m4l25 Truss Analysis Continued

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 
25 
The Direct Stiffness 
Method: Truss Analysis 
(Continued) 
 
Version 2 CE IIT, Kharagpur 
 
Instructional Objectives 
After reading this chapter the student will be able to 
1. Transform member stiffness matrix from local to global co-ordinate system. 
2. Assemble member stiffness matrices to obtain the global stiffness matrix. 
3. Analyse plane truss by the direct stiffness matrix. 
4. Analyse plane truss supported on inclined roller supports. 
 
 
25.1 Introduction 
In the previous lesson, the direct stiffness method as applied to trusses was 
discussed. The transformation of force and displacement from local co-ordinate 
system to global co-ordinate system were accomplished by single transformation 
matrix. Also assembly of the member stiffness matrices was discussed. In this 
lesson few plane trusses are analysed using the direct stiffness method. Also the 
problem of inclined support will be discussed. 
 
Example 25.1 
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be 
constant for all the members. 
 
 
 
Version 2 CE IIT, Kharagpur 
 
 
 
The numbering of joints and members are shown in Fig. 25.1b. Also, the possible 
displacements (degrees of freedom) at each node are indicated. Here lower 
numbers are used to indicate unconstrained degrees of freedom and higher 
numbers are used for constrained degrees of freedom. Thus displacements 6,7 
and 8 are zero due to boundary conditions. 
 
First write down stiffness matrix of each member in global co-ordinate system 
and assemble them to obtain global stiffness matrix. 
 
Element 1: .619.4,60 mL =°=θ Nodal points 4-1 
 
 [ ]
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−−
−−
−−
−−
=
75.0433.075.0433.0
433.025.0433.025.0
75.0433.075.0433.0
433.025.0433.025.0
619.4
1 EAk (1) 
 
Element 2: .00.4,90 mL =°=θ Nodal points 2-1 
 
Version 2 CE IIT, Kharagpur 
 
 [ ]
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−=
1010
0000
1010
0000
0.4
2 EAk (2) 
 
Element 3: .619.4,120 mL =°=θ Nodal points 3-1 
 
[ ]
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−−
−−
−−
−−
=
75.0433.075.0433.0
433.025.0433.025.0
75.0433.075.0433.0
433.025.0433.025.0
619.4
3 EAk (3) 
 
Element 4: .31.2,0 mL =°=θ Nodal points 4-2 
 
 
[ ]
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−
=
0000
0101
0000
0101
31.2
4 EAk (4) 
 
Element 5: .31.2,0 mL =°=θ Nodal points 2-3 
 
[ ]
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−
=
0000
0101
0000
0101
31.2
5 EAk (5) 
 
 
The assembled global stiffness matrix of the truss is of the order 88× . Now 
assemble the global stiffness matrix. Note that the element 111k of the member 
stiffness matrix of truss member 1 goes to location ( )7,7 of global stiffness matrix. 
On the member stiffness matrix the corresponding global degrees of freedom are 
indicated to facilitate assembling. Thus, 
 
Version 2 CE IIT, Kharagpur 
 
[ ]
8
7
6
5
4
3
2
1
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
87654321
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−−
−−−
−−
−−−
−
−−
−−−−
−−−
= EAK 
 
 (6) 
 
Writing the load-displacement relation for the truss, yields 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−−
−−−
−−
−−−
−
−−
−−−−
−−−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
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
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p
 (7) 
 
The displacements 1u to 5u are unknown. The displacements 0876 === uuu . 
 
Also 05321 ==== pppp . But 4 10 kNp = − . 
 
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−−
−
−
−
−
=
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
−
5
4
3
2
1
487.00433.0094.0054.0
025.0025.00
433.00866.000
094.025.00575.00
054.0000108.0
0
10
0
0
0
u
u
u
u
u
EA (8) 
 
Solving which, the unknown displacements are evaluated. Thus, 
 
Version 2 CE IIT, Kharagpur 
 
AE
u
AE
u
AE
u
AE
u
AE
u 334.13;642.74;668.6;64.34;668.6 54321 =−==−== (9) 
Now reactions are evaluated from equation, 
 
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
−
−
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−−
−−−
−−
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
334.13
642.74
668.6
64.34
668.6
1
000162.0094.0
00433.0094.0054.0
094.000162.0094.0
8
7
6
EA
EA
p
p
p
 (10) 
 
 
Thus, 
 
6 7 85.00 kN ; 0 ; 5.00 kNp p p= = = . (11) 
 
Now calculate individual member forces. 
 
