**General definition**

A loaded structure is in*equilibrium*if it does not move as a rigid body. Rigid body movement can either be a translation (movement in a straight line) or a rotation or a combination of both.

For a structure to be in equilibrium, the effect of the loads, which tend to move or rotate the structure, must be balanced by reactive forces (*reactions*) developed at the*supports*upon which the structure is erected.**Plane structures**

A structure which lies within a single plane (i.e. in two dimensions) is a plane structure..**Conditions for equilibrium**

The requirement that a plane structure does not move in*any*direction may be specified by stating that it must not move in any two perpendicular directions. Normally, but not essentially, the two directions are taken to be horizontal and vertical.

The structure will not move in any direction provided that there is no resultant force in that direction.

Thus, for no movement in the horizontal direction the algebraic sum of all the forces acting horizontally must be zero - that is,.

Similarly, for no movement vertically,

The requirement that a structure does not rotate in a plane implies that it can not rotate about any axis at right angles to that plane. thus, there must be no resultant moment of force about any point in the plane. Hence, for no rotation in the plane, the algebraic sum of the moment of all the forces about any point in the plane must be zero - that is,

where the point in the plane about which the moments are taken can be either within or external to the structure.

For complete equilibrium of a plane structure,The the algebraic sum of all the horizontal forces equals 0; The the algebraic sum of all the vertical forces equals 0; The the algebraic sum of all the moments of all the forces about any point equals 0;

These are the three equations for the statical equilibrium of plane structures.

Sufficient supports and corresponding support reactions must be provided to enable the above-listed equations to be satisfied. Three unique equations are involved: thus, the magnitude of the*three*unknown fixing reactions (or fixing moments) may be determined. If a structure is provided with just sufficient supports (no more than three unknown reactions), it may be completely analysed by use of the above equations and is*externally statically determinate*. If there are too many support reactions, a solution is not possible by the use of the above equations alone and the structure is statically indeterminate. If a structure has too few support reactions, it will move as a rigid body.**Supports**(Figure 1.1)**Figure 1.1**Support types

(i) A*roller support*provides one reaction only, of unknown magnitude, and acting at right angles to the direction of motion of the roller. Such a support permits linear movement in one direction only and also permits rotation.

(ii) A*pinned support*provides a reaction of unknown magnitude and unknown direction which may be completely defined by determining the horizontal and the vertical components of reaction. Such a support prevents any linear movement but permits rotation.

(iii) A*fixed support*provides a reaction of unknown magnitude and unknown direction and also a fixing moment (a total of three unknowns). Such a support prevents any linear movement and also prevents any rotation.**Three-pinned plane structures**

If a structure contains an internal pin (hinge)(e.g. as in Figure 1.2), such that one part of the structure can rotate about the pin independently of the rotation of the other part, then one additional unique equation of equilibrium may be written, since the sum of the moments of all the forces on either part of the structure about that pin must be zero. This enables one additional unknown component of support reaction to be determined.**Figure 1.2**Three-pinned arch structure**Space structures**

A three-dimensional structure is a*space structure*.

For a space structure, the sum of all the forces in each of the three mutually perpendicular directions must be zero, and the sum of the moments of all forces about three mutually perpendicular axes (*Z, Y*and*Z*) must be zero. Thus,

The sum of the forces in the *X*direction equals 0;The sum of the forces in the *Y*direction equals 0;The sum of the forces in the *Z*direction equals 0;The sum of the moments about the *X*axis equals 0;The sum of the moments about the *Y*axis equals 0;The sum of the moments about the *Z*axis equals 0;**Mass structures****Figure 1.3**Gravity dam

A mass structure, such as the gravity dam shown in figure 1.3, is one which depends upon its own weight to ensure equilibrium. Thus for equilibrium,the weight ( *W*) of the structure and the vertical components*T*of any loads (_{V}*T*) must be balanced by the vertically upward reaction (*V*) of the ground (or foundation) below the structure;the tendency for the structure to slide horizontally under the action of any horizontal components ( *T*) of load_{H}*T*must be prevented by a reaction (*P*)from the ground behind the structure and/or by a frictional force (*R*) between the structure and the ground beneath it;the *overturning*moment of the loads about a probable centre of rotation (O) must be balanced by the*restoring moment*about the same point due to the self-weight of the structure.

