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Pure BendingStevenVukazich
SanJoseStateUniversity
Consider an Elastic Beam Subjectedto Pure Bending
x
y
π"Beam Cross
Sectionπ΄ π΅
πΆ π·
π π
MML
Lπ΄β² π΅β²
πΆβ² π·β²
πβ² πβ²
y
z x
z
Neutral axisof the beam
Notes
x
y
π"Beam Cross
Sectionπ΄ π΅
πΆ π·
π π
MML
Lπ΄β² π΅β²
πΆβ² π·β²
πβ² πβ²
y
z x
z
1. Beam is prismatic and symmetric about the y axis:2. Material is linear elastic;3. The x axis is attached to the neutral axis of the beam;4. Pure bending (no internal shear) βthe beam deforms in a circular arc.
Study the Geometry of the Deformed Shape of a Small Slice of the Beam
π
π΄
πΆ π·
π
π΅
π
πΆβ² π·β²
π
π΄β² π΅β²
πβ² πβ²y
π"
π"
l
Center of curvature
Radius of curvature
Neutral axis
Note:β’ π΄,π,πΆ,andπ΅,π,π·,arestraightlineswhose
intersectiondefinethecenterofcurvature,O;β’ πΆ,π·, < πΆπ·β’ π΄,π΅, > π΄π΅;β’ π,π, = ππ = π";β’ π" = ππ;β’ π = π β π¦ π
π
Bending Strain of the Horizontal Fibers of the Beam
π
π΄
πΆ π·
π
π΅
π
πΆβ² π·β²
π
π΄β² π΅β²
πβ² πβ²y
π"
π"
l
Radius of curvatureNeutral axis
π" = ππ
π
Length of fiber, l, after deformation
π = π β π¦ π
Original length of all fibers, lO
π =βππ"=π β π"π"
=π β π¦ π β ππ
ππ= β
π¦π
Equilibrium of the Small Segment of Beam
x
ππΉ = πππ΄π
N πππ΄O
= 0
π
π΄
πΆ π·
π
π΅
π
πΆβ² π·β²
π
π΄β² π΅β²
πβ² πβ² y
π"
π"
l
π y
z x
ππ΄
Beam cross sectional area = A
QπΉR
οΏ½
οΏ½
= 0+
QπU
οΏ½
οΏ½
= 0+ βN π¦ππΉO
β π = 0 βN π¦πππ΄O
β π = 0
N ππΉO
= 0
Constitutive Law for Beam Material
Beam is made of linear elastic material
π
π
π = πΈπ
E = Modulus of Elasticity1
Review Relationships from Geometry of Deformation, Equilibrium, and Constitutive Law
π = πΈπ
π = βπ¦π
N πππ΄O
= 0 βN π¦πππ΄O
βπ = 0
π = βπΈπ¦π
2.
3.
1.
Substituting equation 1 into equation 3
Force Equilibrium in x Direction
N πππ΄O
= 0
π = βπΈπ¦π
N βπΈπ¦πππ΄
O= 0
βπΈπN π¦ππ΄O
= 0
y
z
ππ΄
Beam cross sectional area = A
Can only be satisfied if the neutral axis is at the centroid of the beam cross sectionN π¦ππ΄
O= 0
Moment Equilibrium
π = βπΈπ¦π
βN βπ¦πΈπ¦πππ΄
Oβπ = 0
πΈπN π¦Yππ΄O
βπ = 0
y
z
ππ΄
π =πΈπN π¦Yππ΄O
βN π¦πππ΄O
βπ = 0
Moment of Inertia about the centroid of the beam cross section
πΌ = N π¦Yππ΄O
π =πΈπΌπ
Moment-Curvature Relationship
y
ππ΄
Moment of Inertia about the centroid of the beam cross section
πΌ = N π¦Yππ΄O
π =πΈπΌππ
π΄
πΆ π·
π
π΅
π
πΆβ² π·β²
π
π΄β² π΅β²
πβ² πβ²y
π"
π"
l
π
π =1π
Define:π = Curvature of the beam
π =ππΈπΌ
The moment-curvature relationship is the basis of bending deformation theory
Bending Stress Distributiony
π =πΈπΌπ
Neutral axis
zxy
π = βπΈπ¦π
1π=ππΈπΌ π = βπΈπ¦
1π
= βπΈπ¦ππΈπΌ
π = βππ¦πΌ
π¦max = πc1
c2
πmax = βππ_πΌ
πmax =ππYπΌ
π
Summary for Pure Bending of an Elastic Beam
y
z
π = βππ¦πΌ
c1
c2
1. Neutral axis (Ο = 0) is located at the centroid of the beam cross section;
2. Moment-Curvature relationship is basis of bending deformation theory;
3. Bending stress varies linearly over beam cross section and is maximum at the extreme fibers of the beam;
π =ππΈπΌ
πmax =πππΌ
Neutral axisMoment-Curvature relationship
Bending stress distribution