Bending of thin flexible fibre reinforced sheets

Bending of thin flexible fibre reinforced sheets

The International Journal of Cement Composites and Lightweight Concrete, Volume 9, Number4 November 1987 Bending of thin flexible fibre r e i n f o ...

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The International Journal of Cement Composites and Lightweight Concrete, Volume 9, Number4

November 1987

Bending of thin flexible fibre r e i n f o r c e d sheets S. B a r - S h l o m o *

Synopsis It has been found that the bending moment to which a thin fibre reinforced cement slab is exposed may be calculated in most cases by applying the usual bending formulae, based on the assumption that small deflections are obtained during their testing. Some slabs, however, may exhibit in certain conditions, so large deflections just before failure, that the above mentioned assumption and the resulting simple bending formulae are no more applicable. A simple dimensionless correction factor is derived from the theory of bending with large deflections, which, when used for multiplying the maximum bending moment, calculated by the simple small deflection formula, results in values equal to those resulting from the exact method of calculation, in which the actual magnitude of the beam deflection is taken into consideration. In addition, simple formulae are derived for the calculation of the slope at the end of a simply supported beam loaded at its mid-span and of the secant modulus of elasticity of such a beam, both taking into account the non-linear theory of large deflections. Keywords Fibre cement composites, thin sheets, flexural strength tests, structural analysis, large deflection theory, bending moments, error analysis, deflection, modulus of elasticity, non-linear theory, testing, strength of materials

INTRODUCTION The bending strength is often an important quality parameter of relatively thin fibre reinforced cement slabs such as flat or profiled sheets for cladding roofs, facades or ceilings. Very often this bending strength is used as one of the acceptance criteria specified by the relevant product standards. The bending tests are usually carried out by using the slab, or part of it, as a beam resting on two simple supports, fixed at a constant distance, with the load applied at its mid-span. In the calculation of the bending moment and the resulting stress, usually the simple theory of bending is applied, in which the angle ¢o between the tangent to the bent beam and its initial unbent line is assumed to be negligibly small, even at its two simple supports (Figure 1)t. As a result of such an assumption the maximum bending moment in this loading system is given for the point m at the centre of the beam by the equation: Mm or Mm



= FL/4



1- Note that in Figures 1 and 2 the applied toad is assumed to act upwards and the vertical reactions downwards. * Consultant, Fibre Cement Industry, Israel. Received 29 May 1987

Accepted 21 August 1987

© Longman Group UK Ltd 1987 0262-5075/87/09407243/$02.00

However, if the beam resting on two simple supports fixed at a constant distance L = 21 will not remain negligibly curved and the concentrated mid-span load F -- 2P will move upwards byym, the diagram shown in Figure 2 will be obtained. The aim of this paper is to show that even when the Ym deflection in Figure 2 is quite considerable, one still can obtain fairly exact bending moments by applying the simple small deflection equation (1), provided the latter is multiplied by an easily obtainable dimensionless correction factor.

BEAM BENT TO A LARGE DEFLECTION From Figure 2 and by disregarding possible friction between the beam and the supports, one can write for the bending moment at A (x, y): MA = Pa + H h where and

a= 1- x h=ym-y H = Ptan¢o ¢o is the angle between the tangent to the bent beam at each of its supports and the initial straight line of the beam.

Hence at A (and without friction): MA = P (1 - x) + (P (Ym - Y) tan¢o


The maximum bending moment will obviously occur


Bend/ng of thin f/exibl'e fibre reinforced sheets



I i

Figure 1 Bending of a beam with small deflections (Cbo~ O)


mt L= 21




Figure 2 Bending of a beam with large deflections

F= 2P


L= 21. Fig. 2

at the centre of the beam in Figure 2, i.e. at x = y -- 0. Therefore: Mmax = P (1 + ymtan(bo)


Dividing the latter expression by equation (1) one obtains a dimensionless ratio: I~ = 1 + (ym/l)tan(bo where

Mmax = K Mm

(5) (6)

One can see that the actual large deflection bending moment Mmax might be obtained from the simple equation (1), provided the dimensionless factor K from equation (5) were known. In equation (5), the values of the span L = 21 and the mid-span deflection Ym at any applied load F = 2P are easily measurable during the bending tests, and therefore: p. = 1 + 2 (ym/L) tan(bo


and the.only missing parameter to be established is the unknown angle (bo.

CALCULATION OF THE ANGLE (bo The differential equation of the bent beam is given by the following known expression: d ¢/ds = M/El where


s is the length measured along the bent beam, E its modulus of elasticity, and I its second moment of area.

By differentiation of the latter equation in respect to s and considering equation (3), one obtains: d2(b/ds 2 = - k 2 where


k 2 = P/El.

