Post on 21-Feb-2021
transcript
In-Class Activities:
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Today’s Objective : 14th September
To:
a) Understand the context, concept and derivation for deflection and slope.
b) Be able to calculate max and specific values.
DEFLECTION OF BEAMS
Deflections of Beams
Serviceability of beams
Deflection limits for beams appearance (sagging) fitness for purpose (machinery, pipe grades) structural (avoid unintended load paths)
General Guidelines
Limit total deflections to span/250
Different design codes give guidance on the limits for deflections
To limit deflections, we need to be able to calculate them!
250maxLv ≤
deflection (v)
The elastic curve – deformed shape of the beam
The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of the beam is called the elastic curve
v
v
Revision- Geometry of the beam deformations
The deflection, slope and curvature of a beam are related by:Deflection = vSlope = dv/dx = θCurvature = d2v/dx2 = dθ /dx = KThe curvature is the inverse of the radius, K = 1/R
ie. for a segment of the beam dx
dx = Rdθ hence,1/R = dθ /dx
Consider the deflected shape of the beam given below
dθ
dx
R
Geometry of the beam deformations: Strain
At a distance y from the neutral axis (N.A.)
Strain:
LL∆=εdsdsds )'( −=
θθθ
RdRddyR −−
=)(
Ry
−=
Moment-Curvature Relationship
We have shown that the strain ɛ at a distance y from the N.A.
ɛ=-y/R (1)Hooke’s law
ɛ=σ/E (2)From the flexure formula, the stress σ at the point under bending is given by
σ=-My/I (3)Substituting (3) into (2), the strain can be expressed by
ɛ=-My/EI (4)Substituting (4) into (1), the moment-curvature relationship can be found as
1/R=M/EI or K=M/EI
EIM
dxvdK == 2
2Curvature:
∫∫ === Mdx
vdEIdxdvEIEI 2
2
θSlope:
∫∫ ∫∫== Mdx
vdEIEIv 2
2
Deflection:
•We need to be able to define the bending moment as a function of the distance from one end of the beam
•The constants of integration need to be determined by evaluating the slope or deflection at particular points –boundary condition
Slope and Displacement by integration
E – Young’s modulus which is a constant for a given material
I – Moment of inertia computed about the neutral axis
Boundary condition
Roller:
Pin:
Fixed:
At a point on the beam where the value of slope or deflection is known, these values are called boundary conditions. For example:
v =0
v =0
v =0
θ =0
Example 1The simply supported beam supports the uniform distributed loading. Determine the equation of the elastic curve, slope and deflection at point C and the maximum deflection of the beam. E=200GPa, I=50(106)mm4.
Reaction forces
F.B.D:
∑ = :0BM 02/101010010 =××+×− AR+
N500=AR
∑ = :0yF+ 10100×=+ BA RR
N500=BR
Moment FunctionA free-body diagram of the segment and coordinate x is shown as
∑ = :0oM+
05002
100)( =×−×+ xxxxM
xxxM 50050)( 2 +−=
Slope and deflection by integration
xxxMdx
vdEI 50050)( 22
2
+−==
123 250
350 Cxx
dxdvEIEI ++−==θ
According to Moment-Curvature relationship, we have
Integrating twice, we obtain the slope and deflection
Slope:
Deflection: 2134
3250
625 CxCxxEIv +++−=
Applying boundary conditions
2134 00
32500
6250 CCEI +++−=
To determine the constants C1 and C2, the boundary conditions will be applied:
Condition 1:
At point A, x = 0 and v = 0. Thus
02 =C
Condition 2: At point B, x =10m and v = 0.
010103
250106250 1
34 +++−= CEI 7.41661 −=C
Slope and deflection at point CWith the values of C1 and C2, the functions of slope and deflection are
7.41662503
50 23 −+−== xxdxdvEIEIθSlope:
Deflection: xxxEIv 7.41663
250625 34 −+−=
7.4166225023
50 23 −×+×−=CEIθ
3300−=
( ) )10(3.3)10)(10(5010200
33003300 41269
−− −=
×−=−=
EICθ
At Point C, x = 2m
4.773327.416623
2502625 34 −=×−+×−=CEIv m)10(73.7 4−−=Cv
?
The maximum deflection
7.41662503
500 23 −+−= xx
5.0=x
-2073.25.07.41665.03
2505.0625 34
max =×−×+×−=EIv
The maximum deflection occurs at the point where the slope θ=dv/dx =0. Assuming θ=0 in the slope function:
Finding x between 0 and 10m
Thus, the maximum deflection is
m)10(07.22073.2 3max
−−=−
=EI
v
Alternatively, by inspection of the elastic curve, the maximum deflection occurs in the middle where the slope is zero.
Summary of results
xxxEIv 7.41663
250625 34 −+−=
xEI
xEI
xEI
v 7.41663250
625 34 −+−=
)10(3.3 4−−=Cθ
1) Elastic curve:
or
m)10(73.7 4−−=Cv
m)10(07.2 3max
−−=v
5.0=x
2) Slope and deflection at point C
3) The maximum deflection
at
Positive Sign Convention
When applying double integration equations, the positive signs for slope and deflection are
Sign convention-cont.The sign convention is helpful to verify your calculation, for example
vA is negativeθA is negative
vB is negativeθB is positive
Q: By inspection, slope and deflection at point A, B and C are positive or negative under the given coordinate?
