DIRECT STIFFNESS METHOD FOR TRUSSES:

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• CHAPTER 3
• DIRECT STIFFNESS METHOD FOR TRUSSES
• 3.1 INTRODUCTION
• In the previous chapter the procedure for
  obtaining the structure stiffness matrix was
  discussed. The structure stiffness matrix was
  established by the following equation
• K TT kc T -----------(2.21)
• However if a large and complicated structure is
  to be analyzed and if more force components are
  to be included for an element then the size of
  composite stiffness matrix kc and deformation
  transformation matrix T will be increased.
  Therefore the procedure outlined in the preceding
  chapter for formation of structure stiffness
  matrix appears to be inefficient, furthermore
  this procedure is not suitable to automatic
  computation.

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• In this chapter an alternative procedure
  called the direct stiffness method is
  introduced. This procedure provides the basis for
  most computer programs to analyze structures. In
  this method each individual element is treated as
  a structure and structure stiffness matrix is
  obtained for this element using the relationship
• Km TTmkm Tm
  -----------(3.1)
• Where
• Km Structure stiffness matrix of an
  individual element.
• Tm Deformation Transformation matrix of an
  individual element.
• km Member Stiffness matrix of an individual
  element.
• Total structure stiffness matrix can be obtained
  by superimposing the structure stiffness matrices
  of the individual elements..

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• As all members of a truss are not in
  the same direction i.e. inclination of the
  longitudinal axes of the elements varies,
  therefore stiffness matrices are to be
  transformed from element coordinate system to
  structure or global coordinate system. When the
  matrices for all the truss elements have been
  formed then adding or combining together the
  stiffness matrices of the individual elements can
  generate the structure stiffness matrix K for
  the entire structure, because of these
  considerations two systems of coordinates are
  required.
• i) Local or member or element coordinate system
• In this coordinate system x-axis is
  collinear with the longitudinal axis of the
  element or member. Element stiffness is
  calculated with respect to this axis. This system
  is illustrated in figure 3.1

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ii) Structure or global coordinate system A
single coordinate system for the entire structure
is chosen, with respect to which stiffness of all
elements must be written. 3.2 PROCEDURE FOR THE
FORMATION OF TOTAL STRUCTURE STIFFNESS MATRIX FOR
AN ELEMENT USING DIRECT STIFFNESS METHOD
Following is the procedure for the
formation of structure stiffness matrix
i) Formation of the element stiffness matrix
using equation 2.16.
----------- (2.16) ii) Formation of the
deformation transformation matrix T for a
single element ?m Tm ?m
-----------(3.2)
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where ?m Element or member deformation
matrix. ?m Structure deformation matrix
of an element or member Tm Element or
member deformation transformation
matrix. (iii) Formation of a structure stiffness
matrix K or an element using the
relation Km TTmkm Tm
-----------(3.1)
The above mentioned procedure is discussed in
detail in the subsequent discussion. 3.2.1 The
formation of element stiffness matrix in local
co-ordinates It has already been discussed in
the previous chapter. However it is to be noted
that for horizontal members the structure
stiffness matrix and element stiffness matrix are
identical because both member coordinate systems
and structure coordinate systems are identical.
But for inclined members deformation
transformation matrices are to be used because
member coordinate system and structure coordinate
system are different therefore their structure
stiffness matrix and element stiffness matrix
will also be different.
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• 3.2.2 The formation of deformation
  transformation matrix
• As the main difference between the previously
  discussed method and direct stiffness method is
  the formation of the deformation transformation
  matrix. In this article deformation
  transformation matrix for a single element will
  be derived.
• Before the formation of deformation
  transformation matrix following conventions are
  to be established in order to identify joints,
  members, element and structure deformations.
• 1) The member is assigned a direction. An arrow
  is written along the member, with its head
  directed to the far end of the member.
• 2) ?i, ?j, ?k, and ?l, are the x and y structure
  deformations at near and far (tail head) ends
  of the member as shown in figure 3.2. These are
  positive in the right and upward direction.
• 3) ?r and ?s are the element deformations at near
  and far (tail head) ends of the member as shown
  in figure 3.2.
• 4)The member axis (x-axis of member coordinate
  system) makes an angle ?x, with the x-axis of the
  structure coordinate system as shown in figure
  3.2.
• 5)The member axis (x-axis of member coordinate
  system) makes an angle ?y, with the y-axis of the
  structure co-ordinate system as can be seen from
  figure 3.2.
• 6) The cosines of these angles are used in
  subsequent discussion Letters l and m represent
  these respectively.

