By J Prescott

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**Example text**

Any cross section where the inertia of the The line in P^ stress is zero is called Thus axis of the section the the neutral axis of the section in dA Let neutral the ,^-axis is 15. fii; denote the element of area of the cross section at distance z from the axis of 1/, PQ the strip in fig. 15. the total tension across this section Then Fig. 15 is T =fP^dA=fEnzdA ^EafzdA . 61) But, iby the equation for the position of the centre of inertia of an area, JzdA^zA, and in our case^ = o by our hypothesis Consequently T= that the ;r-axis passes through the centre of inertia.

Relations between the elastic constants, •'- E 2n=- n m= I a = — 10 7) m—n 2in E = 7i{^m — n) We shall now turn to some of the simplest solutions of the equations of equilibrium and find the stresses corresponding to them, as well as the body forces X, Y, Z. 32. Homogeneous Assume that the body strain. ax, v--by, w^=c2, where a, b, c, are constants. (3-8) SOME PARTICULAR SOLUTIONS OF THE EQUATIONS dv ^«* r«, dw ^ , Pj ^w 5w 5«; dv dw dw dy dz dx dz By dx = (w — n) {a h -\- c) -\- -\- 2na — n){h — n)(c + a) Pz = {m-\-n)c-^{m — n){a-\-h) =^{m-\-n)a-\- {m 'P 2= {m -\- n)h -\- {m S, • • ^ <3-9) '' Therefore Thus ^ ^' ^ = + * + '''a^ = "'a^ = *'^ = ^^^" 29 -\- c) .

24. that, in although there is a sudden change in the shearing force of amount equal to that load, there passing a load, no sudden change in the bending moment. Thus the bending moment has the same value immediately on is opposite sides of a load. It is the of the bending moment dia- slope gram that 50. makes a sudden change. Beams under Fig. 24 distributed loads. w Let the load per unit length on a beam be at distance some origin taken on the line of centres of the beam. Then the load on dx, and w may or a function of x.