1. MOMENT DISTRIBUTION BY Hardy Cross, 1930
DISPLACEMENT MEHTOD OF ANALYSIS

2. Sign Convention:
Clockwise moment: Positive
Counterclockwise moment: Negative
Fixed-End Moments (FEMS):
MAB
+MBA

3. Far End Fixed
Member Stiffness Factor:

4. Far End Hinged
 A M M’

5. DISTRIBUTION FACTOR (DF)M M1 M2 M3 K1 K2 K3

6. Determine the moment at each end
10 kN
2 kN/m
1.5 kN/m
3.75 m m m m
EI = constant

7. 10 kN
2 kN/m
1.5 kN/m
3.75 m m m m

8. 10 kN
2 kN/m
1.5 kN/m
3.75 m m m m

9. 10 kN
2 kN/m
1.5 kN/m
3.75 m m m m

10. 10 kN
2 kN/m
1.5 kN/m
3.75 m m m m

11. 10 kN 2 kN/m 1.5 kN/m AB 3.75 m 3.75 m 5 m 6.25 m SPAN BC CD L 7.5 m
1/7.5
1/5
1/6.25
JOINT A B C D DF
0.4
0.6
0.556
0.444 1
FEM
-9.375
9.375
-4.167
4.167
-4.883
4.883
BAL CO
5.208
-0.716
-2.083
-3.125
0.398
0.318
-4.883
-1.042
0.199
-1.562
-2.442
0.159
0.199
-4.004
-0.080
-0.119
2.226
1.778
-0.159
-0.040
1.113
-0.06
-0.080
0.889

12. 10 kN 2 kN/m 1.5 kN/m 3.75 m 3.75 m 5 m 6.25 m JOINT A B C D DF 0.4
0.4
0.6
0.556
0.444 1 CO
-0.040
1.113
-0.06
-0.08
0.889
-0.140
-0.889
BAL
-0.445
-0.668
0.078
0.062
-0.223
0.039
-0.334
0.031
-0.779
-0.016
-0.023
0.433
0.346
-0.031
-0.008
0.217
-0.012
0.173
-0.028
-0.087
-0.130
0.016
0.012
-0.173
-0.044
0.008
-0.065
0.006

13. 10 kN 2 kN/m 1.5 kN/m 3.75 m 3.75 m 5 m 6.25 m JOINT A B C D DF 0.4
0.4
0.6
0.556
0.444 1 CO
-0.045
0.008
-0.065
-0.087
0.006
0.152
-0.006
BAL
-0.003
-0.005
0.085
0.067
-0.002
0.043
0.034
-0.017
-0.026
0.003
-0.034
SUM
6.642
-6.642
5.37
-5.391
MOM
-10.75
6.60
-6.60
5.40
-5.40

14. MODIFIED STIFFNESS METHOD
10 kN
2 kN/m
1.5 kN/m
3.75 m m m m
SPAN AB BC CD L
7.5 m
5 m
6.25m K
1/7.5
1/5=0.2
(3/4)1/6.25=0.12
JOINT A B C D DF
0.4
0.6
0.625
0.375 1
FEM
-9.375
9.375
-4.167
4.167
-4.883
4.883
BAL CO
5.208
-0.716
-2.083
-3.125
0.448
0.269
-4.883
-1.042
0.224
-1.563
-2.442
0.224
-4.005
-0.09
-0.134
2.503
1.502
-0.045
1.252
-0.067
PRESENTED BY: Engr. Carol E. Dungca

15. AB SPAN BC CD JOINT A B C D DF 0.4 0.6 0.625 0.375 1 CO -0.45 1.252
0.4
0.6
0.625
0.375 1 CO
-0.45
1.252
-0.067
BAL
-0.501
-0.751
0.042
0.025
-0.251
0.021
-0.376
-0.008
-0.013
0.235
0.141
-0.004
0.118
-0.007
-0.047
-0.071
0.004
0.003
SUM
6.646
-6.646
5.386
-5.386
FEM

