View
52.928
Download
3
Category
Preview:
Citation preview
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
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.
Key Points:
1. Internal bending moment causes beam to deform.
2. For this case, top fibers in compression, bottom in tension.
Bending in beams
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
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
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
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
Bending Stress in beams
Bending Stress in beams
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
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
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
Flexure Formula
€
MI
=ER
=σy
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
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
Section Modulus of symmetrical sections
Source:- http://en.wikipedia.org/wiki/Section_modulus
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.
Composite beams Composite beams consisting of layers with fibers, or rods strategically placed to increase stiffness and strength can be “designed” to resist bending.
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
Recommended