Basic mechanics. Basic principles of statics. Statics is the branch of mechanics that deals with the equilibrium of stationary bodies under the action of forces. Mechanics is that branch of science which deals with the forces and their effects units: The basic quantities or fundamental quantities of mechanics are those. Basic principles: Equivalent force system; Equations of equilibrium; Free R. C. Hibbler, Engineering Mechanics: Principles of Statics and Dynamics, Pearson.
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is concerned with the basic principles of the Universe. ▫ is one of the his Principia. ▫. Today, mechanics is of vital inportance to students from all disciplines. Worked Examples from Introductory Physics. Vol. I: Basic Mechanics. David Murdock. Tenn. Tech. Univ. February 24, FLUID MECHANICS. Fluid statics. Equations of fluid motion. Bernoulli's equation. Conservation of mass. Viscous effects.
The point of application must also be specified. A vector is illustrated by a line, the length of which is proportional to the magnitude on a given scale, and an arrow that shows the direction of the force. Static determinacy If a body is in equilibrium under the action of coplanar forces, the statics equations above must apply. In general, three independent unknowns can be determined from the three equations. Note that if applied and reaction forces are parallel i.
Such systems of forces are said to be statically determinate. Vector addition The sum of two or more vectors is called the resultant. The resultant of two concurrent vectors is obtained by constructing a vector diagram of the two vectors. The vectors to be added are arranged in tip-to-tail fashion. Where three or more vectors are to be added, they can be arranged in the same manner, and this is called a polygon. A line drawn to close the triangle or polygon from start to finishing point forms the resultant vector.
The subtraction of a vector is defined as the addition of the corresponding negative vector. A force can be defined quantitatively as the product of the mass of the body that the force is acting on and the acceleration of the force. The following multiples are often used: Concurrent coplanar forces Forces whose lines of action meet at one point are said to be concurrent.
Coplanar forces lie in the same plane, whereas non-coplanar forces have to be related to a three-dimensional space and require two items of directional data together with the magnitude. Two coplanar non-parallel forces will always be concurrent.
When the resultant of all forces acting on a particle is zero, the particle is in equilibrium, i. The closed triangle or polygon is a graphical expression of the equilibrium of a particle. The equilibrium of a particle to which a single force is applied may be maintained by the application of a second force that is equal in magnitude and direction, but opposite in sense, to the first force. This second force, which restores equilibrium, is called the equilibrant. When a particle is acted upon by two or more forces, the equilibrant has to be equal and opposite to the resultant of the system.
Thus the equilibrant is the vector drawn closing the vector diagram and connecting the finishing point to the starting point. The force effects along these axes are called vector components and are obtained by reversing the vector addition method. This does not necessarily apply for more than three forces. Choose the free body to be used. To draw a free-body diagram: The first step in the solution of a problem should therefore be to draw a free-body diagram. The free-body of a particle is used to represent a point and all forces working on it.
Locate all external forces on the free body and clearly mark their magnitude and direction. Include all dimensions that indicate the location and direction of forces.
T AB N. A Chapter 6 — Basic mechanics N Free body diagram for point A 95 Free-body diagram of a particle A sketch showing the physical conditions of a problem is known as a space diagram. The free-body diagram of a rigid body can be reduced to that of a particle. When solving a problem it is essential to consider all forces acting on the body and to exclude any force that is not directly applied to the body.
Determine the tension in the cable and the compression in the rod. Locate and mark unknown external forces and reactions in the free-body diagram. A free-body diagram of a body is a diagrammatic representation or a sketch of a body in which the body is shown completely separated from all surrounding bodies. This should include the weight of the free body. In general. Equilibrium is also indicated by two sums of moments equal to zero. Reactions Structural components are usually held in equilibrium by being secured to rigid fixing points.
N Rough surfaces F Rough surface is capable of supporting a tangental force as well as a normal reaction. Collinear forces Collinear forces are parallel and concurrent.
Note that. The sum of the forces must be zero for the system to be in equilibrium. They will be in equilibrium if the sum of the forces equals zero and the sum of the moments around a point in the plane equals zero. The forces generated in the supports are called reactions. Resultant reaction is vectorial sum of these two. Resultant of gravitational forces The whole weight of a body can be assumed to act at the centre of gravity of the body for the purpose of determining supporting reactions of a system of forces that are in equilibrium.
The fixing points or Table 6. The support is capable of providing a longitudinal reaction H. N Roller support Reaction is normal to the supporting surface only. A can then be found. This gives the line of action of the reaction at A.
From this point of concurrency. Use these three force directions and the magnitude of RL to draw the force diagram. Determine the supporting reactions graphically: This point is the point of concurrency for the resultant load. A vertical component would have indicated a friction force between the ladder and the wall.
Owing to the roller support reaction RB will be vertical. The end A is pinned to a rigid support.
Therefore the resultant line RL must be extended to intersect the vertical reaction of support B. At the bottom. This in turn enables the force vector diagram to be drawn.
RA As the wall is smooth. Example 6. As the two weight forces in this example have the same line of action. Combine the two applied forces into one and find the line of action. Determine the support reactions at the wall RW and at the ground RG. The determination of the moment of a force in a coplanar system will be simplified if the force and its point of application are resolved into its horizontal and vertical components.
It is common knowledge that a small force can have a large turning effect or leverage. At least one point on the line of action for the other reaction must be known 2 and 3 reduce the number of unknowns related to the equations of equilibrium to an acceptable level. Moments of forces The effect of a force on a rigid body depends on its point of application. In mechanics. The line of action of one of the reactions must be known.
The equilibrant of the downward forces must be equal and opposite to their resultant.
According to the third condition for equilibrium. Take the moments around point A. The reactions RA and RB must both be vertical because there is no horizontal force component. By taking moments around the point where the ladder rests on the ground. All forces apart from the two reactions must be known completely. Using the first condition of equilibrium it can be seen that the horizontal component of RG must be equal but opposite in direction to RW.
The direction of a moment about a point or axis is defined by the direction of the rotation that the force tends to give to the body. A moment has dimensions of force times length Nm. A clockwise moment is usually considered as having a positive sign and an anticlockwise moment a negative sign. This provides a method for calculating the resultant of a system of parallel forces.
The distance d is often called lever arm. The moment of force with a magnitude F about a turning point O is defined as: It can be seen from the figure below that beam A carries the floor loads contributed by half the area between the beams A and B.
In the same way. The magnitude of the force is always indicated. A couple acting on a body produces rotation. Beam C carries the loads contributed by the shaded area M. In calculating reactions. Note that the couple cannot be balanced by a single force. Uniformly distributed loads. Concentrated loads are represented by a single arrow drawn in the direction.
Concentrated loads are those that can be assumed to act at a single point. Irregular loading is difficult to deal with exactly. This technique must not be used for calculation of shear force. Let the load arising from the weight of the floor itself and the weight of any material placed on top of it e.
To produce equilibrium.
These can then be dealt with in mathematical terms using the principle of superposition to estimate the overall combined effect. The loading per metre run can then be used to calculate the required size of the beams. Chapter 6 — Basic mechanics 99 Couples Two equal. For the purpose of calculation. Both the shearing force and the bending moment will be zero between C and B. Bending moment M at any transverse cross-section of a straight beam is the algebraic sum of the moments.
The cantilever must therefore transmit the effect of load W to the support at A by developing resistance on vertical cross-section planes between the load and the support to the load effect called shearing force. To prevent rotation of the beam at the support A. For equilibrium. The shearing force and the bending moment transmitted across the section x-x may be considered as the force and moment respectively that are necessary to maintain equilibrium if a cut is made severing the beam at x-x.
