lecture 2: exercise and solution - wind turbine · pdf filelecture 2: exercise and solution 1...

1Lecture 2: Exercise and Solution

Rectangular steel cross-section, with constant thickness of 0.1 m. Numbers indicate height

1)g

measured in metres. Find end-deflection and compare. Analytical solution for 1) 23.126 mm

1)

1D elements with thickness 0.1 m and varying heights. 10 elements.2)

2D elements with thickness 0.1 m.

20x4 elements.

3)

20x4 elements.

2Lecture 2: Exercise and Solution

Problem Deformation at free end

1) 1 Beam element with constant height equal 0.75 m. 23.221 mm

1) 20 Beam element with constant height equal 0.75 m. 23.221 mm

2) 10 Beam elements with separately constant height. 17.696 mm

1 beam element with variable height going from 1 00 meter at fixed 17 176 mm1 beam element with variable height going from 1.00 meter at fixed end to 0.50 m at free end. Section: Tapered I (1.0;0.1;0.5;0.1;0.1;0.1;0.1)

17.176 mm

3) 4 node elements meshed 4x4 elements (height x length). Load applied at middle node at the free end. Deformation determined at the same node.

15.815 mm

the same node.

3) 4 node elements meshed 4x10 elements (height x length). Load applied at middle node at the free end. Deformation determined at the same node.

16.010 mm

3) 4 node elements meshed 4x20 elements (height x length). Load applied at middle node at the free end Deformation determined at

16.022 mmapplied at middle node at the free end. Deformation determined at the same node.

3Contents

ه Element stiffness matrix for beam element

ه Assembling the global stiffness matrix from the element stiffness matrices

ه How to investigate the design (e.g. the placement of reinforcement) of a concrete beam

ه Exercise: Find the optimal placement of the longitudinal reinforcement in a concrete beam

4Beam Element

General form:

Stiffness matrix: Displacement vector: Load vector:

5Stiffness Matrix for a Plane Beam Element: Axial Deformation

E: Elasticity module [MPa] A: Section area [m2]

I: Bending moment of inertia [m4] L: Beam length [m]g g

6Stiffness Matrix for a Plane Beam Element: Transversal Deformation

From: "Teknisk STÅBI" 18. udgave 1999, p. 108

7Stiffness Matrix for a Plane Beam Element: Transversal Deformation

8Stiffness Matrix for a Plane Beam Element: Rotational Deformation

9Element Stiffness Matrix for a Plane 6DOF Beam Element

10Shape Functions for Beam Element

u: Node displacement or rotation w(x): Section displacement

11Shape Functions for a Beam Element

12Assembling the Global System of Equations

Structural system of fixed beam divided into 3 elements

13Stiffness Matrix for Beam Element

14Assembling Finite Elements into a Structure

15Structural Equation

16Structural Stiffness Matrix

19Different Numbering of Nodes and Degrees of Freedom

20Boundary Conditions

Static boundary condition: f3 f6 f9 f12 are momentsStatic boundary condition: f3, f6, f9, f12 are moments

Kinematic boundary condition:

Note: One boundary condition specified for each DOF

21Changing the Boundary Conditions

Static boundary condition: f3 f6 f9 f12 are momentsStatic boundary condition: f3, f6, f9, f12 are moments

Kinematic boundary condition:

Note: One boundary condition specified for each DOF

22Different Stiffness Matrices

Local element stiffness matrix

ه Independent of element placement in global coordinate system

ه Dependent on material properties (e.g. elasticity modulus, E), section properties (e.g. area, A, moment of inertia, I) and element length, L

Global element stiffness matrix

ه Dependent on element placement in global coordinate system

F d f lti l i l l l t tiff t i ith t f tiه Found from multiplying local element stiffness matrix with transformation matrix

Global system matrixy

ه Dependent on degree-of-freedom numbering

ه Found from assembling global element stiffness matrices according to the degree-of-freedom numbering

23Design of Concrete Structures

ه A concrete structure is not in an elastic state at failure

ه At failure the cross section is cracked and only a part of the section contributesه At failure, the cross section is cracked and only a part of the section contributes to the load-carrying capacity – i.e. the part in compression

ه The elastic section-force distribution gives large values over supportsThe elastic section force distribution gives large values over supports

ه Too expensive to design according to the elastic section forces, because the pressure zone of the cross section becomes very small, hence, large amount of p y greinforcement would be needed

ه Instead a plastic section-force distribution is chosen according to the lower-bound theory

ه Find a statically acceptable and safe section force distribution, then the section design is on the safe side

24Example: Design of a Concrete Beam with Multiple Spans


Example: 2.66 m wide concrete T-beam, preinforced in the longitudinal direction with a maximum of 4 bars at the top and bottom of the section.

