Three Moment Derivation

11

Continuous Beam Objectives:

Derive the Clapeyron’s Clapeyron’s theorem of three moments Analyze continuous beam with different moment of inertia with unyielding supports Analyze the continuous beam with different moment of inertia in different spans along with support settlements using three moment equation.

11.0 11. 0

INTR IN TROD ODUCT UCTION ION

A beam is generally supported on a hinge at one end and a roller bearing at the other end. The reactions are determined by using static equilibrium equations. Such as beam is a statically determinate structure. If the ends of the beam are restrained/clamped/encastre/fixed the n the moments are included at the ends by these restraints and this moments make the structural element to be a statically indeterminate structure or a redundant structure. These restraints make the slope s at the ends zero and hence in a fixed beam, the deflection and slopes are zero at the supports.

A continuous beam is one having more than one span and it is carried by several supports (minimum of three supports). supports). Contin Continuous uous beams are widely used in bridge construction. construction. Consi Consider der a three bay of a building which carries the loads W 1 W 2 and W 3 in two ways. ,

W 1

W 2

FIG. 11a Simply supported beam

FIG. 11b Bending moment diagrams

W 3

Continuous Beam

W 1

W 2

A

•

545

W 3

C

B

D

FIG. 11c Continuous beam

+ +

−

−

+

FIG. 11d Bending moment diagram

If the load is carried by the first case then the reactions of individ ual beams can be obtained by equili equ ilibri brium um equ equati ations ons alo alone. ne. The bea beam m defl deflect ectss in the respecti respective span and does not depend on the influence of adjacent spans. In the second case, the equilibrium equations alone would not be sufficient to determine the end moments. The slope at an interior support B must be same on either side of the support. The magnitude of the slope can be influenced by not only the load on the spans e ither side of it but the entire loads on

the span of the continu continuous ous beam. The redunda redundants nts could could be the rea reacti ctions ons or the bending bending moments moments over ov er the sup support port.. Cla Clapey peyron ron (1857) obtained obtained the com compat patibi ibilit lity y equa equatio tion n in ter term m of the end slo slopes pes of the adj adjace acent nt spans. spans. Thi Thiss equ equati ation on is cal called led theorem theorem of thr three ee moments which contain three of the unknowns. unkno wns. It giv gives es the relat relationshi ionship p betwe between en the loadi loading ng and the moments over three adjacent supports at the same level.

11.1

DERIVATION OF CLAPEYR DERIVA CLAPEYRON’S ON’S THEOREM (THEOREM OF THREE MOMENTS)

of a continuous beam with with two spans. The settlement settlement Figure 11e shows two adjacent spans AB and BC of of the supports are ∆ A ∆ B and ∆C and the deflected deflected shape of the beam is shown in A B C (Fig. 11f). 

,

l 1 A

l 2 B

FIG. 11e

C





Continuous Beam

W 1

W 2

A

•

545

W 3

C

B

D

FIG. 11c Continuous beam

+ +

−

−

+

FIG. 11d Bending moment diagram

If the load is carried by the first case then the reactions of individ ual beams can be obtained by equili equ ilibri brium um equ equati ations ons alo alone. ne. The bea beam m defl deflect ectss in the respecti respective span and does not depend on the influence of adjacent spans. In the second case, the equilibrium equations alone would not be sufficient to determine the end moments. The slope at an interior support B must be same on either side of the support. The magnitude of the slope can be influenced by not only the load on the spans e ither side of it but the entire loads on

the span of the continu continuous ous beam. The redunda redundants nts could could be the rea reacti ctions ons or the bending bending moments moments over ov er the sup support port.. Cla Clapey peyron ron (1857) obtained obtained the com compat patibi ibilit lity y equa equatio tion n in ter term m of the end slo slopes pes of the adj adjace acent nt spans. spans. Thi Thiss equ equati ation on is cal called led theorem theorem of thr three ee moments which contain three of the unknowns. unkno wns. It giv gives es the relat relationshi ionship p betwe between en the loadi loading ng and the moments over three adjacent supports at the same level.

11.1

DERIVATION OF CLAPEYR DERIVA CLAPEYRON’S ON’S THEOREM (THEOREM OF THREE MOMENTS)

of a continuous beam with with two spans. The settlement settlement Figure 11e shows two adjacent spans AB and BC of of the supports are ∆ A ∆ B and ∆C and the deflected deflected shape of the beam is shown in A B C (Fig. 11f). 

,

l 1 A

l 2 B

FIG. 11e

C





Continuous Beam

•

549 547

The primary structure is consisting of simply supported beams with imaginery hinges over each supportt (Fig 11g). Fig 11h shows the simply beam bending suppor bending moment diagrams and Fig 11i shows the support moment diagram for the supports. A compatibility equation is derived based on the fact that the end slopes of adjacent spans are equal in magnitude but opposite in sign. Using Fig 11f and the proper property ty similar triangles triangles GD GD DB ¢ 

∆ B − ∆ A + δ B A

=

l 1 δ B A

i . e.

l 1

+

δ B C

l 2

CF HF B B F ¢ F

==

=



∆C − ∆ B + δ B C

l 2 ∆ A − ∆ B

l 1

+

∆C − ∆ B

(i)

l 2

The displacements are obtained as follows. 11 Ï 11 l 1l 1 11 ¸ B B = ¯ δ A x M l · + + ◊ ◊ l12◊l 22 /l 13/ 3 ˝ A x M l M B l B2 · d A A = E I Ì 11 11 2 A A1 1 3 ++ 2 M E11 I 11 Ó 2 3 2 ˛ 1

B

δC =

E 2 I 2

 

1 2

¯ 2 + M C A2 x C l 2 ·

 

(ii)

1 + M B l 2 · 2l 2 /3 3 2

l 2

Combining the equations (i) and (ii)



M A l 1 l 1 l 2 + 2 M B + E 1 I 1 E 1 I 1 E 2 I 2

= 6



∆ A − ∆ B

l 1

+



∆C − ∆ B

l 2



l 2 A1 x A2 x ¯ 1 ¯ 2 + M C +6 + C E 2 I 2 E 1 I 1 l 1 E 2 I 2 l 1





(iii)

The above equation is called as Clapeyron’s equation of three moments. = constant constant); ); when there are no settl settlement ement of In a simplified form of an uniform beam section ( E ( EI I = supports



A1 x ¯ 1 A 2 x ¯ 2 = + 6 M A l 1 + 2 M B (l 1 + l 2 ) + M C l − C 2 l 1 l 2



(iv)

It is to be mentioned here that x¯ 1 and x¯ 2 are measured outwards in each span from the loads to the ends.

11.1.1 11.1. 1

Procedure Procedur e for Analysi Analysing ng the Continu Continuous ous Beams Beams using Theore Theorem m of Three Moments

(1) Draw simple simple beam moment moment diagram for each span span of the beam. Comput Computee the area of the above diagrams viz, A1 , A2 . . . An and locate the centroid of such diagrams x¯ 1 , x ¯ 2 . . . x ¯ n . It mus mustt be reremembered that the distances x¯ 1 , x¯ 2 . . . x ¯ n are the centroidal distances measured towards the ends of each span as shown in Fig. 11j.

548

•

Basic Structural Analysis A1

A2

−

−

x1

x2

FIG. 11j Simple beam moment diagrams

(2) Identify Identify the support moments moments which are to be determined determined viz, M A , M B and M C C (3) Apply three moment moment equation equation for each pair of spans which which results in an equation equation or equati equations ons which are to be solv solved ed simultaneousl simultaneously y. If the beam is of uniform section section ( E EI I = constant) and no support settlements apply equation (iv) and in case the beam is non-uniform and the support settles/raises apply equation (iii). (4) The solution solution of the equations equations gives the values values of the suppor supportt moments and the bending moment diagram can be drawn. (5) The reactions reactions at the support supportss and the shear force diagram diagram can be obtain obtained ed by using equilibrium equilibrium equations.

