The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is 3/8 in.

A) Determine the maximum shear stress developed in the shaft.

TMAX = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

φ = ?

Tmax = 2.934ksi

φ = 4.58°

Explanation:

Given.

Length = 100 ft

Maximum power = 2590 hp

Angular rotate = 1700 rpm .

Outer diameter of the shaft = 8 in. Wall thickness = 3/8 in.

Calculating angular velocity.

= 1700rev/min * 2πrad/rev * 1min/60s

= 56.67πrad/s

Convert power to ft.lb/s

2590hp = 2590 * 550

= 1424500ft.lb/s

At this point, we calculate torque.

Torque = Power/Angular Velocity

Torque = 1424500/56.67π

Torque = 8000.24lb.ft

Using the torsion formula, we'll calculate maximum stress due to the shear force acting on the body:

Tmax = ½(Tc/J)

Tmax =½ of (8001.24 * 12 * 4 )/ (π/4)(4⁴ - 3.625⁴)

Tmax = ½ * 5868.656298363693

Tmax = 2934.328149181846

Tmax = 2.934ksi

Calculating the angle twist

φ = TL/JG

φ = (8001.24 * 12 * 100 * 12 )/( (π/4)(4⁴ - 3.625⁴)(4⁴ - 3.625⁴)(11)(10^6))

φ = 0.08002 rad -- convert to degrees

φ = 4.58° --- approximated

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )

answer:

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explanation: