Moment Distribution Method - I.K. International Publishing ... The moment distribution method can be...

Objectives:

Defi nition of stiffness, carry over factor, distribution factor. Analysis of continuous beams withoutsupport yielding – Analysis of continuous beams with support yielding – Analysis of portal frames – Naylor’s method of cantilever moment distribution – Analysis of inclined frames – Analysis of Gable frames.

2.1 INTRODUCTIONThe end moments of a redundant framed structure are determined by using the classical methods, viz. Clapeyron’s theorem of three moments, strain energy method and slope defl ection method. These methods of analysis require a solution of set of simultaneous equations. Solving equations is a laborious task if the unknown quantities are more than three in number. In such situations, the moment distribution method developed by Professor Hardy Cross is useful. This method is essentially balancing the moments at a joint or junction. It can be described as a method which gives solution by successive approximations of slope defl ection equations. In the moment distribution method, initially the structure is rigidly fixed at every joint or support. The fi xed end moments are calculated for any loading under consideration. Subsequently, one joint at a time is then released. When the moment is released at the joint, the joint moment becomes unbalanced. The equilibrium at this joint is maintained by distributing the unbalanced moment. This joint is temporarily fi xed again until all other joints have been released and restrained in the new position. This procedure of fi xing the moment and releasing them is repeated several times until the desired accuracy is obtained. The experience of designers points that about fi ve cycles of moment distribution lead to satisfactory converging results. Basically, in the slope defl ection method, the end moments are computed using the slopes and defl ection at the ends. Contrarily in the moment distribution method, as a fi rst step — the slopes at the ends are made zero. This is done by fi xing the joints. Then with successive release and balancing the joint moments, the state of equilibrium is obtained. The release-balance cycles can be carried out using the following theorems.

Moment Distribution Method2

Moment Distribution Method

130 Indeterminate Structural Analysis

In conclusion, when a positive moment M is applied to the hinged end of a beam

a positive moment of 12

Ê ˆÁ ˜Ë ¯

M will be transferred to the fi xed end.

2.2.2 Th eorem 2Consider a two span continuous beam ACB as shown in Fig. 2.2(a). A and B are fi xed supports with a prop at C. A moment is applied at C and it is required to know how much moment is distributed between spans AC and CB. Let this moment M be decomposed and distributed as M1 to CA and M2 to CB as shown in Fig. 2.2(b).

i.e. M1 + M2 = M (1)

l1

I1I2

l2

MCA B

FIG. 2.2(a) Continuous beam ACB with moment M

CA B

M1

M1

M2

2M2

2( )FIG. 2.2(b) Distribution of bending moments

The bending moment diagram is drawn by considering each span AC and CB respectively.

CA B

M1

M1

M2

21

++

−−

( )

M221( )

( )( )( )l13

2l13

2l23

l23

FIG. 2.2(c) Bending moment diagram

As the ends A and B are fi xed; the slope between A and B is zero. That is, the area of the bending moment diagram between A and B is zero.

133Moment Distribution Method

C

A B

A1M1

A2

M2

M1

x3

A3

x2

l13

2l13

x1

( )( )

( )

l2

2

FIG. 2.4(c)

The tangent drawn at A passes through C and B. Hence, from the above fi gure;

Ai xi /EIi = 0

i.e. A1 = (1/2) (l1/3) (M1/2) = 1 1

12M l

x1 = l2 + 11

2 2 ;3 3 3

llÊ ˆ+Á ˜Ë ¯ x1 = 2 189

l l+

A2 = (– 1/2) 1 11 1

23 3

M ll MÊ ˆ = -Á ˜Ë ¯

x2 = l2 + 11 23 3 3

lÊ ˆÊ ˆ Ê ˆÁ ˜ Á ˜ Á ˜Ë ¯ Ë ¯ Ë ¯

; x1 = 2 129

l l+

A3 = (1/2) (l2 )M2 = 2 2

2M l

x3 = 223

l

Thus, Aixi/EI gives

2

1 1 2 2 2

1 2

30

12 3M l l M lEI EI

-+ =

2

1 1 22 2

2 1

33 12

M l lM lEI EI

-=

2 2 1 1

2 1

34

l M M lI I

=


2 1

2 1

34

M Mk k

=

i.e., 22

2 1

(3/4 )kMk k

=

If one end of a member is not fi xed then the “stiffness” of that member should be multiplied by (3/4).

2.2.4 Th eorem 4Consider a fi xed beam AB as shown below. End B has settled by a distance . As the ends are fi xed, there must develop a fi xing moment M at the each end of the beam.

FIG. 2.5(a) Sinking of supports in a fi xed beam

A BA1

A2

l2l2

x2

x1

FIG. 2.5(b) Bending moment diagram

Taking moments about B and using the moment area theorem;

Ai xi /EIi = –

i.e. A1 = – 12 2 4

Mll M-Ê ˆ Ê ˆ =Á ˜ Á ˜Ë ¯ Ë ¯

A2 = 12 2 4

l MlMÊ ˆ Ê ˆ =Á ˜ Á ˜Ë ¯ Ë ¯

x1 = 2 5

2 3 2 6l l lÊ ˆ+ =Á ˜Ë ¯

x2 = 13 2 6

l lÊ ˆ =Á ˜Ë ¯


1 54 6 4 6

Ml l Ml lEI

È ˘Ê ˆ Ê ˆ Ê ˆ Ê ˆ- + = -dÁ ˜ Á ˜ Á ˜ Á ˜Í ˙Ë ¯ Ë ¯ Ë ¯ Ë ¯Î ˚

2 25 124 24Ml Ml

EIÊ ˆ-

+ = -dÁ ˜Ë ¯

2( 5 1)

24 24Ml- +

= - d

M = 26El

ld

i.e., when a fi xed ended beam settles by an amount δ at one end, the moment required to make the ends horizontal = 6EI/l2. The above four theorems can be summarised as (1) When the member is fi xed at one end and a moment is applied at the other end

which is simply supported or hinged, the moment induced at the fi xed end is one half of the applied moment. The induced moment at the fi xed end is in the same direction as the applied moment.

(2) If a moment is applied in a stiff joint of a structure, the moment is resisted by various members in proportion to their respective stiffnesses (i.e., moment of inertia divided by the length). If the stiffness of the member is more; then it resists more bending moment and it absorbs a greater proportion of the applied moment.

(3) While distributing the moments in a rigid joint, if one end of the member is not restrained then its stiffness should be multiplied by (3/4).

(4) In a fi xed beam, if the support settles/subsides/sinks by an amount , the moment required to make the ends horizontal is 6EI/l2.

2.3 BASIC DEFINITIONS OF TERMS IN THE MOMENT DISTRIBUTION METHOD(a) StiffnessRotational stiffness can be defi ned as the moment required to rotate through a unit angle (radian) without translation of either end.

(b) Stiffness Factor (i) It is the moment that must be applied at one end of a constant section member

(which is unyielding supports at both ends) to produce a unit rotation of that end when the other end is fi xed, i.e. k = 4EI/l.

(ii) It is the moment required to rotate the near end of a prismatic member through a unit angle without translation, the far end being hinged is k = 3EI/l.


(c) Carry Over FactorIt is the ratio of induced moment to the applied moment (Theorem 1). The carry over factor is always (1/2) for members of constant moment of inertia (prismatic section). If the end is hinged/pin connected, the carry over factor is zero. It should be mentioned here that carry over factors values differ for non-prismatic members. For non-prismatic beams (beams with variable moment of inertia); the carry over factor is not half and is different for both ends.

(d) Distribution FactorsConsider a frame with members OA, OB, OC and OD rigidly connected at O as shown in Fig. 2.6. Let M be the applied moment at joint O in the clockwise direction. Let the joint rotate through an angle . The members OA,OB,OC and OD also rotate by the same angle θ.

OD

C

A

BM

FIG. 2.6

Let kOA, kOB, kOC and kOD be the stiffness values of the members OA, OB, OC and OD respectively; then

MOA = kOA (i)

MOB = kOB (ii)

MOC = kOC (iii)

MOD = kOD (iv)

A B

MBAMAB = 2EI

L= 4EIL

θA= 1

(a) Beam with far end fi xed


dOB = OBkkS = distribution factor for OB

dOC = OCkkS = distribution factor for OC

dOD = ODkkS

= distribution factor for OD

2.4 SIGN CONVENTIONClockwise moments are considered positive and anticlockwise moments negative.

2.5 BASIC STAGES IN THE MOMENT DISTRIBUTION METHODThe moment distribution method can be illustrated with the following example. It is desired to draw the bending moment diagram by computing the bending moments at salient points of the given beam as shown below.

CC'A

A B'I =

100 kN2.5 m

5 m6 m

50 kN/m

2I 3I

FIG. 2.8 A two span continuous beam

Step 1Determine the distribution factor at each joint A,B and C respectively. The distribution factor of a member is the ratio of the stiffness of the member divided by the total stiffness of all the members meeting at that joint. The distribution factor for the fi xed support A is determined by assuming an imaginary span AA (Fig. 2.8). The fl exural rigidity of AA is infi nity. Hence, the stiffness is infi nite. The stiffness of AB is 2I/5 = 0.4I. Hence, the total stiffness is infi nite. Thus, the distribution factor for AA is (infi nity/infi nity) = 1.0 and for AB the distribution factor kAB = (0.4I/infi nity), i.e. zero. In essence dAA = 1.0 and dAB = 0.0 The distribution factor at the support B is determined as follows. The stiffness of the member BA is (0, 4I). The stiffness of the member BC is taken as three-fourths

of its stiffness (refer Theorem 3). Hence, kBC = 3 34 6

IÊ ˆ Ê ˆÁ ˜ Á ˜Ë ¯ Ë ¯ = 0.375I. The sum of the

stiffness at the joint B is k = (kBA + kBC) = 0.775I. Therefore, the distribution factors are dBA = kBA/(kBA + kBC), i.e. dBA = 0.4I/(0.4I + 0.375I) = 0.52; similarly dBC = kBC/k, i.e. dBC

= (0.375I/0.775I) = 0.48. The sum of the stiffnesses is (0.52 + 0.48) = 1.0. The distribution factor for the simple support at C is determined by extending the span to CC. The rigidity of CC is zero and hence kCC = 0. On the other hand, the


stiffness kCB = 0.375I. The total stiffness at C is kCB + kCC = 0.375I. The distribution factor

dCB = kCB/k = 1.0 and dCC = kCC/k = 0.

The above procedure is summarised in the following table for quick understanding.

Table 2.1 Distribution factor at joint B

Joint Members k = I/l k DF

BA2 0.45l I= 52

B 0.775I

BC 3 34 6

IÊ ˆÁ ˜Ë ¯ = 0.3/5I 0.48

Step 2Imagine all the three joints A, B and C are rigidly fi xed with horizontal tangents. Write down the fi xed end moments for the beam AB as if it were built in at A and B and also for the beam BC as if it were built in B and C.

CA Bl/m

100 kN2.5 m

5 m6 m

50 kN/m

62.562.5 150

FIG. 2.9 Fixed end moments

MFAB = 100 5

62.5 kNm8¥

- = -

MFBA = + 100 5

62.5 kNm8¥

= +

MFBC = 250 6

150.0 kNm12¥

- = -

MFCB = + 250 6

150.0 kNm12¥

= +

Step 3Each joint is released in turn and if B is released it will be out of balance. This unbalanced moment (– 150.0 + 62.5 = 87.5) is shown in Fig. 2.10.


A B C

62.5 kNm 87.5 kNm

150 kNm

FIG. 2.10 Out of balance moments

Step 4A moment is applied at B to balance this joint B and it will distribute itself according to the distribution factors. This is shown in the following fi gure.

A

1 0 1 0

B C

62.5 87.50.480.52

150 kNm

FIG. 2.11 Balancing moments

A B C45.5062.5 0 42.00 –150.00 0 kNm

FIG. 2.12 Distributed moments

Step 5Balance the joint C as its moment is zero (end support C is simple support). By balancing the moment – 150, half of it is carried over to B. By balancing the joint B, half of the moment is carried over to joint A, i.e. half of 45.50 kNm.

Step 6Again the joints become out of balance, and the above procedure is repeated until the moments to be distributed become negligible and can be ignored. This is illustrated in Table 2.2.

Table 2.2 Moment distribution table

Joint A B CMembers AB BA BC CBDF 0 0.52 0.48 1FEMSBal

– 62.50 +62.50+45.50

– 150.00+42.00

+150.00– 150.00

CoBal

+22.75+39.00

– 75.00+36.00

Co +19.50Total – 20.25 +147.00 – 147.00 0.00Nature O


In the above moment distribution table a single vertical line is drawn between the members. Double lines are drawn at the end of each joint. The pure moment diagram can be drawn using the end moments in the moment distribution table. The pure moments are the values just to the left of double line. Thus, MA = – 20.25 kNm, MB = – 147.00 kNm, MC = 0.

CA B

−ve

147.00 kNm

20.25

FIG. 2.13 Pure moment diagram

The simple beam moment diagram is drawn by considering each span separately. The simple beam moment diagram is always positive. While the pure moment diagram is negative. The maximum positive bending moment for span AB is (wl/4) = 100 × 5/4 = 125 kNm. The maximum bending moment for a simple beam of BC is wl2/8 = (50 × 62/8) = 225 kNm.