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l Cos qx and m Cos qy l Cos qx
m Cos qy
The algebraic signs of ?s will be automatically
accounted for the members which are oriented in
other quadrants of X-Y plan.
In order to form deformation transformation
matrix, once again consider the member of a truss
shown in figure 3.1.
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Following four cases are considered to form
deformation transformation matrix. Case-1
Introduction of horizontal deformation to the
structure ?i 1, while far end of the member is
hinged (restrained against movement). From the
geometry of figure 3.3(a) ?r ?i Cos qx 1.
Cos qx Cos qx l ?r l ----------
(3.3) ?s 0 ---------- (3.4) Case-2
Introduction of vertical deformation to the
structure ?j 1, while near end of the member is
hinged (restrained against movement). From the
geometry of figure 3.3 (b) ?r ?j Sin qx ?j
Cos qy 1. Cos qy m ?r m ----------
(3.5) ?s 0 ---------- (3.6) Case-3
Introduction of horizontal deformation to the
structure ?k 1, while near end of the member is
hinged (restrained against movement).From the
geometry of figure 3.3(c) ?s ?k Cos qx 1.
Cos qx Cos qx l ?r 0 ----------
(3.7) ?s l ---------- (3.8)
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• Case-4 Introduction of vertical deformation to
  the structure DL 1, while near end is hinged
• From the geometry of figure 3.3(d)
• ?L 1
• ?r 0 ---------- (3.9)
• ?s ? L
• Cos qy m ---------- (3.10)
• Effect of four structure deformations and two
  member deformations can be written as
• ?r ?i Cos qx ?j Cos qy ?k .0 ? L .0
• ?s ?i .0 ?j .0 ?k Cos qx ? L Cos qy

?r l.?i m.?j 0.?k 0. ? L -----------
(3.11) ?s 0.?i 0.?j l.?k m. ? l
----------- (3.12) Arranging equations
3.11 and 3.12 in matrix from
----------- (3.13)
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Comparing this equation with the following
equation ?m Tm ?m ---------------(3.2
) After comparing equation (3.14) and (3.2)
following deformation transformation matrix is
obtained.
-
---------- (3.14) This matrix Tm transforms
four structure deformation (?i, ?j, ?k, ?L ) into
two element deformation (?r and ?s).
3.2.3 Formation of structure stiffness
matrix Structure Stiffness matrix of an
individual member is first to be transformed from
member to structure coordinates. This can be done
by using equation 3.1. K TTm km
Tm --------(3.1)
----- (3.15)
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------(2.16)
--------------------------------(3.16)
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• 3.2.4 ASSEMBLING OF THE INDIVIDUAL STRUCTURE
  ELEMENT STIFFNESS MATRICES TO FORM TOTAL
  STRUCTURE STIFFNESS MATRIX
• Combining the stiffness matrix of the individual
  members can generate the stiffness matrix of a
  structure However the combining process should be
  carried out by identifying the truss joint so
  that matrix elements associated at particular
  member stiffness matrices are combined. The
  procedure of formation of structure stiffness
  matrix is as follows
• Step 1 Form the individual element stiffness
  matrices for each member.
• Step 2 Form a square matrix, whose order should
  be equal to that of structure deformations.
• Step 3 Place the elements of each individual
  element stiffness matrix framed into structure in
  the corresponding rows and columns of structure
  stiffness matrix of step-2.
• Step 4 If more than one element are to be placed
  in the same location of the structure stiffness
  matrix then those elements will be added.

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Following figure shows the above mentioned
process.
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In the above matrix the element in the second
row and second column is k1k2 where k1 is for
member 1 and k2 is for member 2.This is because
both k1 and k2 have structure coordinates 2-2.
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Illustrative Examples Regarding the Formation of
K Matrix
Example 3.1 Form the structure stiffness matrix
for the following
truss by direct stiffness method.
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Member Length Near End X1 Y1 Near End X1 Y1 Far End X2, Y2 Far End X2, Y2 i j k L lCos?x X2-X1/ Length mCos?y Y2-Y1/ Length
1 5L/6 0 0 L/2 2L/3 5 6 1 2 0.6 0.8
2 5L/6 L/2 2L/3 L 0 1 2 3 4 0.6 -0.8
3 L L 0 0 0 3 4 5 6 -1 0
According to eq. (3.16) we get
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Example 3.2 Form the structure stiffness matrix
of the following truss.
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Member Near End Near End Far End Far End Length lCosqx mCosqy i j k L
Member X1 Y1 X2 Y2 Length lCosqx mCosqy i j k L
1 0 0 L L ?2L 0.707 0.707 7 8 1 2
2 L L L 0 L 0 -1 1 2 5 6
3 L L 2L 0 ?2L 0.707 -0.707 1 2 3 4
So
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Now the matrix can be written as k2
So k k1 k2 k3
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• 3.4 Force transformation matrix
• The axial force (wm) in the members of a truss
  can be found by using the relationship between
  member forces and member deformations equation
  (2.15) and between element and structure
  deformations equation (3.1)
• wm km ?m ------------ (2.15)
• ?m Tm ?m ------------ (3.3)
• Substituting value of ?m from equation 3.1 into
  equation 2.15
• wm km Tm ?m ------------ (3.18)
• Substituting value of km and Tm from equation
  2.16 and 3.15 respectively

