## Image Of Simply Support Beam With Central Point Load Hinge Mechanism

The load deflects zero at the supports and 5 at the centre

### Average displacement of the load 2 4

The Internal Work Done in developing the hinges is found from the product of the moment induced (i.e. Mp) and the amount of rotation (e.g. 0) for each hinge.

Internal Work Done - Moment* Rotation Tor each hinge position

The External Work Done by the applied load system is found from the product of the load and the displacement for each load.

KMcrctnl Work Done «(Load x Displacement) >

[internal Woffc IDone = Externil Warlt Done

Consider the previous Example 8.2 of propped cantilever with a single point load. The hinge positions were identified as occurring at support A, and under the point load at B. Assuming rigid links between the hinges, the collapse mechanism of the beam when the hinges develop can be drawn as shown in Figure 8.10(c).

Figure 8.10

From the deformed shape in Figure 8.10:

From the deformed shape in Figure 8.10:

] nlc i'ii a I Work Hone = Paternal Work Done

Consider the previous Example 8.3 of a propped cantilever with a uniformly distributed load. The hinge positions were identified as occurring at support A, and at a point load 0.4142L from the simple support. Assuming rigid links between the hinges, the deformed shape of the beam when the hinges develop can be drawn as shown in Figure 8.11(c).

Figure 8.11

From the deformed shape in Figure 8.11:

For small values of 9 and/p 5=0.586L0=0.414Lp :. p=1.4150

The load deflects zero at the supports and 5 at a distance 0.414L from support B.

Average displacement of the load 2 2

Internal Work Done=External Work Done

3.415Mp9=0.293wL9

Mp=0.0858wL2 (as before)

8.3.1 Example 8.4: Continuous Beam

A non-uniform, three-span beam is fixed at support A, simply supported on rollers at D, F and G and carries unfactored loads as shown in Figure 8.12. Determine the minimum Mp value required to ensure a minimum load factor equal to 1.7 for any span.

A non-uniform, three-span beam is fixed at support A, simply supported on rollers at D, F and G and carries unfactored loads as shown in Figure 8.12. Determine the minimum Mp value required to ensure a minimum load factor equal to 1.7 for any span.

Figure 8.12

There are a number of possible elementary beam mechanisms and it is necessary to ensure all possibilities have been considered. It is convenient in multi-span beams to consider each span separately and identify the collapse mechanism involving the greatest plastic moment Mp; this is the critical one and results in partial collapse.

The number of elementary independent mechanisms can be determined from evaluating (the number of possible hinge positions—the degree-of-indeterminacy).

Nli ir.hcr cif JwweiwtF-i nrleHefTiinaty: = ({it-1 ■+*■)-*

Number of possible hin^e positions - 7 (cit A, B. C, D. E. F and beL<vwn F find Oj

Number of independent mechanisms =(7-3)=4

(Note: In framed structures combinations of independent mechanisms must also be considered see Section 8.5).

FtKCOfed hndG ( 1.7 * 50) ■ 35.0 kN <1.7 x 25) - 42.5 kN (J.7 x 40}"66.0IN

Plastic analysis 631 Consider span ABCD:

In this span there are four possible hinge positions, however only three are required to induce collapse in the beam. There are two independent collapse mechanisms to consider, they are:

(i) hinges developing at A (moment=2Mp), B (moment=2Mp) and D (moment=Mp)

(ii) hinges developing at A (moment=2Mp), C (moment=2Mp) and D (moment=Mp)

Static Method:

The free bendng moment at B=119.0 kNm The free bendng moment at C=68.0 kNm.

In this span the critical value of Mp=33.06 kNm with hinges developing at A, B and D.

Kinematic Method:

In this span only three hinges are required to induce collapse in the beam. Hinges develop at D (moment=Mp), E (moment=Mp) and F (moment=Mp)

Consider span FG:

In this span only two hinges are required to induce collapse in the beam.

Hinges develop at F (moment=Mp), and between F and G (moment=1.5Mp)

Span FG is effectively a propped cantilever and consequently the position of the hinge under the uniformly distributed load must be calculated. (Note: it is different from Example 8.3 since the plastic moment at each hinge position is not the same).

Examples in structural analysis 634 Equate the Mp values to determine x:

6ftr = Mfi.Sft-<lS^6v + 4.0&jr 2.72.15 + 4S.Mar - 146.«8 =0

619 m

L'ht; reader should ton (im ilic v»iutönkp uclr^the Kniemalie Miiltod.

Span: AHCD 11FF

Min¡inum required value ofM./,"! 33-06 kNni 25-5 ^Nni füralöud f;ic!öröri,7 J"

Actual loud factor if an ,\fr 1 {1.7 k JC.6J V33jM (1.7 « 4S-64W53

46.64 kVm

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