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The bending of Built in Beams, which are fixed at both ends.

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Moment-Area method for Built in Beams.

A Beam is said to be Built-in or 'encastre' when both ends are rigidly fixed so that the slope remains horizontal - It is normal for both ends to be at the same level. It follows from the Moment-Area method (See Bending of Beams Part 3) that since the change of slope from end to end and the intercept z are both zero then

$\sum A=0}$

(1)

and

$\sum{A \bar{x}}=0$

(2)

It is easier to consider the Bending Moment diagram for the Built in Beam in the diagram, as the algebraic sum of two parts. The first considers the beam to be Simply Supported (b) and the second due to the end moments which must be introduced to bring the slopes back to zero (c). The area and end reactions obtained, if freely supported, will be referred to as The Free Moment Diagram and the Free reactions $A_1$ , $R_1$ and $R_2$ .

The Fixing Moments at the ends are $\displaystyle M_a\;\;\;and\;\;\;M_b$ and in order to maintain equilibrium when $\displaystyle M_a\;\;\;and\;\;\;M_b$ are unequal the reactions $\displaystyle R\;=\;\frac{(M_a\;-\;M_b)}{l}$ are introduced. These are upwards on the left-hand end and downwards on the right. Due to $\displaystyle M_a\;\;\;\;M_b\;\;\;\;and\;\;\;\;R$ the Bending Moment at a distance x from the left-hand end is given by:-

$M_x\;=\;-\;M_a\;+\;R\;x\;=\;-\;M_a\;+\;\left\{\frac{M_a\;-\;M_b}{l} \right\}\;x$

(3)

This gives a straight line going from a value of $\displaystyle -\;M_a\;\;\;at\;\;\;x\;=\;0\;\;\;\;\;to\;\;\;\;\;-\;M_b\;\;\;at\;\;\;x\;=\;l$ From this can be drawn the Fixing |Moment Diagram $\displaystyle A_2$ (See d)

For the downwards loads $\displaystyle A_1$ is a positive area (Sagging BM) and $\displaystyle A_2$ is a negative area (Hogging BM). Consequently equations (1) and (2) reduce to:-

$A_1\;=\;A_2$

(4)

$And\;\;\;\;\;\;\;A_1\;\bar{a}_1\;=\;A_2 \;\bar{a}_2\;\;\;\;\;(Numerically)$

(5)

These give rise to two important statements:-

Area of the Free Moment Diagram = Area of the Fixing Moment Diagram

The Moments of Areas of Free and Fixing Diagrams are equal

It may be necessary to break down the areas still further to obtain convenient triangles and parabolas.

These two equations allow $\displaystyle M_a\;\;\;and\;\;\;M_b$ to be found and the total reactions at the ends are :-

$R_a\;=\;R_1\;+\;R\;=\;R_1\;+\;\frac{M_a\;-\;M_b}{l}$

(6)

$And\;\;\;\;\;R_b\;=\;R_2\;-\;R\;=\;R_2\;-\;\frac{M_a\;-\;M_b}{l}$

(7)

The Combines Bending Moment Diagram (e) is the Algebraic sum of the two components.

Five Worked Examples

Example 1

Obtain expressions for the Maximum Bending Moment and deflection of a beam of length l and a flexural rigidity EI. The Beam is fixed horizontally at both ends ( built in) and carries a load W which is (a) Concentrated at mid-span and (b) Uniformly distributed over the whole beam.

(a) Concentrated Load

The Fixing Moment at both ends are equal and the area of the diagram is therefore $M\;l = A_2$

The free Moment diagram is a triangle with a maximum ordinate of $\displaystyle \frac{W\;l}{4}$

$\displaystyle \therefore\;\;\;\;\;\;\;A_1\;=\;\frac{1}{2}\;\left( \frac{W\;l}{4}\right)\;l\;=\;\frac{W\;l^2}{8}$

From Equation (1) $\displaystyle A_1\;=\;A_2$

$\therefore\;\;\;\;\;\;M\;=\;\frac{W\;l}{8}$

(8)

Thus the combined Bending Moment diagram is as shown in the lower diagram. The Maximum Bending Moment is $\displaystyle \therefore\;\;\;\;\;\;M\;=\;\frac{W\;l}{8}$ and occurs at the ends (Hogging) and the Centre (Sagging)

