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Anna71 [15]
3 years ago
14

The equation sigma = My/l is used in the mechanics of materials to determine normal stresses in beams. When this equation is exp

ressed in terms of SI base units. M is in newton-meters (N-m), y is in meters (m), and l is in meters to the fourth power (m^4). What are the SI units of sigma? If M = 2000 N-m v = 0.1 m, and l = 7 times 10^-5 m^4, what is the value of sigma in U.S. Customary base units?
Engineering
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

Part a : The SI unit of σ is Pascal.

Part b : The pressure is 414.28 psi.

Explanation:

Part a

The equation is given as

\sigma =\frac{My}{l}

As per the dimensional analysis

M=[N-m]\\y=[m]\\l=[m^4]

So the equation becomes

\sigma =\frac{[N-m][m]}{[m^4]}\\\sigma =\frac{[N][m^2]}{[m^4]}\\\sigma =\frac{[N]}{[m^{4-2}]}\\\sigma =\frac{[N]}{[m^{2}]}\\

As the dimensions are of pressure so the SI unit of σ is Pascal.

Part b

\sigma =\frac{My}{l}\\\sigma =\frac{2000 \times 0.1}{7 \times 10^{-5}}\\\sigma =2857142.857  \, Pa

Pressure in US customary base units is given in psi so

1 Pa =1.45 \times 10^{-4}\\

So

\sigma =2857142.857  \times 1.45 \times 10^{-4} psi\\\sigma =414.28 psi

So the pressure is 414.28 psi.

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The state of plane strain on an element is:
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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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Distributes a floor load or weight

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