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8/11/2019 Week 4 Lecture - Beam Deflection & Analysis (Sept 15, 2014)
http://slidepdf.com/reader/full/week-4-lecture-beam-deflection-analysis-sept-15-2014 1/29
Aer E 3 : Aerospace Structures Laboratory
Week 4 Lecture:
Linear Analysis of a Beam
by Direct Stiffness Method
September 15, 2014
8/11/2019 Week 4 Lecture - Beam Deflection & Analysis (Sept 15, 2014)
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Aer E 3 : Aerospace Structures Laboratory
2
Beams
In this course, we consider uniform elastic materials
A beam is a long straight structure of elastic members
Subjected to transverse loads and bending moments Small displacement causes small change in member length, and hence
negligible axial force
All loads, reactions and deformations lie in the plane of symmetry of
its cross-section, with all the forces perpendicular to its centroidal axis
Elastic material:
transverse and bending deformations allowed
2
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Aer E 3 : Aerospace Structures Laboratory
Linear Analysis by Direct Stiffness Method
Direct stiffness method is the most popular matrix method for
analyzing discrete structures like trusses, beams and frames
Concept of “unit response”:
for each member of a structure, there is this force per unit
displacement defined as stiffness coefficient, kij, which
represent the force at the location acting in the direction of member end force Q i required, along with other end forces, to
cause a unit value of displacement u j, while all other end
displacements are zero
In stiffness method, these kij’s serve as the universal “scalingfactors” for all members’ forces in a structure
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Aer E 3 : Aerospace Structures Laboratory
Linear Analysis by Direct Stiffness Method (cont’d)
Stiffness method works by first identifying where the displacement
unknowns (called degrees-of-freedom) are located in the frame structure
Second, the structure is conceptually broken down to individual bar
members in their own local coordinates, and a local stiffness matrix is
determined for each member
The global stiffness matrix is then assembled in the global coordinates bysumming up the contributions from individual members. External joint
loads (including the equivalent loads resulted from loads not applied at
joints) are transformed to the global coordinates as well.
Solve for the unknown displacements from the global stiffness matrix
equation
From the now known displacements, reverse the process to calculate the
individual member’s internal forces and displacement as well as stresses
and strains 4
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Aer E 3 : Aerospace Structures Laboratory
Naming and Coordinate Systems Setup
Y
X1 2 3
X-Y: one global coordinate system for entire beam; origin is usuallyset at left end
: joint ID
: member ID
1 2 3 4
5
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Aer E 3 : Aerospace Structures Laboratory
Structure Degree-of-Freedom
Every free joint has two degree-of-freedoms (DOF, allowing vertical
translation and rotation). Constrained joints have reduced DOF,depending on the constraint condition
This beam has 6 DOF’s = 2(2 DOF’s per joint) x 4(4 joints in frame) – 2 x 1(fixed end at joint 1 allows no displacement)
1 2 3 41 2 3
6
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8/11/2019 Week 4 Lecture - Beam Deflection & Analysis (Sept 15, 2014)
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Forces, Displacements and Stiffness
in Local Coordinates
Q 4x1 = k4x4 u4x1
Q : end forces
u: member end displacements
k: local stiffness
y
x
Q 1
Q 2
Q 3
Q 4
u1
u2
deformu3
u4
2
Small displacement causes small
change in member length, and
hence negligible axial force
1 2 3 41 2 3
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Local Stiffnessdue to unit displacement u1
y
x
k11
k21
k31
k41
L
u1
k11
k21
x
M
V (=-k11)
Consider a member section x from left end,
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Local Stiffnessdue to unit displacement u1 (cont’d)
2
2d y M
EI dx=Recall beam deflection eq. . Integrate once and twice to arrive at
Boundary conditions (1) x=0, y=1 and dy/dx=0 and (2) x=L, y=dy/dx=0lead to c1=0 and c2=1 and
2
21 11 1
1,
2
dy xk x k c
dx EI
= − + +
10
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11
Local Stiffnessdue to unit displacement u2
Again use beam deflection eq. . Integrate once and twice to arrive at2
2
d y M
EI dx=
y
xk12
k22
k32
k42
u2
L
b e
Boundary Conditions: At end b: x=0, dy/dx=1 and y=0.
2
22 12 1
1,
2
dy xk x k c
dx EI
= − + +
1 21 0andc c= =
11
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Aer E 3 : Aerospace Structures Laboratory
Local Stiffnessdue to unit displacement u2 (cont’d)
At end e: x=L, dy/dx=0 and y=0 and we obtain
2 3
22 12
10
2 2
L Lk k L
EI
= − + +
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Aer E 3 : Aerospace Structures Laboratory
Local stiffness matrix in its entirety:
Local Stiffness Matrix
Derivations can be done similarly for other cases.
