*Module 4 : Plastic Analysis (1)Dr Yan Zhuge

CIVE3011 Structural Analysis and Computer Applications

*IntroductionElastic analysisStatically determinateanalysed by statics alonestatically indeterminateanalysed by statics and compatibilitybased on the following assumptionsstresses in the material < elastic limitdeflections are small

*Introduction (Cont.)Plastic analysisThis analysis is based on finding the collapse load of the structure. This requires a knowledge of what happens at collapse and how the structure behaves when the stresses in the material exceed the elastic limit

*Introduction (Cont.)This Module: We will develop the principles of plastic theory which are widely used in the analysis and design of steel structures.

*Fundamental concepts of plastic behaviour

syuStrainElastic (slope = E)0.0012syPlastic (slope = 0)0.014suts 0.2Strain hardening (slope 0.04E)FailureStress-strain curve for mild steel

*Collapse of structuresAny structure can be made to collapse by applying a load of sufficient magnitudeA structure may collapse by one or a combination ofelastic bucklinginstability of one or more membersplastic collapseformation of plastic hinges

*Plastic behaviour - beamsNeutral AxisCross sectionWwElastic behaviour

*Bending momentPartially plasticNeutral AxisThe beam is loaded further

*Full plasticityBending momentCross sectionStressStrainecetsysyPlastic hingeCollapse load Wc = WwMpWcL/4 = MpWc = 4Mp/LNeutral AxisFull plasticityThe zone of plasticity will spread from the outer fibres towards the centre of the section

*Full plasticityAt this stage, the section has reached its full moment capacity as there is NO more stress capacity available to develop any additional moment of resistance within the section.What do you think will happen to the beam?The beam will collapse!!

*Direct calculation of the plastic moment rectangular sectionRectangularCross sectionStressStrainetsysybdC = T = 0.5 bd yZP= bd2/4 where ZP is the plastic section modulusZe = bd2/6 where Ze is the elastic section modulusShape factor (ZP / Ze) for a rectangular section = 1.5Neutral AxisTC

*Position of the neutral axis?Equilibrium of the cross-sectionWhat does this equation tell you?The neutral axis is the axis of equal area

*Direct calculation of the plastic moment I sectionShape factor 1.15 (about x-x axis)Shape factor 1.6 (about y-y axis)

*Asymmetric sectionsAsymmetric sections don not yield simultaneously at the top and bottom of the section. The neutral axis moves as yield spreads through the section from the centroid before yield, to the axis that bisects the cross sectional area.Can you identify the position of the plastic neutral axis?20 mm above the base.Calculate the fully plastic moment of the section? Take y = 275 N/mm2

*Asymmetric sections (Cont.)Plastic Section Modulus

*Effect of axial forcePresence of axial force reduce the plastic moment capacity. However, in low-rise structure, axial forces are too small to have significant effect on Mp.

*Effect of shear forceShear forces reduce the plastic moment capacity. However, they cause smaller reduction in Mp than axial forces and need to be considered only when they are exceptionally large.

*Collapse mechanismsThe formation of one plastic hinge will reduce the structure to a mechanismTwo plastic hinges are required to form a mechanismStatically DeterminateStatically Indeterminate

*Number of hinges for collapsesn = r + 1complete collapsewhere n is the number of hingesr is the degree of redundanciesstructure is statically determinaten < r + 1 partial collapsestructure is statically indeterminaten > r + 1over collapse

*Method of virtual work At collapse all deformations of the structure occur at the plastic hingesThe principle of virtual work can be applied to these deformationsBending moment remains constant as the structure deformsAxial load effects (axial shortening, buckling) are ignoredS w. = S MP q over all loads over all plastic hinges w set of external forcesMp internal plastic moment at each plastic hinge - virtual displacement in the direction of the force - corresponding virtual rotation

*Method of virtual work (Cont.)no collapseAdditional hingeAdditional hingeAB, BC deflect separately as rigid body, rotations take place at each plastic hinge positionwcCollapse load

*Procedure for analysisSpecify the collapse mechanismImpose the plastic rotations at the hingesFind the relations among the angles/rotationsWrite the virtual work equation. Express the equation in terms of one unknowns.Solve for collapse load

*Example 1 a fixed beamSpecify the collapse mechanismImpose the plastic rotations at the hinges= a tan q = aq, d = b tan f = bf, f = (a/b)qApply the method of virtual workSw = S MpqWc d = 2 Mp(q + f) Wc aq = 2 Mp(1 + (a/b))q

Wc = 2MpL/ab

*Example 2 - Smallest load will first cause collapse

*Example 3 UDL LoadIWD= 4Mpq EWD=2 x work done by load on AB wcqL2/4 = 4Mpq wc = 16Mp/L2Consider AB and BC separately. A doesnt deflect whilst B deflect q L/2. Hence the average deflection over AB will be q L/4.

*Example 4 UDL loadLocation of plastic hinge?

*Contd.We need to determine the value of x to minimise WcThere are two basic approachesMathematical ie. To minimise WcGraphically - Plot values of Wc against sensible values of x

*Continuous beamEach span may have a different plastic momentAt a support, the plastic hinge forms at the plastic moment of the weaker memberEach span must be checked individually and the span with the lowest collapse load determines the collapse the whole beam

*Example 5 continuous beamDetermine the value of

*Contd.Mechanism 1Wc = WwSame as example 3Sw = S MpqWc =W = 16 Mp/L2

= 16Mp/2x22=2Mp

*Contd.Mechanism 2 The lowest of the two calculated values of is 0.47Mp, corresponding to the second collapse mechanism.

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