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2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)

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By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)

Moment distribution method for 2-D frames using MS-Excel. Let’s solve the solved problem 11.9 from Hibbeler’s book using Excel worksheet.

Following is the schematic of the problem: All members have the same EI

Figure 1. Given Frame with its supports and loadings

DATA:

As the frame has sidesway motion at TWO stories, therefore, the moment distribution analysis will

be applied three times i.e., as once for frame 2(b) then 2(c) and then 2(d) in figure below. Then the

three analysis results will be added together.

Step 1:

Name the Joints in any order you like e.g.as shown in Figure 2(e). Now first the frame will be

assumed to be restrained and solved as shown in Figure 2(b).

Figure 2(e). Named joints of given frame

The correction coefficients C’ and C’’ are then calculated by solving following equation

simultaneously, . These correction factors are then multiplied by the internal

A

B

C D

E

F

20kN

40kN

80kN

5m

5m

7m


2


joint moments found from the moment distribution in Figure 2(c) and 2(d). The resultant moments

are then found by adding these corrected moments to those obtained for the frame in Figure 2(b).

Step 2:

In Excel, write the Joint names in a row covering as many columns as the number of members

connected to that joint e.g., joint E connects three members, as shown in the figure 2. The resulting

excel sheet will look like shown in figure 3.

Figure 3. Joints named in Excel


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Now Member names, depending on the joints they join, are fed in the next row, as shown in Figure

4. Note that the names of the members under a particular joint (e.g., E) starts with the letter

representing the joint (i.e., ED,EB,EF). Also, the members with the same names (e.g., BC and CB) are

written close together for easy crossover, but its not nessessery.

Figure 4. Member names in Excel

Step 3.

Write two more rows below the member names, one for each member’s rigidity-modulus (EI) and

the other for the length. These two rows for the given frame are shown in Figure 5. As the EI of all

the members were same, any arbitrary positive number can be fed in Excel for EI (1000 in this case).


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Figure 5. Member (EI) and Lenghts in Excel

Step 4.

Now Stiffness of all members (K=4EI/L) and the Distribution factors (DF=K/K) are written for all

members as shown in figure 6 and 7, respectively. For Hinge and Fixed supports the DF is directly fed

as 1 and 0, respectively, as hinge can distribute the moment fully (i.e., 100%) and fixed can’t

distribute any moment (i.e., 0%) For instance, in figure 7, for fixed supports A and F, the DF is fed as

0.


5


Figure 6. Calculation of K for member AB

Step 5.

Fixed End Moments (FEM) for all members, due to applied loads are written in the next row. In our

case, we have a member force of 20kN only on one member BE as shown in Figure 8, rest of the

forces are applied at the joints and not on the member, therefore, the FEM will be as shown in

Figure 9, taking Clockwise as positive.


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Figure 7. Calculation of DF for moment in member EF from end E to F

Figure 8. Fixed Ended moments on member BE

B 20kN E

MBE=PL/8=20x7/8=-17.5kNm MEB=-MBE=17.5kNm


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Figure 9. FEM’s in excel

Step 6.

Release all joints. This means that moments on all joints are distributed to other joints connected to

them through members, according to their DF’s. For example, at joint B, the total moment

FEMBA+FEMBE+FEMBC is distributed to the member BA, BE and BC according to their DF’s and of

course with opposite sign.


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Figure 10. Opposite (-ve) moment distributed to member BC after Releasing joint B


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Step 7.

Carry over the distributed moment to the other end of each member. As the prismatic members

carry only half of the distributed moment, the half of the released moment is formulated in the next

row from the relative member. For instance, half carried moment MEB is shown in figure 11.

Figure 11. Half carried moment from end B to end E of member BE


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Step 8.

Now copy the “Release” and “Carry O.” rows and keep pasting them bellow, consecutively, until the

last row of “Carry O.” shows numbers approaching to zero, as shown in figure 12.

Figure 12. Copied rows of “Release” and “Carry O.” (Row 9 and 10)


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Step 9.

Now finally, find the sum of all the moments in each column starting from the FEM as shown in

Figure 13(a). Figure 13(b) shows the moments on the frame

(a)


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(b)

Figure 13. End moments on the frame restrained against sway.

The horizontal reactions for 5m tall stories of the frames, shown in Figure 13(b), can be calculated as

follows:

kNMMMMMMMM

R DEEDFEEFCBBCABBA 80805

)(

5

)(

5

)(

5

)(1

kNMMMM

R DEEDCBBC 40405

)(

5

)(2

Step 10.

Calculating the sway reactions as in Figure 2(c) and 2(d), we get frames as Figure 14(a) and 14(b) ,

respectively.

A

B

C D

E

F 4.03

8.05

14.6

1.06

6.57

1.06

1.06

1.06

6.57

14.6

4.03

8.05

R2

R1


13


(a)

(b)

Figure 14. Sway reactions for two sways of the given frame


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As the storey heights are same we have assumed the end moments to be same (i.e., 100kNm)

otherwise these moments should be calculated using assumed values of in the following formula

for moment due to deflection: .

Now, create two more copies of the excel sheet for solving the two sways, as in Figure 15(a) and

15(b).

(a)


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(b)

Figure 15. PartC and PartD Sheets showing Mom-Dist Analysis of frame in figure 14(a) and (b),

respectively.

Now reactions for the above analysis can be calculated as:


16


(a) (b)

Figure 16. End moments on the frame restrained against sway at (a) D and (b) E.

The horizontal reactions for 5m tall stories of the frames, shown in Figure 16(a & b), can be

calculated as follows:

kNMMMMMMMM

R DEEDFEEFCBBCABBA 1305

)(

5

)(

5

)(

5

)(1

kNMMMM

R DEEDCBBC 8.545

)(

5

)(2

kNMMMMMMMM

R DEEDFEEFCBBCABBA 8.545

)(

5

)(

5

)(

5

)(1

kNMMMM

R DEEDCBBC 4.395

)(

5

)(2

Now solving the equations

using matrices algebra

1

2

1

11

22

''

'

R

R

RR

RR

C

C

53.4

53.2

80

40

8.54130

4.398.54

''

'1

C

C

Figure 17 shows the above calculations in Excel:

R2’’

A

B

C D

E

F 95.9

91.8

8.78

53.9

83.1

53.9

53.9

53.9

83.1

8.78

95.9

91.8

R1’

A

B

C D

E

F 12.8

25.7

27.5

45.1

53.3

45.1

45.1

45.1

53.3

27.5

12.8

25.7

R2’

R1’’


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Figure 17. Finding Correction Coefficients C’ and C’’

Now total moment can be calculated by applying the corrections as follows and shown in figure 18.

Mi=M(part b) + M(part c)x C’ + M(part d)x C’’


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Figure 18. Final Bending Moment Calculations

Figure 19 shows the member-end and fixed-support moments calculated in figure 18.


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Now, finally, the reactions can be calculated and shear force and bending moment diagrams can be

drawn as follows.

A

B

C D

E

F 180

107

132

66.9

24.9

67.1

69.1

69.3

38.1

161

188

123

HF

VF

HA

VA

MA= MF=

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