Structural Analysis

1
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] = [T]T [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.

2
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:
[K]m = [T]T
m[k]m [T]m -----------(3.1)
Where
[K]m = Structure stiffness matrix of an individual element.
[T]m = Deformation Transformation matrix of an
individual element.
[k]m = Member Stiffness matrix of an individual element.
Total structure stiffness matrix can be obtained by superimposing the
structure stiffness matrices of the individual elements..

3
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

4
Element 'B'Element 'A'
A B
Structure or
globel axes
Local or member
or element axes for
element B
Local or member
or element axes
for element 'A'
Figure 3.1

5
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.
 













11
11
L
AE
k m
----------- (2.16)
ii) Formation of the deformation transformation
matrix [T] for a single element:
[]m = [T]m []m -----------(3.2)

6
where
[]m= Element or member deformation matrix.
[]m= Structure deformation matrix of an element or member
[T]m= Element or member deformation transformation matrix.
(iii)Formation of a structure stiffness matrix [K] or an element using the
relation
[K]m = [T]T
m[k]m [T]m -----------(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.

7
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.

8Figure 3.2
Near
end
Far
end
Member
direction

x
y
 (b)
r
s


i
j
k

Near
end
Far
end
Member
direction





(a)
x
y
l
i
j
k
r
s

9
   2
12
2
12
1212
YYXX
XX
L
XX




 = Cos x and m = Cos y
 = Cos x =
m = Cos y =
   2
12
2
12
1212
YYXX
YY
L
YY




The algebraic signs of θ‘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.

10
s=0
r
j
i
s=0
j

r
=1
(b)(a) =1j
rjcosy=mr
cosxl
x
i
i
i
y
x

11

r=0


r=0
y
x
Figure 3.3
(d)(c)
=1k
=1
s1cosy=m
1
l
kk
skcosx=
1
s
s
x
y

12
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 x = 1. Cos x = Cos x = 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 x = j Cos y = 1. Cos y = 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 x = 1. Cos x = Cos x = l
r = 0 ---------- (3.7)
s = l ---------- (3.8)

13
Case-4: Introduction of vertical deformation to the structure L = 1, while
near end is hinged
From the geometry of figure 3.3(d)
L = 1
r = 0 ---------- (3.9)
s =  L
Cos y = m ---------- (3.10)
Effect of four structure deformations and two member deformations can be
written as
r = i Cos x + j Cos y + k .0 +  L .0
s = i .0 + j .0 + k Cos x +  L Cos y
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























L
k
j
i
s
r
m
m


00
00

 ----------- (3.13)

14
Comparing this equation with the following equation
[]m = [T]m []m ---------------(3.2)
After comparing equation (3.14) and (3.2) following deformation
transformation matrix is obtained.
  



m
mT m 

00
00- ---------- (3.14)
This matrix [T]m 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] = [T]T
m [k]m [T]m --------(3.1)
  






ml
ml
mT
00
00
----- (3.15)

15
  





11
11
L
AE
k m
------(2.16)
 
Lkji
m
m
m
m
L
k
j
i
K m







































00
00
11
11
0
0
0
0
 
Lkji
mm
mm
L
AE
m
m
L
k
j
i
K m































0
0
0
0
 
L
k
j
i
Lkji
mmmm
mm
mmmm
mm
L
AE
K m





















22
22
22
22




--------------------------------(3.16)

16
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.

17
1
2 3
2
3
2
1
Structure
stiffness
matrix for
member-2
Structure
stiffness
matrix for
member-1
Structure
stiffness matrix
Following figure shows the above mentioned process.

18
 
2
1
21
kk
kk
K
11
11
1 







  
3
2
32
kk
kk
K
22
22
2 








     21 KKK 
 
3
2
1
kk
kkkk
kk
K
321














22
2211
11
0
0
In the above matrix the element in the second row and
second column is k1+k2 where k1 is for member 1 and k2
is for member 2.This is because both k1 and k2 have
structure coordinates 2-2.

19
Illustrative Examples Regarding the Formation of [K] Matrix:
L/2
(x1,y1) = (0,0)
(x1,y1) = (L/2,2L/3)(x2,y2) = (L/2,2L/3)
(x2,y2) = (L,0)
(x1,y1) = (L,0)(x2,y2) = (0,0)
(a)
(b)
(c)
(Structure forces and deformations)
(Free body diagram indicating structure forces
and coordinates)
2L/3
1 1
2 2
6 4
5 3
6 4
5 3
1
5
6
3
4
2
12
11.5
1
1
2
23
3
L/2
Example 3.1: Form the structure stiffness matrix for the
following truss by direct stiffness method.

