8/1 stress and strain in beams. 8/2 bending beams loads on beam produce stress resultants, v & m...
DESCRIPTION
8/3 TYPES OF BENDING PURE BENDING – FLEXURE UNDER CONSTANT M. i.e. V =0 = dM/dx NON-UNIFORM BENDING – FLEXURE WHEN V NON-ZEROTRANSCRIPT
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8/1 STRESS AND STRAIN IN BEAMS
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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
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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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8/4 PURE BENDINGP P
Pure bending region
x
zy Unit (small) Length
P-P
M
V
P.a
a
Non –uniform bending
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8/5 PURE BENDING
Unit (small) length
Straight lines
MM
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8/6 RADIUS OF CURVATURE
Straight linesremain straight
Planes remain plane
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8/7 CURVATURE = 1/dxdsd .
dxd
1
d
ds
dxx
y
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d= dxL1= (-y)dL1= dx - (y/)dxL1- dx = -(y/)dx
L1
x= -(y/)dx/dx=-.yNORMAL STRAIN
NEUTRAL AXIS
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8/9 LONGITUDINAL STRAIN
1
1 max
1max
y
Strain
x
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8/10 LONGITUDINAL STRESS
y
Strain Stress
Stress v Strain
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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
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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
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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
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8/14 MOMENT-CURVATURE EQN
z
y
c1
c2
dAydAdM x
dAyEydAMAA
x 2..
dAyIA 2
= moment of inertiawrt neutral axis
IEM ..
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8/15 FLEXURE FORMULA FOR BENDING STRESSES
IyMyEx
...
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8/16 MAXIMUM STRESSES
OCCUR AT THE TOP & BOTTOM FACES
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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
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8/18 I & S FOR BEAM
12. 3hbI
2
6. hbS 0h
b
z
y
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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
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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
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8/21 STRESSES CAUSED BY BENDING
q=20kN/mP=50kN
RARB
2.5m 3.5m
h=0.7m
b=0.22m
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8/22 V & M DIAGRAMSV=89.2kN
39.2kN
-10.8kN-80.8kN
M=160.4kNm
V
M
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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