11.2 11.2.1 11.2. 1

APPLICATION OF APPLICATION OF THREE MOMENT EQU EQUA ATION IN CASE OF BEAMS WHEN ONE OR BOTH OF THE ENDS ARE FIXED Propped Pro pped Canti Cantileve leverr Beam

Consider the propped cantilever beam of span AB AB,, which is fixed at A at A and supported on a prop at B . It is subjected to uniformly distributed load over the entire span. The fixed end moment at the support A

can be determined by using theorem of three moments. w/m l ′

A

zero span

B

A

FIG. 11k Propped cantilever beam

As the A is fixed support, extend the beam form A to A of span ‘zero length’ and A is simply supported. 



(wl 2 /8). The centroid (1) The simple beam beam moment diagram diagram is a parabo parabola la with a central ordinate ordinate of ( centroid of this bending moment diagram (symmetrical parabola) is at a distance ‘l /2’ from the supports A and B. wl 2/8

A

B

Continuous Beam

It’s area is A =

wl 2

2  

•

wl 3

(l ) = . 3 8 12 (2) The support support moment diagram diagram is drawn drawn as M A l

FIG. 11m Pure moment diagram

(3) Apply three three moment theorem theorem for the the span AB. 

M A (0) + 2 M A (0 + l ) + 0 = −6 ∴

wl 3

l

12

2

  

M A = − wl 2 /8

(4) The support support reactions reactions are computed computed by drawing the free body diagram diagram as w/m

wl 2/8

A

B l V B

V A

FIG. 11n Free body diagram

∑ V = 0; ∑ M A = 0;

V A + V B = wl wl 2

−

8

+

wl 2

2

−

V B l = = 0

and hence 3wl 8 5wl V A = 8 V B =

(5) Using the reactions, reactions, the shear force force diagram and bending moment moment diagrams are obtained obtained as

( ( 5wl

B

8

( ( 3wl

A

8

FIG. 11o Shear force diagram

549

550

•

Basic Structural Analysis

The point of contraflexure is determined by equating the bending moment expression to zero and

hence 5wl wx2 wl 2 x− − = 0 8 2 8 l 2 + 4 x2 − 5lx = 0 Solving the above equation we get x = l and x = 0.25l The location of maximum positive bending moment from support A is obtained by equating the

shear force to zero.

5wl − wx = 0 8 5l x = 8

At this location, the maximum positive bending moment is obt ained from Max + ve BM =

wl 2

−

M C = −

8

+

 5  5  wl

l

8

8

−

w(5l /8)2

2

wl 2

25wl 2 25wl 2 9wl 2 − + = = 0.07wl 2 8 64 128 128 0.07 wl 2 +ve

0.25 l

A

B C

−

3l /8

( ) wl 2 8

FIG. 11p Bending moment diagram

11.2.2

Beams with Both the Ends Fixed

Consider a beam AB of span l is fixed at both the ends. The beam is carrying a concentrated load of W at a distance of ‘l /3’ from the fixed end A . As the end A is a fixed support, extend this A to A  of span ( l  ) of zero length and is also simply supported at A  . Likewise the end B is extended to B  . The simply supported bending moment diagram is drawn with the maximum ordinate as W × (l /3) × (2l /3) = 2W l /9. l

The centroid of the unsymmetrical triangle is shown in Fig. 11.3 j.

Continuous Beam W

l /3

551

2l /3 l

l′ = O A′

•

l′ = O B

A

B1

FIG. 11q Fixed beam

2Wl /9

FIG. 11r Simple beam moment diagram a

b

CG

( (

( (

l +b

l +a

3

3

FIG. 11s Centroid of an unsymmetrical triangle

The centroid of the simply supported BMD is obtained using the above as from B. The area of the bending moment diagram is

1 2  2

(l )

W l

9

=

W l 2

9

4  l

9

from A and

5  l

9

.

The support moment diagram can be drawn by identifying the support moments as M A and M B . Thus M A

M B l

FIG. 11t Pure moment diagram

Applying three moment theorem for a pair of spans of A AB (Ref Eq (iv)) 

>

M A (0) + 2 M A (0 + l ) + M B (l ) = 0 − 6

W l 2

  5  9

l

9

×

1/l

2 M A + M B = − 0.37 W l

552

•


Considering the next pair of spans ABB





>

M A l + 2 M B (l + 0) + M B (0) = −6

W l 2

  4  l

9

9

M A + 2 M B = −0.296 W l

Thus the support moments are obtained by solving the above equations M A = −0.148 W l M B = −0.074 W l

Free body diagram to determine the reactions 0.148 Wl

W

l /3

0.074 Wl

2l /3

C V B

V A

FIG. 11u

Using the static equilibrium;

∑ V = O;

V A + V B = W

∑ M A = O;

−

0.148 W l + W

 l

3

−

V B l + 0.074 W l = O

V B = 0.26W V A = 0.74W

0.74 W

+

−

0.26 W

FIG. 11v Shearforce diagram 0.0986 Wl +

−

−

−

0.148 Wl

−

FIG. 11w Bending moment diagram

0.074 Wl

Continuous Beam

11.3

•

553

NUMERICAL EXAMPLES ON CONTINUOUS BEAMS

A continuous beam ABC is simply supported at A and C and continuous over support B with AB = 4m and BC = 5m. A uniformly distributed load of 10 kN/m is acting over the beam. The moment of inertia is I throughout the span. Analyse the continuous beam and draw SF D and BMD. EXAMPLE 11.1:

10 kN/m

4m

A

5m

B

C

FIG. 11.1a

20

31.25 kNm

A

A

1

2

A

C

− x

− x

1

2

FIG. 11.1b Simple beam moment diagram

M B

FIG. 11.1c Pure moment diagram

Properties of the simple beam BMD

2 2 × 4 × 20 = 53.33 kNm 3 ¯ 1 = 2m x l 1 = 4m

A1 =

2 2 × 5 × 31.25 = 104.17 kNm 3 x¯ 2 = 2.5m l 2 = 5.0m A2 =

Applying three moment equation for the span ABC >

>

M A l 1 + 2 M B (l 1 + l 2 ) + M C l 2 = −

 ¯ ¯  + 6  53.33 2 104.17

2 M B (4 + 5) = −6

A1 x1

A 2 x2

l 1

l 2

×

4

+

18 M B = −6(26.67 + 52.1) M B = −26.26 kNm.

5

×

2.5



Continuous Beam EXAMPLE 11.2:

Analyse the continuous beam by three moment theorem. Draw SF D and BMD. 10 kN 2

10 kN 4m

D

3

3m E 6 m

6m B

A

C

FIG. 11.2a

S OLUTION

The simple beam moment diagram is drawn as M D = Wab /l =

10 × 2 × 4 = 13.33 kNm 6

M E = W l /4 = 10 ×

6 = 15 kNm 4

15 kNm

13.33

A

D

E

B

C


M B

A

C

B


Properties of the simple beam BMD

1 (6)13.33 = 40.0 2 6+2 = 2.67 m ¯ 1 = x 3 l 1 = 6 m A1 =

•

A2 =

1 × 6 × 15 = 45 2

x¯2 = 3 m l 2 = 6 m

555

560

•

Basic Structural Analysis 50 60 kNm +

2 +

A

B

C


M B

A

C


Properties of the simple beam BMD A1 =

2 2 × 5 × 50 = 167.5 kNm 3

1 2 × 10 × 60 = 300 kNm 2 10 + 6 x2 = = 5.33 m 3

A2 =

x1 = 2.5m l 1 = 5.0m

l 2 = 10 m >

>

5 M A + 2 M B (5 + 10) + 10 M C = −6

 167.5

×

2.5

5.0

300 × 5.33 + 10



30 M B = −6 (83.75 + 159.9) M B = −48.73 kNm

Properties of the simple beam BMD ∑ V = 0; ∑ M = 0;

