Theory and Analysis of Continuous Beams

Continuous beams

March: 2012

THEORY AND ANALYSIS

OF

CONTINUOUS BEAMS

1

Phumlani G. Nkosi

5N A

B

24m

RA

1N/m

C

30m

RB

2N D

6m

RC

0

1

Phumlani G. Nkosi, BSc. (Mathematical and statistical sciences: Mathematical Modelling), BSc Hons (Technology Management), DP( Mechanical engineering), B.Tech. Mechanical Engineering student UNISA

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Continuous beams

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Table of Contents

Introduction............................................................................................................................................3 Analysis of continuous beams ..............................................................................................................3

............................................................................................ ............................................. 4 Claperon’s Theorem of Three Moments ............................................... ............................................................................................................................. ................................................................................. 4 Bending moments ............................................ ...................................................................................................................... ...................................................................... 8 Reactions at the support ................................................ Steps in solving continuous beams........................................................................................................9

...................................................................................................................................... ............................................................................... ........... 12 Analysis .................................................................. Summary .............................................................................................................................................23

.................................................................................................................................. ............................................................................................. .......... 23 References ............................................... ................................................................................................................................. ............................................................................................. ......... 24 APPENDIX A .............................................

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CONTINUOUS BEAMS

Introduction Before we elaborate more on continuous beam let us first understand what is meant by the term beam and bending moments in structural engineering. A beam is a horizontal structural element that is capable of withstanding load primarily by resisting bending, bending moments are the bending forces induced to the material as a results of external load/s and due to its own weight over the span. We are now ready to define a continuous beam; A Continuous beam is beam is a beam that is supported by more than two supports as shown in figure 1.

Figure 1: Continuous beam subjected to point load

These beams (Continuous beams) are statistically indeterminate and indeterminate and are known as Redundant or Indeterminate Structures as they cannot be analysed by making use of basic equilibrium, i.e.

∑  = 0,0, ∑  = 0 ∑ = 0

,

As opposed to determinate structures where by the above conditions are applicable. Continuous are mainly used in high risk structures like bridges and buildings as the supports are strong enough to withstand heavy loads. The following are some of the advantages and disadvantages of continuous beams (just to name the few), Advantage/s: 1. Has more vertical load load capacity – can support a very heavy load/s 2. Deflection at the middle of of the span is minimal as as opposed to simple supported beams. Disadvantage/s: Difficult analysis and design procedure

Analysis of continuous beams As stated from the introduction that continuous beams are difficult to analyse one method (not the only method) of analysing them is by making use of Claperon’s Theorem this Theorem this theorem is also known as Theorem of Three Moments and it is used as follows. Two consecutive spans of the continuous beam are considered at simultaneously, each span is treated individually (see figure 2a) as a simple supported beam with external loads and two end support moments. For each central support, one equation is written in terms of three moments (EQ 1). Thus we get as many equations as there unknowns; each will have only three unknowns. The term three moments refers moments  refers to the unknown moments at the central support and at the two ends of any pair of adjacent spans. The theorem equation is derived by considering the deflection and the

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slope of the beam; we will not derive the theorem, we will just give the question and show how it is applied in determining the bending moments acting on the beam for analysis. (The derivation can be found on Strength of Materials for Technologists by JG Drotsky ). Drotsky  ).

Claperon’s Theorem of Three Moments Consider the continuous beam figure 2(a) carrying a load

ሺሻ

supported at

,

By load  we mean any external force such as Transverse load (concentrated point load or uniform distributed load) and bending moments Assume that the beam: 1. Has Uniform cross section 2. Has Uniform Modulus of rigidity and 3. The supports are at the same level level Therefore the Claperyon’s equation is given by:

Where:

 ஼        ̅   ̅

− × − ૛ ሺ + ሻ− × =૟ ቂ ഥ + ഥ ቃ

1

= The Left hand side (RHS) moment due to the support = The central moment due to the central support = The right hand side (RHS) moment due to the support

= The length of the segment = The length of the segment = The Area of the segment = The Area of the segment

= The length from the left support

to the centroid of area

= The length from the left support

to the centroid of area

 

Bending moments Before continuing with this section it is important that you first go through appendix A for revision on  simple supported beams  Knowing what bending moments are acting on the beam will help us to able to calculate the Area/s as well as the net bending moment for segment/span Before applying the equation of the three moments let us first examine what bending moments the beam on figure 2(a) is subjected to:

