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_{ reduced} <\sigma_{ p} – elastic state (desired),
\sigma_{ reduced} =\sigma_{ p} – the point at which the plasticity occurs,
\sigma_{ reduced} >\sigma_{ 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 =\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 _{reduced}=\sqrt{\delta ^{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 _{p} = 250,00MPa\; -\; yield \; \; point
M = -1760,00kN
T = 880,00kN
J_{y} = 246509,00 cm^{4}
S_{y}^{c} = 3,20 * 30,00 * (31,80 + 1,60) = 3206,00 cm^{3}
z = 318,00mm
b = 15,00mm

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

Let’s start with normal stresses:

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

Contact stresses:

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

Reduced stresses:

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

We have a state impossible!

\sigma_{ reduced} >\sigma_{ 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.