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valentinak56 [21]
2 years ago
14

Derive the expression ε=ln(1+e), where ε is the true strain and e is the engineering strain. Note that this expression is not va

lid after the onset of necking
Engineering
1 answer:
Rudiy272 years ago
6 0

The formula for true strain after derivation from basic terms is; ε_t = In(1 + ε_e)

<h3>How to derive the expression for True Strain?</h3>

Formula for Engineering Stress is;

σ_e = Load/Area

Formula for true stress is;

σ_t = Force/Instantaneous Area

Formula for Engineering Strain is;

ε_e = ΔL/L₀

Formula for true strain is;

dε_t = dL/L

Total true strain is gotten from;

ε_t = ∫(dL/L) between boundaries of L_f and L_o

When we integrate between those boundaries, we have;

ε_t = In[(L₀ + ΔL)/L₀

⇒ ε_t = In[(1+ ΔL/L₀)

⇒ ε_t = In(1 + ε_e)

Read more about True Strain at; brainly.com/question/20717759

#SPJ1

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Read 2 more answers
The state of plane strain on an element is:
balu736 [363]

Answer:

a. ε₁=-0.000317

   ε₂=0.000017

θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2}  \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2}  \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6}  \pm 1.67 \times 10^{-4}

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5

θ= -13.28° and  76.72°

To determine the direction of ε₁ and ε₂

\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2}  + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta  + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2}  + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56)  + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain

\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}

=3.335 *10^-4

\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )

ε(avg) =150 *10^-6

orientation of γmax

tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}

θ = 31.71 or -58.29

To determine the direction of γmax

\gamma _{x'y' }=  - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta  + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }=  - \frac{-300*10^{-6} - \ 0}{2} sin(63.42)  + \frac{150*10^{-6}}{2}cos(63.42)

= 1.67 *10^-4

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A protective equipment which protects workers who are passing by from stray sparks or metal while another worker is welding is: E. Welding Screens.

A wielder refers to an individual who is saddled with responsibility of joining two or more metals together by wielding.

During the process of wielding, sparks and minute metallic objects are produced, which are usually hazardous to both the wielder and other workers within the vicinity.

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However, a protective equipment which protects other workers who are passing by from stray sparks or metallic objects while wielder (worker) is welding is referred to as welding screens.

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Which starting circuit uses fuses, switches, and smaller wires to energize a relay and solenoid?
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Option B (Starter Control Circuit) is the right option.

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