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kiruha [24]
3 years ago
10

Consider a cubic workpiece of rigid perfect plastic material with side length lo. The cube is deformed plastically to the shape

of a rectangular parallelepiped of dimensions l1,l2, and l3. Show that the volume constancy requires that the following expression be satisfied: e1+e2+e3 = 0

Engineering
1 answer:
Taya2010 [7]3 years ago
8 0

Answer:  ε₁+ε₂+ε₃ = 0

Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-

                         l₀l₀l₀=l₁l₂l₃

                        \frac{lo*lo*lo}{l1*l2*l3}=1.0

                      taking natural log on both sides

                              ln(\frac{(lo*lo*lo)}{l1*l2*l3})=ln(1)

Considering the logarithmic Laws of division and multiplication :

                                ln(AB) = ln(A)+ln(B)

                                ln(A/B) = ln(A)-ln(B)

                           ln(\frac{(l1)}{lo})*ln(\frac{(l2)}{lo})*ln(\frac{(l3)}{lo}) = 0

Use the image attached to see the definition of true strain defined as

                         ln(l1/1o)= ε₁

which then proves that ε₁+ε₂+ε₃ = 0

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If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam
Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

\zeta =\dfrac{C}{C_c}

Where C is the damping coefficient and Cc is the critical damping coefficient.

C_c=2\sqrt{mK}

So from above we can say that

\zeta =\dfrac{C}{2\sqrt{mK}}

\zeta \alpha \dfrac{1}{\sqrt K}

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0
3 years ago
The error in the trapezoidal rule is proportional to h^2 True b. False
Lynna [10]

Answer:

b) False

Explanation:

in trapezoidal rule the error is proportional to h^3 and the order of accuracy is proportional to h^3 .

Trapezoidal rule is numerical integration method .Trapezoidal rule is used to find the area of curves.In trapezoidal rule we finds the approximate value of integration.But the real value of integration will not differ to much from the value which finds by using trapezoidal rule.

4 0
3 years ago
Is the flow of power reversible in a worm and wheel gear
nikitadnepr [17]

The flow of power cannot be reversed since the slider could not move the worm gears. Since the input has one continuous tooth and the output has not teeth there is no gear ratio and no change in torque and speed.

8 0
3 years ago
Write the heat equation for each of the following cases:
jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

\dfrac{\partial^2T}{\partial x^2}=  \ 0  \  ;  \ if \  T = f(x)  \\ \\ \dfrac{\partial^2T}{\partial y^2}=  \ 0  \  ;  \ if \  T = f(y)  \\ \\ \dfrac{\partial^2T}{\partial z^2}=  \ 0  \  ;  \ if \  T = f(z)

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}

where;

Q_g = heat generated per unit volume

\alpha = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0

where;

The radial directional term = \dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) and the axial directional term is \dfrac{\partial^2 T}{\partial z^2 }

d) The heat equation for a wire going through a furnace is:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]

since;

the steady-state is zero, Then:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]'

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}

4 0
3 years ago
Everyone why are you reporting my answers i didnt do anything to you
masya89 [10]

Answer:

IDK

Explanation:

same thing is happening to me

5 0
3 years ago
Read 2 more answers
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