160.7.2 Moment Curvature Pure Bending Beam Theory Moment Curvature Pure Bending Beam...Summary for...
Transcript of 160.7.2 Moment Curvature Pure Bending Beam Theory Moment Curvature Pure Bending Beam...Summary for...
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