V A + V B1 = 80

5V A + 49 −

V A = 30.2 kN V B1 = 49.8 kN

(i)

V B2 + V C = 25

(ii)

10V B2 − 25(6) − 49 = 0

2

16(5) =0 2

V B2 = 19.9 kN V C = 5.1 kN

(iii) (iv)

562

•


The simple beam moments are M D = 20 × 102 /8 = 250 kNm M E =

50 × 6 × 2 = 75 kNm 8

250 kNm 75 D

A

C

E

B


Properties of simple beam BMD A1 =

2 2 × 10 × 250 = 1666.7 kNm 3

1 2 × 8 × 75 = 300 kNm 2 8+2 x2 = = 3.33 m 3

A2 =

x1 = 5 m l 1 = 10 m

l 2 = 8.0 m

Since A is fixed imagine a span A A of zero length and A  as simply supported. Apply three moment theorem for the spans A AB. M A (0) + 2 M A (0 + 10) + M B (10) = − 6

 1666.7

×

5

10



+0

20 M A + 10 M B = − 5000 2 M A + M B = − 500

(i)

Apply three moment theorem for the spans ABC. >

M A (10) + 2 M B (10 + 8) + 8 M C = − 6

 1666.7

×

10

5

300 × 3.33 + 8



10 M A + 36 M B = − 6(833.35 + 124.875) 10 M A + 36 M B = − 5749.35 Solving equations (i) and (ii) M A = − 197.6 kNm M B = − 104.8 kNm

(ii)

568 570

•


Free body diagram of spans AB and BC

10 kNm

16 kNm 2

A

4 kN/m

16 kNm

24 kN 2

B

C

6m

B

V C

V B2 V

V

A

B1

FIG. 11.7e

FIG. 11.7d

Static equilibrium of AB

∑ V = 0; ∑ M = 0;

Static equilibrium of BC

V A + V B1 = 24

(i)

V B2 + V C = 24

4V A + 16 − 10 − 48 = 0

(ii)

∑ M B = 0;

V A = 10.5 kN.

16 + 6V C − 4 ×

V B1 = 13.5.

V C = 9.3 kN. V B2 = 14.7 kN.

(iii) 62 2

= 0

10.5 9.3

+

x

+

B

A

C

D

−

−

14.7 kN 4m

13.5

6m

FIG. 11.7f Shear force diagram

The zero shear location in span BC is 14.7 − 4 x = 0 x = 3.67 m . ∴

Maximum +ve BM = 14.7(3.67) − 4(3.67)2 /2 − 16 = 11 kNm 24 10 −

16 kNm

+

11 +

−

C A

B

3.67 m

FIG. 11.7g Bending moment diagram

572 570


•

Solving (i) and (ii); From eq (i); M B = −2 M A and putting in eq (ii) M A − 12 M A = −60 ∴

M A = 5.45 kNm. M B = −10.9 kNm.

Free body diagram of span AB and BC 10.9 kNm

5.45 kNm A

B

B

2m A

V

V C

V

B1

B2

FIG 11.8e

Static equilibrium of span AB

Static equilibrium of span BC

∑ V = 0; =

C

2m

FIG 11.8d

V A + V B1

30 kN/m

10.9 kNm

∑ V = 0 0

(i)

V B2 + V C = 60

∑ M B = 0;

(iii)

∑ M B = 0;

5.45 + 10.9 + 2V A = 0

30 × 22 −10.9 + 2V B2 − 2 V B2 = 35.5 kN V C = 24.5 kN

(ii)

V A = −8.2 kN V B1 = +8.2 kN

=

35.5 kN/m A

C

+

B

−

−

8.2 kN

24.5


10.9 9.55 −

A +

B

D

5.45 kNm


C

0

572

•


Applying three moment theorem for the span ABC M A

5 I

5 6  +  240 12.5.67

+ 2 M B = −6

I

I

−

30 ×

6 1.5 I

360 × 3 + 6 × 1.5 I

×

5 I



5 M A + 18 M B − 120 = −6 (128.16 + 120) 5 M A + 18 M B = −1488.96 + 120 5 M A + 18 M B = −1368.96

(ii)

Solving equations (i) and (ii) M A = −33.76 kNm. M B = −66.67 kNm. Shear forces and moments in members AB and BC . Member AB 80 kN 3m

66.67 kNm

2m D

33.76 kNm

V BA

V

AB

FIG. 11.9d

∑ V = 0; ∑ M B = 0;

V AB + V BA = 80

(i)

5V AB + 66.67 − 33.76 − 80(2) = 0

(ii)

V AB = 25.42 ∴

V BA = 54.58

M D = − 33.76 + 25.42(3) = 42.5 kNm. Member BC 20 kN/m 30 kNm

66.67 kNm 6m V

V

BC

CB

FIG. 11.9e

574

•

Basic Structural Analysis D

A

B

E

C

FIG. 11.9i Elastic curve

A continuous beam ABCD is simply supported at A and continuous over spans B and C . The span AB is 6 m and BC are of length 6 m respectively. An overhang CD is of 1 metre length. A concentrated load of 20 kN is acting at 4 m from support A. An uniformly distributed load of 10 kN/m is acting on the span BC . A concentrated load of 10 kN is acting at D . EXAMPLE 11.10:

20 kN 4m

10 kN/m

E

6m

A

10 kN

2m F 6m

B

I

1m C 2 I

2 I

FIG. 11.10a

The simple beam moments are M E

=

M F

=

M C

=

20 × 4 × 2 26.7 kNm 6 62 10 × 45.0 kNm 8 −10 × 1 −10 kNm =

=

=

45 kNm

26.7

A

E

B

F

FIG. 11.10b Simply suppored BMD

C

D

Continuous Beam

575

M C

M B

A

•

B

C

D


Considering spans ABC

1 2 × 6 × 26.7 = 80 .1kNm 2 6+4 = 3.33 m. x1 = 3 l 1 = 6 m . A1 =

6

>

M A + 2 M B I

Properties the simple beam BMD 2 A2 = × 6 × 45 = 180kNm2 3 x2 = 3 m . l 2 = 6 m .

6 6 

(−10 > )

6 M C + + 2 I I 2 I

= −6

 80.1

×

3.33

6

180 × 3 + 6×2



18 M B − 30 = −6 (44.45 + 45) M B = 28.15 kNm. Shear force and bending moment values for the spans AB and BC 20 kN 4

28.15 kNm 2

A

E V AB

B V BA

FIG. 11.10d

Using equilibrium conditions;

∑ V = 0; ∑ M = 0;

V AB + V BA = 20

6V AB + 28.15 − 20(2) = 0 V AB = 1.98 V BA = 18.02

∴

M E = V AB (4) = 7.9 kNm

(i) (ii)

578 576

• •

Basic Structural Analysis 10 kN/m

28.15 kNm

10 kNm

B

C

6m V

V

BC

CB

FIG. 11.10e

Using equilibrium conditions;

∑ V = 0; ∑ M C = 0

V BC + V CB = 10(6) = 60

(iii)

62 =0 10 − 28.15 + 6 V BC − 10 × 2

(iv)

V BC = 33 kN. V CB = 27 kN.