CA B

− 225 kNm

125

+ +

D E

FIG. 2.14 Simple beam moment diagram

The net bending moment diagram is drawn by superimposing the pure moment diagram on the simple beam moment diagram. Thus, the net moment at D and E are

MD = 125 – 147.00 20.25

2+Ê ˆ

Á ˜Ë ¯ = +41.38 kNm

ME = 225 – 147.00 0.00

2+Ê ˆ

Á ˜Ë ¯ = +151.5 kNm

The fi nal net bending moment diagram is as follows.


CA B−−

151.5 kNm

41.4

++

D E

20.25

147

FIG. 2.15 Bending moment diagram

The net bending moment diagrams are preferable in design offi ces. The elastic curve is drawn using the net bending moment diagram.

A B C

FIG. 2.16 Elastic curve

2.6 NUMERICAL EXAMPLES2.6.1 Analysis of Continuous Beams without Support SettlementExample 2.1 Analyse the continuous beam shown in Fig. 2.17 by the moment distribution method. Draw the bending moment diagram and shear force diagram. The beam is of uniform section.

A B C

6 m

10 kN/m30 kN/m

4 m

FIG. 2.17

SolutionStep 1The distribution factors at joint B are evaluated as follows.


Table 2.3 Distribution factors

Joint Members Relative Stiffness (k) Sum k Distribution Factors (k/ k)BA I/6 = 0.167I 0.47

B 0.3555I

BC 34 4

I¥ = 0.188I 0.53

Step 2 Fixed End Moments

MFAB = 210 6

12- ¥

= – 30 kNm, MFBA = +30 kNm

MFBC = 230 4

12- ¥

= – 40 kNm, MFCB = +40 kNm

Step 3 Moment Distribution TableAs the joint C is a hinged end, the moment is zero. Hence, it is balanced fi rst. Then half of this moment is carried over. Then joint B is balanced. From the joint B, the moment is carried over to A.


Joint A B CMembers AB BA BC CBDF 0 0.47 0.53 1FEMSBal

– 30.00 –

+30.00–

– 40.00 –

+40.00– 40.00

Co – – – 20.00TotalBal

– 30.00–

+30.00+14.10

– 60.00+15.90

0.00–

Co +7.05 – – –Final – 22.95 +44.10 – 44.10 0.00

A B C

22.95 kNm44.1 kNm

D E−−

++

11.48 kNm

37.95 kNm



MD = 210 6 22.95 44.1

11.48 kNm8 2¥ +Ê ˆ- =Á ˜Ë ¯

ME = 230 4 44.1 0

37.95 kNm8 2¥ +Ê ˆ- =Á ˜Ë ¯

Step 4 Shear Force DiagramsEquilibrium of span AB

A B6 m

10 kN/m22.95 kNm 44.1 kNm

FIG. 2.19

V = 0; VAB + VBA = 60 (1)

MA = 0; – 22.95 + 44.1 + 10 × 26

2 – 6VBA = 0 (2)

VBA = 33.5 kN

VAB = 26.5 kN Equilibrium of span BC

B C4 m

30 kN/m44.1 kNm

FIG. 2.20

V = 0; VBC + VCB = 120 (3)

MB = 0; – 44.1 + 30 × 24

2 – 4VCB = 0

VCB = 49 kN

VBC = 71 kN

26.5 kN71 kN

33.5 kN49 kN

x1

x2

x16x24

––

FIG. 2.21 Shear force diagram


From similar triangles

1

1(6 )x

x- =

26.533.5

33.5x1 = 159 – 26.5x1

x1 = 2.65 m

2

2

71(4 ) 49

xx

=-

49x2 = 284 – 71x2

x2 = 2.37 m

Example 2.2 Analyse the continuous beam by moment distribution method. Draw the shear force diagram and bending moment diagram.

A B

EI = Constant

E

F

D

C

3 m 3 380 kN 80 kN 12 kN/m

12 m

FIG. 2.22

SolutionDistribution factors The distribution factors at joint B are obtained as follows.


Joint Members Relative Stiffness (k) Sum k Distribution Factors (k/ k)BA I/9 = 0.111I 0.64

B 0.174IBC 3

4 12I

¥ = 0.063I0.36

Fixed end moments

MFAB = – ¥ ¥

= - = -80 3 6

160 kNm9

Wabl


MFBA = +80 × 3 × 69

= +160 kNm

MFBC = – 12 × 212

12 = – 144 kNm

MFCB = +144 kNm


Joint A B C

Members AB BA BC CB

DF 0 0.64 0.36 1

FEMSBal

– 160.00 +160.00 – 144.00 +144.00– 144.00

Co – 72.00

TotalBal

– 160.00 +160.00+35.80

– 216.00+20.20

0.00

Co +17.90

Total – 142.10 +195.80 – 195.80 0.00

Shear forces Equilibrium of span AB

A

ED

B

3 m 3 m 3 m80 kN 80 kN

195.8 kNm142.1 kNm

FIG. 2.23

V = 0; VAB + VBA = 160 (1)

MA = 0; – 142.1 + 195.8 + 80 × 3 + 80 × 6 – 9VBA = 0 (2)

VBA = 86 kN

VAB = 74 kN

MD = 74(3) – 142.1 = 80 kNm

ME = 86(3) – 195.8 = 62.2 kNm


Equilibrium of span BC

12 kN/m195.8 kNm

12 mB CF

FIG. 2.24

V = 0; VBC + VCB = 144 (3)

MB = 0; – 195.8 + 12 × 212

12 – 12VCB = 0 (4)

VCB = 55.7 kN

VBC = 88.3 kN

MF = 55.7 × 6 – 12 × 26

2 = 118.2 kNm

A B C

142.1 kNm

195.8 kNm

D F

−

8062.2

118.2 kNm

E


74

88.3 kN

(12 − x)6 kN

8655.7 kN

x



From similar ∆’s

55.7(12 ) 88.3

xx

=-

88.3x = 668.4 – 55.7x

x = 4.64 m

Example 2.3 Analyse the continuous beam by the moment distribution method. Draw the shear force diagram and bending moment diagram.

A B ED C

120 kN30 kN/m4 m4 m

8 m

FIG. 2.27

SolutionDistribution factors The distribution factors at joint B are evaluated as follows.


Joint Members Relative Stiffness (k) Sum k Distribution Factors (k/ k)

BA I/8 = 0.125I 0.5

B 0.25I

BC I/8 = 0.125I 0.5

Fixed end moments

MAB = – 30 × 28

12 = – 160 kNm; MBC = –

2120 88¥

= – 120 kNm

MBA = +160 kNm; MCB = + 120 kNm


Moment Distribution


Joint A B C

Members AB BA BC CB

DF 0 0.5 0.5 0

FEMSBal

– 160–

+160– 20

– 120– 20

+120–

Co – 10 – – – 10

Final – 170 +140 – 140 +110

Using the above end moments;

MD = 230 8 170 140

85 kNm8 2¥ +Ê ˆ- =Á ˜Ë ¯

ME = +Ê ˆ¥ - =Á ˜Ë ¯

140 1108120 115 kNm4 2

A B C

170 kNm140 kNm

110 kNm

D

− −−+

E+

85 115


Shear force diagrams Equilibrium of span AB

A B

30 kN/m

8 m

170 140

FIG. 2.29

V = 0; VAB + VBA = 240 (1)


MA = 0; – 170 + 140 + 30 × 28 8 0

2 BAV- =

(2)

VBA = 116.3 kN VAB = 123.7 kN


B

120 kN44

8 m

140 kNm 110 kNm

C

FIG. 2.30

V = 0; VBC + VCB = 120 (3)

MB = 0; – 140 + 110 + 120(4) – 8VCB = 0

VCB = 56.3 kN

VBC = 63.7 kN

A B

63.7 kN

116.356.3 kN

123.7 kN

− −+ +

x1


From similar ∆’s

1

1

123.7(8 ) 116.3

xx

=-

116.3x1 = 989.6 – 123.7x1

x1 = 4.12 m

Example 2.4 Determine the end moments for the continuous beam shown in Fig. 2.32. EI is constant. Use the moment distribution method.


A B C D

22 m 250 kN 50 kN25 kN/m

4 m

75 kN/m

6 m

FIG. 2.32

SolutionDistribution factors The distribution factors at joints B and C are obtained as follows.


Joint Members Relative Stiffness Values (k) k Distribution Factors k/ k

BA I/6 0.50

B 2I/6

BC I/6 0.50

CB I/6 = 0.167 0.40

C 0.417

CD I/4 = 0.25 0.60

Fixed end moments

MFAB = – 25 × 26 75 kNm

12= -

MFBA = +25 × 26 75 kNm

12= +

MFBC = – 50 2 4

66.67 kNm6

¥ ¥= -

MFCB = +50 × 2 × 4 66.67 kNm6

= +

MFCD = – 75 × 42/30 = – 40.00 kNm

MFDC = +75 × 42/20 = +60.00 kNm



Joint A B C D

Members AB BA BC CB CD DC

DF 0 0.5 0.5 0.4 0.6 0

FEMSBal

– 75.00 +75.00– 4.17

– 66.67– 4.16

+66.67– 10.67

– 40.00+16.00

+60.00–

Co

Bal

– 2.10 –

+2.67

– 5.34

+2.67

– 2.08

+0.83

–

+1.25

– 8.00

Co

Bal

+1.34

– 0.21

+0.42

– 0.21

+1.34

– 0.54 – 0.80

+0.63

Co

Bal

– 0.11

+0.13

– 0.27

+0.14

– 0.11

+0.04 +0.07

– 0.40

Co

Bal

+.07

– 0.01

+0.02

– 0.01

+0.07

– 0.03 – 0.04

+0.04

Final – 75.8 +73.41 – 73.41 +55.52 – 55.52 – 52.27

The end moments are

MAB = – 75.8 kNm, MB = – 73.41 kNm

Mc = – 55.52 kNm, Mc = – 52.27 kNm

Example 2.5 Analyse and sketch the bending moment diagram for the beam shown in Fig. 2.33. The values of the second moment area of each span are indicated along the members. Modulus of elasticity is constant.

A B C D

1.2540 kN30 kN/m

5 m6 m5 m

100 kN1.25 2.5 m2.5 m 2.5 m

80 kN

2I 3I 4I

FIG. 2.33


Distribution factors


Joint Members Relative Stiffness Values (k) k Distribution Factors k/ k

BA 2I/5 = 0.4I 0.44

B 0.9I

BC 3I/6 = 0.5I 0.56

CB 3I/6 = 0.5I 0.45

C 1.1I

CD 3 4 0.64 5

I I¥ =0.55

Fixed end moments

MFAB = – 100 × 5 62.5 kNm8

= -

MFBA = +100 × 5 62.5 kNm8

= +

MFBC = – 30 × 26 90.0 kNm

12= -

MFCB = +30 × 26 90.0 kNm

12= +

MFCD = 2 2

2 2

1.25 80 3.75 3.75 40 1.2565.625 kNm

5 5¥ ¥ ¥ ¥

- - = -

MFDC = 2 2

2 2

1.25 40 3.75 3.75 80 1.2546.875 kNm

5 5¥ ¥ ¥ ¥

+ + = +

78.0

135.0 kNm

84.587.5

62.5 kNm

125

54.7




Joint A B C D


DF 0 0.44 0.56 0.45 0.55 1

FEMSBal

– 62.5 +62.50+12.10

– 90.00+15.40

+90.00– 10.97

– 65.63– 16.4

+46.88– 46.88

CoBal

+6.05+2.42

– 5.49+3.07

+7.70– 3.47 – 4.23

CoBal

+1.21+0.77

– 1.74+0.97

+1.54– 0.69 – 0.85

CoBal

+0.39+0.15

– 0.35+0.20

+0.49– 0.22 – 0.27

CoBal

+0.080.05

– 0.11+0.06

+0.10– 0.05 −0.05

CoBal

+0.03+0.01

– 0.03+0.02

+0.03– 0.01 – 0.02

Final – 54.74 +78.0 – 78.00 +84.45 – 84.45 0.00

2.6.2 Analysis of Continuous Beam with Support SettlementThe settlement of supports causes the bending moment in the members. The settlement of support is mainly due to soil subsidence. In a fi xed beam of AB, if the support B is lower by , the induced end moment is – 6EI/l2 where is the settlement of supports. If the support B is higher by , the end moment is +6EI/l2. (Refer Basic Structural Analysis – Chapter 11) The moments are computed using the given loading and due to settlement of supports separately and summed up.

Example 2.6 Analyse the beam shown in Fig. 2.35 by the moment distribution method. Support B sinks by 10 mm. E = 200 kN/mm2. I = 4000 × 104 mm4. Draw BMD and SFD.

A B CE D6 m

2470 kN 10 kN15 kN/m

4 m 1 m

FIG. 2.35


SolutionDistribution factors


Joint Members Relative Stiffness Values (k) k Distribution Factors k/ kBA I/6 = 0.167I 0.47

B 0.355I

BC 34 4

IÊ ˆÁ ˜Ë ¯ = 0.188I 0.53

Fixed moments Due to applied loading

MAB = – 4(70) 2

226

= – 31.11 kNm

MBA = +2(70) 2

246

= +62.22 kNm

MBC = – 15 × 42/12 = – 20.00 kNm

MCB = +15 × 42/12 = +20.00 kNm

MCD = – 10(1) = – 10.00 kNm.

Due to sinking of supports

MAB = MBA = 2

2

6 6 8000 1013.33 kNm

1000 6B

AB

EIl

- D - ¥ ¥= = -

¥

MBC = MCB = 2

2

6 8000 106 30 kNm1000 4

B

CB

EIl

¥ ¥D+ = = +

¥

The actual fi xed end moment, is the addition of moment due to applied loading and due to the settlement of supports.

