Presentation on theme: "Rigid-Frame Structures"— Presentation transcript:
1Rigid-Frame Structures CE Design of Multi-Story StructuresT. B. QuimbyUAA School of EngineeringRigid-Frame Structures
2Rigid-FramesResistance to horizontal loading provided by flexural stiffness of the girders, columns, and connections.Opens up the floor space allowing freedom in space utilization.Economical for buildings up to about 25 stories.Well suited for reinforced concrete construction due to the inherent continuity in the joints.Design of floor system cannot be repetitive since the beams forces are a function of the shear at the level in addition to the normal gravity loads.Gravity loads also resisted by frame action.
3Analysis Procedure Approximate gravity load analysis and design Estimate gravity loads and use approximate analysis to determine member forces. (2 cycle moment dist.)Select beam and column sizes using gravity forces and an allowance for additional member forces due to lateral loads.Approximate lateral load analysisCantilever or Portal methodCheck drift.Resize members based on lateral load analysis and drift analysis.Detailed computer analysis and resize members as appropriate.
4Rigid Frame Behavior under Lateral Loading Accumulated story shear above a given story is resisted by shear in the columns at that story.Points of contraflexure are located an midheight of columns and beams since both types of elements are in double curvature.Deflected shape is predominately in a shear configuration with concavity being upwind.A Flexural component in the deflection results from chord action (axial forces in column).
5Moment Frame BehaviorNote the bending in the typical beam, column, and joint.
6Gravity LoadingContinuity at joints tends to create negative moments at supports and Positive moments at midspan of girders.There are two points of inflection on each girder.Each span is effected by the loading of other spans in the structure.Determination of maximum moments and shears must account for probability of uneven Live Load distribution.
7Muller-Breslau Principle The influence line for a reaction or internal stress is, to some scale, the elastic curve of the structure when deflected by an action similar to the reaction of stress.
8Loading for Maximum Mid- Span Moment Entire spans are loaded, no partial span loading.
10Approximate Analysis under Gravity Loading Code recommended values (See ACI 318 section 8.3, UBC has also adopted these)Limited to spans of approximately equal stiffness and a constant magnitude of uniformly distributed load.Two-cycle moment distribution.Multiple elastic analyses using all the potential load patterns.
112 Cycle Moment Distribution More accurate than code coefficients.May use many different types of loading.Method is quick and easy to implement.Can obtain midspan moments and column forces as well as end moments on beams.See the worked problem in the text.The Portland Cement Association publishes the original paper as a pamphlet entitled “Continuity in concrete building frames”
12Determining Loads on Bents Loads are distributed in relation to the relative rigidity of each bent.Must include torsional effects, if any.Text method allows you to compute the point load on each level of each bent. To has a translational and rotational component.
13The Portal Method Best used on shorter, wider frames. Building whose deflection is primarily racking.height to width ratio not greater than 4:1Highly indeterminate frame is reduced to a statically determinate system by the following assumptions:The points of contraflexure are located at the middle of columns and beams. (locations of zero moment)Horizontal shear at midstory levels is shared between the columns in proportion of the width of aisle each column supports.
14Example Portal Problem (“Fundamental Theory of Structures”, 2nd edition)
15Cantilever MethodUsed in structures for which the flexural component of deflection is more prominent.Up to 35 storiesHeight to width ratios up to 5:1Highly indeterminate frame is reduced to a statically determinate system by the following assumptions:The points of contraflexure are located at the middle of columns and beams. (locations of zero moment)The AXIAL STRESS in a column is in proportion to its distance from the centroid of the column areas.
16Example Cantilever Problem (“Fundamental Theory of Structures”, 2nd edition)
17Approximate Drift Analysis Need to consider the effects of:Joint rotation due to girder flexibilityColumn flexibilityOverall bendingNeed to make some changes to equations at the first level because foundation connections are considered to be either fully rigid or fully pinned.
18Controlling DriftJoint rotation due to girder flexibility is normally the predominate component of drift. Increasing girder stiffness will reduce this component.Column flexibility is the next most predominate component. Increasing column stiffness will reduce this component.A look at the contributions of each component of drift can help decide whether to stiffen the girders or the columns.The text proposes a more analytical method for making decisions.
19Effective Shear Rigidity (GA) (GA) corresponds to the shear rigidity of an analogous shear cantilever of sectional area A and Modulus of rigidity G. (See text figure 7.15)(GA) = Qh/d = Q/fFor the drift calculations, the shear stiffness of a story is given in the text's equation 7.28.If the effective shear rigidity (GA) is known for a particular level, finding the story drift is found by the text's equation 7.29
20Computer AnalysisNewer computer programs (such as ETABS) have made hand methods virtually obsolete.The stiffness based programs inherently take into account the relative stiffnesses of frames when determining bent forces.Member forces are more accurate than from the approximate methods.Deflection outputs simplify the drift analysis.
21Lumped Girder Frame Analysis Lumped Girder FramesCan be used for repetitive floorsDo not lump roof levelDo not lump lower few floorsSee text for equivalent girder and column properties.
22Single-Bay Substitute Frame Single-Bay FramesGood for estimating deflections for stability.Good for dynamic analyses where member forces are not required.Can use for a two stage analysisIge = 1*S(Ig/L)i(Ice)i = .5*S(Ic)i(Ace)i = (2/l2)*S(Acc2)i