Positive Failure Moment Negative Failure Moment

Number of reinforcement bars 4 7 8 2 6

Bars at the top 4 4 4 2 4

Bars at the bottom 0 3 4 0 2

z [mm] 414.3 392.9 388.6 390.8 318.9

Failure moment 229.1 380.1 429.7 108.0 264.5


Firstly, an elastic analysis is made ...


Load case 1

75 kN/m75 kN/m

34 kN/m

El ti t di t ib ti

451 kNm 476 kNm

Elastic moment distribution

98 kNm

198 kNm285 kNm

198 kNm299 kNm


k

Load case 2

75 kNm

34 kN/m

El ti t di t ib tiElastic moment distribution

216 kNm 172 kNm 216 kNm

124 kNm 106 kNm 65 kNm


Charniers are added left and right of the middle supports making the system statically determinate (fulfilling first part of the lower-bound theory)

In Staad.Pro, use the “general/spec” fan, choose “beam” then “release” and release Mz, both at the “start” and “end” location of the beam (two steps!)


ه The chaniers model a failure point. Hence, the moment is known at this point. Over the supports the moment is negative (tension at the top of the beam. Be aware that Staad Pro uses a different sign convention) Hence the maximumaware that Staad.Pro uses a different sign convention). Hence, the maximum negative failure moment in this example is +264.5 kNm (not -264.5 kNm).

ه At the failure points the known moment (failure moment) is added as anه At the failure points the known moment (failure moment) is added as an external moment. In Staad.Pro this is done by choosing “Member Load”, “Concentrated Moment” “GZ” and applying a negative moment at right end of the beam (d1 = L, d2 = 0) and a positive moment at left end of the beam, i.e. e bea (d , d 0) a d a pos e o e a e e d o e bea , e(d1=0, d2 =0). The moment distribution from this load case should give the same sign of the moment at the failure points as the elastic analysis.

ه Then the total moment distribution is found as a sum of the moment distribution from the external load applied to the statically determinate system and the moment distribution from the applied external moments at the failure points.

31

Load case 1Example: Design of a Concrete Beam with Multiple Spans

Moment distribution for actual loading on statically determinate system

216 kNm

Load case 1

216 kNm

486 kNm

662 kNm

467 kNm

264 5 kNm 264 5 kNm

Moment distribution for external applied failure moments

264,5 kNm 264,5 kNm

32

Total moment distribution for load case 1Example: Design of a Concrete Beam with Multiple Spans

264,5 kNm 264,5 kNm98 kNm

Total moment distribution for load case 1

362 kNm 397 kNm 392 kNm397 kNm 392 kNm


264,5 kNm 264,5 kNm 216 kNm


, 216 kNm

108 kNm35,4 kNm 18,9 kNm


Beam 1:

The moment distribution curve is moved 1/2⋅z⋅cotθ incurve is moved 1/2⋅z⋅cotθ in the unfavourable direction. On the safe side cotθ=2.5 (DS411 p. 40)(DS411 p. 40)


Beam 2:


Number of bars at the top

Number of bars at the bottom

37Anchoring Lengths for Reinforcement in Concrete

Anchoring factor, ζ0,3 : smooth reinforcement bars0,8 – 0,9 : rough reinforcement bars

38Output report

Modify report file: " " to " " and insert failure moments at the node pointsModify report file: . to , and insert failure moments at the node points

39Excel for plotting the moment curve

40Moment Curves

60

80

40

0

20

-2 0 2 4 6 8 10 12 14 16 18

-40

-20

-2 0 2 4 6 8 10 12 14 16 18

-60

40

-100

-80

-120

41Today's Exercise

Office house

Beam to be designed

42Today's Exercise

43Today's Exercise

Positive Failure Moment Negative Failure Moment

Number of reinforcement bars 4 6 4 6Number of reinforcement bars 4 6 4 6

Bars at the top 2 2 2 2

Bars at the bottom 2 4 2 4

z [mm] 300 300 300 300z [mm] 300 300 300 300

Failure moment 50 100 50 70

1/2⋅z⋅cotθ=375 mm , fyk = 500 MPa , fck = 25 MPa, y ,Anchoring factor, ζ = 0.8 ⇒ anchoring length = 600 mm

44Today's Exercise

Question

1 Fi d h l f h b di i f1. Find the placement of the bending reinforcement of the beam, i.e. make a moment diagram.

Hint:

Find the moment curves with Staad.Pro and use e.g. Microsoft Excel for drawing the curves.An Excel file with the failure moments has beenAn Excel file with the failure moments has been put on the homepage of the course.

lecture 2: exercise and solution - wind turbine · pdf filelecture 2: exercise and solution 1...

Documents