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This section covers deflection coefficients including deflection due to shear and the use of the graphical methods

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The Method of Deflection Coefficients.

It can be seen that any beam of length l and flexural rigidity EI which carries a load W ( no mattter how it is distributed) will have a maximum deflection of $\displaystyle \frac{W\,l^3}{E\,I}$ where k is a constant which depends upon the type of loading and supports.

The value of k has been found for the standard cases of a cantilever and a simply supported beam ( See Deflection of Beams Part 1 Example 4 and Part 3 Example 1) and the deflection in other cases may frequently be built up by superposition.

\b NOTE The Principle of Superposition

Two types of problems will be solved by this method:-

Example 1

A beam of uniform section and length 2l is simply supported at its ends and by an elastic prop at the centre. If the prop deflects an amount $\displaystyle \alpha$ times the load it carries and the beam carries a total uniformly distributed load W show that the the load carried by the prop $\displaystyle \frac{5\,W}{8\left(1\;+\;+\;6\,E\,I\,\alpha /l^3 \right)}$

$If\;\;\;l\;=\;10\;ft.\;\;\;I\;=\;15\;in^4\;\;\;\;W\;=\;10,000\;lb.\;\;\;\;and\;\;\;\;\alpha \;=\;6.93\times10^{-4}\;in.lb.^{-1}$

(1)

Find the position and value of the maximum Bending Moment. $\displaystyle E\;=\;30\times10^6\;lb.\,in^{-2}$

If P is the load on the prop, then its deflection is $\displaystyle \alpha \;P$ carries a uniformly distributed load of 5 tons over its length of 10 ft. The beam is supported by three vertical steel tie rods each 6 ft. long, one at each end and one in the middle, the end rods having diameters of 1 in. and the centre rod 1.25 in.

Calculate the deflection at the centre of the beam below the end points and the stress in each tie rod. $\displaystyle E = 30\times 10^6 lb. in^{-2}$

Using equation (Part 3 no.19) the downwards deflection due to the load only is :-

$\left(\frac{5}{384} \right)\frac{W(2l)^3}{E\,I}$

(2)

Using equation (15) from Part 3. The upwards deflection due to the prop only is:-

$\frac{P\,(2l)^3}{48\;E\,I}$

(3)

By superposition, the net downwards deflection is given by:-

$\alpha \;P\;=\;\left(\frac{5}{384} \right)\frac{W(2l)^3}{E\,I}\;-\;\frac{P\,(2l)^3}{48\;E\,I}$

(4)

$\therefore\;\;\;\;\;\;P\left(\alpha \;+\;\frac{l^3}{6\;E\.I} \right)\;=\;\left(\frac{5}{48} \right)\left(\frac{W\;l^3}{E\,I} \right)$

(5)

Thus

$P = \frac{5\;W}{8(6\,E\,I,\times \alpha /l^3)}$

(6)

Substituting the numerical values given:-

$P = \frac{5\times10,000}{8[(6\times30\times16^6\times15\times6.93\times10^{-4})/120^3\;+\;1]}$

(7)

$\therefore\;\;\;\;\;P\;=\;3000\;lb.$

(8)

The reaction at the end supportsis given by:-

$R=\frac{1}{2}(10,000 - 3000) = 3,500\;lb.$

(9)

And for x < 10 ft.

$M=3500\;x - \left(\frac{10,000}{20} \right)\left(\frac{x^2}{2} \right)$

(10)

For a maximum $\displaystyle \frac{dM}{dx}\;=\;0$

$\therefore\;\;\;\;\;\;x = \frac{3500}{500}\;=\;7\,ft.$

(11)

And

$\hat{M} = 12,250\;lb.ft.$

(12)

Example 2

A horizontal steel beam, $\displaystyle I\;=\;207\;in^4$ carries a uniformly distributed load of 5 tons over its length of 10 ft. The beam is supported by three vertical steel tie rods ech 6 ft. long, one at each end and one in the middle, the end rods having diameters of 1 in. and the centre rod 1.25 in.

Calculate the deflection at the centre of the beam below the end points and the stress in each tie rod. $\displaystyle E\;=\;30\times10^6\;lb\,in^{-2}$ (U.L.)

Let P tons be the load at the centre of the rod. Then by superposition the following equation can be written:-

Stretch of centre rod - stretch of end rod = Deflection of beam due to load - deflection due to centre rod.

