8/1 STRESS AND STRAIN IN BEAMS

8/2 BENDING BEAMSLOADS ON BEAM PRODUCE STRESS RESULTANTS, V & M

V & M PRODUCE NORMAL STRESSES AND STRAINS IN PURE BENDINGV & M PRODUCE ADDITIONAL SHEAR STRESSES IN NON-UNIFORM BENDING

8/3 TYPES OF BENDINGPURE BENDING FLEXURE UNDER CONSTANT M. i.e. V =0 = dM/dx

NON-UNIFORM BENDING FLEXURE WHEN V NON-ZERO

8/4 PURE BENDING

P-PMVP.aaNon uniform bending

8/5 PURE BENDINGMM

8/6 RADIUS OF CURVATURE

8/7 CURVATURE = 1/ddsdxxy

d= dxL1= (-y)dL1= dx - (y/)dxL1- dx = -(y/)dxL1x= -(y/)dx/dx=-.yNORMAL STRAINNEUTRAL AXIS

8/9 LONGITUDINAL STRAIN

1y

Strain

x

8/10 LONGITUDINAL STRESSyStrainStress

8/11 NORMAL STRESS AND STRAIN

1StrainStressx = -y/ = -.yx = E.xx = -E.y/ = -E..yx

8/12 RELATION BETWEEN & MNEED TO KNOW WHERE NEUTRAL AXIS IS. FIND USING HORIZONTAL FORCE BALANCENEED A RELATION BETWEEN & M. FIND USING MOMENT BALANCE gives THE MOMENT CURVATURE EQUATION & THE FLEXURE FORMULA

8/13 NEUTRAL AXIS PASSES THROUGH CENTROIDFirst moment of areaxMyxzdAyc1c2CompressionTension

8/14 MOMENT-CURVATURE EQNzyc1c2dA= moment of inertiawrt neutral axis

8/15 FLEXURE FORMULA FOR BENDING STRESSES

8/16 MAXIMUM STRESSESOCCUR AT THE TOP & BOTTOM FACES

8/17 MAXIMUM STRESSESS1 = I/c1Section modulus1Myx2C1C2

8/18 I & S FOR BEAM0hbzy

8/19 BEAM DESIGNSELECT SHAPE AND SIZE SO THAT STRESS DOES NOT EXCEED ALLOWCALCULATE REQUIRED S=MMAX/ALLOWCHOOSE LOWEST CROSS SECTION WHICH SATISFIES S

8/20 IDEAL BEAMRECTANGULAR BEAM, SR=bh2/6=Ah/6CYLINDRICAL BEAM, S=0.85.SR

IDEAL BEAM, HALF THE AREA AT h/2IDEAL BEAM, S=3.SRSTANDARD I-BEAM, S=2.SR

8/21 STRESSES CAUSED BY BENDINGq=20kN/mP=50kNRARB2.5m3.5mh=0.7mb=0.22m

8/22 V & M DIAGRAMSV=89.2kN39.2kN-10.8kN-80.8kNM=160.4kNmVM

8/23 MAX FOR BEAM0hbzy

MAXIMUM STRESSES