66 Mechanics of Materials 2 §3.2 possible to arrange for all normal loading situations to be associated with elastic stresses in the beam,the additional strength in the partially plastic condition being used as the safety margin to take account of unexpected load increases. Figure 3.4 shows the way in which moments build up with increasing depth or penetration of yielding and associated radius of curvature as the beam bends. Typical shope factor ⑦7 1.5 肠5 写18 1.0 Stress distributions at various stoges R/RE Fig.3.4.Variation of moment of resistance of beams of various cross-section with depth of plastic penetration and associated radius of curvature. Here the moment M carried by the beam at any particular stage and its associated radius of curvature R are considered as ratios of the values at the maximum elastic or initial yield condition.It will be noticed that at large curvature ratios,i.e.high plastic penetrations, the values of M/ME approach the shape factor of the sections indicated,e.g.1.5 for the rectangular section. Shape factors of other symmetrical sections such as the I-section beam are found as follows (Fig.3.5). Stress distributions 王消 A (a)Eiastic (b)Fully plostic Fig.3.5.Plastic bending of symmetrical (I-section)beam. First determine the value of the maximum elastic moment ME by applying the simple bending theory M o 1 y
66 Mechanics of Materials 2 93.2 possible to arrange for all normal loading situations to be associated with elastic stresses in the beam, the additional strength in the partially plastic condition being used as the safety margin to take account of unexpected load increases. Figure 3.4 shows the way in which moments build up with increasing depth or penetration of yielding and associated radius of curvature as the beam bends. Typical shape factor 17 @ 1.7 1.5 '5 w 1.18 2, z 1.0 Stress distrbutions vorious stages Fig. 3.4. Variation of moment of resistance of beams of various cross-section with depth of plastic penetration and associated radius of curvature. Here the moment M carried by the beam at any particular stage and its associated radius of curvature R are considered as ratios of the values at the maximum elastic or initial yield condition. It will be noticed that at large curvature ratios, i.e. high plastic penetrations, the values of MIME approach the shape factor of the sections indicated, e.g. 1.5 for the rectangular section. Shape factors of other symmetrical sections such as the I-section beam are found as follows (Fig. 3.5). Stress distributions I1 (01 Elasi~c (b) Fully plostlc Fig. 3.5. Plastic bending of symmetrical (I-section) beam. First determine the value of the maximum elastic moment ME by applying the simple bending theory
$3.3 Strains Beyond the Elastic Limit 67 with y the maximum distance from the N.A.(the axis of symmetry passing through the centroid)to an outside fibre and o =oy,the yield stress. Then,in the fully plastic condition,the stress will be uniform across the section at oy and the section can be divided into any convenient number of rectangles of area A and centroid distance h from the neutral axis. Then Mrn=∑(a,Ah (3.5) The shape factor MFp/ME can then be determined. 3.3.Application to I-section beams When the B.M.applied to an I-section beam is just sufficient to initiate yielding in the extreme fibres,the stress distribution is as shown in Fig.3.5(a)and the value of the moment is obtained from the simple bending theory by subtraction of values for convenient rectangles. ol ie. ME- y [BD3 bd312 [1212]D If the moment is then increased to produce full plasticity across the section,i.e.a plastic hinge,the stress distribution is as shown in Fig.3.5(b)and the value of the moment is obtained by applying eqn.(3.3)to the same convenient rectangles considered above. BD2 bd27 MFP=Oy 4-4 The value of the shape factor can then be obtained as the ratio of the above equations MFP/ME.A typical value of shape factor for commercial rolled steel joists is 1.18,thus indi- cating only an 18%increase in"strength"capacity using plastic design procedures compared with the 50%of the simple rectangular section. 3.4.Partially plastic bending of unsymmetrical sections Consider the T-section beam shown in Fig.3.6.Whilst stresses remain within the elastic limit the position of the N.A.can be obtained in the usual way by taking moments of area Stress distributions 门Plostic NA (o)Elastic (b)Plastic Fig.3.6.Plastic bending of unsymmetrical (T-section)beam