[dx/ds - (dy/ds)tan(bo]

From Figure 2: dx/ds = cos(b and dy/ds = sin(b, therefore: d2(b/ds 2 = - k 2 (cos(b + tan(bo sin(b) and by integrating once the latter equation: (d(b/ds)2/2 = - k 2 (sin(b - tan(bo cos(b) + C


where C = 0 because M = 0 at 4) = 4)0. Introducing now ds = dy/sin(b, one obtains eventually the following differential equation of the deflection of the beam: dy = [sin (b/k V 2 (tan(bo cos(b - sin(b) ]d (b


By changing the variable from 4) to ¢, where the latter is defined as follows: cos ~b = V'sin (4)0 - (b) , so that cos ¢o = sV~-E(bo sv

(11 )

it can be shown [1 ] that the y/I ratio can be expressed in terms of elliptic integrals as follows: y _ [ 1/2-sin (boCOS ¢ - cos(bo(b (p, ¢) ] I

[ v ~ - c o s (boCOS¢o + sin(bo(b (p, ¢o) }


Since y = Ymax at x = 0, i.e. at 4) = 0, hence ¢ = q~o and therefore, considering equation (11), one can eventually write for the maximum y/I ratio: y

[ 2~/~]-E(bo - (b (p, ¢o)/tan(bo ]

Tm = [ ~,/2sin(bo /tan(bo + 4) (p, ¢o) ]


Bending of thin flexible fibre rein forced sheets



Ym = Ymax

p = sin (~/4)


¢o = arc tan M(t - sin(bo)/sin(bo "~/2



(b(P, ¢o) = 2 f (1 - p2sin2 t))1/2d 8 - f ¢o

and by introducing the latter expression back into equation (8): M = M2PEI (tan(bo cos(b - sin(b) Since M = Mmax at the centre of the beam span, i.e. at an angle d) = O, therefore: Mmax =


(1 - p2sin28) -1/2d e


and let the two ratios on both sides of equation (13) be defined as RI corresponding to that on the left hand side, and as Rr corresponding to that on the right hand side of the said equation. The left hand side ratio RI -- ym/I is a value known from direct measurements during the testing whereas the right hand side ratio Rr includes two incomplete elliptic integrals of the first and second kinds, given by equation (13c), with their parameters defined by equations (13a)and (13b). It can be seen that the Rr ratio is a function of only the angle 4)0 at the fixed supports of the beam in Figure 2, and its numerical value can therefore be calculated for any given ratio ym/I by simply requiring that Rr = RI. A SIMPLIFIED METHOD FOR THE CALCULATION OF ANGLE 4)o Unfortunately, equation (13c), and therefore also the ratio Rr, has no analytical solution; it can be solved either by trial and error, using tables of elliptical integrals, or numerically, iterating to any desired degree of exactness. Both methods are rather cumbersome and hardly applicable to the calculation of bending moments from results of routine testing of flexible slabs. The author has solved equation (13) and its elliptical integrals by applying Simpson's modified methdd of numerical integration [2] to about 100 values of ym/L ratios, ranging between 0.001 to 0.175, and the corresponding values of the angles 4)0 were obtained by comparing RI to Rr so that the condition of IRI - Rr I <10 -6 would always be fulfilled. The obtained angles 4)0 were then statistically correlated with the corresponding values of the ym/L ratios in the aforementioned range and a polynomial correlation of the third degree between the two was established. The said correlation exhibits a very high coefficient of determination and an extremely low standard error of estimate (r2 = 0.999999 and SEE <0.001 degree). This polynomial equation, in its simplified form, but still very exact when the obtained values of 4)o in degrees are used in equation (1), is as follows:

d)o = 172 (ym/L) - 5 (ym/L) 2 - 184 (ym/L) 3


Equations (14), (7), (6) and (1) enable now a simple calculation of the bending moment of a slab in which the magnitude of the deflection is not disregaraed. CALCULATION OF THE SECANT E-MODULUS AT BENDING WITH LARGE DEFLECTIONS From equation (9) and considering the definition of k 2 one obtains:

d (bids = X/2P (tan(bo cos(b - sin(b)/EI

1/2PEI tan¢o


From equations (4) and (15) one obtains for the angle (bo: tan(bo = [El - Plym - MEt (El - 2Plym)]/py2


Note that the second value of tan(bo might be obtained by putting the sign + in front of the square root. It can be shown, however, that for the larger value of the angle, the beam becomes unstable, i.e. it may slip through between the supports due to the load 2P. Therefore, from equations (4) and (16): Mmax = (EI/ym) -

1/(EI/ym)[(EI/ym) - 2PI]


and from the definition of the factor i~ in equation (6) one obtains from equations (17)and (1): i~=mm(m-2) (18) where m = El/Plym Now, by combining equations (18) with (5) and expressing half the load P and half the bending span l in terms of their full values (P = F/2 and I = W2), one obtains for the secant modulus of elasticity E of the beam bent to a maximum deflection Ym: E = FL 2 [2(ym/L) + 4(ym/L) 2 tan(bo + (1/tan(bo)]161 (19) Since in the latter equation all necessary parameters are measured in every bending test and the angle 4)0 is given by equation (14) for any measured ym/L ratio, the secant modulus of elasticity can easily be established for both small and large beam deflections. (Note that in fact the angle 4)0 can never be equal zero.) DISCUSSION A N D S U M M A R Y The few simple equations which were derived above enable the calculation of the bending moment and of the secant modulus of elasticity of a bent beam using the easily measurable span between the supports and the maximum deflection of the beam at the concentrated load, be it a small or a large one. The angle (bo of the slope at the fixed supports is obtained from equation (14) and the factor i.t from (7) turns the maximum bending moment obtained from the simple small deflection formula into the 'true' one, corresponding to the actually measured deflection (equation (13)) and the secant modulus of elasticity E, corresponding to any deflection span and concentrated load at its mid-span, is given by equation (19). The degree of increase of the bending moment (and to a large extent also of the resulting bending stresses, both tensile and compressive) due to the fact that the size of the actual deflections is taken into consideration can be evaluated from equation (6) as follows:

