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of

Structu

I-,.

,

CHAPTER I DEFLECTIONS OF STRUCTURES

1.1 Introduction A rigid body moves slightly when it is subjected to load or strained due to any other reason such as change in temperature, shrinkage or creep of the material of construction.

The displacements of variou s points from their original positions are called dejlectioTls. In a broader sense, deflections comprise linear displacements as well as angular rotations. Resistance of a structural element to deflection is termed stiffness. Stiffness of a structure is of no less importance than its strength. This is because an excessive deflection of a structure may impair its proper function. For example, excessive deflection of floor beams may cause cracking of the plaster and produce drainage problems and excessive deflection of a frame in a building may damage the exterior curtain walls and inte rior partitions.

I

I

Not only is a knowledge of the deflections important for reasons as those mentioned above, but also because it forms the basis of the analysis of statical1y indeterminate structures. 'When the number of equations of eq uilibrium together with any condition equations available from particular datails of construction (section 2.4" Part I) is not sufficient to determine all the external reactions of a structure and the internal forces and moments in its component members, resort is made to compatibility conditions . These conditions which are based on displacements and rotations provide the additional number of equations required for the analysis. There are several methods for computing deflections. Of these methods" the following are the most useful a nd widely used and therefore will be considered in detail in this chapter. I. The double integration method.

2. The moment-area method.

r ~

3. The elastic-load method. 4 . The conjugate beam method.

2 5. The method of virtual work 6. Graphical methods for truss deflections (Williot and Williot-Mohr diagrams). As will be explained subsequently, some of these methods are only suitable to calculate one particular deflection component at an individual point, while others may be used to calculate deflections at several points simultaneously. Each method nas its limitations, advantages and disadvantages. The student should therefore be thoroughly familiar with as many methods as possible and should understand which method is best suited to a particular problem. For instance, the first four methods are used to determine deflections of structures whose members are mainly subjected to bending strains such as beams and frames. The method of virtual work is used to determine deflections due to all types of strains; bending, axial and shear as well as those due to no-load causes. The graphical methods in (6) are applicable only to trusses. Part I Classical Methods of BeaDl Deflections 1.2 The double integration method It is customary to consider downward deflection at a point on a beam as positive. Thus, if the axis of the beam is taken as the x-axis and the beam deflection as the y co-ordinate as shown in Fig. 1.1 . . then the deflected shape of the beam OT, as it is commonly called, the elastic curve of the beam can be expressed as follows

R +x Elastic curve

+y Fig. I.l From calculus, the expression for the curvature is

-

I

R

=±

~/[l dx 2

+

(~.)2]~ dx \

J.l

3 where R is the radius of curvature at any point on the ;beam, y and

cly

dx are the deHcction and the slope of the elastic curve at that point respect-

)2

ively. In most practical cases the slope cly - is small and hence (cl -Y clx dx is very small and can be neglected in comparison with unity. Thus equation 1.1 reduces to : cl 2y

R

± clx 2

1.2

From equation 8.12 (Part I), the expression relating the curvature to the applied moment is : I

R

M EI

1.3

Combining equations 1.2 and 1.3, cl 2 y _ clx 2 -

M

±

EI

1.4

The solution of the differential equation 1.4 is the classical method of studying beam deflections. If the bending moment M is expressed in terms. of x, then the slope of the elastic curve at any point on the beam is found by integrating equation 1.4 once and the deflection is obtained by integrating it twice. Before doing so, however, it is necessary to define the sign conventions in order to decide whether the positive or negative signs that appear in equation 1.4 may be adopted. Following the usual sign convention for the moment and considering downward deflection positive, it is seen from Fig. 1.1, that with positive M, dy decreases as dx cl 2y x increases and thus - - is negative. Similarly, it may be shown that clx 2 cly cl 2 y. .. with negative M, increases with x and hence dx 2 IS pOSItIve. In dx cl 2 y other words, --2 is always opposite in sign to M. Thus, the negative dx sign is used and equation 1.4 becomes:

M EI

•r

1.5

4 By carrying out the integration of equation 1.5 in two steps, the expressions for slopes and deflectiDns of a beam can be obtained. The resulting constants of integration arc determined from either the displacement boundary conditions or from the requirements of continuity of the elastic curve. Displacement boundary conditions refer to the known conditions of deflection and slope at the supports. For example, the deflection at a roller or a hinge support is zeru, and both deflection and slope at a fixed support are zero. On the other hand, continuity of the elastic curve means that there is no sudden change in eith~r the deflection or slope at any point along the curve . The product' EI of the modulus of elasticity E and the moment of inertia I is called the bemfing rigidity.

1.3 Applications to the douhle integration lIlethod Several of the classical problems of beam deflections will now be solved to illustrate the application of the double integration method. As will be seen later, some of the given problems can be solved in a much simpler manner by other methods. Nevertheless, they arc given on purpose to point out the relative ease or difficulty of the method and show its limitations. Exalllple 1.1 A simply supported beam of constant hending rigidity EI carries a uniformly varying load as shown in Fig. 1.2a. Determine the slope and deflection equations, and find the position and value of the maximum deflection.

(a)

(Ll

O.S19L

I

Fig . 1.2 Solution: The reactions arc first found and are as indicated in Fig. 1.2a.

The bending moment at any section along the beam at a distance x from a is given by :

wLx

wx 3

M - - - - -6L 6

5

Applying equation 1.5, d'y dx 2 = -

I E1

(WLx

-6- -

WX 3) 6L

Integrating twice,

dx

Y

4

-1 (WLx2 --

dy ---

12

E1

= -

wx -

24L

I (wLx3 E1 \

3s -

wx'

)

120L + Ax + B

The two boundary conditions needed to determine the constants A and Bare:

= 0 at

y

o=

x

= 0 and

1 E1 ( 0 -

-

0

0 = - - 1 (WL4 -- E1 36

B dy Therefore - , dx

= o and

A = -

y

+0+

= 0 at B)

x

= L

and

- wL'-+AL+B)

120L

7wL3 360

wx 4 - - 7-WL3) = - - 1 (WLx2 -- - -

12

E1

24 L

360

The maximum deflection,) Yrna;V occurs at the point where t he slope

. dy. oftee h l ashe curve, - - J IS zero. dx dy

-

dx x

4 7 WL3) wx -1 (WLx2 -- - ---

=0

=

E1

12

24 L

360

0.519L.

Substituting this value of x in .the expressIon for y, Ymp

=

0.0065 wL4 E1

The elastic curve of the beam is shown in Fig. 1.2b.

i

,

,6

. batnple 1.2 A cantilever of constant bending rigidity EI carries a

COD-

centrated end load as shown in Fig. 1.3a. Determine the slope and deflection equations, and hence find the slope and deflection at the free end.

Fig. 1.3 Solution: Taking the origin at a, the bending moment at a section at a distance x from a is given by :

M = -

P (L-x)

Applying equation 1.5, dOy -

dx 2

=

P (L-x) EI

Integrating twice,

dy dx y

=.!.... (Lx EI

=

:1

_

(~.

A)

x' + 2

3

- x6

+

Ax

+ B)

The two boundary conditions needed to determine the constants A and B are :

dy

- - - o and y dx

o

-

P

+ A)

(0 - 0

EI OandB = O

A

Therefore, dy =...!:... dx EI y

= 0 at x = 0

=

and

(Lx _ ~) 2

:1 (~2 x; ) _

P 0= (O-O+O+B) EI

7

=

At the free end x

=(

a"

Yb

L then,

dY)

dx

=...!

b

= ~ (L' _ EI

L') = PL' 2 2EI

(L3 _ L3) = PL3

EI

2

3 EI

6

The elastic curve of the beam is shown in Fig. D.3b.

Elaunple 1.3 A simply supported beam of constant bending rigidity EI carries an end moment Mo as shown in Fig. 1.4a. Find the expressions for the slope and deflection. Locate the position of maximum deflection and calculate its value. Also calculate the slopes at the supports.

~)~f--_M.

_ __ L

- ----14\

(b)

Fig. 1.4 Solution: The reactions are first found and are as indicated in Fig. 1.4a. The bending moment at any section at a distanc... x from a is given by :

M=-(Mo- M:) d2 _1_ (M _ M.,x)

Applying equation 1.5, y

=

dx'

EI

L

0

Integrating twice,

~ y

~I

(M.,x - ~~2 + A) 1(M.,x2 M.,x3 =lli -2--61:"""+

=

Ax

)

+B

The constants A and B are found from the two boundary conditions : y

=

0

at x

=

0

and y

=

0

at x

=

L

8 2 I L2 +AL+B) O=-(O-O+O+B)andO= - I (MoL - - _ -M oEI EI 2 6

o

B

MoL A= - - 3

and

_1_ (M.x

dy Therefore, - - _ dx

Mox2 _ 2L

EI

_ _ I (MoX2 _ Y - EI 2

MoL) 3

Mox3 _ MoLX) 6 L 3

The maximum deflection, Ymax ' occurs at the point where the slope of the elastic curve, dy ,is zero. Thus, dx •

dy dx

=

x

=

0 = _ I (M x _ EI 0

Mox2 _ 2L

MoL) 3

0.423 L

Substituting this value into the expression for y, 0.064 MoL2

Ymax = -

EI

The negative sign indicates upward deflection. It will be interesting to compare this value of maximum deflection with the value of the deflection at the centre, Yo . Yo is found by ~ubstitu­ ting L/2 for x in the expression for y.

MoL 2

Yo

= -

l6EI

0.0625 MoL2

=-

EI

It is noticed that in this particular case, the maximum deflection occurs at a point at a distance 0.077 L from the centre, and the difference between the maximum deflection and that at the centre is only about 2%. This is characteristic of simply supported beams';.e. regardless of the load system on the beam. . the maximum deflection is always approximately equal to that at the centre, and never occurs further than 0.0774 L either side of the centre.

The slopes at supports a and b are fo und by substituting x = 0 and

x

L in the expression for the slope. a

•

=

dY)

--

( dx

a

=

MoL

- --

3 E~

The elastic curve of the beam is

and S!l(lWn

a" =

(~) dx

in Fig. l.4b.

MoL b

6EI

9

Example 1.4 A simply supported beam of constant bending rigidity EI carries a concentrated load P as shown in Fig: 1.5. Find the expressions for the slope and deflection. Locate the position of maximum deflection and calculate its value. p

(I

0 0

Pb

b

b

I

Po

L

L

L

Fig. 1.5 Solution: The reactions are first found and are as indicated. Unlike the preceding examples, the bending moment has two different equations to apply to parts ac and be of the beam. Thcrcfore~ moment equations are found for each part and the integrations arc carried o ut separately. Not~ that in writing the moment equations the same co-ordinate axes arc used although a solution is also possible by taking two origins, onc at each end. of the beam, but this may lead to confusion in the signs adopted. Part ac.. x

co

N

l{)

..."

"

0

$1

K

'0

18 mt

~

(t)

w,

18

/

/ I (e)

/

/ / a~

I

I

____

/

I / ~

/ /

____

/ ~

____

~

__

~b

Fig . 1.36 W4

=

21.3 x 0.75

Ws = \V6

=

W7

-

Er, =

8 m 2t.

2 W4

at 1.25 m. from h.

8 m 2 t.

-

18.45 x 0.75

6.9 m 2 t.

2

w.

at 1.75 m. from h.

6.9 m

-

at 1 m.

from h.

at 0.5 m. from h.

2t.

128 X 10 4 X 200 X 10- 4 = 25630

ill 2t.

Using the moment-area method, ab = -

1 Er,

(6 .75

51.75

=25600 ---

+

2 X 7.6

+

2 X 8

+

2 X 6.9)

2.02 X 10- 3 rad. (clockwise)

57 I

Yb =

EI

, I

f

(6.75

,

x 2.75

+ 7.6

X 2.5

+8 87.1

X 100

25600

=

+

7.6 X 2

+8

X 1.75

+ 6.9

X I

+ 6.9

X 1.25

X 0.5)

0.34 em. (downward)

Example 1.29 Find graphically the clastic curve of the beam shown in Fig. 1.37a due to the given case of loading. Indicate the values of the maximum positive and maximum negative deflections. The moments of inertia, in m4., at the various locations are as indicated. E = 210 tfcm2.

(')O~,I~' Ul---

Fig. 1.116

c A L I

Fig. lo1l7

r-------------------------------------------------~/~

f

161

(6) Using the double integration method .. derive the expressions for the slope and the deflection along the beam shown in Fig. 1.117, then find the slopes at the supports and the deflection at the mid point c. (7) , (8) Using the double integration method, derive the expressions for the slope and the deflection along each of the beams in Figs. 1.118 and I. II 9, then find the slopes at the supports and the deflection at point c.

Mo

K;)b "-1. 3

~ 11.3

Fig. 1.118

-#

2L

""3 Fig. 1.119

(9), (10) Using the double integration method, re-solve problems (I) and (5) if the moment of inertia of half the beam nearer to the fixed end is doubled. (1 1)- (16) Using the moment-area method, re-solve problems (I) - (4), (7) and (8). (17) Using the moment-area method" find the rotation and deflection at the free end of the cantilever shown in Fig. 1.120 if EI = 5000 m 2t. What will be the percentage reduction in these values if: a) the moment of inertia of part ab is doubled? b) the moment of inertia of part be is doubled?

!'ig. 1.120

Fig. 1.121

(18) Using the moment-area method find the deflection at mid span of the beam shown in Fig. 1.121 ifE = 200 t/cm 2 and 1= 6x lOS cm 4 • What will this deflection be if the moment of inertia of the middle half of the beam is doubled? (19) Using the moment-area method, calculate the deflection at the mid point of span ab and the deflection at point C of the beam shown in Fig. 1.122 if EI = 7500 m"t.

162 1.SP Ll3

Fig. 1.122

Fig. 1.123

(20) Using the moment-area method, calculate the vertical deflection at point c and the horizontal displacement at point d of the beam shown in Fig. 1.123 if EI is uniform. (21), (22) Using the moment-area method, calculate the absolute displacement at point c of each of the beams shown in Figs. 1.124 and 1.125 if EI = 8000 m 2 t.

Fig. 1.124

Fig. l.l25

(23)-(26) Using the elastic-load method, re-solve problems (3),(4),(7) and (18) (27), (28) Using the elastic -load method, find graphically the slope diagram and the elastic curve of each of the beams in Figs. 1.126 and 1.127 ifE = 2000t/cm 2 and 1= 12500 em'. Check analytically the slope at support b and the deflection at points c and d.

4tl 12 Q

n

c

I--- 4 m

d

--1-.

2

..j...

t

~

3 -
ple 2.30 Calculate the reactions of the frame in problem 2.22 (Fig. 2.39) in the absence of load if the indoor temperature drops 30°C while the outdoor temperature rises] DOC . T he frame has a uniform symmetrical

section 50 em. deep,

81m

at _

I X IO- ' I"C. a nd EI

= 25000

m'.t.

t

b

a

( b) M•.!:).

(a) NrD. F ig. 2. 50

Solution : In general the given frame is statically indeterminate to the second degree. A possible main system is obtained by disconnecting the frame at the intermediate hinge. The two redundants Xl and X 2 will then be the horizontal and vertical reaction components at c. Due to symmetry in the statically indeterminate part of the frame, however, there will be no relative vertical displacement between the displaced ends at c due to temperature change. Hence" X 2 is zero and XI is determined from the condition: ~ct

ad

+

XI ~cl

=

0

has already been calculated in example 2.22 and is given by

519 EI

519 25000

= -- = - - X

100 = 2.076 cm.

264

=

Figs 2.50 a and b show the N , .D. and the M,.D. d ue to X, Referring to these figures and applying equation 1.54,

act =

I X 10-5 X

~o x

2 (

.j73 x .j873) 40 50

x- x -

0.160 -

2(

100 - I

8.54 x 3

2

x

3 +9 +2

I.

10-5

)

x 6 100

0.078 = 0.082 em.

0 .082 Henee,X,= - - - = 2.076

0.04 t.

From statics"

Ya

Yb = 0

X.

0.04 t. (to the left)

Xb

0.04 t. (to the right)

M.

Mb

= -

C.36 m.t.

2.18 Applications to statically indeterJD.inate com.posite structures

The principles used in the analysis of statically indeterminate composite structures are the same as those already considered in regard to trusses, beams and frames. The procedures of determining the degree of in determinancy, choosing the redundants, reducing the structure to a stable and statically determinate main system and setting out a number of deflection equations remain as the basic approach. In computing the displacement coefficients in the deflection equations, however" it is necessary to consider the effect of axial deformation in members which are subject primarily to axial forces on one hand and the bending deformation in members which are mainly subject to bending moments on the other hand. Since no new theory is involved. . the method is best explained through discussions on specific examples.

265

£"an.ple 2.31 Find the forces in the link members. and draw the B.M.D. for beam ab of the trussed beam shown in Fig. 2.51a. For beam ab, A = 120 em 2, I = 30000 em 4 and E = 2000 tfem 2. For link members, A,

= 20 em 2 and E =

2000 t fem 2 .

1.5 tim

(a)~4m

r- ~ 14-J-1 ~:;:S~t:~~ 4

4 --!-- 4

1.5t/m

1I2t

1/2

I~I

':0 48mt

(c) M,., F,.Os.

(b) Mn.• t;;.Os

Fig. 2.51

Solution: The beam is internally statically indeterminate to the first degree. A possible main system may be obtained by cutting member cd at any section along its length. The redundant XI' which is the force in the cut member, may be found from the condition: 8 10 + Xl 8 11 = 0

where 3 refers to the relative displacement at the ends of the cut member. EI = 2000 X 30000 X 10- 4 = 6000 m 2 t .

=

EA, EA

=

2000 X 20

40000 t.

