# RC concrete member behavior

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16-Apr-2015Category

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MEMBER BEHAVIORDuctility, Plastic Hinge, Redistribution

DUCTILITYDuctility is an essential property of structures responding inelastically during severe earthquakes Definition: Ductility defines the ability of a structure and selected structural components to deform beyond the elastic limits without excessive strength and stiffness degradation.

GENERAL DUCTILITY

m = y

Strain Ductility Fundamental source of ductility. Related with the ability of the constituent materials to sustain plastic strains without significant reduction in stress.

= y : imposed total strain y: strain at which plastic action starts

Strain Ductility

0

20

Strain Ductility

>> 20

Strain Ductility It should be noted that significant structural ductility can be achieved only if inelastic strains can be developed over a reasonable length of that member. If inelastic action is limited to a very small length (local ductility) very large strain ductility demands may arise.

Local Ductility DemandPEach has area A and length l

P

Pu Py

2 ~ 1 1 = y u

Local Ductility DemandNote that 1 = y 2 (a simplifying assumption) When the applied load reaches Py, the total deflection at the onset of yielding is: y = n1 + y When the load is increased to Pu, the corresponding deflection is: u = n2 + u Now introducing 1 = y 2 into the equation we have:

Local Ductility Demand( D )Chain u = y n 2 + u = n 1 + y Chain n y + u = (n + 1) y Links Links

Dividing the nominator by y, and simplifying the resulting equation ; we have

( D )Chain

n + ( D )Link = or, (n + 1)

( D )Link = (n + 1) ( D )Chain n

Local Ductility DemandNow, if n=8 and (D)Chain= 2 then required ductility demand on the ductile link will be:

( D )Link = (8 + 1) 2 8 = 10For the same assembly, (D)Chain= 3 is obtained when;

( D )Link = (8 + 1) 3 8 = 19!

Local Ductility DemandIn the construction of the chain, however, if we use n brittle links along with m ductile links, than:

( D )Chain

u n y + m u = = y ( n + m) y Chain

Links

( D )Chain( D )Link

n + m( D )Link = ( n + m)

or,

(n + m) ( D )Chain n = m

Local Ductility DemandNow, getting back to the same example; take n=6 and m=3 (so that the total length of the chain does not change) for (D)Chain= 3, the ductility demand on the ductile links will be:

( D )Link = (3 9 6) / 3 = 7

Remarks: The 1st rehearsal illustrates very large local ductilities are required to provide rather small increases in the overall system ductility. The 2nd rehearsal shows the importance of the length over which the inelastic action develops. As the length over which inelastic action develops decreases, the required local ductility demand increases very rapidly, which further requires materials having very large strain ductility.

Curvature DuctilityThe most common and desirable sources of inelastic structural deformations are rotations in the critical regions. Therefore, it is useful to relate section rotations per unit length to causative bending moments. This can easily be achieved by performing section analysis to obtain moment-curvature relationship of R/C sections.

M

Curvature DuctilityMmax

Idealized (Plastic hinge behavior)0.85Mmax

My MyShifting of neutral axis leads to moment increase

Observed

m = yy y m u

In general for beams My=0.75My and corresponding curvature is y=1.33y For heavily reinforced beams and for column sections a section analysis have to be performed.

PLASTIC HINGEIdealized M- response is actually a typical PLASTIC HINGE response. This definition implies that a plastic hinge is a region where concentrated rotations occur.

PLASTIC HINGE

Displacement DuctilityThe most convenient quantity to evaluate the ductility is displacement. Hence, for the figure below we have:

= y

Where =y+p. y: the yield and p:fully plastic components of the total lateral tip deflection.

Displacement DuctilityFor frames the total deflection used is commonly that at the level of roof.

Of particular interest in design is the ductility associated with the maximum anticipated displacement =m. Equally, if not more important are displacement ductility factors that relates interstory deflections to each other.

Displacement Ductility

While displacement ductilities, in terms of the roof deflection , of the two frames shown above are the same, dramatically different results are obtained when the displacements relevant to the first story are compared. This figure illustrates that displacement ductility capacity of such frames, , will largely be governed by the ability of plastic hinges to be sufficiently ductile (as measured by individual member ductility).

