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6
5
4
3
2
1
6
5
4
3
1875 . 0 375 . 0 0 1875 . 0 0 375 . 0
375 . 0 0 . 1 0 375 . 0 0 5 . 0
0 0 1875 . 0 1875 . 0 375 . 0 375 . 0
1875 . 0 375 . 0 1875 . 0 375 . 0 375 . 0 0
0 0 375 . 0 375 . 0 0 . 1 5 . 0
375 . 0 5 . 0 375 . 0 0 5 . 0 2
67 . 6
0
u
u
u
u
u
u
EI
p
p
p
p
zz
We know that 0
6 5 4 3
= = = = u u u u . Thus solving for unknowns and , yields
1
u
2
u
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2
1
0 . 1 5 . 0
5 . 0 0 . 2
67 . 6
0
u
u
EI
zz
(6)
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67 . 6
0
0 . 2 5 . 0
5 . 0 0 . 1
75 . 1
1
2
1
zz
EI
u
u
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=
62 7
905 1
1
.
.
EI
zz
(7)
Thus displacements are,
zz zz
EI
u
EI
u
62 . 7
and
905 . 1
2 1
=
-
=
The unknown joint loads are given by,
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62 . 7
905 . 1
1
0 375 . 0
0 5 . 0
375 . 0 375 . 0
375 . 0 0
6
5
4
3
zz
zz
EI
EI
p
p
p
p
Page 2
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6
5
4
3
2
1
6
5
4
3
1875 . 0 375 . 0 0 1875 . 0 0 375 . 0
375 . 0 0 . 1 0 375 . 0 0 5 . 0
0 0 1875 . 0 1875 . 0 375 . 0 375 . 0
1875 . 0 375 . 0 1875 . 0 375 . 0 375 . 0 0
0 0 375 . 0 375 . 0 0 . 1 5 . 0
375 . 0 5 . 0 375 . 0 0 5 . 0 2
67 . 6
0
u
u
u
u
u
u
EI
p
p
p
p
zz
We know that 0
6 5 4 3
= = = = u u u u . Thus solving for unknowns and , yields
1
u
2
u
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=
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2
1
0 . 1 5 . 0
5 . 0 0 . 2
67 . 6
0
u
u
EI
zz
(6)
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67 . 6
0
0 . 2 5 . 0
5 . 0 0 . 1
75 . 1
1
2
1
zz
EI
u
u
?
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? -
=
62 7
905 1
1
.
.
EI
zz
(7)
Thus displacements are,
zz zz
EI
u
EI
u
62 . 7
and
905 . 1
2 1
=
-
=
The unknown joint loads are given by,
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62 . 7
905 . 1
1
0 375 . 0
0 5 . 0
375 . 0 375 . 0
375 . 0 0
6
5
4
3
zz
zz
EI
EI
p
p
p
p
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-
-
-
=
714 . 0
95 . 0
14 . 2
857 . 2
(8)
The actual support reactions are,
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286 . 9
62 . 7
86 . 7
857 . 22
714 . 0
95 . 0
14 . 2
857 . 2
10
67 . 6
10
20
6
5
4
3
R
R
R
R
(9)
Member end actions for element 1
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905 . 1
0
0
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
66 . 6
10
66 . 6
10
4
3
2
1
zz
zz
EI
EI
q
q
q
q
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=
565 . 8
714 . 10
707 . 5
285 . 9
(10)
Member end actions for element 2
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62 . 7
0
905 . 1
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
66 . 6
10
66 . 6
10
4
3
2
1
zz
zz
EI
EI
q
q
q
q
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0
856 . 7
565 . 8
14 . 12
Page 3
(5)
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6
5
4
3
2
1
6
5
4
3
1875 . 0 375 . 0 0 1875 . 0 0 375 . 0
375 . 0 0 . 1 0 375 . 0 0 5 . 0
0 0 1875 . 0 1875 . 0 375 . 0 375 . 0
1875 . 0 375 . 0 1875 . 0 375 . 0 375 . 0 0
0 0 375 . 0 375 . 0 0 . 1 5 . 0
375 . 0 5 . 0 375 . 0 0 5 . 0 2
67 . 6
0
u
u
u
u
u
u
EI
p
p
p
p
zz
We know that 0
6 5 4 3
= = = = u u u u . Thus solving for unknowns and , yields
1
u
2
u
?
?
?
?
?
?
?
