Statics deals equilibrium of stationary bodies and bodies moving at constant velocity.
The notes on this page include use of vectors. The necessary background information is provided on page Vectors
Forces
A force identified as a localised vector it requires point of application, magnitude and direction.
The contents of this page are listed below
Basics.
Free body diagramA free body diagram is an extremely useful tool for assessing the interaction of forces on bodies This is essentially a sketch of a body which is in equilibrium and is entirely separate from the surroundings. The only rule for drawing free-body diagrams is to depict all the forces which exist for that object in the given situation. Below is shown a typical 2- dimensional free body diagram of a cantilever beam
Forces & Moments
ForceA force can be represented by a localised vector defined by magnitude, direction and point of application . A number of forces applied to any point can be replaced by a single resultant force using the principles of vector addition as shown on the vector page of this website.. From Newtons's first law it is is known that a particle will remain at rest if all the resultant force on the particle = zero . This called the equilibrium condition.. Using cartesian co-ordinate system this may be states as follow;
Moments
A moment applied to a body creates a tendency for the body to rotate.. The moment of a force about a point equals the product of a distance (say r ) = the lever arm and the force (F) acting perpendicular to the lever arm..
The moment of a force is defined in vector form using determinant algebra (in 2 dimensional and 3 dimensional cases) as
If a number of concurrent forces are applied to a point P and r is the position vector from 0 to P. the moment caused by these forces =
Mo = r x (F1 + F2 + F3...) = r x F1 + r x F2 + r x F3...
That is the moment due to several concurrent forces is equal to the sum of the moments of the indivual forces.Couples
If two forces are equal in magnitude (F), parallel in line of action on opposite in direction they result in a couple. The magnitude of the moment of the couple is the distance between the force (e) multiplied by one force. The direction of the couple is identified by the right hand rule....
Couple = e x F
It can be proved that the moment of the couple is the same magnitude at any location...If a number of forces are applied to a body resulting in a number of moments the moments can be combined algebraically to a resultant moment with a force.. An example below shows how two offset forces and a single moment can be combined to a resultant force and moment..
Rigid Body Equilibrium
Rigid Body EquilibriumA particle is in equilibrium if the resultant force acting upon it is zero... A rigid body is in equilibrium if the resultant force is zero and if the resultant moment = 0...
For the special case of 2 dimensional equilibrium which is most often applied for beams and simple structures the equilbrium equation as below is used...
Structures
Any assemblage of materials whose function is that of supporting loads is a structure.. The term may be applied to a bridge, and aeroplane wing a building or a dam. The component parts of a loaded structure are in a state of stress and the laws which govern the distribution of the stress are used to calculate the design a material to enable the structure to safely support the loads.Structures are classified into two general groups : framed structures and mass structures. The former is based on a number of separate bars or plates pinned, rivetted or welded together as a lattice. These depend upon the geometric properties of the arrangement to withstand the load. Mass structures depend upon the mass of material in the structure to withstand the resistance to the load e.g a masonary damn. The notes following relate only to framed structures.
The notes that follow relate to frameworks or trusses. These are arrangements of bars connected at pin joints which do not transmit moments. The connecting bars are only allowed to transmit tensile forces (Ties) or compressive forces (struts). It is assumed that the struts and ties experience virtually zero deformation.
For a plane frame (2- dimensional) the number of bars (N) required with J joints.
N = 2.J -3
For a space frame (3- dimensional) the number of bars (N) required with J joints.N = 3.J - 6
There are three methods of assessing frameworks- Method of Joints
- Graphical Method
- Method of Sections
The steps in this procedure are listed as follows.
- Label all pin joints joints - A, B , c etc.
- Draw and label a free body diagram for the whole framework.
- Calculate the reaction forces using the three equations for static equilibrium aplied to the whole framework
- Draw a free body diagram for each pin assembly. This may be one free body diagram showing all the pin details
- Apply the two equations of static equilibrium (related to Fx and Fy - no moments at joints ) at each joint to identify the forces in the attached bars..
Note: this method is convenient for simple frames but is very complicated for larger frames and cannot be used for space frames (3 -dimensions). The steps in this procedure are listed as follows.
- Label all pin joints joints - A, B , c etc.
- Draw and label a free body diagram for the whole framework.
- Calculate the reaction forces using the three equations for static equilibrium aplied to the whole framework
- Draw a free body diagram for each pin assembly. This may be one free body diagram showing all the pin details
- Produce a force vector polygon for each joint starting with a joint which has no more than two unknowns
- The force polygons can be combined into a single diagram called a Maxwell diagram
The steps 1,2 & 3 above have been ommited as they are relatively obvious
1) below includes a free body diagram for the each of the joints
2) below is a polygon of forces for joint A. The forces in the bar directions are drawn to scale.
3) below is a polygon of forces for joint B. the forces in the bar directions are drawn to scale.
4) below is a polygon of forces for joint C. The forces in the bar directions are drawn to scale.
5) below is a combined force diagram (Maxwell diagram ) for the truss..
Note: This method can be used to determine the forces in selected bars more rapidly then with the method of joints in some cases.. The steps in this procedure are listed as follows.
- Label all pin joints joints - A, B , c etc.
- Draw and label a free body diagram for the whole framework.
- Calculate the reaction forces using the three equations for static equilibrium aplied to the whole framework
- Draw a free body diagram for a selected part of the frame which can include two joints and breaks the bars under consideration
- Calculate the forces in the bars to one side of the cut using the 3- equations for static equilibrium
Taking moments about joint A eliminates Forces FAI and FAB leaving only the force FJI.
Sum Moments about A to zero.
0 = 100.(1,5) - FJI.(1.5) therefore FJI = +100kN (tension)
Taking moments about joint I outside of the free body, eliminates FAI and FJI leaving only FAB
Sum Moments about I to zero.
0 = 50.(1,5) - 14.(1,8) +33,67.(1,8) + FAB therefore FAB. (1,5) = -73,60kN (tension)
Summing moments vertically to zero eliminates both FJI and Force FAB leaving only FAI
0 = 33,67 -14 - FAI (5 / 7.81) therefore FAI = +30,72 (tension)
FJA = -33.67kN (compression) by expansion