Fixed continuous beam ABCD with AB of UDL 2KN/m of length 6m with I st...
**Continuous Beam Analysis**
To find the fixed end moments of the member BC and CB in the given continuous beam, we can use the method of consistent deformations. This method involves calculating the reactions at the supports and then applying the principle of superposition to determine the moments at the fixed ends.
**Step 1: Calculate Reactions**
To determine the reactions at the supports, we need to consider the loads acting on the beam. In this case, we have a uniformly distributed load (UDL) of 2 kN/m on AB, a point load of 5 kN at 3 m and 2 m on BC, and a point load of 8 kN at 2.5 m on CD.
1. Calculate the reaction at support A:
- The UDL on AB creates a reaction of (2 kN/m) * (6 m) = 12 kN at A.
2. Calculate the reaction at support D:
- The point load of 8 kN at 2.5 m on CD creates a reaction of 8 kN at D.
3. Calculate the reaction at support C:
- The point load of 5 kN at 3 m on BC creates a reaction of 5 kN at C.
**Step 2: Apply Principle of Superposition**
To find the fixed end moments at BC and CB, we can consider each load separately and determine the corresponding moments.
1. UDL on AB:
- The UDL creates a triangular distribution of moment along AB.
- The maximum moment occurs at the fixed end B and can be calculated using the formula:
- Mmax = (wL^2) / 12, where w is the UDL and L is the length of AB.
- Mmax = (2 kN/m) * (6 m)^2 / 12 = 6 kN-m.
- Since the beam is fixed at B, the fixed end moment at BC is equal to the maximum moment at B, which is 6 kN-m.
2. Point load at 3 m on BC:
- The point load creates a concentrated moment at the fixed end C.
- The moment can be calculated using the formula:
- Mc = P * a, where P is the point load and a is the distance from the load to the fixed end.
- Mc = 5 kN * (6 m - 3 m) = 15 kN-m.
- Since the beam is fixed at C, the fixed end moment at BC is equal to the concentrated moment at C, which is 15 kN-m.
3. Point load at 2 m on BC:
- The point load creates a concentrated moment at the fixed end C.
- The moment can be calculated using the formula:
- Mc = P * a, where P is the point load and a is the distance from the load to the fixed end.
- Mc = 5 kN * (6 m - 2 m) = 20 kN-m.
- Since the beam is fixed at C, the fixed end moment at CB is equal to the concentrated moment at C, which is 20 kN-m.
4. Point load at 2.5 m on CD:
- The point load creates a triangular distribution of moment along CD.
- The maximum moment occurs at the
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