By taking Moment-Areas about one end for half the beam, the intercept gives the deflection as:-

$de flection\;\;\;\;\;y\;=\;\frac{[\frac{1}{2}\left(\frac{W\;l}{4} \right)\left(\frac{l}{2} \right)]^{\frac{2}{3}}\;-\;M\left(\frac{l}{2} \right)\frac{l}{4}}{E\;I}$

(9)

$=\;\frac{W\;l^3}{192\;E\;I}$

(10)

(b) Uniform Load

Blank

This time the Free Moment diagram is a parabola and the area is given by:-

$A_1\;=\;\frac{2}{3}\;\left(\frac{w\;l^2}{8} \right)l\;=\;\frac{w\;l^3}{8}$

(11)

The Area of the Fixing Moment diagram $\displaystyle A_2\;is\;M\;l$

Equating the areas of the Free and Fixing Moment diagrams gives:-

$M\;=\;\frac{w\;l^2}{12}$

(12)

This is the Maximum Bending Moment.

As before the intercept about one end gives the deflection

$i.e.\;\;\;\;\;\;\;y\;=\;\frac{[\frac{2}{3}\left(\frac{w\;l^2}{8} \right)\left(\frac{l}{2} \right)]^{\frac{5}{8}}\times \frac{l}{2}}{E\;I}$

(13)

$\therefore\;\;\;\;\;y\;=\;\frac{w\;l^4}{384\;E\;I}$

(14)

Note In comparing this equation with that shown in "Bending of Beams Appendix 3 it must be remembered that w is the weight per unit length and that W is the total weight. i.e. $\displaystyle W\;=\;w\;l$

Example 2

A Beam of span 1 ft. has its ends fixed horizontally at the same level and carries a load W at a distance a ft. from one end and b ft. from the other. Deduce expressions for the fixing moments at the ends. Hence show that, for a distributed load on the same beam the fixing moment at one end is given by

$\int_{0}^{1}{\frac{px\;(l\;-\;x)^2}{l^2}\;dx}$

(15)

Where p is the load per ft. run at a distance x from the end considered.

Apply the above result to find the fixing moments when l = 20 ft. and p varies uniformly from Zero at one end to 2 tons per ft. at the other.

The Free Moment diagram is a triangle of height $\displaystyle \frac{W\;ab}{l}$ and the fixing Moments are $\displaystyle M_a\;\;\;and\;\;\;M_b$

Equating Areas:-

$\frac{1}{2}(M_a\;+\;M_b)\;l\;=\;\frac{1}{2}\left(\frac{W\;ab}{l} \right)\;l$

(16)

$\therefore\;\;\;\;\;\;\;M_a\;+\;M_b\;=\;\left(\frac{W\;ab}{l} \right)$

(17)

By Moment-Areas about the left hand end, splitting each figure into two triangles

$\left(\frac{1}{2}\;M_a\;l\right)\frac{l}{3}\;+\;\left(\frac{1}{2}\;M_b\;l \right)\;\frac{2\;l}{3}\;=\;\left\{\frac{1}{2}\left(\frac{W\;ab}{l} \right)\;a \right\}\;\frac{2a}{3}\;+\;\left\{\frac{1}{2}\left(\frac{W\;ab}{l} \right)\;b\right\}\left(a\;+\;\frac{b}{3} \right)$

(18)

$\therefore\;\;\;\;\;\;(M_a\;+\;2M_b)\;\frac{l^3}{3}\;=\;\frac{2\;W\;a^3b}{3\;l}\;+\;\frac{W\;ab^2}{l}\times\left(a\;+\;\frac{b}{3} \right)$

(19)

$Thus\;\;\;\;\;\;M_a\;+\;2\;M_b\;=\;\left(\frac{W\;ab}{l^3} \right)\left(2a^2\;+\;3ab\;+\;b^2 \right)$

(20)

Subtract equation (17) from (20)

$M_b\;=\;\left(\frac{W\;ab}{l^3} \right)(2a^2\;+\;3ab\;+\;b^2\;-\;l^2)$

(21)

