CHAPTER 2- Analysis of Statically Determinate Structures

7/28/2019 CHAPTER 2- Analysis of Statically Determinate Structures

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Alexandria University Dept. of Marine Engineering & Naval ArchitectureFaculty of Engineering Instructor: Prof. H. W. Leheta

CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATESTRUCTURES

2.1 Idealized structure. Idealization of ship structures

Estimates of the loadings, points of application and material strength of a structuremust be made to analyze a real structure. Therefore, an exact analysis can never becarried out. A structural engineer, or naval architect (in case of ship structures) mustbe able to model or idealize the structure in order to make a force analysis of thestructural members. Structural idealization or mathematical modeling (i.e. replacingthe actual structure with a simple system that is possible to analyze mathematically)involves the following techniques: support connections, idealizing the structure andload modeling.

2.1.1 Support Connections

Structural members are joined together in various ways. Fig. 1.9 shows the mostcommon supports. When selecting a particular support or joint, the engineer must beaware of how the assumptions will affect the performance and whether they arereasonable.

2.1.2 Structure Idealization

A structure must be represented in a simple manner to allow analysis. Structuralcomponents have width and thickness. Lines that are located along the centroids of

the structural components are used to represent them. The sketch of a structureidealized in this manner is usually called a line diagram. For the beam shown in Fig.2.1(a), the connection at point A allows some freedom for rotation and may beidealized as a typical pin support, while the support at B provides a point of smoothcontact and may be idealized as a roller support. Fig. 2.1(b) shows the line diagramof the beam.

Fig. 2.1

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A line diagram for a transverse structure of a ship is shown in Fig. 2.2.

Fig. 2.2

2.1.3 Load Modeling

Forces must also be modeled. The analysis of the loadings should give results thatclosely approximate the actual loadings. Concentrated forces rarely act at a point.

They are distributed over small areas. Distributed forces act over a large area. Theymay be: uniform, triangular, trapezoidal, sinusoidal, etc.

2.2 Equations of equilibrium

From statics, we know that a structure or a structural member is in equilibrium whena balance of forces and moments is maintained. For plane (2-D) structures, thefollowing requirements for equilibrium must be satisfied:

= 0xF

= 0yF (2.1)

= 0zM

That is: the algebraic sums of all forces acting in the x and y directions, respectively,must equal zero. Also, the algebraic sum of the moments of these force componentsabout an axis perpendicular to the x-y plane (the z axis) must equal zero.

Before applying the equations of equilibrium it is necessary to draw a free-bodydiagram of the structure or its members.

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2.3 Free-body diagrams

For a member, it must be isolated from its supports and adjacent members. Allforces and moments that act on the member must be shown, including supportreactions. Forces common to two members act with equal magnitudes but different

directions on the respective free-body diagrams of the members.Ifinternal forces and moments at a point in a member are to be determined, a cut orsection perpendicular to the axis of the member at the point in question is made.Each segment is isolated and equations of equilibrium are applied. Internal loadingsconsist of a normal force N, shear force V, and bending moment M. Fig 2.3(a) showsa beam with two supports acted upon by two concentrated loads. In Fig. 2.3(b), afree-body diagram of the entire beam shows the reaction forces. If we cut the beamat A, we obtain two segments of the beam. Fig. 2.3(c) shows the free-body diagramswith the internal forces now seen. Notice that the internal forces on the twosegments are the same but in opposite direction. The right-hand part pushes the left-hand part down and the left-hand part pushes the right-hand part up.

Free-body diagrams open the way to the analysis of structures.

Fig. 2.3

2.4 Determinacy and stability

DeterminacyWhen all the forces in a structure can be determined solely using the equilibriumequations, the structure is statically determinate. Structures with more unknownforces than equilibrium equations are called statically indeterminate.In order to determine if a structure is statically determinate or indeterminate, a free-body diagram must be drawn for its members or segments and the total number ofunknown reaction forces and moments is compared with the number of equilibriumequations (3 equilibrium equations for each segment). So, if there are n segmentsand rreactions:

,3nr= statically determinate (2.2a),3nr> statically indeterminate (2.2b)

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The following examples show how to classify the determinacy of a structure.

2.4.1 Beams

Example 1

Fig. 2.4a to 2.4d show beams that are to be classified as determinate orindeterminate. Fig. 2.4c and 2.4d show compound beams composed of pinconnected members. The free-body diagrams are shown.

Fig. 2.4

Example 2

This example shows pin-connected structures which are similar to beams. The free-body diagrams are shown in Fig. 2.5 along with the resulting classifications.

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Fig. 2.6

Stability

To ensure equilibrium of a structure, the members must be properly constrained bytheir supports. Two situations may occur:Partial constraints Sometimes there are less reactions than equilibrium equationsand the structure is partially constrained, such as shown in Fig. 2.7. Here horizontalequilibrium is not satisfied and the structure is unstable.

Fig. 2.7

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Improper constraints - If support reactions are concurrent at a point or are allparallel, instability or movement of the structure develops, as shown in Fig. 2.8.

Fig. 2.8

For a 2-D structure having n members or segments with runknown reactions, wehave:

nr 3< unstable (2.3a)nr 3 unstable if reactions are concurrent or parallel (2.3b)

The following example shows how structures are classified as stable or unstable.

Example 4

r= 3, n = 1. The structure is stable and statically determinate.

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r= 8, n = 2. The structure is stable and indeterminate to the second degree.

The structure is unstable since the three reactions are concurrent at B.

The structure is unstable since the three reactions are parallel.

r= 7, n = 3. The structure is unstable, AB can move horizontally.

2.5 Princip le of superposition

The total internal loadings at a point in a structure subjected to several externalloadings can be determined by adding together the internal loadings caused by eachof the external loads acting separately.

For the principle of superposition to apply, the following must be satisfied:

1- The material is linear-elastic, and Hooke's law applies: = E.

2- The geometry of the structure does not undergo significant change(displacements are small)

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Fig. 2.10 represents how this principle is applied to a simply supported beam.

Fig. 2.10

2.6 Application of the equations of equilibrium

Consider the three-member pin-connected frame shown in Fig. 2.11. It is subjectedto loads P1 and P2. There are nine unknowns and nine equations of equilibrium, sothe frame is statically determinate. From the free-body diagram of the entire frame,the three reactionsAx,Ay and Cx can be determined. By analyzing any two of the

remaining members, the other six unknowns may be determined. Applyingequilibrium to the third member, the answers can be checked.

Fig. 2.11

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Example 6

The shown girder supports a boat and is idealized as shown in Fig. 2.14a, where A isa roller and B is a pin. Fig. 2.14b shows the loads acting on the girder: a uniformlydistributed deck load, a horizontal wind force and the weight of the boat which isequal to 23 tonnes. Determine the reactions at the supports. The boat and girder areconsidered as a single system.

Fig. 2.14

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Example 7Determine the horizontal and vertical components of reaction at the pins A, B and Cof the pin-connected frame shown in Fig. 2.15.

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CHAPTER 2- Analysis of Statically Determinate Structures

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