = 31.67 in
hence the maximum spacing is 31.67 in
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Answer:
5.634 N rightwards
Explanation:
qo = - 3 x 10^-7 C
q1 = - 9 x 10^-6 C
q2 = 10 x 10^-6 C
r1 = 7 cm = 0.07 m
r2 = 20 cm = 0.2 m
The force on test charge due to q1 is F1 which is acting towards right
According to the Coulomb's law
F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)
F1 = 4.959 N rightwards
The force on test charge due to q2 is F1 which is acting towards right
According to the Coulomb's law
F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)
F2 = 0.675 N rightwards
Net force on the test charge
F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards
