Physics, 05.09.2019 19:30, pc103919

A5-cm diameter solid shaft is simultaneously subjected to an axial load of 80 kn and a torque of 400 nm.
a) determine the principal stresses at the surface assuming elastic behavior.
b) find the largest shear stress.

Answers: 3

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answered: DAVISJOS00123

principal stress σ = 46.46 MPa and -5.72 MPa

maximum shear stress = 26.08 MPa

Explanation:

given data

diameter = 5 cm = 50 mm

axial load = 80 kN = 80000 N

torque = 400 Nm - 400000 N-mm

to find out

principal stresses and shear stress

solution

we find first axial stress that is force / area

axial stress = 80000 / ( π/4×50² )

axial stress σ = 40.74 MPa

and

shear stress = torque × radius / area

shear stress = 400000 × 25 / ( π/4×50² )

shear stress τ = 16.3 MPa

principal stress will be

principal stress = σ/2 ± $\sqrt{(\sigma/2^{2}+\tau^{2} )}$ ..........1

principal stress σ = 40.74 /2 ± $\sqrt{(40.74/2^{2}+ 16.3^{2} )}$

principal stress σ = 46.46 MPa and -5.72 MPa

and

maximum shear stress is

maximum shear stress = $\sqrt{(\sigma/2^{2}+\tau^{2} )}$ ..........2

maximum shear stress = $\sqrt{(40.74/2^{2}+ 16.3^{2} )}$

maximum shear stress = 26.08 MPa

answered: patrickdolano

principle stress =46.46 MPa, -5.72 MPa

maximu shear stress 26.08 MPa

Explanation:

d = 5 cm = 50 mm

axial load = 80 kN

torque = 400Nm

we know that

axial stress = force/area

$= \frac{80*10^3}{\frac{\pi}{4} 50^2}$

= 40.74 MPa

shear stress $= \frac{ Tr}{J}$

$= \frac{400*10^3 *25mm}{\frac{\pi}{32}50^4}$

= 16.3 MPa

principle stress is given as

$\sigma_p = \frac{\sigma_a}{2} \pm \sqrt{ (\frac{\sigma_a}{2})^2 +z^2}$

$=\frac{40.74}{2} \pm \sqrt{ (\frac{40.74}{2})^2 +16.3^2}$

=46.46 MPa, -5.72 MPa

b) maximu shear stress

$\sigma_p = \sqrt{ (\frac{\sigma_a}{2})^2 +z^2}$

$= \sqrt{ (\frac{40.74}{2})^2 +16.3^2}$

= 26.08 MPa

answered: Guest

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i had a friend that had sleep paralysis and he told me a story one time where he was laying in bed and he couldn't move his body no matter how hard he tried and there was someone standing in the corner of his room coming towards him. he couldn't move and you can't "wake up" because you are already conscious. sometimes sleep paralysis can be triggered by frequent waking up during the night based off of multiple trips to the bathroom, or if you had ever been abused, sleep paralysis can start happening from things like that.

answered: Guest

the answer is b i think

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A5-cm diameter solid shaft is simultaneously subjected to an axial load of 80 kn and a torque of 400...