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The beam is subjected to the bending moments coursed by the external load figure 2(b), the bending moments coursed by the central support figure 2 (c) and bending moments coursed by the support and at the end figure 2 (d) which they are assume to be equal to zero unless the beam is fixed at either both supports (Fixed ends Beam, or at one End (Proped beam see figure 6(a)) of is overhanging see figure 4 Before we can express the above statement mathematically it is important to know that we keep sign conversion the same that is if the beam is compressed at the top (i.e. above neutral axis) and elongated at the bottom (i.e. subjected to tensile force) therefore the bending moments is positive, now we can express the above mathematically: Total bending moments = Bending moment due to load + bending moments due to supports

= ሺሻ+ + + = ሺ ሻ− ሺ + + ሻ = =૙ ∴ = ሺ ሻ−

2

Taking into consideration the sign conversion, the net bending moment is given by 3

For the simple beam with three supports as shown in figure 2 (a), 4 5

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Figure 2: (a) continuous beam supporting a load w(x), (b) Bending moment diagram due to external load w(x), (c) Bending moment diagram due to central support C, (d) Bending moment due to end supports , L and R

Let us now illustrate the above with the following example, figure 3 show the concentrated load P acting on the mid-span of LC and uniform distributed load W acting on the span CR, let LC = a and CR = b. Determine the bending moments. The procedure is as follows: We treat each span individually as simple supported beam with external loads and two end support moments:

The table below show the Areas (A), the centrionds ( Span LC CR

Area A

ଶ

= 8 = 12ଷ

ഥ   ̅ = 2   ̅ = 2

Distance

 ̅

) for each span Comment

Simple supported beam with the load acting at the centre( Appendix A, Figure A2) Simple supported beam with uniform distributed load( Appendix A, Figure A1) Table 1

We can now use the three moment’s equation (EQ1) and substitute the values on table determine the central bending moment. The equation is given by:

1 to

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Continuous beams

From EQ 4:

March: 2012

−  ×  − 2 ஼ሺ  + ሻ −  ×  = 6 ൤   ̅ +   ̅൨ = =0 ଶ8 × 2 12ଷ × 2 ∴−2 ஼ =6 ൦ + ൪ ଶ ଷ −3 ∴ ஼ = ቈ 16 + 24 ቉

From the above we can now able to calculate the net bending moment at the middle of each span, the net bending at the mid-span is the difference between the maximum bending moment for a simple beam (figure 3(b)) and the bending moments due to the supports (see figure 3 (c)) : For the span LC the net bending moment is:

And for the span CR

And for the middle of the beam

௡௘௧ ஼ = 4 − 12 ሺ  + ஼ሻ ∴ ௡௘௧ ஼ = 4 − 2஼

ଶ ௡௘௧ ஼ = 8 − 12 ሺ ஼ + ሻ ∴ ௡௘௧ ஼ = 8ଶ − 2஼

௡௘௧ ஽ = 0 − ஼ ∴ ௡௘௧ ஽ = − ஼

We can now draw the bending moment diagram using the above results see figure 3 (d) or 4(c) for the bending moment diagram

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P W L

C

R

(a) a

b

RL

RC

RR

L =

=

AL

(b)

AR

+

+

ML

MR

(c)

MC =

=

(d)

Figure 3

Reactions at the support Having calculated the bending moments above we can now determine the reactions by either taking moment at the supports or using the relationship between the bending moment and the shear force, we will only used the moment theorem, once again we treat each span individually as simple supported beam with external loads and two end support moments Reaction RL: Consider span LC as a simple supported beam: Treat the moments at L as redundant (i.e. the moment at the support that we are calculation the reaction)

Reaction RC:

෍ ஼ =0  × + ஼= ×2 ∴  = ൬ −2 2 ஼ ൰ 8

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Consider the entire beam; and treat the moments at the support L and C as redundant, therefore

Reaction RD:

෍ ஽ =0  × + ஼ + ஽ = × ቀ2 + ቁ + + 2 ൬ −2 2 ஼ ൰ × + ஼ = × ൬ +2 2 ൰ + 2 ଶ ଶ + 2 ∴ ஼ = × ൬ 2 ൰ + 2 − ൬ −2 2 ஼ ൰ ×

RD is determined b using the equilibrium equation: Thus

෍ ௏ =0 ∴ + ஼+ ஽=0 ∴ ஽=−  − ஼

After determining the reactions the SF and the BM diagrams can be drawn as shown on figure 4(b) and (c)

Steps in solving continuous beams 

Step 1 Three condition need to be taken into account before start analysing the beam, namely: I. Simple supported beam:  If the beam is simple supported at both ends as shown in figure 3(a), set the bending moments at the supports equal to zero and start analysing the beam as explained below. II.

Overhanging beam  If the beam is supporting an overhanging load on either both or one side as shown if figure 5(a), start by taking moments about support/s supporting the overhang load to determine the induced moment at that support/s, see figure 5(b), if it is only one side that is overhanging set the other side bending moment at the support equal to zero and continue analysing the beam as explained below

III.

Fixed ends beam  If the beam is fixed in one or both ends as shown if figure 6(a) first release the fixed end/s as shown

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in figure 6(b) set, the released length equal to zero, hence the area for this span is zero then set the bending moments at the far end supports equal to zero and start analysing the beam as explained below.

P W L

(a)

C

R

a

b

RL

RC

RR

L SF Diagram RL

0 (b)

x - RR BM Diaram Mmax

0 MR

ML

(c) MC

Figure 4: (a) continuous beam, (b) SF diagram, (c) BM diagram

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A

B

w N/m

C

D

(a) a

b

A

B

c

w N/m

C

(b) 2

MC = Wc /2

MA= 0 a

b

Figure 5: (a) Simple beam with over-hanging load, (b) a beam is reduced to simple supported ends with no over-hanging load.

A

B

'C

w N/m

(a) a

A

b

B

w N/m

C

(b)

D MD = 0

MA = 0 a

b

c= 0

Figure 6: (a) continuous beam with the fixed end to the right, (b) The fixed end has been released with the span CD 



Step 2 Treat each span individual as simple supported beam taking load/s acting on the span into consideration. Step 3 Draw a free body diagram (figure 3 (b)) representing the bending moments endued by the applied load/s on the span.

 

 

Step 4 Determine the Area(s) /area moments for the load/s acting on simple supported beam. Step 5 Use the three moment theorem equation (EQ1) by considering the consecutive spans (two spans and three supports) to estimate the bending moments induced by the supports. Step 6 Determine the reactions on each support using the moment’s equilibrium equation. Step 7 Draw the shear force (SF) and the bending moments (BM) diagram.

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Analysis The following examples will better clarify the steps st eps above: Example 1 A 60m long beam supports a uniform distributed loan (u.d.l.) of 1N/m throughout its entire length AD, with AB = 24m, BC = 30m and CD = 6m. In addition a concentrated load of 5 N acts at the mid-point of AB and a load of 2 tons acts at D. See figure 7 Determine the bending moments and the reaction forces ant the supports. Draw the S.F. and the B.M. diagram.

5N A

B

1N/m

2 4m

RA

C

3 0m

RB

2N D

6m

RC Figure 7

Step 1

୅=0 ෍ ஼ =0 6 ሺ1ሻ6 ൬2൰ −2×6+ 6+ሺ ஼ = −2× ஼ = −30

Step 2 Figure 8(a) shows that we have three consecutive spans/segments i.e. AB, BC and CD. For segment: AB Segment AB of a beam is subjected to two loads, i.e. the point load with the magnitude of 5N acting on the mid-span and the u.d.l. of 1Nm over the entire span. BC This segment is only subjected to u.d.l. of 1Nm, CD This segment is subjected to two loads, i.e. the u.d.l and the point load with the magnitude of 2N acting at D

Step 3 Draw a free body diagram representing the bending moments endued by the applied load on the span. See figure 8(b)

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Step 4 After separating and identifying the load/s acting on each span, we can now calculate the area/s under each curve, see Appendix A Note:: Note

If the spun is subjected to two or or more loads therefore the total area moment for span will be sum of the area moments for each load (for this example:

஺஻   ̅஺஻஺஻ = ሺ   ̅ሻ௣௢௜௣௢௜௡௧ ௟௢௔ௗ + ሺ   ̅ ሻ௨ௗ௟௨ௗ௟

Calculation of area moment Span AB For point load: 

