In your question, you said to ignore atmospheric effects. However, the atmosphere also acts as a secondary source of radiation (the infamous greenhouse effect). Even if you neglect eddies, advection, etc. you also need to consider that the atmosphere can absorb radiation and emit it back to the surface, otherwise you would find the average temperature of the surface to be the same as the average temperature of the earth.

Edit: I found a decent approximation (approximation being the key word).

Let $Q_S$ is the incoming solar radiation flux $$Q_s= S(1-\alpha)(\frac{\bar{d}}{d})^2cos(\zeta)\tau_s $$ Where $S$ is the solar constant, $\alpha$ is the albedo, $d$ is the distance of the earth from the sun, $\bar{d}$ is the mean distance of the earth from the sun and $\tau_s$ is the transmissivity of the atmosphere (including any clouds above).

The last factor, $cos(\zeta)$ can be computed by$$cos(\zeta)=sin(\psi)sin(\delta)+cos(\psi)cos(\delta)cos(h)$$ where $$\delta=23.45\times\frac{\pi}{180}cos(\frac{2\pi\times(d-d_{solst})}{d_{year}})$$ Where $d$ is the Julian day, $d_{solst}$ is the julian day of the solstice (173) and $d_{year}$ is the number of days in the year (365.25)

Additionally, $h$ is the local hour, as defined by $$h=\frac{(t_{UTC}-12)\times\pi}{12}+\frac{\lambda\pi}{180}$$ where $\lambda$ is the longitude and $t_{UTC}$ is the time in UTC (in hours).