A mass structure is normally designed so that its weight is greater that the minimum required for equilibrium in order to provide a factor of safety against overturning, where

Similarly, a factor of safety against sliding will be provided, where**Principle of superposition**

If a structure is made of linear elastic material and is loaded by a combination of loads which do not strain the structure beyond the linear elastic range, then the resultant effect of the total load system on the structure is equivalent to the algebraic sum of the effect of each load acting separately.

H |
= | horizontal reactions at a support (kN) |

M |
= | moments taken about a support (kNm) |

V |
= | the vertical reaction at a support (kN) |

Clockwise moments are assumed to be positive.

Forces or components of forces acting to the right are assumed to be positive.

Forces or components of forces acting vertically upwards are assumed to be positive.

The direction of action of known forces is indicated by solid arrowheads on the diagrams.

The assumed direction of action of unknown forces is indicated by open arrowheads on the diagrams.

**General definition**

A pin-jointed frame is a structure constructed from a number of straight members connected together at their ends by frictionless pinned joints. It is assumed that all external loads are applied at the joints of the structure such that the internal members are subjected to axial forces only.**Internal statical determinacy**

A plane frame is internally statically determinate (that is, capable of solutions using the equations of equilibrium) if it satisfies the relationship,

where*M*= number of members and*J*= number of joints. Similarly, for a space frame in equilibrium,**Positioning of members**

Not only must the frame satisfy the relationship given above, but also, to ensure stability, the members must be correctly positioned in the frame. In general, a frame built up as a series of triangles will be stable. Fore example, in Figure 2.0 frame (a) is stable and frame (b) is unstable (the right-hand panel is free to collapse as indicated in Figure 2.1(c)).**Figure 2.1****Method of resolution at joints**

To determine the forces in all the members of a frame, it is, in general, best to use the method of*resolution at joints*. At any point in a plane frame at which there are no more that*two*unknown forces, write down two equations of equilibrium obtained by resolving all forces at that joint in two mutually perpendicular directions. Solve these equations for the two unknown forces. Proceed systematically through the frame until all the forces are known. In the case of a space frame, three equations of equilibrium may be written down, thus, joints joint with three unknown forces may be solved.

Usually forces will be resolved in the vertical and horizontal directions (*ΣV = 0*and*ΣH = 0*).However, it may be convenient in some problems to resolve in two other mutually perpendicular directions. In many cases forces can be determined by*inspection*of the direction of members and forces acting at a particular joint.

External support reactions are usually determined before internal forces are calculated. However, depending on the geometry of the frame and the position of the supports, this may not always be necessary.**Method of sections**

To calculate the value of a force in only one of a few members in a frame, use the*method of sections*. Cut the frame, as indicated in Figure 2.2, by a section through the member under consideration and no more that two other members in which the member forces are unknown. Both parts of the structure can then be treated as structures in equilibrium and either part can then be solved by resolving forces or taking moments of forces.**Figure 2.2**Frame cut for solution by method of sections

Graphical methods of solution, although largely superseded by modern computer methods, may be useful to solve frames of complicated geometry. Commencing at a joint at which there are no more than two unknown forces, construct a force polygon for all the internal and external forces acting at that joint. The magnitude and the direction of the unknown forces can then be scaled from the polygon. Proceed systematically through the frame, constructing a force polygon at each joint, to produce one composite force diagram for the whole structure.

F |
= | the internal force in a member (kN) |

V |
= | the vertical component of reaction at the support (kN) |

H |
= | horizontal component of reaction at the support (kN) |

R |
= | the reaction at a support (kN) |

θ |
= | the acute angle between an inclined member and the horizontal |

Tensile forces are assumed to be positive.

Clockwise moments are assumed to be positive.

Forces or components of forces acting vertically upwards are assumed to be positive.

Forces or components of forces acting to the right are assumed to be positive.

In the various methods of analysis employed all unknown internal forces in the members of frames are assumed to be tensile. If the result of a calculation is a positive answer, then the initial assumption was correct, the member is in tension and it is a *tie*. If the answer is negative, the initial assumption was incorrect, the member must be in compression and it is a *strut*.