20
Member Length Near
End
X1 Y1
Far End
X2, Y2
i j k L l=Cosx =
X2-X1/
Length
m=Cos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
 
l
k
j
i
mlmmlm
lmllml
mlmmlm
lmllml
lkji
L
AE
K m













22
22
22
22
According to eq. (3.16) we get

21
 
2
1
6
5
768.0576.0768.0576.0
576.0432.0576.0432.0
768.0576.0768.0576.0
576.0432.0576.0432.0
2165
1 










L
AE
K
 
4
3
2
1
768.0576.0768.0576.0
576.0432.0576.0432.0
768.0576.0768.0576.0
576.0432.0576.0432.0
4321
2 










L
AE
K
 
6
5
4
3
0000
000.1000.1
0000
000.1000.1
6543











L
AE
3K

22
 
6
5
4
3
2
1
768.0576.0000.0000.0768.0576.0
576.0432.1000.0000.1576.0432.0
000.0000.0768.0576.0768.0576.0
000.0000.1576.0432.1576.0432.0
768.0576.0768.0576.0536.1000.0
576.0432.0576.0432.0000.0864.0
654321



















L
AE
K

23
Example 3.2: Form the structure stiffness matrix of the following truss.
L L
P
1
2
3
4
5
6
7
8
(L,L)
(0,0)
(L,0)
(2L,0)
(L,L)
(L,L)
y
x
1
2
32 2
1
1
3
4
6
5
7
8
1
2
Free body diagram showing structure forces and
coordinates
Structure forces and deformations
Structure to be analysed
L
1
2
3

24
Member Near End Far End Length =Cosx m=Cosy i j k L
X1 Y1 X2 Y2
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
 













22
22
22
22
mlmmlm
lmllml
mlmmlm
lmllml
L
AE
mK
So

25
 
2
1
8
7
0.50.5.5.5
0.50.5.5.5
.5.50.50.5
.5.50.50.5
2187
1













2L
AE
K
 
6
5
2
1
1010
0000
1010
0000
6521
2











L
AE
K  
4
3
2
1
.5.5.5.5
.5.5.5.5
.5.5.5.5
.5.5.5.5
3
4321













2L
AE
K
Now the matrix can be written as [k]2:
 
6
5
2
1
414.10414.10
0000
414.10414.10
0000
6521
2
2











L
AE
K  
2
1
8
7
5.5.5.5.
5.5.5.5.
5.5.5.5.
5.5.5.5.
2187
2
1













L
AE
K
So [k] = [k]1 + [k]2 + [k]3

26
 
8
7
6
5
4
3
2
1
5.5.00005.5.
5.5.00005.5.
00414.1000414.10
00000000
00005.5.5.5.
00005.5.5.5.
5.5.414.105.5.414.20
5.5.005.05.01
87654321
2
























L
AE
K

27
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)]
[w]m= [k]m []m ------------ (2.15)
[]m = [T]m []m ------------ (3.3)
Substituting value of []m from equation 3.1 into equation
2.15
[w]m = [k]m [T]m []m ------------ (3.18)
Substituting value of [k]m and [T]m from equation 2.16 and
3.15 respectively
 




























L
k
j
i
mm
mm
L
AE
mw
11
11



28



























L
k
j
i
mm
mm
L
AE
ws
rw


In above Equations, wr= Force at near end
See figure below:
ws = Force at far end
As wr = -ws so:
   

















l
k
j
i
s
mlml
L
AE
w -------------- (3.19)
Sign conventions
wr ws

29
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














k
u
u
k
KK
KK
W
W
2221
1211
----------- (3.20)

30
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 [WK]=[K11]-1 [K11][U]
[u] = [K11]-1 [Wk] ----------- (3.25)
Substituting value u from equation 3.25 into equation 3.24
[Wu] = [K21] [K11]-1 [Wk] ----------- (3.26)

31
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.

