09thick Beams

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Thick curved beams

The cross-sectional depth of a thick curved

beam is comparable to its radius of

curvature and this gives rise to stressconcentration. It is required to ascertain

the variation of stress across the cross-

section. Exact analysis is complex but the

simple approximation which follows -

based on plane cross-sections remaining

plane under load - is adequate for most

practical beams typified by the curved portion of the G-clamp illustrated.

The generalised sector of a thick curved beam with centre of curvature at O and small

angular extent appears below. The radii of interest are identified in the LH sketch.Two radii are particularly significant - the mean (or centroidal) radius rc, and the mean

reciprocal radius rd. They are defined by :

( 1) A * rc = ri

rodA * r ; A / rd = ri

rodA / r

Like riand ro, both rcand rdare intrinsic to the known geometry ( b = f(r) ) of the

cross-section.

Since the beam is curved, the radius defining the unstressed neutral axis, rn, 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.

These positive conventions must be adhered to for the theory developed

Page 1 of 5DANotes: Misc strength topics: Thick curved beams

8/22/2008http://www.mech.uwa.edu.au/DANotes/MST/thick/thick.html


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below to be applicable.

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 'under load in the RH sketch. Assuming that plane cross-sections remain planeduring loading, the increase in length of the typical fibre at radius r is proportional to

the distance from the unstrained neutral axis, ( r-rn). Since the original length of the

fibre is proportional to the radius r, then :fibre strain = extension/free length ( r - rn )/r = 1 - rn /r

Presuming elastic behaviour, the stress in the fibre at radius r is proportional to this

strain, so

( 2) 1 - rn /r = m( 1 - rn /r ) where mand rnare constants, as yet

unknown.

This establishes theformof 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 dA = m

A( 1 - rn /r ) dA = mA ( 1 - rn /rd )

M = A r dA = mA( 1 - rn /r ) r dA = mA ( rc- rn )

Solving these for the two unknown constants :-

( 2) rn = rd( M - F rc ) / ( M - F rd ) ; m = ( M - F rd ) / A ( rc- rd )

It is noticed these two constants - and hence the stress distribution (2) - depend onlyupon the loading ( F & M) and the cross-sectional geometry ( A, rc& rd ).

Two particular locations of the neutral axis occur when

M = 0 whereupon rn= rc, and

F = 0 then rn= rd- the pure bending case most commonly encountered in the

literature.

Summarising - application of the foregoing theory involves the following steps : compute the salient radii rcand rdfrom ( 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).




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The conspicuous radii, rc& rd, for common symmetric cross-sections are found from

( 1) as follows.

Trapezoid- defined by dimensions ri , ro , bi and boas sketched.

The variables r and dA must be expressed as functions of the

same variable before the integrals in ( 1) can be evaluated. Theradius 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 = + r where the constants and are evaluated at ri and ro :

= ( biro- bori ) / ( ro- ri ) ; = ( bo- bi ) / ( ro- ri )

dA = b dr = ( + r ) dr and therefore, from (1) :-

A = AdA = riro( + r ) dr = 1/2 ( ro- ri ) ( bi- bo )

A*rc = AdA*r = ri

ro( + r ) r dr = 1/6 ( ro- ri ) [ bo( 2ro+ ri ) + bi( 2ri+ ro ) ]

A/rd = AdA/r = riro( /r + ) dr = bo- bi+ [ ( biro- bori )/( ro- ri ) ] . ln ( ro /ri )

Rectangle- particularising the trapezoidal results with bo= bi= b ( constant ) :-

A = b ( ro- ri ) ; A*rc =1/2b ( ro

2- ri2 ) ; A/rd = b ln ( ro /ri )

Circle- integrating as above :- 2 rc = ro+ ri ; 2rd = ro+

ri

Built-up sectionsmay 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 = j=1( A )j ; A*rc= j=1( A*r )j ; A/rd=

j=1( A/r )j

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

A = bi ( rm- ri ) + bo ( ro- rm ) using ( 1a)

A*rc =1/2[ bi ( rm

2- ri2 ) + bo ( ro

2- rm2 ) ] whence rcfor the section . . .

A/rd = bi ln ( rm /ri ) + bo ln ( ro /rm ) . . . and rd

element widthinner

radius

outer

radiusA A*rc A/rd

1 bi ri r bi ( rm- ri ) 1/2 bi ( r

2

- ri2

) bi ln ( rm/ri )2 bo r ro bo ( ro- rm ) 1/2 bo ( ro

2- rm2 ) bo ln ( ro/rm )

total j=1( A )j j=1( A*r )j j=1( A/r )j




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EXAMPLE

The G-clamp illustrated above is made from a steel whose design stress is 50 MPa.

It is supposed to sustain a force, P, of 750 N. Can it ?

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 ri= 20, rm= 30, ro= 50, bo= 15, bo= 5

mm :A = 15 ( 30 -20) +5 ( 50 -30) = 250 mm2 watch units!

A*rc = [ 15 ( 302 -202 ) +5 ( 502 -302 ) ]/2 = 7750 mm3 whence rc = 31 mm

A/rd = 15 ln 30/20 + 5 ln 50/30 = 8.636 mm whence rd = 28.95

mm

The stress resultants, F and M, are shown on the free body in the

conventional positive sensesused in developing the theory above.

For equilibrium of the free body, F = P and M = -Pa. Inserting these

into ( 3) gives :-

rn = rd( a + rc )/( a + rd ) = 28.95 ( 40 + 31 )/( 40 + 28.95 ) =

29.81 mmm= -P ( a + rd )/A ( rc- rd ) = -750 ( 40+28.95)/250( 31-28.95) = -101 MPa

The stress distribution is thus = -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 = d+ 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




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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 ?

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Copyright 1999-2005 Douglas Wright, [email protected]

last updated December 2005


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09thick Beams

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