93.3 Strains Beyond the Elastic Limit 67 with y the maximum distance from the N.A. (the axis of symmetry passing through the centroid) to an outside fibre and (T = oY, the yield stress. Then, in the fully plastic condition, the stress will be uniform across the section at oY and the section can be divided into any convenient number of rectangles of area A and centroid distance h from the neutral axis. Then MFP = x(cyA)h (3.5) The shape factor MFp/ME can then be determined. 33. Application to I-section beams When the B.M. applied to an I-section beam is just sufficient to initiate yielding in the extreme fibres, the stress distribution is as shown in Fig. 3.5(a) and the value of the moment is obtained from the simple bending theory by subtraction of values for convenient rectangles. (TI Y i.e. ME = - BD3 bd3 2 =uy [T - -4 5 MFP ‘Cy [T - -4 If the moment is then increased to produce full plasticity across the section, i.e. a plastic hinge, the stress distribution is as shown in Fig. 33b) and the value of the moment is obtained by applying eqn. (3.3) to the same convenient rectangles considered above. BD2 bd2 The value of the shape factor can then be obtained as the ratio of the above equations MFP/ME. A typical value of shape factor for commercial rolled steel joists is 1.18, thus indicating only an 18% increase in “strength” capacity using plastic design procedures compared with the 50% of the simple rectangular section. 3.4. Partially plastic bending of unsymmetrical sections Consider the T-section beam shown in Fig. 3.6. Whilst stresses remain within the elastic limit the position of the N.A. can be obtained in the usual way by taking moments of area Fig. 3.6. Plastic bending of unsymmetrical (T-section) beam
68 Mechanics of Materials 2 §3.4 about some convenient axis as described in Chapter 4.7A typical position of the elastic N.A.is shown in the figure.Application of the simple blending theory about the N.A.will then yield the value of ME as described in the previous paragraph. Whatever the state of the section,be it elastic,partially plastic or fully plastic,equilibrium of forces must always be maintained,i.e.at any section the tensile forces on one side of the N.A.must equal the compressive forces on the other side. ∑stress×area above N.A.=∑stress×area below N.A. In the fully plastic condition,therefore,when the stress is equal throughout the section, the above equation reduces to ∑areas above N.A.=∑areas below N.A. (3.6) and in the special case shown in Fig.3.5 the N.A.will have moved to a position coincident with the lower edge of the flange.Whilst this position is peculiar to the particular geometry chosen for this section it is true to say that for all unsymmetrical sections the N.A.will move from its normal position when the section is completely elastic as plastic penetration proceeds.In the ultimate stage when a plastic hinge has been formed the N.A.will be positioned such that eqn.(3.6)applies,or,often more conveniently, area above or below N.A.total area (3.7) In the partially plastic state,as shown in Fig.3.7,the N.A.position is again determined by applying equilibrium conditions to the forces above and below the N.A.The section is divided into convenient parts,each subjected to a force average stress x area,as indicated,then F1+F2=F3+F4 (3.8) Yielded area Fig.3.7.Partially plastic bending of unsymmetrical section beam. and this is an equation in terms of a single unknown yp,which can then be determined,as can the independent values of F1,F2,F3 and F4. The sum of the moments of these forces about the N.A.then yields the value of the partially plastic moment MPP.Example 3.2 describes the procedure in detail. EJ.Hearn,Mechanics of Materials 1,Butterworth-Heinemann,1997