2000 X 120

240000 t.

Figs. 2.51 band c show the forces and moments in the chosen main .ystem due to the applied loads and due to the case of loading X, = I. Referring to these figures and using the method of virtual work,

266

=J

=_2 (~X

M,M.dl EI 960 = - - - X 100 6000

J =~(~X

EI

5 8 x · 48 X8

3

= -16cm.

J

2

M2 dl F2L N dl '+1:'+' ~ EA, EA

8" --

EI 3

3 2 )+ _ 1

EA,

[4(~)2X5+IX8]+ _ 1 (16X~X~) 8 EA 2 2

4) 48 15.81 ( 6000 + 40000 + 240000 100

=

= 0.8 + 0.0395 + 0.0017 = 0.8412 em. Hence, X,

-16 =- --- =

0.8412

19.0 t.

(tension)

The forces in the link members are readily obtainable from the relationship: F = F,X,. Similarly, the final B.M.D. is obtained from the superposition relationship: M = Mo + MIX, . The result is shown in Figs. 2.52 a and b respectively.

9mt

Fig. 2.52 It would be interesting to compare the negative moment at e in the given trussed beam with that in a similar two-span continuous beam where the moment at the central support is given by : M

=- -- =wL2

1.5 X 8"

8

8

=-

12 m.t.

Alternative solution : The beam may be reduced to a stable and statically determinate main system by inserting a hinge in the beam at e. The redundant X'I is then the bending moment at e and is found from the condition: where

a.

aeo

+

X'I

act =

0

refers to the relative rotation at e.

267 X;=1 m!

X;=O

8

lifo

1i6J ~'Il"

.~

1m!

+

+

12

12m! (a) Mo.,No.'~.Os.

(b) Ml.,Nr'~ .Os. Fig. 2.53

Figs. 2.53 a and b show the forces and moments in the chosen main system due to the applied loads and due to the case of loading X', = I. Referring to these figures and using the method of virtual work" a",

f J

M,Modl EI

1:

=

J

M,Modl EI

=_

+ 1:

~ (~ X 8 EI 3

F,FoL EA,

X 12 X

+

3

EA

~) = 6000 64 X

2

L [~X l6X8+4(~X

F,FoL = EA, EA,

SN,Nodl

24

100 = _

10) 5] = 84.3340000 X 100 =

1.067 0.211

N,Nodl I ( I ) 21.33 X 100 EA = EA 8 X X 16 = 240000 = 0.009

6

a.o = -

a.,

=

M~dl EI +

J

Midl ~(~ X I -~-= EI EI 3

J

FiL 1: EA

,+

2) =

0.847

N:dl EA

16xlOO =0.089 3 X 6000

FiL 1: - = _ I [ 4 (5) - 25 + EA, EA, 24

J a.,

J

1.067 + 0.211 + 0.009 = -

(

~) 3

2

8

]

= 1.756 X 100 = 0.0044 40000

[(~)"

N:dl =_1 X 16] = 4 X 100 =0.000 EA EA 6 9 X 240000 = 0.089 + 0.0044 + 0.000 = 0.0934 - 0.847 X', = = 9 m.t. 0.0934

+

The positive sign indicates that the moment at e acts in the assumed

268 direction. This is consistent with the value obtained from the first, and obviously simpler, main system.

Example 2.32 A cantilever ab is braced by a system of link members aa shown in Fig. 2.54 a. Draw the S.F. and B.M.Ds. for beam ab if E = 2000 tlcm 2, I for beam = 20000 cm 4 and L/A for link members = 18 em-I. Neglect axial deformation in the beam.

36 61 ~o~

f~

61

____~c~______~b

1.5

L::------e: t.:.- 2 m _I-'d_

o

(a)

2

2mt

6

1187t

)(,=1

8.85 14.7

.4/3

8.85

(d)

(c) M, .. F,.Os

121

61

18.3ml

12

+ + 2.8

2.8

5.7 (llS.M.O

(e) S.F.O

Fig. 2.54

Solution: The structure is statically indeterminate to the first degree. A possible main system may be obtained by cutting link member cd at any section along its length. The redundant XI' which is the force in this member, is then obtained from the condition: 3 10 + XI 8 11 = o.

269 Figs. 2.54 band e show the forces and moments due to the applied loads and due to the case ofloading X, = I. Referring to these figures: and using the method of virtual work,

~'o

S

= =

8"

M,MOdl EI

112

EI =

_ M:dI

-EI =

I [2X2 =

2

EI

2 X

(12 + 36) 2

3 X 12 +

112 X 100 4000 = 2.S em.'

+

1: F:L

EA

~In X3X2'+~X2']+~~J(ir+Gr+12] 10.67 X 100 + -IS 4000 2000

Hence, X,

2.S

= - -- = 0.316

50

X -

9

= 0.266

+ 0.050

= 0.316 em.

S.S5t. i.e. S.S5t. (tension)

The free body diagram together with the S.F. and B.M.Ds. are shown 10

Figs. 2.54 d,e and f.

Alternative solution:

+

Q

6/5mt

e

.31

-415

0

(b) M,., 'i.Ds.

(a) M M,Modl EI

lOOx2 ( 20000 3

x-

2

24

x6 2

x 2+ -

x3

2

2·

3

x

2-24

x

6.7

2

x 6.72 X 9 X

I)

= -1.632 em.

8"

=J

M~dl EI

1

+

X

_ :I..:OO .:......:X ..:.....:2:...,( · - 6 X 22 20000 3' I X 1200 40000

L

EA

+

6.72 +- x2 2) 3

0.17

+ 0.03

= 0.2 em. Hence, X', = -

-1.632

-....:..:.:..:.::.

8.16 t.

0.2

which is identical to the corresponding

value obtained in the first

solution. It is interesting to note that if the deformation in the tie is neglected the value of the redundant will be, X', = 1.632/0.17 = 9.6 t. witb an error of about 18%.

Example 2.34 Find the reactions and the force in the tie cd ofthe frame shown in Fig. :1.58 a. For tie EA = 10000 t. and for frame EI = 20000 m 2t.

Solution : The frame is statically indeterminate to the second degree. A possible main system may be obtained by replacing the hinge support at b by a roller and cutting the tie at a section x along its length. The redundants Xl and X 2 are therefore the horizontal reaction component at b and the force in the tie respectively. These may be found from the simultaneous solution of the two superposition equations :

8. 0 8xo

+ X, 0., + X 2 8. 2 + X, 8x ' + X 2 tx2

0

- 0

r

I

273

, 3

1c 6m

La I-

b 4 -I-- 6 m ---I-- 4 -'

(al

.-------,3 m t

(clM,.D. Fig. 2.58 The Mo.D., the M,.D. and the M 2.D. a re shown in Figs. 2.58 b, c and d respectively. Referring to these figures and using the method of virtual work. .

8. 0

=

J

M,Mo dl

El

= _ _1_( El

5 X 40 (6 + 2) 2 + 40 X 8 X 9)

2

4480

El 8.,

I

I

d

8. 2

=s =S -

I

8,0

El

= -

1 ( -6- (6 2 ) 2 + - 5 3 3

El

(6 2 + 9 2 + 6 x 9) 2

+ 9 X 8 X 9) =

1362

El M,M2dl

_1 ( 5 X 3

El

El

2

(6

+ 2) 2 +

3

X

8 X 9)

336

=f -

M'dl I

El M2MO dl

El 1360

El

=

-

- 1(5

El

X 40

2

X 2 X 2 + 40 X 8 x 3)

274

336 EI

- - - from Maxwell's theorem

=J

+

(2-3 (3)2X2 +8X3X3)

IxL= _

EA

EI

+

I X 16 102 16 EA = Ef+EA

134

EI Substituting these values in the superposition equations,

+

-

4480

1362 XI + 336 X,

0

-

1360 + 336 XI + 134 X,

0

Solving these two equations simultaneously,

XI

= 2.04 t.

and

X2

= 5.03

t.

The remaining reactions are found from statics and are as follows:

Xa = 2.04 t. (to the right) and Ya = Y b = IO t. (upward)

2.19 Deflections of statically indetern1inate structures In the deve10pment of the method of virtual work for deflection com~ putations, there has been no restrictions on whether the structure was statically determinate or indeterminate.

It follows that the method of

virtual work is equally applicable to the determination of the deflection of statically indeterminate structures. Consider for example the statically indeterminate beam shown in Fig. 2 .59 a, and let it be required to calculate the vertical deflection at

point c. Under the given load, the Mo.D. will be as shown in Fig. 2.59 b. The first step for the calculation of the deflection at c is to place a unit load there and find the corresponding MI.D . This has been done and the result is presented in Fig. 2.59 c. Following the usual procedure of graphical multiplication~

!

r

275

w tim (a)

1I2

wL2J12

I

I

I

Ll2~"

I

c

+ wL2/B

1t (c)M,.O .

LIB

+ Ll4

Ll2

1t (d)M;O

~

_ _ _ _...:: ::"L ----- -j

1t

+

Ll4 Fig. 2.59

j

-

_2_[ EI

wL2 12

X~(~-~) 2

(5

S

S

2 L +"3 X"'2 X- s- a X4"- a ] wL2

L

L)

wL 4 384EI is rather The proced ure describ ed above, althoug h quite permiss ible" minate compli cated as it involve s the analysi s of the given statical ly indeter load unit a beam twice; once under the given load and anothe r under r method " placed at the point where the deflecti on is require d. Anothe ements displac and a much simple r one, is sugges ted by the fact that the corresp onof the given statical ly indeter minate beam are identic al to their applied the by ding values in any chosen main system when acted upon momen ts.. loads and the redund ant reactio ns, and Of interna l forces or may be they simulta neously . Theref ore, once the redund ants are found,

276 considered as external loads on the chosen main system and the required deflection is found in the usual manner by placing a unit load on the chosen main system. To generalize, any main system other than that used in the analysis of the statically indeterminate structure may serve the purpose. This is because while the final B.M.D. under the given case of IOl.ding is unique, the main system used in the analysis is arbitrarily chosen. As an application to the above statement, consider the main system

shown in Fig. 2.59 d. __ 1_ ( wL2 EI 12 wL2

x

f

L

x-x X -

8

3 8

X

L

2

L

4

3

2

---X-

2

L)

2

=

wL4 384EI

Consider next the main system shown in Fig. 2.59 e.

8e

=

M·;Modl EI

_2_(..!. x EI 3

~

X

wL2 8

2

__l_ x ~ x~x 2

2

4

x .2.. 8

X~ 4

wL')= wL4 12 384 EI

This demonstrates that on calculating the displacements of statically indeterminate structures, the M1.D. of any statically determinate main

system may replace the corresponding B.M.D. in the original statically indeterminate structure.

Esunple 2.35 Calculate the horizontal displacement at the eaves of the frame shown in Fig. 2.41a if it has a uniform bending rigidity EI = 2000 m 2 t.

I

1.1!6

J '-- 8 m--l-- 4 -l- 4 -< (b)M, D.

Fig. 2.60

(

I r

277 Solution: The final B.M.D. of the frame under the given case ofloading has already been found in example. 2.24 (Fig. 2.41). This is considered as the Mo.D. and is reproduced in Fig. 2.60 a. The best main system for the calculation of the horizontal displacement at the eaves is obtained by removing the support at c and replacing the hinge at b by a roller. The Mt.D: associated with this main system due to a unit horizontal load at ,

d is shown in Fig. 2.60 b. Using the method of virtual work,

8

d

=

S

M,Mo dl = ' _1_ ( _ 3.89x6 x...!.x 6 _ 3.89X5 5 x EI EI 2 3 2 +5.33X5 x4 _5.57X5 xl +5.33X5 2 2 2

29.28 =---= EI

X2)

29.28 - - X 100=-1.46cm. i.e. 1.46 cm.(to the left) 2000

Example 2.36 C~lculate the vertical displacement at joint c of the truss shown in Fig. 2. 21. L/A = 8 cm- 1 and E = 2000 t/cm 2 . S:.lution : Furces Fo in the truss m:::mbers under the given case of loading are readily obtainable from example 2.10 (Fig. 2.21), and are as given in the second column of Table 2.7. For c3.lculating the deflection at point cJa unit downward vertical

load is placed at c. The corresponding F I-forces in the given statically indeterminate truss are easily deduced from the Fo·forces; F, = Fo/24. 8 - - - (83.09) =0.332 cm. (downward) 2000 A main system obtained by either removing the support at g or cutting member kd ,Jay be conveniently used for the calculation of 8c' The forces F'l in the members df the chosen main system due to a unit vertical load at c are as listed in the fourth column of Table 2.7. The product F,'F. is listed in column 5 of the table.

8

2000

(83.24)= 0.332 cm. (downward)

which is identical to the value obtained above by applying a unit load to the original statically indeterminate truss.

Member

F.

F,F.

F,'

F,' F.

Member

ah hi ij jk ab be cd bh bi ie ej jd kd

- 7.8 -9.2 -18.4 + 6.7 + 4.6 +13 .8 + 5.1 + 7.7 - 7.7 + 7.7 +22.3 -17.7 - 7.5

2.54 3.:4 14.10 1.87 0.88 7.93 1.08 2.47 2.47 2.47 20.80 13.05 2.28

-5/12 -1/2 -I 0 +1/4 +3/4 +1/2 +5/12 -5/12 +5/12 +5/6 -5/6 0

+ 3.24 + 4.60 + 18.40 0 + 1.1 5 + 10.35 + 2.55 + 3.20 + 3.20 + 3.20 + 18.60 +14.75

gn nm ml Ik gf fe ed fn fm me el

Id

F. +2 .3 +2.8 +5.6 +6.7 -1.4 --4.2 -6.9 -2.3 +2.3 -2.3 +2.3 +2.3

F,F. +0. 22 +0.33 +1.30 +1.87 +0 .08 +0.73 +1.98 +0.22 +0.22 +0.22 +0.22 +0 .22 83,09

Table 2.7

F'

,

F,'F.

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 83 .24

279 2.20 Influence lines for silDple statically indeternUuate structures

f-

I,

,

I

t

!

i I

lt

!

~

l

Similar to any load function in statically determinate structures, the influence line for a redundant in statically indeterminate structures may be obtained by successive placing of a unit load at each possible position on the structure and computing the value of the redundant under consideration corresponding to each position of the unit load. These values form the ordinates to the required influence line. Although this is quite permissible, it is very tedious as it involves the analysis of the given statically indeterminate structure several times; once for each position of the unit load. Another and a much simpler method of constructing the influence lines for redundants follows from Maxwell's theorem of reciprocal deflections. This is explained in the following. Consider the restrained beam shown in Fig. 2.61a and let it be required

(al

~a I

(bll.l.6b

L

I

(cll.L. Xl

Fig. 2.61 to construct the influence line for the reaction at b. If the support at b is removed and a unit downward load is applied there) then as has been demonstrated in section 2.17} the resulting elastic curve which is shown in Fig. 2.61b is the influence line for the deflection at b. Now} if a unit load acts vertically downwards at an arbitrarily chosen point then, by definition} the ordinate to the influence line at this point will represent the deflection produced at b. For example} a unit load at c will produce a deflection 8bc at b. The corresponding redundant reaction is obtained from the well known deflection equation: Xl = _

8bc

S.,

280 where 8b1 is the vertical deflection at b due to a unit vertical load applied there, which is also the ordinate 8bb ,tO the influence line for Sb . Since point c W3.S arbitrarily chosen, it follows that the influence line for the reaction at b is obtained by dividing the ordinates to the influence line for the deflection at b by 8bb • This influence line is shown in

Fig. 2.6Ic. As a second example, consider the two-hinged truss shown in Fig. 2.62a, aqd let the influence line for the redundant horizontal reaction

(a)

(c) I. L. X k-----.c:-~/--::/'1

b'rresponding displacement diagram drawn with 25d~-values and assuming member 5 - 6 to remain fixed in direction is shown in Fig. 2 .65 c. Indicated on this diagram are the displacements at the panel points, which represent the ordinates of the influence line for the relative displacement at the cut section. The influence line for X is obtained by dividing these ordinates by S11 which is the relative displacement at the cut section due to X = 1. This is approximately equal to the relative displacement between joints 5 and 12 which, by measure· m~nt from the displacement diagram, is equal to 0.42cm. The influence line for X, which is the required influence line for the force in member 5-12, is thus as shown in Fig. 2.65 d. The influence line for the force in member 1·4 shown in Fig. 2.65e is obtained from the relationship: LL. F = LL. Fo + F, (LL. X); F, = - 3.35. The same influence line drawn on a horizontal base-line is shown in Fig. 2.65 f.

Example 2.40 If load is transmitted to the frame shown in Fig. 2.66 a at panel points spaced at 3 m, construct the influence lines for the horizontal reaction at b and the bending moment at section n at the middle of cd . EI = 20000 m"t.

)

---------

i I

t

287

r

c

n

d

3

La

6m

I

9m (a)

4

6mt

3

1

10.5

13.5

I

16.5

I (b)M, .D.

16.5 0.20

+ 13."1------~0

(d)l.L. 6 b

0.34cm 10.

(e) l.L. X

0.252 t

-

2.25 ........ .........

.

\

0.38 Fig. 2.66

288 Solution : The given frame is statically indeterminate to the first degree. A possible main system may be obtained by replacing the hinge at b by a rollerJ and the redundant X is then the horizontal reaction component at h. The influence line for X is obtained from the relationship: LL. X= - (LL. 3 b) I 3bl ; 3bl is the horizontal displacement at b due to a unit load applied there.

3bl is nIcuIated with reference to Fig.2.66 b which shows the B.M.D. due to a unit load at h.