The assumed direction of action of unknown forces is indicated by open arrowheads on the diagrams of the frame.

The direction of action of known forces is indicated by solid arrowheads on the diagrams of the frames.

**General definition of shearing force**

The shearing force at any section in a horizontal beam is the algebraic sum of all the vertical forces to the left, or to the right, of that section.**General definition of bending moment**

The bending moment at any section in a beam is the algebraic sum of the moments about that section of all the forces to the left, or to the right, of that section.**Shearing force diagram**

A shearing force diagram is a graph showing the variation of the shearing force along the length of the beam or other structural member.

Shearing force diagrams resulting from the action of concentrated loads on horizontal beams consist of horizontal straight lines with vertical jumps at load positions.

Shearing force diagrams resulting from the actions of uniformly distributed loads on horizontal beams consist of sloping straight lines.**Bending moment diagram**

A bending moment diagram is a graph showing the variation of the bending moment along the length of a beam or other structural member.

Bending moment diagrams resulting from the action of concentrated point loads consist of straight lines with changes of direction at the points of application of the loads.

Bending moment diagrams resulting from the action of uniformly distributed loads consist of parabolic curves.

At the point of application of a couple to a horizontal beam, there will be a vertical step in the bending moment diagram.

The bending moment at an internal pin or a pinned support is zero.**Points of contraflexure**

The direction of curvature changes at a point of contraflexure - that is, at a point where the bending moment is zero.**Standard formulae for maximum bending moment**

If a single concentrated vertical load*W*acts at the mid-point of a simply supported beam of span*L*, then the maximum bending moment is at the point of application of the load and is of magnitude*WL/4*.

If a uniformly distributed load*w*per unit length acts along the entire length of a simply supported beam of span*L*, then the maximum bending moment is at mid-span and is of magnitude*wL*.^{2}/8

H |
= | horizontal reactions at the support (kN) |

V |
= | vertical reactions at the support (kN) |

B.M. |
= | bending moment (kNm) |

S.F. |
= | shear force (kN) |

The shearing force at a section in a beam is taken to be positive if the resultant vertical loading on the part of the beam to the left of that section tends to move that part of the beam vertically upwards.

The bending moment in a beam is taken to be positive if it causes the beam to sag.

Bending moment diagrams are plotted such that the positive bending moments are on the tensile side of the beam or other structural member.

**Direct (normal) stress**

A force which acts normal to a surface causes stress which also acts normal to that surface. Provided that the force passes through the centroid of the surface, then the stress will be uniform over the whole surface. This type of stress is known as a*direct*stress (σ). The direct stress can be calculated from the expression**Direct strain**

A body subject to a direct stress will deform and be in a state of*strain*. The direct strain (*ε*), which is measured in the same direct as the direct stress(*σ*), is given by the expression

where both*δL*and*L*are measured in the direction of the applied normal force (*P*).**Stress-strain relationship**

The direct stress (σ) and the direct strain (ε) are related by Young's modulus of elasticity (*E*) in the expression**Lateral strain**

The straining of any elastic material will give rise to a change in the lateral dimensions and, hence, to*lateral strain*in all directions at right angles to the direction of the applied force.**Poisson's ratio**

For any elastic material, the direct (longitudinal) strain and the indirect lateral strain are related by Poisson's ratio (*v*), which is given by the expression**Stress in thin-walled cylinders**

Thin-walled cylinders, such as pressure vessels, are subject to direct stresses along their length normal to all cross-sections of the cylinder. They are also subject to stresses which act circumferentially around the cylinder. These circumferential stresses are called*hoop stresses*. Although there are standard formulae for the calculation of both the longitudinal and hoop stresses, they are best calculated by the application of fundamental principles as illustrated by the examples in this chapter.

A |
= | area (mm^{2}) |

E |
= | Young's modulus of elasticity (kN/mm^{2}) |

L |
= | length (m) |

P |
= | force (kN) |

ϵ |
= | strain (strain is dimensionless and therefore has no units) |

ν |
= | Poisson's ratio (dimensionless) |

σ |
= | stress (N/mm^{2}) |

Tensile stress and tensile strain are both taken as positive.

**Bending stress**

The effect of a bending moment applied to a cross-section of a beam is to induce a state of stress across that section. These stresses are known as bending stresses and they act normally to the plane of the cross-section.

