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Shear on Beam Sections.-In the Fig. 371 the rectangular element abcd on the unstrained beam becomes a'b'c'd' when the beam is bent, and the element has undergone a shear. The total shear force on any vertical section W, and, assuming for the present that the shear stress is evenly distributed over the whole section, the mean shear W

stress

=

1=3 where A = the area
A'

of the section; or we may write it

W

B. H

But we have shown (p.

313) that the shear stress along

FIG. 371.

any two parallel sides of a rectangular element is equal to

the shear stress along the other two

parallel sides, hence the shear stress

along cd is also equal to

W

A

The shear on vertical planes tends to make the various parts of the beam slide downwards as shown in Fig. 372, a, but the shear on the horizontal planes tends to make the parts of the beam slide as in Fig. 372, b. This action may be illustrated by bending some thin strips of wood, when it will be found that they slide over one another in the manner shown. often fail in this manner when tested.

In the paragraph above we assumed that the shear stress was evenly distributed over the section; this, however, is far from being the case, for the shearing force at any part of a beam section is the algebraic sum of the shearing forces acting on either side of that part of the section (see p. 397). We will now work out one or

NA

У

FIG. 372.

a

Solid timber beams

B

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two cases to show the distribution of the shear on a beam

section by a graphical method, and afterwards find a mathematical expression for the same.

In Fig. 373 the distribution of stress is shown by the width of the modulus figure. Divide the figure up as shown into strips, and construct a figure at the side on the base-line aa, the ordinates of which represent the shear at that part of the section, i.c. the sum of the forces acting to either side of it, thus

The shear at I is zero

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2 is proportional to the area of the strip between

1 and 2 = cd on a given scale.

3 is proportional to the area of the strip between 1 and 3 = ef on a given scale.

4 is proportional to the area of the strip between 1 and 4 = gh on a given scale.

5 is proportional to the area of the strip between I and 5 ij on a given scale.

=

6 is proportional to the area of the strip between I and 6 = kl on a given scale.

Let the width of the modulus figure at any point distant y from the neutral axis = b; then

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=

Thus the shear curve is a parabola, as the ordinates ƒƒ1, etc., vary as y2; hence the maximum ordinate kl (mean ordinate) (see p. 30), or the maximum shear on the section is of the mean shear.

In Figs. 374, 375 similar curves are constructed for a circular and for an I section.

It will be observed that in the I section nearly all the shear is taken by the web; hence it is usual, in designing plate girders of this section, to assume that the whole of the shear is taken by the web. The outer line in Fig. 375 shows the total shear and the inner figure the intensity of shear at the different layers. The shear at any section (Fig. 376) ab , and the intensity of

=

W

2

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W

area of the web, or the intensity of shear =

But the

2ht

intensity of shear stress on aa, the intensity of shear stress

FIG. 374

Mean

heights

FIG. 375.

on ab, hence the intensity of the shear stress between the web

and flange is also

=

W 2ht

We shall make use of this when

working out the requisite spacing for the rivets in the angles between the flanges and the web of a plate girder.

In all the above cases it should be noticed that the shear stress is a maximum at the neutral plane, and the total shear

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there is equal to the total direct tension or compression acting above or below it.

We will now get out an expression for the shear at any part of a beam section.

We have shown a circular section, but the argument will be seen to apply equally to any section.

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breadth of the section at a distance y from the neutral axis ;

stress on the skin of the beam distant Y from the neutral axis ;

stress at the plane b distant y from the neutral axis; bending moment on the section cd;

moment of inertia of the section;

shear force on section cd.

The area of the strip distant y

from the neutral axis

= b.dy

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Substituting the value of ƒ in the above, we have

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M

the total force acting on the strip=b.y.dy

But M = S/ (see p. 399)

Substituting in the equation above, we have

I

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Dividing by the area of the plane, viz. b. 7, we get—

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the maximum value at the neutral axis,} =
when b = B, and y = o

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the mean intensity of shear stress =

where A = the area of the section.

The ratio of the maximum
intensity to the mean

S

A

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K =

IB

S

Y

b.y.dy

The value of K is easily obtained by this expression for geometrical figures, but for such sections as tram-rails a graphic solution must be resorted to.

The value of

O

b.y.dy is the sum of the

dy

FIG. 377.

moments of all the small areas b.dy about the N.A., between the limits of y = o, i.e. starting from the N.A., and y = Y, which is the moment of the shaded area A, about the N.A., viz. A,Ye, where Y is the distance of the centre of gravity of the shaded area from the N.A.

C

Since the neutral axis passes through the centre of gravity of the section, the above quantity will be the same whether the moments be taken above or below the N.A.

Deflection due to Shear.-The shearing action of a load on a beam causes the deflection to be greater than it otherwise would be. In the figure the beam is supposed to be subject

to shear only, which deflects it an amount = x.

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W

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f.

(approx. in this case)

FIG. 378.

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where K is the ratio of the maximum to the

A '

mean shear stress (see last paragraph); A = sectional area of beam. Hence

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