Simplification using Newton's second law I am not sure if my simplification works in this problem:
Problem:
I have a beam which is strap around with cargo straps. First picture presents section through second picture. So applying Newtons second law:
$$F=G=\frac{m_{beam} g}{2} $$
so the F is the force in the cargo straps. Therefore I can calculate which king of cargo straps I need to use. Is this simplification legit or is it not approved method?
Black box is the beam and red lines are the straps. Green arrow is $G$
and red arrows are the $F$.


 A: The forces you have calculated are acceptable. However there are other things you will need to consider from an engineering point of view.
Not so important:
Because the straps are not directly connected to beams, rather the beam is resting on it. The beam will also deflect due to self weight. To be safe I would also check the slope of the beam at the ends given they are free to rotate.
$$\theta=\frac{MgL^{2}}{24EI}$$
where $E$ is the young's modulus of elasticity and $I$ is the moment of inertia.
You don't want the angle to be too large otherwise the strap may slide up the beam. If the angle is large, you will need to check the friction between the strap and the beam given the load Mg/2.
Important:
You will also need to check the internal capacity of the beam itself to make sure it can support it's own weight. The maximum shear force in the beam is Mg/2, you will need to check if the material can handle this. In addition, the beam bends when it is held from the straps.
You will get a maximum bending moment in the middle of the beam of:
$$M_{b}=\frac{MgL}{8}$$
The maximum stress that arises from carrying the beam is $f$:
$$f=\frac{M_{b}D}{2I}$$
where $D$ is the depth of your beam and $I$ is the moment of inertia.
If the stress you calculate is larger than the tensile or compressive strength (e.g. for steel $f=300MPa$) of your beam then your beam may get damaged during transport.
Also initial loading may be uneven, use straps that can take the whole load individually.
A: In addition to the other answers, another aspect that may be worth considering is how the sharp edges of the beam produces addition stress on the straps. If the lower corners of the beam are sharp enough, the straps could fail at a much lower load than you'd otherwise predict.
In practice, the beam cross section either has rounded corners, or would be surrounded by something with rounded corners, so you might get something that resembles this:

where $F = \frac{1}{2} m_{\text{beam}} g$. Assuming the case where the beam is static, the tension of the parts of the straps above the beam is $T=F/2$. In fact, the tension is $T=F/2$ everywhere in the straps, provided we make the simplifying assumption that we can ignore friction.
Since the strap is thin and flexible, we assume it can only carry tension, and not shear force or bending moments (like in beams). As a result, this means that the beam infact applies no force to the horizontal section of the straps below (as the horizontal strap carries only tension, which is horizontal, it cannot remain in equilibrium if the beam applies any vertical force onto it). This then means that all the weight of the beams is applied at the rounded corners.
Since the beam applies only force to the corner, and equivalent loading situation on the straps can be expressed as two cylinders with forces at 45 degrees.

Why are the forces at 45 degrees? This becomes clear if we consider a free-body diagram of one of the corners, rotated round to show the symmetry of the loading:

Clearly, the cylinder applies a pressure onto the strap, and we can calculate the pressure the cylinder exerts on the strap, which is equal to
$$p = \frac{T}{BR}$$
where $T$ is strap tension, $B$ is the width of the strap, and $R$ is the cylinder radius. From this, we can see that the pressure goes up if the cylinder radius decreases (i.e. the corners become sharper)
So, ignoring the effect of corner sharpness for a moment, we might say the strap fails if the tensile stress $\sigma$ exceeds a max stress $\sigma_{\text{max}}$. Stress is related to tension by
$$\sigma = \frac{T}{Bt}$$
where $t$ is the strap thickness, so that $Bt$ is the strap cross-sectional area. And so, noting that $F = 2T$ and $F = \frac{1}{2}m_\text{beam}g$, we might say that the maximum beam mass that could be supported is
$$m_{\text{max}} = \frac{4Bt}{g}\sigma_\text{max}$$
If we now consider the effects due to corner sharpness, it's a bit more complicated how much stress it takes for the strap to fail, since we have stresses in two directions (stress due to tension along the length of the strap, and stress due to the corner pressure perpendicular to the length of the strap).
A way to deal with this is to determine an equivalent stress $\sigma_\text{eqv}$ so that failure occurs when $\sigma_\text{eqv} = \sigma_\text{max}$ (see von Mises yield criterion and Tresca yield criterion for examples. Which yield criterion is appropriate will depend on the material). As a crude but simple approach, let's define the equivalent stress as
$$\sigma_\text{eqv} = \sigma_\text{tension} + \sigma_\text{pressure} = \frac{T}{B}\left(\frac{1}{t} + \frac{1}{R}\right)$$
Then, we find that the maximum mass before failure occurs is now
$$m_{\text{max}} = \left(1+\frac{t}{R}\right)^{-1}\frac{4Bt}{g}\sigma_\text{max}$$
So, in conclusion,
we estimate that the effect of the sharp corners of the beam reduce the strength of the straps by a factor of $1+\frac{t}{R}$, and so you should scale the design strength of the straps accordingly!
