Thick curved beams

The generalised sector of a thick curved beam with centre of curvature at O and small angular extent Φ appears below. The radii of interest are identified in the LH sketch. Two radii are particularly significant - the mean (or centroidal) radius r_c, and the mean reciprocal radius r_d. They are defined by :

( 1 )       A * r_c   =   ∫ _ri^r_o dA * r   ;           A / r_d   =   ∫ _ri^r_o dA / r

Like   r_i and r_o, both   r_c and r_d are intrinsic to the known geometry ( b = f(r) ) of the cross-section.
Since the beam is curved, the radius defining the unstressed neutral axis, r_n, is not immediately obvious as it would be in pure bending of a straight beam - it depends on the load and is unknown initially. The RH sketch shows the assumed positive senses of the two components of the resultant of stresses acting on the RH cross-section :

• a tensile force, F, which is chosen to pass through the centre of curvature, and
• a corresponding bending moment, M, which tends to sharpen the curvature.

In the normal situation the resultants may be ascertained easily since they equilibrate other loads, which are usually known. Shear is neglected as in simple bending of a straight beam.
Application of M increases the curvature, so the angle subtended by the sector increases to Φ' under load in the RH sketch. Assuming that plane cross-sections remain plane during loading, the increase in length of the typical fibre at radius r is proportional to the distance from the unstrained neutral axis, ( r-r_n). Since the original length of the fibre is proportional to the radius r, then :
      fibre strain   =   extension/free length   ∝   ( r - r_n )/r   =   1 - r_n /r
Presuming elastic behaviour, the stress σ in the fibre at radius r is proportional to this strain, so

( 2 )       σ   ∝   1 - r_n /r   =   σ_m ( 1 - r_n /r )      where   σ_m and r_n are constants, as yet unknown.

This establishes the form of the stress variation across the cross-section; it is clear that the distribution is non-linear, giving rise to high stress magnitudes at small radii - that is, stress is concentrated where curvature is sharpest where r becomes small. Determination of the two constants follows from the necessity for the resultants of the stress distribution ( 2) to be the known F and M noted above. Thus, for equivalence of the distribution and the resultants, using ( 1) and ( 2) :-
      F     =   ∫ ^A σ dA     =   σ_m ∫ ^A ( 1 - r_n /r ) dA       =   σ_m A ( 1 - r_n /r_d )
      M   =   ∫ ^A σ r dA   =   σ_m ∫ ^A ( 1 - r_n /r ) r dA   =   σ_m A ( r_c - r_n )
Solving these for the two unknown constants :-

( 2 )       r_n   =   r_d ( M - F r_c ) / ( M - F r_d )   ;           σ_m   =   ( M - F r_d ) / A ( r_c - r_d )

It is noticed these two constants - and hence the stress distribution ( 2) - depend only upon the loading ( F & M) and the cross-sectional geometry ( A, r_c & r_d ).
Two particular locations of the neutral axis occur when
      M   =   0     whereupon   r_n = r_c, and
      F     =   0     then   r_n = r_d - the pure bending case most commonly encountered in the literature.

Summarising - application of the foregoing theory involves the following steps :

• compute the salient radii   r_c and r_d from ( 1) and known cross-sectional geometry (see below),
• determine the stress resultants ( F & M) necessary to equilibrate other given loads, with positive senses to the convention noted above,
• evaluate the constants from ( 3), hence completely define the stress distribution by ( 2).

The conspicuous radii, r_c & r_d, for common symmetric cross-sections are found from ( 1) as follows.

Trapezoid - defined by dimensions   r_i , r_o , b_i and b_o as sketched.

The variables   r and dA must be expressed as functions of the same variable before the integrals in ( 1) can be evaluated. The radius   r itself is the logical choice for independent variable, since, from the sketch   dA = b.dr, where   b is evidently linearly dependent on r, say
      b   =   α + β r     where the constants   α and β are evaluated at   r_i and r_o :
      α   =   ( b_i r_o - b_o r_i ) / ( r_o - r_i )   ;       β   =   ( b_o - b_i ) / ( r_o - r_i )

∴ dA   =   b dr   =   ( α + β r ) dr     and therefore, from ( 1) :-
      A   =   ∫ ^A dA       =   ∫_ri^ro ( α + β r ) dr     =   1/2 ( r_o - r_i ) ( b_i - b_o )
  A*r_c =   ∫ ^A dA*r   =   ∫_ri^ro ( α + β r ) r dr  =   1/6 ( r_o - r_i ) [ b_o( 2r_o + r_i ) + b_i( 2r_i + r_o ) ]
  A/r_d =   ∫ ^A dA/r   =   ∫_ri^ro ( α/r + β ) dr   =   b_o - b_i + [ ( b_ir_o - b_or_i )/( r_o - r_i ) ] . ln ( r_o /r_i )

