The deflection for the 1st element of the beam can be written in terms of the shape functions as,
‫ݒ = )ݏ(ݒ‬ଵܰଵ(‫ )ݏ‬+ ߠଵܰଶ(‫ )ݏ‬+ ‫ݒ‬ଶܰଷ(‫ )ݏ‬+ ߠଶܰସ(‫ )ݏ‬െ െ െ െ െ െ െ െ െ (2.1) The deflection for the 2nd element of the beam can be written in terms of the shape functions as,
0
=
0.5
െ
െ
െ
െ
െ
െ
െ
െ
െ
െ
െ
െ
െ
െ
െ
െ(4)
‫(ݒ‬0.5) = ‫ݒ‬ଵܰଵ(0.5) + ߠଵܰଶ(0.5) + ‫ݒ‬ଶܰଷ(0.5) + ߠଶܰସ(0.5) = 0.01ܰଷ(0.5) = 0.0417(3 × 0.5ଶ െ 2 × 0.5ଷ) = 0.021݉
From (3.1),
ߠ(0.5)
‫ݏ‬
=
‫ ݔ‬െ ‫ݔ‬ଶ ‫(ܮ‬ଶ)
=
‫ݔ‬
‫ݒ‬ଶ = 0.0417 ݉ ܽ݊݀ ߠଶ = 0 ‫݀ܽݎ‬, which are the deflection and slope at the midpoint of the beam, that is at x=1m.
The deflections and slopes at points in between nodes can be interpolated using the shape functions.
Solution:
Element 1
Element 2
(a) Given, ‫(ܮ‬ଵ) = ‫(ܮ‬ଶ) = ‫ = ܮ‬1݉; ‫ = ܫܧ‬1000ܰ݉; ‫ =of length L, the structural stiffness matrix is defined as,
=
1 ‫(ܮ‬ଶ)
(‫ݒ‬ଶ
݀ܰଵ ݀‫ݏ‬
+
ߠଶ
݀ܰଶ ݀‫ݏ‬
+
‫ݒ‬ଷ
݀ܰଷ ݀‫ݏ‬
+ ߠଷ ݀݀ܰ‫ݏ‬ସ)
െ െ െ െ(3.2)
The point x=0.5m is in the 1st element.
From (2.1),
‫ݏ‬
=
‫ ݔ‬െ ‫ݔ‬ଵ ‫(ܮ‬ଵ)
=
0.5 െ 1
ܰଶ(‫(ܮ = )ݏ‬ଵ)(‫ ݏ‬െ 2‫ݏ‬ଶ + ‫ݏ‬ଷ) െ െ െ െ(ܾ) ܰସ(‫(ܮ = )ݏ‬ଵ)(െ‫ݏ‬ଶ + ‫ݏ‬ଷ) െ െ െ െ െ (݀)
݀‫ ݏ݀ )ݏ(ݒ݀ )ݏ(ݒ‬1 ݀‫)ݏ(ݒ‬ ߠ(‫(ܮ = ݔ݀ ݏ݀ = ݔ݀ = )ݏ‬ଵ) ݀‫ݏ‬
=
1 ‫(ܮ‬ଵ)
(‫ݒ‬ଵ
݀ܰଵ ݀‫ݏ‬
+
ߠଵ
݀ܰଶ ݀‫ݏ‬
+
‫ݒ‬ଶ
݀ܰଷ ݀‫ݏ‬
+ ߠଶ ݀݀ܰ‫ݏ‬ସ)
െെ
െ െ(3.1)
The slope for 2nd element of a beam is,
Deflection and slope at x=0.5m:
݀‫ ݏ݀ )ݏ(ݒ݀ )ݏ(ݒ‬1 ݀‫)ݏ(ݒ‬ ߠ(‫(ܮ = ݔ݀ ݏ݀ = ݔ݀ = )ݏ‬ଶ) ݀‫ݏ‬
=
1 ‫(ܮ‬ଵ)
݀‫(ݒ‬0.5) ݀‫ݏ‬
=
1 ‫(ܮ‬ଵ)
(‫ݒ‬ଵ
݀ܰଵ(0.5) ݀‫ݏ‬
+
ߠଵ
݀ܰଶ(0.5) ݀‫ݏ‬
+
‫ݒ‬ଶ
݀ܰଷ(0.5) ݀‫ݏ‬
+
ߠଶ
݀ܰସ(0.5) ݀‫ݏ‬
=
1 ‫(ܮ‬ଵ)
(‫ݒ‬ଶ
݀ܰଷ݀(‫ݏ‬0.5))
=
1
×
0.0417
×
(6
×
0.5
െ
6
×
0.5ଶ)
×
(െ6
×
0.5
+
6
×
0.5ଶ)
=
െ0.063
‫݀ܽݎ‬
(a) Deflection and slopes ‫ = ݔ ݐܣ‬0.5݉ ‫(ݒ‬0.5) = 0.021݉ ߠ(0.5) = 0.063 ‫݀ܽݎ‬
‫ = ݔ ݐܣ‬1݉ ‫(ݒ‬1) = 0.0417 ݉
ߠ(1) = 0 ‫݀ܽݎ‬
‫ = ݔ ݐܣ‬1.5݉ ‫(ݒ‬1.5) = 0.021݉ ߠ(1.5) = −0.063 ‫݀ܽݎ‬
=
‫ܫܧ‬ ሾ‫(ܮ‬ଶ)ሿଷ
