This problem boils down to solving the issue of bending a simple curved rod. Curved rods mean rods which already in an unformed state have a curved linear axis (Zielnica 1996). We shall limit the following considerations to rods of symmetrical cross sections relative to the plane of the rod’s axis. Such a rod is shown in Fig. 6.27.
Fig. 6.27 A rod of a symmetrical cross section relative to the plane of the rod’s axis: a load state, b state of stress in the elementary section of a rod 
The state of stress prevailing in a rod can be analysed on an infinitely small element (Fig. 6.27), which is cut out from the rod, leading two planes perpendicular to its axis, passing through the centre of curvature and inclined towards each other at an angle d$>. As a result of bending the rod by moment M, the transverse cross section вв rotates in relation to the cross section aa by the angle Ad$>, taking the position of в’в’. Some of the fibres are subject to lengthening. Only the length of the fibre 00′ passing through the intersection point of the cross sections is not changed. This fibre belongs to a layer that is neutral to the element. It should be noted here that the length of particular fibres of an unformed element is not created equal. Therefore, the adopted Bernoulli’s principle and Hooke’s law are preserved only if the neutral layer 00′ is situated at a certain distance c from the axis of the rod. Any fibre, distant by yo from the neutral layer, changes its length by the section BB’ = yo Ad$>. The relative lengthening of the fibre is determined by the equation:
where p indicates the distance from the centre of the curvature of the rod axis to the considered fibre. The coordinates of normal stresses in the considered fibre of the element—in accordance with Hooke’s law—shall be as follows:
j, yoDdu
ax = exE = E. (6.35)
PdU
Let us next consider the balance of forces acting on the cut element (Fig. 6.28). Balance will be maintained if the following conditions are met: EX = 0, EM = 0. Condition EX can be written as follows:
axdA = 0.
A
Fig. 6.28 The balance of forces acting on the sheared element of the rod
By substituting ax with the equation:
we obtain:
^ E / — dA = 0:
du J P
A
Because the equation:
The second of these conditions of balance, i. e. 1M relative to the neutral axis, is written as follows:
After converting the equation and using in it the predefined relations, we obtain:
(6.46)
In the above equation, the expression
У0 dA,
has a value equal to zero. Therefore, we can write:
whereby
j yodA = S, (6.50)
A
indicates the static moment of the field of transverse cross section counted with respect to the neutral axis. Because the distance from the centre of gravity of the cross section to the neutral axis was marked previously by c, the static moment S can also be presented as follows:
S = A • c.
By comparing the previous equation, we obtain:
A
Adra M = EAc, du 
(6.53) 
which gives: 

Adu M du EAc 
(6.54) 
Therefore, the stress function is presented as follows: 

yo M Г x. q Ac 
(6.55) 
By substituting the subsequent equations, 

P = r + У, 
(6. 56) 
c = r – Го, 
(6.57) 
The equation of the moment M can be now written as follows: 
we obtain the equation written in the main axes xy of the cross section: 
which can be used in calculations of the normal stresses’ values in curved rods, bent in the curvature plane. The individual symbols in the equation mean:
M bending moment acting in the discussed cross section of the rod,
A area of the cross section of the rod, r radius of the curvature of the rod axis,
c distance of the neutral axis of the cross section of the rod to its centre of gravity, y coordinate of the point, in which we calculate the value of stress.