3bl =

M dl = 1 (3_ (3)2 +-(3 9 2 + _1_

f

2

EI

EI

3

62 + 3

3

X

6)+ 6- (6)2) 3

270 270 == - - X 100 = 1.35 cm. EI 20000

The LL. for 3b is obtained graphically in the manner described in section 1.8 and the result is shown in Fig. 2.66 d. The scales used are: L.S., I cm = 200 cm, E.L.S., 1 cm. = 10 m 2t and ELS., I cm. = 8000 m 2t. The displacement scale is thus made up as follows: Icm =

200 X 10 8000

= 0.25 cm.

Thus.. by measurement, the ocdinates of the jnfluence line for Bb are as indicated in Fig. 2.66 d. The ordinates of the influence line for X are obtained by dividing the corresponding ordinates of the influence line for 3b by 3bh where 3bl = 1.35. The r esult'is shown in Fig. 2.66 e. The influence line for M n, shown in Fig. 2.66 f, is obtained from the relationship: LL. M = I.L. Mo + MI (LL.x); MI = - 4.5. The same influence line drawn on a horizontal base-line is shown in Fig. 2.66 g.

Exan>ple 2.41 If load is transmitted to the arch with a tie shown in Fig. 2.67a at panel points a-b, construct the influence line for the bending moment at section n at the middle of panel cd. For arch EI = 40000 m 2t and for tie A = 100 cm. 2 and E = 2100 tlcm 2. Solution: The given structure is statically indeterminate to the first degree. A possible main system may be obtained by cutting the tie at any section along its l ongth , and the redundant then the force in the tie. The influence line for X is obtained by dividing the ordinates of the influence line for 3b by 3 11 , where 3 JI is the relative displacement at the cut section due to X = I. The LL. for 3b has been found in example 1.77 (Fig. 1.108 e) and is re-produced in Fig. 2.67 b.

Xes

289

~ jm~.

(a)

6X4=24m~

I

0.88

0.96 em.

0.52

(e) I.L. X

/

4.5 /

+

I. L. Mo -4.5(I.L.X )

~,..",.---,-===

(d) I. L. Mnf----'_ _ _ _ _ _ _~

Fig. 2.67 Referring to Fig. 1.108 b which shows the M,.D.,

S

_JM,2 dl

I xL EI+EA

11-

2 X 100 (5.3 4.47 4.04 (3.5)2 + (3.5 2 + 5.5 2 + 3.5X5.5) + 40000 3 3 3

= --

I x24xlO0 =1.236 + 0.011=1.247 cm. 2100 X 100 The LL. for X is thus as shown in Fig. 2.67 c. The positive sign indicates that X acts as assumed in Fig. 1.108b. ( ' (5.5

2

+

2

6 + 5.5 xg)

)

+

The LL. for M., shown in Fig. 2.67 d, is obtained from the relationship: I.L. M = I.L. Mo M, (LL.X); M, = - 4.5. The same influence line drawn on a horizontal base-line is shown in Fig. 2.67c.

+

290 R"n"ple 2.42 If load is transmitted to the trussed beam shown in Fig. 2.68a at panel points spaced at 3 m, construct the influence line for the moment at point e. For beam I = 200000 em" and A = 300 cm 2 , for linIc members A = 30 em 2 and E = 2000 t/cm 2 . (a)

(b)M,,'i. ""lE"--L--t----'-------.------=~ ~"."
.L---------=::""i (b) BM.D 4.6mt

~6

~92

tC-0-18--0-76! rO-9-"---0.-9!

.

kl.76

~-0.4-9--0-.4"'9l

----v----

1.721"

l. 43

t

0.49

076 t

0.49

+

+

0.94

0.941

.-------------------------

(e)

Fig. 3.18

(d)S.F'D

343 Solution: There afe two unknown support moments. These may be determined from the two equations of three moments for supports 1 and 2 . . Since the bending rigidity is constant throughout the beam, form 3.13 of the equation of three moments is used. Equation of three moments for support 1,

Mo

1.0,

2 M, (6 + 8) + M2 X 8 = 6 X 4000 (

~.

!

8, - 80 8, - 82) + .:.!...-~ L o, LI2

1.5 - 0 1.5 - 0) 600 + 800

28 M, + 8 M2 = 105

(i)

Equation of three moments for support 2,

-0, -0,)

82 82 M, LI2 + 2 M2 (LI2 + L 23 )+ M, L23=6EI ( - - + - LI2 L23

M, X 8 + 2 M2 (8 + 6) =6x4000 (

+

8 M,

28 M, = -

0 -

1.5

800

+

0-0) 600 (ii)

45

A simultaneous solution of equations (i) and (ii) leads to : M, =

+ 4.6 m.t. and M2 = -

2.92 m.t.

The B.M.D. is thus as shown in Fig. 3.18 b. The reactions are found in the usual manner. Referring to Fig. 3.18 c, 4.6 = - = 0.78 t. (upward) 6 4.6

6

r !,

I

+ 2 M, (Lo, + L 12 )+M2 LI2=6 EI (

4.6

+ 2.92 8

+ 2.92 + -2.92 = ---'--8 6

4.5

Y,=

2.92 6

-

0.78 -

0.94

0.94 = 1.72 t. (downward)

+ 0.49

1.43 t. (upward)

0.49 t. (downward)

The S.F.D. is thus as shown in Fig . 3.18 d. A solution to the same problem, assuming EI = 8000 m 2 t., will lead to :

344 M,

+ 9.2

=

m.t. and M2

= -

5.84 m.t.

This shows that the greater the value of EI or, to put it In general terms, the greater the EI/L ratio for the spans, the greater are the moments produced in the beam due to differential settlement.

ExalDple 3.13 For the continuous beam shown in Fig. 3.19 a J draw the B.M. and S.F.Ds. if support 1 settles by 2 em., support 2 by 1 em., and the end supports remain undisplaccd. The beam is made of steel (E = 2000 t/em 2 ), and its cross-section throughout is a S.LE. No. 38 (I = 24000 cm 4 ).

a

2

8m

6

3

( a)

6 0..2

k----------:::::o-""""'===~___l (b)B.M.O.

+ 2.2 mt

-:jY

~2tr 0..367

0..367

a3t

io.3

~ 0..0.33

(el 0..033

'- - - - - y - - - - '

'------v----'

0..333

0. .667 0..367

0..367

+

0..0.33

0..0.33

(d) S.F.O

j 0..31

0..3

Fig. 3.19

Solution: In order that the resulting moments may be expressed in m-t, units, the bending rigidity of the beam must be in m-t units.

EI

=

2000 X 24000 X 10- 4

=

4800 m 2t •

f

~

,

345 Equation of three moments for support 1.1

MoLo,

+ 2 M, (Lo, + L 12)+M2L 12 = 6 EI 3, (

30 + 3, L-

Lo,

2 M, (6 + 8) + M2 X 8

= 6 X 4800 (2 - 0 + 2 600

7 M, + 2 M2 =

32)

'2

1)

800

15

(i )

Equation of three moments for support 2,

I

M, L'2 + 2 M2 (L 12 + L 2,) + M,L21 = 6 EI(32 - 3, + 32 - 3,) L'2 L 2,

t

M, X 8 + 2 M2 (8 + 6)

,,

~

i

!

I-

t

=

I - 2 + I6iiO - 0) 6 X 4800 800 (

2 M, + 7 M2 = 3

(ii)

A simultaneous solution of equations (i) and (ii) leads to : M,

=

+ 2.2 m.t. and M2

= -

0.2 m.t.

The B.M.D. is as shown in Fig. 3.19 h. The reactions are calculated in the usual manner. Referring to Fig. 3.19 c,

2.2

= -6 = Y,

=

0.367 t. (upward)

2.2 (2.2 + - 68

0.2)

= -0.367-0.3 = 0.667 t. (downward

_ 2.2 + 0.2 + 0.2 = 0.3 + 0.033 8 6 Y, = -

0.2 -

6

= 0.033 t. (downward)

The S.F.D. is as shown in Fig. 3.19 d.

=

0.333 t. (upward)

)

346 Esam.ple 3.14 For the continuous beam shown in Fig. 3.20 a, draw the B.M. and S.F.Ds. due to the given loads. If support 2 settles by 1.2 cm., calculate the percentage increase in the B.M . at support I. EI = 5000 m 2t.

(a}

(b)

12

12

12ml

12 6.1mt

6

+

+

(c)B.M.D.

+ 6mt

l.St/m 6

4

6

~1.28 4.n

U8t t 034 7.28 3.34

----v-

10.62 t 4.72 t

+

(d)

2)

t

134

7.28

2

6

---v--

4.661

+

~t

0.34ll~ 2.66

t

0.34 2.66

+ 2.66

2

2

(e)S.F.Q

4

41

Fig. 3.20

Solution: The simple B.M.D. is shown in Fig. 3.20 b. The elastic reactions due to the applied loads are calculated with reference to this figure and are as follows ;

r, r,

2 I 12 + 4 ="3X4X I2 + -i X 2 X 12 =32 +48=80m 2 t. = 48 + (I8 X 2 + 36 X 5) X 1/9= 48 + 24 = 72 m 2 t.

(

,

r

I

l

347 Effect of applied loads only. Equation of three moments for support 1,

MoLo, 2 M,

+ 2 M, (Lo, + L 12 ) + (8 + 12) + M2 X 12 = +

10 M,

3 M2 = -

M2L12

= -

6"

6 X 80

-

120

(i )

Equation of three moments for support 2,

M,L 12

+

2 M2 (L'2

12 M,

+

2 M2 (12

2 M,

+

+

+

7 M2 = -

L 2,)

9) -

+

M,L 2,

= -

6 X 9 = -

6 '2

6 X 72

63

(ii)

Solving equations (i) and (ii) simultaneously,

M, = -

10.2 m.t. and M2 = -

6.1 m.t.

Effect of settlement in the absence of load.

Equation of three moments for support 1,

2 M, X 20 40 M,

+

+

12 M2 = 6 X 5000 (

12 M2 = -

a ;;0~·2)

30

(iii)

Equation of three moments for support 2,

12 M,

+

2 X 21 M2

12 M,

+

42 M2 = 70

=

1.2 6 X 5000 ( 1200

+

1.2 ) 900 (iv)

A simultaneous ~olution of equations (iii) and (iv) leads to :

M,

= -

1.36 m.t. and M2

=

2.03 m .t.

Percentage increase in the moment at support

1.36

= 10.2 --

X 100

=

13.4%

348 3.5 Advantages and disadvantages of continuous beaDls

A comparison between the simple beam B.M.Ds. on one hand" and the final B.M.Ds. of the continuous beams in example 3.l - 3.ll on the other hand, shows that generally the greatest numerical value of the bending moment in a continuous beam is less than that for the same beam if it were composed of a series of simple spans. Since the design is normally governed by the greatest value of the bending moment along the beam., a continuous...,beam system leads to more economical design. Further, the aver~ge numerical value of the bending moment in a continuous beam is less than that in the corresponding series of simple beams) and also the bending moment changes sign along the beam. Since the deflection is mainly the integrated effect of the bending strains, a continu6us beam system leads to a stiffer structure. On the other hand, as illustrated by examples 3.l2, 3.l3 and 3.l4 continuous beams may be subjected to large additional moments if any support settlement takes place whereas statically determinate beams are not affected by such support movements. The moment due to settlement may be large enough to change the sign of the bending moments due to the given loading in some parts of the beam; a serious matter no doubt for reinforced concrete structures where the tension side decides where the main reinforcement is to be placed. In a loaded continuous beam, two points of zero moment, called points of inflection or contraflexture, usually occur in each span. If at these

(al

w

(b)

t---

(el

L

t,

L

J

L

I" " r' l' r ,",i I

w

I--

L

I

t~~8'8' L' L

L

Fig. 3.2l

r ------------------------------------------

---l

l

349 points proper hinges are inserted, a statically determinate beam, commonly called cantilever or Gerber beam, results. The B.M.D. of this heam win not be different from that of the corresponding continuous beam. This is illustrated in Fig. 3.21. Fig. 3.21a shows the B.M.D. of the continuous beam shown in Fig. 3.21b, and also for the cantilever beam shown in Fig. 3.2Ic, which is identical to the previous one except in having two hinges in the central span at the locations of the points of inflection.

3.6 Application of the !Dethod of the eqaation of three !DODlents to statically indetenninate fralBes The equation of three moments can be applied to the analysis of rigidjointed frames provided that : (I) joint translation is not allowed, (2) no more than two members are rigidly connected to a single joint. Fig. 3.22 shows examples of frames satisfying these two conditions Note that translation of the joints is generally prevented by virtue of the disposition of the supports or due to complete symmetry in the frame and loading.

c

b

c

b

"

(a)

(b)

b

a

d'

c' (e)

c

f (d)

Fig. 3.22

d

350 In Fig. 3.22 a, joint b can translate neither vertically nor horizontally due to the support conditions of the frame, and twa members only meet at joint h . In Fig. 3.22 b, neither joint b nor joint c can translate horizon.. tally due to the complete symmetry in both frame and loading. Further, two members only meet at each of joints band c. In Fig. 3.22 c, ron of joints b, c, d, and e can translate horizontally due to the complete symmetry in both frame and loading. On the other hand, three members meet at joints c and d. Nevertheless the equation of three moments can still be applied as members ee' and dd' are pin-connected to. the continuous member bed e. Also, it may be argued that the equation of three moments is inapplicable to the frame shown in Fig. 3.22d on the basis that three members are rigidly connected to joint b. A little thought, however, will show that overhang bd is statically determinate and all it has to do with the analysis is to. provide a moment of known value and 3ense at joint h. The procedure of analysis is essentially the same as that outlined in section 3.2 in regard to beams. Equations of three moments corresponding to the number of intermediate joints, a fixed support being considered·as one, are established. These equations when solved simultaneously, give the unknown joint moments. Once the joint moments are determined, it will be an easy matter to calculate the reactions and draw the N.F. and S.F.Ds. by considering the equilibrium of individual members as free bodies. The final B.M.D. may be obtained in the same manner but, easier still, it may be obtained by superimposing the indeterminate moments to the simple moments. The application of the equation of three moments to the analysis of statically indeterminate frames will be illustrated by a number of examples.

351

ExalDple 3. IS Usi ng the equation of t h ree mom en ts, const r uct t he N .F.,S. F . and B.M.Ds . for t he fr a m e shown i n Fig . 3 .23 a. T he rel at ivr moment s of i nert ia a re as ind icated.

4\

211m

i

t

I

b

e

21

%1

6m

l

a

b

I-

12mt

6m--l

6

( b)

( a)

I

4t 211m

+

r

14 t

30ml

I

3t

~

141

(e) B M.D.

r

14

141

14

3t

3

(d) Free body diagram

+ 3

1

3

2t

31

L-1I4

( ~)

N.F.D.

(IlS.F.D. Fig . 3.23

Solution: There are two unknown moments at joints band c, but from symmetry Mb = Me and one equation of three moments will be sufficient to determine the unknown joint moments . The simple B.M.D. is shown in Fig. 3.23 h. For ease of calculating the elastic reaction, it is presented in parts.

352 Considering I

Ir,

+(~

r. =

X 6 X 36

+

+

X 6 X 12) = 90

Equation of three moments for joint b,

M

•

Lab 1

+

2 M (Lab + L b,) b 1 2

+

6)

2 Mb (6

+

Mb = M, = -

Mb

+

M L b, = '2

6 = -

X

6

X

90

18. m.t.

The final B.M.D. may be constructed now and will be as shown in Fig. 3.23 c. The free body diagram for members ab and bc are shown in Fig. 3.23 d . From this diagram, the N.F. and S.F.Ds. shown in Figs.

3.23 e and f Can be easily established. E",aD1ple 3.16 Using the equation of three moments, draw the N.F., S.F. and B.M.Ds. for the frame shown in Fig. 3.24a ifit has constant EI.

Solution : There are two unknown joint moments Mb and Me- These may be determined from the equations of three moments for joints band c. The simple B.M.D. is shown in Fig. 3.24 b. Referring to this figure,

3

6

X

2 -

2 3

X

+ -32

4

X

4

=

24

X

9

+ 64 =

73

24 = 64

X

Equation of three moments for joint b,

MaLa.

+

2 Mb (6

+

Lbo)

+

+ 8) + 8 M,

-

6

2 Mb (La.

+ 4 M,

14 Mb

=

-

M,L., X

-

6 rb

73

(i)

219

Equation of three moments for joint c,

+ 2 M, + 16 M,

MbLb'

(LbO

8 M.

=

-

-

48

Mb

+ 2 M,

=

+

L,o)

+

MoL,o

6 X 64 lii)

Solving equations (i) and (ii) simultaneously, Mb = - 1O.25m.t. and ~ = - 18.88 m.t. The final B.M.D. is found by superimposing the simple bending moments in Fig. 3.24h to the indeterminate moments

353 311m

c "

b~O 9 + t 64

4t

+ 6ml

~===-

24mt

Q

__ 8m

( b)

( a)

18.88 18.88mt 311m

3:12 13.061

0.875

(e) B.M.D.

10.921 10.92t 1 . 2

3.721

3.72

3.72

t

0.28

10.92

(e) N.F.D.

028

13.08

.72

(f) SFD

F ig. 3.24 t bus found. The result is shown in Fig. 3.24 c. The free-body diagram _ of the frame is shown in Fig. 3.24 d. From this diagram, it will be an easy matter to construct the N .F. and S.F.Ds. shown in Figs. 3.24 e and f respectively .