Member 1: mLml 619.4;866.0;50.0 === . 
 
{ } [ ]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=
2
1
8
7
1 619.4
'
u
u
u
u
mlmlAEp 
 
{ } [ ]1 6.6671' 0.5 0.866 5.77 kN34.644.619
AEp
AE
⎧ ⎫= − − =⎨ ⎬−⎩ ⎭ (12) 
 
Member 2: mLml 0.4;0.1;0 === . 
 
{ } [ ]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=
2
1
4
3
1 0.4
'
u
u
u
u
mlmlAEp 
 
{ } [ ]1 74.6421' 1 1 10.0 kN34.644.619
AEp
AE
−⎧ ⎫= − = −⎨ ⎬−⎩ ⎭ (13) 
 
Member 3: mLml 619.4;866.0;50.0 ==−= . 
 
 
Version 2 CE IIT, Kharagpur 
 
{ } [ ]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=
2
1
6
5
1 619.4
'
u
u
u
u
mlmlAEp 
 
{ } [ ]1
13.334
1' 0.5 0.5 0.866 6.667 5.77 kN
4.619
34.64
AEp
AE
⎧ ⎫⎪ ⎪= − − =⎨ ⎬⎪ ⎪−⎩ ⎭
 (14) 
 
Member 4: mLml 0.31.2;0;0.1 === . 
 
{ } [ ]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=
4
3
8
7
1 31.2
'
u
u
u
u
mlmlAEp 
 
{ } [ ]1 01' 1 1 2.88 kN6.6672.31
AEp
AE
⎧ ⎫= − = −⎨ ⎬⎩ ⎭ (15) 
 
Member 5: mLml 0.31.2;0;0.1 === . 
 
{ } [ ]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=
6
5
4
3
1 31.2
'
u
u
u
u
mlmlAEp 
 
{ } [ ]1 6.6671' 1 1 2.88 kN13.3342.31
AEp
AE
⎧ ⎫= − = −⎨ ⎬⎩ ⎭ (16) 
 
Example 25.2 
Determine the forces in the truss shown in Fig. 25.2a by the direct stiffness 
method. Assume that all members have the same axial rigidity. 
 
Version 2 CE IIT, Kharagpur 
 
 
Version 2 CE IIT, Kharagpur 
 
 
The joint and member numbers are indicated in Fig. 25.2b. The possible degree 
of freedom are also shown in Fig. 25.2b. In the given problem 21 ,uu and 3u 
represent unconstrained degrees of freedom and 087654 ===== uuuuu due 
to boundary condition. First let us generate stiffness matrix for each of the sixmembers in global co-ordinate system. 
 
Element 1: .00.5,0 mL =°=θ Nodal points 2-1 
 
 [ ]
2
1
4
3
0000
0101
0000
0101
0.5
2143
1
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−
= EAk (1) 
 
Element 2: .00.5,90 mL =°=θ Nodal points 4-1 
 
Version 2 CE IIT, Kharagpur 
 
 [ ]
2
1
8
7
1010
0000
1010
0000
0.5
2187
2
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−= EAk (2) 
 
Element 3: .00.5,0 mL =°=θ Nodal points 3-4 
 
 [ ]
8
7
6
5
0000
0101
0000
0101
0.5
8765
3
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−
= EAk (3) 
 
 
Element 4: .00.5,90 mL =°=θ Nodal points 3-2 
 
 [ ]
4
3
6
5
1010
0000
1010
0000
0.5
4365
4
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−= EAk (4) 
 
Element 5: .07.7,45 mL =°=θ Nodal points 3-1 
 
 [ ]
2
1
6
5
5.05.05.05.0
5.05.05.05.0
5.05.05.05.0
5.05.05.05.0
07.7
2165
5
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−−
−−
−−
−−
= EAk (5) 
 
Element 6: .07.7,135 mL =°=θ Nodal points 4-2 
 
Version 2 CE IIT, Kharagpur 
 
 [ ]
4
3
8
7
5.05.05.05.0
5.05.05.05.0
5.05.05.05.0
5.05.05.05.0
07.7
4387
6
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−−
−−
−−
−−
= EAk (6) 
 
