Lecture 10 bending stresses in beams
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Transcript of Lecture 10 bending stresses in beams
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Unit 2- Stresses in Beams
Lecture -1 – Review of shear force and bending moment diagram
Lecture -2 – Bending stresses in beams
Lecture -3 – Shear stresses in beams
Lecture -4- Deflection in beams
Lecture -5 – Torsion in solid and hollow shafts.
Topics Covered
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Theory of simple bending (assumptions)
Material of beam is homogenous and isotropic => constant E in all direction
Young’s modulus is constant in compression and tension => to simplify analysis
Transverse section which are plane before bending before bending remain plain after bending. => Eliminate effects of strains in other direction (next slide)
Beam is initially straight and all longitudinal filaments bend in circular arcs => simplify calculations
Radius of curvature is large compared with dimension of cross sections => simplify calculations
Each layer of the beam is free to expand or contract => Otherwise they will generate additional internal stresses.
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Key Points:
1. Internal bending moment causes beam to deform.
2. For this case, top fibers in compression, bottom in tension.
Bending in beams
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Key Points:
1. Neutral surface – no change in length.
2. Neutral Axis – Line of intersection of neutral surface with the transverse section.
3. All cross-sections remain plane and perpendicular to longitudinal axis.
Bending in beams
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Key Points:
1. Bending moment causes beam to deform.
2. X = longitudinal axis
3. Y = axis of symmetry
4. Neutral surface – does not undergo a change in length
Bending in beams
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P A B
RB RA M M
Radius of Curvature, R
Deflected Shape
Consider the simply supported beam below:
M M
What stresses are generated within, due to bending?
Bending Stress in beams
Neutral Surface
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M=Bending Moment
M M
Beam
σx (Tension)
σx (Compression)
σx=0
(i) Bending Moment, M (ii) Geometry of Cross-section
σx is NOT UNIFORM through the section depth
σx DEPENDS ON:
Axial Stress Due to Bending:
stress generated due to bending:
Neutral Surface
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Bending Stress in beams
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Bending Stress in beams
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Stresses due to bending
N’ N’
R
E F
A’ C’
B’ D’
Strain in layer EF
€
=yR
€
E =Stress_ in _ the_ layer _ EFStrain _ in _ the_ layer_ EF
E =σyR⎛
⎝ ⎜
⎞
⎠ ⎟
€
σy
=ER
€
σy
=ER
€
σ =ERy
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Neutral axis
N A
y dy
dA force on the layer=stress on layer*area of layer
€
=σ × dA
=ER× y × dA
Total force on the beam section
€
=ER× y × dA∫
=ER
y × dA∫
For equilibrium forces should be 0
€
y × dA = 0∫Neutral axis coincides with the geometrical axis
Stress diagram
x
σx
M M
σx
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Moment of resistance
N A
y dy
dA force on the layer=stress on layer*area of layer
€
=σ × dA
=ER× y × dA
Moment of this force about NA
€
=ER× y × dA × y
=ER× y 2 × dA
Total moment M=
€
ER× y 2 × dA =
ER∫ y 2∫ × dA
Stress diagram
€
y 2 × dA∫ = I
€
M =ERI⇒ M
I=ER
x
σx
M M
σx
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Flexure Formula
€
MI
=ER
=σy
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Beam subjected to 2 BM
In this case beam is subjected to moments in two directions y and z. The total moment will be a resultant of these 2 moments.
You can apply principle of superposition to calculate stresses. (topic covered in unit 1).
Resultant moments and stresses
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Section Modulus
Section modulus is defined as ratio of moment of inertia about the neutral axis to the distance of the outermost layer from the neutral axis
€
Z =Iymax
MI
=σy
MI
=σmaxymax
M =σmaxIymax
M =σmaxZ
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Section Modulus of symmetrical sections
Source:- http://en.wikipedia.org/wiki/Section_modulus
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Section Modulus of unsymmetrical sections
In case of symmetrical section neutral axis passes through geometrical center of the section. But in case of unsymmetrical section such as L and T neutral axis does not pass through geometrical center.
The value of y for the outermost layer of the section from neutral axis will not be same.
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Composite beams Composite beams consisting of layers with fibers, or rods strategically placed to increase stiffness and strength can be “designed” to resist bending.
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Composite beams
b
t t
y
d
€
σ1E1
=σ2E2
σ1 =E1E2σ2
= mσ2 m=modular ratio
€
M =σyI
M = M1 + M2
=σ1yI1 +
σ2yI2
=σ2ymI1 + I2[ ]
Equivalent I (moment of inertia)=
€
mI1 + I2