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Beam Deflection
BIOE 3200 - Fall 2015Watching stuff break
Learning Objectives
ο‚— Define beam deflection (Ξ΄) and identify
the factors that affect it
ο‚— Determine deflection and slope in
beams in bending using double
integration method
BIOE 3200 - Fall 2015
Deflection in beams
Deflection is
deformation
from original
position in y
direction
BIOE 3200 - Fall 2015
Recall relationships between shear
force, bending moment and normal
stresses
BIOE 3200 - Fall 2015
To balance forces within
beam cross section:
P = Οƒx 𝑑𝐴
V =  π‘₯𝑦 𝑑𝐴
𝑀 𝑧 = βˆ’ Οƒx 𝑦 𝑑𝐴
 π‘₯𝑦
Bending moments balance normal
stresses across area of cross
section. This is how we relate
applied loads and deformation
(stress and strain).
BIOE 3200 - Fall 2015
Simplify: 𝑀𝑧 = 𝐸
𝑑2 𝑒 π‘œπ‘¦(π‘₯)
𝑑π‘₯2 𝑦2
𝑑𝐴
Recall that 𝐼x = 𝑦2
𝑑𝐴
So:
where u0y(x) is displacement in y direction
General formula for beam
deflection involves double
integration of bending moment
equation
ο‚— Governing equation for deflection:
𝑑2Ξ΄
𝑑π‘₯2 =
𝑀(π‘₯)
𝐸𝐼
, where deflection Ξ΄(x) = uoy(x)
ο‚— Solve double integral to get equation for Ξ΄(x)
(elastic curve), or the deflected shape
ο‚— Shape of beam determined by change in
load over length of beam
BIOE 3200 - Fall 2015
How to calculate beam deflection
using double integration method
BIOE 3200 - Fall 2015
Ξ΄(x)
EI = Flexural Rigidity
ο‚— Governing equation:
𝑑2Ξ΄
𝑑π‘₯2 =
𝑀
𝐸𝐼
𝐸𝐼
𝑑δ
𝑑π‘₯
= 0
π‘₯
𝑀 π‘₯ 𝑑π‘₯ + 𝐢1 Note:
𝑑δ
𝑑π‘₯
= tan ΞΈ β‰ˆ ΞΈ(x)
οƒ  𝐸𝐼θ(x) = 0
π‘₯
𝑀 π‘₯ 𝑑π‘₯ + 𝐢1
𝐸𝐼δ = 0
π‘₯
[ 0
π‘₯
𝑀 π‘₯ 𝑑π‘₯ + 𝐢1 ] dx + 𝐢2
𝐸𝐼δ = 0
π‘₯
𝑑π‘₯ 0
π‘₯
𝑀 π‘₯ 𝑑π‘₯ + 𝐢1x + 𝐢2 - general formula for
beam deflection
Sign conventions
for beam
deflection
ο‚— X and Y axes: positive to the right and
upward, respectively
ο‚— Deflection Ξ΄(x): positive upward
ο‚— Slope of deflection at any point
𝑑δ
𝑑π‘₯
and
angle of rotation ΞΈ(x): positive when CCW
with respect to x-axis
ο‚— Curvature (K) and bending moment (M):
positive when concave up (beam is
smiling) BIOE 3200 - Fall 2015
Ξ΄(x)
What affects deflection?