There are eight possible global degrees of freedom for the truss shown in the 
figure. Hence the global stiffness matrix is of the order ( 88× ). On the member 
stiffness matrix, the corresponding global degrees of freedom are indicated to 
facilitate assembly. Thus the global stiffness matrix is, 
 
 
[ ]
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−−−
−−−
−−−
−−−
−−−
−−−
−−−
−−−
=
271.0071.000071.0071.020.00
071.0271.0020.0071.0071.000
00271.0071.020.00071.0071.0
020.0071.0271.000071.0071.0
071.0071.020.0071.0271.0071.000
071.0071.000071.0271.0020.0
20.00071.0071.000271.0071.0
00071.0071.002.0071.0271.0
AEK (7) 
 
The force-displacement relation for the truss is, 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−−−
−−−
−−−
−−−
−−−
−−−
−−−
−−−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
271.0071.000071.0071.020.00
071.0271.0020.0071.0071.000
00271.0071.020.00071.0071.0
020.0071.0271.000071.0071.0
071.0071.020.0071.0271.0071.000
071.0071.000071.0271.0020.0
20.00071.0071.000271.0071.0
00071.0071.002.0071.0271.0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p
 
 (8) 
 
The displacements 21,uu and 3u are unknowns. 
Here, 1 2 35 kN ; 10 ; 0p p p= = − = and 087654 ===== uuuuu . 
Version 2 CE IIT, Kharagpur 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−−−
−−−
−−−
−−−
−−−
−−−
−−−
−−−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
0
0
0
0
0
271.0071.000071.0071.020.00
071.0271.0020.0071.0071.000
00271.0071.020.00071.0071.0
020.0071.0271.000071.0071.0
071.0071.020.0071.0271.0071.000
071.0071.000071.0271.0020.0
20.00071.0071.000271.0071.0
00071.0071.002.0071.0271.0
0
10
5
3
2
1
8
7
6
5
4
u
u
u
EA
p
p
p
p
p
(9) 
 
Thus, 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−
−
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
−
3
2
1
271.0020.0
0271.0071.0
20.0071.0271.0
0
10
5
u
u
u
 (10) 
 
Solving which, yields 
 
AE
u
AE
u
AE
u 825.53;97.55;855.72 321 =−== 
 
Now reactions are evaluated from the equation, 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−
−−
−−
−
=
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
3
2
1
8
7
6
5
4
071.020.00
071.000
0071.0071.0
0071.0071.0
071.000
u
u
u
p
p
p
p
p
 (11) 
 
4 5 6 7 83.80 kN ; 1.19 kN ; 1.19 kN ; 3.8 0 ; 15.00 kNp p p p kN p= − = − = − = =
 
In the next step evaluate forces in members. 
 
Element 1: .00.5,0 mL =°=θ Nodal points 2-1 
 
Version 2 CE IIT, Kharagpur 
 
 { } [ ]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=
2
1
4
3
1 0.5
'
u
u
u
u
mlmlAEp 
 
{ } [ ]1 53.8251' 1 1 3.80 kN72.8555.0
AEp
AE
⎧ ⎫= − = −⎨ ⎬⎩ ⎭ (12) 
 
Element 2: .00.5,90 mL =°=θ Nodal points 4-1 
 
{ } [ ]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=
2
1
8
7
1 5
'
u
u
u
u
mlmlAEp 
 
{ } [ ]1 01' 1 1 11.19kN55.975
AEp
AE
⎧ ⎫= − =⎨ ⎬−⎩ ⎭ (13) 
 
Element 3: .00.5,0 mL =°=θ Nodal points 3-4 
 
{ } [ ]{ } 000
5
'1 == AEp (14) 
 
Element 4: .00.5,90 mL =°=θ Nodal points 3-2 
 
 { } [ ]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=
4
3
6
5
1 5
'
u
u
u
u
mlmlAEp 
{ } [ ] { }1 1' 0 53.825 05
AEp
AE
= = (15) 
 
Element 5: .07.7,45 mL =°=θ Nodal points 3-1 
 
{ } [ ]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=
2
1
6
5
1 07.7
'
u
u
u
u
mlmlAEp 
Version 2 CE IIT, Kharagpur 
 