Substituting given values:-

$\frac{p\times72}{\left(\frac{\pi }{4} \right)\left(\frac{5}{4} \right)^2E}\;-\;\frac{(5\;-\;P)\times72}{2\left(\frac{\pi }{4} \right)E}\;=\;\frac{5\times5\times120^3}{384\,E\times207}\;-\;\frac{P\times120^3}{48\,E\times207}$

(13)

$\therefore\;\;\;\;\;5.87\,P\;-\;229\;+\;4.58\,P\;=\;543\;-\;174\,P$

(14)

$Thus\;\;\;\;\;\;P\;=\;\frac{772}{278.5}\;=\;2.77\,tons$

(15)

The stress in the centre rod $\displaystyle =\;\frac{2.77}{\left(\frac{\pi }{4} \right)\left(\frac{5}{4} \right)^2}\;=\;2.26\;tons\:in^{-2}$

Stress in an end rod $\displaystyle =\;\frac{(5\;-\;2.77)}{2\times\pi /4}\;=\;1.43\;tons\;in^{-2}$

Thedeflection of the centre of the rod relative to the ends is given by the difference of stretches of the tie rods. This equals:-

$\frac{2.26\times72\;-\;1.43\times72}{30\times10^6}\;\times2240\;\;\;\;\;\;\;\;NB \;\;\;(2240 lb.\;=\;1\;ton)$

(16)

$=\;0.0045\;in.$

(17)

Deflection due to Shear

It can be shown that the shear stress set up in the transverse section of a beam and the accompanying shear strain will cause a distortion of the cross-section and since the shear stress varies from zero at the extreme fibres to a maximum at the neutral axis, cross sections can no longer remain plane after bending.

In fact the "warping" will be of the form shown in the diagram. The left-hand view being for positive shear and the right-hand for negative shear. These strains are incompatible with the theory of pure bending but nevertheless a good approximation in deflection can be obtained by strain energy methods. It should also be noted that the shear distribution near to the application of a concentrated load must differ considerably from that given by the theory since there can be no sudden change of shear strain from one type to the other as would be implied for a simply supported beam with a central load.

Strain Energy due t0 Shear $\displaystyle = \left(\frac{s^2}{2C} \right)\times Volume$

And for the whole beam

$U_s = \left(\frac{1}{2}C \right)\int \int s^2\;Da.dx$

(18)

Where dA is an element of cross-section and dx an element of the length.

The integration can only be performed for a particular cross-section over which the variation of x is known and rectangular and I-sections will be calculated.

Rectangular Section

It can be shown that using equation (18)

$U_s=\frac{1}{2C}\int \left[\int_{-\frac{d}{2}}^{\frac{d}{2}}{\frac{36\,F^2}{b^2\,d^2}\left(\frac{d^4}{16} - \frac{d^2\;y^2}{2} + y^4 \right)b\;dy} \right]\;dx$

(19)

$=\frac{1}{2C}\;\int\;\frac{36\;f^2}{b\;d^6}\left[\frac{d^4\;y}{16} - \frac{d^2\,y^3}{6} + \frac{y^5}{5} \right]_{-\frac{d}{2}} ^{\frac{d}{2}}\;dx$

(20)

$=\frac{18}{C\,b\,d}\int F^2\times 2\left(\frac{1}{32} - \frac{1}{48} + \frac{1}{160} \right)\;dx$

(21)

$=\frac{3}{5}\,C\,b\,d\int_{0}^{1}{F^2\;.dx}$

(22)

For a Cantilever with a load of W at the free end W = F

Thus from equation (22)

$U_s\;=\;\frac{3\;W^2\;l}{5\;C\,b\,d}$

(23)

If $\displaystyle \delta _s$ is the deflection due to shear Then

$U_s\;=\;\frac{1}{2}\;W\;\delta _s$

(24)

For a Cantilever with a uniformly distributed Load.

