6.

A solid round bar with a diameter of 2.32 in has a groove cut to a diameter of 2.09 in, with a radius of 0.117 in. The bar is no

Alenkinab [10]

Answer:

nf=1.11 (Goodman criterion)

ny=2.41 (factor of safety for fatigue)

Explanation:

From the table A-20 Deterministic ASTM minimum tensile and yield strengths for HR and CD steels, we have approximately:

Sut=120 kpsi

Sy=66 kpsi

Due Sut<1400 Mpa, the endurance limit is:

$Se=0.5Sut=0.5*120=60 kpsi$

The surface condition modification factor is:

$ka=a(Sut)^{b}=2.7(120)^{-0.265} =0.759$

The effective diameter is:

$de=0.37d=0.37*2.09=0.7733 in$

The size factor is:

$kb=0.879de^{-0.107} =0.879(0.7733)^{-0.107} =0.9$

The endurance limit at critical location is:

$See=ka*kb*kc*kd*kd*ke*kf*Se=0.759*0.9*1*60=40.986 kpsi$

$\frac{D}{d}=\frac{2.32}{2.09} =1.11\\\frac{r}{d}=\frac{0.117}{2.09} =0.056$

From Figure A-15-15 chart, the Kf = 2.1

The notch sensitivity is:

$\sqrt{a}=0.246-3.08(10^{-3})Sut+1.51(10^{-5})Sut^{2}-2.67(10^{-8})Sut^{3}$

$\sqrt{a}=0.246-3.08(10^{-3})(120)+1.51(10^{-5})(120^{2})-2.67(10^{-8})(120^{3})=0.048$

$q=\frac{1}{1+\frac{\sqrt{a} }{\sqrt{r} } }=\frac{1}{1+\frac{0.048}{\sqrt{0.117} } }=0.877$

The fatigue stress is:

$Kf=1+q(Kt-1)=1+0.877(2.1-1)=1.96$

The moment of inertia is:

$I=\frac{\pi }{64} d^{4}=\frac{\pi }{64} (2.09^{4})=0.936 in^{4}$

The maximum stress is:

$omax=\frac{M*c}{I}=\frac{25000*\frac{2.09}{2} }{0.936} =27911.32 psi=27.911 kpsi$

The mean stress is:

$om=Kf\frac{omax+omin}{2} =1.96\frac{27.911+0}{2}=27.35 kpsi$

The alternate stress is:

$oa=Kf|\frac{omax-omin}{2}|=27.35 kpsi$

The fatigue factor using Goodman is:

$\frac{1}{nf}=\frac{oa}{See}+\frac{om}{Sut}\\\frac{1}{nf}=\frac{27.35}{40.986}+\frac{27.35}{120}$

nf=1.11

the factor of safety is:

$ny=\frac{Sy}{omax}=\frac{66}{27.35} =2.41$

4 0

4 years ago

A 1,040 N force is recorded on a hemispherical vane as it redirects a 2.5 cm- blade diameter water jet through a 180 angle. Dete

Alex777 [14]

This question is incomplete, the complete question is;

A 1,040 N force is recorded on a hemispherical vane as it redirects a 2.5 cm- blade diameter water jet through a 180 angle.

Determine the velocity of the flowing water jet if the blade is assumed to be frictionless.

Answer: the velocity of the flowing water jet is 32.55 m/s assuming the blade is frictionless

Explanation:

Given that;

Force Ft = 1040 N

diameter d = 2.5 cm = 0.025 m

we know that; force acting on Hemispherical plate is;

Ft = 2δav²

where

a is area = π/4(0.025)²

δ is density of water = 1000 kg/m³

v is velocity = ?

now we substitute

1040 = 2 × 1000 × (π/4(0.025)²) × v²

1040 = 0.9817v²

v² = 1040 / 0.9817

v² = 1059.3867

v = √1059.3867

v = 32.5482 ≈ 32.55 m/s

Therefore the velocity of the flowing water jet is 32.55 m/s assuming the blade is frictionless

8 0

3 years ago

Determine the complex power, apparent power, average power absorbed, reactive power, and power factor (including whether it is l

LenKa [72]

Answer:

A) complex power = apparent power = ( 108.253 + 62.5 i ) VA

active power = 108.25 watts

reactive power = 62.5 VAR'S

power factor = cos ∅ = cos 30° = 0.866 ( lagging )

also ; current lags voltage by 30°

B) your question is not well written hence no answer

Explanation:

A) v(t) = 100 cos (377t - 30° ) v

Vrms = $\frac{100}{\sqrt{2} }$ ∠ -30°

I(t) = 2.5cos(377t- 60°) A

Irms = $\frac{2.5}{\sqrt{2} }$ ∠ -60°

determine complex power apparent power . ...... power factor

Note : complex power = apparent power

=( Irms ) * Vrms

= ( $\frac{2.5}{\sqrt{2} }$ ∠ -60° ) * ( $\frac{100}{\sqrt{2} }$ ∠ -30° )

= 125 ∠ 30°

= ( 108.253 + 62.5 i ) VA ( complex power )

active power = 108.25 watts

reactive power = 62.5 VAR'S

power factor = cos ∅ = cos 30° = 0.866 ( lagging )

current lags voltage by 30°

4 0

3 years ago

What is the chord length of an airplane called?

aliya0001 [1]

Answer:

The distance from the leading edge to the trailing edge is called the chord, denoted by the symbol c. The ends of the wing are called the wing tips, and the distance from one wing tip to the other is called the span, given the symbol s

3 0

3 years ago

A three-phase voltage source with a terminal voltage of 22kV is connected to a three-phase transformer rated 5MVA 22kV/220V. The

amid [387]

Answer:

A) See Attachment B) See Attachment C) 186.153

Explanation:

C) Per phase Voltage= 220/∛3

= 6.037

S=V²/Z

= (6.037)²/(0.01+i0.05)

= 62051.282-i310256.41

Active power losses per phase= 62.051kW

total Active power losses= 62.051×3

=186.153kW

6 0

4 years ago