6161103 2.7 position vectors
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The document discusses position vectors in 3D coordinate systems. It defines a position vector r as a fixed vector that locates a point P(x,y,z) in space relative to another point, such as the origin O. The components of r are the distances from O to P along the x, y, and z axes. As an example, it shows calculating the position vector and length of an elastic rubber band attached between points A and B to determine the band's direction from A toward B.
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6161103 8.9 chapter summary and review
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6161103 10.8 mohr’s circle for moments of inertia
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6161103 10.6 inertia for an area
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The document summarizes key concepts about internal loadings and shear and moment diagrams in structural members. It discusses how to determine the normal force, shear force, and bending moment at a cross section using a method of sections. It also describes how to construct and plot shear and moment diagrams as functions of location along the member length using equilibrium equations. Graphical methods for establishing shear and moment diagrams using relationships between loadings, shear, and moment are also presented. Finally, techniques for analyzing cables subjected to concentrated and distributed forces are summarized.
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6161103 10.7 moments of inertia for an area about inclined axes
This document discusses calculating moments of inertia for an area about inclined axes. It provides transformation equations to relate moments of inertia with respect to x-y axes to moments with respect to inclined u-v axes. It also describes how to determine the principal axes, which are the orientations that produce the maximum and minimum moments of inertia. An example is provided to illustrate finding the principal moments of inertia for a beam cross-section.
6161103 2.8 force vector directed along a line
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6161103 2.6 addition and subtraction of cartesian vectors
6161103 2.6 addition and subtraction of cartesian vectors
6161103 10.5 moments of inertia for composite areas
6161103 10.5 moments of inertia for composite areas
6161103 11.3 principle of virtual work for a system of connected rigid bodies
6161103 11.3 principle of virtual work for a system of connected rigid bodies
6161103 10.7 moments of inertia for an area about inclined axes
6161103 10.7 moments of inertia for an area about inclined axes
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Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
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To find the unknown parameters such as reaction forces and moments induced by the body.
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Chapter 12 Section 12.1 Three-Dimensional Coordinate Sys
Chapter 12
Section 12.1: Three-Dimensional Coordinate Systems
We locate a point on a number line as one coordinate, in the plane as an ordered pair, and in
space as an ordered triple. So we call number line as one dimensional, plane as two
dimensional, and space as three dimensional co – ordinate system.
In three dimensional, there is origin (0, 0, 0) and there are three axes – x -, y - , and z – axis. X –
and y – axes are horizontal and z – axis is vertical. These three axes divide the space into eight
equal parts, called the octants. In addition, these three axes divide the space into three
coordinate planes.
– The xy-plane contains the x- and y-axes. The equation is z = 0.
– The yz-plane contains the y- and z-axes. The equation is x = 0.
– The xz-plane contains the x- and z-axes. The equation is y = 0.
If P is any point in space, let:
– a be the (directed) distance from the yz-plane to P.
– b be the distance from the xz-plane to P.
– c be the distance from the xy-plane to P.
Then the point P by the ordered triple of real numbers (a, b, c), where a, b, and c are the
coordinates of P.
– a is the x-coordinate.
– b is the y-coordinate.
– c is the z-coordinate.
– Thus, to locate a point (a, b, c) in space, start from the origin (0, 0, 0) and move a
units along the x-axis. Then, move b units parallel to the y-axis. Finally, move c
units parallel to the z-axis.
The three dimensional Cartesian co – ordinate system follows the right hand rule.
Examples:
Plot the points (2,3,4), (2, -3, 4), (-2, -3, 4), (2, -3, -4), and (-2, -3, -4).
The Cartesian product x x = {(x, y, z) | x, y, z in } is the set of all ordered triples of
real numbers and is denoted by 3 .
Note:
1. In 2 – dimension, an equation in x and y represents a curve in the plane 2 . In 3 –
dimension, an equation in x, y, and z represents a surface in space 3 .
2. When we see an equation, we must understand from the context that it is a curve in the
plane or a surface in space. For example, y = 5 is a line in 2 �but it is a plane in 3 �
������
3. in space, if k, l, & m are constants, then
– x = k represents a plane parallel to the yz-plane ( a vertical plane).
– y = k is a plane parallel to the xz-plane ( a vertical plane).
– z = k is a plane parallel to the xy-plane ( a horizontal plane).
– x = k & y = l is a line.
– x = k & z = m is a line.
– y = l & z = m is a line.
– x = k, y = l and z = m is a point.
Examples: Describe and sketch y = x in 3
Example:
Solve:
Which of the points P(6, 2, 3), Q(-5, -1, 4), and R(0, 3, 8) is closest to the xz – plane? Which point
lies in the yz – plane?
Distance between two points in space:
We simply extend the formula from 2 to . 3 . The distance |p1 p2 | between the points
P1(x1,y1, z1) and P2(x2, y2, z2) is: 2 2 21 2 2 1 ...
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2. Common vector operations include addition, subtraction, and determining the resultant or sum of multiple vectors.
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6161103 11.7 stability of equilibrium
The document discusses the stability of equilibrium for mechanical systems. It defines three types of equilibrium:
1) Stable equilibrium occurs when a small displacement causes the system to return to its original position with potential energy at a minimum.
2) Neutral equilibrium occurs when a small displacement does not change the potential energy, which remains constant.
3) Unstable equilibrium occurs when a small displacement causes the system to move farther away from the original position with potential energy at a maximum.
The stability of equilibrium depends on the potential energy function and its derivatives evaluated at the equilibrium position. A system is stable if the second derivative of the potential energy is positive and unstable if it is negative.