MFAB = – 31.11 – 13.33 = – 44.44 kNm

MFBA = +62.22 – 13.33 = +48.89 kNm

MFBC = – 20.00 + 30.00 = +10.00 kNm

MFCB = +20.00 + 30.00 = +50.00 kNm

MFCD = – 10.00 kNm



Joint A B CMembers AB BA BC CB CDDF 0 0.47 0.53 1 –

FEMSBal

– 44.44 +48.89– 27.61

+10.00– 31.22

+50.00– 40.00

– 10.00

CoBal

– 13.84+9.40

– 20.00+10.60

Co +4.70

Final – 53.58 +30.62 – 30.62 +10.00 – 10.00

−53.58

93.33 kNm

−30.6230 kNm

−10

A E B C D


Shear forces Equilibrium of span AB

A B

53.58 kNm 30.62 kNm70 kN4 2

FIG. 2.37

V = 0; VAB + VBA = 70 (1)

M = 0; – 53.58 + 30.62 + 70(4) – 6VBA = 0 (2)

VBA = 42.84 kN

VAB = 27.16 kN


B

30.62 kNm

C

10 kNm

4 m

15 kN/m

FIG. 2.38

V = 0; VBC + VCB = 60 (3)

MB = 0; – 30.62 + 10 + 15 × 2

C4 4 02 BV- =

VCB = 24.85 kN

VBC = 35.15 kN

35.15

42.84 kN

10 kN

24.85 kN

27.16

A B (4 − x)

x

DC


The location of zero shear force is located from similar triangles as

24.854 35.15

xx

=-

35.15x = 99.4 – 24.85x

x = 1.65 m

Example 2.7 Analyse the continuous beam by the moment distribution. The supports B and C settle by 8 mm and 4 mm respectively. EI = 30000 kNm2. Sketch the SFD and BMD.

BDAE6 m4 m

2 m 2 m240 kN 40 kN/m

C

FIG. 2.40




Joint Members Relative Stiff Values (k) k k/ kBA I/4 = 0.25I 0.67

B 0.375I

BC3 0.1254 6

I I¥ = 0.33

Fixed end moments MFAB = MAB + MAB where MAB is the fi xed end moment due to loading and MAB is the fi xed end moment due to settlement of supports. Due to applied loading

MAB = – 240 × 4 120 kNm8

= -

MBA = + 240 × 4 120 kNm8

= +

MBC = – 40 × 26 120 kNm

12= -

MCB = + 40 × 26 120 kNm

12= +

Due to settlement of supports

A

B'

C

CB

' Δ2 = 4 mm

θB = −lBC

Δ2θA = lAB

Δ1

Δ1 = 8 mm

4 m 6 m

+

FIG. 2.41 Settlement diagram

MAB = MBA = 12 21

6 8 16 3000 90 kNm1000 4

EIl

- D= - ¥ ¥ ¥ = -

MBC = MCB = 22 22

6 ( ) 4 16 3000 20 kNm1000 6

EIl

- - D= + ¥ ¥ ¥ = +


Hence, MFAB = – 120 – 90 = 210 kNm

MFBA = +120 – 90 = +30 kNm

MFBC = – 120 + 20 = – 100 kNm

MFCB = +120 + 20 = + 140 kNm

Moment distribution table


Joint A B C

Members AB BA BC CB

DF 0 0.67 0.33 1

FEMSBal

– 210.00 +30.00+46.90

– 100.00+23.10

+140.00– 140.00

CoBal

+23.45 –+46.90

– 70.00+23.10

–

Co +23.45

Total – 163.10 – 123.8 – 123.80 0.00

Equilibrium of span AB

2 m

4 m

240 kN

163.10

2 m123.8 kNm

VBAVAB D

FIG. 2.42

V = 0; VAB + VBA = 240 (1)

MA = 0; – 163.10 + 123.8 + 240(2) – 4VBA = 0 (2)

VBA = 110.2 kN

VAB = 129.8 kN

MD = – 163.10 + 129.8(2) = 96.5 kNm



6 m

123.8 kNm

VCBVBC E

40 kN/m

FIG. 2.43

V = 0; VBC + VCB = 240 (3)

MB = 0; – 123.8 – 6VCB + 40 × 26 0

2=

VCB = 99.4 kN

VBC = 140.6 kN

ME = – 123.8 + 140.6 × 3 – 40 × 23 118 kNm

2=

96.5

163.10 123.80

118 kNm

E CDBA


The reader can locate the point of contrafl exure and the maximum positive bending moment using basics.

129.8 kN 140.6

110.2 kN 99.4 kN

A D (6 x1)x1

FIG. 2.45


The location of zero shear force can be obtained from similar triangles,

1

1

99.4(6 ) 140.6

xx

=-

and hence x1 = 2.5 m

Example 2.8 Analyse the continuous beam by the moment distribution method. Sketch the shear force and bending moment diagrams. Relative to the support A, the support B sinks by 1 mm and the support C raises by 1/2 mm, EI = 30000 kNm2.

4 m 4 m80 kN

2 EI EI

DA B C

20 kN/m

6 m

FIG. 2.46



Joint Members Relative Stiffness Values (k) k k/ k

BA28I

0.67

B 3I/8

BC 34 6 8

I I¥ = 0.33

Fixed end moments Due to loading

MAB = – 80 × 8 80 kNm8

= -

MBA = + 80 kNm

MBC = – 20 × 26 60 kNm

12= -

MCB = +60 kNm


Due to settlement of supports

FIG. 2.47 Settlement angles diagram (R = ∆/l)

MAB = 12 21

6 (2 ) 6 30000 12 5.63 kNm

1000 8E I

l- D ¥ ¥

= - ¥ = -¥

MBA = 121

6 (2 )5.63 kNm

E Il

- D= -

MBC = 22 22

6 ( ) 6 30000 1.57.5 kNm

1000 6EI

l- - D ¥ ¥

= + =¥

It should be noted that the support B settles by 1 mm. In span AB, B settles and hence AB rotates in the clockwise direction. Therefore, (1/l1) is positive. In span BC, the support C raises by 1/2 mm from the original level AC. The total displacement is 1(1/2) mm with respect to the displaced position B. BC displaces to BC. Hence, the rotation angle (2/l2 ) is negative since the span BC rotates in the anticlockwise direction with respect to B. The fi xed end moment is the sum of the moments due to applied loading and due to the settlement of supports. MFAB = – 80 – 5.63 = – 85.63 kNm

MFBA = + 80 – 5.63 = + 74.37 kNm

MFBC = – 60 + 7.50 = – 52.50 kNm

MFCB = + 60 + 7.50 = + 67.50 kNm


Joint A B C

Members AB BA BC CBDF 0 0.67 0.33 1FEMSBal

– 85.63 +74.37– 14.65

– 52.50– 7.22

+67.50−67.50

CoBal

– 7.33+22.61

– 33.75+11.14

Co +11.31Final – 81.65 +82.33 – 82.33 0.00


Shear force diagrams Equilibrium of span AB

81.65 kNm 82.33 kNm80 kN

A BD

4 4 m

FIG. 2.48

V = 0, VAB + VBA = 80 (1)

MA = 0; – 81.65 + 82.33 + 80(4) – 8VBA = 0 (2)

VBA = 40.1 kN

VAB = 39.9 kN

MD = – 81.65 + 39.9 × 4 = 77.95 kNm


B C

82.3320 kN/m

6 m

FIG. 2.49

V = 0; VBC + VCB = 120 (3)

MB = 0; – 82.33 + 20 × 26 6 0

2 CBV- =

VCB = 46.3 kN

VBC = 73.7 kN The zero shear force location in BC is obtained from

46.36 73.7

xx

=-

73.7x = 277.8 – 46.3x

x = 2.32 m


CA BD

39.9 kN

73.7 kN

46.3 kN

40.1

x


Absolute maximum BM is 46.3 × 2.32 – 20 × 22.32 53.6 kNm

2=

C

A B

− −

+ +

D

81.65 kNm

77.95 kNm

53.6 kNm

2.32 m

88.33 kNm


Example 2.9 A continuous beam is loaded as shown in Fig. 2.52. During loading the support B sinks by 10 mm. Determine the bending moments at the supports. Sketch the BMD. Given that I = 1600(10)4 mm4; E = 200 kN/mm2. Use moment distribution method. Draw SFD also.

AB

C

8 m 4 m

3 kN/m 8 kN

2I I

2 2

FIG. 2.52




Joint Members Relative Stiffness Values (k) kSkk

BA 2 0.258I I= 0.57

B 0.438I

BC 3 0.1884 4

I I¥ = 0.43

Fixed end moments Due to loading

MFAB = – 3 × 28 16 kNm;

12= - MFBA = + 16 kNm

MFBC = – 8 × 4 4 kNm;8

= -

MFCB = + 4 kNm

Due to Settlement

A C8 m

10 mm

2 mθ

B

FIG. 2.53 Settlement diagram

MFAB = 11 12 2

1

6 10 16 2 3200 ( / is positive)1000 8

EI ll

D- = - ¥ ¥ ¥ ¥ D∵

MFAB = – 6 kNm

MFBA = +6 kNm

MFBA = 22 22 2

2

6 3200 ( 10)6 12 kNm ( / is negative)1000 4

EI ll

¥ ¥ -D- = - = + D

¥∵

MFCB = + 12 kNm

It is to be mentioned here that B settles by 10 mm with respect to A. The settlement angle is positive as AB rotates in clockwise direction with respect to A. The settlement


angle for CB is negative as CB rotates in the anticlockwise direction. With respect to C. Thus, the fi xed end moments at the joint is the sum of the moments due to loading and the settlement. They are as follows. Fixed end moments MFAB = – 16 – 6 = – 22 kNm

MFBA = 16 – 6 = 10 kNm

MFBC = – 4 + 12 = 8 kNm

MFCB = + 4 + 12 = 16 kNm


Joint A B C

Members AB BA BC CB

DF 0 0.57 0.43 1

FEMSBal

– 22.00–

+10.00–

+8.00–

+16.00– 16.00

Co – – – 8.00 –

TotalBal

– 22.00–

+10.00– 5.70

0.00– 4.30

0.00–

Co – 2.85 – – –

Final – 24.85 +4.30 – 4.30 0.00

Shear forces Free body diagram

A B24.85 4.30

3 kN/m

8 m

FIG. 2.54(a)

V = 0; VAB + VBA = 24 (1)

MA = 0; – 24.85 + 4.30 + 3 × 28 8 0

2 BAV- =

VBA = 9.4 kN

VAB = 14.6 kN


MD = – 24.85 + 14.6 × 4 – 23 4

9.55 kNm2¥

=

B E C

4.3 kNm8 kN

2 m 2 m

FIG. 2.54(b)

V = 0; VBC + VCB = 8

MB = 0; – 4.3 + 8(2) – 4VCB = 0

VCB = 2.9 kN

VBC = 5.1 kN

ME = 2.9(2) = 5.8 kNm

CA BD

14.6 kN

9.4

5.1 kN

2.9 kN

E


− −

+ +

24.85 kNm4.30 kNm

9.555.8 kNm



2.6.3 Analysis of Beams with Variable Moment of InertiaExample 2.10 A horizontal beam ACB, three metres long is fi xed at both ends A and B. They are at the same level. The member has a change of section at the centre of the span C such that the second moments of area are I for a distance AC and 2I for distance CB. A concentrated load of 500 kN acts at the midpoint C. Determine the fi xing end moments at A and B.

A

B

1.5 m

500 kN

1.5 m

C

I

2I

FIG. 2.57

Solution

A B

C '

Δ6EIΔ1.52

6EIΔ1.52

6EΔ1.52

(2I)

(2I)

Δ1.52

6E

FIG. 2.58

Applying the moment distribution method,


Joint Members Relative Stiffness (k) Sum k Distribution Factor (k/ k)

CA I/1.5 0.33

C 2I

CB 2I/1.5 0.67

The point C is displaced vertically by an amount ∆ to the point C keeping the section at C clamped (i.e., preventing any rotation). If an arbitrary end moment MCA is given an arbitrary value of 100, the fi xed end moment of MCB will be 200.



Joint A C BMembers AC CA CB BC

DF 1 0.33 0.67 1

FEMSBal

– 100 – 100– 33

+200– 67

+200

Co – 16.5 – 33.5

Final – 116.5 – 133 +133 +166.5

Equilibrium of span AC

A C

116.5 kNm

1.5 m

133 kNm

VAC VCA

FIG. 2.59

MC = 0;

– 116.5 – 133 + 1.5VAC = 0

VAC = 166.33 kN

Equilibrium of span CB

C B

133 kNm

1.5 m

167.5 kNm

VCB VBC

FIG. 2.60

Taking moment about C;

133 + 167.5 – 1.5VBC = 0

VBC = 200.33 kN

Resolving the forces vertically

= 166.33 + 200.33 = 366.66 kN


i.e., if the concentrated load applied at the centre is 366.66 kN then this will yield the moments given in the table. But the applied load is 500 kN. Hence, the moments at A and B are obtained as

MA = – 116.5 × 500 158.9 kNm

366.66= -

MB = +167.5 × 500 228.4 kNm

366.66= +

Example 2.11 A vertical column of 8 m height is to carry a crane girder load of 50 kN applied at an eccentricity of 0.2 m. Calculate the moments at A and B due to this load assuming both ends are fi xed.

Solution: Evaluate the distribution factor at C.