33 1.98

10 kN +

– A

+

B

6m

6m

–

C

D

–

27 18.02


7.9 +

A

26 28.15

+ 10 kNm

− B

− E


C

D


30 kN 4

40 kN

2

2

20 kN

2 F

E A

B

2 m D C

FIG. 11.11a

S OLUTION

The simple beam moments at E and F are M E = M F =

Wab l W l

4

30 × 4 × 2 = 40 kNm 6 40 × 4 = 40 kNm 4

=

=

40

40 kNm

+

+

E

B

F

C

D

FIG. 11.11b Simply supported beam BMD

M

M B

C

−

A

577

Analyse the continuous beam shown in figure by three moment theorem. Draw

SF D & BMD.

A

•

B

C


Properties of simply supported beam BMD 1 1 A1 = × 6 × 40 = 120 kNm2 A2 = × 4 × 40 = 80 kNm2 2 2 6+4 x1 = x2 = 2 m . = 3.33 3 l 1 = 6.00 l 2 = 4.0 m.

D

578

•


Applying three moment theorem for spans AB & BC

1

3.33 80(2) × 6 × 40 × 6 M A+ 2 M B (6 + 4) + 4 M C = −6 + 2 6.00 4 >

>



20 M B − 160 = −6 (66.6 + 40) 20 M B = −479.6 M B = −23.98 kNm Free Body diagrams 40 kN

30 kN 4m

2m

2m

24 kNm

2m F

E V A

V B2

V B1

V

C

FIG. 11.11e

FIG. 11.11d

∑ V = 0; ∑ M B = 0;

40 kNm

V A + V B1 = 30

(i)

6V A + 24 − 30(2) = 0

(ii)

∑ V = 0; ∑ M B = 0;

V A = 6 kN V B1 = 24 kN

∴

V B2 + V C = 40

(iii)

40 + 40(2) − 24 − 4V C = 0

V C = 24 kN V B2 = 16 kN

20 16

+ 6

+

+ 2m

A

4m

E

2m B

2m

–

C

F

–

24 kN

24


2m D

(iv)

Continuous Beam Continuous Beam 40

40 −24

E

581 579

40 kNm

−

+

A

•

F

B

C

D

FIG. 11.11g Bending moment diagram EXAMPLE 11.12:

Draw the shear force diagram and bending moment diagram for the beam shown

in figure. 10 kN/m A'

3m

A

6m

B

C

FIG. 11.12a

45 kNm +

11.25

+

A

B

C

FIG. 11.12b Simply supported beam BMD

S OLUTION As the end A is fixed, imagine an imaginery span A A of zero length with no load and A is simply supported.

Considering the span A  AB

  +  0

>

M A (0) + 2 M A (0 + 3) + 3 M B = − 6

   

2 × 3 × 11.25 1.5 3 3

6 M A + 3 M B = − 67.5

(i)

Considering the span ABC >

3 M A + 2 M B (3 + 6) + 6 M C = − 6



22.5 × 1.5 3

+

180 × 3 6



3 M A + 18 M B = − 6 (11.25 + 90) 3 M A + 18 M B = − 607.5

(ii)

582 580

•


Solving (i) & (ii) 34.77 kNm. M A = 6.14 kNm. M B =

−

The BMD is drawn using the above end moments as 45 kNm 11.25 34.77 +

A

+

− C

B

+

6.14 kNm FIG. 11.12c Bending

moment diagram

Shear force and BM values for spans AB and BC Static equilibrium of AB 6.14 kNm

10 kN/m

34.77 kNm B

3m

A V AB

V BA FIG. 11.12d

∑ V = 0; ∑ M

B

=

0;

V AB + V BA =

30

32 34.77 + 6.14 + 3V AB − 10 × 2

(i) =

0

(ii)

1.36 kN. V BA = 28.64 kN. V AB =


10 kN/m

34.77 kNm B

C

6m x

V BC

2

V

CB

FIG. 11.12e

Continuous Beam

∑ V = 0; ∑ M C = 0;

V BC + V CB = 10(6) = 60 −

34.77 + 6V BC

62 − 10 × = 0 2

Maximum positive BM is span AB The location of zero shear force in AB zone is 1.36 − 10 x1 = 0. x1 = 0.135 m

35.8 1.36

x D x

B

2

C

E

1

28.64

24.2 kN


34.77 kNm 6.28 29.3 +

A

+

−

B

583 581

(iii)

V BC = 35.8 kN . V CB = 24.2 kN .

A

•

+

C

E

2.42 m 6.14


M X1X1 = 6.19 + 1.35 (0.135) − 10 (0.135)2 /2 = 6.28 kNm.

582

•


Maximum positive BM in span BC

The location of zero shear force in BC zone is 24.2 − 10 x2 = 0 x2 = 2.42 m M ×2×2 = 24.2(2.42) − 10(2.42)2 /2

= 29.3 kNm.

A continuous beam ABCD is of uniform section. It is fixed at A , simply supported at B and C and CD is an overhang. AB = BC = 5 m and CD = 2 m. If a concentrated load of 30 kN acts at D , determine the moments and reactions at A , B and C . Sketch the shear force and bending moment diagram and mark in the salient values. EXAMPLE 11.13:

30 kN

5m A'

A

5m B

2m C

D

FIG. 11.13a

S OLUTION As the end A is fixed imagine a imaginery span A  A of zero length and A is simply supported. Apply three moment theorem for the spans A  AB >

M A (0) + 2 M A (0 + 5) + 5 M B = − 6(0 + 0)

10 M A + 5 M B = 0 2 M A + M B = 0

(i)

Apply three moment theorem for the spans ABC 5 M A + 2 M B (5 + 5) − 60(5) = − 6(0 + 0) 5 M A + 20 M B = 300 Solving (i) and (ii) M A = −8.57 kNm. M B = +17.14 kNm.

(ii)

Continuous Beam

•

583

Shear force and BM values for spans AB and BC Span AB 8.57 kNm

17.14 kNm

A

B

5m V

V

AB

BA

FIG. 11.13b

∑ V = 0; ∑ M B = 0;

V AB + V BA = 0

(i)

5V AB − 8.57 − 17.14 = 0 V AB = 5.14 kN. V BA = −5.14 kN.

Span BC 17.14 kNm

60 kNm

B

C

5m V BC

V CB

FIG. 11.13c

∑ V = 0; ∑ M = 0;

V BC + V CB

=

0

60.0 + 17.14 + 5V BC = 0 V BC = −15.43 kN. V CB = +15.43 kN. 30 kN

5.14 +

+

A

B

C

−

15.43

FIG. 11.13d Shear force diagram

D

584

•

Basic Structural Analysis 60 8.57

A

B

C

D

17.10 kNm

FIG. 11.13e Bending moment diagram

Analyse the continuous beam by the theorem of three moments. Draw neat sketches of SF D and BMD. Clearly indicate all the salient values. EXAMPLE 11.14:

20 KN/m

16 kN

2m

40 kN

2m

2m

4m A'

4m

A

C

D

E

FIG. 11.14a

S OLUTION

The simple beam moments are M B = M D =

wl 2

8 wl

4

+

=

42 4 = 20 × + 16 × 4 8 4

wl

40 ×

4 4

=

B

56 kNm

40 kNm

40

A

=

40

C

D


E

Continuous Beam

•

585

16

A

E

C

B

FIG. 11.14c Simple beam moment diagram

Properties of simple beam BMD 2 1 × 4 × 40 + × 4 × 16 = 138.67 3 2 l 1 = 4 m

1 2 × 4 × 40 = 80 kNm 2 x2 = 2 m

A1 =

A2 =

x1 = 2 m

l 2 = 4 m

Applying three moment theorem for span A  AC



M A l 1 + 2 M A l 1 + l 2



+ M C l 2 = −

 6

A1 x1 l 1

+

A 2 x2 l 2

2 M A (4) + 4 M C = − 6 × 138.67 ×

2 4

 

