8/1 STRESS AND STRAIN IN BEAMS. 8/2 BENDING BEAMS LOADS ON BEAM PRODUCE STRESS RESULTANTS, V & M V &...

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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

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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

• 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 BENDINGP P

Pure bending region

x

zy Unit (small) Length

P-P

M

V

P.a

a

Non –uniform bending

8/5 PURE BENDING

Unit (small) length

Straight lines

MM

8/6 RADIUS OF CURVATURE

Straight linesremain straight

Planes remain plane

8/7 CURVATURE = 1/dxdsd .

dxd

1

d

ds

dxx

y

d= dxL1= (-y)dL1= dx - (y/)dxL1- dx = -(y/)dx

L1

x= -(y/)dx/dx=-.yNORMAL STRAIN

NEUTRAL AXIS

8/9 LONGITUDINAL STRAIN

1

1 max

1max

y

Strain

x

8/10 LONGITUDINAL STRESS

y

Strain Stress

Stress v Strain

8/11 NORMAL STRESS AND STRAIN

1

1 max

1max

y

Strain Stress

x = -y/ = -.y

x = E.x

x = -E.y/ = -E..y

x

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

AA

x dAyEdA 0... 0. A

dAy

First moment of area

x

M

y

x z

dA

y

c1

c2

Compression

Tension

8/14 MOMENT-CURVATURE EQN

z

y

c1

c2

dAydAdM x

dAyEydAMAA

x 2..

dAyIA 2

= moment of inertiawrt neutral axis

IEM ..

8/15 FLEXURE FORMULA FOR BENDING STRESSES

IyMyEx

...

8/16 MAXIMUM STRESSES

OCCUR AT THE TOP & BOTTOM FACES

8/17 MAXIMUM STRESSES

S1 = I/c1

Section modulus

1

M

y

x

2

C1

C2

1

1

11

.SM

IcM

2

22

.SM

IcM

8/18 I & S FOR BEAM

12. 3hbI

2

6. hbS 0h

b

z

y

8/19 BEAM DESIGN

• SELECT SHAPE AND SIZE SO THAT STRESS DOES NOT EXCEED ALLOW

• CALCULATE REQUIRED S=MMAX/ALLOW

• CHOOSE LOWEST CROSS SECTION WHICH SATISFIES S

8/20 IDEAL BEAM

• RECTANGULAR BEAM, SR=bh2/6=Ah/6

• CYLINDRICAL BEAM, S=0.85.SR

• IDEAL BEAM, HALF THE AREA AT h/2

• IDEAL BEAM, S=3.SR

• STANDARD I-BEAM, S=2.SR

8/21 STRESSES CAUSED BY BENDING

q=20kN/mP=50kN

RARB

2.5m 3.5m

h=0.7m

b=0.22m

8/22 V & M DIAGRAMSV=89.2kN

39.2kN

-10.8kN-80.8kN

M=160.4kNm

V

M

8/23 MAX FOR BEAM

32

2 018.06

7.022.06. mxhbS

0h

b

z

y

MPamkNm

SM

C 9.8018.0

4.1603

max

MPaT 9.8

MAXIMUM STRESSES