MISCELLANEOUS STRENGTH TOPICS

• bending of beams which are initiallly not straight but deliberately curved,
• asymmetric (or 'unsymmetrical') bending where we introduce a matrix approach which will be later found indispensable in the analysis of fillet welded joints,
• detailed stress concentration in way of elastic contacting bodies such as the round pin and hole which in the previous chapter we were obliged to approximate using projected areas.

Castigliano's theorem

The load G may be a translational force in which case the displacement δ_G is linear; alternatively the load G may be a rotational moment or torque in which case the displacement δ_G also is rotational. The displacement and the load are reckoned positive in the same sense - eg. NNE or clockwise as shown.

For a system in which bending strain only is significant, that is for a beam-system :
                δ_G   =   ^∂/_∂G [ ∫_length ( M²/2EI ) ds ]       or, changing the order of operations

( ii)         δ_G   =   ∫_length ^M/_EI.^∂M/_∂G .ds       where s is the distance along the beam.

The deflection under a particular load is thus found by expressing M (and EI if it's not constant) algebraically in terms of distance along the beam, s, then applying ( ii). A couple of examples follow showing this applied to straight beams, however Castigliano comes into his own when the deflections of curved beams are sought.

Use Castigliano in bending ( ii) with Q as the load and EI constant :
            δ_G   =   ∂U/∂Q   =   ¹/_EI ∫₀^L   M ^∂M/_∂Q   ds     where   M = function(s)

There are two spans, ie. two regions of integration. From a free body of each, the bending moment may be found in terms of the known external loads :
(a)       M   =   Qs ;                              ∂M/∂Q = s ;     0 <= s <= L/2
(b)       M   =   Qs + P( s - L/2) ;     ∂M/∂Q = s ;     L/2 <= s <= L

So   EI δ_Q   =   ∫₀^L/2 Qs.s ds   + ∫_L/2^L [ Qs + P( s - L/2) ].s ds   =   ( Q/3 + 5P/48 ) L³

The displacement of the cantilever under the load Q is therefore ( Q/3 + 5P/48 ) L³/EI

If a displacement is needed for which there is no corresponding actual load, then a fictitious dummy load must be applied to the system. The value of such a dummy load is zero, but it must be included algebraicallly in order to deduce partial derivatives ( for use in ( ii) ) - for although a dummy load is actually zero, partial derivatives involving it generally are not zero. Once these partial derivatives have been ascertained, the dummy load is set to zero when working out the integral in ( ii).

Since the rotational displacement of the end at Q is required here, the dummy rotational load M_o is defined at that end as shown. This external moment will appear on each of the free bodies of the previous example, so that the bending moments at s become :
(a)       M   =   M_o + Qs ;                              ∂M/∂M_o = 1 ;     0 <= s <= L/2
(b)       M   =   M_o + Qs + P( s - L/2) ;     ∂M/∂M_o= 1 ;     L/2 <= s <= L

Having evaluated the partial derivative of the bending moments with respect to the dummy load, the dummy load is zeroed in the energy integral. So, with   M_o = 0 :

            δ_Mo   =   ∂U/∂M_o   =   ¹/_EI ∫₀^L   M   ^∂M/_∂Mo   ds

                            Do not confuse the bending moment M with the external load M_o

So   EI δ_Mo   =   ∫₀^L/2 Qs.1 ds   + ∫_L/2^L [ Qs + P( s - L/2) ].1 ds   =   ( Q/2 + P/8 ) L²

The slope of the cantilever at the load Q is therefore ( Q/2 + P/8 ) L²/EI

Thin curved beams

• A thin curved beam is characterised by a beam depth which is small compared to the radius of curvature, and as a result curvature effects and stress concentration in the cross-section are negligible, leading to stresses which are essentially linear like those of a straight beam. With stress concentration not an issue, attention is focused on beam deflections, since the familiar formulae used for straight beam deflections cannot be used for curved beams.
• A thick curved beam is characterised by a beam depth which is of the same order as the radius of curvature and so stresses are non-linear, increasing towards the inside of the bend where curvature effects are more pronounced. We examine this stress concentration later, however let's first look at thin curved beams.

A thin curved beam is typified by the quadrantal cantilever illustrated. Its radius is R and it is subjected to two components of end load, V & H. The loaded end's deflection is sought.

Castigliano's theorem may be applied directly, although the angle θ is usually a more convenient independent variable than the distance s when it comes to curved beams. Thus in the sketch, at the general free body's cut end defined by θ, the bending moment required for equilibrium is :
M = HR ( 1 - cosθ ) + VR sinθ.
The strain energy of direct tensile and shear forces is negligible in the presence of bending, so the forces need not be evaluated.
Clearly in ( ii)   s = Rθ   so that all terms are functions of θ, enabling the integral to be evaluated.

A proving ring of radius R and rectangular cross-section b*d is subjected to a force, F, as shown. The elastic modulus and design stress of the material are E and S respectively. Determine :
    - the ring deflection under the load,
    - ditto, transverse to the load, and
    - the maximum allowable load on the ring.

Consider only one quadrant since the system is symmetric - say the top left quadrant. This is modelled as a cantilever ( which automatically preserves the lower end vertical ) and loaded with -

• a vertical force V equalling half the total load F (the top RH quadrant supports the other half)
• a dummy load, H, which is necessary for determining the requested horizontal deflection, and
• an external moment, M_o. This is not a dummy load. If there were no moment at the end then the end would rotate as shown at (c). But the end remains horizontal due to symmetry of the complete ring, so there must be a moment M_o exerted by the RH mirror quadrant to ensure compatibility δ_Mo = 0 as shown at (d).
  M_o is in fact a bending moment in the complete ring, however it shows up as an (unknown) external moment on the FBD of the LH quadrant.

Inserting these into ( ii) with EI constant gives the various deflection components as :-
      δ_V   =   ¹/_EI ∫₀^L   M ^∂M/_∂V ds   =   ¹/_EI ∫₀^π/2( VRsinθ - M_o).Rsinθ.R dθ         = (^π/₄VR - M_o) R²/EI

      δ_H   =   ¹/_EI ∫₀^L   M ^∂M/_∂H ds   =   ¹/_EI ∫₀^π/2( VRsinθ - M_o).R(1-cosθ).R dθ = ( VR/2 - (^π/₂-1)M_o) R²/EI

      δ_Mo =   ¹/_EI ∫₀^L   M ^∂M/_∂Mods  =   ¹/_EI ∫₀^π/2( VRsinθ - M_o).(-1).R dθ             = (-VR + ^π/₂ M_o) R/EI

But there is no end rotation, so setting   δ_Mo = 0 in the last equation resolves the indeterminacy, giving   M_o = ²/_πVR.
Substituting this value of M_o into the first two equations gives :   δ_V = (^π/₄ - ²/_π) VR³/EI   and   δ_H = (²/_π -¹/₂) VR³/EI.

So, for the complete ring, since   V = F/2 then :
    diametral deflection, coaxial with load   = 2 δ_V =   (^π/₄ - ²/_π) FR³/EI
    diametral deflection, transverse to load = 2 δ_H =   (²/_π - ¹/₂ ) FR³/EI.

From the bending moment equation above, the maximum moment occurs at θ = 0 and is   M_max = M_o = ²/_πVR   from above.
Since bending is the only significant load building block, the maximum stress is
    σ   =   M_max y_max/I   =   (²/_πVR )( d/2 ) /( bd³/12 ).

The maximum allowable load on the ring occurs when σ reaches the design stress, S, and is therefore F_max = πbd²S / 6R.