This are eccentrically loaded bolts in which the plane of loading is parallel to the bolt plane.When deriving an equation to this particular problem almost all books start by assuming that $$\frac{F_{1}}{r_{1}}=\frac{F_{2}}{r_{2}}=\frac{F_{3}}{r_{3}}=...=c$$ So, how can one make such assumptions and have a correct answer.Meaning, wouldn't a different assumption cause different results? Is there some type physical meaning to the assumption made?

I too had the same doubt but here is what I make sense of it.

$$I*\alpha=P*e$$ ; α=angular acceleration

$$I= m_1*l_1^2 +m_2*l_2^2+m_3*l_3^2+m_4*l_4^2$$

$$α=P*e/(m*(l_1^2+l_2^2+l_3^2+l_4^2))$$ -1

Angular acceleration of all mass particles in the system is same. Know let us assume that there was no external Torque ($$P*e$$) and the bolts produced the same reaction Moment. Then each bolts contribution for the net Moment must be in such that its angular acceleration about CG is same.

So,

$$F_1/(l_1*m_1) = F_2/(l_2*m_2)=F_3/(l_3*m_3)=F_4/(l_4*m_4)= α$$

General assumption is all bolts are same. So, $$m_1=m_2=m_3=m_4=m$$

$$F_1/l_1 = F_2/l_2 =F_3/l_3 =F_4/l_4 = m*α$$ -2

From 1 & 2

$$F_1/l_1 = F_2/l_2 =F_3/l_3 =F_4/l_4 = α = P*e/(l_1^2+l_2^2+l_3^2+l_4^2)$$

The last equation is given in many textbooks. So I assume the approach must be true.