Boundary condition for a symmetry structure
Due to symmetry, only half of the structure needs to be calculated for slope and deflection. In such cases, the slope at the symmetry axis is zero. This boundary condition will greatly simplify our calculation.
Symmetry condition:
2at0 Lxdxdv ==
Previous example using the symmetry condition
2134 00
32500
6250 CCEI +++−=
xxxMdx
vdEI 50050)( 22
2
+−==
123 52505
3500 CEI +×+×−=
123 250
350 Cxx
dxdvEIEI ++−==θ
2134
3250
625 CxCxxEIv +++−=
From previous slides, we obtained
Boundary conditions:1) At point A, x = 0 and v = 0.
02 =∴C
2) When x = 5m and θ = 0.
xxxEIv 7.41663
250625 34 −+−=
7.41661 −=∴C
Function of elastic curve:
-Same to the previous result
Continuity conditionsIf a single x coordinate cannot be used to express the equation for the slope or deflection, then continuity conditions must be used to evaluate some of integration constants.
Continuity conditions – cont.Each coordinate is only valid within the regions
and ax ≤≤ 10 )(2 baxa +≤≤
Continuity conditions require
)()( 21 avav =)()( 21 aa θθ =
Each coordinate is only valid within the regions
and ax ≤≤ 10 bx ≤≤ 20Continuity conditions require
)()( 21 avav =)()( 21 aa θθ −=
Example 2
Determine the equations of the elastic curve for the beam using x1 and x2coordinates. Specify the slope at A and the maximum deflection. EI is constant.
Reaction Forces
∑ = :0AM+ 03
23
=×−×−×LPLPLRB
PRB =
∑ = :0yF+ 0=−−+ PPRR BA
PRA =
Moment Functions30 1 Lx ≤≤When
∑ = :0M+
0)( 11 =− PxxM
11)( PxxM =
23 2 LxL ≤≤When
∑ = :0M+
0)3
()( 222 =−+−LxPPxxM
3)( 2
PLxM =
Slope and deflection by integrationThus, for 30 1 Lx ≤≤
23 2 LxL ≤≤
1121
12
)( PxxMdx
vdEI ==
3)( 22
2
22 PLxM
dxvdEI ==
211311 6
CxCxPEIv ++=
For
423222 6
CxCxPLEIv ++=
Note: There are four constants of integration: C1, C2, C3 and C4. To determine the above four constants, four additional equations are needed by applying boundary conditions
------(1)
----------(2)
------(3)
--------(4)
12111
1
1
2)( CxPxEI
dxdvEI +== θ
32222
2
3)( CxPLxEI
dxdvEI +== θ
Boundary conditions
Because only half of the structure is considered, so only the boundary condition at point A will be applied, that is,
When 01 =x , 01 =v
213 00
60 CCPEI +×+×=
So 02 =C ------------------------------------------(5)
. From eqn.(2)
Boundary condition due to symmetry
Due to the symmetry of the structure, we know
when 2/2 Lx = , 02
2 =dxdv
From eqn.(3),
3230 CLPLEI +×=
6
2
3PLC −= ------------------------------------------(6)
Continuity conditions
From eqn.(1) and eqn.(3), slope continuity condition is
when 3/21 Lxx == , continuity conditions will be applied
)3
()3
( 21LEILEI θθ =
63332
2
1
2 PLLPLCLP−×=+
×
9
2
1PLC −=
From eqn.(2) and eqn.(4), deflection continuity condition is
4
2223
36363936CLPLLPLLPLLP
+×−
×=×−
×
)3
()3
( 21LEIvLEIv =
162
3
4PLC =
------------------------------------------(7)
-----(8)
Slope and deflection
)30( 1 Lx ≤≤
)23( 2 LxL ≤≤
)30( 1 Lx ≤≤
)23( 2 LxL ≤≤
1
2311 96
xPLxPEIv −=
16266
3
2
2222
PLxPLxPLEIv +−=
92)(
22111
PLxPxEI −=θ
63)(
2
222PLxPLxEI −=θ
Slope:
Deflection:
Slope at point A:99
02
222 PLPLPEI A −=−×=θ
EIPL
A 9
2
−=θ
1622626
322
maxPLLPLLPLEIv +
×−
×=
The max v occurs at x2=L/2
EIPLv
64823 3
max −=
Steps for DOUBLE INTEGRATION
Select appropriate coordinator(s) and establish the equation(s) for M(x)
Integrate to get slope, EIθ. Integrate again to get deflection, EIv. Apply the boundary conditions and continuity
conditions to θ and v to evaluate the integration constants.
)(xM
Why is deflection calculation important?
a) Deflection is another important set off calculation in structural analysis.
b) Loads create bending and bending causes deflection
c) Deflection is an important service condition
d) Deflection is not important at all
What does dv/dx represent?
a) Deflectionb) The elastic curvec) The curvatured) The slope
What is the purpose of boundary conditions?
a) Allows to calculate or eliminate constants
b) Defines the equations at boundariesc) Is a part of all equationsd) Not useful at all