A load $\displaystyle w\;\delta x$ acting on a length $\displaystyle \delta x$ (situated at a distance x from the fixed end) will produce a deflection due to shear oat this point of $\displaystyle \frac{\,w\,\delta x.x}{5\;C\,b\,d}$ . For this load alone the distortion produced is indicted in the diagram and is uniform shear force over the length x abd is zero over the rest of the beam (l - x). Hence the total deflection due to shear for all the distributed load is given by:-

$\int_{0}^{1}{\frac{6\,w\,x\,dx}{5\;C\,b\,d}}\;=\;\frac{3\;w\,l^2}{5\;C\,b\,d}$

(25)

For a Simply Supported Beam with Central Load W

$F\;=\;\pm \frac{W}{2}$

(26)

Thus by using equation (22) again

$U_s\;=\;\int_{0}^{1}{\frac{3(W^2/4)}{5\;C\,b\,d}.dx}$

(27)

$\;\frac{3W^2\;l}{20\;C\,b\,d}$

(28)

But since

$U_s\;=\;\frac{1}{2}\;W\;\delta _s$

(29)

By substitution in equation (28 )

$\delta _s\;=\;\frac{3\;W\;l}{10\;C\,b\,d}$

(30)

The simplified deflection is as shown in the upper diagram and since the shearing force is constant over each half, this case is equivalent to a cantilever of length l/2 carrying an end load of W/2

If the load is not centrally applied but divides the length into $\displaystyle l_1\;\;and\;\;l_2$ Then we can treat either end as a cantilever with an end load equal to the reaction on that side.

$Hence\;\;\;\;\;\;\;\delta _s\;=\;\frac{^\;(Wl_2/l_1)l_1}{5\;C\,b\,d}$

(31)

$=\;\frac{^\;W\;l_1\,l_2}{5\;C\,b\,d\,l}$

(32)

Blank

A Simply supported Beam with A Uniformly Distributed Load

Considering a load

$w\.\delta x$

(33)

only at a distance x from one end $\displaystyle \left(x\,<\,\frac{l}{2}\right)$ the deflection at the load will be:-

$\frac{6\,w\,\delta x\;(l\;-\;x)\,x}{5\;C\,b\,d\,l}$

(34)

Note this has already been proved in equation (30)

By proportion the deflection at the centre of the beam :-

$\frac{6\,w\,\delta x\;(l\;-\;x)\,x}{5\;C\,b\,d\,l}\left(\frac{l/2}{l\;-\;x} \right)$

(35)

Then the total central deflection due to shear is:-

$=\;\int_{o}^{\frac{l}{2}}{\frac{3\,w\,x\,dx}{5\;C\,b\,d}}}\;=\;\frac{3\;w\;l^2}{20\;C\;b\,d}$

(36)

I-section

The shear force is treated as being uniformly distributed over the web area. Thus $\displaystyle s\;=\;\frac{F}{b\,d}\;\;\;and\;\;\;\int dA\;=\;bd$ and using equation (18)

$\therefore\;\;\;\;\;\;U_s\;=\;\left(\frac{1}{2}C \right)\int \left(\frac{F^2}{b^2\;d^2} \right)b\;d.dx$

(37)

$=\;\frac{(\int F^2\;dx)}{2\;C\,b\,d}$

(38)

By similar methods to those used fro a rectangular section the deflections due to shear may be obtained as follows:-

Cantilever with end load $\displaystyle \delta _s\;=\;\frac{W\;l}{C\;b\,d}$

Cantilever with distributed load $\displaystyle \delta _s\;=\;\frac{W\;l}{2\;C\;b\,d}$

Simply supported beam with central load $\displaystyle \delta _s\;=\;\frac{W\;l}{4\;C\;b\,d}$
Simply supported beam with distributed load $\displaystyle \delta _s\;=\;\frac{W\;l}{8\;C\;b\,d}$

The Strain Energy method known as "Castigliano's Theorem" ( See Bending of Curved Bars) may be used where a number of loads exist concurrently or to find the distributed load by imposing a concentrated load at a deflection point and latter giving it a value of zero. i.e. $\displaystyle \delta _s\;=\;\left( \frac{\partial U_s}{\partial P}\ \right)_{p\;=\;0}$

Example 3

For a given cantilever of rectangular cross-section, length l and depth d show that if $\displaystyle \delta \;\;and\;\;\delta _s$ are the deflections due to shear and Bending due to a concentrated load at the free end, $\displaystyle \frac{\delta }{\delta _s}\;=\;k\,.\,\left(\frac{d}{l} \right)^2$ and find the value for k for steel. $\displaystyle E\;=\;30\times10^6\;lb.in^{-2}\;\;;\;\;C\;=\;11.5\times10^6\;lb.in.^{-2}$ .