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=
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2
1
0 . 1 5 . 0
5 . 0 0 . 2
67 . 6
0
u
u
EI
zz
(6)
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=
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67 . 6
0
0 . 2 5 . 0
5 . 0 0 . 1
75 . 1
1
2
1
zz
EI
u
u
?
?
?
?
?
?
?
?
?
? -
=
62 7
905 1
1
.
.
EI
zz
(7)
Thus displacements are,
zz zz
EI
u
EI
u
62 . 7
and
905 . 1
2 1
=
-
=
The unknown joint loads are given by,
?
?
?
?
?
?
?
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?
? -
?
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=
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?
62 . 7
905 . 1
1
0 375 . 0
0 5 . 0
375 . 0 375 . 0
375 . 0 0
6
5
4
3
zz
zz
EI
EI
p
p
p
p
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
=
714 . 0
95 . 0
14 . 2
857 . 2
(8)
The actual support reactions are,
?
?
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?
?
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?
?
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?
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-
=
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286 . 9
62 . 7
86 . 7
857 . 22
714 . 0
95 . 0
14 . 2
857 . 2
10
67 . 6
10
20
6
5
4
3
R
R
R
R
(9)
Member end actions for element 1
?
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-
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-
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+
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=
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905 . 1
0
0
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
66 . 6
10
66 . 6
10
4
3
2
1
zz
zz
EI
EI
q
q
q
q
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
565 . 8
714 . 10
707 . 5
285 . 9
(10)
Member end actions for element 2
?
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?
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-
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?
62 . 7
0
905 . 1
0
1
0 . 1 375 . 0 5 . 0 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
5 . 0 375 . 0 0 . 1 375 . 0
375 . 0 1875 . 0 375 . 0 1875 . 0
66 . 6
10
66 . 6
10
4
3
2
1
zz
zz
EI
EI
q
q
q
q
?
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?
=
0
856 . 7
565 . 8
14 . 12
Summary
In the previous lesson the beam element stiffness matrix is derived from
fundamentals. Assembling member stiffness matrices, the global stiffness matrix
is generated. The procedure to impose boundary conditions on the load-
displacement relation is discussed. In this lesson, a few continuous beam
problems are analysed by the direct stiffness method.
Read More
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34 videos|140 docs|31 tests
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FAQs on The Direct Stiffness Method: Beams - 5 - Structural Analysis - Civil Engineering (CE)
| 1. What is the direct stiffness method for beams? |  |
Ans. The direct stiffness method is a numerical technique used to analyze the behavior of beams. It involves breaking down the beam into smaller elements and determining the stiffness matrix for each element. These stiffness matrices are then assembled to form the global stiffness matrix, which can be used to solve for displacements, forces, and moments in the beam.
| 2. How does the direct stiffness method handle boundary conditions? | |
Ans. The direct stiffness method handles boundary conditions by applying constraints or known displacements at the ends of the beam. These constraints are incorporated into the global stiffness matrix through the use of boundary condition equations. By solving the resulting system of equations, the method can determine the unknown displacements, forces, and moments at any point along the beam.
| 3. What are the advantages of using the direct stiffness method for beam analysis? | |
Ans. The direct stiffness method offers several advantages for beam analysis. Firstly, it allows for the analysis of complex beam structures with varying material properties, cross-sections, and loading conditions. Secondly, it provides accurate results by considering the stiffness of each individual element, rather than assuming a constant stiffness throughout the beam. Lastly, it can handle both linear and nonlinear behavior, making it versatile for a wide range of applications.
| 4. What are the limitations of the direct stiffness method for beam analysis? | |
Ans. While the direct stiffness method is a powerful tool for beam analysis, it does have some limitations. One limitation is that it assumes linear elastic behavior, which may not accurately represent the actual behavior of some materials. Additionally, it requires dividing the beam into discrete elements, which can be time-consuming for complex structures. Lastly, the method may become computationally expensive for large-scale analyses due to the need to solve a system of equations.
| 5. How does the direct stiffness method differ from other methods of beam analysis? | |
Ans. The direct stiffness method differs from other methods of beam analysis, such as the flexibility method or finite element method, in its approach to solving the beam equations. While the flexibility method focuses on determining the flexibility matrix of each element, the direct stiffness method calculates the stiffness matrix. The finite element method, on the other hand, uses a similar approach but allows for more complex element shapes and behaviors. Each method has its own advantages and limitations, making them suitable for different types of beam analysis problems.
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