354

Example 3.17 For the frame of constant EI shown in Fig. 3.25 a, calculate the reactions and draw the B.M.D. 1 tim

loSt

2~

4.51

c

d

I 4m b

LL-l-Q

0_'_ _

12 m

----1-1~J

Ca)

I , +

8.4mt

,1

I 10.5t (e) B.M 0

(bJ Fig. 3.25

Solution: There are four unknown joint moments :tvl a , Me,M d and Mb

Mb and Me = Md and one is left with two unknown moments only which may be determined from the equations of three moments for joints a and c. but from symmetry', Ma

=

Referring to Fig. 3.25 b which shows the simple R.M.D., the elastic reactions are

rcc =

calculated

i(1

as follows:

X 12 X 18)

I

r·

I

,

,

355 rc =

rei

+ rer

=

3

+

72 =

75 m 2t.

Equation of three moments for joint a,

MoLo. + 2 Ma (Loa + Lac) + MeL.e 2M.x6+6Me~-6x6

(i)

2 M a+ M e = - 6 Equation of three moments for joint c,

MaLae + 2 Me (Lac + Led) + MdLed 6 Ma + 2 Me (6 + 12) + 12 Me M. + 8 Mc

~

=

-

-

6 rc

6 X 75

= - 75

(ii)

Solving equations (i) and (ii) simultaneously,

M.

~

Mb

~

+ 1.8 m.t. and Mc

~

Md

= - 9.6 m.t.

The final B.M.D. is obtained in the usual manner and is as shown in Fig. 3.25 c. Indicated on the same figure are the reactions at support a.

Exam.ple 3.18 Using the equation of three moments, construct the B.M.D. for the frame shown in Fig. 3.26 a if the moment of inertia varies as indicated. Solution: Overhangs eb and cf are statically determinate and all they have to do with the analysis of the statically indeterminate bent abed is to provide moments = 2 X 3 = 6 m.t. acting as indicated in Fig. 3.26 h. These moments may be considered to act either at the ends of member bc or members ab and cd acco rding to whether the main system considered is obtained by introducing hinges at the top of members ab and cd or at the ends of member bc. The corresponding Mo.Ds. are shown in Figs. 3.26 c and d. Taking I = Ta

rb

=

It' and considering the main system in Fig. 3.26 c ..

0

~( ~

X 6 X 36 -

6 X 6)

Equation of three moments for joint a ..

~

54

356 21

2t

2t 2tfm

fe

b

6m

1

e

21

d

a

1=3

"3J

~

12m {al 6

(bl b 6

e

e

+

+

36mt

36mt

(el

(d)

18

"-;)"'"18-6:~' hl

12

+ 18 mt

12

(e l

(g)B.M.D.

(I) Fig. 3.26

2Ma +Mb "",O

(i)

Equation of three moments for joint h,

Ma Lab

+ 2 Mb (Lab + LbC) + Mc Lbc =

lab

6 Ma

Ma

+

lab

2 Mb (6

+ 5Mb

=

-

+

Ibc

6)

+

_

6 rb

Ibc

Mb X 6 = -

54

6 X 54

(ii)

Solving equations (i) and (ii) simultaneously, Ma

=

6 m.t. and Mb

= -

12 m.t.

The final moments at joint b are obtained by co-mbining the indeterminate moments thus found and the moments in the chosen main system. The resulting free-body diagram of joint b will be as shown in Fig. 3.26e.

357 Considering next the main system in Fig. 3.26 d,

ra

I 6 X 6 =-- x - = 6

2

3

2) (2 = ( 6X6 - 2X31+ 3 x 6 X 3 6)1 2=84 Equation of three moments for joint a,

Mo Loa + 2 Ma (Lo, + Lab) + Mb Lab = lOa

lOa

lab

6 ra

lab

2Ma X6+Mb X 6 = - 6 x 6 2 Ma + Mb = - 6

(iii)

Equation of three moments for joint b,

6 Ma + 2 Mb X 12 + Mb X 6 = -

6 X 84 (iv)

M. + 5Mb = -84 Solving equations (iii) and (iv) simultaneously, M. = + 6 m.t . and Mb = -

18 m .t.

The final moments at joint b will thus be as shown in Fig. 3.26f, which are identical to those obtained previously. The final B.M.D. is obtained in the usual manner and is as shown in Fig. 3.26 g.

358 EXAMPLES TO BE WORKED OUT (1)-(5) Draw the B.M. and S.F.Ds. for each of the continuous beams shown in Figs. 3.27-3.31. EI ~ constant.

I--- L. m --t--- 2 --+- 2 Fig. 3.27

211m

Xliii'!' 1..--_ 6m Fig. 3.28 211m

~IIIII!I

I--- 6m Fig. 3.29

it

~ " ,n'~" , I - - 6m

Fig. 3.30

---'!

Fig. 3.31

(6)-(16) Dnw the B.M. and S.F.Ds. for each of the continuous beams shown in Figs. 3.32-3.42. EI constant.

1.6 rim

Fig . 3.32

Fig . 3.33

359

Fig. 3.34

ILt

~Q I- 3

JLt

12t _

c

Ii

-L3m~

3 -l- 3 Fig. 3.35

-'§zJ

2-1 Fig. 3.36 1.5 11m

c

8 m ---l ~

-1-_ _

Fig. 3.37

'II •

181

Qi£:~t('1'1 Illb~

l- 2 -+-- L m

611 161

c~

---I- 2 -t-- Z -I- 2 -4. 2 -J

Fig. 3.38 1.5 t

%

~

2tl

12t

it

1-2-1-2+2 I

Bm

Lomt

~3

__ 3

:tt/m

~I"I!"I"~ I

8m Fig.3 .40

~.-l 11J

Fig. 3.39

. E: ~

Ifm 2m

O.Bt/m I I I I I I I I I I I I I

I

360

Fig. 3.41

Fig. 3.42 (17) The three-span continuous beam shown in Fig. 3.43 has a total length of 3L. Determine the span lengths L 1, L2 and L} in terms of L so that

M.

=

Mb = M

=

-

wLI2

-- .

12

c

EI = constant.

(18) Determine the bending moments at the supports of the continuous beam shown in Fig. 3.44 if: (a) end a is free, (b) end a is simply supported, and (c) end a is fixed. EI = constant.

Q

L ..

iC I

lSl/m

!!"Ii

Bm

~ I

Fig. 3.44 (19) - (24) Re-solve problems 6,7,8,10,12 and 14 if the moment of inertia varies from span to span as follows : Problem 6 (Fig. 3.32), lab: Ibc : Icd = 2 : 1.2 : I Pwblems 7 and 8 (Figs. 3.33 and 3.34) , lab: Ibc : ICd = 1.5 : 2 : I Problems 10 and 12 (Figs. 3.36 and 3.38), lab: Ibc = 1.5 : 1 Problem 14 (Fig. 3.40), lab: Ibc : Icd =_ 1 : 2 : I

rr

I I

361 (25) - (31) Determine the bending moments at the supports of the beams in problems 1,2,6,7,8,9 and 11 in the absence of the applied loads due to the following support movements: Problem Problem Problem Problem Problem Problem

1 2 6 7 8 9

(Fig. (Fig. (Fig. (Fig. (Fig. (Fig. Problem II (Fig.

3.27) , 3.28), 3.32), 3.33), 3.34), 3.35), 3.37),

Bb Bb Bb

2.4 cm. 2 cm. 2 em and Be Be = 4cm. Be 3 cm. Bb 2.5 cm. Bb 3.6 cm.

El = EI = 0.5 em. EI EI El El

5000 ill 2t. 3000 m 2t.

EI = 6000 5000 4000 4000

4000m 2t. m 2 t. m 2 t. m 2 t. m 2 t.

•

(32) The continuous beam in Fig. 3.41 has uniform I = 9&00 cm 4 and E = 2000 tfcm 2. What will be the percentage change in the max;mum shearing force and support moment if support c settles by 0.75 em ? (33)-(43) U sing the equation of three moments, draw the N.F.,S.F. and B.M.Ds. for the frames shown in Figs. 3.45-3.55. The moment of i.nertia varies as indicated.

41

r

L r===::i:1'i:"'mC=II;:J 2

21

61

61

___11m

CIL

6m

4

L

1

1

L " - - 4x3 ,, 12m ------l Fig. 3.45

1.5t1m

t-.-_ _ 12m

Fig. 3.46 1.51/m

41

21

lIS

1

I

4S

r

8m Fig. 3.47

Fig. 3.48

,

2r

362 1.5 II m

!

[i ::::'r:~ :l: ::::I

1.51

4

1

L

1.51

I

I

1.51/m

~4m - l -

311m

I--

4 -I

Fig. 3.49

3x4=12m - _

Fig. 3.50

lOt

21

21 2t1m

6m

1

L..--

9

_-'--_ 3x4:12m

I

1 211m

-.I. 9--,

Fig. 3.51

2t

4t

41

21

r

6m

L

I.t

8m

L.!-

211m

~8m - - I

1.5 11m

L3-l-- 4x3=12m --+-

~

Fig. 3.53

Fig. 3.52

61

I

,

3

t

1.st/m

41 3

3m

~2

3d:!t Fig. 3.54

m------J

~

8m Fig. 3.55

;f

.,-

CHAPTER 4 THE METHODS OF ELASTIC CENTRE AND COLUMN ANALOGY Part I

The Method of Elastic Centre

4.1 Introduction The method of elastic centre which was suggested by the German engineer Carl Culman stems from the method of consistent deformations and provides a simplified solution to a special type of structures. Its application is limited to three times statically indeterminate single span frames or arches and closed frames or rings. Although it is applicable to all frames of the abovementioned type, yet it is best suited to the analysis of frames which have an axis of symmetry such as those shown in Fig. 4.1.

r

t

rI

,-

I,,

"

(a)

"

"

"

(b)

"

(e)

"

I

!

I...

r

I

(d)

'.i. .>.. (e)

~

(f)

Fig. 4.1 The philosophy 'lfthe method lies in the fact that in general the analysis of frames which are statically indeterminate to the third degree requires the simultaneous solution of three equations if the method of consistent deformations is used. Each of these equations generally contains all the three unknown redundants. The method of elastic centre provides a means 363

364

by which the three unknown redundants may be found by individual solution, rather than by simultaneous solution, of three equations. As will be seen subsequently, this has the advantage of reducing the number of displacement components to be calculated in addition to saving in the time required for the solution of the deflection equations involved. 4.2 Development of the method of elastic centre

The method may be developed with reference to the fixed arch shown in Fig. 4.2 a.

(b)

(ai

Fig. 4.2

A possible main system is shown in Fig. 4.2 b and the reclunrlants ,X, Y and M are found from the usual deflection equations, 8. = 8,0 + X 8~l + Y 8'2 + M 8., = 0

4.1

Y. = Y,o + X Y.l + Y Y.2 + My., = 0

4.2

a.

=

a.o

+X

+ Y n a2 + M

aaI

aa]

= 0

4.3

Using the method of virtual work for the determination of the displacement coefficients and noting that the moments produced by unit values of X, Y and M are in succession, ·M l = y,M2 = x and M, = I, equations 4.1, 4.2 and 4.3 may be re-stated as :

S

MlMOdl+XSM~dl -+ EI EI

= 0

... 4.4

yS M~dl

= 0

... 4.5

M'Modl+xS~+ YS~+MSMldl=O EI EI EI EI

... 4.6

S

M2Modi + X { EI

S

...'!.!

ySXYdl + MS y EI EI

J

xydl + EI

EI

+ MS Xdl EI

It may he noticed that dl/EI appears in every term of the deflection

r---------------------

365 equations 4.4-4.6. Now, if an element dllong is imagined to have a width

equal to I/EI, then dl (I/EI) will represent an imaginary area called the elastic area. The centroid of this area is called the elastic centre and is denoted by c.

J

XYdl = ~

In view of this,

S

Ei=

r

xydA =1"" JYdl EI = J ydA = 8. and fXdl JXdA = 8y 4.7

Introducing the terms in equation 4.7 into equations 4.4-4.6,

S

S

J

M,Modl + X M!dl + I Y + 8 M = 0 EI lEI""

...

4.8

M 2 M o dl+ 1 x+ySMidl +8 M=O EI xy EI y

...

4.9

S

M3

MSM~EIdl

M O dl + 8 X + 8 Y + EI . ,

= 0

••. 4.10

The simplification obtained is based on the idea of connecting the assumed free end of the structure to the elastic centre c by a rigid (undeformable). arm. Fig. 4.2 c shows such an arm, and also three new redundants X., X 2 and X3 acting at c and replacing those at a. Obviously X., X 2 and Xl are statically equivalent to X, Y and M, and their magnitude should

be such that to keep c in an identical state to that of a. Thus, referring to the new rectangular axes wi th c as origin,

... 4.11 1'e

aeo

+ XI

1'el

+ X2

1'e2

+ X3

1'e3

= 0

... 4.12

de.O

+ XI

act

+ X2

de2

+ X3

a e3 = 0

... 4.13

Equations 4.11-4.13 may be expressed in a manner similar to that used

in equations 4.8-4.10. Thus, M,M O dl + X EI

+ 8. X3 = 0

... 4.14

M 2Mo di + I X + X JM!dl + 8 X = 0 EI "" I 2 EI y 3

... 4.15

S S

IJfM!EIdl + I "" X

.Jf M 3EIM .. dl +8. X ,+ 8

X y

2

+

X

2

SM~dl EI

3

=0

... 4.16

366 Since by difinition of the centroid Sx = Sy = 0, equations 4.14, 4.15 and 4.16 reduce to : MIModl + X SM;dl + I X = 0 EI lEI xy2

S

M2 S

M di o + I X EI xy

I

... 4.17

dl + X JMi = 0 2 EI

... 4.18

S

M3Modl+ X SMjdl = 0 EI 3 EI

... 4.19

Equation 4.19 yields directly the value of X3 and the values of X 2 and Xl are obtained from the simultaneous solution of equations 4.17 and 4.18. If further either axis x or axis y is an axis of symmetry, I:w:.y = 0 and equations 4.17 and 4.18 reduce to:

MIMOdl + X f M;dl = 0 X EI I EI ' I

S

J

=_SMIModI/J EI

M;dl EI

and

(M2Modl +X (Midi = EI 2 EI

J

J

0 X =_fM2Mo dl jjMi dl ' 2 EI EI

J

4.3 Applications to the method of elastic centre Example 4.1 Using the method ofelastic centre, calculate the reactions and draw the B.M.D· for the frame shown in Fig.4.3a (Note that this frame has been previously analysed by the method of consistent deformations in example 2.27 (Fig. 2.24).

Solution: The elastic centre will obviously be on the vertical line midway between the two columns. Considering I = If.' the distance ween the elastic centre and the top member will be :

= 2 m.

2 X 6 X 3

y =

2

X

6

+ 0.5

X

y bet-

12

A possible main system is obtained by removing the restraints at a. A rigid arm is attached to the frame at a and extended to c where the redundants XI' X 2 and Xl are assumed to act in the directions shown in Fig. 4.3 b.

I

I· 367

I

12mt

I l

21

6m I

1 b

Q

"-

12

12 m ----l

I...---

(b)

(a)

(c)Mo·D

n:T'~ '" ~ \:1tLl ~,]j 4

4

6

(d)M,.D.

X2"lt 6 (e)M 2·D.

Fig. 4.3

Referring to Figs. 4.3 c-f, which show the Mo,D., the MI.D., the M 2 .D. and the M,.D.,

~

= cO

?!..

SMIModl = _I (4 X 6 X 12 _ 2 X 6 X 12) = EI EI, 2 2 EI,

~c1 SM~t =

=

I 2 E - -I J 2 : 6 (4 + 22_ 2 .X 4)+iXI2X 2X2]

72 EI,

XI = -

YeO

Ye2

=

~eO = ~el

SM2M odl EI

-

72 72

=

rM~dl = _ 1_

=

JEI

=

. -It., I.e. It. (to the left)

432 _1_ (12 X 6 X 6) = EI , EI,

EI,

[2 X 6 X 6 X 6 +

~ X~(62+ 6 2 -

23

504 EI,

X2 = -

Yeo = Ye2

-

432 6 6 = - - t., i.e - t . (downward) 504 7 .7

-

6X6)]

368

"co "c3

=

=

SM

3M"dl =

EI

JM~dl EI

_1_ (12 x 6 x 1) EJ.

72 EIr 18

= _1_ (2 x 6 x 1 x 1 +! x 12 x 1 Xl=EJ. ) EIr 2

X 3 = - -"co = -

-72

18

u e]

. 4 m.t. (antI·cI--'-'~) = - 4 m .t., I.e. ut,;AWlse 12m!

4! 3.14

1.14 3417

(dB.M.D.

(b)

(a)

Fig. 4.4 X" X 2 and X. are shown in their proper sense in Fig. 4.4a. Remembering that a force is equivalent to a force and a couple, the redundant reac>tion components at a are easily found as follows : X. = 1 t. (to the left), Y. = 6/7 t. (downward) and M. = 4 1 X 4 - 6/7 X 6 = 2.86 m.t (anticIockwise) The remaining reaction components may be found from statics and are as indicated in Fig. 4.4 b. The final B.M.D. is shown in Fig. 4.4c. Note that this diagram is identical to that previously obtained by the application of the method of consistent deformations.

+

L

2m .---

41

-v--

T--'-t-- ."C

--,,.-'

r,eu

~

(b)t.\,.D.

(a)

6mt

6

tl Jlrll

6 (d)MxD.

6 Fig. 4.5

12

2m!

2

J~~l 4

(e)M\D

4

l

369 Alternative solution : The frame can he equally solved hy cutting the top member at a section taken at its mid point, and by assuming that two rigid arms are attached to the two parts of the frame and extended to the elastic centre c as shown in Fig. 4.5a. The Mo.D. due to the applied loads and the B.M.Ds. due to unit values of Xl, X 2 and X3 are shown in sue.. cession in Figs. 4.5 hoe. Referring to these figures,

E..