16. AB SPAN BC CD L 6 m 9 m 4m K 1/6 (3/4)(1/9)=1/12 JOINT A B C D DF 2/3
10kN/m
30 kN
A m B m C m D
SPAN AB BC CD L
6 m
9 m 4m K
1/6
(3/4)(1/9)=1/12
JOINT A B C D DF
2/3
1/3 1
FEM
-67.5
67.5
-120
BAL CO
SUM
-67.5
-52.5 45
22.5
52.5
22.5
26.25
26.25
-17.5
-8.75
-8.75
13.75
27.5
-27.5
120
-120
13.75
27.5
-27.5
120
-120

17. Determine the member end moment for the three span continuous beam shown due to the uniformly distributed load and due to the support settlement of 15 mm at B, 36 mm at C, and 18 mm at D. E = 200 GOPa, I = 1705 x 106 mm4
32 kN/m
A m B m C m D

18. 32 kN/m
0.015 m
0.036 m
0.018 m
0.021 m
32 kN/m ∆ ∆

19. ∆ ∆
PRESENTED BY: Engr. Carol E. Dungca

20. ∆ ∆
PRESENTED BY: Engr. Carol E. Dungca

21. SPAN AB BC CD K ¾(1/5) = 0.15 1/5 = 0.2 ¾(1/5) JOINT A B C D DF 1
0.429
0.571
FEM(LOAD)
FEM(SETTLE)
-66.67
66.67
FEM(total)
BAL CO
70.096
71.257
8.728
11.616
87.846
58.462
5.808
58.462
5.808
-3.316
-2.492
PRESENTED BY: Engr. Carol E. Dungca

22. SPAN AB BC CD K ¾(1/5) = 0.15 1/5 = 0.2 ¾(1/5) JOINT A B C D DF 1
0.429
0.571
BAL CO
-1.658
0.711
0.947
9.531
7.160
4.766
0.474
-2.045
-2.721
-0.271
-0.203
-0.136
-1.379
0.058
0.078
0.787
0.592
-3.316
-2.492
PRESENTED BY: Engr. Carol E. Dungca

23. SPAN AB BC CD K ¾(1/5) = 0.15 1/5 = 0.2 ¾(1/5) JOINT A B C D DF 1
0.429
0.571
BAL
0.058
0.078
0.787
0.592 CO
0.394
0.039
-0.169
-0.225
-0.022
-0.017
-0.011
-0.113
0.005
0.006
0.065
0.048
0.033
-0.014
-0.019
SUM/FEM
424.62
PRESENTED BY: Engr. Carol E. Dungca

24. 10 kN 5kN/m 50 kN m JOINT A B C D MEMBER AB BA BC CB CD DC K 0.188
A EI B
8m C m D
4m m
3EI EI
JOINT A B C D
MEMBER AB BA BC CB CD DC K
(3/4)(2/8)= 3/16= 0.188
0.188
0.375
0.281 DF 1
0.334
0.666
0.572
0.428
JT. COUPLE CO
FEM
-10 10
26.667
BAL 50
16.7
33.300
16.650
16.650 10
5.567
11.100
-9.524
-7.126 5
-4.762
5.550
PRESENTED BY: Engr. Carol E. Dungca

25. JOINT A B C D MEMBER AB BA BC CB CD DC DF 1 0.334 0.666 0.572 0.428 CO
-4.762
5.550
0.238
-7.784
BAL
-0.079
-0.159
4.452
3.332
2.226
-0.080
-0.743
-1.483
0.046
0.034
0.023
-0.741
-0.008
-0.015
0.424
0.317
FEM
36.437
13.563
43.444
PRESENTED BY: Engr. Carol E. Dungca

26. Determine the internal moment acting at each joint
Determine the internal moment acting at each joint. Assume A,D, and E are pinned and B and C are fixed joints. E = 29 x 103 ksi
PRESENTED BY: Engr. Carol E. Dungca

27. JOINT A B C E D AB BA BD BC CB CE EC DB K 0.125 0.047 0.056 0.078 DF 1
MEMBER AB BA BD BC CB CE EC DB K
0.125
0.047
0.056
0.078 DF 1
0.548
0.206
0.246
0.418
0.582
FEM
-12 12
-108
108
-30 30
BAL CO
-96 78 12
52.608
19.776
23.616
-30 6
-15
11.808
-3.192
5.645
2.122
2.534
1.334
1.858
0.667
1.267
0.667
1.267
-0.366
-0.137
-0.164
-0.530
-0.737
75.887
21.761
89.275
PRESENTED BY: Engr. Carol E. Dungca