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3 years ago

SPELL HELLO PLZ I NEED HELP BIG TIME

Engineering

2 answers:

QveST [7]3 years ago

8 0

Answer:

hello

Explanation:

jeka57 [31]3 years ago

8 0

Answer:

HELLO

Explanation:

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What math subjects are going to be seen in Industrial Engineering?

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Answer:

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3 years ago

Read 2 more answers

In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri

ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

Here

$\sigma_x$ is the normal stress which is 1750 psf
$\sigma_y$ is 0
$\tau_{xy}$ is the shear stress which is 800 psf

So the formula becomes

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}$

Similarly, the minimum normal stress is given as

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}$

The maximum shear stress is given as

$\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}$

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