Structural Concrete - Hassoun

16.14 Moment Redistribution of Maximum Negative or Positive Moments in Continuous Beams 599
Example 16.8
Determine the maximum elastic moments at the supports and midspans of the continuous beam of four
equal spans shown in Fig. 16.32a. The beam has a uniform section and carries a uniform dead load
of 8 K/ft and a live load of 6 K/ft. Assume 10% maximum redistribution of moments and consider the
following two cases: (1) When the live load is placed on alternate spans, calculate the maximum positive
moments within the spans, and (2) when the live load is placed on adjacent spans, calculate the maximum
negative moments at the supports.
Solution
1. The beam has a uniform moment of inertia I and has the same E; thus, EI is constant. The
three-moment equation to analyze the beam and for a constant EI is
M A L 1 + 2M b (L 1 + L 2 )+M c L 2 =− w 1 L3 1
4
Because the spans are equal,
− w 2 L3 2
4
M A + 4M B + M C =− L2
4 (w 1 + w 2 ) (16.26)
In this example M A = M E = 0. Six different cases of loading will be considered, as shown in
Fig. 16.31:
Case 1. Dead load is placed on the whole beam ABCDE (Fig. 16.32b).
Case 2. Live load is placed on AB and CD for maximum positive moments within AB and CD
(Fig. 16.32c).
Case 3. Similar to Case 2 for beams BC and DE (Fig. 16.32d).
Case 4. Live load is placed on AB, BC,andDE for a maximum negative moment at B (Fig. 16.32e).
Case 5. Live load is placed on spans CD and DE (Fig. 16.32f).
Case 6. Live load is placed on BC and CD for a maximum negative moment at C (Fig. 16.32g).
2. Case 1. Apply Eq. 16.26 to the beam segments ABC, BCD, andCDE, respectively:
Solve the three equations to get
For a 10% reduction in moments,
4M B + M C =− (20)2 (8 + 8) =−1600 K ⋅ ft
4
M B + 4M C + M D =−1600 K ⋅ ft
M C + 4M D =−1600 K ⋅ ft
M B = M D =−342.8K⋅ ft and M C =−228.6K⋅ ft
M ′ B = M′ D
= 0.9(−342.8) =−308.5K⋅ ft
= 0.9(−228.6) =−205.7K⋅ ft
M ′ C
The corresponding midspan moments are
Span AB = DE = w D L2
8
Span BC = CD = w D L2
8
+ 1 2 M B = 8(20)2
8
− 1 × 308.5 = 245.8K⋅ ft
2
− 1 8(20)2
(308.5 + 205.7) = − 257.1 = 142.9K⋅ ft
2 8