m4l26 Temperature and Errors

Prévia do material em texto

Version 2 CE IIT, Kharagpur 
 
 
 
 
 
 
 
 
 
Module 
4 
 
Analysis of Statically 
Indeterminate 
Structures by the Direct 
Stiffness Method 
Version 2 CE IIT, Kharagpur 
 
 
 
 
 
 
Lesson 
26 
The Direct Stiffness 
Method: Temperature 
Changes and 
Fabrication Errors in 
Truss Analysis 
 
 
Version 2 CE IIT, Kharagpur 
 
Instructional Objectives 
After reading this chapter the student will be able to 
1. Compute stresses developed in the truss members due to temperature 
changes. 
2. Compute stresses developed in truss members due to fabrication members. 
3. Compute reactions in plane truss due to temperature changes and fabrication 
errors. 
 
 
26.1 Introduction 
In the last four lessons, the direct stiffness method as applied to the truss 
analysis was discussed. Assembly of member stiffness matrices, imposition of 
boundary conditions, and the problem of inclined supports were discussed. Due 
to the change in temperature the truss members either expand or shrink. 
However, in the case of statically indeterminate trusses, the length of the 
members is prevented from either expansion or contraction. Thus, the stresses 
are developed in the members due to changes in temperature. Similarly the error 
in fabricating truss members also produces additional stresses in the trusses. 
Both these effects can be easily accounted for in the stiffness analysis. 
 
 
26.2 Temperature Effects and Fabrication Errors 
 
Version 2 CE IIT, Kharagpur 
 
 
Consider truss member of length L, area of cross section A as shown in 
Fig.26.1.The change in length lΔ is given by 
 
TLl Δ=Δ α (26.1) 
 
where α is the coefficient of thermal expansion of the material considered. If the 
member is not allowed to change its length (as in the case of statically 
indeterminate truss) the change in temperature will induce additional forces in the 
member. As the truss element is a one dimensional element in the local 
coordinate system, the thermal load can be easily calculated in global co-
ordinate system by 
 
1( )tp AE L′ = Δ (26.2a) 
 
2( )tp AE L′ = − Δ (26.2b) 
 
or 
 
( ){ }
⎭⎬
⎫
⎩⎨
⎧
−
+Δ=
1
1' LAEp t (26.3) 
 
The equation (26.3) can also be used to calculate forces developed in the truss 
member in the local coordinate system due to fabrication error. LΔ will be 
considered positive if the member is too long. The forces in the local coordinate 
system can be transformed to global coordinate system by using the equation, 
 
( )
( )
( )
( )
( )( ) ⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
t
t
t
t
t
t
p
p
p
p
p
p
'
2
'
1
4
3
2
1
sin0
cos0
0sin
0cos
θ
θ
θ
θ
 (26.4a) 
 
where ( ) ( )tt pp 21 , and ( ) ( )tt pp 43 , are the forces in the global coordinate system at 
nodes 1 and 2 of the truss member respectively Using equation (26.3), the 
equation (26.4a) may be written as, 
 
1
2
3
4
( ) cos
( ) sin
( ) cos
( ) sin
t
t
t
t
p
p
AE L
p
p
θ
θ
θ
θ
⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪= Δ⎨ ⎬ ⎨ ⎬−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎩ ⎭⎩ ⎭
 (26.4b) 
Version 2 CE IIT, Kharagpur 
 
 
The force displacement equation for the entire truss may be written as, 
 { } [ ]{ } { }tpukp )(+= (26.5) 
 
where ,{ }p is the vector of external joint loads applied on the truss and ( ){ }tp is the 
vector of joint loads developed in the truss due to change in 
temperature/fabrication error of one or more members. As pointed out earlier. in 
the truss analysis, some joint displacements are known due to boundary 
conditions and some joint loads are known as they are applied 
externally.Thus,one could partition the above equation as, 
 
[ ] [ ]
[ ] [ ]
{ }
{ }
( )
( ) ⎪⎭
⎪⎬⎫⎪⎩
⎪⎨⎧+⎭⎬
⎫
⎩⎨
⎧⎥⎦
⎤⎢⎣
⎡=
⎭⎬
⎫
⎩⎨
⎧
tu
tk
k
u
u
k
p
p
u
u
kk
kk
p
p
2221
1211 (26.6) 
 
where subscript u is used to denote unknown quantities and subscript k is used 
to denote known quantities of forces and displacements. Expanding equation 
(26.6), 
 { } [ ]{ } [ ]{ } ( ){ }tkkuk pukukp ++= 1211 (26.7a) 
{ } [ ]{ } [ ]{ } ( ){ }21 22u u k u tp k u k u p= + + (26.7b) 
 