Relationship b/w and

Given the curvature distributions at the yield and at the maximum response, the required ductility can be obtained by integration

m m ( x)dx K1m = = = = K y y ( x)dx K 2 y

Relationship b/w and Yield Displacement: Actual curvature diagram is nolinear. However, by adopting a linear moment diagram as shown, e, the yield displacement can be calculated as (by applying moment-area theorem):

y =

y L23

(b) Maximum Displacement: The curvature distribution at the maximum displacement corresponds to a maximum curvature at the base of the cantilever.

Relationship b/w and (b) Maximum Displacement (cont) The extent over which plasticity spreads marks the length of the plastic hinge developing at the column end. For convenience in calculation, an equivalent plastic hinge length, lp, is defined and over this length a plastic curvature p=m-e is assumed (to be equal to m-y) lp is chosen such that the plastic displacement at the top of the cantilever, p, predicted by the simplified approach is the same as that derived from the actual curvature distribution.

Relationship b/w and Thus, the plastic rotations occurring within the plastic hinge is:

p=plp = (m-y) lpAssuming that p is concentrated at the mid-height of the plastic hinge, we have:

p=p (L-0.5lp)= (m-y)lp(L-0.5lp)

Relationship b/w and Thus the ductility factor is:p y + p = = 1+ = y y y which may be written as

= 1 +

p y

( = 1+

m

y )l p (L 0.5l p )

y L2

3

Relationship b/w and Therefore:

lp lp = 1 + 3( 1) L 0.5 L L or conversely

= 1 +

( 1)

lp lp 3 1 0.5 L L

Example 1 Consider that the cantilever has a plastic hinge length, approximately equal to 1/10 of its total length, for a displacement ductility factor of 6 what is the required curvature ductility? = 1 +

( 1)lp lp 3 L 0.5 L L

== 1 +

(6 1) 1 1 3 1 0.5 10 10

= 17.5!

As will usually be the case, the curvature ductility factor is much larger than the displacement ductility factor.

PLASTIC HINGE LENGTH Theoretical values for the equivalent plastic hinge length lp, based on integration of the curvature distribution for typical members, would make lp proportional to L. Such values however do not match well with the experimentally measured lengths.Reasons: 1) Theoretical M- diagram ends abruptly at the base of the cantilever, while steel tensile stains continue, due to anchorage slip, for some depth into the footing. The elongation of bars beyond the theoretical base leads to additional rotations and tip deflection. This phenomenon is referred to as tensile strain penetration or bond slip.

PLASTIC HINGE LENGTHReasons: 2) The spread of plasticity resulting from inclined flexure-shear cracking: Remember that inclined cracks result in steel stresses, therefore strains, some distance above the base (usually proportional to d) being higher than predicted by the bending moment at that level.

PLASTIC HINGE LENGTHGood estimates of the plastic hinge length are defined by: Paulay: lp = 0.08L + 0.022dbfyk 0.044dbfyk Here db is the bar diameter and fyk is the characteristic yield strength of the longitudinal steel, measured in MPa.

PLASTIC HINGE LENGTHMattock and Corley: lp = 0.5d + 0.05zHere d is the effective depth of the section and z is the length of the shear span.

b f yk cu = 0.003 + 0.02 + 140 z

2

Here b is the section width is the volumetric ratio of transverse and compression steel and fyk is steel yield strength in MPa. z being the shear span is the distance between the critical section and the nearest inflection point.

Example 2Given: 400 mm x 400 mm square section cantilever column S420 820 longitudinal bars L = 3.0 m Paulay: lp = 0.08 x 3,000 + 0.0022 x 20 x 420 = 425 mm For S220, we have lp = 406 mm

Example 2Given: 400 mm x 400 mm square section cantilever column S420 820 longitudinal bars , d = 360 mm L = 3.0 m Mattock lp = 0.5d + 0.02z = 0.5 x 360 + 0.02 x 3000 = 240 mm Experimental results show scattered values of lp ranging fom d/2 to h for building columns.

Example 3 Let us retry Example 1 with revised value of the plastic hinge length. L = 4 m, h = 0.8 m, 828, fyk = 300 MPa Displacement ductility factor = 6lp = 0.08 x 4,000 + 0.0022 x 28 x 300 = 505 mm = 1 +3

( 1)lp l 1 0.5 p L L

= 1+

(6 1) 0.505 0.505 1 0.5 3 4,000 4,000

= 15.1