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Transformation Between Global and
Local Coordinate Systems
Here we only consider beam in horizontal position, so
Member local coordinates ~ member global coordinates
We will need rotation later in frame analysis
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Aer E 3 : Aerospace Structures Laboratory
Local Fixed-End External Forces
External forces can be applied not only at joints but also at members.
The latter, called fixed-end forces Q f , are conceptually generated as the
member were fixed at ends. Equivalent forces at the joints (resulted
from the reactions at fixed ends) are then added to the member’s end
forces
W
Q = ku + Q f
equivalent loads added
to member’s end forces Q f3Q f1
Q f2 Q f4
reactions at the fixed
ends become equivalent
loads at joints
external loads applied to member
member ends fixed and reactions
to the external loads generatedW
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Aer E 3 : Aerospace Structures Laboratory
Local Fixed-End External Forces (cont’d)
Source: reference 1
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Aer E 3 : Aerospace Structures Laboratory
Assembly of the Structure Stiffness Matrix
Using Member Code Numbers
A beam with loads at joints and at members
2 DOF’s at joint 2: 2 unknown
displacements d’s and 2external joint loads P’s
(resulting from all applied loads)
Corresponding FBD
Source: reference 1
Only 1 DOF at joint 3 (since it
has support from below): 1unknown rotation d3 and 1
external joint load P3 (resulting
from all applied loads)
17
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Aer E 3 : Aerospace Structures Laboratory
Assembly of the Structure Stiffness Matrix
Using Member Code Numbers (cont’d)
Matching member end displacements u’s with joint displacements d’s
Breaking down to FBDs of individual members or joints
Superscripts (1), (2) or
(3) denote member
Source: reference 1
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Super i denotes member
(= 1, 2 or 3)
Expanding the equations for member
end forces (shown on the right) yields:
source: reference 1
Assembly of the Structure Stiffness Matrix
Using Member Code Numbers (cont’d)
see pages 8&15
Plugging these into the equation for P, from the previous slide,
gives us the desired structure stiffness relationships
19
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P=Sd + Pf
and
with global beam Stiffness matrix S
Sum of equivalent joint loads
In matrix form
source: reference 1
Sum of stiffness of both
members at the free joints
Assembly of the Structure Stiffness Matrix
Using Member Code Numbers (cont’d)
20
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Aer E 3 : Aerospace Structures Laboratory
Solving Beam Stiffness Assembly
Unknown displacement d can be solved from the beam stiffness
matrix equation P = Sd + Pf .
From d, we can get u by matching member end displacements.
From Q=ku+Q f to solve for Q , both in local coordinates.
Given the four components of Q we can further express the shear
forces and bending moments as functions of position along the
member. Member’s stress and strain can then be calculated from
these forces and moments
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A Worked Example1000lb
1250ft-lb
60ft40ft
E=29000ksi, A=11.8in2
, I=310in4
60ft40ft
1
2
3
4
5
6
Degrees of Freedom: Unknown Displacements 3,4
1 2
1 2 3
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A Worked Example (cont’d)member 1:
tracking with node number
corresponds to unknown d’s
L = 40’*12 in/ft = 480in, E=29000ksi, A=11.8in2, I=310in4
1 2 3 4
1
2
3
4
1 2 3 4
1
2
3
4
Warning: here only finite digits are displayed 23
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A Worked Example (cont’d)member 2:
tracking with node number
corresponds to unknown d’s
L = 60’*12 in/ft = 720in, E=29000ksi, A=11.8in2, I=310in4
3 4 5 6
3
4
5
6
3 4 5 6
3
4
5
6
Warning: here only finite digits are displayed 24
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A Worked Example (cont’d)
Assembly of Global Beam Stiffness Matrix
The rows and columns corresponding to the known displacementsare crossed out here to show us the structure stiffness matrix S.
1 2 3 4 5 6
12
34
56
corresponds to unknown d’s
K =
25
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A Worked Example (cont’d)
No fixed-end forces: Pf = 0
Warning: here only finite
digits are displayed
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A Worked Example (cont’d)
Now that the unknown displacements have been found, we simply plug
these values back into F-F f =KU to find the end reaction forces in the
global system
=
The big K from page 25
U: no other joint
can move except3&4 (page 22)
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Lab and Homework this Week:
Lab 3 – Beam Deflection and Analysis
Prelabs due in lab sessions Thursday Sept. 18 or Tuesday, Sept. 23
Homework 3 due online 11:59pm, Wednesday, Sept. 24
lab report due online 11:59pm, Friday, Sept. 26 (Thursday sections)
and Sunday, Sept. 28 (Tuesday sections) 28
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Reference
Aslam Kassimali, Matrix Analysis of Structures,
second edition, Cengage Learning, 2012, Ch. 5
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