Effect of Axial Load on Plastic Moment

Plastic Analysis of BeamsFailure (Collapse) Mechanisms

the failure mechanism describes the state in which a beam acts as a mechanism at the point of failure in this state we have sections of the beam rotating with respect to each other about plastic hinges a statically determinate beam will fail completely when a single plastic hinge is formed but indeterminate beams may have to develop several before collapse occurs as shown below e.g. consider the indeterminate beam below

the elastic bending moment diagram

the moment at the support is larger so it reaches a value equal to the plastic moment Mp first and a plastic hinge is formed as load is further increased the beam now acts as if it is simply supported until plastic moment reached at B and another hinge is formed beam can take NO MORE load

the corresponding bending moment diagram is

moment distribution occurs in indeterminate beams i.e. when support A reaches Mp then moment stays constant (at plastic hinge) and bending increases at B instead

Remember: Yield condition bending moment at any point must not exceed plastic moment at that point

moment redistribution:- causes change in shape of bending moment diagram (elastic vs plastic)- increases ultimate strength of structure (as one section fails then other sections have to carry extra load)

ultimate load can be found using statics:

Considering BC

and, therefore

plastic analysis of indeterminate simpler than elastic analysis since we dont have to use elastic bending moment diagram at all we simply find positions of plastic hinges and calculate using statics

NB: -ultimate load not affected by imperfections and sinking supports - cannot use superposition- loads assumed to be applied simultaneously and ratio between them remains constant

Virtual Work Method

considering the propped cantilever again

at collapse the beam is in equilibrium with plastic hinge at A and B lets say that AB rotates by a small amount ; BC also rotates by the same amount

vertical displacement at B is L/2 (small angles) angle at B is 2 reactions at A and C do not displace vertically the only force doing work is WU internal forces doing work are MP at A and B (resist rotation)

and as before

since plastic hinges form at regions of peak bending moment it follows that for beams with a series of point loads, hinges will form under the loads

e.g. calculate the minimum value of w required to cause collapse

largest moments occur at A and some point C between A and B rotation of AB and BC are and respectively vertical deflection of C is

total load on AC is wx acting at x/2 and will be displaced a distance d/2 total load on CB is w(L-x) and its centroid also displaced d/2 using virtual work

NB: B is a hinge- rotates- no plastic hinge formation combining equations above

finally

equation yields more than one value of we need to find minimum value of w

which reduces to

solving this quadratic and ignoring the negative value of x gives

substituting value of x back into expression for w gives

now use lower bound theorem to check we have critical mechanism taking moments about Awhich gives

taking moments about B

vertical equilibrium is satisfied

now considering RHS and taking moments about C

substituting for RB and w gives

same result from considering LHS load satisfies vertical and moment equilibrium bending moment at any distance x1 from B

and

from which we get

substituting for RB, x1 and w into

givestherefore yield criterion is met

so the yield mechanism assumed is correct.