Hence find the least value of l/d if the deflection due to shear is not to exceed 1% of the total. (U.L.)

It has been shown that:-

$\delta _s\;=\;\frac{6\;W\;l}{5\;C\;b\,d}$

(39)

$\delta _b\;=\;\frac{W\;l^3}{3\;E\,I}\;=\;\frac{\;W\;l^3}{E\;b\;d^3}$

(40)

For a rectangular Section:-

$\therefore\;\;\;\;\;\frac{\delta _s}{\delta _b}\;=\;\frac{6}{5\times4}\times\frac{E}{C}\times\left(\frac{d}{l} \right)^2\;=\;k\left(\frac{d}{l} \right)^2$

(41)

$\therefore\;\;\;\;\;k\;=\;\left(\frac{3}{10} \right)\left(\frac{E}{c} \right)\;=\;\left(\frac{3}{10} \right)\left(\frac{30}{11.5} \right)\;=\;0.783$

(42)

$If\;\;\;\;\frac{\delta _s}{\delta _p\;+\;\delta _s}\;=\;0.01$

(43)

Then

$\frac{\delta_s}{\delta _P}\;=\;\frac{0.01}{0.99}\;=\;0.783\;\left(\frac{d}{l} \right)^2$

(44)

$\therefore\;\;\;\;\;The\;least\;value\;of\;\frac{l}{d}\;=\;\sqrt{0.783\times99}\;=\;8.8$

(45)

Example 4

A 10 in. by 6 in. R.S.J. with web 0.4 in. , flanges 0.7 in/ thick acts as a horizontal cantilever 12 ft. long and carries a load of 2 tons at 6 ft. from the end and assuming that the shear force is carried by the web and is uniformly distributed, calculate the deflection at the end.

$E\;=\;12,800\;tons\;in^{-2}\;\;and\;\;C\;=\;5000\;tons\;in^{-2}$

(46)

(U.L.)

$I\;=\;\frac{6\times10^3\;-\;5.6\times8.63^3}{12}\;=\;204\;in.$

(47)

Using the Moment of Area method ( See Deflection of Beams Part 3) the end deflection due to bending is given by:-

$\left(\frac{1}{2}\times12\times6 \right)\frac{10\times12^3}{12,800\times204}\;=\;0.238\;in.$

(48)

The Deflection at the load due to Shear is:-

$\frac{W\;l}{C\;b\,d}\;=\;\frac{2\times6\times12}{5000\times0.4\times8.6}\;=\;0.0084\;in.$

(49)

Since the shear force is zero beyond the load there is no further deflection due to shear between the load and the end of the beam.

Thus the deflection at the end of the beam is the result of both bending and shear and is:-

$0.238\;+\;0.0084\;=\;0.2464\;in.$

(50)

Deflection by Graphical Method

It was shown in the pages on "Shearing force and Bending Moment" that a Funicular Polygon could be used to perform a double integration of the load curve and this would produce the Bending Moment diagram. Since $\displaystyle \frac{d^2y}{dx^2}\;=\;\frac{M}{E\,I}$ it follows that a double integration of the Bending Moment curve will produce the Deflection Curve

If EI is constant, draw the B.M. diagram and divide it into a number of strips of width $\displaystyle \delta x$ . Now draw a vertical line to represent the areas $\displaystyle M\delta x$ and join this to a pole O on the right of the line. Proceed in the normal way to draw the funicular polygon which will be a series of straight lines to be smoothed out into a curve.The vertical ordinates on this diagram represent deflection and it will usually be necessary to slew the diagram through an angle in order to produce a horizontal base ( eg fro a simply supported beam)

If the scales are $\displaystyle 1\;in.\;=\;s_1\;lb.in^2\;\;\;"M\;\delta x"\;units,\;1\,in.\;=\;s_2\;in.\;length,\;and 1\;in.\;=\;s_3\;"EIy"\;units$ then the distance h is given by $\displaystyle \frac{s_3}{s_1\;s_2}\;in.$

If then the Deflection scale required is $\displaystyle 1\;in.\;=\;s_4\,in.\;=\;\frac{s_3}{EI}\;\;h\;=\;\frac{EI\;s_4}{s_1\;s_2}$