ο‚— Bending moment
β—¦ Magnitude and type of loading
β—¦ Span (length) of beam
β—¦ Beam type (simply supported, cantilever)
ο‚— Material properties of beam (E)
ο‚— Shape of beam (Moment of Intertia I)
BIOE 3200 - Fall 2015
How to complete double
integration
𝐸𝐼δ = 0
π‘₯
𝑑π‘₯ 0
π‘₯
𝑀 π‘₯ 𝑑π‘₯ + 𝐢1x + 𝐢2
ο‚— Find C1 and C2 from boundary conditions
(supports)
ο‚— Example: cantilever beam with load at free
end
BIOE 3200 - Fall 2015
Using boundary conditions to
calculate deflections in beams
ο‚— Other examples of boundary
conditions:
BIOE 3200 - Fall 2015
Pulling it all together
ο‚— Relation of the deflection Ξ΄ with beam
loading quantities V, M and load
ο‚— Deflection = Ξ΄
ο‚— Slope = d Ξ΄ / dx
ο‚— Moment M(x) = EI
𝑑2Ξ΄
𝑑π‘₯2
ο‚— Shear V(x) = - dM/dx = - EI
𝑑3Ξ΄
𝑑π‘₯3 (for
constant EI)
ο‚— Load w(x) = dV/dx = - EI
𝑑4Ξ΄
𝑑π‘₯4 (for constant
EI)
BIOE 3200 - Fall 2015
Load, moment, deformation and
slope can all be sketched for a
beam
BIOE 3200 - Fall 2015
Procedure for calculating
deflection by integration method
ο‚— Select interval(s) of the beam to be used,
and set coordinate system with origin at
one end of the interval; set range of x
values for that interval
ο‚— List boundary conditions at boundaries of
interval (these will be integration
constants)
ο‚— Calculate bending moment M(x)
(function of x for each interval) and set it
equal to EI
𝑑2Ξ΄
𝑑π‘₯2
ο‚— Solve differential equation (double
integration) and solve using known
integration constants BIOE 3200 - Fall 2015
Typical deflection equation
Simply supported beam under uniform
constant load:
Ξ΄x =
𝑀 π‘₯
24 𝐸 𝐼
(𝑙3
βˆ’ 2 𝑙 π‘₯2
+ π‘₯3
)(at any point x)
BIOE 3200 - Fall 2015
Loa
d
Material
Property
Shape
Propert
y
L
Ξ”max
Span
Examples of deflection
formulae
FBD for simply supported beam under
constant uniform load:
ο‚— Ξ΄max =
5 𝑀 𝑙4
384 𝐸 𝐼
(at midpoint)
ο‚— Ξ΄x =
𝑀 π‘₯
24 𝐸 𝐼
(𝑙3
βˆ’ 2 𝑙 π‘₯2
+ π‘₯3
)(at any point x)
BIOE 3200 - Fall 2015
Examples of deflection
formulae
ο‚— Simply supported beam, point load at
midspan
ο‚— Ξ΄max =
𝑃 𝐿 π‘œ
3
48 𝐸 𝐼
(at point of load)
ο‚— Ξ΄x =
𝑃 π‘₯
48 𝐸 𝐼
(3 𝐿 π‘œ
2
βˆ’ 4 π‘₯2
) (where x <
𝐿 π‘œ
2
; symmetric about midspan)
BIOE 3200 - Fall 2015
x
Examples of deflection
formulae
ο‚— Cantilever beam loaded at free end
ο‚— Ξ΄max =
𝑃 𝐿3
3 𝐸 𝐼
(at free end)
ο‚— Ξ΄x =
𝑃 π‘₯
2
6 𝐸 𝐼
(3 𝐿 - π‘₯)(everywhere else)
BIOE 3200 - Fall 2015
P

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8 beam deflection

  • 1. Beam Deflection BIOE 3200 - Fall 2015Watching stuff break
  • 2. Learning Objectives ο‚— Define beam deflection (Ξ΄) and identify the factors that affect it ο‚— Determine deflection and slope in beams in bending using double integration method BIOE 3200 - Fall 2015
  • 3. Deflection in beams Deflection is deformation from original position in y direction BIOE 3200 - Fall 2015
  • 4. Recall relationships between shear force, bending moment and normal stresses BIOE 3200 - Fall 2015 To balance forces within beam cross section: P = Οƒx 𝑑𝐴 V =  π‘₯𝑦 𝑑𝐴 𝑀 𝑧 = βˆ’ Οƒx 𝑦 𝑑𝐴  π‘₯𝑦
  • 5. Bending moments balance normal stresses across area of cross section. This is how we relate applied loads and deformation (stress and strain). BIOE 3200 - Fall 2015 Simplify: 𝑀𝑧 = 𝐸 𝑑2 𝑒 π‘œπ‘¦(π‘₯) 𝑑π‘₯2 𝑦2 𝑑𝐴 Recall that 𝐼x = 𝑦2 𝑑𝐴 So: where u0y(x) is displacement in y direction
  • 6. General formula for beam deflection involves double integration of bending moment equation ο‚— Governing equation for deflection: 𝑑2Ξ΄ 𝑑π‘₯2 = 𝑀(π‘₯) 𝐸𝐼 , where deflection Ξ΄(x) = uoy(x) ο‚— Solve double integral to get equation for Ξ΄(x) (elastic curve), or the deflected shape ο‚— Shape of beam determined by change in load over length of beam BIOE 3200 - Fall 2015