஺஻   ̅஺஻஺஻ = ሺ   ̅஺஻஺஻ሻ௣௢௜௣௢௜௡௧ ௟௢௔ௗ + ሺ   ̅஺஻஺஻ሻ௨ௗ௟௨ௗ௟

Method 1

ሺ   ̅ሻ௣௢௜௣௢௜௡௧ ௟௢௔ௗ = 8ଶ஺஻ × ஺஻2 ଶ 5×24 ሺ   ̅ሻ௣௢௜௣௢௜௡௧ ௟௢௔ௗ = 8 × 242 =4320 ℎ2 = ሺ   ̅ሻ௣௢௜௣௢௜௡௧ ௟௢௔ௗ = ൬12 × 4 ൰ × 23 + ൬12 × 4 ൰ × ൬3 + ൰ ℎ : = 2 = ஺஻ ሺ   ̅ሻ௣௢௜௣௢௜௡௧ ௟௢௔ௗ = ൭12 × 12×12×൬൬5×244 ൰൱×൬2×123 ൰ + ൭12 ×12×൬5×244 ൰൱×൬123 +12൰ ሺ   ̅ሻ௣௢௜௣௢௜௡௧ ௟௢௔ௗ = 4320 ଷ ሺ   ̅ሻ௨ௗ௟௨ௗ௟ = 12஺஻ × 2 ሺ   ̅ሻ௨ௗ௟௨ௗ௟ = 1×2412 ଷ × 242 ሺ   ̅ሻ௨ௗ௟௨ௗ௟ = 1382 138244 ஺஻   ̅஺஻஺஻ = 4320 + 13824 181444 ஺஻   ̅஺஻஺஻ = 1814

Method 2

For u.dl. load 

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Span BC

 ஻஼ ̅஻஼஻஼ = 12ଷ஻஼ × 2஻஼  ஻஼ ̅஻஼஻஼ = 1×3012 ଷ × 302 337500 ஻஼   ̅஻஼஻஼ = 3375

Step 5 Applying Equation 1 to the span A B C 

− ஺ × ஺஻ − 2 ஻ሺ ஺஻ + ஻஼ሻ − ஼ × ஻஼ = 6 ൤ ஺஻஺஻  ̅஺஻஺஻ + ஻஼஻஼  ̅஻஼஻஼൨ 1811424 + 3375030 ൨ 0×2 0× 24− 2 ஻−ሺ54ሻ51084ሻ − ሺ−+3090090ሻ ×03=0 6= ×1×6 1൤18114 ஻∴ =96.2 881 ஻ The Bending Moment at mid-point of AB 

ଶ ஺஻ = + ௡௘௧ ஺஻ 4 8 ஺஻ − 2஻ ଶ 5×24 1×24 = + ௡௘௧ ஺஻ 4 8 − 96.22 ∴ ௡௘௧ ஻஼ =53.9 The Bending Moment at the mid-point of BC 

ଶ ௡௘௧ ஻஼ = 8 ஻஼ − 12 ሺ ஻ + ஼ሻ ଶ 1×30 1 ሺ96.96.2+30ሻ = − +30ሻ ௡௘௧ ஻஼

8 2 ∴ ௡௘௧ ஻஼ = 4949..4

Step 6

Determine the reactions on each support using the moment’s equilibrium equation.

For span AB 

෍ ஻ =0 14

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1×24×൬ 4×൬242 ൰ ஼ ×24+ ஻ =5×൬242 ൰+ 1×2 2 ஺ = ൬288+60−96. 24 ൰=10.5 For span ABC 

෍ ஼ =0 54 × ஺ +30× ஻ × + ஼ =+5൬242 +30൰+1×54×൬542 ൰ 54 × 10.5 + 30 ஻ +30=5×42+54 ×27 ஻ = ൬1668−597 30 ൰ ஻ =35.7 ෍ ௏ =0 2+ 1×660 ஺ + ஻ + ஼ = 5 + 2+1× ஼ = 67 − 10.5 − 35 ∴ ஼ = 20.8

Step 7 Draw the SF and the BM diagrams figure 8 (c) and (d) From the Shear Force diagram at the maximum bending moment occurs at:

And at

12 =10. 5 ൬ ଵ 10.5+1.5൰ ଵ =10.5 ଶ =17. 2൬17.230+12.8൰ ଶ =17.2 15

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The maximum bending moment is given by the area under the curve from the shear force diagram

௠௔௫ = ஺ + 12 ஺=0

At x = 10.5m the maximum bending moment is

ℎ ∴ ௠௔௫ = 12 ×10.5 ×10.5 =55.3 ௠௔௫ = 12 ℎ − ஻ ∴ ௠௔௫ = 12 × 1717..2 × 1717..2 − 9696..2 = 5151..72 ሺ 52 ሻ

At x = 17.2m the maximum bending moment is:

The diagram for the SF and BM is shown below:

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5N A

B

1N/m

24m

C

30m

RA

2N D

6m

RB

RC

L

1

(b)