68 Mechanics of Materials 2 g3.4 about some convenient axis as described in Chapter 4.t A typical position of the elastic N.A. is shown in the figure. Application of the simple blending theory about the N.A. will then yield the value of ME as described in the previous paragraph. Whatever the state of the section, be it elastic, partially plastic or fully plastic, equilibrium of forces must always be maintained, i.e. at any section the tensile forces on one side of the N.A. must equal the compressive forces on the other side. 1 stress x area above N.A. = 1 stress x area below N.A. In the fully plastic condition, therefore, when the stress is equal throughout the section, the above equation reduces to areas above N.A. = areas below N.A. (3.6) and in the special case shown in Fig. 3.5 the N.A. will have moved to a position coincident with the lower edge of the flange. Whilst this position is peculiar to the particular geometry chosen for this section it is true to say that for all unsymmetrical sections the N.A. will move from its normal position when the section is completely elastic as plastic penetration proceeds. In the ultimate stage when a plastic hinge has been formed the N.A. will be positioned such that eqn. (3.6) applies, or, often more conveniently, area above or below N.A. = total area (3.7) In the partially plastic state, as shown in Fig. 3.7, the N.A. position is again determined by applying equilibrium conditions to the forces above and below the N.A. The section is divided into convenient parts, each subjected to a force = average stress x area, as indicated, then -- Yielded ----- =* Fig. 3.7. Partially plastic bending of unsymmetrical section beam. and this is an equation in terms of a single unknown 7,. which can then be determined, as can the independent values of F 1, F2, F3 and F4. The sum of the moments of these forces about the N.A. then yields the value of the partially plastic moment Mpp. Example 3.2 describes the procedure in detail. E.J. Hearn, Mechanics of Materials 1, Butterworth-Heinemann, 1991
$3.5 Strains Beyond the Elastic Limit 69 3.5.Shape factor-unsymmetrical sections Whereas with symmetrical sections the position of the N.A.remains constant as the axis of symmetry through the centroid,in the case of unsymmetrical sections additional work is required to take account of the movement of the N.A.position.However,having deter- mined the position of the N.A.in the fully plastic condition using eqn.(3.6)or(3.7),the procedure outlined in $3.2 can then be followed to evaluate shape factors of unsymmetrical sections-see Example 3.2. 3.6.Deflections of partially plastic beams Deflections of partially plastic beams are normally calculated on the assumption that the yielded areas,having yielded,offer no resistance to bending.Deflections are calculated therefore on the basis of the elastic core only,i.e.by application of simple bending theory and/or the standard deflection equations of Chapter 5f to the elastic material only.Because the second moment of area I of the central core is proportional to the fourth power of d, and I appears in the denominator of deflection formulae,deflections increase rapidly as d approaches zero,i.e.as full plasticity is approached. If an experiment is carried out to measure the deflection of beams as loading,and hence B.M.,is increased,the deflection graph for simply supported end conditions will appear as shown in Fig.3.8.Whilst the beam is elastic the graph remains linear.The initiation of yielding in the outer fibres of the beam is indicated by a slight change in slope,and when plastic penetration approaches the centre of the section deflections increase rapidly for very small increases in load.For rectangular sections the ratio MFP/ME will be 1.5 as determined theoretically above. Theoretical collapse lood Theoretical initial yield/ _ading condition Deflection Fig.3.8.Typical load-deflection curve for plastic bending. 3.7.Length of yielded area in beams Consider a simply supported beam of rectangular section carrying a central concentrated load W.The B.M.diagram will be as shown in Fig.3.9 with a maximum value of WL/4 at EJ.Hearn,Mechanics of Materials 1.Butterworth-Heinemann,1997
