Effect of axial load on plastic moment analysis of framesPlastic Analysis of Structures1when a beam section subjected to plastic moment only stress distribution is as shown

presence of axial load causes additional stresses BUT still cannot exceed y

Effect of Axial Load on Plastic Moment

stress distribution in the presence of axial load

region of section under compressive stress increases while that under tensile stress decreasesA modified stress distribution (for bending) results

Section is doubly symmetric so we can assume that area under action of compressive stresses equal to area under action of tensile stressesN.A. is in original positionBy inspection we can see that the plastic moment has been reducedReduced plastic moment MP,R given byZa is the plastic section modulus for the area where the axial load actsRemember the plastic modulus is given by

Such that

force on the area where the axial load acts isRearranging the term gives

And if we substitute this back into the expression for plastic modulus we get

Reduced plastic moment MP,R becomes

Now let the mean axial stress over the entire cross-section due to P be aWe can write

Reduced plastic moment becomes

Reduced plastic modulus ZP,R can be expressed as

K constant; depends on beam section geometry

n ratio of mean axial stress to yield stress

Expression for MP,R and ZP,R valid as long as N.A. lies in web

NB design of beams subjected to both bending and axial load must also take into account local and global stability

Analysis similar to that of beams i.e. assume collapse mechanism and calculate collapse load using statics or virtual workHowever frame collapse mechanisms have 2 componentsBeam mechanismSway mechanism

Remember sway arises due to asymmetry of loading or supportsAnalysis will be illustrated by example

Plastic Analysis of Frames

E.g. Determine the collapse load W of the fame. The plastic moment for all members is 200kNm. Also calculate support reactions at collapse.Both frame and loading asymmetric therefore sway will occur.Bending moment diagram & 3 possible failure mechanisms are shown below.

Notice the hinge cancellation at B in the combined mechanism this is due to the fact that the moment due to vertical load opposes that due to horizontal load- moment at B smallest

consider member BD only under action of vertical loadsuppose BC given a rotation ; CB also rotates by (due to symmetry) angle at C is 2 remember we are assuming small angles therefore tan

consider BCtan = x/2 = so, x = 2

therefore vertical distance travelled by load W is 2

Beam mechanism

Using virtual work & equating external work done by load to internal work done by plastic moment at hinges

from which we get

Sway mechanism Now consider whole frame under action of horizontal load AB rotated by angle consider AB: tan = = x/4so, x = 4 this is the lateral sway of frame consider DE:tan = = x/2 but x = 4 = 4/2 = 2 so ED rotates through angle 2 using virtual work again

or

Combined Mechanism Now consider whole frame under action of both loads and apply virtual work equation- remember no hinge at B

and

We could get same result by beam mechanism + sway mechanism contribution from hinge at B

i.e.

contribution from hinge B is circled add the expressions and subtract circled quantities

To get critical mechanism compare the 3 answers

the lowest is for the combined mechanism

so

use statics to find reactionse.g. for reactions at E, take moments about D where internal moment is MP

and resolving horizontally

now take moments about A to get vertical reactions

resolving vertically

but what happens when the members have different plastic moments?

say member BCD now has a plastic moment 2MP while AB and DE have plastic moments MP

the 3 collapse mechanisms are shown below

vertical members are weaker so hinges form as shown

beam mechanism and

but what happens when the members have different plastic moments?

sway mechanism

so

combined mechanism

from which we get

Find the collapse load W if the collapse mechanism is as shown.

Portal frames with pitched roofs

use concept of instantaneous centres

BC rotates by Assumptions: C moves perpendicular to BC to C DE rotates about E such that D moves horizontally to D this implies that CD rotates about instantaneous centre IWhere I is intersection of BC and ED produced

IC and ID rotate through same angle

considering the triangles highlighted in red

now considering triangles highlighted in green

considering triangles highlighted in blue

example has no sway as load is symmetrical if a horizontal load is applied, other failure mechanisms possible as shown