0.2 m

50 kN2 m

6 m

A

B

C

FIG. 2.61


Joint Members Relative Stiffness (k) k (k/ k)

CA I/6 0.25

C 0.667I

CB I/2 0.75


Taking moment about C for the portion AC,

12 + 8 – 6RA = 0

RA = 3.33 kN

Resolving all the forces horizontally

RA + RB – Y = 0

3.33 + 18 – Y = 0

Y = 21.33 kN

The sway moments corresponding to the sway force 5 kN is obtained by multiplying the moments in the moment distribution table as

MAC = 5 8 1.88 kNm

21.33Ê ˆ =Á ˜Ë ¯

MCA = 5 12 2.81 kNm

21.33Ê ˆ =Á ˜Ë ¯

MCB = 5 12 2.81 kNm

21.33Ê ˆ ¥ - = -Á ˜Ë ¯

MBC = 5 24 5.63 kNm

21.33Ê ˆ ¥ - = -Á ˜Ë ¯

Then the fi nal moments is the sum of sway and nonsway moments as

MAC = MAC + MAC = 1.25 + 1.88 = 3.13 kNm

MBC = MBC + MBC = 3.75 – 5.63 = – 1.88 kNm

2.7 ANALYSIS OF RECTILINEAR FRAMESExample 2.12 Analyse the frame shown in Fig. 2.65 by moment distribution method. Draw the bending moment diagram.

FIG. 2.64

B

A

RA

RB

8.00

24.00

2 m

6 m

C − Y12.00

12.00


B C2 m

2 m

1 m

20 kN/m

EI

EI2

A

FIG. 2.65



Joint Members Relative Stiffness Values (I/l) k k/ k

BA I/2 0.5B I

BC3 2 /24 3

I IÊ ˆ =Á ˜Ë ¯ 0.5

Fixed end moments

MBC = 2 22 2

2 20 0

( ) (20 ) (3 )13.33 kNm

3xwdx l x x dx x

l- -

= =Ú Ú

MCB = 2 22 2

20 0

( ) (3 ) 20 8.89 kNm9

x xI x wdx x dxl

- = - =Ú Ú Nonsway Analysis


Joint A B CMembers AB BA BC CBDF 0 0.5 0.5 1.0FEMSBal +6.66

– 13.33+6.67

+8.89– 8.89

CoBal

+3.33+2.22

– 4.44+2.22

Co +1.11+4.44 +8.88 – 8.88 0.00


MBC = – 8.88 + 50(0.107) = – 3.530 kNm MCB = 0

BC

3.53 20 kN/m

1 m2 mD

FIG. 2.67

MB = 0

3VC + 3.53 = 220 2

2¥

VC = 12.157 kN

MD = 12.157(1) = 12.157 kNm

V = 0

VB + VC = 2(20)

VB = 40 – 12.157 = 27.843 kN (SF)X = 27.843 – 20x = 0

x = 1.392 m

Maximum +ve BM = 27.843(1.392) – 220(1.392) 3.53

2-

ME = 15.851 kNm

B CE D3.585

3.585

3.585 kNm

12.157 kNm15.851 kNm

1.392 m

−

−

+ve

AFIG. 2.68 Bending moment diagram


Example 2.13 Analyse the frame shown in Fig. 2.69 by the moment distribution method. Draw the bending moment diagram.

E AG

B C

F

D

3 m2 m

3 m 3 m1 m

10 kN/m

25 kN

40 kN10 kN

I

2I

2I

3 m

FIG. 2.69



Joint Members Relative Stiffness Values (I/l) k (k/ k)

BA 34 3 4

I I¥ = 0.26

B BD (2I/5) 0.98 0.41

BC 2I/6 0.33

Fixed end moments

MFAE = +10 × 1 = 10 kNm

MFAB = – 10 × 23 7.5 kNm

12= -

MFBA = +10 × 23 7.5 kNm

12= +

MFBC = – 40 × 6 30.0 kNm8

= -


MFCB = + 40 × 6 30.0 kNm8

= +

MFBD = +2 × 25 × 32/52 = +18.0 kNm

MFBE = – 3 × 25 × 22/52 = – 12.0 kNm


Joint A B C D

Members AE AB BA BD BC CB DB

DF – 1.00 0.26 0.41 0.33 0.00 0.00

FEMSBal

+10.00 – 7.50– 2.50

+7.50+1.17

+18.00+1.85

– 30.00+1.48

+30.00 – 12.00

CoBal

– 1.25+0.33 +0.51 +0.41

+0.74 +0.93

Co 0.21

Total +10.00 – 10.00 +7.75 +20.36 – 28.11 +30.95 – 11.07

31.25

11.07

20.36 7.75

28.11 kNm 30.95 kNm

60–10.00

CGBA

F

D



Example 2.14 Analyse the frame by the moment distribution method. Draw the bending moment diagram.

A E

D

BC3

3 m

3 m

2100 kN 20 kN/m

2I I

I

FIG. 2.71



Joint Members Relative Stiffness (k) k DF = k/ k

BA 2I/5 = 0.4I 0.41

B BD I/3 = 0.33I 0.98I 0.34

BC 3 ( /3) 0.254

I I= 0.25

Fixed end moments

MFAB = – 2 × 100 × 2

23 72 kNm5

= -

MFBA = +3 × 100 × 2

22 48 kNm5

= +

MFBC = – 20 × = -23 15 kNm

12

MFCB = +20 × 23 15 kNm

12= +



Joint A B C DMembers AB BA BD BC CB DBDF 0 0.41 0.34 0.25 1 0FEMSBal

– 72.00–

+48.00– 13.53

–– 11.22

– 15.00– 8.25

+15.00– 15.00

CoBal

– 6.77+3.08 +2.55

– 7.50+1.88

– 5.61

Co +1.54 +1.28Total – 77.23 +37.55 – 8.67 – 28.87 0.00 – 4.33


Example 2.15 Analyse the frame by the moment distribution method. Draw the bending moment diagram.

B

D

E

C

15 m

15 m

15 m

7 m1 kN/m8 kN

A

I

I I

I

15 m

21

FIG. 2.73




Joint Members Relative Stiffness (k) k k/ k

BBA

BC

I/15

I/152I/15

0.5

0.5

CCBCECD

I/15I/15I/15

3I/150.330.330.33

Fixed end moments

MFBC = – 1 × 152/12 = – 18.75 kNm

MFCB = +1 × 152/12 = +18.75 kNm

MFCD = – 8 × 15/8 = – 15.00 kNm

MFDC = +8 × 15/8 = +15.00 kNm


Joint A B C D

Members AB BA BC CB CE CD DC EC

Distribution Factors 0 0.5 0.5 0.33 0.33 0.33 0 0FEMSBalance +9.38

– 18.75+9.37

+18.75– 1.25 – 1.25

– 15.00– 1.25

+15.00

Co +4.69 – 0.63 +4.69 – 0.63 – 0.63

Bal +0.32 +0.31 – 1.56 – 1.57 – 1.56

Co +0.16 – 0.78 +0.16 – 0.78 – 0.79

Bal +0.39 +0.39 – 0.05 – 0.05 – 0.06

Co +0.20 – 0.03 +0.20 – 0.03 – 0.03

Bal +0.02 +0.01 – 0.07 – 0.06 – 0.06

Final

Moments +5.05 +10.11 – 10.11 +20.87 – 2.93 – 17.93 +13.56 – 1.45


Table 2.34 Distribution factors for symmetrical case

Joint Members Relative Stiffness k DF = k/ k

BA 5I

0.8

B 0.25I

BC 1 0.052 10

I IÊ ˆ =Á ˜Ë ¯ 0.2

Fixed end moment

MFBC = (10) 1.25

8 8WWl W- = - = -


Joint A BMembers AB BA BCDF 0 0.8 0.2FEMBalCo +0.5W

+1.00W– 1.25W+0.25W

Final +0.5W +1.00W – 1.00W

Thus, from the above moment distribution table, joint moment 1.00W = 300

W = 300 kN

Hence, MAB = 0.5 × 300 = 150 kNm

450

150 150

300

300

300

300 kNm

A

B C

D

Elastic Curve

FIG 2.76 Bending moment diagram FIG 2.77



Joint A B

Members AB BA BC

DF 0.0 0.5 0.5

FEMSBalance +15.00

– 30.00+15.00

Co +7.50

Final Moments +7.50 +15.00 – 15.00

30

15 kNm15 kNm

7.5 kNm 7.5 kNm

−−− −

+

+ +


Example 2.18 The culvert shown in Fig. 2.81 is of constant section throughout and the top beam is subjected to a central concentrated load of 25 kN. Assume the base pressure is uniform throughout and analyse the box culvert. Draw the bending moment diagram.

5 m5 m

10 m

25 kN

2.5 kN/m

DA

B C

FIG. 2.81


Solution: The above frame is symmetrical and symmetrically loaded. This can be analysed using moment distribution in a straightforward and in a simpler manner by taking advantage of symmetry. The distribution factors are worked out by taking the central member BC as half of its value of stiffness. The process of moment distribution is carried out by considering only half of the frame.


Table 2.37

Joint Members Relative Stiffness Values k k/ k

BA 0.110I I= 0.67

B 0.15I

BC1 0.052 10

I IÊ ˆ =Á ˜Ë ¯ 0.33

AB 0.110I I= 0.67

A 0.15I

AD1 0.052 10

I IÊ ˆ =Á ˜Ë ¯ 0.33

Fixed end moments

MFBC = 1025 31.25 kNm

8 8Wl- = - ¥ = -

MFAD = 2 2102.5 20.83 kNm

12 12wl+ = + ¥ = +

Moment Distribution Due to symmetry analyse half of the frame. The joints A, B, C and D are rigid. This frame is recognised as a continuous frame. It implies that when joint moments are balanced, it is being carried over to the neighbouring joints. This carry over moment is again balanced, the process is continued and all the joints are balanced. Hence, the joint moments are

MA = MD = 9.06 kNm

MB = MC = 16.95 kNm


9.06

9.06 kNm

22.19

45.55

16.95 16.95 kNm

DA

B

− −

−−

− −

+

+

C


2.9 ANALYSIS OF UNSYMMETRICAL FRAMESIn the previous article, the frame was symmetrical and was symmetrically loaded. Half frame technique was used by taking beam stiffness as half of its original stiffness. It indicates that sways do not occur in the following cases (i) a structure which is held against lateral movement (Ref. Example 2.15) (ii) where both geometry of the frame and the loading are symmetrical (Ref. Examples 2.16 - 2.18).


Joint A B

Members AD AB BA BC

DF 0.33 0.67 0.67 0.33

FEMSBal

+20.83– 6.94 – 13.89 +20.83

– 31.25+10.42

CoBal – 3.47

+10.42– 6.95

– 6.95+4.63 +2.32

CoBal – 0.77

+2.32– 1.55

– 3.48+2.32 +1.16

CoBal – 0.38

+1.16– 0.78

– 0.78+0.52 +0.26

CoBal – 0.17

+0.26– 0.08

– 0.39+0.26 +0.13

CoBal – 0.04

+0.13– 0.09

– 0.04+0.03 +.01

Final 9.06 – 9.06 16.95 – 16.95


In reality, the frames are unsymmetrical and the loading could also be unsymmetrical. In such frames, there will be rotation of joints as well as the lateral displacement joints. The lateral displacement causes the frame to sway either to the right or to the left with respect to the centre line of the frame. The sway (side sway) is caused by (i) Portal frames with unequal columns (Fig. 2.83a) (ii) Portal frames whose columns are of different moment of inertia (Fig. 2.83b) (iii) Unsymmetrical loading on the beams (Fig. 2.83c) (iv) Lateral loading (Fig. 2.83d)

(a)

I1

I2I3

Δ Δ Δ Δ

(b)

I1

I2

I3

(d)

I1

I2

P

I3

ΔΔΔ

Δ

(c)

I

2 I

W

I

FIG. 2.83 Examples of sway frames

Analysis of sway frames is done in the following way. (1) The moment distribution was carried out by assuming that the joints do not get

displaced. The moments obtained from the moment distribution table are called nonsway moments.

(2) The horizontal reactions at the base of the columns are found out which helps in fi nding the net out of balance force. A prop is assumed to act opposite to the direction of the above force.

(3) Allow the frame to sway in the direction of sway force which is equal and opposite to the prop force (which acts along the axis of the beam level). Let the actual prop force be X.

(4) Perform the sway moment distribution, by assuming arbitrary moments as per the joint moment ratios.


(5) Calculate the displacement force (Y) from the sway moment distribution. (6) The correction factor/sway factor k = X/Y. The correction factor gives the

direction of sway of the frame. If the value of k is positive, then the frame sways in the direction of the sway solution. If the value of k is negative, then the frame sways in the direction opposite to that of assumed sway.

(7) The fi nal end moments are obtained as

Final end moments = Nonsway moments +‘Correction factor’ × Sway moments Expressing mathematically;

MAB = MAB + k MAB

where MAB is the end moment of the member AB, MAB is the moment obtained from nonsway moment distribution, MAB is the computed moment obtained from sway moment distribution k is the correction factor.

