8/1 STRESS AND STRAIN IN BEAMS
8/2 BENDING BEAMS LOADS ON BEAM PRODUCE STRESS RESULTANTS, V & M V & M PRODUCE NORMAL STRESSES AND STRAINS IN PURE BENDING V & M PRODUCE ADDITIONAL SHEAR STRESSES IN NON-UNIFORM BENDING
8/3 TYPES OF BENDING PURE BENDING – FLEXURE UNDER CONSTANT M. i.e. V =0 = dM/dx NON-UNIFORM BENDING – FLEXURE WHEN V NON-ZERO
8/4 PURE BENDING PP Pure bending region x z y Unit (small) Length P -P M V P.a a Non –uniform bending
8/5 PURE BENDING Unit (small) length Straight lines M M
8/6 RADIUS OF CURVATURE Straight lines remain straight Planes remain plane
8/7 CURVATURE = 1/ dd ds dx x y
d = dx L 1 = ( -y)d L 1 = dx - (y/ )dx L 1 - dx = -(y/ )dx L1L1 x = -(y/ )dx/dx=- .y NORMAL STRAIN NEUTRAL AXIS
8/9 LONGITUDINAL STRAIN 1 y Strain xx
8/10 LONGITUDINAL STRESS y StrainStress Stress v Strain
8/11 NORMAL STRESS AND STRAIN 1 y StrainStress x = -y/ = - .y x = E. x x = -E.y/ = -E. .y xx
8/12 RELATION BETWEEN & M NEED TO KNOW WHERE NEUTRAL AXIS IS. FIND USING HORIZONTAL FORCE BALANCE NEED A RELATION BETWEEN & M. FIND USING MOMENT BALANCE – gives THE MOMENT CURVATURE EQUATION & THE FLEXURE FORMULA
8/13 NEUTRAL AXIS PASSES THROUGH CENTROID First moment of area xx M y x z dA y c1c1 c2c2 Compression Tension
8/14 MOMENT-CURVATURE EQN z y c1c1 c2c2 dA = moment of inertia wrt neutral axis
8/15 FLEXURE FORMULA FOR BENDING STRESSES
8/16 MAXIMUM STRESSES OCCUR AT THE TOP & BOTTOM FACES
8/17 MAXIMUM STRESSES S 1 = I/c 1 Section modulus 11 M y x 22 C1C1 C2C2
8/18 I & S FOR BEAM 0 h b z y
8/19 BEAM DESIGN SELECT SHAPE AND SIZE SO THAT STRESS DOES NOT EXCEED ALLOW CALCULATE REQUIRED S=M MAX / ALLOW CHOOSE LOWEST CROSS SECTION WHICH SATISFIES S
8/20 IDEAL BEAM RECTANGULAR BEAM, S R =bh 2 /6=Ah/6 CYLINDRICAL BEAM, S=0.85.S R IDEAL BEAM, HALF THE AREA AT h/2 IDEAL BEAM, S=3.S R STANDARD I-BEAM, S=2.S R
8/21 STRESSES CAUSED BY BENDING q=20kN/m P=50kN RARA RBRB 2.5m 3.5m h=0.7m b=0.22m
8/22 V & M DIAGRAMS V=89.2kN 39.2kN -10.8kN -80.8kN M=160.4kNm V M
8/23 MAX FOR BEAM 0 h b z y MAXIMUM STRESSES