2

3

MB

17.2 10.5

SF DIAGRAM '8

F 

'2

0 '1.5 '6.5 x1 = 10.5

x2 = 17.2 12.8

'18.5 '55.1

(d)

'52

53.9 '49.4

BM DIAGRAM

0 M

'30

'96.2

Figure 8: (a) continuous beam, (b) free body diagram for BM induced by external load, (c) SH diagram, (d) BM diagram

The bending moment diagram can also be drawn as shown below, figure 9

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BM Diagram 49.4 96.2 53.9

M 30

0 Figure 9: BM Diagram

Example 2 This question is taken directly from UNISA past exam paper (January/February 2010)

Solution Let:

஺=0

Areas Consider span AB

Consider the span BC

஽ = −10 ×1= −10

=஺஻ 12ଷ = 1 ×125ଷ = 12512 ஺஻   ̅ = 12512 × 52 = 62524   ̅஻஼஻஼ ≠ ଶ

Since the concentrated does not act at the mid-span:

, the figure below represent the BM

diagram induced by the concentrated point load, figure 10

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1

L1 =2m

2

L2=1m

Figure 10: BM diagram for the point load

ℎ= × ଵ× ଶ Where:

= ଵ+ ଶ

Method 1:

Moments about B

Moments about C

Method 2:

Moments about B:

40 ∴ ℎ = 20×2×1 = 3 3 = ஺஻ = 12 × ℎ = 12 × 3 × 403 = 20 × ሺ +3 ଵሻ =20 × 53 = 1003 ஺஻   ̅஻ = × ሺ +3 ଶሻ =20 × 43 = 803 ஺஻   ̅஻ = 1 = 12 × ℎ = 12 × 2 × 403 = 403 2 = 12 × ℎ = 12 × 1 × 403 = 203 ஺஻   ̅஻ = 403 × 23 × 2 + 303 ൬13 ×1+2൰= 1003 19

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Moments about C:

஺஻   ̅஼ = 403 ൬13 ×2+1൰+ 203 × 23 × 1 = 803 = ૚ + ૛ = ૝૙૜ + ૛૙૜ = ૛૙ ଷ ଷ 3 × 4 ஼஽ = = = 16

Note: Consider the span CD

12 412 16 × 2 = 32 ஼஽   ̅ = 16× − ஺ × ஺஻ − 2 ஻ሺ ஺஻ + ஻஼ሻ − ஼ × ஻஼ = 6 ൤ ஺஻஺஻  ̅஺ + ஻஼஻஼  ̅஻൨ 0×5− 2 ஻ሺ8ሻ + ஼ ×3= ×3 = 6 ൤62562425 × 15 + 1003 × 13൨ −16 ஻ − 3 ஼ =97.2 ஼ =−ሺ32.64+5.33 ஻ሻ − ஻ × ஻஼ − 2 ஼ሺ ஻஼ + ஼஽ሻ − ஽ × ஼஽ = 6 ൤ ஻஼஻஼  ̅஼ + ஼஽஼஽  ̅஽൨ −3 ஻ − 2 ஼ሺ7ሻ + 10×10×44 = 6 ൤80830 × 15 + 32× 14൨ −3 ஻ − 14 ஼ =101.33−40 −3 ஻ − 14ሾ14ሾ−ሺ32.32.64+5.33 ஻ሻሿ =101.33−40 9 6 ∴ ஻ = ൬61.33−456. 52 71.62ሺ ൰=−5. ∴ ஼ =32.64−5.33ሺ3 −5.−5.52ሻ2ሻ = −3.22

Consider span AB and BC:

Consider span BC and CD:

Reactions: Consider span AB:

Consider span ABC:

෍ ஻ =0 5 ஺ × 5 + ஻ = 1×5×൬2൰ ∴ ஺ = ൬12.5−5.5 52൰=1. 4 20

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෍ ஼ =0 5 ஺ × 8 + ஻ × 3 + ஼ = 1×5×൬2 +3൰+20×1 ∴ ஻ = ൬47.5−14.3 42൰=11.03 Consider span ABCD:

And

஽

෍ ஽ =0 +7൰+20×ሺሺ1 + 4ሻ +3×4× 42 ஺ ×12+ ஻ × 7 + ஼ × 4 + ஽ = 1×5×൬52 +7൰+20× 1 ∴ ஼ = ൬171.5−104. 4 ൰=16.85 ෍ ௏ =0 1×5+220+3 0+3×4+ ×4+110 ஺ + ஻ + ஼ + ஽ = 1×5+ ∴ ஽ =17.72