The loading shape is triangular and is the result of such actions as the pressure of water on retaining walls and dams. The primary load effect that a beam is designed to resist is that of bending moments but. Shear force V is the algebraic sum of all the transverse forces acting to the left or to the right of the chosen section.
The free-body diagrams of the two portions of the beam are shown in D. Consider the cantilever AB shown in A. The bending effect of the load will cause the beam to deform as in C. Failure to transmit the shearing force at any given section.
X R Shear-force SF and bending-moment BM diagrams Representative diagrams of the distribution of shearing forces and bending moments are often required at several stages in the design process. In the case of uniformly distributed loads. Where reinforcement may be required in certain types of beam. This is referred to Definitions Shear force Q is the algebraic sum of all the transverse forces acting to the left or to the right of the chosen section.
Thus it will be seen that uniform loads cause gradual and uniform change of shear. The bending-moment diagram gives useful information on the deflected shape of the beam. Maximum bending moment values will occur where the shear force is zero or where it changes sign. The shear-force diagram will provide useful information about the bending moment at any point. For a beam or part of a beam carrying a UDL only.
This point is referred to as point of inflexion. Table 6. These tend to make the beam concave upwards and are called sagging bending moments. Over a part of the span for which SF is zero. Bending moment M at any transverse cross section of a straight beam is the algebraic sum of the moments.
If the moment is anticlockwise on the left and clockwise on the right. These graphical representations provide useful information regarding: It is usual to construct these diagrams in sets of three. The bending moment is considered positive if the resultant moment is clockwise on the left and anticlockwise on the right.
Some rules for drawing shear-force and bendingmoment diagrams are: When the load is uniformly distributed. The most likely section where a beam may fail in shear or in bending.
If a beam is subjected to two or more different systems of loading. In the absence of distributed loads. At the point where the shear-force diagram passes through zero i.
These diagrams are obtained by plotting graphs with the beams as the base and the values of the particular effect as ordinates. Shearing forces. At a point where the bending-moment diagram passes through zero. Consider a section just to the right of D.
Section just to the left of G. Variation of shear under a distributed load must be linear. Section just to the left of F. Consider a section just to the right of E.
First use the condition for equilibrium of moments about a point: Consider a section through the beam just to the left of D. Determine the reactions at the supports and draw the shear-force and bendingmoment diagram.
Section just to the right of F. The following example demonstrates the construction of diagrams representing shearing forces and bending moments.
As we approach the right-hand end of the beam we find the mathematics easier to consider on the right-hand side of any section. Critical points are: Calculate the values of the shear force to the left and to the right of all critical points. These two transverse sections are the two most likely points for failure in shear. One is at E and the other is between H and G.
Calculate values of the bending moment at all critical points. It is the effect that one load would have on the bent shape at the chosen point that determines the sign. When designing beams in materials such as concrete. The bending moment at G is obviously zero 5. Critical points for bending moment are: Chapter 6 — Basic mechanics The shear-force diagram in the example has two points where the shear force is zero.
At point H we have the maximum bending moment: For the calculation of primary stresses. The force can be a pure tension conventionally designated positive.
It will read a value of zero after 2. Equilibrium of Forces 6. Centre of Gravity 7. Moment of Intertia 8. Principles of Friction 9. Applications of Friction Principles of Lifting Machines Simple Lifting Machine Support Reactions Analysis of Perfect Frames Analytical Method Analysis of Perfect Frames Graphical Method Equilibrium of strings Virtual Work Linear Motion Motion Under Variable Acceleration Relative Velocity Projectiles Motion of Rotation Combined Motion of Rotation and Translation Simple Harmonic Motion Laws of Motion Motion of Connected Bodies Helical Springs and Pendulums Collision of Elastic Bodies Motion Along a Circular Path Balancing of Rotating Masses Work, Power and Energy Motion of Vehicles Transmission of Power by Belt and Ropes Transmission of Power by Gear Trains Hydrostatics Equilibrium of Floating Bodies.
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