# Stress reduced by hypotheses Hubert-M-H

A word of introduction to the guide. At the beginning, as usual, a bit of theory, I will explain the H – M – H hypothesis (plasticity condition), then I will give the formulas and, for example, I calculate the stresses reduced in the beam.

Hypothesis Hubert-Mises-Hencky (plasticity condition) – one of the performance hypotheses, in which the energy of the form deformation is compared, namely:
the material passes into a plastic state at a given point, when the energy density of the form strain (energy of deviators) reaches a certain limit value, characteristic for a given material.
Stress state as in the beam:
$\sigma_{&space;reduced}&space;<\sigma_{&space;p}$ – elastic state (desired),
$\sigma_{&space;reduced}&space;=\sigma_{&space;p}$ – the point at which the plasticity occurs,
$\sigma_{&space;reduced}&space;>\sigma_{&space;p}$– impossible state; before, plasticity (destruction) occurred.

Now I will quote the necessary formulas to calculate the reduced stresses. There are exactly three formulas: the first is for normal stresses, the next for tangential stresses, and the last for calculating the reduced stresses.

Pattern for normal stresses.

$\delta&space;=\frac{M*z}{I_{y}}$

where:
M – this is the value of the bending moment in the place of the calculated reduced stress
z – distance of the point to the center of gravity
$I_{y}$ – moment of inertia relative to the horizontal axis

Pattern for tangential stresses.

$T=\frac{T*S_{y}^{c}}{J_{y}*b}$

where:
$T$ – value of cutting force at the place of calculated reduced stress
$S_{y}^{c}$– static moment at point C. We calculate it as follows:
area over the calculated point, multiply the distance of the center of gravity of the surface area over the point to the center of gravity of the whole figure.
$J_{y}$ – moment of inertia relative to the horizontal axis
$b$ – the width of the field on which the point is located

The last formula for calculating the reduced stresses.

$\sigma&space;_{reduced}=\sqrt{\delta&space;^{2}+3*T^{2}}$

where:
$\sigma$ – calculated normal stresses
$T$ – calculated tangential stresses

We will now go on to calculate the example to better understand how to use the above formulas. We have a fixed beam with an I-section.

Description:
Dimensions are in m. A rectangular distributed force acts on the beam. Suppose we count the stresses reduced at the point located in the place of restraint.

Data:
$\sigma&space;_{p}&space;=&space;250,00MPa\;&space;-\;&space;yield&space;\;&space;\;&space;point$
$M&space;=&space;-1760,00kN$
$T&space;=&space;880,00kN$
$J_{y}&space;=&space;246509,00&space;cm^{4}$
$S_{y}^{c}&space;=&space;3,20&space;*&space;30,00&space;*&space;(31,80&space;+&space;1,60)&space;=&space;3206,00&space;cm^{3}$
$z&space;=&space;318,00mm$
$b&space;=&space;15,00mm$

Let’s remember about units to convert to the same when calculating.
We substitute our data for the patterns that appear higher.

$\sigma=\frac{M&space;*&space;z&space;}{Jy}&space;=&space;\frac{-1760,00&space;*&space;0,318}{246509&space;*&space;10^{-8}&space;}&space;=&space;227,042MPa$

Contact stresses:

$\tau&space;=&space;\frac{880,00&space;*&space;3206&space;*&space;10^{-6}}{246509&space;*&space;10^{-8}&space;*&space;0,015}&space;=&space;76,299MPa$

Reduced stresses:

$\sigma&space;_{reduced}&space;=&space;\sqrt{\sigma^{2}&space;+&space;3&space;*&space;\tau^{2}}=&space;\sqrt{(227,04&space;+&space;3&space;*&space;76,29)}&space;=&space;262,70&space;MPa&space;>&space;\sigma&space;_{p}&space;(250,00MPa)$

We have a state impossible!

$\sigma_{&space;reduced}&space;>\sigma_{&space;p}$ – impossible state; before, plasticity (destruction) occurred.

This is enough of the reduced stress. Once again, remember to run calculations in units of the same degree!
I invite you to the next guides, if everything is clear or to previous ones, if there are any ambiguities.