8 M A + 4 M C = − 416.01

(i)

Applying theorem of three moments for the spans ACE >

M A (4) + 2 M C (4 + 4) + M E (4) = − 6

 138.67

×

2

4

80 × 2 + 4



4 M A + 16 M C = − 6(109.335) = − 656.01

(ii)

M A = − 36 kNm M C = − 32 kNm

Solving Equations (i) and (ii)

Free body diagram 36 kNm

2m

16 kN

20 kN/m

32 kNm

A

B

A

V

C1

C

FIG. 11.14d

40 kN 2m 4m

4m V

32 kNm

V E

V C2 C

D

FIG. 11.14e

E

588 586


•

Static equilibrium of AC

Static equilibrium of CE

∑ V = 0;

∑ V = 0;

V A + V C 1 = 16 + 4(20) = 96

∑ M A = 0 −

V C 2 + V E = 40

(i)

36 + 32 + 4V A − 16(2) − 20 ×

4

∑ M E = 0;

2

2

(iii)

=0

32 + 4V C 2 − 40(2) = 0

(ii)

(iv)

−

V A = 49 kN V C 1 = 47 kN

∴

V C 2 = 28 kN V E = 12 kN

49 28 +

9

+

E

D

B A

−

C

−

7 47 kN

12

FIG. 11.14f Shear force Diagrams

56

40 kNm

+

36

32

−

−

A

B

+

C

D

E


EXAMPLE

11.15:

Sketch the BMD for the continuous beam shown in figure. 60 kN 1m

3m 4m

O A'

30 kN

20 kN/m

A

D

3 I

B

FIG. 11.15a

E 4 I

1m C

F

Continuous Beam

587

40

45

A

•

D

B

E

C

FIG. 11.15b Simple beam moment diagram M

B

M C

M

A

A

B

C


S OLUTION

Properties of the simple beam BMD 1 2 × 4 × 45 = 90 kNm 2 4+1 = 1.67 m x1 = 3 l 1 = 4 m A1 =

A2 =

2 2 × 4 × 40 = 106.7 kNm 3

x2 = 2 m l 2 = 4 m

Since A is fixed assume an imaginery span of A A of zero length with no loading. Assume A  as simply supported. Apply three moment equation for the span A AB, M A (0) + 2 M A

      0 + 4  +  4  = 6  0 + 90 M B

3 I

−

3 I

4

×

×

8 M A + 4 M B = − 315

 2.33 

3 I

(i)

Applying three moment theorem for the spans AB and BC ; M A

4 3 I

+ 2 M B

4



 





4 4 90 × 1.67 106.7 × 2 + + M C = − 6 + . 3 I 4 I 4 I 4 × 3 I 4 × 4 I

1.33 M A + 2 M B (1.33 + 1.0) − 30 = − 6 (12.525 + 13.338) 1.33 M A + 4.66 M B = 30 − (25.863)6 1.33 M A + 4.66 M B = − 125.18 Solving (i) and (ii); M A = −30.3kNm. M B = −18.1kNm.

(ii)

588 590


•

Free body diagrams of span AB and BC 60 kN

30.3 kNm

1m

18.1 kNm

3m

30 kNm

B B

A

B1

C

4m

V B2

V

V A

V C

FIG 11.15e

FIG 11.15d

Static equilibrium of AB


∑ V = 0;

∑ V = 0;

V A + V B = 60

V B2 + V C = 80

(i)

∑ M B = 0; −

20 kN/m

18.1 kNm

(iii)

∑ M B = 0 42 −18.10 + 30 + 4V = 0 B2 − 20 × 2 V B2 = 37.02 kN.

30.3 + 18.1 + 4V A − 60(3) = 0

V A = 48.05 kN

(ii)

V B2 = 11.95 kN

V C = 42.98 kN.

48.05 30 37.02

+

+

+

A

D

C

E

B

−

F

−

1.85 11.95

42.98 kN


+

30.3

18.1

+

15.95 30.0 kNm

45.0 −

−

−

A

D

B

E


C

F

Continuous Beam

•

589

Analyse the continuous beam by three moment theorem. E is constant. Draw the bending moment diagram. EXAMPLE 11.16:

20 kN/m

80 kN

2m

1m 4 m F

3m E

A'

3 I

3 I

A

2 I

B

1m

D

2 I

C

FIG. 11.16a

40 kNm

53.33

A

E

B

F

C

D

FIG. 11.16b Simple beam BMD M B

10 kNm M A B

A

D

C


S OLUTION As the end A is fixed assume an imaginery span A of zero length with no load and A is simply supported;

Apply three moment theorem for spans A  AB >

M A

0 3 I

+ 2 M A

  0 3

>

I

    3 3 80 + + = 6 0+ 3 3 3 I

M B

I

−

1.33 × 3 I

×



2 M A + M B = − 70.93

(i)

Applying three moment theorem for spans ABC M A

3 3 I

+ 2 M B

3

 4

4 + 3 I 2 I

−

10

2 I

= − 6

 80

1.67 106.67 × 2 + 3 × 3 I 2 I × 4 ×



M A + 6 M B − 20 = − 6(14.84 + 26.67) M A + 6 M B = − 249.06

(ii)

590

•


Solving (i) and (ii) M A = −16.05 kNm. M B = −38.84 kNm. Free Body diagrams of span AB and BC

16.05 kNm

80 kN 2m E

V

A

B1

4m

V B2

V C

FIG. 11.16e

FIG. 11.16d

∑ V = 0;

∑ V = 0;

V A + V B1 = 80

(i)

∑ M B = 0 −

10 kNm

38.84 kNm

1m

V

20 kN/m

38.84 kNm

V B2 + V c = 80

(iii)

∑ M C = 0

16.05 + 38.84 − 80(1) + 3V A = 0

(ii)

20 × 42 10 − 38.84 + 4V B2 − =0 2

V A = +19.07 kN

V B2 = 47.21 kN

V B1 = +60.93 kN

∴

V C = 32.79 kN

M E = −17.86 + 20.88(2)

= 23.9 kNm The location of zero shear in zone BC is obtained from 47.21 − 20 x = 0 x = 2.36 m ∴

2.362 Max +ve BM = −38.84 + 47.21 × 2.36 − 20 × 2 = 16.88 kNm

At the midspan of BC ; 22 = 15.58 kNm M F = −38.84 + 47.21 × 2 − 20 × 2

(iv)

Continuous Beam Continuous Beam

593 591

•

19.07 47.21 20 + + A

E

+ C

B

−

D

−

60.93

33.79


53.33

15.58 38.84

+

+

16.05

10 kNm −

−

− 2m

A

E

1m

B

2.32 m

C

D

1m


A continuous beam ABC is fixed at A and C . It is continuous over a simple support B. Span AB is 5 m while BC span is 6 m. It is subjected to a concentrated load of 60 kN at 3 m from A and the span BC is subjected to uniformly distributed load of 10 kN/m. The ratio o f flexural rigidity of span BC to BA is 1.5. Sketch the shear force and bending moment diagram. Use Clapeyron’s theorem of three moments. EXAMPLE 11.17:

60 kN 3m

10 kN/m

2m 6m

5m A'

