4 Effect of Axial Load & Frames

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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 crosssection 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