2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
1
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-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
2
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
3
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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).
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
4
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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.
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
5
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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.
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
6
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
7
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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.
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
8
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
Figure 10. Opposite (-ve) moment distributed to member BC after Releasing joint B
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
9
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
10
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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)
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
11
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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)
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
12
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
(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
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
13
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
(a)
(b)
Figure 14. Sway reactions for two sways of the given frame
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
14
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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)
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
15
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
(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:
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
16
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
(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’’
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
17
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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’’
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
18
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
Figure 18. Final Bending Moment Calculations
Figure 19 shows the member-end and fixed-support moments calculated in figure 18.
2-D Frame with sway (Moment Distribution Method) in Excel. CE-5110 (Adv. Struct. Anal. I)
19
By Dr. M Adil M.Sc Fall 2013 (modified 06-10-2013)
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=