Current rating for busbars
A very good source of information with regards to copper is ... the Copper Alliance's documentation relating to ratings and sizing.

Your example is a common case for busbars and a convenient method to use is sizing by specific heat capacity: the amount of energy needed to raise a mass of material's temperature by a given amount.


  \$Q=S\cdot m\cdot t \$
  
  \$t_r = \frac{Q}{S\cdot m}\$
  
  \$Q\$ = amount of heat energy added (J)
  \$S\$ = specific heat capacity of the material (J/kg/K or J/kg/&deg;C)
  \$m\$ = mass of the bar (kg)
  \$t_r\$ = temperature rise (K or &deg;C)


Where the energy dissipated in the bar \$Q\$ is:  


  \$Q = P\cdot T\$
  
  P = power dissipated (W)
  T = Time the power is dissipated over (s)


Thus:


  \$\ \ \ \ \ \ \ t_r = \frac{P \cdot T}{S \cdot m}\$
  \$\therefore \Delta t_r = \frac{P}{S\cdot m} \$


The power dissipated is the ohmic loss (copper loss):


  \$P = I^2 R\$


Or, using the characteristics (resistivity and dimensions) of the material:


  \$P= I^2 \frac{\rho l}{A}\$
  
  \$\rho\$ = resistivity of the material (&ohm;&middot;m)
  \$l\$ = length of the bar (m)
  \$A\$ = cross-sectional area of the bar (m2) 


And so:


  \$\Delta t_r = \frac{I^2 \cdot \rho \cdot l}{A \cdot S \cdot D \cdot l \cdot A}\$  
  
  \$\ \ \ \ \ \ =\frac{I^2 \cdot \rho}{A^2 \cdot S \cdot D}\$
  
  \$\ \ \ \ \ \ = \frac{\rho}{S \cdot D} \cdot (\frac{I}{A})^2 \$
  
  \$D\$ = Density (kg&middot;m&minus;3)


Essentially these are the steps I go through (originally sourced from the copper alliance 15years ago) to provide a 1st pass sizing &amp; suitability. The entire document contains useful information as well pertaining to copper for current carrying