Instantaneous Centre of Rotation

In more complex frames it is convenient to use the 'instantaneous centre of rotation method' when developing a collapse mechanism. The technique is explained below in relation to a simple rectangular portal frame and subsequently in Example 8.7.

Consider the asymmetric rectangular frame shown in Figure 8.25 in which there are two independent mechanisms, one beam and one sway. The frame requires three hinges to cause collapse. Both mechanisms can combine to produce a collapse mechanism with hinges developing at A, C and D. In this mechanism there are three rigid-links, AB'C', C'D' and D'E as shown.

|40kN 1-IOfcN

|40kN 1-IOfcN

Figure 8.25

The centre-of-rotation for link AB'C' is at A and the remote end C moves in a direction perpendicular to line AC shown. The centre-of-rotation for link D'E is at E and the remote end D moves in a direction perpendicular to line ED shown.

In the case of link C'D', the centre-of-rotation must be determined by considering the direction of movement of each end. C' moves in a direction perpendicular to AC and consequently the centre-of-rotation must line on an extension of this line. Similarly, it must also lie on a line perpendicular to the movement of D, i.e. on an extension of ED. This construction is shown in Figure 8.26(a). The position of this centre-of-rotation is known as the instantaneous centre-of-rotation and occurs at the instant of collapse.

The work equations can be developed and the required Mp value determined by considering the rotation of the hinges and the displacements of the loads. Consider the geometry shown in Figure 8.26(b) and equate the displacements in terms of 0, p and a as follows:

The horizonta I di spk&emcnl U D' & - 3.0/3 =6.00 .*. p = 2,0 0

The vertical di splaccni enl CC vtniiaj - 2.0 a - 4.0 f? .*. a = 2,00

(Note: equating the horizontal displacement of point C will give the same result, i.e.

§C,horizontal=3.0a= 6.00) The rotation at the hinge at A=a=2.00 Note: no internal work is done at support E Internal Work Done=External Work Done

A4f («) + 2.0A/r {0+a) + A/p {0+ $ - (10,0 x $>) + {40,0 x Sc, ^) Mp {2.00)+2Mfr {0+ + >Vfp {&+ = (10.0 x +<40-0 x 4.00)

The reader should confirm that this is the critical value by calculating the reactions and checking that the bending moment on the frame does not exceed the appropriate Mp value for any member. (Note: In the case of member BCD this is equal to 2.0Mp=40 kNm).

Was this article helpful?

+1 0

Post a comment