If the known displacement vector { } { }0=ku then using equation (26.2a) the 
unknown displacements can be calculated as 
 
{ } [ ] { } ( ){ }( )tkku ppku −= −111 (26.8a) 
If { } 0≠ku then 
{ } [ ] { } [ ]{ } ( ){ }( )tkkkuu pukpku −−= − 121 (26.8b) 
 
After evaluating unknown displacements, the unknown force vectors are 
calculated using equation (26.7b).After evaluating displacements, the member 
forces in the local coordinate system for each member are evaluated by, 
 { } [ ][ ]{ } { }tpuTkp ′+′=′ (26.9a) 
or 
( )( ) ⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧+
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
−
−=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
t
t
p
p
v
u
v
u
L
AE
p
p
'
2
'
1
2
2
1
1
'
2
'
1
sincos00
00sincos
11
11
θθ
θθ
 
Version 2 CE IIT, Kharagpur 
 
Expanding the above equation, yields 
 
{ } { } LAE
v
u
v
u
L
AEp Δ+
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=′
2
2
1
1
1 sincossincos θθθθ (26.10a) 
And, 
 
{ } { }
1
1
2
2
2
cos sin cos sin
u
vAEp AE L
uL
v
θ θ θ θ
⎧ ⎫⎪ ⎪⎪ ⎪′ = − − − Δ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭
 (26.10b) 
Few problems are solved to illustrate the application of the above procedure to 
calculate thermal effects /fabrication errors in the truss analysis:- 
 
Example 26.1 
Analyze the truss shown in Fig.26.2a, if the temperature of the member (2) is 
raised by Co40 .The sectional areas of members in square centimeters are 
shown in the figure. Assume 25 /102 mmNE ×= and 1/ 75,000α = per Co . 
 
 
Version 2 CE IIT, Kharagpur 
 
 
The numbering of joints and members are shown in Fig.26.2b. The possible 
global displacement degrees of freedom are also shown in the figure. Note that 
lower numbers are used to indicate unconstrained degrees of freedom. From the 
figure it is obvious that the displacements 0876543 ====== uuuuuu due to 
boundary conditions. 
The temperature of the member (2) has been raised by Co40 . Thus, 
 
TLL Δ=Δ α 
( )( ) 3102627.24023
75000
1 −×==ΔL m (1) 
 
The forces in member (2) due to rise in temperature in global coordinate system 
can be calculated using equation (26.4b).Thus, 
 
( )
( )
( ) ⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
−Δ=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
θ
θ
θ
θ
sin
cos
sin
cos)(
2
1
6
5
LAE
p
p
p
p
t
t
t
t
 (2) 
 
For member (2), 
 
242 102020 mcmA −×== and o45=θ 
Version 2 CE IIT, Kharagpur 
 
( )
( )
( )
( )
5
6 4 11 3 3
1
2
1
2
1
220 10 2 10 2.2627 10 /10
1
2
1
2
t
t
t
t
p
p
p
p
− −
⎧ ⎫⎪ ⎪⎪ ⎪⎧ ⎫ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪= × × × × ×⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎪ ⎪⎪ ⎪−⎪ ⎪⎩ ⎭
 (3) 
 
 
( )
( )
( )
( )
5
6
1
2
1
1
150.82
1
1
t
t
t
t
p
p
kN
p
p
⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎩ ⎭⎩ ⎭
 (4) 
 
In the next step, write stiffness matrix of each member in global coordinate 
system and assemblethem to obtain global stiffness matrix 
 
Element (1): 240 1015,3,0 mAmL −×===θ ,nodal points 4-1 
 
4 11
'
3
1 0 1 0
0 0 0 015 10 2 10
1 0 1 03 10
0 0 0 0
k
−
−⎡ ⎤⎢ ⎥× × × ⎢ ⎥⎡ ⎤ =⎣ ⎦ ⎢ ⎥−× ⎢ ⎥⎣ ⎦
 (5) 
 