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In above Equations, wr Force at near end
See figure below
ws Force at far end
As wr -ws so
-------------- (3.19)

• Sign conventions
• wr ws

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• If
• wr is positive then member is in compression.
• ws is negative then member is in compression.
• wr is negative then member is in tension.
• ws is positive then member is in tension.
• 3.5 ANALYSIS OF TRUSSES USING DIRECT STIFFNESS
  METHOD
• Basing on the derivations in the preceding
  sections of this chapter a truss can be
  completely analyzed. The analysis comprises of
  determining.
• i) Joint deformations.
• ii) Support reactions.
• iii) Internal member forces.
• As the first step in the analysis is the
  determination of unknown joint deformation. Using
  the equations can do this.
• W K ?
• The matrices W, K and ? can be divided in
  submatrices in the following form

----------- (3.20)
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• Where
• Wk Known values of loads at joints.
• Wu Unknown support reaction.
• ?u Unknown joint deformation.
• ?k Known deformations, generally zero due to
  support conditions.
• K11, K12, K21, K22 are the
  sub-matrices of K
• Expanding equation 3.20
• Wk K11 ?u K12 ?k ----------- (3.21)
• Wu K21 ?u K22 ?k ----------- (3.22)
• If the supports do not move, then ?k 0
  therefore equation 3.21 3.22 can be written as
• Wk K11 ?u ----------- (3.23)
• Wu K21 ?u ----------- (3.24)
• By pre-multiplying equation 3.23 by K11-1
  following equation is obtained
• K11-1 WKK11-1
  K11DU
• ?u K11-1 Wk
  ----------- (3.25)
• Substituting value ?u from equation 3.25 into
  equation 3.24
• Wu K21 K11-1 Wk
  ----------- (3.26)

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• Using equation 3.25, 3.26 and 3.19, joint
  deformations, support reactions and internal
  member forces can be determined respectively.
• As this method does not depend upon degree of
  indeterminacy so it can be used for determinate
  as well as indeterminate structures.
• Using the basic concepts as discussed in the
  previous pages, following are the necessary steps
  for the analysis of the truss using stiffness
  method.
• 1-Identify the separate elements of the structure
  numerically and specify near end and far end of
  the member by directing an arrow along the length
  of the member with head directed to the far end
  as shown in the fig. 3.2
• 2-Establish the x,y structure co-ordinate
  system. Origin be located at one of the joints.
  Identify all nodal co-ordinates by numbers and
  specify two different numbers for each joint (one
  for x and one for y). First number the joints
  with unknown displacements.
• 3-Form structure stiffness matrix for each
  element using equation 3.16.
• 4-Form the total structure stiffness matrix by
  superposition of the element stiffness matrices.
• 5-Get values of unknown displacements using
  equation 3.25.
• 6-Determine support reaction using equation 3.26.
• 7-Compute element or member forces using equation
  3.19.

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3.6- Illustrative Examples Regarding Complete
Analysis of Trusses Example 3.3 Solve truss in
example 3.1 to find member forces.
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The K matrix for the truss as formed in example
3.1 is
Using the relation W K ? we get
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From equation (3.25) ?u K11-1 Wk
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Using the relation Wu K21 ?u
Now the member forces can be found using the
relation
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kips
kips
kips
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Final Results sketch
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Example 3.4
Solve the truss in example 3.2 to find member
forces.
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The K matrix for the truss as formed in example
3.2 is as follows
Now as W k?
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As ?k 0 or ?3, ?4, ?5, ?6, ?7,?8, all are zero
due to supports and using
Wk k11 ?u we get
Which gives
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W1 0 W2 P Also using Wu k21 ?u we
have
Or

Or
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Or
Or
Now Wu K21 ?u
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W3 0.207 P W4 -0.207 P W5 0 W6 -0.5858
P W7 -0.207 P W8 -0.207 P
Now as
So
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So, for member 1
Or ws 0.707 x 0.5858 P w1 0.414 P Similarly,
for member 2
Or w2 1 x 0.5858 P Or w2 0.5858 P
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For member 3
Or w3 0.707 x 0.5858 P w3 0.414 P
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w1 0.414 P w2 0.586 P w3 0.414 P
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Example 3.5 Analyze the truss shown in the
figure.

A and E are constant for each member.
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Member Length l m i j k l
1 10 -0.6 -0.8 7 8 1 2
2 11.34 -0.7071 0.7071 1 2 3 4
3 8 1 0 3 4 5 6
4 6 1 0 5 6 7 8
5 8 0 1 1 2 5 6
Using the properties given in the above table we
can find the structures stiffness matrices for
each element as follows.
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Using equation K K1K2K3K4K5 we
get the structure stiffness matrix of (8x8)
dimensions. Partitioning this matrix with respect
to known and unknown deformations we get K11
and K12 portions as follows.
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Using equation DuK111Wk we get the
Putting the above value in equation
WuK21Du
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Element forces can be calculated using equation
as follows.
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