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mina [271]
2 years ago
7

A 20cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter of the bar is 0.490 at this load, de

termine: I. the engineering stress and strain, and [2] II. the true stress and strain
Engineering
1 answer:
Margaret [11]2 years ago
4 0

If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.

<h3>What is strain?</h3>

Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.

  1. L = 20 cm d x 1 = 0.21 cm
  2. dx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3
  3. (assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333
  4. b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333
  5. = 1561.84 MPa.

Read more about the diameter :

brainly.com/question/358744

#SPJ1

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The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevat
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Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}

Properties: The density of water \delta = 1000 kg/m^3

Conversions:

165 \  ft \  to \  meters  = 50 m  \\ \\7000 \ lbm/s \  to  \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \  to \  kilowatt = 1166 kw \\ \\

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

e_{mech_{in}} - e_{mech_{out}} = gh - 0

Then;

gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}

gh = 0.491 kJ/kg

\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg

= 1559 kW

Therefore; the overall efficiency is:

\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}

= \dfrac{1166 \ kW}{1559 \ kW}

= 0.75

= 75%

b) mechanical efficiency of the turbine:

\eta_{turbine- generator} = \eta_{turbine}\times   \eta_{generator}

thus;

\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%

6 0
3 years ago
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