32
3.6- Illustrative Examples Regarding Complete Analysis of Trusses:
Example 3.3: Solve truss in example 3.1 to find member forces.
L/2
(x1,y1) = (0,0)
(x1,y1) = (L/2,2L/3)(x2,y2) = (L/2,2L/3)
(x2,y2) = (L,0)
(x1,y1) = (L,0)(x2,y2) = (0,0)
(a)
(b)
(c)
(Free body diagram indicating structure forces and
coordinates)
2L/3
1 1
2 2
6 4
5 3
6 4
5 3
1
5
6
3
4
2
12
11.5
1
1
2
23
3

33
The [K] matrix for the truss as formed in example 3.1 is:
 
6
5
4
3
2
1
768.0576.0000.0000.0768.0576.0
576.0432.1000.0000.1576.0432.0
000.0000.0768.0576.0768.0576.0
000.0000.1576.0432.1576.0432.0
768.0576.0768.0576.0536.1000.0
576.0432.0576.0432.0000.0864.0
654321



















L
AE
K
Using the relation [W] = [K] [] we get




















k
u
u
k
KK
KK
W
W
2221
1211

34



































































6
5
4
3
2
1
654321
6
5
4
3
2
1
768.0576.0000.0000.0768.0576.0
576.0432.1000.0000.1576.0432.0
000.0000.0768.0576.0768.0576.0
000.0000.1576.0432.1576.0432.0
768.0576.0768.0576.0536.1000.0
576.0432.0576.0432.0000.0864.0
L
AE
W
W
W
W
W
W
From equation (3.25) [u] = [K11]-1 [Wk]





















2
1
536.1000.0
000.0864.0
2
1
1
W
W
AE
L































12
5.11
651.0000.0
000.0157.1
2
1
AE
L














812.7
31.13
2
1
AE
L

35
Using the relation:
[Wu] = [K21] [u]






































812.7
310.13
768.0576.0
576.0432.0
768.0576.0
576.0432.0
6
5
4
3
W
W
W
W






























1.66
1.25
13.66
10.25
6W
5W
4W
3W
Now the member forces can be found using the relation:

36
   

















L
k
j
i
mm
L
AE
wr 
    09.2
812.7
31.13
00.0
00.0
8.06.08.06.0
5
6
1 














AE
L
L
AE
w kips
    09.17
00.0
00.0
812.7
31.13
8.06.08.06.0
5
6
2 














AE
L
L
AE
w kips
    00.0
00.0
00.0
00.0
00.0
0.00.10.00.13 













AE
L
L
AE
w
kips

37
11.5k
12k
1.25k 10.25k
13.66k
17.09kip
2.09kips
1.66k
0 kips
Final Results & sketch

38
Example 3.4:
L L
P
1
2
3
4
5
6
7
8
(L,L)
(0,0)
(L,0)
(2L,0)
(L,L)
(L,L)
y
x
1
2
32 2
1
1
3
4
6
5
7
8
1
2
(Structure to be analysed)
L
1
2
3
(Free body diagram showing structure forces and
coordinates)
Solve the truss in example 3.2 to find member forces.

39
The [K] matrix for the truss as formed in example 3.2 is as follows:
 
8
7
6
5
4
3
2
1
5.5.00005.5.
5.5.00005.5.
00414.1000414.10
00000000
00005.5.5.5.
00005.5.5.5.
5.5.414.105.5.414.20
5.5.005.05.01
87654321
2


























L
AE
K
Now as W = k

40




















































































8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
5.5.00005.5.
5.5.00005.5.
00414.1000414.10
00000000
00005.5.5.5.
00005.5.5.5.
5.5.414.105.5.414.20
5.5.005.05.01
87654321
2L
AE
W
W
W
W
W
W
W
W




















k
u
u
k
kk
kk
W
W
2221
1211
As k = 0 or 3, 4, 5, 6, 7,8, all are zero due to supports and using
[Wk] = [k11] [u] we get
W
W
AE
L
1
2
1
22
1 0
0 2 414





 











.


Which gives

41
W1 = 0
W2 = P
Also using [Wu] = [k21] [u] we have
0
2
1 0
0 2 414
1
2P
AE
L





 











.


Or





















PAE
L 0
414.20
012
1
2
1













2 1
2 414
2 414 0
0 1
0L
AE P
.
.
.
Or

42


1
2
2
2 414
0




 






L
AE P.
Or
2
2
2 414

LP
AE.
Or
2 05858 .
PL
AE
Now Wu = [K21] [u]
W
W
W
W
W
W
AE
L
3
4
5
6
7
8
1
22
5 5
5 5
0 0
0 1414
5 5
5 5
























 
 


























. .
. .
.
. .
. .


