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Engineering › Materials › Beams ›

Built in Beams II

Shear and Deflection formulae for Built in and Continuous Beams

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Contents

Fixed At Both Ends. Uniform Load. Total Load W
Fixed At Both Ends. Load At Centre.
Fixed At Both Ends. Load At Any Point.
Continuous Beam With Two Equal Spans. Uniform Load
A Continuous Beam With Two Unequal Spans And Unequal Uniform Loads.
A Continuous Beam With Two Unequal Spans With Two Unequal Loads At Any Point On Each
Page Comments

Fixed At Both Ends. Uniform Load. Total Load W

The stress at any Point : $S = \displaystyle\frac{W\;l}{2\;Z}\left\{\displaystyle\frac{1}{6} - \displaystyle\frac{x}{l} + \left(\displaystyle\frac{x}{l} \right)^2\right\}$

The Maximum Stress is at the ends and is $\hat{S} = \displaystyle\frac{W\;l}{12\;Z}$

The Stress is zero at $\displaystyle x = 0.7887\;l$ and $x = 0.2113\;l$

The Greatest negative Stress is at the centre and is $\displaystyle S_c\;= - \frac{W\;l}{24\;Z}$

The Deflection at any Point is given by $y = \displaystyle\frac{W\;x^2}{24\;E\:I\;l}\;\;(l - x)^2$

The Maximum Deflection is at the Centre and is $\hat{y} = \displaystyle\frac{W\;l^3}{384\;E\;I}$

Deflection is a term that is used to describe the degree to which a structural element is displaced under a load.

Fixed At Both Ends. Load At Centre.

The Stress between each end and the Load: $S = \displaystyle\frac{W}{2\;Z}\;\;\left(\displaystyle\frac{1}{4}\;l - x \right)$

The Stress at each end is : $S_e = \displaystyle\frac{W\;l}{8\;Z}$

The Stress at the middle is: $S_m\;= - \displaystyle\frac{W\;l}{8\;Z}$

These are the maximum Stresses and are equal and opposite.

The stress is zero at $\displaystyle x = \frac{1}{4}\;l$

The Deflection at any point is given by: $y = \displaystyle\frac{W\;x^2}{48\;E\;I}\;\;(3l - 4x)$

The Maximum Deflection is at the Load and is: $\displaystyle \hat{y} = \frac{W\;l^3}{192\;E\;I}$

Fixed At Both Ends. Load At Any Point.

The Stress at any Point within the segment of length

$S_a = \frac{W\;b^2}{Z\;l^3}\;\;\left(al - x[l + 2a] \right)$

The Stress at any point within the segment of length

$S_b = \frac{W\;a^2}{Z\;l^3}\;\;\left(bl - v[l + 2b] \right)$

The Stress at the end next to segment of length

: $S = \displaystyle\frac{W\;a\;b^2}{Z\;l^2}$

The Stress at the end next to segment of length

: $S = \displaystyle\frac{W\;a^2\;b}{Z\;l^2}$

The Maximum Stress is at the end next to the shorter segment.

The Stress is Zero for: $x = \displaystyle\frac{a\;l}{l + 2\;a}$ and $v = \displaystyle\frac{b\;l}{l + 2\;b}$

The Greatest negative stress is at the Load and is given by: $S\;= - \displaystyle\frac{2\;W\;a^2\;b^2}{Z\;l^3}$

The Deflection for the segment of length

is given by:

$y = \frac{W\;x^2\;b^2}{6\;E\;I\;l^3}\;\;2a(l - x) + l(a - x)$

The Deflection for the segment of length

is given by:

$y = \frac{W\;v^2\;a^2}{6\;E\;I\;l^3}\;\;2b(l - v) + l(b - v)$

The deflection at the Load is: $y = \displaystyle\frac{W\;a^3\;b^3}{3\;E\;I\;l^3}$

Let

be the length of the longer segment and

the shorter one.