൦െ612
4 െ6
െ6 12
െ26൪ = 1000 ൦െ612
4 െ6
െ6 12
െ26൪
6 2 െ6 4
6 2 െ6 4
Assembling the element stiffness matrices above, we obtain the following matrix equation,
12 6‫ ܮ‬െ12 6‫ܮ‬
ሾ݇ሿ
=
‫ܫܧ‬ ‫ܮ‬ଷ
൦െ61‫ܮ‬2
4‫ܮ‬ଶ െ6‫ܮ‬
െ6‫ܮ‬ 12
െ2‫ܮ‬6ଶ‫ܮ‬൪
6‫ ܮ‬2‫ܮ‬ଶ െ6‫ ܮ‬4‫ܮ‬ଶ
The element stiffness matrix for element 1 is:
12 6 െ12 6
12 6 െ12 6
桁架单元例子 MATLAB 1
Problem 1: Consider the clamped-clamped beam shown below. Assume there are no axial forces acting on the beam. Use two elements to solve the problem. (a) Determine the deflection and slope at x = 0.5, 1 and 1.5 m; (b) Draw the bending moment and shear force diagrams for the entire beam; (c) What are the support reactions? (d) Use the beam element shape functions to plot the deflected shape of the beam. Use EI = 1,000 Nm, L = 1 m, and F = 1,000 N.
(b) The bending moment and shear force diagrams for the entire beam: The bending moment can be found from,
‫݀ ܫܧ‬ଶ‫ݒ‬ ‫ܮ = )ݏ(ܯ‬ଶ ቆ݀‫ݏ‬ଶቇ
For the 1st element:
We have to express these moments in terms of x to plot the results. For the first element,
For the second element,
‫ݏ‬
=
‫ ݔ‬െ ‫ݔ‬ଵ ‫(ܮ‬ଵ)
=
‫ݔ‬
െ 1
0
;
⇒ ‫ ݔ = ݏ‬െ െ െ െ െ െ െ (8)
݀ܰଷ ݀‫ݏ‬
+
ߠଷ
݀݀ܰ‫ݏ‬ସ൰൱
= 1000ሾ‫ݒ‬ଶ(െ6 + 12‫)ݏ‬+ ߠଶ(െ4 + 6‫(ܮ)ݏ‬ଶ) + ‫ݒ‬ଷ(6 െ 12‫)ݏ‬+ ߠଷ(െ2 + 6‫(ܮ)ݏ‬ଶ)ሿ = 1000ሾ0.0417(െ6 + 12‫ )ݏ‬+ 0 + 0 + 0ሿ
= െ250.2 + 500.4‫ ݉ܰ ݏ‬െ െ െ െ െ െ െ െ െ െ(7)
From (3.2),
ߠ(1.5)
=
1 ‫(ܮ‬ଶ)
݀‫(ݒ‬0.5) ݀‫ݏ‬
=
1 ‫(ܮ‬ଶ)
(‫ݒ‬ଶ
݀ܰଵ(0.5) ݀‫ݏ‬
+
ߠଶ
݀ܰଶ(0.5) ݀‫ݏ‬
+
‫ݒ‬ଷ
݀ܰଷ(0.5) ݀‫ݏ‬
+
ߠଷ
݀ܰସ(0.5) ݀‫ݏ‬
=
1 ‫ܮ‬
(‫ݒ‬ଶ
݀ܰଵ݀(‫ݏ‬0.5))
=
1
×
0.0417
‫ ۏ‬0 0 6 2 െ6 4 ‫ ۏ ے‬0 ‫ܥ ۏ ے‬ଷ ‫ے‬
Striking out the rows and columns of zero elements, (1) reduces to,
1000 ቂ204 80ቃ ቂ‫ߠݒ‬ଶଶቃ = ቂ10000ቃ
Solving the above set of equations we get,
‫(ܯ‬ଵ)(‫)ݏ‬
=
‫ܫܧ‬ ሾ‫(ܮ‬ଵ)ሿଶ
݀ଶ‫ݒ‬ ݀‫ݏ‬ଶ
=
1000
݀ ൭݀‫ݏ‬
൬‫ݒ‬ଵ
݀ܰଵ ݀‫ݏ‬
+
ߠଵ
݀ܰଶ ݀‫ݏ‬
+
‫ݒ‬ଶ
݀ܰଷ ݀‫ݏ‬
+
ߠଶ
݀݀ܰ‫ݏ‬ସ൰൱