Rectangle - particularising the trapezoidal results with   b_o = b_i = b ( constant ) :-

  A   =   b ( r_o - r_i )   ;       A*r_c   =   ¹/₂ b ( r_o² - r_i² )   ;       A/r_d  =   b ln ( r_o /r_i )

Circle - integrating as above :-     2 r_c   =   r_o + r_i  ;       2 √r_d   =   √r_o + √r_i

Built-up sections may be analysed by dividing them into a number of simple shapes for each of which the integrals in ( 1) are calculable, then applying ( 1) in discrete form :-

( 1a )       A   =   Σ_j=1 ( δA )_j   ;       A*r_c =   Σ_j=1 ( δA*r )_j   ;       A/r_d =   Σ_j=1 ( δA/r )_j

Consider the T-section shown, consisting of two rectangular elements :-

element	width	inner radius	outer radius	δA	δA*rc	δA/rd
1	bi	ri	rm	bi ( rm - ri )	1/2 bi ( rm2 - ri2 )	bi ln ( rm/ri )
2	bo	rm	ro	bo ( ro - rm )	1/2 bo ( ro2 - rm2 )	bo ln ( ro/rm )
total				Σj=1 ( δA )j	Σj=1 ( δA*r )j	Σj=1 ( δA/r )j

∴     A   =   b_i ( r_m - r_i ) + b_o ( r_o - r_m )         using ( 1a)
  A*r_c   =   ¹/₂ [ b_i ( r_m² - r_i² ) + b_o ( r_o² - r_m² ) ]       whence   r_c for the section . . .
  A/r_d   =   b_i ln ( r_m /r_i ) + b_o ln ( r_o /r_m )     . . . and   r_d

The most heavily stressed cross-section is assumed to be just to the left of A-A as shown, since

• it is the furthest cross-section from the load, so it is the cross-section subjected to the largest bending stresses, and
• it lies in a 'thick curved beam' portion of the clamp and so is subjected to stress concentration.

Section properties :
From the above T-section results with   r_i = 20,     r_m = 30,     r_o = 50,     b_o = 15,     b_o = 5 mm :
      A     =     15 ( 30 -20) +5 ( 50 -30)                       =   250 mm²            watch units!
  A*r_c   =   [ 15 ( 302 -202 ) +5 ( 502 -302 ) ]/2   =   7750 mm³       whence   r_c   =   31 mm
  A/r_d   =     15 ln 30/20 + 5 ln 50/30                     =   8.636 mm         whence   r_d   =   28.95 mm

The stress resultants, F and M, are shown on the free body in the conventional positive senses used in developing the theory above. For equilibrium of the free body,   F = P and   M = -Pa. Inserting these into ( 3) gives :-
    r_n   =   r_d ( a + r_c )/( a + r_d )       =   28.95 ( 40 + 31 )/( 40 + 28.95 )   =   29.81 mm
    σ_m =   -P ( a + r_d )/A ( r_c - r_d ) = -750 ( 40+28.95)/250( 31-28.95) = -101 MPa
The stress distribution is thus     σ = -101 ( 1 - 29.81/r ) MPa     where   r is in mm.

The stress distribution is plotted and yields extreme values of 49 MPa tensile at the inside (20 mm radius), and 41 MPa compressive at the outside. As the stress nowhere exceeds 50 MPa, the claimed capacity seems vindicated.
But, for the straight portion :-
      I   =   15 ( 10³/12 +10*6² ) + 5 ( 20³/12 +20*9² )   =   18083 mm⁴, so
      σ   =   σ_d + σ_b   =   P/A + My/I
            =   750/250 - 750(40+31).( r -31)/18083   =   3 - 2.945 ( r -31) MPa

This stress distribution also is plotted and indicates a tensile stress of 35 MPa at the inside, and 53 MPa compressive at the outside. The clamp is therefore some 6 % under-designed and unsafe. But it is physically impossible for two distinct stress distributions to occur at an infinitessimal distance apart across the interface A-A. The actual distribution at A-A will therefore lie somewhere between those found - however this does not invalidate the conclusions regarding clamp inadequacy . . . . why ?