# Example 81 Encastre Beam

An encastre beam is 8.0 m long and supports an unfactored load of 40 kN/m as shown in Figure 8.3. Assuming that the yield stress py=460 N/mm2 and a load factor X=1.7, determine the required plastic moment of resistance and plastic section modulus.

Figure 8.3 Solution:

The number of hinges required to induce collapse=(lD+1)=3 (see Figure 8.1)

The possible hinge positions are at the supports A and B and within the region of a distributed load since these are the positions where the maximum bending moments occur. Superimpose the fixed and free bending moment diagrams:

Figure 8.4

The beam has two redundancies (ignoring horizontal components of reaction) therefore a minimum of three hinges must develop to create a mechanism. Since the beam is uniform, at failure all values of the bending moment at the hinge positions must be equal to the plastic moment of resistance and cannot be exceeded anywhere:

It is evident from the above that all three conditions in Section 8.1.2 are satisfied and consequently the Mp value calculated for the required collapse load is true to achieve a load factor of 1.7

It is evident from the above that all three conditions in Section 8.1.2 are satisfied and consequently the Mp value calculated for the required collapse load is true to achieve a load factor of 1.7

8.2.2 Example 8.2: Propped Cantilever 1

A propped cantilever is 6.0 m long and supports a collapse load of 24 kN as shown in Figure 8.5. Determine the required plastic moment of resistance Mp.

A propped cantilever is 6.0 m long and supports a collapse load of 24 kN as shown in Figure 8.5. Determine the required plastic moment of resistance Mp.

Figure 8.5 Solution:

The number of hinges required to induce collapse=(lD+1)=2 (see Figure 8.1)

The possible hinge positions are at the support A and under the point load since these are the positions where the maximum bending moments occur.

The support reactions for the free bending diagram are: Va=8.0 kN and Vc=16.0 kN

The maximum free bending moment at Mfree,C,=(8.0x4.0)=32.0 kNm

The bending moment at B due to the fixed moment= -[M1x(2.0x6.0)] =-0.333Mi kNm

Figure 8.6

The beam has one redundancy (ignoring horizontal components of reaction) therefore a minimum of two hinges must develop to create a mechanism. Since the beam is uniform, at failure all values of the bending moment at the hinge positions must be equal to the plastic moment of resistance and cannot be exceeded anywhere:

A/, H and (Ml + 0.333M) = (A/p + MttfiQ = 1333MV - 320 Hit required plastic moment of resistance Mt ■ (32,0/1.333) = 24,0 kN nt

As in Example 8.1 all three conditions in Section 8.1.2 are satisfied and consequently the true value of Mp has been calculated for the given collapse load.