$3.5 Strains Beyond the Elastic Limit 69 35. Shape factor - unsymmetrical sections Whereas with symmetrical sections the position of the N.A. remains constant as the axis of symmetry through the centroid, in the case of unsymmetrical sections additional work is required to take account of the movement of the N.A. position. However, having determined the position of the N.A. in the fully plastic condition using eqn. (3.6) or (3.7), the procedure outlined in $3.2 can then be followed to evaluate shape factors of unsymmetrical sections - see Example 3.2. 3.6. Deflections of partially plastic beams Deflections of partially plastic beams are normally calculated on the assumption that the yielded areas, having yielded, offer no resistance to bending. Deflections are calculated therefore on the basis of the elastic core only, i.e. by application of simple bending theory t and/or the standard deflection equations of Chapter 5 to the elastic material only. Because the second moment of area I of the central cors is proportional to the fourth power of d, and I appears in the denominator of deflection formulae, deflections increase rapidly as d approaches zero, i.e. as full plasticity is approached. If an experiment is carried out to measure the deflection of beams as loading, and hence B.M., is increased, the deflection graph for simply supported end conditions will appear as shown in Fig. 3.8. Whilst the beam is elastic the graph remains linear. The initiation of yielding in the outer fibres of the beam is indicated by a slight change in slope, and whtn plastic penetration approaches the centre of the section deflections increase rapidly for very small increases in load. For rectangular sections the ratio MFP/ME will be 1.5 as determined theoretically above. P J x Theoretical collapse load Theoret lcal w -2oding condition L Deflection / Fig. 3.8. Typical load-deflection curve for plastic bending 3.7. Length of yielded area in beams Consider a simply supported beam of rectangular section carrying a central concentrated load W. The B.M. diagram will be as shown in Fig. 3.9 with a maximum value of WL/4 at E.J. Hearn, Mechanics of Materials I, Butterworth-Heinemann, 1997
70 Mechanics of Materials 2 $3.7 W/2 一L/2一儿/2 w/2 B.M.Diagram WL/4 Fig.3.9. the centre.If loading is increased,yielding will commence therefore at the central section when (WL/4)=(BD2/6)oy and will gradually penetrate from the outside fibres towards the N.A.As this proceeds with further increase in loads,the B.M.at points away from the centre will also increase,and in some other positions near the centre it will also reach the value required to produce the initial yielding,namely BD2/6.Thus,when full plasticity is achieved at the central section with a load Wp,there will be some other positions on either side of the centre,distance x from the supports,where yielding has just commenced at the outer fibres;between these two positions the beam will be in some elastic-plastic state.Now at distance x from the supports: 2 2WpL B.M.=w,2=MFP=54 L x=3 The central third of the beam span will be affected therefore by plastic yielding to some depth.At any general section within this part of the beam distance x'from the supports the B.M.will be given by xBa (3D2-d2] BM.=W2=12 (1) BD WpL Now since 0,=w4 0y= BD2 Therefore substituting in (1), x' w,2= 23D2-4 pL BD2 (3D-)L 6D2 L d21 术= 2 3D2 This is the equation of a parabola with x'=L/2 when d =0 (i.e.fully plastic section)
70 Mechanics of Materials 2 03.7 w/2 F-% L/2 & L/2 w/2 Fig. 3.9. the centre. If loading is increased, yielding will commence therefore at the central section when (WL/4) = (BD2/6)o, and will gradually penetrate from the outside fibres towards the N.A. As this proceeds with further increase in loads, the B.M. at points away from the centre will also increase, and in some other positions near the centre it will also reach the value required to produce the initial yielding, namely BD20,/6. Thus, when full plasticity is achieved at the central section with a load W,, there will be some other positions on either side of the centre, distance x from the supports, where yielding has just commenced at the outer fibres; between these two positions the beam will be in some elastic-plastic state. Now at distance x from the supports: x2 2 W,L B.M. = W,- = -MFp = -- 23 34 L 3 .. x=- The central third of the beam span will be affected therefore by plastic yielding to some depth. At any general section within this part of the beam distance x’ from the supports the B .M. will be given by Now since XI Bo B.M. = W,- = ‘[3D2 - d2] 2 12 L WPL --(Tv=wp- -(T -- BD~ 4 4 ’- BD2 Therefore substituting in (l), W,,- x’ B WPL = -[3D2 - d2]- 2 12 BD2 L I (3D2 -d2) 6D2 x= 2 This is the equation of a parabola with XI = L/2 when d = 0 (i.e. fully plastic section)