A = 100 (Minas- Mm)/Mm -- lOOMm (IJ.- 1)/Mm = 100(11- 1), where A is expressed in per cent.


Bending of thin flexible fibre reinforced sheets








/ t

0.1 005 0.04 0.03







i I




0.02 i i


0.01 Fig 3






Figure 3 The difference in % between the true bending moment and the one calculated by assuming small beam deflection versus deflection to span ratio

Figure 3 is obtained when z& in per cent is drawn against the relative beam deflection ym/L. One can see from that figure that the bending moment, calculated according to the theory of large deflections, will exceed by about 1 per cent the moment obtained from the usual small deflection assumptions only when the relative maximum deflection ym/L will be slightly bigger than

0.04. If the bending span in standard tests is L = 150mm (often used for testing slates), the deflection which would result in such a difference of 1% is about Ym = 6mm; and if the span is L = 1150mm (usual for corrugated sheets) then Ym = 4 6 r a m In practical terms, therefore, and in most cases of bending flexible sheets, the usual formulae based on the small deflection theory will provide sufficiently precise results. There might, however, be cases when the bending test results of thin fibre cement sheets must be treated by applying the large deflection theory, i.e. the above mentioned few simple formulae. As an illustration of such a case consider, for instance, the following actual bending test results obtained on some 80 x 200ram fibre reinforced cement slabs, bent on a span of 150ram. These specimens were flat slabs, referred to as product No. 8391, tested recently, and marked as '0-specimens' [3]. Readings of deflections and load were recorded at the limit of proportionality (LOP) and at failure. The results are given in the following Table 1. (The weight of the tested thin slabs is not considered in the calculations). In the example shown in the table above one can see that whereas at the LOP the bending moments, corrected according to the large deflection theory, are by only about 0.001 to 0.007 per cent bigger than those corresponding to small deflections, those at failure are by 8.65% to 15.5% higher. If, for example, the modulus of rupture (MOR) of specimen No. 1, calculated according to the small deflection theory is about 19.68 N/mm 2, in actual fact it is 21.73N/mm 2 if the large ultimate deflection of 19.5mm at a bending span of 150mm is not disregarded. This is a non-negligible difference of more than 10%. In specimen No. 4 the said difference is about 15.5%. Similar results are obtained [3] from bending tests carried out on corrugated fibre reinforced sheets, except that in their case the differences obtained from applying the small or large deflection theory are rather small. It can be shown, for instance, that the ultimate bending moment corresponding to the large deflection theory will be by only 0.95% bigger than that obtained from the small deflection theory, even when the ultimate deflection at bending on a 1150mm long span was as large as 45.6mm. From the test results listed in Table 1 one may

Table 1 Bending test results - large deflection theory

Specimen No. 1

2 3 4



At failure

Thickness t mm

Ym mm


¢o degrees

& %

Ym mm


Oo degrees

& %

4.9 5.0 5.0 5.1

0.49 0.41 0.50 0.19

60.48 50.96 53.77 25.11

0.56 0.47 0.57 0.22

0.006 0.004 0.007 0.001

195 17.8 19.7 23.6

168 171 167 135

21.87 20.03 22.09 26.22

10.44 8.65 10.66 15.50

Bending of thin flexible fibre reinforced sheets


Table 2 Secant moduli of elasticity calculated by equation (19) At LOP Specimen No.

I mm 4

Ym mm

F kN

¢o degrees

E kN/mm 2

1 2 3 4

784.33 833.33 833.33 884.34

0.49 0.41 0.50 0.19

0.06048 0.05096 0.05377 0.02511

0.56 0.47 0.57 0.22

11.1 10.5 9.1 10.5

* Length of span L = 150mm, width of slabs B = 80mm

calculate the secant E-moduli of the slabs by applying equation (19). The results of such a calculation are given in Table 2. It is worth noting that the values of the secant E-moduli arrived at in Table 2 are almost identical with those indicated by the computer attached to the testing machine, and which were calculated from the test results by applying an entirely different method.

REFERENCES 1. Frisch-Fay, R. 'Flexible Bars', Butterworths, London, 1962. 2. Heilborn, J. 'Science and Engineering Programs', Apple II Edition. 3. Ametex Ltd. Private communication Ametex Laboratory Testing Report, Niederurnen, Switzerland, 1987.