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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.

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Presentation on theme: "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."— Presentation transcript:

1 8/1 STRESS AND STRAIN IN BEAMS

2 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

3 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

4 8/4 PURE BENDING PP Pure bending region x z y Unit (small) Length P -P M V P.a a Non –uniform bending

5 8/5 PURE BENDING Unit (small) length Straight lines M M

6 8/6 RADIUS OF CURVATURE  Straight lines remain straight Planes remain plane 

7 8/7 CURVATURE  = 1/   dd ds dx x y

8  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

9 8/9 LONGITUDINAL STRAIN 1 y Strain xx

10 8/10 LONGITUDINAL STRESS y StrainStress   Stress v Strain

11 8/11 NORMAL STRESS AND STRAIN 1 y StrainStress  x = -y/  = - .y  x = E.  x  x = -E.y/  = -E. .y xx

12 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

13 8/13 NEUTRAL AXIS PASSES THROUGH CENTROID First moment of area xx M y x z dA y c1c1 c2c2 Compression Tension

14 8/14 MOMENT-CURVATURE EQN z y c1c1 c2c2 dA = moment of inertia wrt neutral axis

15 8/15 FLEXURE FORMULA FOR BENDING STRESSES

16 8/16 MAXIMUM STRESSES OCCUR AT THE TOP & BOTTOM FACES

17 8/17 MAXIMUM STRESSES S 1 = I/c 1 Section modulus 11 M y x 22 C1C1 C2C2

18 8/18 I & S FOR BEAM 0 h b z y

19 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

20 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

21 8/21 STRESSES CAUSED BY BENDING q=20kN/m P=50kN RARA RBRB 2.5m 3.5m h=0.7m b=0.22m

22 8/22 V & M DIAGRAMS V=89.2kN 39.2kN -10.8kN -80.8kN M=160.4kNm V M

23 8/23  MAX FOR BEAM 0 h b z y MAXIMUM STRESSES


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