PLASTIC BENDING OF BEAMS

STRUCTURAL MATERIALS MAY BE IDEALIZED AS ELASTIC-PLASTIC MATERIALS AS DISCUSSED EARLIER. SUCH A MATERIAL FOLLOWS HOOKE’S LAW UP TO THE YIELD STRESS AND THEN YIELDS PLASTICALLY UNDER CONSTANT LOAD. STRUCTURAL STEELS MAY BE IDEALIZED AS ELASTIC-PLASTIC MATERIALS BECAUSE SUCH MATERIALS HAVE SHARPLY DEFINED YIELD POINTS AND UNDERGO LARGE STRAINS DURING YIELDING. IN THIS PARTICULAR CASE EFFECTS OF STRAIN-HARDENING ARE DISREGARED BECAUSE IT IS SAFE AS STRAIN HARDENING PROVIDES AN INCREASE IN THE STRENGTH. NOW CONSIDER A BEAM OF SUCH MATERIAL SUBJECTED TO PURE BENDING DUE TO APPLICATION OF COUPLE OF BENDING MOMENTS. IF THE VALUE OF APLLIED BENDING MOMENT IS SMALL, THE MAXIMUM STRESSES IN THE BEAM ARE LESS THAN YIELD STRESS, THE STRESS DISTRIBUTION WOULD BE SAME AS DISCUSSED EARLIER. UNDER SUCH CONDITION NEUTRAL AXIS PASSES THROUGH THE CENTROID OF THE CROSS SECTION AND THE SAME EQUATIONS DISCUSSED EARLIER WOULD BE USED TO CALCULATE NORMAL STRESS (My/I) AND CURVATURE (-M/EI). SUCH RESESTULTS WOULD BE VALID TILL THE STRESS FARTHEST FROM THE NEUTRAL AXIS REACHES THE YIELD STRESS. NOW THE CORRESPONDING MOMENT ACTING ON THE BEAM IS CALLED THE YIELD MOMENT “My”, AND IS GIVEN AS

My = σy I / c = σy S

NOW IF THE BENDING MOMENT IS INCREASED ABOVE THE YIELD MOMENT, THE STRAINS AT THE EXTREME POINTS OF THE X-SECTION WILL CONTINUE TO INCREASE AND MAXIMUM STRAIN WOULD EXCEED THE YIELD STRAIN. HOWEVER, BECAUSE OF PERFECTLY PLASTIC YIELDING, THE MAXIMUM STRESSES WOULD REMAIN CONSTANT AND EQUAL TO YIELD STRESS. THE STRESS DISTRIBUTION BE AS SHOWN IN FIG. C. THE OUTER REGION OF THE BEAM WILL BECOME PLASTIC WHILE A CENTRAL CORE STILL REMAINS ELASTIC. AS THE BENDING MOMENT IS FURTHER INCREASED THE PLASTIC REGION EXTENDS FARTHER INWARD TOWADS THE NEUTRAL AXIS TILL A COMPLETE PLASTIC REGION IS REACHED AS SHOWN IN FIG. D.

THE STRAINS IN THE EXTREME FIBRES AT THIS STAGE ARE ABOUT 10 15 TIMES GREATER THAN THE YIELD STRAIN AND ELASTIC CORE IS ALMOST DISAPPEARED. NOW TILL THIS STAGE THE BEAM WOULD HAVE REACHED ITS ULTIMATE MOMENT-RESISTING CAPACITY. IN THIS CONDITION THE STRESS DISTRIBUTION CAN BE IDEALIZED AS CONSISTING OF TWO RECTANGULAR PARTS. THE BENDING MOMENT CORRESPONDING TO THIS IDEALIZED STRESS DISTRIBUTION IS KNOWN AS THE PLASTIC MOMENT “Mp”, AND IT REPRESENTS THE MAXIMUM MOMENT THAT THE BEAM OF ELASTIC-PLASTIC MATERIAL CAN SUSTAIN. OF COURSE THE DETERMINATION OF PLASTIC MOMENT IS OF GREAT IMPORTANCE AS IT IS THE LIMITING OR MAXIMUM MOMENT OF THE BEAM. IN ORDER TO FIND THE RELATIONSHIP TO CALCULATE “Mp”, FIRST WE LOCATE THE NEUTRAL AXIS OF THE X-SECTION SO THAT IT CAN EASILY BE DETERMINED WHERE THE COMPRESSIVE AND TENSION REGIONS EXIST. AS CAN BE OBSERVED FROM THE FIGURE THAT ABOVE THE NEUTRAL AXIS, EVERY ELEMENT IN THE X-SECTION HAS A COMPRESSIVE STRESS EQUAL TO σy, AND BELOW THE NEUTRAL AXIS STRESS IS IN TENSION. NOW TOTAL TENSILE AND COMPRESSIVE FORCES, “T” AND “C”, BELOW AND ABOVE THE NEUTRAL AXIS OF THE X-SECTION ARE σyA1 and σyA2.