Member (2): 241020,23,45 mAmL −×=== oθ , nodal points 3-1 
 
[ ]
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−−
−−
−−
−−
×××=
−
5.05.05.05.0
5.05.05.05.0
5.05.05.05.0
5.05.05.05.0
23
1021020 1142k (6) 
 
Member (3): ,30,1015,90 24 mLmA =×== −oθ nodal points 2-1 
 
Version 2 CE IIT, Kharagpur 
 
4 11
3
3 3
0 0 0 0
0 1 0 115 10 2 10
0 0 0 03 x 10 10
0 1 0 1
k
−
⎡ ⎤⎢ ⎥−× × × ⎢ ⎥⎡ ⎤ =⎣ ⎦ ⎢ ⎥× ⎢ ⎥−⎣ ⎦
 (7) 
 
The global stiffness matrix is of the order 88× ,assembling the three member 
stiffness matrices, one gets 
 
[ ]
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−
−−
−−
−
−−−
−−−
=
00000000
010000000100
0014.4714.470014.4714.47
0014.4714.470014.4714.47
000010001000
00000000
0014.4714.47100014.14714.47
010014.4714.470014.4714.147
103k (8) 
 
Writing the load displacement equation for the truss 
 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−
−−
−−
−
−−−
−−−
00000000
010000000100
0014.4714.470014.4714.47
0014.4714.470014.4714.47
000010001000
00000000
0014.4714.47100014.14714.47
010014.4714.470014.4714.147
103
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
0
0
1
1
0
0
1
1
640 
(9) 
 
In the present case, the displacements 1u and 2u are not known. All other 
displacements are zero. Also 021 == pp (as no joint loads are applied).Thus, 
 
Version 2 CE IIT, Kharagpur 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
8
7
6
5
4
3
2
1
p
p
p
p
p
p
p
p
=
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−
−−
−−
−
−−−
−−−
=
00000000
010000000100
0014.4714.470014.4714.47
0014.4714.470014.4714.47
000010001000
00000000
0014.4714.47100014.14714.47
010014.4714.470014.4714.147
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
+
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
0
0
1
1
0
0
1
1
640 (10) 
 
Thus unknown displacements are 
 
1
1
3
2
147.14 47.14 0 11 ( 150.82 )
47.14 147.14 0 110
u
u
− −⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫= −⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ −⎣ ⎦ ⎩ ⎭ ⎩ ⎭⎩ ⎭ (11) 
 
4
1
4
2
7.763 10
7.763 10
u m
u m
−
−
= ×
= × 
 
Now reactions are calculated as 
 
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
+
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
+
⎭⎬
⎫
⎩⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−
−−
−
=
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
0
0
1
1
0
0
640
0
0
0
0
0
0
000000
01000000
0014.4714.4700
0014.4714.4700
00001000
000000
00
0100
14.4714.47
14.4714.47
1000
00
10
2
13
8
7
6
5
4
3
u
u
p
p
p
p
p
p
 
 
3
4
5
6
7
8
0
77.63
77.63
77.63
77.63
0
p
p
p
p
p
p
⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭
kN (12) 
 
Version 2 CE IIT, Kharagpur 
 
 
 
The support reactions are shown in Fig.26.2c.The member forces can be easily 
calculated from reactions. The member end forces can also be calculated by 
using equation (26.10a) and (26.10b). For example, for member (1), 
o0=θ 
100103'2 ×=p [ ]0101− 4
4
0
0
7.763 10
7.763 10
−
−
⎧ ⎫⎪ ⎪⎪ ⎪⎨ ⎬×⎪ ⎪⎪ ⎪×⎩ ⎭
 (13) 
 = 77.763 kN. Thus the member (1) is in tension. 
 
Member (2) 
o45=θ 
281.94103'2 ×=p [-0.707 -0.707 0.707 0.707]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
×
×
−
−
3
3
102942.3
102942.3
0
0
 
kN. 78.1092 −=′p 
 
Thus member (2) is in compression 
 
 
 
Version 2 CE IIT, Kharagpur 
 
Example 26.2 
Analyze the truss shown in Fig.26.3a, if the member BC is made 0.01m too short 
before placing it in the truss. Assume AE=300 kN for all members. 
 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Solution 
A similar truss with different boundary conditions has already been solved in 
example 25.1. For the sake of completeness the member of nodes and members 
are shown in Fig.26.3b.The displacements 6543 ,,, uuuu , 7u and 8u are zero due to 
boundary conditions. For the present problem the unconstrained degrees of 
freedom are 1u and 2u .The assembled stiffness matrix is of the order 88× and is 
available in example 25.1. 
In the given problem the member (2) is short by 0.01m.The forces developed in 
member (2) in the global coordinate system due to fabrication error is 
 