The Maximum Deflection is in the longer Segment and occurs at $\displaystyle v = \frac{2\;b\;l}{l + 2\;b}$

$\hat{y} = \frac{2\;W\;a^2\;b\63}{3\;E\;I\;(l + 2\;b)}$

Continuous Beam With Two Equal Spans. Uniform Load

A beam is a horizontal structural element that is capable of withstanding load primarily by resisting bending. The bending force induced into the material of the beam as a result of the external loads, own weight, span and external reactions to these loads is called a bending moment.

The Stress at any Point is given by: $s = \displaystyle\frac{W\;(l - x)}{2\;Z\;l}\;\;\left(\displaystyle\frac{1}{4}l - x \right)$

The Maximum Stress is at Point

and is: $\hat{S} = \displaystyle\frac{W\;l}{8\;Z}$

The Stress is zero at $\displaystyle x = \frac{1}{4}l$

The greatest negative Stress is at $\displaystyle x = \frac{5}{8}\;l$ and is: $y\;= - \displaystyle\frac{9}{128}\;\;\displaystyle\frac{W\;l}{Z}$

The Deflection at any Point is given by: $y = \displaystyle\frac{W\;x^2\;(l - x)}{48\;E\;I\;l}\;\;(3l - 2x)$

The Maximum Deflection is at $\displaystyle x = 0.5785\;l$ and is given by: $\hat{y} = \displaystyle\frac{W\;l^3}{185\;E\;I}$

The Deflection at the centre of each Span is: ${y} = \displaystyle\frac{W\;l^3}{192\;E\;I}$

The Deflection at the point of greatest negative Stress is ${y} = \displaystyle\frac{W\;l^3}{187\;E\;I}$
And is at $x = \displaystyle\frac{5}{8}\;l$

A Continuous Beam With Two Unequal Spans And Unequal Uniform Loads.

Between $\displaystyle R_1$ and

The Stress is given by: $S = \displaystyle\frac{l_1 - x}{Z}\left\{\displaystyle\frac{(l_1 - x)\;W_1 }{2\;l_1} - R_1\right\}$

Between $\;R_2$ and

The Stress is given by: $S = \displaystyle\frac{l_2 - u}{Z}\;\left\{\displaystyle\frac{(l_2 - u)\;W_2}{2\;l_2} - R_2\right\}$

Stress at the Support

is: $S_R = \displaystyle\frac{W_1\;l_1^2 + W_2\;l_2^2}{8\;Z\;(l_1 + l_2)}$

The greatest Stress in the first span is at: $x = \displaystyle\frac{l_1}{W_1}\;\;(W_1 - R_1)$

And is $\displaystyle \frac{R_1^2\;l_1}{2\;Z\;W_1}$

The greatest Stress in the second Span is at: $u = \displaystyle\frac{l_2}{W_2}\;\;(W_2 - R_2)$

And is $\displaystyle - \frac{R_2^2\;l_2}{2\;Z\;W_2}$

The Deflection between $R_1 \;and\; R$ is given by:

$y = \frac{x(l_1 - x)}{24\;E\;I}\left\{(2l_1 - x)(4R_1 - W_1) - \frac{W_1\;(l_1 - x)^2}{l_1} \right\}$

The Deflection between $R_2\; and \;R$ is given by:

$y = \frac{u(l_2 - u)}{24\;E\;I}\left\{(2l_2 - u)(4R_2 - W_2) - \frac{W_2\;(l_2 - u)^2}{l_2} \right\}$

The above example is so complicated that convenient general expressions for the maximum deflections cannot be obtained

Between point

and the Load the Stress at any Point is: $S = \displaystyle\frac{W}{16\;Z}\;\;(3\;l - 11\;x)$

Between Point

and the Load the stress at any Point $S\;= - \displaystyle\frac{5}{16}\;\;\displaystyle\frac{W\;v}{Z}$

The Maximum Stress at Point

$\hat{S} = \displaystyle\frac{3}{16}\;\;\displaystyle\frac{W\;l}{Z}$

The Stress is Zero at: $x = \displaystyle\frac{3}{11}\;l$

The greatest negative Stress is at the centre of the Span and is:

$\hat{S}_{-ve}\;= - \frac{5}{32}\;\;\frac{W\;l}{Z}$

Between Point

and the Load the Deflection at any point is:

$y = \frac{W\;x^2}{96\;E\;I}\;\;(9\;l - 11\;x)$

Between Point

and the Load the Deflection at any point is:

$y = \frac{W\;v}{96\;E\;I}\;\;(3\;l^2 - 5\;v^2)$

The Maximum deflection is at $\displaystyle v = 0.4472$ and is: $\hat{y} = \displaystyle\frac{W\;l^3}{107.33\;E\;I}$

The Deflection at the Load is: $y_w = \displaystyle\frac{7}{768}\;\;\displaystyle\frac{W\;l^3}{E\;I}$

A Continuous Beam With Two Unequal Spans With Two Unequal Loads At Any Point On Each

Between $\displaystyle R_1$ and

the Stress is: $S\;= - \displaystyle\frac{w\;r_1}{Z}$

Between $\displaystyle R$ and

the Stress is: $S = \displaystyle\frac{I}{l_1\;Z}\;\;[m(l_1 - u) - W_1\;a_1\;u]$

Between $\displaystyle R$ and

the Stress is: $S = \displaystyle\frac{I}{l_2\;Z}\;\;[m(l_2 - x) - W_2\;a_2\;x]$

Between $\displaystyle R_2$ and

the Stress is: $S\;= - \displaystyle\frac{v\;r_2}{Z}$

Stress at the Load: $S\;= - \displaystyle\frac{a_1\;r_1}{Z}$

Where

$m = \frac{1}{2(l_1 + l_2)}\;\;\left(\frac{w_1\;a_1\;b_1}{l_1}\;(l_1 + a_1) + \frac{w_2\;a_2\;b_2}{l_2}\;\;(l_2 + a_2) \right)$

Stress at Support

: $S_R = \displaystyle\frac{m}{Z}$

Stress at load

: $S\;= - \displaystyle\frac{a_2\;r_2}{Z}$

The greatest of these is the maximum Stress.

The Deflection between $\displaystyle R_1$ and

$y = \frac{w}{6\;E\;I}\;\left\{(l_1 - w)(l_1 + w)r_1 - \frac{W_1\;b_1^3}{l_1} \right\}$

The Deflection between $\displaystyle R$ and

$y = \frac{u}{6\;E\;I\;l_1}\;\;[W_1\;a_1\;b_1(l_1 + a_1) - W_1\;a_1\;u^2 - m(2\;l_1 - u)(l_1 - u)]$

The Deflection between $\displaystyle R$ and

$y = \frac{x}{6\;E\;I\;l_1}\;\;[W_2\;a_2\;b_2(l_2 + a_2) - W_2\;a_2\;x^2 - m(2\;l_2 - x)(l_2 - x)]$

The Deflection between $\displaystyle R_2$ and

$y = \frac{v}{6\;E\;I}\;\left\{(l_2 - v)(l_2 + v)r_2 - \frac{W_2\;b_2^3}{l_2} \right\}$

The Deflection at Load $\displaystyle W_1$ : $y = \displaystyle\frac{a_1\;b_1}{6\;E\;I\;}\;[2a_1\;b_1\;W_1 - m(l_1 + a_1)]$

The Deflection at Load $\displaystyle W_2$ : $y = \displaystyle\frac{a_2\;b_2}{6\;E\;I\;}\;[2a_2\;b_2\;W_2 - m(l_2 + a_2)]$

Page Comments

$levettp\′s Photo$

levettp

16 Sep 10, 10:10AM

Whats the units of measure for the variables?

$EngNoob\′s Photo$

EngNoob

17 Nov 08, 11:28AM

(7 replies)

Z Reference

Anyone please advise on what Z is in the equation? thanks.

$CodeCogs\′s Photo$

CodeCogs

18 Nov 08, 5:58PM

Z is the section modulus. It is equal to

$Z=\frac{I}{y}$

(1)

where

I is the second moment of inertia
y is the distance between the axes of I and Z.

