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Engineering › Materials › Beams ›

Built in Beams

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

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Contents

Moment-area Method For Built In Beams.
Macaulay Method
Page Comments

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

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

and

The Fixing Moments at the ends are $\displaystyle M_a$ and

and in order to maintain equilibrium when $\displaystyle M_a$ and

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 $\displaystyle M_a$ ,

and

the Bending Moment at a distance

from the left-hand end is given by:

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

This gives a straight line going from a value of $\displaystyle - M_a$ at

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

) and $\displaystyle A_2$ is a negative area (Hogging

). Consequently equations (1) and (2) reduce to:

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

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

to be found and the total reactions at the ends are :

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

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

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

Example:

[imperial]

Example - Example 1

Problem

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.
b) Uniformly distributed over the whole beam.

Workings

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) $\displaystyle A_1 = A_2$

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

Thus the combined Bending Moment diagram is as shown in the lower diagram. The Maximum Bending Moment is $\displaystyle M = \frac{W\;l}{8}$ 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:

$y = \frac{[\displaystyle\frac{1}{2}\left(\displaystyle\frac{W\;l}{4} \right)\left(\displaystyle\frac{l}{2} \right)]^{\frac{2}{3}} - M\left(\displaystyle\frac{l}{2} \right)\displaystyle\frac{l}{4}}{E\;I}= \frac{W\;l^3}{192\;E\;I}$

b) Uniform Load

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

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

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

Equating the areas of the Free and Fixing Moment diagrams gives $M = \displaystyle\frac{w\;l^2}{12}$

This is the Maximum Bending Moment.

As before the intercept about one end gives the deflection

i.e.

$y = \frac{[\displaystyle\frac{2}{3}\left(\displaystyle\frac{w\;l^2}{8} \right)\left(\displaystyle\frac{l}{2} \right)]^{\frac{5}{8}}\times \displaystyle\frac{l}{2}}{E\;I}= \frac{w\;l^4}{384\;E\;I}$

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. $\displaystyle W = w\;l$

Solution

a) $y = \displaystyle\frac{[\displaystyle\frac{1}{2}\left(\displaystyle\frac{W\;l}{4} \right)\left(\displaystyle\frac{l}{2} \right)]^{\frac{2}{3}} - M\left(\displaystyle\frac{l}{2} \right)\displaystyle\frac{l}{4}}{E\;I}= \frac{W\;l^3}{192\;E\;I}$
b) $y = \displaystyle\frac{[\displaystyle\frac{2}{3}\left(\displaystyle\frac{w\;l^2}{8} \right)\left(\displaystyle\frac{l}{2} \right)]^{\frac{5}{8}}\times \displaystyle\frac{l}{2}}{E\;I}= \frac{w\;l^4}{384\;E\;I}$

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.

Example:

[imperial]

Example - Example 1

Problem

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 $\displaystyle\frac{3}{4} ton/ft$ 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.

Workings

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

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

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

Integrating,

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

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

Integrating,

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

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

Also when $x = 20$ $\displaystyle\frac{dy}{dx} = 0$ and $y = 0$

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

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

(1)

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

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

(2)

Subtracting equation (2) from (1)

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

Thus, $R_a = 7.97\;tons$

From equation (1)

$M_a = 28.45\;tons-ft.$

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

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

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

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

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.

$\therefore\;\;\;\;\;E\;I\;\frac{dy}{dx}\;= - 28.45x + 7.97\times20 - \frac{x^3}{8} + \frac{(x - 10)^3}{8} = 0$

$x^2(3.98 - 3.75) - x(28.75 - 37.5) - 125 = 0$

$0.235x^2 + 9.05x - 125 = 0$

Solving this quadratic gives:

$x = 10.8\;ft.$

Substituting this value in the deflection equation gives:-

$E\;I\;y\;= - \frac{28.45\times10.8^2}{2} + \frac{7.97\times10.8^3}{6} - \frac{10.8^4}{32} + 0.8^432\;= - 414\;tons-ft.$

Thus the maximum deflection is given by:

$\hat{y} = \frac{414\times 12^3}{13200\times 500} = 0.109\;in.$

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