$But\;since\;\;l\;=\;(a\;+\;b)\;\;\;\;M_b\;=\;\left(\frac{W\;ab}{l^3} \right)\;\;(a^2\;+\;ab)$

(22)

$\therefore\;\;\;\;\;\;\;\;M_b\;=\;\left(\frac{W\;ab}{l^3} \right)\;a\;(a\;+\;b)\;=\;\frac{W\;a^2\;b}{l^2}$

(23)

From Equation (17)

$M_a\;=\;\frac{W\;ab}{l}\;-\;\frac{W\;a^2b}{l^2}\;=\;\frac{W\;ab^2}{l^2}$

(24)

Using this equation for a distributed load. The Fixing Moment $\displaystyle \delta \;M_a$ due to the load $\displaystyle p\;\delta x$ on a short length at a distance x from the end $\displaystyle =\;p\;\delta x\times (l\;-\;x)^2$ .

Integrating for the whole beam

$M_a\;=\;\int_{0}^{l}{\frac{px(l\;-\;x)}{l}\;dx}$

(25)

$p\;=\;\frac{2x}{20}\;\;\;tons\; per ft.\;\;\;and\;\;\; l\;=\;20\;ft.$

(26)

$\therefore\;\;\;\;\;M_a\;=\;\int_{0}^{20}{\frac{2x}{20}\times \frac{x\;(20\;-\;x)^2}{20^2}}\;dx$

(27)

$=\;\frac{1}{4000}\int_{0}^{20}{(400x^2\;-\;40x^3\;+x^4)\;dx}$

(28)

$\therefore\;\;\;\;\;\;\;M_a\;=\;26\;\frac{2}{3}\;\;tons-ft.$

(29)

Blank

$\therefore\;\;\;\;\;M_b\;=\;\int_{0}^{20}{\frac{2x}{20}\times \frac{x^2\;(20\;-\;x)^2}{20^2}}\;dx$

(30)

$=\;\frac{1}{4000}\int_{0}^{20}{(20x^3\;-x^4)\;dx}$

(31)

$\therefore\;\;\;\;\;\;\;M_b\;=\;40\;\;tons-ft.$

(32)

It can be seen from the last two examples that for standard cases the maximum Bending Moment occurs at one of the fixed ends. More complicated loadings may be built up by superposition as will be seen in the next Example. It can be assumed that, For any combination of downward loads the Maximum Bending Moment is given by the greater fixing Moment

Example 3

A built-in beam of span 12 ft. carries a uniformly distributed load of half a ton per foot over its whole length. There are also concentrated loads of 2 tons at 3 ft. and 3 tons at 8 ft. from one end. If the Bending Stress is limited to 6 tons/sq.in, calculate the section Modulus required and sketch the Bending Moment diagram.

For each concentrated load $\displaystyle M_a\;=\;\frac{W\;a^2b}{l^2}\;\;\;\;and\;\;\;\;M_b\;=\;\frac{W\;ab^2}{l^2}$ (see eqwuations (23) and(24)) and for the distributed load

$M\;=\;\frac{wl^2}{12}$

(33)

By combination

$Total\;\;\;\;M_a\;=\;\frac{2\times 3\times9^2}{12^2}\;+\;\frac{3\times 8\times 4^2}{12^2}\;+\;\frac{12^2}{2\times 12}$

(34)

$\therefore\;\;\;\;\;\;M_a\;=\;10.04\;tons-ft.$

(35)

$Total\;\;\;\;M_b\;=\;\frac{2\times 3^2\times9}{12^2}\;+\;\frac{3\times 8^2\times 4}{12^2}\;+\;\frac{12^2}{2\times 12}$

(36)

$\therefore\;\;\;\;\;\;M_b\;=\;12.46\;tons-ft.$

(37)

The maximum Bending Moment is thus 12.46 tons-ft. = fZ

$\therefore\;\;\;\;\;\;\;The\;Section\;Modulus\;=\;12.46\times 12 \div 6 \;\;\;\;\;\; ( note\;change\;from\;ft.\;to\;in.)$

(38)

$\therefore\;\;\;\;\;\;\;Z\;=\;24.92\;in^3$

(39)

The following combined Bending Moment diagram has been built up from its component parts and the main values are shown.