A

C

B

EI

1.5 EI

FIG. 11.17a

S OLUTION The simple beam moments are

M D = M E =

Wab 60 × 3 × 2 = = 72 kNm l 5 wl 2 8

=

10 × 62 8

= 45 kNm

C

592


•

72

A

45 kNm

B

D

E

C

FIG. 11.17b Simple beam BMD M B M A

M

C

A

B

C


Since A is fixed imagine a span of zero length A A with no load and A is simply supported. Apply three moment theorem for the spans A  AB

Properties of the simple beam BMD 1 × 5 × 72 = 180 2 5+2 x2 = = 2.33 3 l 2 = 5.0

A1 = 0

A2 =

x1 = 0 l 1 = 0 M A

 l 1 I 1

+ 2 M A



l 1 I 1

  5 5

l 2 + I 2

2 M A

+ M B

+ M B

I

l 2 I 2

I

 6  180

=−

A1 x1 l 1

= −6

Properties of the simple beam BMD A1 = 180 kNm2

5+3 3 l 1 = 5 m

x1 =

=

2.67 m

A2 =

2 2 × 6 × 45 = 180 kNm 3

x2 = 3 m l 2 = 6 m

×

5 × I

10 M A + 5 M B = −503.28 Apply three moment theorem for the spans ABC

  2.33

A2 x2 + l 2

(i)

Continuous Beam

M A

5 I

+ 2 M B

5

6 6 180 × 2.67 180 × 3 + + M C = − 6 + I 1.5 I 1.5 I 5 6 × 1.5



 



•

593



5 M A + 18 M B + 4 M C = − 6 (96.12 + 60) 5 M A + 18 M B + 4 M C = − 936.72

(ii)

Applying three moment theorem BCC  As the end C is fixed imagine a span CC  of zero length and C  is simply supported M B

6 1.5 I

+ 2 M C

>

6 



+ 0 + M C

1.5 I

0 1.5 I

= − 6

 180

3 +0 6 × 1.5



×

4 M B + 8 M C = − 360

(iii)

Solving equations (i), (ii) and (iii) M A = 31.62 kNm. M B = 37.4 kNm. M C = 26.29 kNm. Shear force and bending moment values for the spans AB and BC respectively.

Span AB 31.62 kNm A

60 kN 3m

2m D

V AB

37.4 kNm B V BA

FIG. 11.17d

∑ V = 0; ∑ M B = 0;

V AB + V BA = 60

(i)

5V AB − 60(2) − 31.62 + 37.4 = 0

(ii)

V AB = 22.84 kN V BA = 37.16 kN Span BC M D = 22.84(3) − 31.62 = 36.9 kNm 37.4 kNm

10 kN/m

26.29 kNm

B

C V BC

6m

E

FIG. 11.17e

V CB

596 594

•


∑ V = 0;

V BC + V CB =

60

(iii)

62 − 37.4 + 26.29 + 6V = 0 BC − 10 × 2

∑ M = 0; C

V BC = 31.85 kN V CB = 28.15 kN

32 M E = 31.85(3) − 37.4 − 10 × = 13.15 kNm 2

31.85 22.84 +

+

F

A

D

C

B

−

−

37.16 28.15 kN 2.815


The location of zero shear in span CB is obtained by equating the shear force equation to zero as

(SF ) xx = 28.15 − 10 x = 0 x = 2.815 m M F

= 28.15(2.815) − 10(2.815)2 /2 − 26.29 = 13.2 kNm

+

37.49

37.4

+

13.15

31.62

26.29 kNm −

−

A

D

B

F


C


•

595

A continuous beam ABCD is of uniform section as shown in figure. EI is constant.

Draw the SF D and BMD 10 kN/m

6m E

A

B

6m F

6m G

C

D

FIG. 11.18a

S OLUTION

The simple beam moments are 10 × 62 M E = M F = M G = = 45 kNm 8 45

45

A

45 kNm

B

C

D

FIG. 11.18b Simple beam BMD M

M C

B

A

D B

C


Considering spans ABC A1 =

2 2 × 6 × 45 = 180 KNm 3

x1 = 3 m >

6 M A + 2 M B (6 + 6) + 6 M C = −6

 180

×

6

3

180 × 3 + 6

24 M B + 6 M C = −6(90 + 90) = −1080

 (1)

596

•


Considering span BCD >

6 M B + 2 M C (6 + 6) + 6 M D = −6

 180

×

6

3

180 × 3 + 6



6 M B + 24 M C = −1080

(2)

Solving Equations (1) and (2) M B = −36 kNm M C = −36 kNm Shear force and bending moment values in the spans ABC, BCD

Consider span AB 10 kN/m 36 kNm A

B

6m

V AB

V BA

FIG. 11.18d

∑ V = 0; ∑ M B = 0; ∴

V AB + V BA = 6(10) = 60

(i)

62 6V AB + 36 − 10 × =0 2

(ii)

V AB = 24 kN V BA = 36 kN

Consider span BC 10 kN/m

36 kNm

36 kNm

B

C

6m

V BC

V CB

FIG. 11.18e

∑ V = 0; ∑ M C = 0;

V BC + V CB = 60

(iii)

62 6V BC − 36 + 36 − 10 × =0 2

(iv)

V BC = 30 kN V CB = 30 kN ,


•

599 597

Span CD 10 kN/m

36 kNm

D

C

6m

V CD

V DC

FIG. 11.18f

∑ V = 0;

V CD + V DC = 10 × 6 = 60

∑ M D = 0;

62 6V CD − 36 − 10 × = 0 2

(v)

V CD = 36 kN V DC = 24 kN 24

30

+

36

+

E

A

−

x

+

F

B

−

G

C

D

−

3m

1

36 6m

30

24 kN

6m

6m 2.4

FIG. 11.18g Shear force diagram

The location of zero shear is calculated as 24 − 10 x1 = 0 x1 = 2.4 m M E = 24(2.4) − 10(2.4)2 /2 = 28.8 kNm M F = 30(3) − 36 − 10 × 32 /2 = 9.0 kNm M G = 24 (2.4) − 10(2.4)2 /2 = 28.8 kNm

28.8

36

+

9 36

28.8 kNm

+ −

A

E

B

+

−

F

C

FIG. 11.18h Bending moment diagram

G

D

598


•

EXAMPLE 11.19:

Analyse the continuous beam by three moment theorem. Also draw SFD and BMD. 25 kN/m

50 kN/m

2.8 m

3m E

A

15 kN/m

4m

F

B

G

C

D

FIG. 11.19a 49 30 28.1

A

E

B

F

C

G

D

FIG. 11.19b Simply supported BMD M B

A

M C

B

C

D


S OLUTION

Properties of the simple beam BMD A1 =

2 2 × 3 × 28.1 = 56 .2 kNm 3 x1 = 1.5 m

A2 =

2 2 × 2.8 × 49 = 91.47 kNm 3 x2 = 1.4 m

l 1 = 3 m

A3 = 80 kNm2 x3 = 2 m

l 2 = 2.8 m

l 3 = 4 m

Applying three moment theorem for spans ABC >

M A (3) + 2 M B (3 + 2.8) + 2.8 M C = −6

 56.2

×

3

1.5

91.47 × 1.4 + 2.8



11.6 M B + 2.8 M C = −6(28.1 + 45.74) 11.6 M B + 2.8 M C = −443

(i)

Continuous Continuous Beam Beam

• •

599 601

Applying three moment theorem for spans BCD >

2.8 M B + 2 M C (2.8 + 4) + 4 M D = −6



91.47 × 1.4 80 × 2 + 2.8 4



2.8 M B + 13.6 M C = −6 (45.74 + 40) 2.8 M B + 13.6 M C = −514.44

(ii)

