Structural Analysis & Design

UNIT 22: STRUCTURAL BEHAVIOUR AND DETAILING

Learning outcomesLO1- Ability to determine the properties of structural materials

LO2- Understanding of fundamental structural conceptsLO3- Analysis of statically determinate structures

LO4- Design of structural elementsLO5- Detailing of structural elements

LinksUnit 26 : Properties and performance of construction materials

Unit 34 : Structural analysis and design

04/22/23

Prepared by: Eng. Chamil Duminda Mahagamage B.Sc.Eng (Hons), C Eng, MIE(SL)

1International College of Business and Technology

M/601/1282 - Lecture Note: 03

04/22/23

Prepared by: Eng. Chamil Duminda Mahagamage B.Sc.Eng (Hons), C Eng, MIE(SL) 2International College of Business and Technology


Analysis of Structures

Analytical MethodsThere are three approaches to the analysis

1. Mechanic of Materials (Strength of materials)2. Elastic theory (Special case of the more general field of continuum mechanics)3. Finite element

Mechanic of Material ApproachAssumptionsa) The materials in question are elastic, that stress is related linearly to strainb) Material (but not the structure) behaves identically regardless of direction of

the applied load.c) All deformations are small.d) Beams are long relative to their depth.

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Statically determinate and indeterminate structuresIf for the stable structure it is possible to find the internal forces in all the

members constituting the structure and supporting reactions at all the supports provided from statical equations of equilibrium only, the structure is said to be statically determinate.

System for which the principles of statical equilibrium are insufficient to determine support reactions and/or internal force distributions, i.e. there are greater number of unknowns than the number of equations of statical equilibrium, are known as statically indeterminate or hyperstatic systems.Structural systems may be

1. Externally indeterminate but internally determinate2. Externally determinate but internally indeterminate3. Externally and internally indeterminate4. Externally and internally determinate

** A system which is externally and internally determinate is said to be determinate system.

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Statical Determinacy in framesLet: m= total number of members in a truss j= number of connections

Minimum number of members required for the geometric or configuration stability m = 2j-3

If m< 2j-3 structure is unstable m> 2j-3 structure with redundant members

** The expression serves as an indicator whether or not the internal forces in a structure can be calculated by the equations of statics. Since all the j joints of the structure are in equilibrium and there are two equilibrium equations involving the summation of forces in X- and Y- directions, namely ΣFx=0 and ΣFy=0 at each joint, 2j equations of static equilibrium are available for the entire structure to compute the support reactions and internal forces in all the members.

Let: m= unknown member forces r= unknown support reactions

For the adequacy of the number of the available equilibrium equations to compute the support reactions and internal forces in all the members,

2j=(m+r) or m=2j-r - statically determinate

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If 2j < m+r, there are more unknowns than the number of equilibrium equations, the structure is statically indeterminate. The degree of indeterminacy n=(m+r)-2j

If 2j > m+r, there are more equilibrium equations available than the number of unknowns, such a structure is a mechanism and always unstable. The structure does not have unique solution. Existence of more than one solution indicates instability

Ex:1 Determine whether the trusses shown in following figures are stable. If stable, then find whether they are statically determinate or indeterminate.

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Sign ConventionUse normal sign convention adopted for a 3-dimensional right-handed system of

Cartesian or rectangular coordinate axes OX, OY and OZ with origin O on the extreme left of the structure. The forces measured from the origin towards positive directions of axes are always positive. Rotational moments expressed in vector form pointing towards positive directions of the axes are positive. Thus, moments that tend to produce counterclockwise rotations are considered positive and those tend to produce clockwise rotations are considered negative.

Free - Body Diagrams For static analysis of bodies subjected to external loads, analytical diagrams that

illustrate the force systems acting on the objects are called equilibrium or free-body diagrams. Using equilibrium concepts, the numerical values of reactions that occur at supports and hence internal forces, i.e. axial forces, shear forces and bending moments can be determined. The major application of equilibrium analysis is in the evaluation of reactions and internal forces by representing an object by a series of free body diagrams.

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Shear Force and Bending MomentRelationships among loading, shear force, and bending moment

A small element of the beam shown in figure above (left) is taken at a distance x from end 1. The forces acting on the element are shown in figure above (right).

Resolving forces vertically,V = (V+δV)+wδx and δV/δx = -w

Limiting condition dV/dx = -w indicates that the slope of the shear force diagram, at any section,

equals the intensity of loading at that section.

