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!