Solving (i) and (ii) M B = −30.58 kNm, M C = −31.53 kNm Free body diagrams of AB , BC and CD 30.58 kNm

25 kN/m

V

3m

30.58 kNm

31.53 kNm

2.8 m

V

V

50 kN/m

15 kN/m

4.0 m

V C

V C

B2

31.53 kNm

1

1

V D

B1

A

FIG. 11.19f

FIG. 11.19e

FIG. 11.19d

Static equilibrium of spans AB , BC and CD ∑ V = 0;

∑ V = 0;

V A + V B1 = 75

∑ V = 0;

V B2 + V C1 = 140

(i)

∑ M B = 0;

∑ M C = 0;

25 2 ×3 = 0 3V A + 30.58 − 2

V A = 27.3 kN V B1 = 47.7 kN

V C = 70.34 kN

+

+

47.7

V D = 22.12 kN

+

x −

V C2 = 151.53/4 = 37.88 kN

37.88

69.66

27.3

1

42 −31.53 + 4V =0 C 2 − 15 × 2

31.53 − 30.58 + 2.8V B2

(ii)

2

(v)

∑ M D = 0;

2.82 −50 × = 0 (iv) 2 V B2 = 69.66 kN

x

V C2 + V D = 60

(iii)

x −

3

70.34


−

22.12 kN

(vi)

600 602

• •


The locations of shear forces in zones 27.3 − 25 x1 = 0

AB, BC and CD are

69.66 − 50 x2 = 0

x1 = 1.09 m

37.88 − 15 x3 = 0

x2 = 1.39 m

M 1 = 27.3(1.09) − 25 × 1.09

2

x3 = 2 .52 m

/2 = 14.9 kNm

M 2 =

−

30.58 + 69.66(1.39) − 50 × 1.392 /2 = 17.94 kNm

M 3 =

−

31.53 + 37.88(2.52) − 15 × 2.522 /2 = 16.3 kNm

14.9

30.58

+

17.94

31.53

+

28.1

−

1.09

16.3

+

30 kNm

−

2.52 m

1.39

FIG. 11.19h Bending

moment diagram

Analyse the continuous beam by theorem of three moments and draw SFD and BMD. EI is constant. EXAMPLE

11.20:

10 kN

15 kN

5 kN/m

2m

3m

4m A

E

5m

6m F

B

2m

C

G

D

FIG. 11.20a

10

A

22.5 kNm

B FIG. 11.20b Simple

beam BMD for span ABC

C

Continuous Beam

•

601

22.5 18 kNm

B

C

D

FIG. 11.20c Simple beam BMD for span BCD

M C M

B

FIG. 11.20d Pure moment diagram

S OLUTION

Referring to Fig. 11.20 b Properties of simple beam BMD 1 2 × 10 × 4 = 20 kNm 2 x1 = 2 m

A1 =

2 2 × 6 × 22.5 = 90 kNm 3 x2 = 3 m

A2 =

l 2 = 6 m

l 1 = 4 m Applying three moment theorem for spans ABC, >

4 M A+ 2 M B (4 + 6) + 6 M C = −6

 20

×

4

2

90 × 3 + 6



20 M B + 6 M C = −6(10 + 45) 20 M B + 6 M C = −330

(i)

Referring to Fig. 11.20 c Properties of simple beam BMD A1 =

2 2 × 6 × 22.5 = 90 kNm 3

x1 = 3 m l 1 = 6 m

1 2 × 5 × 18 = 45 kNm 2 5+2 x2 = = 2.33 m 3 l 2 = 5 m

A2 =

602

•


Applying three moment theorem for spans BCD Considering span BCD >

6 M B + 2 M C (6 + 5) + 5 M D= −6

 90

×

6

3

45 × 2.33 + 5



6 M B + 22 M C = −6(45 + 20.97) 6 M B + 22 M C = −395.82

(ii)

Solving (i) and (ii) M B = 12.09 kNm. M C = 14.69 kNm. Shear force and bending moment values for spans AB, BC and CD. 10 kN 2m

A

12.09 kNm

2m

B

E V AB

V BA

FIG. 11.20e

∑ V = 0; ∑ M B = 0;

V AB + V BA = 10

(i)

4V AB + 12.09 − 10(2) = 0

(ii)

Solving (i) and (ii) V AB = 1.98 kN V BA = 8.02 kN M E = 1.98(2) = 3.96 kNm Span BC 12.09 kNm

5 kN/m 14.69 kNm C

B

6m

V BC

V CB

FIG. 11.20f

∑ V = 0; ∑ M B = 0;

V BC + V CB = 6(5) = 30 kN

5 × 62 6V BC + 14.69 − 12.09 − = 0 2 V BC = 14.56 kN V CB = 15.44 kN

(iii)


•

603 605

The location of shear force is zero is found out as 14.56 − 5 x = 0 x = 2.91 m Hence Max +ve BM = 14.56(2.91) − 12.09 − 5 ×

2.912 2

= 9.11 kNm

Span CD 15 kN 14.69 kNm 3m

2m

C

D

V CD

V CD

FIG. 11.20g

∑ V = 0; ∑ M D = 0;

V CD + V DC = 15

(iv)

5V CD − 14.69 − 15(2) = 0 V CD = 8.94 kN V DC = 6.06 kN 8.94

14.56

1.98 +

A

+

+

E

B

D

C −

−

−

6.06 kN 8.02

14.69

FIG. 11.20h Shear force diagram

12.09

+

9.11

15.44 kNm

3.96

−

A

+

+

E

B

FIG. 11.20i Bending moment diagram

12.12

−

C

D

604

•


A continuous beam ABCD is simply supported at A and D. It is continuous over supports B and C . AB = BC = CD = 4 m. EI is constant. It is subjected to uniformly distributed load of 8 kN/m over the span BC . Draw the shear force diagram and bending moment diagram. EXAMPLE 11.21:

8 kN/m

4m A

4m

4m E

B

C

D

FIG. 11.21a

S OLUTION

The simple beam moment 8 × 42 M E = = 16 kNm 8 16 kNm

A

B

E

C

D

FIG. 11.21b Simple beam bending moment diagram

A

M B

M

B

C

C

D


Consider span ABC

Applying three moment theorem; 2 4 × 16 × 2 4 M A + 2 M B (4 + 4) + 4 M C = −6 0 + × 3 4





16 M B + 4 M C = −128

(i)

Consider span BCD

2

4 × 16 × 2 > × +0 4 M B + 2 M C (4 + 4) + 4 M D= −6 3 4 4 M B + 16 M C = −128

 (ii)

Continuous Beam

•

605

Solving (i) and (ii) M B = −6 4 kNm M C = −6 4 kNm .

.

Shear force and bending moment values of spans AB BC and CD span AB ,

6.4 kNm A

B

4m V AB

V BA

FIG. 11.21d

∑ V = 0; ∑ M B = 0;

V AB + V BA = 0

(i)

4V AB + 6 4 = 0

(ii)

.

V AB = −1 6 kN V BA = +1 6 kN .

.

span BC 6.4 kNm

8 kN/m

B

6.4 kNm C

4m

V BC

V CB

FIG. 11.21e

∑ V = 0; ∑ M C = 0;

V BC + V CB = 8(4) = 32

(iii)

8 × 42 − 6 4 + 6 4 + 4V =0 BC − 2

(iv)

.

.