V

V+δV

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Alternatively, sincedV = -wdx ʃdV = ʃ-wdx, and

V2-V1 = ʃx1

x2 -wdx where,

V1= Shear force in the beam at x=x1, V2 = shear force in the beam at x=x2 and the change in shear force between the two sections equals the area of the load intensity diagram between the two sections.

Taking moments about the lower right corner of the element gives the expressionM = (M+δM)- Vδx + w (δx )2/2

Neglecting the small value (δx )2 , δM/δx = V

The limiting condition isdM/dx = V indicates that the slope of the bending moment diagram at any

section equals the shear force at that section.

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It is possible for the total work done by the forces to be zero even though the particle is not in equilibrium if the virtual displacement is taken to be in a direction perpendicular to their resultant, R. We cannot, therefore, state the converse of the above principle unless we specify that the total work done must be zero for any arbitrary displacement. Thus;

A particle in equilibrium under the action of a system of forces if the total work done by the forces is zero for ant virtual displacement of the particle.

Note-: The ∆v is a purely imaginary displacement and is not related in anyway to the possible displacement of the particle under the action of the forces F. ∆v has been introduced purely as a device for setting up the work-equilibrium relationship. The forces, F, therefore remain unchanged in magnitude and direction during this imaginary displacement; this would not be the case if the displacement were real.

Principle of Virtual work for a Rigid Body

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Consider the rigid body shown in figure above, which is acted upon by a system of external forces, F1, F2,…., Fk,…, Fr. These external forces will induce internal forces in the body, which may be regarded as comprising an infinite number of particles; on adjacent particles, such as A1 and A2, these internal forces will be equal and opposite, in other words self-equilibrating. Suppose now that the rigid body is given a small, imaginary, that is virtual displacement, ∆v (or a rotation or a combination of both), in some specified direction. The external and internal forces then do virtual work and the total virtual work done , W t, is the sum of the virtual work, We, done by the external forces and the virtual work, W i, done by the internal forces.

Wt = We + Wi

Since the body is rigid, all the particles in the body move through the same displacement, ∆v, so that the virtual work done on all the particles is numerically the same. However, for a pair of adjacent particles, such as A1 and A2 in above figure, the self equilibrating forces are in opposite directions, which means that the work done on A1 is opposite in sign to the work done on A2. Thus the sum of the virtual work done on A1 and A2 is zero. The argument can be extended to the infinite number of pairs of particles in the body from which we conclude that the internal virtual work produced by a virtual displacement in a rigid body is zero.

Wt = We

Since the body is rigid and the internal virtual work is therefore zero, we may regard the body as a large particle. It follows that if the body is in equilibrium under the action of set of forces, F1, F2,…., Fk,…, Fr , the total virtual work done by the external forces during an arbitrary virtual displacement of the body is zero.

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Virtual work in a deformable bodyIn structural analysis we are not generally concerned with forces acting on a rigid body. Structures and structural members deform under load, which means that if we assign a virtual displacement to a particular point in a structure, not all points in the structure will suffer the same virtual displacement as would be the case if the structure were rigid. This means that the virtual work produced by the internal forces is not zero as it is in the rigid body case, since the virtual work produced by the self-equilibrating forces on adjacent particles does not cancel out. The total virtual work produced by applying a virtual displacement to a deformable body acted upon by a system of external forces is Wt = We + Wi

If the body is in equilibrium under the action of the external force system then every particle in the body is also in equilibrium. Therefore, from the principle of virtual work, the virtual work done by the forces acting on the particle is zero irrespective of whether the forces are external or internal. It follows that, since the virtual work is zero for all particles in the body, it is zero for the complete and We + Wi = 0

Note that in the above argument only the conditions of equilibrium and the concept of work are employed. Thus the above equation does not require the deformable body to be linearly elastic (i.e.it need not obey Hooke’s law) so that the principle of virtual work may be applied to any body or structure that is rigid, elastic or plastic. The principle does require that displacements, whether real or imaginary, must be small, so that we may assume that external and internal forces are unchanged in magnitude and direction during the displacements.

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In addition the virtual displacements must be compatible with the geometry of the structure and the constraints that are applied, such as those at a support .

ProblemsCalculate the support reactions in the simply supported beam shown in figures below.

1. 3.

2. 4.

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Structural Analysis & Design