( )
( )
( )
( )
( )
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
−
−=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
θ
θ
θ
θ
sin
cos
sin
cos
4
01.0
02
01
04
03
AE
p
p
p
p
 
 =
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
75.0
0
75.0
0
kN (1) 
Now force-displacement relations for the truss are 
 
Version 2 CE IIT, Kharagpur 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−+
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−−
−−−
−−
−−−
−
−−
−−−−
−−−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
0
0
0
0
75.0
0
75.0
0
162.00934.00000162.0094.0
0934.0487.0000433.0094.0054.0
00162.0094.000162.0094.0
00094.0487.00433.0094.0054.0
000025.0025.00
0433.00433.00866.000
162.0094.0162.0094.025.00575.00
094.0054.0094.0054.0000108.0
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
AE
p
p
p
p
p
p
p
p
 
 (2) 
 
Note that 0876543 ====== uuuuuu 
Thus, solving 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎭⎬
⎫
⎩⎨
⎧−
⎭⎬
⎫
⎩⎨
⎧⎥⎦
⎤⎢⎣
⎡=
⎭⎬
⎫
⎩⎨
⎧ −
75.0
0
0
0
575.00
0108.01
1
2
1
AEu
u
 (3) 
01 =u 
and, mu 32 103478.4 −×−= (4) 
 
Reactions are calculated as, 
 
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
−
+
⎭⎬
⎫
⎩⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−−
−−
−
−
−
=
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
0
0
0
0
75.0
0
162.0094.0
094.0054.0
162.0094.0
094.0054.0
25.00
00
2
1
8
7
6
5
4
3
u
u
AE
p
p
p
p
p
p
 (5) 
 
 
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
−
−
=
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
211.0
123.0
211.0
123.0
424.0
0
8
7
6
5
4
3
p
p
p
p
p
p
 (6) 
 
The reactions and member forces are shown in Fig.26.3c. The memberforces 
can also be calculated by equation (26.10a) and (26.10b). For example, for 
member (2), 
 
o90=θ 
Version 2 CE IIT, Kharagpur 
 
4
300'
2 =p [0 -1 0 1] L
LAE
u
u
u
u
Δ−
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
2
1
4
3
 
 
 ( ) ( )
4
01.0300103478.4
4
300 3 −−×−= − 
 
 kN 424.04239.0 ≅= (7) 
 
Example 26.3 
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the 
member BC is raised by Co40 and member BD is raised by Co50 .Assume 
AE=300KN for all members and 
75000
1=α per o C. 
 
 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Solution 
For this problem assembled stiffness matrix is available in Fig.26.4b.The joints 
and members are numbered as shown in Fig.26.4b. In the given problem 
4321 ,,, uuuu and 5u represent unconstrained degrees of freedom. Due to support 
conditions, 0876 === uuu . 
 
The temperature of the member (2) is raised by o50 C.Thus, 
 
mTLL 32 10333.3505
75000
1 −×=××=Δ=Δ α (1) 
 
The forces are developed in member (2), as it was prevented from expansion. 
 ( )
( )
( )
( ) ⎪
⎪
⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
−××=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
θ
θ
θ
θ
sin
cos
sin
cos
10333.3300 3
2
1
8
7
f
f
f
f
p
p
p
p
 
Version 2 CE IIT, Kharagpur 
 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
=
1
0
1
0
 (2) 
 
The displacement of the member (5) was raised by Co40 . Thus, 
 
mTLL 35 10771.34025
000,75
1 −×=××=Δ=Δ α 
 
The forces developed in member (5) as it was not allowed to expand is 
 
( )
( )
( )
( ) ⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
−××=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
707.0
707.0
707.0
707.0
10771.3300 3
8
7
6
5
t
t
t
t
p
p
p
p
 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
−=
1
1
1
1
8.0 (3) 
 
The global force vector due to thermal load is 
 
( )
( )
( )
( )
( )
( )
( )
( ) ⎪
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
=
⎪⎪
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎪⎪
⎨
⎧
1
0
8.0
8.0
0
0
8.1
8.0
8
7
6
5
4
3
2
1
t
t
t
t
t
t
t
t
p
p
p
p
p
p
p
p
 (4) 
 
Writing the load-displacement relation for the entire truss is given below. 
 