{ } [ ]1 72.8551' 0.707 0.707 1.688 kN55.977.07
AEp
AE
⎧ ⎫= − − = −⎨ ⎬−⎩ ⎭ (16) 
 
Element 6: .07.7,135 mL =°=θ Nodal points 4-2 
 
 { } [ ]
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−−=
4
3
8
7
1 07.7
'
u
u
u
u
mlmlAEp 
 
{ } [ ] { }1 1' 0.707 53.825 5.38 kN7.07
AEp
AE
= = (17) 
 
 
25.2 Inclined supports 
Sometimes the truss is supported on a roller placed on an oblique plane (vide 
Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is 
zero. ..ei displacement along "y is zero in the present case. 
 
 
Version 2 CE IIT, Kharagpur 
 
 
 
If the stiffness matrix of the entire truss is formulated in global co-ordinate system 
then the displacements along y are not zero at the oblique support. So, a special 
procedure has to be adopted for incorporating the inclined support in the analysis 
of truss just described. One way to handle inclined support is to replace the 
inclined support by a member having large cross sectional area as shown in Fig. 
25.3b but having the length comparable with other members meeting at that joint. 
The inclined member is so placed that its centroidal axis is perpendicular to the 
inclined plane. Since the area of cross section of this new member is very high, it 
does not allow any displacement along its centroidal axis of the joint A . Another 
method of incorporating inclined support in the analysis is to suitably modify the 
member stiffness matrix of all the members meeting at the inclined support. 
 
 
Version 2 CE IIT, Kharagpur 
 
Consider a truss member as shown in Fig. 25.4. The nodes are numbered as 1 
and 2. At 2, it is connected to a inclined support. Let '' yx be the local co-ordinate 
axes of the member. At node 1, the global co-ordinate system xy is also shown. 
At node 2, consider nodal co-ordinate system as "" yx , where "y is perpendicular 
to oblique support. Let 1'u and 2'u be the displacements of nodes 1 and 2 in the 
local co-ordinate system. Let 11 ,vu be the nodal displacements of node 1 in 
global co-ordinate system xy . Let 2 2" , "u v be the nodal displacements along "x -
and "y - arein the local co-ordinate system "" yx at node 2. Then from Fig. 25.4, 
 
xx vuu θθ sincos' 111 += 
 
"2"22 sin"cos"' xx vuu θθ += (25.1) 
 
This may be written as 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎦
⎤⎢⎣
⎡=
⎭⎬
⎫
⎩⎨
⎧
2
2
1
1
""2
1
"
"sincos
00
00
sincos
'
'
v
u
v
u
u
u
xx
xx
θθ
θθ
 (25.2) 
Denoting "" sin";cos";sin;cos xxxx mlml θθθθ ==== 
 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎦
⎤⎢⎣
⎡=
⎭⎬
⎫
⎩⎨
⎧
2
2
1
1
2
1
"
"""
00
00'
'
v
u
v
u
ml
ml
u
u
 (25.3a) 
 
 or { } [ ]{ }uTu '' = 
 
where [ ]'T is the displacement transformation matrix. 
 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Similarly referring to Fig. 25.5, the force 1'p has components along x and 
y axes. Hence 
 
xpp θcos'11 = (25.4a) 
 
xpp θsin'12 = (25.4b) 
 
Version 2 CE IIT, Kharagpur 
 
Similarly, at node 2, the force 2'p has components along "x and "y axes. 
3 2" ' cos xp p θ ′′= (25.5a) 
 
4 2" ' sin xp p θ ′′= (25.5b) 
 
The relation between forces in the global and local co-ordinate system may be 
written as, 
 
1
2 1
3 2
4
0cos
0 'sin
cos" '0
sin" 0
x
x
x
x
p
p p
p p
p
θ
θ
θ
θ
⎡ ⎤⎧ ⎫ ⎢ ⎥⎪ ⎪ ⎧ ⎫⎪ ⎪ ⎢ ⎥=⎨ ⎬ ⎨ ⎬′′⎢ ⎥ ⎩ ⎭⎪ ⎪ ⎢ ⎥⎪ ⎪ ′′⎢ ⎥⎩ ⎭ ⎣ ⎦
 (25.6) 
 
{ } [ ] { }'' pTp T= (25.7) 
 