Bending stresses vary linearly across the section, with maximum value (compression or tension) at the outer fibres of the beam and with zero value at the level of the*neutral axis*.**Neutral axis**

The*neutral axis*is the axis of the cross-section of a beam at which both the bending strain and bending stress are zero. The neutral axis passes through the centroid of the cross-section.**Equations of bending**

The fundamental equations that govern the bending of a beam or other structural member are given by:where σ = the bending stress in a layer of fibres distance *y*from the neutral axis;*M*= the bending moment at the section under consideration; *I*= the second moment of area of the cross-section taken about the neutral axis; *E*= Young's modulus of elasticity; and *R*= the radius of curvature at the section under consideration. **Second moment of area**

The*second moment of area*of a cross-section about the neutral axis is a geometrical property of that cross-section and can be obtained from tables of standard values or can be calculated.

For a rectangular section of breadth*b*and depth*d*, the second moment of area (*I*) of that section about an axis through the centroid is given by:_{CC}**Parallel axis theorem**

The second moment of area (*I*) of a cross-section about an_{XX}*X-X*axis which is parallel to and at a distance*h*from an axis through its own centroid is given bywhere *I*_{CC}= the second moment of area about the centroidal axis; and *A*= the area of the cross-section. **Elastic section modulus**

The elastic section modulus (*Z*) of a section is the second moment of area (*I*) about the axis of bending, divided by the distance (*y*) from the neutral axis to the furthermost fibre of the section. That is,_{max}

If the neutral axis is not at the mid-height of a section, that section will have two section moduli - one relating to the compression face and one relating to the tension face.**Moment of resistance**

The moment of resistance (*M*) of a beam at a particular section is given bywhere *σ*_{max}= the maximum permissible bending stress; and *Z*= the elastic section modulus of that section.

A |
= | area (mm^{2}) |

h |
= | the distance between an axis through the centroid of a section and any parallel axis (mm) |

I_{CC} |
= | the second moment of area of a section about an axis through the centroid of its own cross-sectional area (mm^{4} or cm^{4}) |

I_{XX} |
= | the second moment of area of a section about an X-X axis (mm^{4} or cm^{4}) |

σ |
= | bending stress (N/mm^{2}) |

y |
= | the distance from the neutral axis to the level at which the bending stress is being calculated (mm) |

y_{max} |
= | the distance from the neutral axis to the top or bottom face of the beam (mm) |

Z |
= | elastic section modulus of a section about an X-X axis (mm^{3} or cm^{3}) |

**Single-axis bending**

The maximum stress in a section subject to direct axial loading together with an externally applied moment acting about either*X-X*or the*Y-Y*axis are given by

or

respectively, whereA = the area of the cross-section under load; *P*= the direct axial load; *M*_{X}= the externally applied moment about the *X-X*axis;*M*_{Y}= the externally applied moment about the *Y-Y*axis;*Z*_{X}= the elastic section modulus of the section about the *X-X*axis; and*Z*_{Y}= the elastic section modulus of the section about the *Y-Y*axis.**Biaxial bending**

The maximum stresses in a section subject to direct axial loading together with externally applied moments acting about two mutually perpendicular axes (*X-X*and*Y-Y*) are given by**Eccentrically loaded sections**

If a load*P*acts at an eccentricity*e*from the centroidal axis of a section, then that loading is equivalent:- A direct load
*P*acting axially through the centroid of the section, together with - a moment of value
*P × e*acting about the axis of bending which passes through the centroid of the section.

- A direct load

A |
= | the area of the cross-section under load (mm^{2}) |

e_{x} |
= | the eccentricity of the load measured perpendicular to the Y axis (mm) |

e_{y} |
= | the eccentricity of the load measured perpendicular to the X axis (mm) |

M_{X} |
= | the external applied moment about the X-X axis (kNm) |

M_{Y} |
= | the external applied moment about the Y-Y axis (kNm) |

P |
= | the direct axial load (kN) |

Z_{X} |
= | the elastic modulus of the section about the X-X axis (mm^{3}) |

Z_{Y} |
= | the elastic modulus of the section about the Y-Y axis (mm^{3}) |