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scoundrel [369]

3 years ago

9

For an isotropic material, E and ν are often chosen as the two independent engineering constants. There are other elastic consta

nts: the shear modulus G, the bulk modulus K, and the Lames’ constants, μ and λ.
Suppose that G = E/[2(1+v)].
Prove that μ and λ, and K, satisfy the following relations:

μ = G = 3K(1-2v)/[2(1+v)], λ = Ev/ [(1+v)(1-2v)] = 2vG/ (1-2v),
K = E/ 3(1-2v) = λ+(2/3)μ

To prove it, apply pure shear stress τ, and connect it to pure shear strain γ by G, ie. τ = Gγ. Then in the 45° orientation, consider its normal stresses σ, and σ: are related to its normal strains ε, and e, through E and v in Hooke's law. We can easily establish this relation.

1 answer:

pav-90 [236]3 years ago

8 0

Answer:

khgy

Explanation:

nbv

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Example – a 100 kW, 60 Hz, 1175 rpm motor is coupled to a flywheel through a gearbox • the kinetic energy of the revolving compo

rjkz [21]

Answer:

1200KJ

Explanation:

The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.

P (rotor-loss) = 3 x K.E

P = 3 x 300 = 900 KJ

After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;

KE = 300 KJ

Since it is in opposite direction, it will also add up to rotor loss

P ( rotor loss ) = 900 + 300 = 1200 KJ

7 0

3 years ago

Which option identifies the type of device the engineer will develop in the following scenario?

Stells [14]

It would be actuator

4 0

2 years ago

Read 2 more answers

A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s

sergiy2304 [10]

Answer:

$D_o=11.9inch$

Explanation:

From the question we are told that:

Thickness $T=0.5$

Internal Pressure $P=2.2Ksi$

Shear stress $\sigma=12ksi$

Elastic modulus $\gamma= 35000$

Generally the equation for shear stress is mathematically given by

$\sigma=\frac{P*r_1}{2*t}$

Where

r_i=internal Radius

Therefore

$12=\frac{2.2*r_1}{2*0.5}$

$r_i=5.45$

Generally

$r_o=r_1+t$

$r_o=5.45+0.5$

$r_o=5.95$

Generally the equation for outer diameter is mathematically given by

$D_o=2r_o$

$D_o=11.9inch$

Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

$D_o=11.9inch$

7 0

3 years ago

Determine the deflection at the center of the beam. Express your answer in terms of some or all of the variables LLL, EEE, III,

Rom4ik [11]

Answer:

See explanations for step by step procedures to get answer.

Explanation:

Given that;

Determine the deflection at the center of the beam. Express your answer in terms of some or all of the variables LLL, EEE, III, and M0M0M_0. Enter positive value if the deflection is upward and negative value if the deflection is downward.

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3 years ago

Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.

nikitadnepr [17]

Answer:

a) $m_e= 3.05 Kg$

b) $\rho=1072.3kg/m^3$

c) $m_e= 3.05 Kg$

Explanation:

From the question we are told that:

Beaker Mass $m_b=1.20$

Liquid Mass $m_l=1.85$

Balance D:

Mass $m_d=3.10$

Balance E:

Mass $m_e=7.50$

Volume $v=4.15*10^{-3}m^3$

a)

Generally the equation for Liquid's density is mathematically given by

$m_e=m_b+m_l+(\rho*v)$

$\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}$

$\rho=1072.3kg/m^3$

b)

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

$m_d=m- (\rho *v )$

$m=3.10+ (1072.30 *4.15*10^{-3}m^3 )$

$m=18.10kg$

c)

Generally the equation for E's Reading at A pulled is mathematically given by

$m_e=m_b+m_l$

$m_e = 1.20 + 1.85$

$m_e= 3.05 Kg$

6 0

3 years ago