Using displacement and force transformation matrices, the stiffness matrix for 
member having inclined support is obtained. 
 [ ] [ ] [ ][ ]''' TkTk T= 
 
[ ] ⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
−
−
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
=
""
00
0011
11
"0
"0
0
0
ml
ml
L
AE
m
l
m
l
k (25.8) 
 
Simplifying, 
 
[ ]
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢
⎣
⎡
−−
−−
−−
−−
=
2
2
2
2
"""""
"""""
""
""
mmlmmlm
mllmlll
mmmlmlm
lmlllml
L
EAk (25.9) 
 
If we use this stiffness matrix, then it is easy to incorporate the condition of zero 
displacement perpendicular to the inclined support in the stiffness matrix. This is 
shown by a simple example. 
 
 
Version 2 CE IIT, Kharagpur 
 
Example 25.3 
Analyse the truss shown in Fig. 25.6a by stiffness method. Assume axial rigidity 
EA to be constant for all members. 
 
 
Version 2 CE IIT, Kharagpur 
 
 
 
The nodes and members are numbered in Fig. 25.6b. The global co-ordinate 
axes are shown at node 3. At node 2, roller is supported on inclined support. 
Hence it is required to use nodal co-ordinates "" yx − at node 2 so that 4u could be 
set to zero. All the possible displacement degrees of freedom are also shown in 
the figure. In the first step calculate member stiffness matrix. 
 
Member 1: .00.5,87.6,13.143 " mLxx =°=°= θθ Nodal points 1-2 
 12.0";993.0";6.0;80.0 ===−= mlml . 
 
 [ ]
4
3
2
1
014.0119.0072.0096.0
119.0986.0596.0794.0
072.0596.036.048.0
096.0794.048.064.0
0.5
4321
1
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−
−−−
−
= EAk (1) 
Version 2 CE IIT, Kharagpur 
 
 
 
 
Member 2: .00.4,30,0 " mLxx =°=°= θθ Nodal points 2-3 
 50.0";866.0";0;1 ==== mlml . 
 
 
 
Version 2 CE IIT, Kharagpur 
 
 [ ]
4
3
6
5
014.0119.0072.0096.0
119.0986.0596.0794.0
072.0596.036.048.0
096.0794.048.064.0
0.4
4365
2
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−
−−−
−
= EAk (2) 
 
Member 3: .00.3,90 mLx =°=θ , 1;0 == ml Nodal points 3-1 
 . 
 [ ]
2
1
6
5
1010
0000
1010
0000
0.3
2165
3
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−= EAk (3) 
 
 
For the present problem, the global stiffness matrix is of the order ( )66× . The 
global stiffness matrix for the entire truss is. 
 
[ ]
6
5
4
3
2
1
333.0000333.00
025.0125.0217.000
0125.0065.0132.0014.0.019.0
0217.0132.0385.0119.0159.0
333.00014.0119.0405.0096.0
00019.0159.0096.0128.0
654321
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−
−
−−
−−−−
−
= EAk (4) 
 
Writing load-displacement equation for the truss for unconstrained degrees of 
freedom, 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−
−−
−
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−
3
2
1
385.0119.0159.0
119.0405.0096.0
159.0096.0128.0
0
5
5
u
u
u
 (5) 
 
Solving , 
 
AE
u
AE
u
AE
u 12.33;728.3;408.77 321 ==−= (6) 
 
Version 2 CE IIT, Kharagpur 
 
Now reactions are evaluated from the equation 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−
−=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
12.33
728.3
40.77
1
0333.00
217.000
132.0014.0.019.0
6
5
4
AE
AE
p
p
p
 (6) 
 
4 5 62.85 kN ; 7.19 kN ; 1.24 kNp p p= = − = − 
 
 
Summary 
Sometimes the truss is supported on a roller placed on an oblique plane. In such 
situations, the direct stiffness method as discussed in the previous lesson needs 
to be properly modified to make the displacement perpendicular to the roller 
support as zero. In the present approach, the inclined support is handled in the 
analysis by suitably modifying the member stiffness matrices of all members 
meeting at the inclined support. A few problems are solved to illustrate the 
procedure. 
 
	Analysis of StaticallyIndeterminateStructures by the DirectStiffness Method
	The Direct StiffnessMethod: Truss Analysis(Continued)
	Instructional Objectives
	25.1 Introduction
	25.2 Inclined supports
	Summary