2.9.1 Joint Moment RatiosThe various moment ratios for rectangular frames are given below. Portal frames with both ends fi xed

I1

I3L

W

D

C

A

B

I2H2

H1

FIG. 2.84(a)

D

C

A

B Δ Δ

FIG. 2.84(b)


MAB = MBA = 121

6EIH

D-

and MCD = MDC = 222

6EIH

D-

2

1 22

2 1

AB

DC

M I HM I H

=

Portal frame with end fi xed and other hinged

I1

I3

L

D

C

A

B

I2H2

H1

FIG. 2.85(a)

D

C

A

B Δ Δ

FIG. 2.85(b)

MAB = 0 ( Hinged)

MBA = 121

3EIH

D-

MDC = MCD = 222

6EIH

D-


Nonsway Analysis


Joint A B C D


DF 0 0.45 0.55 0.55 0.45 0

FEMSBal +12.96

– 28.8015.84

+19.20– 10.56 – 8.64

CoBal

6.48+2.38

– 5.28+2.90

+7.92– 4.36 – 3.56

– 4.32

CoBal

1.19+0.98

– 2.18+1.20

+1.45– 0.80 – 0.65

– 1.78

CoBal

0.49+0.18

– 0.40+0.22

+0.60– 0.33 – 0.27

– 0.33

CoBal

0.09+0.08

– 0.17+0.09

+0.11– 0.06 – 0.05

– 0.14

CoBal

0.04+0.01

– 0.03+0.02

+0.05– 0.03 – 0.02

– 0.03

Total 8.29 16.59 – 16.59 13.19 – 13.19 – 6.60

MAB MBA MBC MCB MCD MDC

Sway Analysis The assumed sway moments are in the joint ratio

2

2

6 /36 /3

BA

CD

EIMM EI

- D=

- D

i.e. MBA : MCD = – 1 : – 1

The moments are arbitrarily assumed in the columns as – 100 kNm. Final moments These are obtained by adding the nonsway and sway moments as

MAB = 8.29 – 82.33 k

MBA = 16.59 – 64.73 k

MBC = – 16.59 + 64.73 k

MCB = 13.19 + 64.73 k

MCD = – 13.19 – 64.73 k

MDC = – 6.60 – 82.33 k


D

C

A 8.006.89 kNm

B

15.5014.29 kNm

32.89


D

C

A

B


Example 2.20 Analyse the given frame and draw the bending moment diagram

2I

D

C

A

BI

2I

3 6 m100 kN

50 kN4

4

FIG. 2.90



Table 2.42

Joint Members Relative Stiff Values (I/l) k k/ k

BA 2I/8 = 0.25I 0.7B 0.36I

BC I/9 = 0.11I 0.3CB I/9 = 0.11I 0.3

C 0.36ICD 2I/8 = 0.25I 0.7

Fixed end moments

MFAB = 850 50 kNm

8 8Wl- = - ¥ = -

MFBA = 850 50 kNm

8 8Wl+ = + ¥ = +

MFBC = 2 2

2 26100 3 133.33 kNm9

Wabl

- = - ¥ ¥ = -

MFCB = 2

22 2

6100 3 66.67 kNm9

Wa bl

+ = + ¥ ¥ = +

Nonsway Analysis


Joint A B C DMembers AB BA BC CB CD DCDF 0 0.70 0.30 0.30 0.70 0FEMS Bal

– 50.00 +50.00 +58.33

– 133.33+25.00

+66.67– 20.00 – 46.67

CoBal

+29.20+7.00

– 10.00 +3.00

+12.50– 3.75 – 8.75

– 23.34

CoBal

+3.50+1.32

– 1.88+0.56

+1.50– 0.45 – 1.05

– 4.38

CoBal

+0.66+0.16

– 0.23+0.07

+0.28– 0.08 – 0.20

– 0.53

CoBal

+0.08+0.03

– 0.04+0.01

+0.04– 0.01 – 0.03

– 0.10

– 16.56 +116.84 – 116.84 56.70 – 56.70 – 28.35



Sway Analysis


Joint A B C D


DF 0 0.70 0.30 0.30 0.70 0

FEMSBal

+100.00 +100.00– 70.00 – 30.00 – 30.00

+100.00– 70.00

+100.00

COBal

– 35.00+10.50

– 15.00+4.50

– 15.00+4.50 +10.50

– 35.00

CoBal

+5.25– 1.57

+2.25– 0.68

+2.25– 0.68 – 1.57

+5.25

COBal

– 0.79+0.24

– 0.34+0.10

– 0.34+0.10 +0.24

– 0.79

COBal

+0.12– 0.04

+0.05– 0.01

+0.05– 0.02 – 0.03

+0.12

Final 69.58 39.13 – 39.13 – 39.13 +39.13 +69.58


The fi nal moments are the addition of nonsway moments and a constant times sway moments; for example, MAB = MAB + k MAB. Therefore, End moments

MAB = – 16.56 + 69.58 k

MBA = +116.84 + 39.13 k

MBC = – 116.84 – 39.13 k

MCB = +56.70 – 39.13 k

MCD = – 56.70 + 39.13 k

MDC = – 28.35 + 69.58 k

The value of k is determined from the column shear condition as follows. Column shear condition Considering the free body diagram of column AB and the value of HA isdetermined as


23.096 – 27.18 k = 50

k = – 0.99 Hence, Final moments MAB = – 16.56 + 69.58(– 0.99) = – 85.4 kNm

MBA = +116.84 + 39.13(– 0.99) = +78.1 kNm

MBC = – 116.84 – 39.13(– 0.99) = – 78.1 kNm

MCB = +56.70 – 39.13(– 0.99) = +95.4 kNm

MCD = – 56.70 + 39.13(– 0.99) = – 95.4 kNm

MDC = – 28.35 + 69.58(– 0.99) = – 97.2 kNm

D

C

A 97.285.4

78.1B

+ 95.4 kNm

200

+

− −

100


Example 2.21 Analyse the portal frame shown in Fig. 2.94 by the moment distribution method. Sketch the bending moment diagram.

3 m

60 kN/m

4 m

A

B C

D

I

I2I

FIG. 2.94



Table 2.45

Joint Members Relative Stiffness Values k k/ k

BBA

BC

2I/4

I/30.83I

0.60

0.40

CCB

CD

I/3

I/40.58I

0.57

0.33

Fixed end moments

MFBC = – 60 × = -23 45 kNm

12

MFCB = +60 × 23 45 kNm

12= +

Though the boundary conditions and the loading are symmetrical, the frame sways as the columns AB and CD are having different moments of inertia of the cross–section. Hence, we have to do nonsway and sway analysis to determine the joint moments. Nonsway Analysis


Joint A B C DMembers AB BA BC CB CD DCDF 0 0.60 0.40 0.57 0.33 0FEMSBal +27.00

– 45.00+18.00

+45.00– 25.65 – 19.35

CoBal

+13.50+7.70

– 12.83+5.13

+9.00– 5.13 – 3.87

– 9.68

CoBal

+3.85+1.54

– 2.56+1.02

+2.56– 1.43 – 1.13

– 1.94

CoBal

+0.77+0.43

– 0.72+0.29

+0.51– 0.29 – 0.22

– 0.57

CoBal

+0.22+0.09

– 0.15+0.06

+0.15– 0.08 – 0.06

– 0.11

CoBal

+0.05+0.02

– 0.04+0.02

+0.03– 0.02 – 0.01

– 0.03

Final +18.39 +36.78 – 36.78 +24.65 – 24.65 – 12.33



Column shear condition In the given portal frame, the columns are fi xed and there is no horizontal load acting externally. Hence, the horizontal equilibrium gives

MAB + MBA + MCD + MDC = 0

(18.39 + 36.78 – 24.65 – 12.33) – (14.57 + 9.09 + 7.26 + 8.63) k = 0

18.19 – 39.55 k = 0

k = 0.46

Substituting this value of k in the above equations; End moments

MAB = 18.39 – 14.57(0.46) = 11.7 kNm

MBA = 36.78 – 9.09(0.46) = 32.6 kNm

MBC = – 36.78 + 9.09(0.46) = – 32.6 kNm

MCB = 24.65 + 7.26(0.46) = +28.0 kNm

MCD = – 24.65 – 7.26(0.46) = – 28.0 kNm

MDC = – 12.33 – 8.63(0.46) = +16.3 kNm

D

C

A 16.311.7

B32.6

28.0 kNm

37.2



D

C

A

B


Example 2.22 Analyse the portal frame shown in Fig. 2.97 by the moment distribution method. Draw the bending moment diagram.

2EI

D

C

A

B

EIEI

6 3 3 m

6 m

60 kN 60 kN

E F

FIG. 2.97

Solution: The frame is subjected to asymmetrical loading. Hence, nonsway and sway analyses were carried out separately and the results are summed up. Distribution factors

Table 2.48

Joint Members Relative Stiff Values (k) k DF

BBA

BC

I/6

2I/12 = I/6I/3

0.5

0.5

CCB

CD

2I/12 = I/6

I/6I/3

0.5

0.5


Fixed end moments

MFBC = 22

2 2

60 9 31260 123.75 kNm8 8 12

Wl Wabl

¥ ¥- - = - ¥ - = -

MFCB = 22

2 2

60 3 91260 191.25 kNm8 8 12

Wl Wa bl

¥ ¥+ + = ¥ + = -

Nonsway Analysis


Joint A B C D


DF 0.0 0.5 0.5 0.5 0.5 0.0

FEMSBal +61.88

– 123.75+61.87

+191.25– 95.63 – 95.62

CoBal

+30.94+23.91

– 47.82+23.91

+30.94– 15.47 – 15.47

– 47.81

CoBal

+11.96+3.87

– 7.74+3.87

+11.96– 5.98 – 5.98

– 7.74

CoBal

+1.94+1.50

2.99+1.50

+1.94– 0.97 – 0.97

– 2.99

CoBal

+0.75+0.24

– 0.49+0.25

+0.75– 0.37 – 0.38

– 0.49

CoBal

+0.12+0.10

– 0.19+0.09

+0.13– 0.06 – 0.07

– 0.19

Final 45.71 91.50 – 91.50 118.49 – 118.49 – 59.22


Sway moments The sway moments are assumed in the joint ratio of

2

2

6 /6 16 /6 1

BA

DC

EIMM EI

- D -= =

- D -

MBA : MDC = – 100.00 : – 100.00


Example 2.23 Analyse the given frame by the moment distribution method. Draw the bending moment diagram and shear force diagram.

1.5I

D

C

A

B2I

I

3 m2 m

3 m

6 m

80 kN

10 kN/m

20 kN

FIG. 2.100 Distribution factors

Table 2.51

Joint Members Relative Stiffness k k/ k

BA 1.5 0.256

I I= 0.38

B 0.65I

BC2 0.45I I= 0.62

CB 2 0.45I I= 0.55

C 0.73I

CD 0.333I I= 0.45

Fixed end moments

MFAB = – 10 × 26 30 kNm

12= -

MFBA = +10 × 26 30 kNm

12= +

MFBC = 2

2

2(90)3 64.8 kNm5

- = -


MFCB = +3(90) 2

23 43.2 kNm5

= +

Nonsway Analysis


Joint A B C DMembers AB BA BC CB CD DCDF 0 0.38 0.62 0.55 0.45 0

FEMSBal

– 30.00 +30.00+13.22

– 64.80+21.58

+43.20– 23.76

–– 19.44

–

CoBal

+6.61+4.51

– 11.88+7.37

+10.79– 5.93 – 4.86

– 9.72

CoBal

+2.26+1.13

– 2.97+1.84

3.69– 2.03 – 1.66

– 2.43

CoBal

+0.57+0.39

– 1.02+0.63

+0.92– 0.51 – 0.41

– 0.83

CoBal

+0.20+0.10

– 0.26+0.16

0.34– 0.19 – 0.15

– 0.21

CoBal

+0.05+0.04

– 0.10+0.06

+0.08– 0.04 – 0.04

– 0.08

Final – 20.31 +49.39 – 49.39 +26.56 – 26.56 – 12.44


Sway Analysis

MBA : MCD = – 6E(1.5I) 2 2: 6 ( )

6 3E ID D-

MBA : MCD = – 15.00 : – 40.00 kNm

Correction factor MAB = – 20.31 – 14.15 k

MBA = +49.39 – 13.34 k

MBC = – 49.39 + 13.34 k

MCB = +26.56 + 22.63 k

MCD = – 26.56 – 22.63 k

MDC = – 12.44 – 31.30 k


The value of k is determined from the column shear condition as Column shear condition MB = 0

MAB + MBA + 6HA = 10 × 26

2 and hence

HA = 30 – ( )

6AB BAM M+

Table 2.53 Moment distribution ta ble

Joint A B C D


DF 0 0.38 0.62 0.55 0.45 0

FEMSBal

– 15.00 – 15.00+5.70

–+9.30

–+22.00

– 40.00+18.00

– 40.00

CoBal

+2.85– 4.18

+11.00– 6.82

+4.65– 2.56 – 2.09

+9.00

CoBal

– 2.09+0.49

– 1.28+0.79

– 3.41+1.88 +1.53

– 1.05

CoBal

+0.25– 0.36

+0.94– 0.58

+0.40– 0.22 – 0.18

+0.77

CoBal

– 0.18+0.04

– 0.11+0.07

– 0.29+0.16 +0.13

– 0.09

CoBal

+0.02– 0.03

+0.08– 0.05

+0.04– 0.02 – 0.02

+0.07

Final – 14.15 – 13.34 +13.34 +22.63 – 22.63 – 31.30


FIG. 2.101

BMBA

MAB

VA

HA

6 m

20

10 kNm


MC = 0

MCD + MDC + 3HD = 0

HD = ( )

3CD DCM M+

-

HA + HD = 6(10) + 20

( ) ( )30 80

6 3AB BA CD DCM M M M+ +

- - =

Substituting the values of MAB, MBA, MCD and MDC from the above equations; and solving

k = 1.843

Final moments

MAB = – 20.31 – 14.15(1.843) = – 46.4

MBA = +49.39 – 13.34(1.843) = +24.8

MBC = – 49.39 + 13.34(1.843) = – 24.8

MCB = +26.56 + 22.63(1.843) = +68.3

MCD = – 26.56 – 22.63(1.843) = – 68.3

MDC = – 12.44 – 31.30(1.843) = – 70.1

FIG. 2.102

CMCD

MDC

VD

HD

3 m


D

C

A

70.1 kNm

46.4

B

24.8 kNm68.3 kNm

69.8

9.4

108


46.4 kN

46.4 kN

44.7

56.4

33.6

45.3 kN


Example 2.24 Analyse the rigid portal frame by moment distribution method and hence draw the bending moment diagram (Fig. 2.105).