The effects of complete and perfect end fixings are to reduce the maximum Bending Moment ( and hence the Stress) and to reduce the deflection, as may be appreciated from the previous examples. In Practice it is almost impossible to ensure no change of slope at the ends, so that usually the degree of fixing is both imperfect and indeterminate. A rotation of the ends proportional may be allowed for and this is shown in the next example where the stiffness of the built in ends has been estimated empirically.

A further practical problem is the danger of settlement of one end relative to the other. This produces an appreciable change in the value of the fixing Moment.

Example 4

Blank

A rung of a vertical ladder is in a horizontal plane and has the form of three sides of rectangle, the short sides of length b and the long side 4 b. The rung is made of steel of circular section and the shrt sides are securely built in to the vertical sides of the ladder. If a vertical load W is carried in the middle of the long side. find the twisting Moment on each of the short sided in terms of W and b

$E\;=\;30\times 10^6\;lb.\;in^{-2}\;\;\;\;\;\;and\;\;\;\;\;\;C\;=\;11.5\times 10^6\;lb.in^{_2}$

(40)

Let t be the twisting Moment on each of the short sides. This then acts as a Bending Moment on each end of the long side and if $\displaystyle \theta$ is the angle of twist of the short sides it is also the angle of slope of the long side.

For the Twisting of a short side:-

$\frac{T}{J}\;=\;\frac{C\;\theta }{b}$

(41)

$Or\;\;\;\;\;\;\theta \;=\;\frac{T\;b}{C\;J}$

(42)

Treating the long side as a centrally loaded beam with incomplete fixing moments T. The increase in slope from end to end is $\displaystyle \frac{\sum{A}}{E\;I}$

$\therefore\;\;\;\;\;\;2\;\theta \;=\;\frac{\frac{1}{2}\times\frac{W}{4}\times4\;b\times4\;b\;-\;t\times 4\;b}{E\;I}$

(43)

$Thus\;\;\;\;\;\;\theta \;=\;\frac{W\;b^2\;-\;2\;T\;b}{E\;I}$

(44)

Equating (42) and (44) and noting that J = 2I

$\frac{T\;b}{2\;C\;I}\;=\;\frac{W\;b^2\;-\;2\;T\;b}{E\;I}$

(45)

From Which:-

$T\;=\;\frac{2\;C\;W\;b}{E}\;-\;\frac{4\;C\;T}{E}$

(46)

$\therefore\;\;\;\;\;T\;=\;\frac{2\;\frac{C}{E}\;W\;b}{1\;+\;\frac{4\;C}{E}}\;=\;\frac{(23/30)\;W\;b}{1\;+\;(46/30)}\;=\;0.303\;W\;b$

(47)

Example 5

Find an expression for the change of Fixing Moments and the End Reactions when one end of a built in Beam of span l sinks by an amount u below the other. The ends remain horizontal.

If M is the change in Fixing Moment it must be hogging at one end and sagging at the other. The change in end reactions R must be given by:-

$R\;=\;\frac{2\;M}{l}$

(48)

The Bending Moment is shown in the diagram.

$u\;=\;\sum{\frac{A\;\bar{x}}{E\;I}}\;=\;\frac{(\frac{1}{2}M\timesl/2)\frac{5}{6}\;l\;-(\frac{1}{2}\;M\;l/2)\frac{l}{6}}{E\;I}$

(49)

$=\;\frac{M\;l^2}{6\;E\;I}\;=\;\frac{R\;l^3}{12\;E\;I}\;\;\;\;\;\;since\lR\;=\;\frac{2M}{l}$

(50)

$\therefore\;\;\;\;\;\;M\;=\;\frac{6\;E\;I\;u}{l^2}$

(51)

$And\;\;\;\;\;\;R\;=\;\frac{12\;E\;I\;u}{l^3}$

(52)

Alternatively this question can be solved by treating it as two Cantilevers of length l/2 carrying end loads of R.

$Hence\;\;\;\;\;\;u\;=\;\frac{2\times R(l/2)^3}{3\;E\;I}\;=\;\frac{R\;l^3}{12\;E\;I}$

(53)

$And\;\;\;\;\;\;M\;=\;\frac{R\times l}{2}\;\;\;\;\;as\;before$

(54)

Macaulay Method

When the Bending Moment diagram does not lend itself to simplification into convenient areas it may be quicker to use the calculus method. (See Bending of Beams Part 3). This also has the advantage of giving directly the Fixing Moments and end Reactions and enables the maximum deflection to be found.