Z is usually calculated on the surface of a beam, so for a rectangular box, y=half its height.

(I think the pages need updating to explain this.)

$EngNoob\′s Photo$

EngNoob

20 Nov 08, 9:37AM

Hey

Can you confirm this please?

Z = I / y

I = BD^3 / 12

Y-rectange = d / 2

Y is described as the distance from the neutral axis, but distance to where? top or bottom?

Can this be calculated as

Z = BD^2 / 6 ?????

Thanks for any more light you can shed on this.

I also want to add that this site is one of the best, if not the best resource i have came upon whilst studying Mechanical Engineering, its excellent, and i will be passing the link on to fellow students and lectureres.

$lodmore\′s Photo$

lodmore

21 Nov 08, 11:17AM

I entered the original reference page and I would have answered your query earlier but my Broadband went down yet again!

I am sure you know the equation $\displaystyle \frac{f}{y}\;=\;\frac{M}{I}\;=\;\frac{E}{R}$ well neglect the right hand fraction and re-arrange the other two so:-

$f\;=\;y\times \frac{M}{I}$

(1)

If you look at this equation you can see that for any given cross section y mas a maximum value as it is the distance from the Neutral Axis to the "outside" of the section. Likewise I is a Constant for that section being bent in that plane. Hence I can re-write the above equation as:-

$\hat{f}\;=\;M\times \frac{\hat{y}}{I}\;=\;\frac{M}{Z}$

(2)

Actually it might be better to write this as:-

$M\;=\;Z\times \hat{f}$

(3)

M in this equation is the Maximum Bending moment that can be carried by the given section for a given maximum Stress. It is called the Moment of Resistance

Now look at your Post of the 20Th Nov.

Y is actually the distance from the neutral axis to where ever you want to know the stress. Of course this is usually the "outside" of the section because that is where the stress is a maximum. You ask whether you should measure upwards or downwards. Well that depends upon what you want. On a simply supported beam subjected to a downwards load the upper surfaces goes into Compression and the lower Tension and the important one depends upon the relative maximum compressive and Tensile stresses.

You asked about a rectangular section and the neutral axis is central. Had it been a "T" cross section the neutral axis would no longer be central and its position would need to be calculated.

In the particular example you gave you are correct. I hope that I have sorted out your problems.

$EngNoob\′s Photo$

EngNoob

21 Nov 08, 3:09PM

Hey

Sorted them out, and thanks for that additional info.

I am now trying to figure out how to calculate using two loading points on a fixed end beam and having little, well no luck. From reading i can see that maybe you would use superpossition, but not 100% sure.

I am trying to figure out the Macaulay Method too, but because my maths isnt that good i am making no progress their. Rather than give up, it be good if some one can identify the maths required to solve so i can learn them .

thanks

$lodmore\′s Photo$

lodmore

21 Nov 08, 5:34PM

May I suggest that you study example 3 in "The Bending Of Beams Part 5" The beam in the example is both built in and carries two loads. However it also carries a uniform load which you will have to remove!

I am afraid that you will have to learn some Mathematics if you want to be an engineer. As a help we have some useful pages on Integration; Differential equations. Trigonometry and Coordinate (Analytical) Geometry. They all have worked examples which should help you

$beamy\′s Photo$

beamy

7 Aug 10, 5:00PM

I do not understand why the modular ratio m is introduced in the reaction loads in the case of the 'Continuous Beam With Two Unequal Spans With Two Unequal Loads At Any Point On Each'.

$beamy\′s Photo$

beamy

30 Dec 10, 10:11AM

In the section 'Fixed at both ends. Load at any point', the deflection at the load is given as y = Wa^3b^2/6EIL^3. However, where load is in the centre, i.e. subsituting a = L/2 and b = L/2, the load deflection in the equation becomes:

W = W(0.5L)^3(L - 0.5L)^2/6EIL^3

= WL^3L^2/192EIL^3

= WL^5/192EIL^3

= WL^2/192EI

which is not correct, because it should be WL^3/192EI

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