Structural Concrete - Hassoun

602 Chapter 16 Continuous Beams and Frames
3. Case 2. Apply Eq. 16.26 to ABC, BCD,andCDE, respectively:
Solve the three equations to get
4M B + M C =− (20)2 (6) =−600 K ⋅ ft
4
M B + 4M C + M D =−600 K ⋅ ft
M C + 4M D =−600 K ⋅ ft
M B = M D =−129.6K⋅ ft
The corresponding elastic midspan moments are
Beam AB = w L L2
8
M C =−86.4K⋅ ft
+ M B
2 = 6(20)2 − 129.6 =+235.2K⋅ ft
8 2
BC = 0 − 1 (129.6 + 86.4) =−108 K ⋅ ft
2
CD = w LL 2
8
− 1 6(20)2
(129.6 + 86.4) = − 108 =+192 K ⋅ ft
2 8
DE = 0 − 1 × 129.6 =−64.8K⋅ ft
2
To reduce the positive span moment, increase the support moments by 10% and calculate the
corresponding positive span moments. The resulting positive moment must be at least 90% of the
first calculated moments given previously.
M ′ B = M′ D
= 1.1(−129.6) =−142.6K⋅ ft
= 1.1(−86.4) =−95.0K⋅ ft
M ′ C
The corresponding midspan moments are
Beam AB = w L L2
8
+ M′ B
2 = 6(20)2 − 142.6 =+228.7K⋅ ft
8 2
BC =− 1 (142.6 + 95) =−118.8K⋅ ft
2
CD = w L L2
8
+ 1 2 (M′ C + M′ 6(20)2
D
)= − 1 (95 + 142.6) =181.2K⋅ ft
8 2
DE =− 1 × 142.6 =−71.3K⋅ ft
2
4. Case 3. This case is similar to Case 2, and the moments are shown in Fig. 16.32d.
5. Case 4. Consider the spans AB, BC, andDE loaded with live load to determine the maximum
negative moment at support B:
4M B + M C =− w LL 2
=− 6(20)2 =−1200 K ⋅ ft
2 2
M B + 4M C + M D =− w L L2
4
M C + 4M D =− 6(20)2
4
=− 6(20)2
4
=−600 K ⋅ ft
=−600 K ⋅ ft