Worked Example

Example 6

A Beam of uniform Section is built in at each end and has a span of 20 ft. It carries a uniformly distributed load of 3/4 ton/ft/ on the left hand half together with a 12 ton load at 25 ft. from the left hand end. Find the end reactions and Fixing Moments and the magnitude and position of the maximum deflection.

$E\;=\;30\times 10^6lbs.in^{-2}\;\;\;\;and I\;=\;500in.^4$

(55)

Taking the Origin at the left hand end and let the Fixing Moments be $\displaystyle M_a\;\;and\;\;M_b$ . The Reactions $\displaystyle R_a\;\;and\;\;R_b$

$Then\;\;\;\;E\;I\;\frac{d^2y}{dx^2}\;=\;-\;M_a\;+\;R_ax\;-\;\frac{3}{4}\times \frac{x^2}{2}\;+\;\frac{3}{4}\frac{(x\;-\;10)^2}{2}\;-\;12(x\;-\;15)$

(56)

Integrating

$E\;I\;\frac{dy}{dx}\;=\;-\;M_ax\;+\;R_a\frac{x^2}{2}\;-\;\frac{x^3}{8}\;+\;\frac{(x\;-\;10)^3}{8}\;-\;6(x\;-\;15)^2\;+\;A$

(57)

$When \;\;\;\;x\;=\;0\;\;\;\;\;\frac{dy}{dx}\;=\;0\;\;\;\;\;\;\therefore\;A\;=\;0$

(58)

Integrating

$E\;I\;y\;=\;-\;M_a\frac{x^2}{2}\;+\;R_a\frac{x^3}{6}\;-\;\frac{x^4}{32}\;+\;\frac{(x\;-\;10)^4}{8}\;-\;2(x\;-\;15)\;+\;B$

(59)

$When\;\;\;\;\;x\;=\;0\;\;\;\;y\;=\;0\;\;\;\;\;\;\therefore B\;=\;0$

(60)

$Also\;\;\;When\;\;\;x\;=\;20\;\;\;\frac{dy}{dx}\;=\;0\;\;\;\;and\;\;\;\;y\;=\;0$

(61)

$\therefore\;\;\;\;-\;M_a\times20\;+\;R_a\times \frac{20^2}{2}\;-\;\frac{20^3}{8}\;+\;\frac{10^3}{8}\;-6\times 5^2\;=\;0$

(62)

$\therefore\;\;\;\;\;\;10\;R_a\;-\;M_a\;=\;\frac{205}{4}$

(63)

$\therefore\;\;\;\;-\;M_a\times\frac{20^2}{2}\;+\;R_a\times \frac{20^3}{6}\;-\;\frac{20^4}{32}\;+\;\frac{10^4}{32}\;-2\times 5^3\;=\;0$

(64)

$\therefore\;\;\;\;\;\;\frac{20}{3}\;R_a\;-\;M_a\;=\;\frac{395}{16}$

(65)

Subtracting equation (65) from (63)

$\frac{10}{3}R_a\;=\;\frac{425}{16}$

(66)

$Thus\;\;\;\;\;\;R_a\;=\;7.97\;tons$

(67)

From equation (63)

$M_a\;=\;28.45\;tons-ft.$

(68)

$But\;\;\;\;\;\;R_a\;+\;R_b\;=\;Total\;downwards\;Load\;=\;19.5\;tons$

(69)

$\therefore\;\;\;\;\;\;R_b\;=\;11.53\;tons$

(70)

$And\;\;\;\;\;-\;M_b\;=\;Value\;of\;B.M.\;at\;x\;=\;20$

(71)

$=\;-\;28.45\;+\;7.97\times 20\;-\;\frac{3\times 20^2}{8}\;+\;\frac{3\times 10^2}{8}\;-\;12\times 5\;=\;-41.55\;tons-ft.$

(72)

Since the concentrated load is greater than the total concentrated load and acts at an equal distance from the nearest end, it may be deduced that zero slope occurs at a value between 10 and 15 ft.