SolutionDistribution factorsTable 2.54

Joint Members Relative Stiffness (I/l) k k/ k

BBA

BC

2I/5

2I/4

0.9I0.44

0.56

CCB

CD

2I/4

3 1.54 3

I¥

0.875I0.57

0.43


Column shear equation

MB = 0

MAB + MBA + 5HA – 20 × 25 0

2=

HA = 50 – 0.2(MAB + MBA)

MC = 0

MCD + 3HD = 0

HD = – 0.33MCD

H = 0; HA + HD = 20(5)

50 − 0.2(MAB + MBA) – 0.33 MCD = 100 (1)

FIG. 2.106

BMBA

MAB

A HA

5 m20 kN/m

FIG. 2.107

CMCD

VD

VC

D HD

HC

3 m


The end moments are obtained using the sway and nonsway moment distribution table as

MAB = – 45.49 – 86.00 k

MBA = +34.05 – 64.09 k

MBC = – 34.05 + 64.09 k

MCB = 6.50 + 44.11 k

MCD = – 6.50 – 44.11 k MDC = 0.00

Substituting the above end moments in Eq. (1);

k = 1.022

Hence, the end moments are obtained by back substitution as

MAB = – 133.4 kNm; MBA = – 31.5 kNm; MBC = +31.5 kNm;

MCB = +51.6 kNm; MCD = – 51.6 kNm; MDC = 0

133.4 kNm

51.6 kNm40

62.5


Example 2.25 The frame shown in Fig. 2.109 is hinged at A. The end D is fi xed and the joints B and C are rigid. Column CD is subjected to horizontal loading of 30 kN/m. A concentrated load of 90 kN acts on BC at 1 m from B. Analyse the frame and sketch the BMD?


D

C

A

B

1.5I

I

I

1 m1 m

34 m

90 kN

30 kN/m

FIG. 2.109


Table 2.57

Joint Members Relative stiffness k k k/ k

B

BA

BC

¥ =3 1.5 0.3754 3

l I

I/2 = 0.5I0.875I

0.43

0.57

CCB

CD

I/2 = 0.5I

I/4 = 0.25I0.75I

0.67

0.33

Fixed end moments

MFBC = – 90 × 2 22.5 kNm8

= -

MFCB = +90 × 2 22.5 kNm8

= +

MFCD = – 30 × 24 40 kNm

12= -

MFDC = +30 × 24 40 kNm

12= +


Nonsway Analysis


Joint A B C D


DF 1 0.43 0.57 0.67 0.33 0

FEMSBal +9.68

– 22.50+12.82

+22.50+11.73

– 40.00+5.77

+40.00

CoBal – 2.52

+5.87– 3.35

+6.41– 4.30 – 2.11

+2.89

CoBal +0.92

– 2.15+1.23

– 1.68+1.12 +0.56

– 1.06

CoBal – 0.24

+0.56– 0.32

+0.62– 0.42 – 0.20

+0.28

CoBal +0.09

– 0.21+0.12

– 0.16+0.11 +0.05

– 0.10

CoBal – 0.03

+0.06– 0.03

+0.06– 0.04 – 0.02

+0.03

Final 0 +7.90 – 7.90 +35.95 – 35.95 +42.04

Members MAB MBA MBC MCB MCD MDC

Sway AnalysisThe sway moments are assumed in the following ratios

2

2

3 (1.5 ) /36 ( ) /4

BA

CD

E IMM E I

D=

D

43

BA

CD

MM

=

MBA : MCD = +40 : +30 kNm

Final moments MAB = 0

MBA = 7.90 + 25.80 k

MBC = – 7.90 – 25.80 k

MCB = +35.95 – 23.22 k MCD = – 35.95 + 23.22 k

MDC = +42.04 + 26.58 k


Equilibrium of column CD

MC = 0;

MCD + MDC + 30 × 24 4 0

2 DH- =

HD = 60 + 1 ( )4 CD DCM M+

Considering the horizontal equilibrium of the whole structure;

HA + HD = 4(30)

i.e. ++ + =

( )60 120

3 4CD DCBA M MM

Substituting the values of the moments from the fi nal moment equations;

k = 2.65 The end moments were calculated as

MAB = 0

MBA = 7.90 + 25.80 × 2.65 = 76.3 kNm

MBC = – 7.90 – 25.80 × 2.65 = −76.3 kNm

MCB = +35.95 – 23.22 × 2.65 = – 25.6 kNm

MCD = – 35.95 + 23.22 × 2.65 = +25.6 kNm

MDC = +42.04 + 26.58 × 2.65 = +112.5 kNm

FIG. 2.111

CMCD

MDC

HDD

HC

4 m 30 kN/m


FIG. 2.114

BMBA

MAB

HAA

1 m

3 m20 kN/m

Sway Analysis


Joint A B C D


DF 0 0.5 0.5 0.5 0.5 0

FEMSBal

– 100.00 – 100.00+50.00 +50.00 +50.00

– 100.00+50.00

– 100.00

CoBal

+25.00

– 12.50

+25.00

– 12.50

+25.00

– 12.50 – 12.50

+25.00

CoBal

– 6.25+3.12

– 6.25+3.13

– 6.25+3.13 +3.12

– 6.25

CoBal

+1.56

– 0.78

+1.56

– 0.78

+1.57

– 0.79 – 0.78

+1.56

CoBal

– 0.39+0.20

– 0.40+0.20

– 0.39+0.20 +0.20

– 0.69

CoBal

+0.10

– 0.05

+0.10

– 0.05

+0.10

– 0.05 – 0.05

+0.10

Final – 79.98 – 60.01 +60.01 60.02 – 60.02 – 79.98


Column shear condition Equilibrium of column AB

MB = 0;

MAB + MBA – 20 × 23 4 0

2 AH+ =

HA = 90 ( )

4AB BAM M- +


MC = 0;

MCD + MDC + 4HD = 0

HD = ( )

4CD DCM M- +

Considering the horizontal equilibrium Substituting the values in terms of moments and simplifying HA + HD = 20 × 3 MAB + MBA + MCD + MDC = – 150

(– 26.41 + 11.81 + 3.38 + 1.68) – (79.98 + 60.01 + 60.02 + 79.98)k = – 150 k = 0.502 End moments MAB = – 26.41 – 79.98(0.502) = 66.6 kNm MBA = 11.81 – 60.01(0.502) = −18.3 kNm MBC = – 11.81 + 60.01(0.502) = +18.3 kNm MCB = – 3.38 + 60.02(0.502) = +26.8 kNm MCD = +3.38 – 60.02(0.502) = – 26.8 kNm MDC = 1.68 – 79.98(0.502) = – 38.5 kNm

D

C

A 38.566.6

B

18.3

26.8 kNm

26.8 kNm18.3


FIG. 2.115


Example 2.27 Analyse the rigid frame shown in Fig. 2.117 by the moment distribution method. Draw the bending moment diagram.

D

C

A

B

1.5EI

EIEI4 m

6 m

5 m

100 kN

FIG. 2.117

Solution: In the above problem, a lack of symmetry makes the frame to sway to the right. In the sway analysis, it is assumed that the joints B and C have no movement of rotation but there is only lateral translation. In the fi rst instant, an external force necessary to prevent the lateral translation is assumed. The moments at the supports and joints are computed for the above external force. This is followed by a redistribution of moments allowing for the side sway by removing the assumed external force. The frame is analysed by considering the sway moments only. Distribution factors

Table 2.63

Joint B CMembers BA BC CB CD

k3 0.1884 4

I IÊ ˆ =Á ˜Ë ¯1.5 0.25

6I I=

1.5 0.256

I I= 0.205I I=

k 0.438I 0.45I

DF = k/k 0.43 0.57 0.56 0.44

Sway moments The sway moments are assumed in the following ratio

2

2

3 /46 /5

BA

CD

EIMM EI

- D=

- D

MBA : MCD = – 25.00 : – 32.00 kNm


Moment distribution table

Table 2.64

Joint A B C D


DF 0 0.43 0.57 0.56 0.44 1

FEMSBal

0 – 25.00+10.75 +14.25 +17.92

– 32.00+14.08

– 32.00

CoBal – 3.85

+8.96– 5.11

+7.13– 4.00 – 3.13

+7.04

CoBal +0.86

– 2.00+1.14

– 2.56+1.43 +1.13

– 1.57

CoBal – 0.31

+0.72– 0.41

+0.72– 0.40 – 0.32

+0.57

CoBal +0.08

– 0.20+0.12

– 0.21+0.12 +0.09

– 0.16

CoBal – 0.03

+0.06– 0.03

+0.06– 0.03 – 0.03

+0.05

Final – 17.50 +17.50 +20.18 – 20.18 – 26.07

Moment 0

Column shear condition

MB = 0 MC = 0

4HA + MBA = 0 MCD + MDC + 5HD = 0

HA = 4

BAM-

HD =

( )5

CD DCM M- +

FIG. 2.118 FIG. 2.119


Considering the horizontal equilibrium, substituting the values of moments, HA and HD are calculated as

HA = ( 17.50)

4.38 kN4

- -=

HD = 1

( 20.18 26.07) 9.25 kN5

-- - =

The addition of HA and HD gives the lateral load which produces the assumed moments. Thus, HA + HD gives 13.63 kN but the applied load is 100 kN. Hence, the end moments in the table are to be multiplied by the ratio (100/13.63 = 7.34). The fi nal moments are

MAB = 0 × 7.34 = 0

MBA = – 17.50 × 7.34 = – 128.5 kNm

MBC = +17.50 × 7.34 = +128.5 kNm

MCB = +20.18 × 7.34 = +148.1 kNm

MCD = – 20.18 × 7.34 = – 148.1 kNm

MDC = – 26.07 × 7.34 = – 191.4 kNm

D

C

A 191.4

B

148.1 kNm

148.1 kNm128.5

128.5

−−

+

+


D

C

A

B θθ



Example 2.28 Analyse the frame by moment distribution method and draw the shear force diagram and bending moment diagram.

FIG. 2.122


Table 2.65

Joint Members Relative Stiff (I/l) k k/ k

BBA

BC

I/4

2I/43I/4

0.33

0.67

CCB

CD

2I/4

3 0.1884 4

I IÊ ˆ =Á ˜Ë ¯

0.688I0.73

0.27

Sway moments

The sway moments are assumed in the following ratio:

2

2

6 /4 23 /4 1

BA

CD

EIMM EI

- D -= =

- D -

Hence, MBA : MCD = – 20 : – 10 kNm


Table 2.66 Sway moment distribution table

Joint A B C D


DF 1 0.33 0.67 0.73 0.27 0

FEMSBal

– 20.00 – 20.00+6.60 +13.40 +7.30

– 10.00+2.70

CoBal

+3.30– 1.21

+3.65– 2.44

+6.70– 4.89 – 1.81

CoBal

– 0.61+0.81

– 2.45+1.64

– 1.22+0.89 +0.33

CoBal

+0.41– 0.15

+0.45– 0.30

+0.82– 0.60 – 0.22

CoBal

– 0.08+0.10

– 0.30+0.20

– 0.15+0.11 +0.04

CoBal

+0.05– 0.02

+0.06– 0.04

+0.10– 0.07 – 0.03

– 16.93 – 13.87 +13.87 8.99 – 8.99 0.00



MB = 0 MC = 0

MAB + MBA + 4HA = 0 MCD + 4HD = 0

HA = 1 ( )4 AB BAM M- + HD =

4CDM

FIG. 2.123

BMBA

A

4 m

MAB

HA

HB

VA

FIG. 2.124

C MCD

D

4 m

HC

HD

VC

VD


Example 2.29 Analyse the frame shown in Fig. 2.127 by the moment distribution method. Draw the bending moment diagram.