Version 2 CE IIT, Kharagpur 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
+
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−−−
−−−
−−−
−−−
−−
−−−
−−−
−−−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
1
0
8.0
8.0
0
0
8.1
8.0
271.0071.000071.0071.02.00
071.0271.002.0071.0071.000
00271.0071.02.00071.0071.0
02.0071.0271.000071.0071.0
071.0071.02.00129.0071.000
071.0071.000071.0271.0020.0
2.00071.0071.000271.0071.0
00071.0071.0020.0071.0271.0
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
AE
p
p
p
p
p
p
p
p
(5) 
 
In the above problem 087654321 ======== pppppppp and 
0876 === uuu . 
 
Thus solving for unknown displacements, 
 
⎟⎟
⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜⎜
⎜
⎝
⎛
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
−
−
−
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−−
−
−−
−
−−
=
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
8.0
0
0
8.1
8.0
0
0
0
0
0
271.000071.0071.0
0129.0071.000
0071.0271.0020.0
071.000271.0071.0
071.002.0071.0271.0
1
5
4
3
2
1
AE
u
u
u
u
u
 (5) 
 
Solving equation (5), the unknown displacements are calculated as 
 
mu
umumumu
0013.0
0,0005.0,0020.0,0013.0
5
4321
−=
=−===
 (6) 
 
Now, reactions are computed as, 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
+
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−−
−−
−−−
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
1
0
8.0
0071.0071.02.00
2.0071.0071.000
071.02.00071.0071.0
5
4
3
2
1
8
7
6
u
u
u
u
u
p
p
p
 (7) 
 
All reactions are zero as truss is externally determinate and hence change in 
temperature does not induce any reaction. Now member forces are calculated by 
using equation (26.10b) 
 
Member (1): L=5m, o0=θ 
 
Version 2 CE IIT, Kharagpur 
 
5
'
2
AEp = [-1 0 1 0]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
2
1
4
3
u
u
u
u
 (8) 
 
='2p 0.1080 Kn 
 
Member 2: L=5m, o90=θ ,nodal points 4-1 
 
5
'
2
AEp = [0 -1 0 1] 5
2
1
8
7
10771.3300 −××−
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
u
u
u
u
 (9) 
 
 =0.1087 kN 
 
Member (3): L=5m, o0=θ ,nodal points 3-4 
 
5
300'
2 =p [-1 0 1 0]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
8
7
6
5
u
u
u
u
 (10) 
 
 =0.0780kN 
 
Member (4): ,5,90 mL == oθ nodal points 3-2 
 
5
300'
2 =p [0 -1 0 1]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
4
3
6
5
u
u
u
u
=0 (11) 
 
Member (5): 25,45 == Loθ ,nodal points 3-1 
 
25
300'
2 =p [-0.707 -0.707 0.707 0.707] 3
2
1
6
5
10333.3300 −××−
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
u
u
u
u
 (12) 
 
 =-0.8619 kN 
 
Member (6) : 25,135 == Loθ ,nodal points 4-2 
Version 2 CE IIT, Kharagpur 
 
25
300'
2 =p [0.707 -0.707 -0.707 0.707]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
4
3
8
7
u
u
u
u
= 0.0150 kN. (13) 
 
 
Summary 
In the last four lessons, the direct stiffness method as applied to the truss 
analysis was discussed. Assembly of member stiffness matrices, imposition of 
boundary conditions, and the problem of inclined supports were discussed. Due 
to the change in temperature the truss members either expand or shrink. 
However, in the case of statically indeterminate trusses, the length of the 
members is prevented from either expansion or contraction. Thus, the stresses 
are developed in the members due to changes in temperature. Similarly the 
errors in fabricating truss members also produce additional stresses in the 
trusses. In this lesson, these effects are accounted for in the stiffness analysis. A 
couple of problems are solved. 
	Analysis of StaticallyIndeterminateStructures by the DirectStiffness Method
	The Direct StiffnessMethod: TemperatureChanges andFabrication Errors inTruss Analysis
	Instructional Objectives
	26.1 Introduction
	26.2 Temperature Effects and Fabrication Errors
	Summary