  • 7. How to calculate beam deflection using double integration method BIOE 3200 - Fall 2015 Ξ΄(x) EI = Flexural Rigidity ο‚— Governing equation: 𝑑2Ξ΄ 𝑑π‘₯2 = 𝑀 𝐸𝐼 𝐸𝐼 𝑑δ 𝑑π‘₯ = 0 π‘₯ 𝑀 π‘₯ 𝑑π‘₯ + 𝐢1 Note: 𝑑δ 𝑑π‘₯ = tan ΞΈ β‰ˆ ΞΈ(x) οƒ  𝐸𝐼θ(x) = 0 π‘₯ 𝑀 π‘₯ 𝑑π‘₯ + 𝐢1 𝐸𝐼δ = 0 π‘₯ [ 0 π‘₯ 𝑀 π‘₯ 𝑑π‘₯ + 𝐢1 ] dx + 𝐢2 𝐸𝐼δ = 0 π‘₯ 𝑑π‘₯ 0 π‘₯ 𝑀 π‘₯ 𝑑π‘₯ + 𝐢1x + 𝐢2 - general formula for beam deflection
  • 8. Sign conventions for beam deflection ο‚— X and Y axes: positive to the right and upward, respectively ο‚— Deflection Ξ΄(x): positive upward ο‚— Slope of deflection at any point 𝑑δ 𝑑π‘₯ and angle of rotation ΞΈ(x): positive when CCW with respect to x-axis ο‚— Curvature (K) and bending moment (M): positive when concave up (beam is smiling) BIOE 3200 - Fall 2015 Ξ΄(x)
  • 9. What affects deflection? ο‚— Bending moment β—¦ Magnitude and type of loading β—¦ Span (length) of beam β—¦ Beam type (simply supported, cantilever) ο‚— Material properties of beam (E) ο‚— Shape of beam (Moment of Intertia I) BIOE 3200 - Fall 2015
  • 10. How to complete double integration 𝐸𝐼δ = 0 π‘₯ 𝑑π‘₯ 0 π‘₯ 𝑀 π‘₯ 𝑑π‘₯ + 𝐢1x + 𝐢2 ο‚— Find C1 and C2 from boundary conditions (supports) ο‚— Example: cantilever beam with load at free end BIOE 3200 - Fall 2015
  • 11. Using boundary conditions to calculate deflections in beams ο‚— Other examples of boundary conditions: BIOE 3200 - Fall 2015
  • 12. Pulling it all together ο‚— Relation of the deflection Ξ΄ with beam loading quantities V, M and load ο‚— Deflection = Ξ΄ ο‚— Slope = d Ξ΄ / dx ο‚— Moment M(x) = EI 𝑑2Ξ΄ 𝑑π‘₯2 ο‚— Shear V(x) = - dM/dx = - EI 𝑑3Ξ΄ 𝑑π‘₯3 (for constant EI) ο‚— Load w(x) = dV/dx = - EI 𝑑4Ξ΄ 𝑑π‘₯4 (for constant EI) BIOE 3200 - Fall 2015
  • 13. Load, moment, deformation and slope can all be sketched for a beam BIOE 3200 - Fall 2015
  • 14. Procedure for calculating deflection by integration method ο‚— Select interval(s) of the beam to be used, and set coordinate system with origin at one end of the interval; set range of x values for that interval ο‚— List boundary conditions at boundaries of interval (these will be integration constants) ο‚— Calculate bending moment M(x) (function of x for each interval) and set it equal to EI 𝑑2Ξ΄ 𝑑π‘₯2 ο‚— Solve differential equation (double integration) and solve using known integration constants BIOE 3200 - Fall 2015
  • 15. Typical deflection equation Simply supported beam under uniform constant load: Ξ΄x = 𝑀 π‘₯ 24 𝐸 𝐼 (𝑙3 βˆ’ 2 𝑙 π‘₯2 + π‘₯3 )(at any point x) BIOE 3200 - Fall 2015 Loa d Material Property Shape Propert y L Ξ”max Span
  • 16. Examples of deflection formulae FBD for simply supported beam under constant uniform load: ο‚— Ξ΄max = 5 𝑀 𝑙4 384 𝐸 𝐼 (at midpoint) ο‚— Ξ΄x = 𝑀 π‘₯ 24 𝐸 𝐼 (𝑙3 βˆ’ 2 𝑙 π‘₯2 + π‘₯3 )(at any point x) BIOE 3200 - Fall 2015
  • 17. Examples of deflection formulae ο‚— Simply supported beam, point load at midspan ο‚— Ξ΄max = 𝑃 𝐿 π‘œ 3 48 𝐸 𝐼 (at point of load) ο‚— Ξ΄x = 𝑃 π‘₯ 48 𝐸 𝐼 (3 𝐿 π‘œ 2 βˆ’ 4 π‘₯2 ) (where x < 𝐿 π‘œ 2 ; symmetric about midspan) BIOE 3200 - Fall 2015 x
  • 18. Examples of deflection formulae ο‚— Cantilever beam loaded at free end ο‚— Ξ΄max = 𝑃 𝐿3 3 𝐸 𝐼 (at free end) ο‚— Ξ΄x = 𝑃 π‘₯ 2 6 𝐸 𝐼 (3 𝐿 - π‘₯)(everywhere else) BIOE 3200 - Fall 2015 P

Editor's Notes

  1. Long bone curvature results in bending when compressive loads applied; convex surface under tension, convex surface under compression.