D

C

A

B

3I

I I

4 m

4

5 m

20 kN50 kN/m

40 kN

FIG. 2.127


Table 2.67

Joint Members Relative Stiffness (I/l) k k/ k

B

BA

BC

3 1 .0944 8

Ê ˆ =Á ˜Ë ¯

3 0.65I I=

0.694I

0.14

0.86

CCB

CD

3 0.65I I=

Ê ˆ =Á ˜Ë ¯3 .0944 8

I I

0.694I0.86

0.14

Fixed end moments

MFAB = – 40 × 8 40.0 kNm8

= -

MFBA = +40.0 kNm

MFBC = 250 5

104.17 kNm12

- ¥= -

MFCB = +50 × 25 104.17 kNm

12= +


End moments

MAB = 0

MBA = 75.18 – 90.27 k

MBC = – 75.18 + 90.27 k

MCB = +21.07 + 90.27 k

MCD = – 21.07 – 90.27 k

MDC = 0

Nonsway Analysis

Table 2.68

Joint A B C D


DF 1 0.14 0.86 0.86 0.14 1

FEMSBal

– 40.00+40.00

+40.00+8.98

– 104.17+55.19

+104.17– 89.58

0– 14.59

0

CoBal

+20.00+3.47

– 44.79+21.32

+27.60– 23.74 – 3.86

CoBal +1.66

– 11.87+10.21

+10.66– 9.16 – 1.50

CoBal

0+0.64

– 4.58+3.94

+5.11– 4.40 – 0.71

0

CoBal +0.31

– 2.20+1.89

+1.97– 1.69 – 0.28

CoBal +0.12

– 0.85+0.73

+0.95– 0.82 – 0.13

75.18 – 75.18 +21.07 – 21.07



Hence, MBA : MCD = – 100.00 : – 100.00

20 – 0.125MBA – 0.125MCD = 60

20 – 0.125(75.18 – 90.27 k) – 0.125(– 21.07 – 90.27 k) = 60

Solving for k = 2.07

Substituting the values of k by back substitution:

MAB = 0, MBA = – 111.86, MBC = +111.86,

MCB = 207.9, MCD = – 207.9 kNm, MDC = 0


Joint A B C D


DF 1 0.14 0.86 0.86 0.14 1

FEMSBal

0 – 100.00+14.00

0+86.00

0+86.00

– 100.00+14.00

0

CoBal – 6.02

+43.00– 36.98

+43.00– 36.98 – 6.02

CoBal +2.59

– 18.49+15.90

– 18.49+15.90 +2.59

CoBal – 1.11

+7.95– 6.84

+7.95– 6.84 – 1.11

CoBal +0.48

– 3.42+2.94

– 3.42+2.94 +0.48

CoBal – 0.21

+1.47– 1.26

+1.47– 1.26 – 0.21

0 – 90.27 +90.27 +90.27 – 90.27 0



FIG. 2.133

Decomposition of the structure taking advantage of symmetry. Fixed end moments in symmetrical case

MFBC = – 20 × 2 × 3 24 kNm5

= -

MFCB = +20 × 2 × 3 24 kNm5

= +

Table 2.71 Moment distribution for symmetric case

Joint A B

Members AB BA BC

DF 0 0.62 0.38

FEMBal +14.88

– 24.00+9.12

Co 7.44

Final 7.44 14.88 – 14.88

Skew–symmetric case The cantilever stiffness for the column AB is taken as (I/3). While the skew stiffness of the beam BC is taken as 6(2I/5). The derivation of the cantilever stiffness and skew stiffness are available in standard references.


FIG. 2.134

A

B

20 kN2 m

(a) Symmetric case (b) Skew symmetric case

21 2I

5( ) I3

A

B

20 kN2 m

2I5( )

I3

6Axisofsymmetry

Axisofsymmetry

Table 2.72 Distribution factors for skews symmetric case

Joint Members Relative Stiffness k DF = k/ k

BA I/3 = 0.33I 0.12

B 2.73I

BC 6(2I)/5 = 2.4I 0.88

Fixed end moments in skew symmetric case

MFBC = 2 2

2 2

2 20 3 3 20 24.8 kNm

5 5¥ ¥ ¥ ¥

- + = -

MFCB = – 4.8 kNm

Table 2.73 Moment distribution for skew symmetric case

Joint A BMembers AB BA BC

0 0.12 0.88FEMS – 4.80Bal +0.58 +4.22Co – 0.58Final – 0.58 +0.5 – 0.58


FIG

. 2.1

37

Dec

ompo

sitio

n of

giv

en lo

ad s

yste

m t

o sy

mm

etri

c an

d sk

ew s

ymm

etri

c lo

adin

g sy

stem

s

C

AD

B6

63

36

m6

33

++

=EI

EIEI

EIEI

EI

2EI

2EI

2EI

6 m

33 EI

EI

2EI

6060

6030

3030

30 k

N

Cas

e (i)

Cas

e (ii

)C

ase

(iii)


Table 2.76

Joint A B

Members AB BA BC

DF 0.0 0.67 0.33

FEMSBal +45.0

– 67.5+22.5

CO +22.5

Total 22.5 45.0 – 45.0

Case (iii) Skew symmetric caseThe distribution factors at joint B are calculated by considering the cantilever stiffness and taking the skew stiffness of the beam. MFBC = (3 × 30 × 92/122 – 9 × 30 × 32/122) = + 33.75 kNm. The cantilever stiffness is taken as (I/6) and for the beam (6 × 2I/12). Hence,

dBA : dBC = (I/6) : (6 × 2I/12), i.e. dBA : dBC = 1 : 6

i.e. dBA = (1/7) and dBC = (6/7).

Table 2.77

Joint A B

Members AB BA BC

DF 0 0.14 0.86

FEMSBal – 4.72

+33.75– 29.03

CO +4.72

Total +4.72 – 4.72 +4.72

The fi nal moments are obtained by adding:


Case (i) 30.00 60.00 – 60.00 +60.00 – 60.00 – 30.00

Case (ii) 22.50 45.00 – 45.00 +45.00 – 45.00 – 22.50

Case (iii) 4.72 – 4.72 +4.72 +4.72 – 4.72 +4.72

Total 57.22 100.28 – 100.28 109.72 – 109.72 – 47.78

The above moments tally with the moments obtained by classical moment distribution (Ref. Ex. 2.22). The Naylor’s method is very much useful if the reader identifi es the application of the same to specifi c problems.


The distribution factors are obtained as follows.

B

BA BC

2I

4I

4I

0.5 0.5

Table 2.78 Member end moments for symmetrical loading

Members AB BA BC

DF 0 0.5 0.5

FEMSBal

– 26.67 +26.67– 13.33 – 13.34

CO – 6.67

Bal 0.00 0.00 0.00

Total – 33.34 13.34 – 13.34

The analysis for antisymmetrical distribution is carried out using cantilever moment distribution. The stiffness for the column is taken as (EI/4) and for the beam it is taken as six times the original stiffness. Hence, 6E(2I)/4 = EI. The fi xed end moments for the columns are taken as

MFAB = 2 2 2420 106.67 kNm

12 2 2 3 3wl wl l wlÊ ˆ- - = - = - ¥ = -Á ˜Ë ¯

MFBA = 22 2 20 4

53.33 kNm12 2 2 6 6wl wl l wl ¥Ê ˆ+ - = - = - = -Á ˜Ë ¯

The carryover factor to the column is (– 1) in antisymmetrical loading. The distribution factors for the antisymmetrical loading is

k

k

DF = kkS


B

BA BC

4I

46(2I)

0.077 0.923

13

1 12

Table 2.79 Member end moments for symmetrical loading

Members AB BA BC

DF 0 0.77 0.923

FEMSBal

– 106.67 – 53.33+4.17 +49.16

CO 4.17

Bal 0.00 0.00 0.00

Final – 110.84 – 49.16 +49.16

The end moments of the members were obtained by adding the moments due to symmetric loading and antisymmetric loading.


Symmetric Loading – 33.34 +13.34 – 13.34 +13.34 – 13.34 +33.34

Antisymmetric Loading – 110.84 – 49.16 +49.16 +49.16 – 49.16 – 110.84

Total – 144.18 – 35.82 +35.82 +62.5 – 62.50 – 77.50

Joint Memberkkk

kkS


C

A

B

E

F

18.47

18.47 18.47 kNm

66.07

66.0711.53 54.56 kNm

54.56

65.4465.44 kNm


2.11 ANALYSIS OF FRAMES WITH INCLINED LEGSExample 2.34 Analyse the rigid frame shown in fi gure by the moment distribution method and draw the BM diagram Support A is hinged and support D is fi xed support.

I

2I

D

C

A

B

2I

6 m 3 m

4 m

10 kN

O

FIG. 2.143

SolutionDetermination of sway (∆) of members


FIG. 2.144 Sway displacement diagram

Δ

'

C C1

C

O

FIG. 2.145

Consider the above triangle, CC = ∆ cot = 0.75∆ CC = ∆ cosec = 1.25∆ Sway of AB = 1 = ∆AB = BB = +∆

Sway of BC = 2 = ∆BC = – CC = – ∆ cot = – 0.75∆ Sway of CD = 3 = ∆CD = +CC = ∆ cosec = +1.25∆

Sway fi xed end moments MFAB = 0

MFAB = – 3EI11/l21 = 2 33 /4

16EI EI- D = - D

MFBC = MFCB = – 6EI22/l22 = 26 (2 )( 0.75 )/6

4EIE I D- - D =

MFCD = MFDC = – 6EI33/l23 = 2 36 (2 )( 0.25 )/5

5EIE I D- - D =


MFBA : MFBC : MFCD = 3 1 3: :

16 4 5- -

– 15 : +20 : – 48

Relative Stiffness k Values and Distribution Factors


Joint Members Relative Stiffness Values I/l k DF = k/ k

BA3 34 4 16

I IÊ ˆ =Á ˜Ë ¯ 0.36

B = 0.1875I = 0.5175I

BC2 0.336I I= 0.64

CB 2 0.336I I= 0.45

C 0.73I

CD 2 0.45I I= 0.55

Table 2.83 Sway moment distribution table

Joint A B C DMembers AB BA BC CB CD DCDF 1 0.36 0.64 0.45 0.55 0FEMSBal

– 15.00– 1.80

+20.00– 3.20

+20.00+12.60

– 48.00+15.40

– 48.00

COBal – 2.27

+6.30– 4.03

– 1.60+0.72 +0.88

+7.70

COBal +0.13

+0.36– 0.23

– 2.02+0.91 +1.11

+0.44

COBal – 0.17

+0.46– 0.29

– 0.12+0.05 +0.07

+0.56

COBal – 0.01

+0.03– 0.02

– 0.15+0.07 +0.08

+0.04

COBal – 0.01

+0.04– 0.03

– 0.01+0.005 +0.005

+0.04

– 19.39 19.39 30.46 – 30.46 – 39.22


10 kN

D

B C

C

C1

A

3 mΔ Δ

'


Example 2.35 Determine the end moments and draw the bending moment diagram.

6 kN/m

12 kN

2 m

C

D

B

A

3 m

1.5 m

1.5 m

FIG. 2.149

Solution: In inclined frames, the sway of the members is to be determined from geometry. Let the column AB sway by an amount ∆. The beam member BC moves horizontally to BC as shown in Fig. 2.150. The displacement CC is related to sway ∆ and the angle of the inclined member CD. Thus, using the geometry of the defl ected shape the sway of the members are related. Fixed end moments due to loading

MFAB = – 12 × 3 4.5 kNm;8

= -

MFBA = +12 × 310 = +4.5 kNm

MFBC = – 6 23 4.5 kNm

12¥ = -

MFCB = +6 × 23 4.5 kNm

12= +


Determination of sway (∆) of the member

6 kN/m

C

CB

C1

D

RCD

RAB

B

A

Δ Δ

''

φ

FIG. 2.150 Sway displacement diagram

C

C1C

'

φ

FIG. 2.151

From ∆CC1C; cot = 1C C ¢D

C1C = cot

cosec = CC ¢

D

CC = cosec

Referring to the above given frame

cot = 2/3 = 0.67

cosec = 13/3 1.20=

Sway of AB = ∆

Sway of BC = – 0.67∆

Sway of CD = +1.20∆


Sway moment distribution The sway moments are assumed as

MFAB = MFBA = 12 21

6 6 0.6673

EI EI EIl

D- = - = - D

MFBC = MFCB = 2 22 22

6 ( 0.7 )6 0.4473

EIEI EIl

- DD- = = + D

MFCD = MFDC = 3 323

6 (1.2 )6 0.55413

EIEI EIl

DD- = - = - D

MFBA : MFBC : MFCD = – 0.667EI∆ : +0.447EI∆ : – 0.554EI∆

= – 66.70 : +44.70 – 55.40


Joint A B C D


0 0.5 0.5 0.55 0.45 0

FEMSBal

– 66.70 – 66.70+11.00

+44.70+11.00

+44.70+5.89

– 55.40+4.82

– 55.40

COBal

+5.50– 1.48

+2.95– 1.48

+5.50– 3.03 – 2.47

+2.41

COBal

– 0.74+0.76

– 1.52+0.76

– 0.74+0.41 +0.33

– 1.24

COBal

+0.38– 0.10

+0.21– 0.11

+0.38– 0.21 – 0.17

+0.17

COBal

– 0.05+0.06

– 0.11+0.05

– 0.06+0.03 +0.03

– 0.09

COBal

+0.03– 0.01

+0.02– 0.01

+0.03– 0.02 – 0.01

+0.02

Total – 61.58 – 56.47 +56.47 +52.88 – 52.88 – 54.13




Table 2.85

Joint Members Relative Stiff Values (k) k k/ k

BBA

BC

I/3

I/32I/3

0.5

0.5

CCB

CD

I/3

/ 3I2I/3

0.55

0.45

Table 2.86 Nonsway member distribution table

Joint A B C D


DF 0 0.5 0.5 0.55 0.45 0

FEMSBal

– 4.50 +4.500

– 4.500

+4.50– 2.48

0– 2.02

0

COBal

0+0.62

– 1.24+0.62

00 0

– 1.01

COBal

+0.310

00

+0.31– 0.17 – 0.14

0

COBal

0+0.05

– 0.09+0.04

00 0

– 0.07

COBal

+0.030

00

+0.02– 0.01 – 0.01

0

Total – 4.16MAB

+5.17MBA

– 5.17MBC

2.17MCB

– 2.17MCD

– 108MDC

– 61.58MAB

– 56.47MBA

+56.47MBC

52.88MCB

– 52.88MCD

– 54.13MDC

End moments The end moments are the sum of the moments of the nonsway and sway moments.