3 = JMIModl = _I [(2 X 6 _ '1: X 6) 12] = _ co EI EI, 2 2 EI,

I , 2X 6 2 = JMfdl = _ 1- [ -3(4 +22_2 X 4) + 2 x2xI2x2 EI EI, J = -

72

EI, 3eo 72 = = 3d 72

XI = -

+

It.

Yeo = fM2Modl = _ _ 1- (12 X 6 X 6) = _ 432 ill EI, EI,

Yc2 =

f~dl = EI

E.

_ 1- [2 X 6X6X6+!.. X (6 2 + 6 2 _6 X 6)] EI, 2 3

504

EI,

X2

= - -Yeo =

a

= JM1Modl = _ 1- (12 X 6 X I)

Ye2

EI

cO

ac3 =

X,

f

"C"" _

M~dl

432

504

=

+ 6- t. 7

72

EI ,

EI,

~

18

EI

= _1- (2 X I X 6 X I + X 12 X I X I = ) EI, EIr 2

aeo

= _..I!..

e,

= -

4 m.t.

18

which are consistent with the values obtained previously.

370

EsaJnpJe 4.2 Using the method of elastic centre, construct the B.M.D. of the frame shown in Fig. 4.6a if EI = constant.

31 211m

r

e

1

5m

La

7m

L:~6m~~ii (a)

31

x=3m 211m

2

32

(b)

(e)Mo·D.

5 (d)M,.D.

1ml

3 (e) Mz.D.

(t)MJ.D.

(g) B.M.D.

Fig. 4.6 Solution: The elastic centre is determined first. This is done with respect to the beam and the right column as follows:

x = 5.38

X 7 + 6 X 3 _ 55.7 '" 3 m. 5.38 6 7 - 18.38 .

+ +

371 /

y=

5.38 x 2.5 + 7 x 3.5 5.38 + 6 + 7

=

37.97 '" 2 m. 18.38

The frame is cut at a section just to th e right of corner b and two rigid arms connect the two cut ends of the frame to the elastic centre as shown

in Fig. 4.6 b. The Mo.D. due to the applied loads and the B.M.Ds. due to unit values of the redundants X h X 2 and X3 are shown in succession

in Figs. 4.6 c-f. Seo

+

X I Sci

Yeo

+ +

XI Yel

ncO

The deflection equations will be :

+ +

X3 acl =

X 2 Se2

=

Xl Yc2 =

0 0

0

Using the method of virtual work and referring to Fig. 4.6,

213.6 EI Sci

=

SMi dl = _I [5.38 (2 2 EI EI 3

+

+ 37-

=~ [2

Sel _ SMIM2 dl EI

+3 X6 2

EI

32

(2 2

X 5.38(3 2

X 2 _ 3 X6 X 2 2

2 X 3)

_

+ 52 _

+ ~)

+

+

6 X 2 X 2 81.1

2 X 5) ] = - EI

3 X 5.38(3 2

3

+ ~ X2) 3

5 X 7 X 3 _ 2 X 7 X 3] 2 2

16.4 EI 'Yeo

=

SM 2 M odl EI

=

2- [ 6

EI

X 5.38 (3 2

(4~ X 3 'Yet

=

16.4

4~

+~ 3

X 2)

+~

X 3)- 32 X 7 X 3].

8c2 = - - from Maxwell's theorem EI

X 6 X 36

3

710 =-ru

372 M2dl 1 [5.38 6 _2_",,_ - - (3"+5"+3x5)+- (32+32-3X3) EI EI 3 3 169

J

Yc2

a

+ 7 X 3 X 3] = -EI-

SM M

I 3 Odl. 1 [5.38 X 6 X 1+.3 X6X36XI+32X7Xl] EI EI 2

=

co

312.1

--EI

a I c

SMi

dl EI

=

= -..r.. (5.38 X 1 X 1 + 6 X 1 X I + 7 X I X 1) EI

18.38 EI Substituting these values in the deflection equations,

-

213.6 + 81.1 XI + 16.4 X 2 = 0

(i )

-

710

( ii)

+ 16.4X I + 169X2

=

0

312.1 + 18.38 X3 = 0 F rom

.

equatlOn

( ... ) X III

~

3

(iii)

312.1 = -18.38

17 m.t.

Solving equations (i) and (ii) simultaneously, XI

~

1.8 t.

and

X2

~

4 t.

The reactions may now be calculated and the final B.M.D. drawn and will be as shown in Fig. 4.6g.

I

373

I

Part 2

l-

The Method of Colwnn Analogy

F !

II

4.4 Introduction The cJlumn analogy is a special method of analysis applicable to frames of the one-Io:>p type and statically indeterminate up to and including the third degree. Fig. 4.7 shows examples of such frames. The me~hod is based on the analogy between stresses in an eccentrically loaded short column and the redundant moments in the frame, and hence the name.

Fig. 4.7

4.5 DevelopDlent of the Dlethod of colUDlD analogy The method will be developed with reference to Fig. 4.8.

y b,p:JITIrr:qc

-A-Ilumn and following the sign convention used in reg ard to the general equation of stress (section 8.11, Part 1) with which the student is already familiar.

Compressive stress will then correspond

to negative Mi (moment causing tension in the outer fibers), and conversly tensile stress to positive moment (moment cau--_ _ 12m ( a)

2. 31 (c)

( d ) B.M.D

Fig. 4.12 Solution: Two different main systems, corresponding to the concentrated

and uniformly distributed load systems.. are considered . The first is obtained by removing the hinge support at d and the second by replacing the fixed support at a by a roller. The resulting Mo-diagram is shown in

Fig. 4.12b. Properties of the section of the analogous column,

384 there Since suppor t d is hinged , an infinite area will be concen trated analothe of ~cction The co and hence the centroid coinsides with d. A = momen ts of gous column , assumi ng EI = 1, is shown in Fig. 4.12c. The throug h d inertia and produc t of inertia with referen ce to x and y axes arc to be found next.

Ix =

C_·5_X_0_;_~_X_7_.5_2 + 3.75

+ Iy = 3.75

122

X

X 122

+ (3

Ia. = 3.75 (- 12) (2.25)

+ 12

X 2.25 2)

(I

X 0.25 X 6 2 2

X6 X6 12

+6

X 32) = 216.7 .

) X 6 2 = 684

12

+3

+

3 (- 6) (6)

+ 6 to)

(3) = -

202.5

Straini ng actions ,

N,

= ( 15

2= 28.125 t

X 7.5) I 2

N2 = ( 2/3 X 12 X 72) 1/4 = 144 ~

Mx

=

28.125

+

My

=

144

6-

X

144 X 6 = 835.87 5 28.125

X

12

=

...t

526.5 ~+-

Indeter minate momen ts,

_ N M,_

A

+

o+

+

MyI" --- M"I XY

I"Iy ---

2

IXY

x

+

MXIy -

I"Iy ---

MyIXY 2

Y

IXY

(526.5) (216. 7) ~ (- 835.87 5) ( - 202.5) 216.7 X 684 - (- 202.5) 2 (- 835.87 5) (684) -(526 .5) (- 202.5) 216.7 X 684 - (- 202.5) 2

Mb = -

0.515( - 12)-4 .338(- 1.5) - 12.687 m.t 0.515{ - 12)-4. 338(6 ) - -19 .848 m.t

Me = -

0.515(0 ) -4.33 8 (6) - - 26.028 m.t

M. = -

Final momen ts, M.= - 15 + 12.687 - - 2.313m .t

Mb=

Me =

0- 19.848 - -19.84 8m.t 0 - 26.028 - - 26.028 m.t

The final B.M.D . is thus as shown in Fig. 4.12d.

x

y - -0.51 5x-4. 338y

I:-

•

385 I!.aJnple 4.7 Using the method of column analogy, construct the B.M.D. for the frame shown in Fig. 4~13a. The moment of inertia varies as indicated. 21

e

21 3m

1

31

a

1

4m

x

L .!: I

3 _____ 3m

(a) M - 6.57 58 Y Tl2

(b) ""00

=14.22

·Ib9

:1 (e)

(d)BM.O.

Fig. 4.13

Solution: A convenient main system is obtained by releasing the right support. The resulting Mo.D. is thus as shown in Fig. 4.13b. The section

of the analogous column assuming EI = I is shown in Fig. 4.13c. Referring to this figure,

+ 3 X 1/2 + 5 X 1/3 = 3 + 1.5 + 1.67 = X 1.5 + 1.67 X 4.5 = 1.58 m.

A = 1 X 3 X=

y =

1.5

6.17

6.17

3 X 1.5

+ 1.67 6.17

X 2

1.27 m.

Before proceeding any further" it will be helpful to discuss a particular problem which is encountered quite often. This is the evaluation of the moments and pr.oduct of inertia of a long and narrow rectangular section with respect to axes which are at an angle with the rectangle. '

r

y

/-l~-'---.

v

Fig. 4.14 Referring to Fig. 4.14, L/2

I. 2J

b dl (I sin 8)'

=

=

2b sin2 8 L 3 /24

o

•

Av' 1=• 12

."

i-.3:;

L/2

:r,. = 2J bdl (I cos 8)' = 2b cos' 8 L 3/24 o Ah' 12

:r,.=-

•.. 4.36

L/2

I., 2J

b dl (I sin 8) (I cos 8)

=

=

2 b sin 8 cos 8 L 3 /24

o I.,.

Avh 12

=-

••• 4.37

Applying equations 4.35, 4.36 and 4.37, the moments and product of inertia of the section of the analogous col umn are as follows :

.

I

387

=

1,.

=

3 X 32 + 3 12

3

X

1.5

1.58 2 +

X

0.23 2 + 1.5

32

X

12

1.67 X 4 2 1.272 + --,-12 + 1.67 X 0.73 2 = 7.94

X

+ 1.5

X

2 1.67 X 3 2 0.08 + - -12

+ 1.67 =

I.,.

X

1.67

3 (- 0.23) (- 1.58)+ 1.5 (- 0.08) (1.27)

2.92' = 24.12 3 12

X

X

4

+ 1.67 (2.92) (- 0.73) = - 4.29

N

Straining actions, 3 X 6

- - x l = 9t..+9

2

-

=

9 X 0.73 9 X 1.58

=

=

6.57

+

t. -

6.57

14.22_... -14.22

Indeterminate moments,

9

=-+ 9.17

(-14.22) (7.9'1) - (- 6.57) (- 4.29) x 7.94 X 24.12 - (- 4.29)"

(- 6.57) (24.12) -(-14.22) (- 4.29) + 7.94 X 24.12-(-4.29)2 Y + 1.4586 - 0.8152 x - 1.2681 Y Mia = 1.4586 - 0.8152 (-

1.58) -

1.2681 ( -

1.58) -

1.2681 (1.27)

Mlb

=

1.4586 - 0.8152 ( -

Mi.

=

1.4586 -

M ..

=

1.458'; - 0.8152 (4.42) -

0.8152 (1.42) -

1.2681 (1.27)

1.73) = 4.94 m.t.

=

=-

1.2681 ( - 2.73)

=

1.14 m.t. 1.31 m.t. 1.31 m.t.

F.na1 moments, The final moments at all the joints are the same as the indeterminate

moments except at a where M. = - 6 + 4.94 B.M.D is thus as shown in Fig. 4.13 d.

=-

1.06 m.t. The final

388

&alllple 4.8 The frame shown in Fig. 4.1Sa has stepped columns. Draw the B.M.D. for the frame under the given crane loads if the moment of inertia varies as indicated. d

31

9

mt

3

9

1--_12m _ _

(bl"'\,.D.

(al ,,\,407 y

3EI,11

.~~____~==~!J8mt

•

2.06

.09

3.4

0.54

(dl B.M.D.

(el Fig. 4.15

SoIutioll : A convenient main system is obtained by cutting member cd at any section along its length. The resulting Mo.D. is thus as shown in Fig.4.15b. The section of the analogous column assuming 3EI = I is

shown in Fig. 4.15c.

A

=

Y

=

I,

=

12 X I

Referring to this figure,

+ 2 (3

+2

X 3

2 (3 X 3 X 1.5

+6

54.

X 6) = 54

X 2 X 6)

= 3.17

3 X 33 (12 X I) 3.172 + 2 ( '--'1"-2- + 9 X 1.67 2

+ 3

I X 12 =----+ 2 (9 12

X

62

+ 12

+

2

X

63

12

12 X 2.83 2) = 448

X 6 2 ) = 1656

389 Straining actions,

NI

=

Nz =

=

(3 X 6 X 2) (9

X

6

X

2) =

+ 36 + 108

N

=

36

+ 108 = + 144

M.

=

(36

+ 108) 2.83 = 407.f

My =

(108 -

36) 6 = 432-++

Indeterminate moments,

N

My

M.

A

M; =

±

144

1. Y ± I; x

407 432 448 Y ± 1656 x

+ 54 ±

=

M;

=

2.67 ± 0.9085 Y± 0.2609 x

M.

=

2.67

+ 0.9085 (5.83) + 0.2609 (6)

=

9.54 m.t.

M, = 2,67 - 0.9085 (0.17)

+

=

4.09 m.t .

¥. =

+ 0.2609 (6) =

1.36 m.t.

2.67 -

0.9085 (3.17)

0.2609 (6)

Md = 2.67 - 0.9085 (3.17) - 0.2609 (6) = -

Mr = Mb

=

2.67 2.67

+

1.78 m.t.

0.9085 (0.17) -

0.2609 (6)

= 0.94 m.t.

0.9085 (5.83) -

0.2609 (6)

=

6.40 m.t.

Final moments,

M,

=

9.54 -

M. (belo:w)

=

9.00

= 0.54 m.t.

4.09 -

=-

4.91 m.t.

3.00 = -

2.06 m.t.

9.00

M. (above) = 4.09 m.t.

Me

=

Md

=-

1.36 m.t. 1.78 m.t.

Mr (above)

=

0.94 m.t.

Mr (below) = 0.94 -

Mb = 6.40 -

3

= 3.40 m.t.

The final B.M.D. is thus as shown in Fig. 4.15 d.

,

I.

I

390 Ezaxnple 4.9 Using the method of column analogy, construct the B.M.D. for the frame shown in Fig. 4.16a. The moment of inertia varies as indicated. 31

b

1.21

(c)

(d)B.M.D.

Fig. 4.16 Solution : A convenient main system is obtained by releasing support d. The resulting Mo.D. is thus as shown in Fig. 4.16b. The section of the analogous column assuming EI = I is shown in Fig. 4.16c. Referring to this figure, A

= 2 (5 X I) + 6 X 2 (5x I x2) 15

y

=

5 X 6

12

2

5 = 15 6

4 3

2 [5 X 3 2] + 2 ---'-:-1-2:"" +5 (4.5) = 225 ,-

Straining actions,

N

=+ (

12 X 2

5) 1=+30

391

M,: =

30 x 5

=

-++

150

Indeterminate moments"

M,

=

~ ± M. y ± M,x Ix

1\

30

=

M.

15 ±

=

2 ± 1.5y ± 0.66 x

=

2

Mb

+

2 -

Me = 2 -

M.

If

40 X 3 150 80 Y ± 225 x

=

2

+

!) +

0.66 (6)

+

10 m.t.

0.66 (3) =

+

2

1.5( ; ) -

0.66 (3) -

-

2 m.t.

l.5( :) -

0.66 (6) -

+2

1.5(

1.5( : )

+

m.t.

m.t.

Final moments,

The final moments of all the joints are the same as the indeterminate moments except at joint a where M. = 12 + \0 = - 2 m.t. The final B.M.D. is thus as shown in Fig. 4.l6d.

392

Ezample 4.10 Using the method of column analogy, construct the B.M.D. for the frame of constant EI shown in Fig. 4.17a.

211m

LCi1: 1D ::ITII:JIITIII~i:r:TIID

64mt

64'---~~~--~16~

3m

fa

e

8

6 b

&4

48

--1-4-1

.....Lam (a)

'b) M•. D

fM x=4852.8 41.39 ml

x My =768

Y

2f.2~5~m~~~~~~--..-....,

r

3.75

+-.

.

.

6t6m-l

I N2:384

I-

N31

N,=384

EI=l

25.54

30.69 (d) B.M.D.

(e)

Fig. 4.17 Solution: A main system is obtained by disconnecting the hinge at c. The resulting Mo.D. is shown in Fig. 4.17b. The section of the analogous column and the corresponding straining actions is shown in Fig. 4.17c. It should be noted that since an infinite area is concentrated at the location of the hinge" the centroid coinsides with c. Also, since"the indeterminate bent is symmetrical, x and yare the principal axes; IXY = O. Properties of the section of the analogous column,

A I X 63

2 --( 12 Iy

=

2 (6

X

+6

82

X

+ 8.55 12X

Straining actions"

NJ

=

(64

X

62

6) = 384

+ 8.55212X 8

2

+ 8.55

32

X

+ 8.55 4

2)

X

)

1.5 2 = 519.2

1132.4

393

N2

=

= M. = M, = N.

N3 = (0.33 X 64 X 8.55) 1 = 182.4 (48 X 6) 1 = 288 (384 + 288) 6 + 2 X 182.4 X 2.25 = 4852.8t (384 - 288) 8 = 768-++

Indeterminate moments,

M.

M,

= -Ix

M.

=

Md

=

M,

= =

M,.

y

M,

+ -Iy

9.346 9.346 9.346 9.346

X X X X

x

4852.8

+ 0.678 + 0.678

9 3 39-

768

=-y +- x = 9.346 Y + 0.678 x 519.2 1132.4 X X 0.678 X 0.678 X

8 8 8 8

= 89.54m.t. =

33.46 m.t.

= 22.61 m.t. = 78.69 m.t.