  2. Unloaded beam: neutral axis is level Applied load: neutral axis bends and takes on curved shape Distance d (also delta; maximum deflection is delta max) is distance between unloaded and loaded neutral axis (NA). Delta max occurs at mid-span of simply supported beam, or at free end of cantilever beam
  3. The axial stressΒ Οƒx produces a moment about the z-axis that must be equivalent to the moment resultantΒ Mz. In other words, Οƒx balances moment Mz, and Mz balances Οƒx
  4. Recall that stress = strain x Young’s modulus (ο³ο€½πœ–E) Note that we must integrate Mz over the cross-section A,Β Β which lies in the y-zΒ plane. Assume that Young's modulusΒ EΒ is a constant over the cross-section, and thusΒ EΒ may be taken outside the integral. Since uoy(x)Β is not a function ofΒ yΒ orΒ z, it may also be taken outside the integral. Replacing strain and E for stress, moment equation includes E and second derivative of displacement term u(x). This is also called delta (Ξ΄) Iz = Second moment of area Note EI = flexural rigidity; material and shape property of beam, assumed to be constant throughout beam (unless otherwise noted)
  5. The equation for beam deflection is an ordinary second order differential equation. It defines the transverse displacement in terms of the bending moment. We know that bending moment Mz is a function of position x along the beam, and we have already learned how to determine the equation for Mz. So deflection can be determined by double integrating the equation for M. Note: EI = flexural rigidity
  6. Integrate 2nd order differential equation twice to get expression for Ξ΄(x) ΞΈ(x) – approximate slope at location Q along the beam for small deflections, because d𝛅 dx = tan ΞΈ β‰ˆ ΞΈ(x) for small deflections Beam deflection defined between two points; red curve represents the deflected shape of the beam. Can express elastic curve as a function of x – get equation of line, can get deflection at any point on the beam. Governing equation: M = complete equation for moment distribution along a beam; Differential equation can be integrated in each particular case to find the deflection delta, provided by bending moment M and flexural rigidity EI are known.
  7. Beam between two points; red curve represented the deflected shape of the beam. Can express elastic curve as a function of x – get equation of line, can get deflection at any point on the beam. Governing equation: M = complete equation for moment distribution along a beam; Differential equation can be integrated in each particular case to find the deflection delta, provided by bending moment M and flexural rigidity EI are known. EI = flexural rigidity
  8. Higher magnitude, more deflection; Uniform load produces less deflection than single point (same equivalent resultant) Same load on a longer beam will be greater than on a shorter beam Stiffer beam (higher E) will deflect less Shape of beam – mainly cross section (rectangular, circular, I-shape)
  9. At points A and B for simply supported beam, deflection = 0 For cantilever, slope and deflection at A = 0 Use these boundary conditions to calculate C1 and C2 (2 constants, 2 boundary conditions) (1) the x and y axes are positive to the right and upward, respectively; (2) The deflection Ξ½ is positive upward; (3) The slope dΞ½ dx and angle of rotation ΞΈ are positive when counterclockwise with respect to the positive x axis; (4) The curvature k is positive when the beam is bent concave upward; and (5) the bending moment M is positive when it produces compression in the upper part of the beam.
  10. For more complex loading scenarios where equation for moment cannot be defined for entire x, cut beam into sections where M can be represented by one function – concentrated loads, discontinuity in distributed loads
  11. Sign convention: (1) the x and y axes are positive to the right and upward, respectively; (2) The deflection Ξ½ is positive upward; (3) The slope dΞ½ dx and angle of rotation ΞΈ are positive when counterclockwise with respect to the positive x axis; (4) The curvature k is positive when the beam is bent concave upward; and (5) the bending moment M is positive when it produces compression in the upper part of the beam.
  12. Example (W is a distributed load): At any distance x from left to right (except midspan) Delta max is at mid-span
  13. Delta max is at mid-span
  14. Delta x left or right of midspan, deflection will be less than delta max (deflection at midspan)
  15. All these equations have the same elements: load, geometry, material properties and shape properties