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Andrews [41]
3 years ago
7

9. An elevator on a construction site is being operated at rated capacity of 6 tons, and is supported by two standard steel cabl

es, each having a cross-section area of 1.25-in.2 and an effective modulus of elasticity of 12 * 106 psi. As the elevator is descending at a constant 500 ft/min, an accident causes the top of the cable, 200 ft above the elevator, to stop suddenly. (a) Calculate the Impact Factor. (b) Estimate the maximum elongation and maximum tensile stress developed in cable. (c) Estimate the maximum elongation and maximum tensile stress developed in cable if the cable length was 100 ft. (d) Comment on the effect of changing cable length on maximum tensile stress developed in the cable.
Engineering
1 answer:
aleksklad [387]3 years ago
8 0
Woah that is a longgg question
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The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pu
marishachu [46]

The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

3 0
3 years ago
Note that common skills are listed toward the top, and less common skills are listed toward the bottom.
Sladkaya [172]

Answer:

BDEG

Explanation:

got it right on the test on edge because i used my b r a i n

5 0
3 years ago
What are Tresca and Von Mises yield criteria?
elena-s [515]

Answer

For isotropic material plastic yielding depends upon magnitude of the principle stress not on the direction.

Tresca and Von Mises yield criteria are the yield model which is widely used.

The Tresca yield criterion stated that yielding will occur in a material only when the greatest maximum shear stress reaches a critical value.

max{|σ₁ - σ₂|,|σ₂ - σ₃|,|σ₃ - σ₁|} = σ_f

under plane stress condition

  |σ₁ - σ₂| = σ_f

The Von mises yielding criteria stated that the yielding will occur when elastic energy of distortion reaches critical value.

σ₁² - σ₁ σ₂ + σ₂² =  σ²_f

5 0
3 years ago
Estimate the endurance strength, Se, of a 37.5-mm- diameter rod of AISI 1040 steel having a machined finish and heat-treated to
7nadin3 [17]

Answer:

endurance length is 236.64 MPa

Explanation:

data given:

d = 37.5 mm

Sut = 760MPa

endurance limit is

Se = 0.5 Sut

   = 0.5*760 = 380 MPa

surface factor is

Ka = a*Sut^b

where

Sut is ultimate strength

for AISI 1040 STEEL

a = 4.51, b = -0.265

Ka = 4.51*380^{-0.265}

Ka = 0.93

size factor is given as

Kb =1.29 d^{-0.17}

Kb = 0.669

Se = Sut *Ka*Kb

    = 380*0.669*0.93

Se = 236.64 MPa

therefore endurance length is 236.64 MPa

4 0
3 years ago
The component of a regenerative vapor power cycle that permits only liquid to pass through to a region of lower pressure is a?
Zepler [3.9K]

The component of a regenerative vapor power cycle that permits only liquid to pass through to a region of lower pressure is a Valve/trap.

<h3>What is vapor?</h3>
  • In physics, a substance in the gas phase at a temperature lower than its critical temperature is referred to as a vapor or vapor.
  • This means that the vapor can be condensed to a liquid by increasing the pressure on it without decreasing the temperature.
  • An aerosol is distinct from a vapor.
  • A suspension of minute liquid, solid, or both types of particles in a gas is known as an aerosol.
<h3>Why does vapor form?</h3>
  • Evaporation or sublimation are two processes that can be used to create it.
  • Unlike clouds, fog, or mist, which are only suspended drops of liquid water in the atmosphere, watevaporur is a gas and cannot be seen.
  • In the atmosphere, water vapor frequently exists below the boiling point.

Learn more about vapor here:

brainly.com/question/14578189

#SPJ4

3 0
1 year ago
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