“A1” AND “A2” ARE THE X-SECTIONAL AREAS BELOW AND ABOVE THE NEUTRAL AXIS. AS PER FIRST EQUATION OF STATICS (∫ σ dA = 0),

T – C = 0 OR A1 = A2

¨AS THE TOTAL AREA OF THE X-SECTION IS EQUAL TO “A”, THEREFORE,

A = A1 + A2, AND A1 = A2 = A/2

IT MEANS THAT THE NEUTRAL AXIS WOULD DIVIDE THE X-SECTION INTO TWO SECTIONS OF EQUAL AREAS. NOW THE PLASTIC MOMENT “Mp” CAN BE FOUND BY INTEGRATING THE SECOND EQUATION OF STATICS (∫ σ y dA = M) OR BY SIMPLY TAKING THE MOMENTS ABOUT THE NEUTRAL AXIS OF THE TWO FORCES. THUS

Mp = Ty1 + Cy2

HERE “y1” AND “y2” ARE THE DISTANCES FROM THE NEUTRAL AXIS TO THE CENTROIDS “c1” AND “c2” OF THE AREAS “A1” AND “A2”.

NOW REPLACING T AND C BY “σy A/2”, FOLLOWING EQUATION IS OBTAINED:

Mp = σyA (y1 + y2) / 2

NOW THE PROCEDURE TO FIND “Mp” IS TO DIVIDE THE X-SECTIONS OF THE BEAM INTO TWO EQUAL AREAS; TO LOCATE THE CENTROID OF EACH HALF AND THEN USE THE ABOVE EQUATION. THIS EXPRESSION CAN BE REWRITTEN AS

Mp = σy Z

WHERE

Z = A (y1 + y2) / 2

AND “Z” IS CALLED THE PLASTIC MODULUS OF THE X-SECTION. THE PLASTIC MODULUS MAY BE INTERPRETED GEOMETRICALLY AS THE FIRST MOMENT OF THE AREA OF THE X-SECTION ABOVE THE NEUTRAL AXIS PLUS THE FIRST MOEMENT OF AREA BELOW THE NEUTRAL AXIS. THE RATIO OF THE PLASTIC MOMENT FOR A BEAM TO ITS YIELD MOMENT IS SOLEY A FUNCTION OF THE SHAPE OF THE X-SECTION, AND IS CALLED THE SHAPE FACTOR “f”:

f = Mp / My = Z / S

NOW FOR A RECTANULAR X-SECTION WITH “b” AND “h” THE PLASTIC MODULUS IS GIVEN AS

Z = bh / 2 (h/4 + h/4) = bh²/4

AS THE SECTION MODULUS WITH RESPECT TO YIELD IS EQUAL TO S = bh²/6, THE SHAPE FACTOR FOR A RECTANGULAR BEAM IS GIVEN AS

f = 3/2

IT MEANS THAT PLASTIC MOMENT FOR A RECTANGULAR BEAM IS 50% GREATER THAN THE YIELD MOMENT. THE SHAPE FACTOR “f” FOR WIDE-FLANGE BEAMS IS TYPICALLY IN THE RANGE OF 1.1 TO 1.2 DEPENDING UPON THE PROPORTIONS OF THE CROSS SECTION.