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:
Explanation:
We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.
<h3>1. Force of 1st Object </h3>First, we can find the force of the first, 8 kilogram object.
According to Newton's Second Law of Motion, force is the product of mass and acceleration.
The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.
- m= 8 kg
- a= 1.6 m/s²
Substitute these values into the formula.
Multiply.
Now, use the force we just calculated to complete the second part of the problem. We use the same formula:
This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.
- F= 12.8 kg *m/s²
- m= 2 kg
Substitute the values into the formula.
Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.
The units of kilograms cancel.
The acceleration is 6.4 meters per square second.