Good question. What we're looking for is the semi's optimum load configuration (let's call it B).
Stuff that seem important.
1. Max load, M
2. Max space, S
3. Maximum safe load, A
4. Average volume of objects to be loaded, v
5. Average weight of objects to be loaded, w
6. Quantity of objects to be loaded, q
7. Minimum economically viable load, m (weight is what consumes fuel, there's no point huring a semi if a van can do it)
8. Density of loaded objects (weight per unit volume), d
We can see that ...
a) [imath]qv \leq S[/imath]
b) [imath]m < qw \leq A[/imath]
c) Total volume is determined by density of load. [imath]\frac{A}{d} = T_v[/imath], Total volume of load. [imath]q = \frac{T_v}{v}[/imath]
d) You may find that space restrictions may be more important than weight restrictions i.e. the semi will fill up at weights less than full weight capacity, but going by what you see in India and Pakistan, it's the opposite i.e. the semi can carry more than there's space on it. That is to say [imath]qm << A[/imath] and [imath]qv \geq S[/imath], Also, for semis used for industrial purposes, e.g. transporting metallic objects e.g. turbines, these being denser material, the situation maybe reversed.
A typical loading routine could start off with [imath]A/d = T_v[/imath]. We want to maximize load, keeping safety a priority ([imath]money \propto weight[/imath]). Then we find out q, that we do by [imath]T_v/v = q[/imath]. Load on ...
Remember if v is disproportionately large, q will be low and if q is low [imath]qw < m[/imath].