Final moments,

= Md = -

+ +

64 89.54 = 25.54 m.t. 64 33.46 = - 30.54 m.t. M, (above) = - 64 + 22.61 = - 41.39 m.t. M. (below) = - 48 22.61 = - 25.39 m.t. Mb = - 48 78.69 = 30.69 ITLt. M.

+

+

The final B.M.D. is thus as shown in Fig. 4.17d. 4.8 Application of the D1ethod of colUIDD analogy to statically iadeterDlinate beams

The method of column analogy is applicable to the analysis of single span beams of the type shown in Fig. 4.18 whether they are prismatic or nonprismatic.

Fig. 4.18 The procedure of analysis is essentially the same as that outlined in section 4.4 in regard to frames. In this case, however, the section of the

analogous column is a narrow strip of zero moment of inertia about the and hence equation 4.33 reduces to :

x~a](is,

M,

N

M,

= A+lx y

... 4.38

and the final moment is again determined from the superposition equation, M = ~ M,

+

394 E>cunple 4.11 Determine the end moments for the beam shown in Fig. 4.19a if the moment of inertia varies as indicated.

(a)

~~a 1--::2!-r------: II~b~ -;--1-';--

LJ4 -l-LJ4 -+--LJ4 --t- LJ4

(b)"\,.D.

PU4 (e)

Fig. 4.19 Solution: A possible main system is obtained by replacing the fixed ends by simple supports. The corresponding Mo.D. is shown in Fig. 4.19b. Assuming EI = I, the section of the analogous column will be as shown in Fig.4.16c. Referring to this figure, A = 2 (0.25 L X I N

+ 0.25 L

X 0.5) = 0.75 L

I PL L ) (PL PL L) I] 5 PL2 =2 [( "2Xa x 4" 1+ 4" x SX"4 ="64.1.

"2

Since here My = 0, the moments at the fixed ends are given by :

5PL2/ 5 PL M. =Mb=-M 0.75L=-~ which are the final moments since the Mo-moments at points a and bare both zero. It should be remembered that the main system considered was arbitrarily chosen. Any other main system would have yielded the same result for the final moments, but the main system considered is the easiest as it takes advantage ofthe symmetry ofloading on the section of the analogous column.

.~

395 EzaJnple 4.12 Find the moment at end a and the reaction at support b of the beam shown in Fig. 4.20a if the moment ofinertia varies as indicated.

wllm

(a)

a

21 LJ2

1 LJ2

(b)~.D.

y EI=1

112

Ie)

y Fig. 4.20 Solution : A possible main system is obtained by replacing the fixed support at a by a hinge. The corresponding Mo.D. is shown in Fig. 4.20b. Since the beam is simply supported at b, the section ofthe analogous column will have an infinite area concentrated at b and its centroid will coinside . with b as shown in Fig. 4.20c. Referring to this figure" A

=

'"

= 0.5 (0.5 L) 3 + (0.5 X 0.5 L) (0.75 L) 2 + 1 (0.5 L),/12 12 + (I X 0.5 L) (0.25 L)2 = =_1 2

N2

(.! 3

X ~ X WL2) = wL3 2 8 48

2

=

I (

3

L

X

WL2)

'2 X ""8

= WL3(-2 X.!:.. X ~) + 48 8 2 2

=

wL3 24

J. J.

WL3(~ X.!:..) = 24

8

2

7wL4 ++256

7wL4 L 7wL2 = - 256 X 3 L3/16=-

48

which is the final moment since the Mo-moment at a is zero. 7wL2 wL2 17wL ~Ma=0=LXYb+4i\-2 'Yb = 4 i \ '

3L3 16

396 4,9 Determination or stiffness and carry-over factors Perhaps the most important application of the method of column analogy is the determination of the stiffness and carry-over factors of nonprismatic members. A knowledge of these values is essential to the moment distribution method of analysis which will be presented in chapter 6. At this stage it will be advantageous to define these new terms. Stiffness : The bending stiffness at end a of member ab, which is usually denoted by Kabj is the moment required to produce a unit rotation at end a while end b is supported in some particular manner; usually fixed. Carry-over factor: The carry-over factor from end a to end b of member ab, which is usually denoted by Cab' is the ratio of the moment induced at end b to the moment applied at end a. The stiffness and carry-over factors at an end of a member depend on the shape of the member and the restraining conditions at its other end. Consider a prismatic member ab such as that shown in Fig. 4.21, and let it be required to determine the stiffness Kab and the carry-over factor Cab' If a moment Mab is applied at end a causing this end to rotate' an angle aa = I then by definition Mab = K ab • Also, as a result of M ab, some moment Mba will be induced at end b preventing this end from rotation, and again by defini tion Mba = Cab K ab •

(al

(bl

Fig. 4.21 In order to determine Kab and Cab the usual procedure would be to load the section of the analogous column by the moments on the beam. These moments should satisfy the following displacement boundary conditions :

397

(I) The net area of the (M/EI)-diagram between a and b should be equal to aa since ab = O. (2) The moment of the (M/EI) - diagram between a and b about a should he equal to zero since there is no relative displacement between the two ends. A concentrated load = I rad. applied at the rotated end a satisfies both these conditions,and can therefore replace the actual elastic loading on the beam. Consequently, whenever the stiffness and the carry-over factors are required, an equivalent concentrated load of a unit value, is applied to the section of the analogous column at the point corresponding to the rotated end of the beam. This rule applies regardless of the shape of the beam.

Referring to Fig. 4.21 b which shows the section of the analogous column corresponding to the prismatic beam considered" K

-f

ab -

N Mx X L/2) --+- -I+ (IL3/12EI A I - L/EI

a -

(L/2)

The stiffness of a prismatic member is thus given by : Kab =

... 4.39

4 EI/L

I

(I X L/2)

L/EI

L /12 EI

Kab Cab = fb = - -

Substituting for Cab --

3

(L/2) = -

2 EI/L

KabJ

I

-2

-

The negative sign results from the sign convention used in the method of column analogy. The carry-over factor from one end to the other end of a prismatic member is given by : C

I

a

b

= -2 .

...

4.4{)

The student is advised to memorize the stiffness and ~rry-over factors of prismatic members derived above as they will be needed later in regard to the method of moment distribution. ,

398

Esanlp le 4.13 Calcula te the stiffness factors at ends a and b and the carryover factors from a to b and from b to a of the nonpri smatic beam shown in Fig.4.2 2a.

Solutio n: The section of the analog ous column is shown in Fig. 4.22h. Referr ing to this figure ..

0.8 L (1/2EI) + 0.2 L (I/EI) = 0.6 L/EI (0.4 L/EI) 0.4 L + (0.2 L/EI) 0.9 L x = = 0.567 L 0.6 L/EI 1 1 0.4 L 1 Iy = 2EI (0.8 L)3 X 12 + EI (0.167 L)2 + EI (0.2 L3)

A

=

X

+ 0.2

1 12

L (0.333 L) 2 = 0.055 L3/EI EI

To determ ine Kab and Cab' place a unit load at a,

N K.b

=

f. =

Mx

A +1

=

EI (1 0.6 L +

X

0.567 L) (0.567 L) EI 0.055 L3

= 7.5 EI/L

EI K b C b= fb = - - a a 0.6 L Cab

(I X 0.567 L) (0.433 L) EI _

0.055 L3

-

-

2.8 E

IfL

= - 2.8/7.5 = - 0.37

To determ ine Kba and Cba' place a unit load at b,

N

K ..

=

~ _ (I. X 0.433 L) (0.433 L) EI = _ 2.8 EI/L 0.6 L 0.055 L 3 -2.8/5 .1 = -0.55

=

a

=

EI (1 X 0.433 L) (0.433 L) EI 0.6 L + 0.055 L3

5.1 ElfL

= f

Kba Cab

C ..

Mx

= fb =A: +1 =

EXAMPLES TO BE WORKED OUT (1)-(8) Using the method of elastic centre, construct the B.M.D for each of the frames shown in Figs. 3.52,3.53 and 4.23-4.28. The of inertia moment varies as indicated

8t

r

-E

4m

4m

Ul

~3m

3

----I

7-

Sm Fig. 4.24

Fig. 4.23

2t1m

f 4m

t

21

31 21

21 21

4m

1 LI Fig. 4.25

111m

31 __ 2..L-Sm Fig. 4.26

~2...J

400 (9)-(14) Using the method of column analogy, r.e-solve prt>blems 1-8 (Figs. 3.52. 3.53 and 4.23-4.28). (15) - (18) Using the method of column ilJlalogy qmsttuct the , B.M.D. for each of the frames in Figs. 3.47, 3.48, 3.54 and 3.55. (19)-(26) Using the method of column analogy, ennstruct the B.M. and S.F.Ds. for each of the frames shown in Figs. 4.29- 4.36. The moment of inertia varies as indicated.

L

t

4

4

L-""____

L4 -L--- 12

~~ ~j

m

II ",:" ~ 2t1m

L- 3 - L - 6 m ---l.- 3 -I

FigA.29

Fig. 4.30

iii

311m

16

I

6m---l-~;f

4m

~2i L3

Fig. 4.32

Fig. 4.31

,11

-,

21

3m 2t

2

211m

21

6

I

J

8m Fig. 4.33

Fig. 4.34

i

4

J

401 211m I I .. I "

8t

I I i I I I

2~/.m

8t

1

21

21

6m

..J 2 1-" --I-4m

t

4-1-4

Fig. 4.36

Fig. 4.35

(27)-(30) Using the method of column analogy, determine the moments and stiffnesses at ends a and b, and the carry-over factors from a to b and from b to a for each of the beams shown in Figs. 4.37 - 4.40.

~Q J

6mt

21

4m

O~ b~ I

~Q

~

I

6m

t

b~

I 3---1

Fig. 4.38

Fig. 4.37

F=::=l==~~·1.6m

/1

1- 3 -J,.....: 6m-l--3 .. Fig. 4.39

151

l~! 14 t b~l

r I--

~~8

IS

4x3 =12m

Fig. 4.40

--J"

I

!•

r

f

!

403

CHAPTER 5 THE SLOPE - DEFLECTION METHOD

5.1 Force and displaceD1ent D1ethods of analysis In all the methods of elastic structural analysis, two basic conditions must be fulfilled. These are (1) Static equilibrium. (2) Compatibility of deformations. Two main methods of analysis are possible. In the first, static equilibrium is always satisfied and equations are established to express the second condition; compatibility of deformations in the structure. Since these equations use forces as unknowns. . the method is referred to as the force method. The method of consistent deformations presented in chapter 2 and the method of the equation of three moment.s presented in chapter 3 are examples of the force method. In the second method,compatibility of deformations in the structure is always maintained and equations are established to express the first condition;static equilibrium of the structure. Since these equations use displacements as unknowns} the method is referred to as the displacement method. The slope-deflection method to be considered in this chapter and the moment distribution method to be considered in the following chapter are examples of the displacement method.

5.2 Degree of freedoD1 The number of possible joint rotations and independent joint translations in a structure is called the degrees of freedom of the structure. There are three types of degrees of freedom (1) The degree of freedom in rotation The degree of freedom in rotation of a beam or a frame is equal to the number of possible joint rotations. It follows that if:, sr = degree of freedom in rotation}

j

number of joints including supports}

f

number of fixed supports,

s, = j - f

5.1

404 (2) The degree of freedom in translation The degree offreedom in translation of a frame is equal to the number of independent joint translations which can be given to the frame. Since in general each joint has two joint translations" the total number of possible joint translations = 2 j. Since, on the other hand, each fixed or hinged support prevents two of these translations, and each roller support or connecting member prevents one of these translations, the total number of the available translational restraints is (2f 2h r m), where

+

f

number of fixed supports"

h

number of hinged supports,

r

number of roller supports,

m

number of members.

+ +

The degree of freedom in translation, sp is thus given by : St

=

2j -

(2f

+ 2h + r + m)

5.2

(3) Combined degree offreedom The combined degree of freedom of a frame" s, is the sum of its degrees of freedom in rotation and translation. Thus,

5.3 As opposed to the degree of statical indeterminancy, n, which refers to the number of redundant forces, s is sometimes called the degree of kinematic indetenninancy. If s = 0, the structure is said to be kinematically determinate. The fixed beam is an example of a kinematically determinate structure.

5.3 Outlines of the slope-deflection

D1eth~d

The slope·deflection method is applicable to the analysis of statically indeterminate beams and frames. It is particularly useful for the analysis of highly statically indeterminate structures which generally have a low degree of kinematic indeterminancy.

~

I

I

II

It

!

I

405 ~~a~

________~b~________-TC________~d

e

f Fig. 5.1

This is demonstrated py the frame shown in Fig. 5.1. The frame is nine times statically indeterminate. This calls for the simultaneous solution of nine equations if the force method of analysis were to be used. On the other hand, the frame has only two unknown rotations.. ab and ac, and is thus kinematically indeterminate to the second degree. This calls for the simultaneous solution of two equations only if the slope-deflection method is used. The general procedure for the analysis is outlined below: (1) The degree of freedom in rotation and translation are found separately. This is done either by inspection or by the application of equations 5.1 and 5.2. (2) The moments acting on the ends of individual members of the structu(e are expressed in terms of the rotations of the joints at the ends of the member, the rotation of the member itself (if any) and the load acting on the member. This is done by the slope-deflection equations to he derived in section 5.5.

(3) The conditions of static equilibrium provide as many equations as there are joint and member rotations. When solved simultaneously, these equations give the unknown rotations . (4) Back-substitution of the thus-determined rotations into the slopedeflection equations yields the moments acting on the ends of individual members.

5.4 Sign conventions Before proceeding to derive the basic slope-deflection equations, it is necessary to establish specified sign conventions for the moments and rotations.

406 Moments acting on the ends of members, joint and member rotations are considered positive when occuriDg in a clockwise direction. For example, a fixed beam acted upon by a downward load will have negative and positive moments at its left and right ends respectively. Also, if the right end of a member sinks with respect to the left end, the member rotation is positive.

This sign convention, though suitable for considering joint equilibrium" contradicts the graphical presentation of the bending moments along the length of a member where moments producing tension in the lower fibers of the member are considered positive and vice versa. Nevertheless, once the end moments are obtained it is quite easy to change over to the usual graphical convention for plotting B.M.Ds.. This is done by drawing freebody diagrams for individual members with the end moments indicated according to the sign convention used in the slope-deflection method. 5.5 Derivation of the slope-deflection equations Fig. 5.2a shows a single member ab of a continuous beam or a frame . As a result of the loads applied to the structure, the member considered will assume the given general deformed shape. Ends a and b have been caused to rotate through angles aa and ab and the axis of the member has rotated an angle.p = t. /L as a result of end b settling an amount t.

Fig. 5.2

relative to end a. It is now required to find expressions for the final moments M .b and Mb • acting on ends a and b in terms of the rotations aa and Ub at joints a and b, the rotation of the member, ifJab and the loads acting on the member.

407 First assume, as in Fig. 5.2h.. that ends a and b are fixed, i.e. with ~ero rotation at the ends. The loads on the member produce the fixed-end momeu'ts MPab and MFba indicated in Fig. 5.2h in their assumed positive direction. These fixed-end moments must be corrected to allow for the end rotations a, and ab and the member rotation ",. The effects of these rotation. will be found separately and the final result will then be obtained by superposition. Il,l Fig. 5.2c the tangent to the elastic curve at a is rotated by a moment M'a while end b is kept fixed. Using the second moment-area theorem it may be easily shown that M' b = t M' a' and from the first momentarea theorem, a. = M' aL/4EI. H'ence,

M'. = 4 EI a./L and M'b = 2 EI a./L. Similarly, with reference to Fig. 5.2d it may be shown that ;

In Fig. 5.2e, end b is shown displaced relative to end a by an amount!::i while the tangents to the elastic curve at a and b are kept fixed against rotation. The end moments Mill a and M'" b may again be expressed in terms of the member rotation ,p using the moment-area theorems. Thus,

Mm.

= Mm

b

= - 6 EI !::ilL' = - 6 EI of/L.

Ifnow, for each end ofthe member all the end moments corresponding to the various deformations are added to the fixed-end moments, the following expressions are obtained :

... 5.4

Equations 5.4 are the basic slope-deflection equations. If there is no relative displacement between the ends of the member, of = 0 and these equations reduce to :

.. . 5.5 Mba = MF b.

+ 2 EI/L (2 ab + a. )

:408 It should be noted that the coefficient (2EIJL) appearing in the slopedeflection equations-is generally different for each member in the structure. If the (2EIJL)-values for all the members are made n times smaller (or larger), the effect will only be to make all the rotations n times larger (or 3.p) and (2 EI/L) smaller) while the products (2 EI/L) (2 (2 a. a. - 3.p) in equations 5.4 or (2 EI/L) (2 a. a.) and (2 EI/L) (2 ab + aa) in equations 5.5 remain unchanged, and consequently the values of the end moments remain 'also unchanged. Thus., if the actual values of the rotations are not of interest, as is usually the case.f the relative values of (2 EI/L) rather than their absolute values may be used in the slope-deflection equations. If the relative values of (2 EI/L) are denoted by K, the slope-deflection equations in 5.4 and 5.5 may be restated -a s :

a. + a. -

+

M •• = MF••

+ K •• (2

a. + a.

+

5.6 -3.p)

and

...

5.7

5.6 Fixed-end DlODlents MFab and MFba appearing in the slope-deflection equations are the moments that develop at the ends ofa member assuming it to be restrained against translation and rotation as in the case of a fixed-end beam. Expressions for these moments due to common cases of loading are given in Fig. 5.3. Expression for other cases of loading may be found from the first principles or from the given expressions using the principle of superposition. It should be remembered that the given expressions ar~ for prismatic members.