Maximum bending moment: From the SF diagram the maximum bending moments occurs at:

And at

5 6൰=1. 4 ଵ =1. 4 ൬1.4+3. ଶ =4. 2 ൬124 ൰=1. 4 3

The SF and the BM diagram is shown below figure11 (c) and (d) From here one can easily calculate the maximum bending moments when V =0 and the resultant moments at the mid-span AB and CD. From the results above one can also draw the bending moments like one in figure 9 for the this problem

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20KN

10KN 1m

(a)

A

1KN/m

B

5m

(b)

C

1m 3KN/m

3m

D

E

4m

0 MB MC MD 10

SF DIAGRAM

7.43 4.28

F

(c)

0

1.4

1.4m

1.43m 3.6 7.72

12.57

9.36

M BM DIAGRAM

0.98

(d) 0

0.4

0.37

0.61

3.2

5.5

10

Figure 11: (a) continuous beam, (b) free body diagram, (c) SF Diagram, (d) BM Diagram

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Summary Continuous beams have been employed for ages as they are most reliable for the civil and engineering structures with high level of risk to collapse e.g. bridges, their complex structure make them difficult to analyse during the design phase, one way of analysing them is by making use of Claperon’s Theorem of Three Moments, whereby for each inter mediate support an equation is developed in terms of three moments, thus we get as many equation as there are unknowns in order to analysis the beam, this method was demonstrated in this paper by making used of examples.

References

ሾ1ሿ ሾ2ሿ ሾ3ሿ 4

JG Drotsky, Strenght of materials for technologist. 1997. Tech Books

H.P. Gavin, The three-moment equation for continuous beam analysis, Uncertainty, Design and Optimization. CE130L. Spring 2009.Duke University. Dr. Amlan K. Sengupta, Prof. Devdas Menon, Pre-stressed Concrete Structures, Indiana Institute of Technology Madras Kharagpur, The three moment’s equation for continuous beams analysis, Version 2 IIT

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APPENDIX A Revision on simple supported beams Figure A1: Simple beam uniform distributed load W

L R

R L/2

L/2 SF Diagram

Vmax = R = WL/2

BM Diagram

2

Mmax = WL /8

The area under the curve for a bending moment is calculated as follows: Consider the diagram below: W

L x y y x dx

The bending moment at any point along the span is given by:

௫= Consider the strip for the above figure: The Area (Audl) of a strip is given by:

The total area is:

= 2ሺ−ሻ

௨ௗ௟ = ℎ =  ்௢௧௔௟ = න଴ 2 ሺ − ሻ 24

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ଶ   = න − න ்௢௧௔௟ ଴ 2 ଴ 2 ଶ ଷ ்௢௧௔௟ = ቤ 4 − 6 ቤ 00 ଷ ଷ 3 − 2 = ்௢௧௔௟ 12 ∴ ்௢௧௔௟ = 12ଷ  ̅ = 2

The centriod:

Figure A2: Simple beam concentrated load at the centre P

L/2

L/2

R

R

L

SFD

Vmax = R = P/2

BMD Mmax = PL/4

The area under the curve for a bending moment is calculated as follows: Consider the diagram below:

x y y1

y2 y x dx

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Continuous beams

March: 2012

From the above figure it can be seen that are two curves, thus:

ଵ= 2

0≤ ≤ 2 ଶ = ሺ 2− ሻ 2 ≤ ≤

And

The Area (Apl) of a strip is given by:

The total area is:

௣௟ =

= ଵ +ଶ ଶ  = න + න ்௢௧௔௟ ଴ ଵ ଶ ଶ ଶ  ்௢௧௔௟ = න଴ 2 + නଶ ሺ 2− ሻ

ଶ ଶ 0 2 = ቤ + ฬ − ்௢௧௔௟ 4 0 2 4 ቤ 0 0 2 2 ଶ ଶ ଶ ଶ ଶ = + − − + ்௢௧௔௟ 16 2 4 4 16 ଶ ∴ ்௢௧௔௟ = 8 = 12 × ℎ ଶ 1 ∴ = 2× × 4 = 8

The area can also be calculated from the bending moment diagram as follows:

The centriod:

 ̅ = 2 Figure A2: Simple beam concentrated load at any point

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Continuous beams

March: 2012

P

a

b L

V1 = Pb/L

V2 = Pa/L

Mmax = Pab/L a

b L

The area for r the above figure can be calculated as follows:

The centriod:

= 12 × ℎ = 12 × × = 2   ̅ = ሺ +3 ሻ

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