W
W
W
W
W
W
P
3
4
5
6
7
8
1
2 414
5 5
5 5
0 0
0 1414
5 5
5 5
0
























 
 


























.
. .
. .
.
. .
. .

43
W
W
W
W
W
W
P
P
P
P
P
3
4
5
6
7
8
1
2 414
05
05
0
1414
05
05













































.
.
.
.
.
.
W3 = 0.207 P
W4 = -0.207 P
W5 = 0
W6 = -0.5858 P
W7 = -0.207 P
W8 = -0.207 P
Now as






























l
k
j
i
s
r
mlml
mlml
L
AE
w
w
So
 
2
1
8
7

















l
k
j
s
i
mlml
L
AE
w

44
So, for member # 1:
 





















AE
L
P
L
AE
wsw
5858.0
0
0
0
707.0707.0707.0707.01
Or
ws= 0.707 x 0.5858 P
w1= 0.414 P
Similarly, for member # 2:
 
6
5
2
1
0
0
5858.
0
10102














 AE
PL
L
AE
wws
Or w2 = 1 x 0.5858 P
Or w2 = 0.5858 P

45
For member # 3:
 
4
3
2
1

















l
k
j
s
i
mlml
L
AE
W
 















0
0
5858.
0
707.707.707.707.3 AE
PL
L
AE
w
Or
w3= 0.707 x 0.5858 P
w3= 0.414 P

46
P
0.207P 0.59P 0.207P
0.207P
0.207P
0.414P 0.414P
0.586P
w1 = 0.414 P
w2 = 0.586 P
w3 = 0.414 P

47
Example 3.5 Analyze the truss shown in the figure.
5k
2kW1,1
W7,7
W8,8
W5,5
W6,6
W3,3
W4,4
W2,2
12
3
4
5
8' 6'
8'
Truss to be analysedStructure forces and
displacements
A and E are constant for each member.

48
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.

49
Element forces and displacements
w , 5
w , 4
5
4
w , 
w , 
2
3 3
w , 9 9
6
w , 
3
6
w , 
5
w , 10
7
10
7
1
2 2
w , 
w , 
4
8
1
8
1

50
6
5
4
3
0000
0125.00125.0
0000
0125.00125.0
k
6543
4
3
2
1
044.0044.0044.0044.0
044.0044.0044.0044.0
044.0044.0044.0044.0
044.0044.0044.0044.0
k
4321
2
1
8
7
064.0048.0064.0048.0
048.0036.0048.0036.0
064.0048.0064.0048.0
048.0036.0048.0036.0
k
2187
3
2
1

















































51
8
7
4
3
125.00125.00
0000
125.00125.00
0000
k
8743
8
7
6
5
0000
0167.00167.0
0000
0167.00167.0
k
8765
5
4






























Using equation [K] = [K1]+[K2]+[K3]+[K4]+[K5] 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.

52




























665.9
983.62
2
5
233.0004.0
004.008.0
1
2
1
 
2
1
233.0004.0
004.008.0
21
11 





K
 
8
7
6
5
4
3
064.0048.0
048.0036.0
125.00
00
044.0044.0
044.0004.0
21
12

























K
Using equation [Du]=[K11]–1[Wk] we get the
Putting the above value in equation [Wu]=[K21][Du]

53












































































405.2
803.1
208.1
0
197.3
197.3
665.9
983.62
064.0048.0
048.0036.0
125.00
00
044.0044.0
044.0004.0
6
5
4
3
2
1
W
W
W
W
W
W
Element forces can be calculated using equation as follows.

54
4.54 kips
3.197 K
3.197 K
1.208kips
-3.006kips
2 K
5 K
1.803 K
2.405 K
0 K
1.208 K
 
 
 
 
  kipw
kipw
kipw
kipw
kipw
208.1
0
0
665.9
983.62
1010
8
1
0
0
0
0
0
0101
6
1
0
0
0
0
0
0101
8
1
540.4
0
0
665.9
983.62
707.0707.0707.0707.0
314.11
1
006.3
665.9
983.62
0
0
8.06.08.06.0
10
1
5
4
3
2
1










































































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