409

M~b

wI! -12

a~

f

-7

~

I---

r

8

I----

Pa(L-a) L

Mb(2a-b)

t3 'tEl (12-11) d

I

a

~

PL

r

L12

I--

t=-

a

12

I

~

Pba 2

~

PL

~

Pa(L-a) L

~

Ma(2b-a)

t3

I

8

L12 ----'

r

~f-a-.J

wI!

/.

b

I

PI

~

~b

I

L

!

z Pab

Mba

w I I

I

r

F

Loading Case

L a ---
, ""'Z7

I

5614 2

3'i"-37!..!2,--;----,3"1.372 (d)S.F.O.

/?\-

~

~

6.386

h--.

1c::=+:C=l-I___+I""-2+-=:::::::::"....~=-_-f2=+=:i2t

3.8'"'2"'"8----:3~.828 Fig. 5.5

~14

I

412 Solution : Since there is no joint translation, form 5.7 of the slope-deflection equations may be used. The relative stiffnesses and fixed-end moments are as follows :

K'b : Kbc = 1/6 : 1/6 = I : I MFab = -

PL/8 = -

7.2

MPbC = - wL2/12 = - 2

X

6/8 = -

X 6 2112

5.4 m.t., MF b• = + 5.4 m.t.

--: -

6 m.t., MFcb = + 6 m.t.

Here aa is known to be zero and there are two---unknown a. - values; and ac . The slope-deflection equations are:

Mab Mba Mbc Mcb

= = = =

+ +

5.4 5.4 6 6

+ + + +

I (2 aa + I (2 ab + I (2 ab + I (2 a c. +

Q.b

ab)' aa), a c)' ab)'

M.b = - 5.4 + ab Mba = + 5.4 + 2 ab Mbc = - 6 + 2 ab + a c MCb == + 6 + ab + 2 a c (Note that a slope-deflection equation needs not be written for cantilever cd as it is statically determinate. Considering a section just to the right of c, Mcd = - 2 X 2 = - 4 m.t. The negative sign for the moment is required since the moment acting on the end of member cd. is anticlockwise . With two unknown rotations, two equations of statics are required. These are provided by the conditions of equilibrium at joints band c.

At joint b, At joint c,

Mba + Mbc = 0 Mcb - 4 = 0

It is of particular importance to note that all the unknown moments should be assumed positive while any known moment should be substituted for in the equilibrium equation with its known value and sense of direction.

Substituting from the slope-deflection equations into the joint equilibrium equations,

+

(i ) (ii)

4 ab a c - 0.6 = 0 ab+2ac+2 =0 Solving equations (i) and (ii) simultaneously, ab = 0.457 and a c = - 1.23 Substituting these values in the slope-deflection equations.. Mab = -

5.4 + (0.457) = - 4.943 m.t.

M.,. = + 5.4 + 2 (0.457) = 6.314 m.t. Mb< = - 6 + 2 (0.457) + (- 1.23) = - 6.314 m.t. Mcb = + 6 + (0.457) + 2 (- 1.23)

= 4.000 m.t.

I

413 Having thus determined the end moments, the free-body diagrams for all the members may be drawn as shown in Fig. 5.5b. The correspond· ing B.M. and S.F.Ds. are obtained in the usual manner and are as shown in Figs. 5.5c and d respectively.

Example 5.3 AU'lly,e the continuous beam shown in Fig. 5.6a if the moment of inertia varies along the beam as indicated.

(a)

'@

(b)

2t1m

i [ I I

l'

4.66

19t

e

:a

41

t

d

I 1.21 I I I ~ 2 - l - - 4 ___ 2 -1--2 -I-- 2-1

r '~---L..--~~I~ 805mt

t

7.34 6.31t

269 5.03

80~_

2.97

- ____68 .1 mt

12

(e ) 8.M.D. ~=---:-h;--7--~--E-""7,.L.--=~~=---=-=t:~":>!

+

4.66 t (d)S.F.D.

~

6.31

6.311

5.~

2.69

2.69

~

~

2.~

2.97

7.34 Fig. 5.6 Solution: The relative stiffnesses and fixed-end momen ts are as follows:

K'b : K"" : Kcd = 1/6 : 1.21/6 : 1/6 = 5 : 6 : 5 MFab

= - wL2/12 = -

M"bC =

-

Pba 2/L2

MFcb

=

MPcd

= -

M"dc =

Pab 2/L2

+

=

2 X 6 2112

=-

6 m.t.,

9 X 2 X 4 2 /6 2 = -

9 X 4 X 22/6 2 =

Pa (L - a)/L 5.33 m.t.

= -

=

-

+

8 m.t.

4 m.t.

4 X 2 X 416 = -

5.33 m.t.,

414

The slope-deflection equations are : Mab Mba M bo Mob Mod Mdo

= = = = = =

+ + +

6 + 5 (2 aa 6 + 5 (2 ab 8 + 6 (2 ab 4 + 6 (2 a o 5.33 + 5 (2 5.33 + 5 (2

+ + + + ao ad

ab)

Mab

aa)

Mba

a o)

, M bo

ab)

,Mob

+ ad) , + a o) ,

Mod Mdo

= = = = = =

+ + +

6 + IO aa + 5 ab 6 + 5 aa + IO ab 8 + 12 a b + 6 a~ 4 + 6 ab + 12 a o 5.33 + IO a o + 5 5.33 + 5 a o + IO

ad ad

It should be noted that in keeping with the general development of the slope-deflection equations, the fixed-end moments MFab and MFde must be included although it is known that the final moments at .a and bare

zero. Later, Mab of statics.

=

0 and Mdc

=

0 provide two of the required equations

Inspection of the above slope-deflection equations shows that there are four unknown rotations; aa ,ab , a c and ado Therefore, four equations of statics are required. These are provided by the conditions that the sum of the moments at each of joints a,b,c and d is zero. Thus, M'b

= 0,

Mba

+

Mho

= 0,

Mob

+

Mod

= 0 and

Mdo

=

O.

Substituting from the slope-deflection equations into the joint equilibrium equations,

10

a. +

5

ab

5a.+ 22a.+ 6 6

a. +

"'0

22 "0 5 ""

=6

(i )

=6

( ii) (iii) (iv)

+5 +

.... = 1.33 10 "II = - 5.33

Solving equations (i) - flV) simultaneuusIy, a. = 0.663,

"'" = - 0.126,

Bo

=

0.2.45 and

Substituting these values into the slope-deflection

ad

=-

o.W

equatiens~

Mab

= - 6 + 10 (0.663) + 5 (- 0.126) = -

6 + 6.63 - 0.63 = 0

Mba

= + 6 + 5 (0.663) + 10 (-0.126) = + 6

+ 3.31-1.26 =8.05 m.t.

M bo = - 8 + 12 (-0.126) + 6 (0.245) = = -8.05 m.t.

8-

1.512 + 1.47

Mob

= + 4 + 6 (-0.126)+12 (0.245) =+4--D.756+2.94D = 6.ISm.t.

Mod

= - 5.33 + 10 (0.245) + 5 (- 0.66) = - 5.33 + 2.45 = -6.18 m.t.

Mdo

= +5.33 + 5 (0.245) + 10 (-0.66)

= + 5.33

3.30

+ 1.23 - 6.6

=

0

415 Having thus determined the end moments, the free-body diagrams for all the members are obtained and the B.M. and S.F.Ds. are constructed in the usual manner. The result is shown in Figs. 5.6 b-d. It should be noted that in this problem there are two unknown moments only although by the slope-deflection method four unknown angles are involved. It follows that a solution by the force method will be simpler than that worked out above.

Esa.nple 5.4 Analyse the beam shown in Fig. 5.7a in the absence of loads for the following simultaneous support movements: A downward movement of 0.3 em. in addition to a clockwise rotation of 0.001 rad. at support a, a downward movement of 1.2 em. at support b, and a downward movement of 0.6 em. at support c. EI = 5000 m 2t. (a)

1°

I--- Sm

(b)

c

b

£

S

32

1}

q?2ml 1.191

d

t

1.19 0.66

0.66

3.22ml (c) B.M.D.

+

3.93

1.19 1

1.19 +

(d} 5.F.o.

0.66

0.66

Fig. 5.7 Solution: Since there are finite joint and member rotations, form 5.4 of the slope-deflection equations should be used with the absolute (2EI/L)values. Further, as there are no loads all the fixed-end moments are zero. In this case it is known that: 1.2-0.3

a.

= + 0.001, "'ab= - 600 - = 0.0015 and "'be = --

1.2-D.6

6

00

= -- 0.001

.-----------------------------------------------------------------

416 The slope-deflection equations are thus as folleWs :

+ a" -

Ma. = 2 (5000/6) (2 X 0.001 M •• = 2 (5000/6) (2

a. +

0.001 -

+ a, + 3 2 (5000/6) (2 a.,. + a" + 3

3 X 0.0015)

3 X 0.0015)

M., = 2 (5000/6) (2 ".

X

0.001)

Me. =

X

0.001}

There are two unknown rotatioIlS'; Q.iJ..and -a.c ."l'hese'" may be found from the conditions of equilibrium at joints band c. Thus~

and Substituting from the slope-deflection equations into the above two equilibrium equations" (i) 4a.+ a,-0.5 X 10- 3 =0

a. + 2 a, + 3

X 10- 3

=

(ii)

0

Solving equations (i) and (ii) simultaneously,

a. =

0.57

X

10- 3 rad. and

a, = -

1.79

X

10- 3 rad.

Substituting these values into the slopeMdeflectioD equations,

+

Mab = 2 (5000/6) (2 X 10- 3 0.57 X 10- 3 _ 4.5 X HI-3) = - 3.22 m.t. M b• = 2 (5000/6) (2 X 0.57 X 10- 3 + 1 X 10- 3 - 4.5 X JO-3) = - 3.93 m.t. M b, = 2 (5000/6) (2 X 0.57 X 10- 3 - 1.79 X to..- 3 + 3 X JO-3) = 3.93 m.t.

M •• = 2 (5000/6) (- 2 X 1.79 X 10- 3 + 0.57 X 10- 3 + 3 X t!r"3}=

n

In this type of problem, particular care should be taken with regard to the proper signs of the known a and of angles and also the .....ls used. H~ving

determined the end moments, the free-body diagrams for individual members are obtained and the B.M. and S.F.I:>$.. are constructed as shown in Figs. 5.7 bod.

Example 5.5 For the restrained beam shown in Fig. 5.8, determine the fixing moment at end a ifsupport b settles 8 per unit load and EI = constant.

r

I--

P

I

U2

---4-1_ Fig. 5.8

lI2

I

f

I

~

I

I

~

417 Solution ;~b= 8xY. = 8 (Pf2 +M..JL). Henc.e,~ = 8 (Pf2 + Ma.{L){L. Since there is finite rotation of member ab, form 5.4 of the sIope-deHeetion equations should be used. Thus, PL/8 + (2 EI{L) [a. - 3 8 {P/2 + Ma.{L){L] PL/8 + 2 EI aofL - 6 EI 8 {P/2 + M • .fL){L2

1.1•• = = -

M'a

=

+ PL/8 + (2 EI{L) [2 ao ~3 8 {P/2 + M •.fL){L] + 4 EI a.fL - 6 EI 8 {P/2 + M ••{L){L2

= + PL/8

There is

dition ; 1.1.. 2EI a,JL

=

ODe

unknown rotation, aa,-

This is obtained from the con-

+ 4 EI a.{L - 6 EI 8 {P/2 + Mab{L){L2 or, 3EI 8 {P/2 + Ma.{L){L2- PL/16 = 0 = PL/8

Snb'tituting this value inti> the first slope-deHection equation, 1.1.. = -

(PL/16

+ EI 8 pf2L2) / {1/3 + EI 8 {L3)

If support b does not settie, 8 = 0 and 1.1•• = - 3PL/ 16 which is the known value for a restrained beam subject to a ceotral coz:centrated load P. It is also noticed that sinking of support b reduces the reaction at b and increases the end moment at a. 5.8 AppW:ations to static:ally indeterminate fraJnes with traDslation

DO

joint

Frames that do not involve transIation of the joints may be solved using form 5.5. or 5.7 of the sIope-deHeetion equations. In this case, as for statically indeterminate beams, there will be as many joint equilibrium equations as there are unknown joint rotations. Mter solvirg thefe m3'nt is distributed to members ab and be according to the distribution factors at joint h. The distributed m:>rnents (D. Mts.) are indicated on the second line. (6) A line is drawn bolow the distributed moments indicating that the preceding moments are in equiJibrium. (7) Hdf the distributed moments at one end of a member is carried ov.::r to its other end. These cl.rried over moments (C.O. Mts.) are written on a new line.

Steps 3,5 and 6 form a conplete cycle of mo:>ment distribution.

Genera~ly

the curied over moments start a new cycle and as many cycles as neces-

sary for a rCl.son'1.blc degree of accuracy are carried out. It should always be remembered that the last step in a cycle is to balance joints which can rotate and carryover moments to joints which cannot. (8) The final moments acting on the ends of the V'arious members are obt3.ined by summing up the moments in the respective columns. fhese moments are indicated on the last line. (9) Having thus determined the moments acting on the ends of the various members, it will be an easy matter to draw the required B.M.D. Esan1ple 6.2 Analyse the continuous beam shown in Fig. 6.4 by moment distribution.

EI =

constant.

Solution: The analysis of the beam is presented in Fig. 6.4.' In the previous example, one joint only was free to rotate. Thus, one cycle oflocking and releasing the joint was ad~quate'Tor a complete and exact solution. In this example joints band c are free to rotate. Mter the first cycle, the carried over moments at joints band c are out of balance. These moments can be looked upon as a new set of fheed-end moments and ,distributed in the usual manner during a second cycle, and so on. In this example

,-

:

!

f

455 four cycles are carried out. It is seen that the carried over moments in the four th cycle have become so small that they do not deserve further dis tribu tion.

9mt

~a1-4 C;~" 2

111m

CD

CD

--l

K

I

D.Fs.

13

F.E.MIsn.Mt!.. C.O.Mts. D.Mts.

C.O.MIs.

n. Mts. C.O.Mls. D.Mts. C.O.Mts.

0 ~

J.

_0_ .0.67 _0._ t 0.11 _0_ ,0..0.2 ~

Final Mts. .1.80

I

9mt

I !

I

12m

"~2:24 d~ -1

CD

113 -12

,12

LL-

~ ~

13 1/3

,2 - 0.67 .0.33 - D.lt .0.06 - 0..0.2

0 - 2 .1.33 .. 0.67 ( J - 0.33 .0.22 .0.11 0 - 0..0.6 .0.04 ~

0 0 1.33 0 -0.22

0.

-4 0 - 0..67

0.

2 3

- 0..11

~o.D4

0.

4

-0..0.2

=

.9.59 -9.59

-9.59 -9.59

-t.8D

Fig. 6.4 The symmetry in the beam and the loading suggests a simpler solution. utilizing modified stiffness for member be as explained in section 6.3. This simplified solu tion is presented in the following example.

Example 6,3 Analyse the symmetrical beam shown in Fig. 6.5 using modified stiffness for the central span. EI = constant. I

~a

9mt

EA

1-4 ---12 I

CD CD

K

K' D.Fs. F.E.Mts.

t

1tIm I 12' m

! I ! I

9mt

~3 ·d~ 121-4--J

CD

-tX1:@

CD CD

+___-J

t:tQ:[';- ---1)0p.S!fD"f.2f-----_ _ -3 0. -11

D.Mts. ~ C.O.Mts. .4.8 Final Mts.

_,9_ .6 _. 1_4

":1.i

Fig. 6.5 Solution : By using modified stiffness for the central span, the carry over moments across this span are actually eleminated. Also, from symmetry of beam and loading the moments on each side of the centre line are

r ,.

,

,

456 symmetrical. Thus, it is possible to work with only one half of the beam as shown in Fig. 6.5. The student is advised to compare the relative ease of the solution pre. sented above to that presented in the previous example for the same problem. He is recommended to adopt the later solution based on the use of modified stiffness.

Example 6.4 Analyse the continuous beam shown in Fig. 6.6. The moment of inertia varies as indicated.

311m

~__~____~~~IoID~C 21 -=t::'-- - 6 m K O.Fs.

F.E.Mls D.Mts e.O,MIS. D,Mls e.O.MIS D.Mls C.O.MIS. O.Mls.

C.D.Mls. O.Mls. C.O.Mls,

O.Mls C.O.MIs.

lol

-6

0 .0.5 _0_ .075 0 .0. 08 _0_ .0.12

0 .. 0.01

0 .0,02 ~

FlOal Mts

4.52

1

r1/3 2131 .6

-9 .2 - 4.5

.1 - 0.1.5 ~ 0 -0.5 .0.16 .0.34 0 _0.75 .0.25 .0.50 -0.08 0 .0.02 .0.06 -0.12 0 .0.04 .o.oa

=

CD

IT .S -S ~

C yel.1

-=L-

2

.1.5 _1.5 .0.171-

~_ .0.25 1

.•~ 0.03 1~

3

4

5 6

~

,.6.97 -8.S7

0

Fig. 6.6 Solution: Following normal moment distribution procedure, the analysis will be as shown in Fig. 6.6. It should be particularly noted that the simply supported end c, as any other joint, is first locked such that the fixed-end moments are deVeloped in member be. The fixed-end moment at c is next eleminated during the course of moment distribution. It should also be noted that the distribution factor at the simply supported end c = l. This will be obvious if member bc is looked at as being connected to another member of zero stiffness. Although the above procedure is quite basic, it is seen that convergence to a solution is rather slow. (Six cycles were carried out before the carried over moments have become too small to influence the final result). This is because the hinged end c of member bc keeps sending back relatively

457 large carried over momen ts to joint h. Anothe r solutio n based on the knowle dge that the final momen t at the hinged end must be zero and utilizin g modified stiffness for member be is possible. This solutio n is presen ted in the followi ng examp le.