V BC = 16 kN V CB = 16 kN span CD 6.4 kNm D

C

4m V CD

V CD

FIG. 11.21f

606 608


•

∑ V = 0; ∑ M = 0; D

V CD + V DC = −

0

6.4 + 4V CD

(v)

=

0

(vi)

1.6 kN V DC = −1.6 kN V CD =

16

1.6 kN

+

+

B A

C

−

D

−

−1.6

16


16 kNm +

6.4

6.4

−

−

A

B

C

D

FIG. 11.21h Bending moment diagram EXAMPLE

11.22:

Analyse the beam shown in figure by SFD and BMD. EI is constant. 10 kN/m

6m A

6m B

6m

C

D

6m

E

FIG. 11.22a

45

45

A

B

C

FIG. 11.22b Simple beam BMD

D

E

Continuous Beam M

M

M

B

C

D

B

A

C

•

607

D

E


Properties of simple beam BMD A1 =

2 × 6 × 45 = 180 3 x1 = 3 m

A2 =

2 2 × 6 × 45 = 180 KNm 3 x2 = 3 m

l 1 = 6 m

l 2 = 6 m

Applying 3 moment theorem for the spans ABC

180 × 3 > 6 M A+ 2 M B (6 + 6) + 6 M C = −6 0 + 6





24 M B + 6 M C = −540

(i)

Applying 3 moment theorem for the spans BCD

6 M B + 2 M C (6 + 6) + 6 M D = −6

 180

×

3

6

180 × 3 + 6



6 M B + 24 M C + 6 M D = −6(180) = −1080 M B + 4 M C + M D = −180

(ii)

Applying 3 moment theorem for spans CDE

6 M C + 2 M D (6 + 6) + 6 M E = −6

 180

×

6

3

+0



6 M C + 24 M D = −540

(iii)

solving equations (i), (ii) and (iii) M B = −12.86 kNm. M C = −38.57 kNm. M D = −12.86 kNm. Free body diagram of AB, BC, CD and DE span AB B

A

6m V AB

V BA

FIG. 11.22d

12.86 kNm

608

•


∑ V = 0; ∑ M B = 0;

V AB + V BA = 0

(i)

6V AB + 12.86 = 0

(ii)

V AB = −2.14 kN V BA = +2.14 kN. span BC 12.86 kNm

10 kN/m

38.57 kNm C

B

6m

V BC

V CB

FIG. 11.22e

∑ V = 0; ∑ M C = 0;

V BC + V CB = 60

(iii)

10 × 62 − 12.86 + 38.57 + 6V =0 BC − 2

(iv)

V BC = 25.72 kN V CB = 34.28 kN span CD 38.57 kNm

10 kN/m

12.86 kNm

C

D

6m

V

CD

V

DC

FIG. 11.22f

∑ V = 0; ∑ M D = 0;

V CD + V DC = 6(10) = 60 kN

(v)

62 12.86 − 38.57 + 6V CD − 10 × =0 2

(vi)

V CD = 34.28 kN V DC = 25.72 kN span DE 12.86 kNm

V DE

6m

FIG. 11.22g

V ED

610

•


S OLUTION

M B = −20(1) = −20 kNm

The simple beam moments are 20 × 42 M F = = 40 kNm 8 60 × 4 M G = = 60 kNm 4 60 kNm

40 kNm

B

C

F

D

G

E

FIG. 11.23b (b) Simple beam BMD M D

M C

20

B

C

D

E

FIG. 11.23c (c) Pure moment diagram

Apply 3 moment theorem for the spans BCD



2 2 −20(3) + 2 M C (3 + 4) + M ( ) = − + × × × 4 6 0 4 40 D 3 4 −



60 + 14 M C + 4 M D = −320 14 M C + 4 M D = −260

(i)

Apply 3 moment theorem for the spans CDE

2 1 2 × 4 × 40 × M C (4) + 2 M D (4 + 4) + 4 M E = −6 + × 4 × 60 × 3 4 2 4 >

2

4 M C + 16 M D = −6(53.33 + 60) = −680 Solving (i) and (ii) M C = −6.92 kNm M D = −40.77 kNm

 (ii)

Continuous Beam



+ 6 EI

δ A

L1

+

δC

L2

613

•



A1 x1 = 960

2 × 6 × 180 × 3 = 2160 3 960 2160 240 120 + + 6 EI + Substituting, M A × 4 + 2 M B (4 + 6) = −6 4 6 EI × 4 EI × 6 A2 x2 =





4 M A + 20 M B = −3120 M A = −163.33 kNm M B = −123.33 kNm

Solving (1) and (2),

2m

40 kN/m

123.33 kNm

2m

A

B

130 kN

B

110 kN

6m

140.56 kN

140.56

A

C

B

94.44

110

FIG. 11.25e Shear force diagram

240 163.33 kNm −

+

123.33 + −

FIG. 11.25f Bending moment diagram

O C

94.44 kN

FIG. 11.25d Free body diagram of spans AB and BC

130 kN

(2)

hogging BM.

240 kN 163.33 kNm



M A + 5 M B = −780

→





180

618 616

•

Basic Structural Analysis 100 kN

30 kN/m

4m A

6m

8m

C

B

EI

2.5 EI

Ans: R A = 45kN,

R B = 165.5kN,

RC = 69.5kN

(11.3) A continuous is supported andand loaded as shown in Figure. If ABCD continuous beam beamof ofuniform uniformsection section is supported loaded as shown in gure. ABCD If the support bymm, 10 mm, determine the resultants and moments at the supports. B sinks B sinks the support by 10 determine the resultants and questions at the supports. 2

Assume E = 2(10)5 N/mm ; I = 6(10)7 mm4 40 kN 10 kN/m

20 kNm

3m A

B

4m

C 1m

6m

D

EI = Constant

Ans: V AB =

+16.5 kN ,

M B =

−

V BA =

+23.5,

V BC =

+19,

V CB =

+21.0

14 kNm

(11.4) Determine the reactions at A , B and C of the continuous beam shown in figure. 8 kN

3 kN/m

3m

1m A

5m

4m B

I

C

1.25I

Ans: V AB = 6.75 kN, M A =

V BA = 1 .25,

3.31 kNm,

−

M B =

V BC = 6 .31;

3.87,

−

M CB =

V CB = 8 .69

+7.44

Continuous Beam

•

617

(11.5) Analyse the continuous beam shown in Figure and determine the reactions 80 kN

40 kN

50 kN/m 1m

2m

2m

4m A

4m

4m

2I

B

1.5I

2m

D

I

C

Ans:

V AB = 41 68 kN V BA = 38 32 .

,

V CD = 28 03 .

V DC = 11 97

,

.

M B = −52 93 .

.

V BC = 102 88 V CB = 97 12

,

.

,

.

M A = −17 98 kNm

,

.

,

M C = −41 42 M D = −9 29 kNm

,

.

,

.

(11.6) Analyse three span continuous beam by three moment theorems. Draw the BMD and shear force diagram. Determine the end moments and reactions E I is constant. 50 kN

50

75 kN/m

25 kN/m 2m A

6m

2 m 6m

B

4m

C

D

EI = Constant

Ans:

(i) R A = 75 39 kN .

R B = 127 59 kN

,

.

M A = −75 78 kNm .

,

,

RC = 97 85 kN .

M B = −73 44 kNm .

,

,

R D = 99 17 kN .

M C = −55 55 kNm .

,

M d = −55 2 kNm .

.

(11.7) Analyse and draw BMD and SFD for the beam shown in Figure. The values of second moment area of each span are indicated along the members. Modulus of elasticity is constant. 100 kN 30 kN/m 2.5 m

2.5 m 2I

1.25 m

3I

A

B

80 kN 2.5 m

40 kN 1.25 m

4I

6m

D

C

Ans:

M A = −56 02 kNm .

,

M B = −75 47 kNm .

,

M C = −94 3 kNm .

,

M D = 0

Three Moment Derivation

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