MAB = – 4.16 – 61.58 k

MBA = +5.17 – 56.47 k

MBC = – 5.17 + 56.47 k

MCB = +2.17 + 52.88 k


MCD = – 2.17 – 52.88 k

MDC = – 1.08 – 54.13 k


FIG. 2.152(a)

BMBA

SBA

A

12 kN

MAB

SAB

NA

NB

1.5 m

1.5 m

FIG. 2.152(b) FIG. 2.152(c)


FIG. 2.156

cosec = 5/4 = 1.25

cot = 3/4 = 0.75

From the displacement diagram

cot = 1B B¢D

B1B = cot

cosec = 1

BBBB

¢

BB = BB1 cosec

BB = 1.25∆

CC = 1.25∆

Vertical displacement between B and C

= BB1 + C1C = 2cot = 1.5∆

Sway of AB = AB = BB = cosec = +1.25∆

Sway of BC = ∆BC = – 1.5∆

Sway of CD = ∆CD = cosec = +1.25∆

Nonsway Analysis

MFBC = – 100 × = -8 100 kNm8

MFCB = +100 × 8 100 kNm8

= +


A

MAB

S ABNA

MBA

M CD

MDC

SCD

SDC

SBA

NBNB

ND

4 m

5 m

3 m

5 m

25 kN

B

D

C

(a) (b)

FIG. 2.157

Considering the above free body diagram,

MB = 0;

MAB + MBA + 5 SAB = 0

i.e. SAB = ( )

5AB BAM M+

-

(1)

MC = 0;

MCD + MDC – 5 SDC = 0

SDC = ( )

5CD DCM M+

- (2)

On simplifying

(MAB + MDC) + 1.75(MBA + MCD) + 100 = 0

The values of the moments are obtained from the above nonsway moment distribution and sway moment distribution case as

{(38.27 – 26.39 k) – (38.27 + 24.78 k)} + 1.75{(76.55 – 22.74 k) + (– 76.55 – 24.78 k)} + 100 = 0


– 51.17 k – 83.16 k + 100 = 0

k = 0.74

Final moments MAB = 38.27 – 26.39 k(0.74) = 18.74 kNm

MBA = 76.55 – 22.74 k(0.74) = 59.72 kNm

MBC = – 76.55 + 22.74(0.74) = – 59.72 kNm

MCB = +76.55 + 24.78(0.74) = 94.88 kNm

MCD = – 76.55 – 24.78(0.74) = – 94.88 kNm

MDC = – 38.27 – 27.40(0.74) = – 58.55 kNm

CB

59.7294.88 kNm122.7

18.74 kNm 58.7A D


ME = 8 1100 (59.72 94.88) 122.7 kNm4 2

¥ - + =


2.12 ANALYSIS OF GABLE FRAMESExample 2.37 Analyse the rigid frame shown in Fig. 2.161 by the moment distribution method and sketch the bending moment diagram.


C

DB

A E

200 kN

3 m

8 m

4 m4 m

1.5I1.5I

II

φ

FIG. 2.161

SolutionSway (∆) values of the members The Gable frame sways equally on both sides (either side) at B and D by virtue of symmetry on the frame and the loading on YY axis.

Sway of AB = ∆AB = – BB = – ∆ (negative sign because AB sways in theanticlockwise direction with respect to AB)

Sway of BC = ∆BC = +C1C = +cosec

∆BC = 53

+ D

Sway of CD = ∆CD = – C2C = – cosec

∆CD = 53

- D

Sway of DE = ∆DE = DD

∆DE = +∆


FIG. 2.162 Sway diagram FIG. 2.163

Sway moment values(Fixed end moments due to sway)

MFAB = MFBA = 1 12 2

6 6 ( )8 64 64EI EIEI EI- D - D

- = - = - = +

MFBC = MFCB = 2 22 22

6 (1.5 ) (5/3 )6 155 25

EEI EIl

- D DD D- = - = -

MFCD = MFDC = 3 32 23

6 (1.5 ) ( 5/3 )6 155 25

E ZEI EIl

- - DD D- = - = +

MFDE = MFED = 4 42 24

66 68 64EIEI EI

l- DD D= - = -

6 /64 10

15 / 25 64FAB

FBC

EIMM EI

+ D += =

- D -

Assume the sway moments in the above proportion as follows:

MFAB = MFBA = +10.00 kNm

MFBC = MFCB = – 64.00 kNm

MFCD = MFDC = +64.00 kNm

MFDE = MFED = – 10.00 kNm

Distribution factorsDue to symmetry, consider joints B and C only.


C

B

MBC

200 kN

VBC

HCB

MCB

VCB

4 m

B

AMAB

100 kN

100 kN

HBC

MBA

HAB

HBA

8 m 3 m

Column AB Member BC

FIG. 2.164(b) FIG. 2.164(c)

In Fig. 2.164(b);

MB = 0;

– 8HAB + MBA + MAB = 0

HAB = ( )

8AB BAM M+

HBC = HBA = HAB = (MAB + MBA)/8

VAB = VBC = VBA = 100 kN

Referring to Fig. 2.164(c) and taking moment about (c),

4 VBC – 3 HBC + MBC + MCB = 0

Substituting the value of HBC in terms of moments,

( )4(100) 3 ( ) 0

8AB BA

BC CBM M

M M+

- + + =

3200 – 3(MAB + MBA) + 8(MBC + MCB) = 0

Substituting the end moments from moment distribution table with a multiplier ‘k’;

3200 – 3(18.1 k + 26.2 k) + 8(– 26.2 k – 45.1 k) = 0

k = 4.55


Solution: The Gable frame sways equally on either side B and D by virtue of symmetry of the frame and loading.

A

C

C

E

D DBB

Sway Diagram

θ

θ θ

θ

ΔΔ

ΔC1 C2

′′′

FIG. 2.167 Sway diagram

FIG. 2.168

In Fig. 2.166 length BC = 2 26 3 6.71 m+ =

cosec θ = 6.71/3 = 2.237

In Fig 2.167 cosec θ = 1C C ¢D

C1C = cosec θ = 2.237∆

In Fig. 2.167 Sway of AB = BB = – ∆

Sway of BC =C1C = +2.237∆

Sway of CD =C2C = – 2.237∆

Sway of ED = DD = +∆


Fixed end moments

Due to loading

6 m

3 m

B

C

25 kN/m

FIG. 2.169

MFAB = MFBA = MFDE = MFED = 0

MFBC = – 25 × 26 75 kNm

12 FCDM= - =

MFCB = +25 × 26 75 kNm

12 FDCM= + =

Sway moment values

MFAB = MFBA = 1 12 2

6 (2 )( ) 126 36

E IEI EI-D- = - =

MFBC = MFCB = 2 22 22

40.276 (3 )(2.237 )66.71 45

EIE IEIl

- DDD- = - =

15

40.27FAB

FBC

MM

+=

-

Assume the sway moments in the above proportion as

MFAB = MFBA = +15.00 kNm

MFBC = MFCB = – 40.27 kNm

Distribution factors Due to symmetry, consider joints B and C only.


6 m

3 m

HCB

MCB

MBC

BHBC

VBC = 150 kN

C25 kN/m

FIG. 2.171

MC = 0

150(6) + MBC + MCB – 3HBC – 25 × 26 0

2=

As HBC = HBA = HAB and replacing in terms of moments as above.

150 × 6 + MBC + MCB – 3 ( ) 450 06 AB BAM M+ - =

900 + (64.13 – 58.94 k) – 0.5(48.38 + 46.3 k) – 450 = 0

489.94 – 82.09 k = 0

k = 5.97

End momentsThe end moments are obtained by substituting the value of k in the above equations as

MAB = 16.13 + 20.43(5.97) = 138.1 kNm

MBA = 32.25 + 25.87(5.97) = 186.7 kNm

MBC = – 32.25 – 25.87(5.97) = – 186.7 kNm

MCB = +96.38 – 33.07(5.97) = – 101.0 kNm

HAB = 54.13 kN6

AB BAM M+=

From geometry; cos θ = 6/6.71 = 0.894 and

sin θ = 3/6.71 = 0.447


25 kN/m

6 m

150 kN186.7

186.7109.9

54.13

101.0

54.13 kN

24.2

24.2

101.048.39

101.0

115.44

(6.71 − x)

x

C

θ

θ

SFD

FIG. 2.172 Bending moment diagram (BMD)

54.13 kN

54.13 kN

−

SFD

(b)

186.7 kNm

138.1 kNm

−

+

BMD

(c)

6 m

138.11 kNm

54.13 kN

54.13

186.67

B

A

(a)

FIG. 2.173

Referring to shear force diagram,

109.96.71 24.2

xx

=-

x = 5.5 m

186.67


MFCD = MFDC = 2 22 22

6 6 34 8

EI EI EIl

D D- = - = D

MEF = MFE = 3 32 23

6 6 63 9

EI EI EIl

D D D- = - =

MFBC = MFCB = 0

MFCE = MFEC = 0

MFBA : MFCD : MFEF = 12 3 6: :

25 8 9EI EI EID D D- - -

MFAB : MFCD : MFEF = – 0.48∆ : – 0.375∆ : – 0.67∆

Hence, MFAB : MFCD : MFEF = – 48.00 : – 37.50 : – 66.7 kNm

54

A

B C

D

E

F

3I I

I

2I

2I

Δ Δ Δ

FIG. 2.177


Table 2.93

Joint Members Relative Stiffness Values (k) I/l k/ k

BBA

BC

2I/5

2I/54I/5

0.50

0.50

CCBCDCE

2I/5I/4I/4

0.9I0.440.280.28

EEC

EF

I/4

I/30.58I

0.43

0.57



MB = 0

MAB + MBA + 5HA = 0

HA = – 0.2(MAB + MBA)

MC = 0

MCD + MDC + 4HD = 0

HD = – 0.25(MCD + MDC)

ME = 0

MEF + MFE + 3HF = 0

HF = – 0.33(MEF + MFE)

Considering the horizontal equilibrium 80 – HB – HD – HF = 0

80 + 0.2(MAB + MBA) + 0.25(MCD + MDC) + 0.33(MEF + MFE) = 0

FIG. 2.178

5 m

MBA

MAB

HA

VA

B

A

FIG. 2.179

4 m

MCD

MDC

HD

HC

VD

VC

C

D

FIG. 2.180

3 m

MEF

MFE

HF

E

F


80 + {0.2(– 36.66 – 25.36) + 0.25(– 34.09 – 35.77) + 0.33(– 29.68 – 48.18)}k = 0

k = 1.44 Final momentsThe moments obtained at the end of moment distribution table is multiplied by k and hence

MAB = – 52.8 kNm, MBA = – 36.5 kNm, MBC = – 36.5 kNm

MCB = +24.1 kNm, MCD = – 49.1 kNm, MDC = – 51.5 kNm

MCE = +35.4 kNm, MEC = +42.7 kNm, MEF = – 42.7 kNm

MFE = – 69.4 kNm

35.4 kNm

42.7 kNm

69.4

51.5

52.79

36.5 ECB

A

D

F


REVIEW QUESTIONSRemembrance 2.1. Who has developed the moment distribution method? 2.2. Is the moment distribution a stiffness method or fl exibility method? 2.3. List the advantages of the moment distribution method? 2.4. List the important steps in the moment distribution method? 2.5. Does the axial deformation considered in the development of the moment

distribution method? 2.6. Defi ne rotational stiffness? 2.7. Explain distribution factor? 2.8. Defi ne carry over factor? 2.9. Which moment distribution is preferable for symmetric frames subjected to lateral

loads as storey heights?


2.10. What are the other names for cantilever moment distribution method? 2.11. What are the advantages of Naylor’s method? 2.12. What is the rotational stiffness of a cantilever? 2.13. What is the magnitude of a stiffness of a member under antisymmetric bending

in Naylor’s method? 2.14. List the values of symmetric stiffness factor, skew symmetric stiffness factor and

cantilever stiffness factor?

Understanding 2.1. In a member AB, if a moment of 10 kNm is applied at A, what is the moment

carried over to the fi xed end B? 2.2. What is sinking of supports? What is its effect on the end moments of the

member? 2.3. Calculate the MFBA and MFBC for the beam shown in fi gure below due to sinking

of support by 2 mm. E = 6000 N/mm2 and I = 1.6(10)8 mm4.

2 m

B CA

4 m

EI 2EI

2.4. Calculate the MFBC and MFCB for the beam shown in fi gure below EI = 10 × 1011 Nmm2. Support B and C sinks by 2 mm and 3 mm respectively.

3 m 3 mB DA

4 mC

2.5. Is it possible to determine the beam defl ections in a continuous beam by moment distribution method?

2.6. List the reasons for sidesway of the portal frame. 2.7. What is balancing at a joint? 2.8. While applying the moment distribution method, a designer remembers that

“nothing comes back from the fi xed end”. Justify. 2.9. What we can understand from the computed sway correction factor, i.e., if it is

positive what does it indicate as well as if it is negative what does it point out? 2.10. Why sway correction is required when we analyse unsymmetrical frames? 2.11. Does the carry over factor is half for non-prismatic members? 2.12. When a symmetrical structure is subjected to symmetrical loading, it does not

sway. Does this statement applicable to Gable frames? If not, why?

Moment Distribution Method - I.K. International Publishing ... The moment distribution method can be...

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