Examp le 5.6 Analys e the beam shown in Fig. 6.7 using modifie d stiffness for span be. The momen t of inertia varies as indicat ed.

~o

18t

I--- 3

£'

K' D.Fs.

iol

F.E.Mt5.

-6 ~

C.O.M's. final t-IIts. _4.5

~

21 6m

';

CD

G) G)

K

D.Mt!..

3

311m 1 1 1 1 1

-fl(2=@ 0.4 0.6 .6 ·13 5 ~ • 4.5 ~

~

.9 - 9

Fig. 6.7 Solutio n: The analysi s of the beam is presen ted in Fig. 6.7. It should be noted that a modifie d stiffness of 3/4 X 2 = 1.5 is used for membe r bc to allow for the hinged end c. It should also be noted that a modifie d fixedend momen t at joint b ( - 13.5 m.t.) is used to accoun t for the hinged end c. This modifie d fixed-e nd momen t may be calcula ted indepe ndently by momen t distribution as follows : Assum ing both ends of membe r be to be fixed, wL2 3x6 2 Mbo = - - - = - - - = - 9 m.t. and Mob = + 9 m.t. 12 12 Now Mob cannot exist and should therefo re be elemin ated by applyin g an equal and opposit e momen t = - 9 m.t. The carried over momen t from c to b

= ~ (- 9) = -

4.5 m .t. Thus, the final momen t at c = 0, as it 2 should be, and the modified fixed-e nd momen t at b = 9 - 4.5 = - 13.5 m.t. as indicat ed. The studen t is advised to compar e the relative ease of the solutio n present ed above to that present ed in the previou s exampl e for the same problem . He is recomm ended to adopt the later solutio n based on the use of modifie d stiflhes s.

458 Es:unple 6.6 Analyse the continuous beam shown in Fig. 6.8. EI = constant. (This problem has been previously solved by the method of the equation of three moments in ExamFle 3.5-Fig. 3.10).

~a

r

1-- 2 --I- 2 K K'

D.Fs.

101

Rele-oslng joint c F.E.Mts. D.Mts. C.O.Mt~

-4

~

~-

Finol Mts. - 2.83

2t/m

~

! ! ! !

6m

I

0)

CD

CD

fX2=@ fill

1m!

~

l~J 2...J

11 tal

-6 - 15 .4 -7.5 .. 2.33 .1.17

.6 -3 -3 0

.6.33 -6.33

-;) ::)

Fig. 6 .8 Solution: The analysis of the beam is presented in Fig. 6.B. The only new point in this example is the manner of ha ndling the overhanging end. The overhanging end is similar to the simply supported end in Example 6.5, but in determining the modified fixed-end moment at end b of member be, it must be remembered that now the moment at c has a fixed value 1.5 X 2 = 3 m.t. However, instead of independently calculating the modified fixed-end moment at end b (allowing for the simple support at c and the applied loads on be including a clockwise moment of 3 m.t. at c) , it is easier to first release joint c and allow it to rotate under the unbalanced moment there and then proceed with the moment distribution. Further, si.nce a modified stiffness has been used for member bc, no moments are carried over from b to c. 6.9 Applications to statically indeterminate frantes with translation

DO

joint

The anl.ly-3ls of statically indeterminate frames with no joint translation follows: exactly the same procedure presented in the preceding section dealing with beams. In the case of frames, however, systematic arrangem ent o f th: calculations becomes m ore important. Many arrangements h ave b~en suggested. . but the author prefers the arrangement indicated in Fig. 6.9.

r- --------------------------------------------_________

459 In this arrang ement distrib ution factors and momen ts are written paralle l to the membe r to which they belong . The values related to one end of a membe r are written next to this end on the clockw ise side of the

00 c~

Fig. 6.9 membe r. Furthe r, the calcula ted values are written outwar d from the line diagram of the frame. This arrang ement is adopte d in the followi ng exampl es and the studen ts are advised to do the same in order to reduce the possibi lity of error. Examp le 6.7 Analys e the frame shown in Fig. 6.lOa. The momen t of inertia varies as indicat ed. (This problem has been previou sly solved by the slope-d eflectio n method in Examp le 5.6 - Fig. 5.10) b

r

1.5t1m

5t

¢.

c

21

5

d

4

l ,

-1--4 m

----I

"

(a)

(b) Fig. 6.10

Solutio n: The analysi s is present ed in Fig. 6.lOb. A modifie d stiffness is used for membe r be to allow for symme try. The relati'le stiffnes ses are thus as follows :

460

K'b : Kbc

= ; : }(

2:)

= 8 :5

= K/'f. K The distribu tion factors are calcula ted from the relation sbip, D and are' as follows : 5 5 8 8 Db = - - = - a n d D bc = - - =13 5 8 13 5 8 •

+

+

The fixed-end momen t at end b of membe r bc is : M' = _ b

c

(WL 2 12

+ PL) 8

= _ (1.5 X 8 12

2

+5

X 8) = _ 13 m.t. 8

it has been Note that since a modified stiffhess for member be was used) possible to work with only one half of the frame. constan t. Examp le 6.8 Analys e the frame shown in Fig. 6.lla. EI = method n (This proble m has been previou sly solved by the slope·d eflectio in Examp le 5.7 - Fig.5.11). .18.875

:-r.m -0- ' .16

3 tim

r

c~

b

3

41

CD 3

La

8m

(bl

(al

Fig. 6.11 d stiffness Solutio n: The analysis is presen ted in Fig. 6.lIb. A modifie stifrelative The is used for membe r ab to allow for the hinged end at a. fnesses are thus as follows : K.b: Kbc =+(+ ):-f = I: I DbC = 0.5. The distribution factors at joint b are thus: Dba = 0.5 and allow for A modified fixed-e nd momen t is used at end b of membe r ab to

I

I

I

L

II

461 . 4x6 th e hinged end at a, M~ = + - + - 1(4-X6) - = .' 8 2 8 wL2 3 x8 2 M"bc = = - -12 - = - 8 m.t. and ~b =

12

+ 4.5m.t.

+ 8 m.t.

Example 6.9

Analyse the frame shown in Fig. 6.12a. The moment of inertia yuies as indicated. (This problem has been previously solved by the slope-deflection method in Example 5.8-Fig. 5.12)

12m (a)

o - 18

..9-

~ I

~~:::~ NO.4 o ci ~o d_ I

I

8

_4

.. 1.6

- 2 _0_ .0.32

r:;, -0.64

CD

0.0.6 -0.32

....lL.- 0.16

:iii.48

-10.56

o

I

/

/"'1-4

ooh·o~o~ ~

(b) Fig. 6.12 Solution: The analysis is presented in Fig. 6.12h. A modified stiffness is used for member ce to allow for the hinged ::nd at c. The relative stiffnesses are thus as follows :

K.b:K",,:Kbd:K, e

= ~:~ : 12 8

1.51: 6

~(.!..) = 4:4:2:1 4 6

These values are shown encircled. The distribution factors are calculated

from the relationship: D = K/I:. K and

are as follows :

462 4

At joint b, Dba and Dbd

=

4+4+2

2 4+4+2

=

=

4

+

4 4

+

2 = 0.4

0.2

_4_ = 0.8 and Dec = _1_ = 0.2 4+1 4+1 0 since the overhanging member has zero stiffness.

At joint c, Dcb Note that Dee

= 0.4, Db< =

=

The fixed-end moments are: 1.5 X 122

12

18 m.t. and M~a =

2 1.5 X 8 = _ B m.t. and M~b = 12

+

+

18 m.t.

8 m.t.

It should be noted that over only three cycles of moment distribution, the largest moment agrees with its corresponding value obtained by the slope-deflection method to within 0.2%.

6.1 0 Applications to fraD1es with a single degree of freedom in translation In the preceding section, the application of the moment distribution method has been restricted to frames with no joint translation. As explained in sections 3.6 and 5.9, joint translation is generally prevented due to either complete symmetry in frame and loading or the disposition of the supports. Before proceeding with the analysis of frames with joint translation" it will be necessary to develop expressions for the moments developed at the ends of a prismatic member due to a relative joint translation while both ends are being held against rotation.

Fig. 6.13

i

I-

, I

r

!

463

It is assumed that end b is forced through some displacement!:::. relative to end a, while both ends are being held against rotation. The relationship between the fixedend moments developed at ends a and b on one hand and the relative displacement between the two ends on the other hand may be easily found from the moment-area theorems or more readily from the slope-deflection equations: Fig. 6.13a shows a fixed-end prismatic beam abo

M,b =

M~b +

2 :1 (2 a,

+

ab

Mba =

M~, +

2:1 (2

+

a, _

ab

_

3 b.) L

... 6.21

3:)

... 6.22

Since no load is applied between a and b, M: b = M~a = O. since both ends are held against rotation, U a = Ub = O. Thus,

Mb = _ 6EIb. a L2

.. . 6.23

If one end is hinged as shown in Fig. 6.13 h, the modified fixed-cnd moment at the other end may be found by moment distribution as follows :"

Assuming both ends to be fixed~ then according to equation 6.23~

-

6 Elb.

L2 Now if end b is hinged~ by applying a moment = from b to a =

+ -I

2

Mba

+

cannot exist and should be eleminated

6 EI b. at b. L2

(6 EI b.) = L2

+

3 EI L2

The carried over moment f\,

.

Thus.. the modified

fixed-end moment at end a allowing for the hinged support at b is given by :

-

3 El b. L2

... 6.24

In order to illustrate the method of handling joint

I

simple frame shown in Fig. 6.14 is considered.

.L -

systems shown in Fig. 6.14h and c .

I

Also,

translation~

the

The applied loads shown in Fig. 6.14a are equivalent to the two load The first load system consists of the applied loads in addition to some artificial restraining force Ha

preventing~

the lateral deflection or sway of

464 the frame. With joint translation thus prevented, the resulting non-sway moments may be determined by moment distribution as for any structure involving no joint translation. Next the artificial restraining force H . may be calculated by considering the equilibrium of the frame.

b

C

+ d

(al

z

(c)

(bl

Fig.6.14 The second load system consists of a single force equal in magnitude and oDoosite in direction to Hao Once the as~ociat(d moments, whic.h will be referred to as the actual sway mument.s , are found the final moments will be given by : Final moments = Non-sway moments

+

Actual sway moments

At this stage" the student is supposed to be capable of determining the non-sway moments and the problem thus reduces to finding the actual sway moments. A direct solution for the actual sway moments is not possible. However they may be found indirectly by first determining sway 11lIJ17lenls emsi')tent with some force He. having the same line of action as Hal and then finding the actual sway moments by direct proportion. The procedure is outlined belo (I) Joints band c are forced through some displacement I:; while being held against rotation. The corresponding fixed-end moments developed at the ends of the columns can be determined from equation 6.23. These moments are proportional to

(~~)

.

Since

I:;

is arbitrarily

chosen, any set of numerical values for the fixed-end moments may be chosen so long as they are proportional to their respective values of (I

1:;)

LZ •

(2) The frame is held against any further translation while joints b and

465 c are allowed to rotate. The resulting sway moments are determined by moment distribution as for zny structUJe involvir g r.o joint trznslaticn.

r

(3) The ""Iue of the force He c;>.Using the chosen displacement !!, and consistent with the sway moments calculated in step 2 is determined by considering the equilibrium of the frame. (4) Since the artificial restraining force cannot exist~ the actual sway moments should correspond to a force which when added algcbraicly to the artificial restraining force the latter is cancelled out. This condition is expressed mathematically as ;

H. ±

xJf"

=

... 6.25

0

where x is a factor to be applied to the sway moments determined in step 2 to obtain the actual sway moments" Actual sway moments

x (sway moments)

=

It remains to mention that the positive sign in equation 6.25 is used if both Ha and He have the same sense and vice versa.

The procedure outlined above will be illustrated by a number of examples.

ExaJnple 6.10 Analyse the frame shown in Fig . 6.15a. EI = constant. (This problem has been previously solved by the slope-deflection method in Example 5.10 - Fig. 5.16) Solution: The analysis is presented in steps. (1) The relative sti1f'nesses and distribution factors arc evaluated.

K'b : K"" : Ked = ] - : 6

~ : 2. 12

4

=

2 : I :3

These stiflnesses are shown encircled in Fig. 6.15 b. At joint b, Dba

2 At joint c, Deb

2

2

+I

3

= -- = =

3 3 = --- = -

I

--- I + 3

I

and Db< = - - - = 2 + I 3

4

I

+

3

4

These distribution factors are shown in the boxes attached to the line

diagram of the fr ame in Fig. 6.15 b . (2) Assuming that the frame is pr evented from sway by a horizontal force H., the fixed-end moments due to the applied loads are calculated.

rp::r:::o:=r::.iI::i:I:r::r===ll 2 tim

3

4m

61

12m (a)

.20.82

:orr .. 0.5

:o:sr .. 3.25

:--s.24 114

I

:::I~ I ~ ~ 113 mOONO:~N_24 "';..

..

....

N~ ~

.. 6.5 ~

I

I

I

I

~ o~ol~ ~ I

..

0

oool",ol~ol~l~

• 1

0:0:4 \V

Ha·0.991!

I

I

I

IMI' d ;;;"

.-0 ..

I

....

(b) Non-5way MIs.

.6.6

:aT _0.4 -0.4 .1.7 .5.S

o

ul ~~

I'" I'"

·0.....: 0

I"

I

tD~ ..

I

0 ~ 113 +3.3

N

:rr-0.9

or:~ol~ol~ol~l~ .

-0.2 .0.07

:sm l~ 0

I

I:l> 1:gIUl

0 0 MODCOf'-o

I"

•

I

..

I

(c) Sway MIs. _ 0.89 .0.68 -0.68

0.991 I

.. 0.89

.1.96 .t.Ol

(d)Aclual Sway MI5.

Fig. 6.15

•

I

I I

467

I

6 x 6 -B- = -

-

2

122 = 12

X

r

I

4.5 m.t. and MFb, --

+ 4 •5 m .t.

24m.t. and MF'b = + 24 m.t.

(3) The fixed-end moments in step 2 are distributed in the usual manner to obtain the non-sway moments shown in Fig. 6.15b. (4) The artificial restraining force Ha is calculated by considering the eq uili bri urn of the frame as follows !

3.13+ 19 .77 ----'----'----'- 6

3

=

Xd = 10.41 + 20.82 4

O. BI 7t . _

/

7.B08 t. +-

L X= Ha + 6 + 0.BI7 .L 7.808,

H, = 0.991 t. _

(5) The frame is forced through some displacement b. and the corresponding fixed-end moments are calculated . M=b

F

=

Mba = -

6 E1 b. 62 ~ say -

10

6 E1 b. M~d = M~c = _~-' say-22.5

(6) The fixed-end moments in step 5 are distributed in the usual manner to obtain the sway moments shown in Fig. 6.I5e. (7) The consistent straining force H, is calculated by considering the equilibrium of the frame as follows: Xa

x.

7.54 + 5.07 = 2.lt . +6 = _14_..:..c 55'-+ .:......::6~.6 = 5.29 t.+4

LX = 0=H,-2.1-5.29,

He = 7.39 t._

(8) From equation 6.25, the multiplying factor x is obtained, 0.991 + 7.39 x = 0,

x = -0.135

(9) Factor x is applied to the sway moments in Fig. 6.15 c to obtain the actual sway moments shown in Fig. 6.15 d. (10) The final moments are obtained by adding the non-sway moments in Fig. 6.l5b to the actual sway moments in Fig. 6.15d. These will be as follows:

468

+ = +

Mab =

3.13

M b•

19.77

M b, = -

M'b

=

+ 1.01 = + 4.14 m.t. + 0.68 = + 20.45 m.t. 0.68 = -

19.77 -

+ 20.82 -

M,d = -

20.82

M d, = -

10.41

=

0.89

+ +

20.45 m.t.

+ 19.93 m.t.

0.89 = -

19.93 m.t.

1.96 = -

8.45 m.t.

These moments are very nearly the same as those previously obta ir.cd by the slope-deflection method.

Exatnple 6.11 Analyse the frame shown in Fig. 6.16a. The moment of inertia varies as indicated. (This problem has been previously solved by the slope-deflection method in Example 5.9-Fig. 5.15) Solution: The analysis is presented in steps.

(l) The relative st:& lcsses and distribution factors arc evaluated. K'b : K b, : K,d

=

-.!. : ~ : ~ 8

6

6

= I :3:I

These stifFnesses are shown encircled in Fig. 6.16b. At joint b" Dba =

I

-- -

1+3 At joint c,

DCb

I

-

3

3

+3

4

3

3

1+3

4

and Dbc = - - = _.-

4

= - - = - and

I

Note that the overhang has zero stiffness. The distribution factors are

shown in the boxes attached to the line diagram of the frame in Fig. 6.16b.

(2) Assuming that the frame is prevented from sway by a horizontal force RaJ the fixed-end moments due t? the applied loads are calculated.

M'be

10 X 3.2 X 4.8 2

----;:--- = 82

, l O x 4.8 X 3.22 M'b = + Moe = -

10 X 2 =

-

+

11.52 m.t.

7.68m.t.

20 m.t.

(3) These moments are distributed to obtain the non-sway moments

shown in Fig. 6.16 b.

469

101

c

e

6m I

a

d

"3.2---,-" ' I.-

"J

4.8m----l- 2

(a)

I

I

I l - I. 62 • 1. 22 • 0.65 ~ 0.49

.

,

I t J'

~

2r~ ,

r

F

0

(tl) Non-sway Mt5.

I

I

'" IN 1"'cf1 c:

- , 96I~ ~ ~