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Marrrta [24]
3 years ago
5

Write the definitions for engineering stress, true stress, engineering strain, and true strain for loading along a single axis.

A commercially pure copper wire originally 10.00 m long is pulled until its final length is 10.10 m. It is annealed, and then pulled again to a final length of 10.20 m. What is the engineering strain associated with each of the two steps in the process? What is the true strain for each step? What are the total engineering and true strains for the combined steps?
Engineering
1 answer:
shusha [124]3 years ago
5 0

Answer and Explanation:

Engineering Stress of a material is defined as the applied load or force divided by the original cross sectional Area of the material.

σ(engineering) = F/(Ao)

True Stress is defined as the applied load or force divided by the actual cross-sectional area (the changing area with respect to time) of the material at that point in time. It's an instantaneous stress.

σ(true) = F/A

Engineering strain is a measure of how much a material deforms under a particular load. It is the amount of deformation in the direction of the applied force divided by the initial length of the material.

ε(engineering) = Δl/lo

True Strain measures instantaneous deformation. It is obtained mathematically by integrating strain over small time periods and summing them up. Hence,

ε(true) = In (lf/lo)

The calculations,

First step, 10m to 10.1m, Δl = 0.1m, lf = 10.1m, lo = 10m

ε(engineering) = 0.1/10 = 0.01

ε(true) = In (10.1/10) = 0.00995

Second step, 10.1m to 10.2m, Δl = 0.1m, lf = 10.2m, lo = 10.1m

ε(engineering) = 0.1/10.1 = 0.0099

ε(true) = In (10.2/10.1) = 0.00985

Overall, 10m to 10.2m, Δl = 0.2m, lf = 10.2m, lo = 10m

ε(engineering) = 0.2/10 = 0.02

ε(true) = In (10.2/10) = 0.0198

QED!

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Answer:

a. \frac{-kdT(0)}{dx} =q_{0}=5000W/m^2

b.833.3(0.006-x)+112

c. 117 deg C

Explanation:

Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conductivity of k 5 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron,

<u>Assumption</u>

Heat conduction is steady state and unidimensional  2. thermal conductivity is constant. Heat supplied is not in the plate

4. we disregard heat loss

Heat flux=heat/area

\alpha/A=800W/160*10^-4

with direction to the surface been in the x direction,

the mathematical expression will be

\frac{d^2T}{dx^2}=0..............1

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from fourier law, for conductivity

T(L)=T2=112C

b. integrating equation 1 twice we have\dT/dx=c1

T(x)=C1x+C2

C1 and C2 are arbitrary constant

at x=0 the boundary conditions become

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C1=-(qo/k)

at x=L          

=T(L)=C1L+C2=T2

C2=T2-cL1

C2=T2+qoL/k

Juxtaposing C1 and C2 into the general equation , we have

T(x)=-qo/k+T2+qoL/k=qo(L-k)/k+T2

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Answer:

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Required

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U = N/m^2

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