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KiRa [710]
3 years ago
5

A cylindrical specimen of this alloy 12.7 mm in diameter and 250 mm long is stressed in tension and found to elongate 7.6 mm. On

the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why.
Engineering
1 answer:
Margarita [4]3 years ago
5 0

Answer:

Condition 2 is true.

\epsilon_{test}>\epsilon_{yield}

we can not calculate the load with given information.

According to above values, Condition 2 is satisfied so we can not find the load with given information because material deformation lies in plastic region.

Explanation:

In order to check whether we can find the load or not we have to check the following conditions:

Condition 1:

\epsilon_{test}

If this condition is true then we can calculate the load.

Condition 2:

\epsilon_{test}>\epsilon_{yield}

If this condition is true then we can not calculate the load with given information.

Calculating \epsilon_{test}:

\epsilon_{test}=\frac{Elongation}{original\ length}

\epsilon_{test}=\frac{7.6\ mm}{250\ mm}\\ \epsilon_{test}=0.0304

Calculating \epsilon_{yield}:

\epsilon_{yield}=\frac{Yield\ Stress}{Elastic\ modulus}\\ \epsilon_{yield}=\frac{280\ Mpa}{105*10^3\ MPa} \\\epsilon_{yield}=2.67*10^{-3}=0.00267

Hence:

Condition 2 is true.

\epsilon_{test}>\epsilon_{yield}

According to above values, Condition 2 is satisfied so we can not find the load with given information because material deformation lies in plastic region.

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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

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Answer:

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The entropy is equal to:

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Answer:

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3 years ago
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