2. Learning Objectives
Define beam deflection (δ) and identify
the factors that affect it
Determine deflection and slope in
beams in bending using double
integration method
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 = σx 𝑑𝐴
V = 𝑥𝑦 𝑑𝐴
𝑀 𝑧 = − σx 𝑦 𝑑𝐴
𝑥𝑦
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 𝑢 𝑜𝑦(𝑥)
𝑑𝑥2 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
Governing equation for deflection:
𝑑2δ
𝑑𝑥2 =
𝑀(𝑥)
𝐸𝐼
, where deflection δ(x) = uoy(x)
Solve double integral to get equation for δ(x)
(elastic curve), or the deflected shape
Shape of beam determined by change in
load over length of beam
BIOE 3200 - Fall 2015
8. Sign conventions
for beam
deflection
X and Y axes: positive to the right and
upward, respectively
Deflection δ(x): positive upward
Slope of deflection at any point
𝑑δ
𝑑𝑥
and
angle of rotation θ(x): positive when CCW
with respect to x-axis
Curvature (K) and bending moment (M):
positive when concave up (beam is
smiling) BIOE 3200 - Fall 2015
δ(x)
9. What affects deflection?
Bending moment
◦ Magnitude and type of loading
◦ Span (length) of beam
◦ Beam type (simply supported, cantilever)
Material properties of beam (E)
Shape of beam (Moment of Intertia I)
BIOE 3200 - Fall 2015
10. How to complete double
integration
𝐸𝐼δ = 0
𝑥
𝑑𝑥 0
𝑥
𝑀 𝑥 𝑑𝑥 + 𝐶1x + 𝐶2
Find C1 and C2 from boundary conditions
(supports)
Example: cantilever beam with load at free
end
BIOE 3200 - Fall 2015
11. Using boundary conditions to
calculate deflections in beams
Other examples of boundary
conditions:
BIOE 3200 - Fall 2015
12. Pulling it all together
Relation of the deflection δ with beam
loading quantities V, M and load
Deflection = δ
Slope = d δ / dx
Moment M(x) = EI
𝑑2δ
𝑑𝑥2
Shear V(x) = - dM/dx = - EI
𝑑3δ
𝑑𝑥3 (for
constant EI)
Load w(x) = dV/dx = - EI
𝑑4δ
𝑑𝑥4 (for constant
EI)
BIOE 3200 - Fall 2015
14. Procedure for calculating
deflection by integration method
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
List boundary conditions at boundaries of
interval (these will be integration
constants)
Calculate bending moment M(x)
(function of x for each interval) and set it
equal to EI
𝑑2δ
𝑑𝑥2
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:
δx =
𝑤 𝑥
24 𝐸 𝐼
(𝑙3
− 2 𝑙 𝑥2
+ 𝑥3
)(at any point x)
BIOE 3200 - Fall 2015
Loa
d
Material
Property
Shape
Propert
y
L
Δmax
Span
16. Examples of deflection
formulae
FBD for simply supported beam under
constant uniform load:
δmax =
5 𝑤 𝑙4
384 𝐸 𝐼
(at midpoint)
δx =
𝑤 𝑥
24 𝐸 𝐼
(𝑙3
− 2 𝑙 𝑥2
+ 𝑥3
)(at any point x)
BIOE 3200 - Fall 2015
17. Examples of deflection
formulae
Simply supported beam, point load at
midspan
δmax =
𝑃 𝐿 𝑜
3
48 𝐸 𝐼
(at point of load)
δx =
𝑃 𝑥
48 𝐸 𝐼
(3 𝐿 𝑜
2
− 4 𝑥2
) (where x <
𝐿 𝑜
2
; symmetric about midspan)
BIOE 3200 - Fall 2015
x
18. Examples of deflection
formulae
Cantilever beam loaded at free end
δmax =
𝑃 𝐿3
3 𝐸 𝐼
(at free end)
δx =
𝑃 𝑥
2
6 𝐸 𝐼
(3 𝐿 - 𝑥)(everywhere else)
BIOE 3200 - Fall 2015
P
Editor's Notes
Long bone curvature results in bending when compressive loads applied; convex surface under tension, convex surface under compression.
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
The axial stress σx produces a moment about the z-axis that must be equivalent to the moment resultant Mz.
In other words, σx balances moment Mz, and Mz balances σx
Recall that stress = strain x Young’s modulus (𝜖E)
Note that we must integrate Mz over the cross-section A, which lies in the y-z plane. Assume that Young's modulus E is a constant over the cross-section, and thus E may be taken outside the integral. Since uoy(x) is not a function of y or 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 (δ)
Iz = Second moment of area
Note EI = flexural rigidity; material and shape property of beam, assumed to be constant throughout beam (unless otherwise noted)
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
Integrate 2nd order differential equation twice to get expression for δ(x)
θ(x) – approximate slope at location Q along the beam for small deflections, because d𝛅 dx = tan θ ≈ θ(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.
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
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)
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 ν is positive upward;
(3) The slope dν dx and angle of rotation θ 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.
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
Sign convention:
(1) the x and y axes are positive to the right and upward, respectively;
(2) The deflection ν is positive upward;
(3) The slope dν dx and angle of rotation θ 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.
Example (W is a distributed load):
At any distance x from left to right (except midspan)
Delta max is at mid-span
Delta max is at mid-span
Delta x left or right of midspan, deflection will be less than delta max (deflection at midspan)
All these equations have the same elements: load, geometry, material properties and shape properties