Structural Concrete - Hassoun

16.14 Moment Redistribution of Maximum Negative or Positive Moments in Continuous Beams 603
Solve the three equations to get
For 10% reduction in moment at support B,
M ′ B
The corresponding midspan moments are
Beam AB = w L L2
8
BC = w LL 2
M C =−42.9K⋅ ft
M B =−289.3K⋅ ft
M D =−139.3K⋅ ft
= 0.9 ×(−289.3) =−260.4K⋅ ft
8
+ M B
2 = 6(20)2 − 260.4 = 169.8K⋅ ft
8 2
− 1 (260.4 + 42.9) =148.4K⋅ ft
2
CD =− 1 (42.9 + 139.3) =−91.1K⋅ ft
2
DE = 300 − 1 × 139.3 =+230.4K⋅ ft
2
6. Case 5. This is similar to Case 4, except that one end span is not loaded to produce maximum
positive moment at support B (or support D for similar loading). The bending moment diagrams
are shown in Fig. 16.32f.
7. Case 6. Consider the spans BC and CD loaded with live load to determine the maximum negative
moment at support C:
4M B + M C = w L L2
4
=−600 K ⋅ ft
M B + 4M C + M D =− w L L2
2
=−1200 K ⋅ ft
Solve the three equations to get
For 10% reduction in support moments,
M ′ C
M C + 4M D =− w L L2
4
M C =−257.2K⋅ ft
M B = M D =−85.7K⋅ ft
=−600 K ⋅ ft
= 0.9 ×(−257.2) =−231.5K⋅ ft
M ′ B = M′ C
= 0.9 ×(−85.7) =−77.2K⋅ ft
The corresponding midspan moments are
Beam AB = DE =− 77.2 =−38.6K⋅ ft
2
BC = CD = w L L2
8
− 1 (231.5 + 77.2) =6(20)2 − 154.3 = 145.7K⋅ ft
2 8