6161103 10.9 mass moment of inertia
Mass moment of inertia measures an object's resistance to angular acceleration and is defined as the integral of the second moment of mass elements about an axis. It depends on the axis chosen and the distribution of mass. For composite objects, the total mass moment of inertia is found by summing the contributions of each component. The parallel axis theorem allows calculating mass moments of inertia about parallel axes.
6161103 10.4 moments of inertia for an area by integration
This document discusses calculating moments of inertia for planar areas using integration. It describes:
1) Choosing a differential element for integration that has size in only one direction to simplify the calculation.
2) The procedure involves specifying a rectangular differential element and orienting it parallel or perpendicular to the axis of rotation.
3) Moments of inertia are calculated through single or double integration, depending on whether the element has thickness in one or two directions.
6161103 9.2 center of gravity and center of mass and centroid for a body
This document discusses the center of gravity, center of mass, and centroid of rigid bodies. It defines these terms and presents methods to calculate them using integrals of differential elements. Examples are provided to demonstrate calculating the centroid for areas and lines using appropriate coordinate systems and differential elements. Centroids are found by taking moments of these elements and the document outlines the general procedure to perform these calculations.
6161103 9.6 fluid pressure
1) Pascal's law states that pressure in a fluid at rest acts equally in all directions. The pressure (ρ) at a point depends on the specific weight (γ) or density (ρ) of the fluid and the depth (z) of the point, where ρ = γz or ρgz.
2) For a submerged plate, the pressure at different points depends on the depth of each point. The resultant force on the plate acts through its center of pressure, not the plate's centroid.
3) For curved or variable width plates, integration is used to determine the total pressure distribution and locate the center of pressure.
6161103 9.3 composite bodies
This document discusses analyzing the center of gravity of composite bodies made up of simpler shapes. It provides the procedure for doing so which includes: [1] Dividing the body into composite parts; [2] Determining the coordinates of each part's center of gravity; [3] Using equations to calculate the total center of gravity by taking weight-weighted sums of the coordinate positions. Examples are provided to demonstrate locating the center of gravity for areas and volumes made of multiple components.
6161103 8.4 frictional forces on screws
Screws are commonly used as fasteners to transmit power or motion between machine parts. A screw can be thought of as an inclined plane wrapped around a cylinder. When a screw is rotated, the nut moves along the inclined plane of the screw thread. The distance the nut moves in one revolution is called the lead of the screw. Large axial loads on a screw generate significant frictional forces on the threads. The moment required to turn a screw under an axial load depends on the frictional forces and can be calculated using force and moment equilibrium equations. A self-locking screw will remain locked in place through frictional forces alone when the applied moment is removed.
6161103 8.3 wedges
Wedges are simple machines that transform applied forces into larger forces directed at approximately right angles. They are used to give small adjustments to heavy loads by applying a smaller force. To analyze wedges, force and moment equilibrium equations are used along with frictional force equations, with unknown normal and frictional forces. Wedges can be self-locking if the frictional forces alone are enough to hold the load without any applied force needed.
6161103 8.2 problems involving dry friction
This document discusses problems involving dry friction. It describes three types of friction problems: equilibrium, impending motion at all points, and impending motion at some points. Examples are provided to demonstrate how to set up and solve static friction problems using free body diagrams, equilibrium equations, and frictional equations. The key steps are to determine the number of unknowns and equations, draw free body diagrams, apply the appropriate equilibrium and frictional equations, and solve the system of equations.
6161103 7.3 relations between distributed load, shear and moment
1) The document discusses the relationships between distributed loads, shear forces, and bending moments in beams.
2) It shows that the slope of the shear diagram is equal to the distributed load intensity, and the slope of the moment diagram is equal to the shear force.
3) The change in shear force between two points is equal to the negative area under the distributed load diagram between those points. The change in bending moment is equal to the area under the shear diagram.
6161103 7.2 shear and moment equations and diagrams
1) Beams are structural members designed to support loads perpendicular to their axes. Simply supported beams are pinned at one end and roller supported at the other, while cantilevered beams are fixed at one end and free at the other.
2) Internal shear forces (V) and bending moments (M) must be determined for beam design. V and M diagrams graphically display these values and can be discontinuous where loads change.
3) The procedure involves determining support reactions, then calculating V and M values along the beam using the method of sections to draw the diagrams.
6161103 7.1 internal forces developed in structural members
This document discusses how to determine the internal forces in structural members using the method of sections. It explains that an imaginary cut is made through the member to separate it into segments. The internal forces acting on the cut section become external forces on the free body diagrams of each segment. Equilibrium equations are then used to calculate the normal force, shear force, bending moment, and torsion at the cut section. Several examples are provided to demonstrate how to set up and solve for the internal forces using this method.
6161103 7.4 cables
1) Cables are used in structures like suspension bridges to support and transmit loads. They are considered flexible and inextensible.
2) Cables taking concentrated loads form straight line segments with constant tension. Cables under distributed loads form parabolic shapes with tension varying along the cable length.
3) When a cable's own weight is considered, its deflection curve is defined by hyperbolic functions with tension also varying along the cable.
6161103 6.6 frames and machines
Frames are stationary structures that support loads, while machines contain moving parts to transmit and alter forces. Free-body diagrams are used to analyze frames and machines by showing all forces on each member. Two-force members have equal and opposite internal forces. Equations of equilibrium for each member are applied to determine unknown forces at pins and supports. Common forces between connected members are internal and not shown on individual free-body diagrams.
6161103 6.5 space trusses
A space truss is a 3D structure formed by joining members at their ends. The simplest space truss is a tetrahedron formed from six members. Additional members would be redundant. Space truss members can be treated as two-force members if loading is applied at the joints, which use ball and socket connections. The method of joints and method of sections can be used to analyze space trusses by applying equilibrium equations at each joint or section. The example problem shows using these methods to determine member forces in a space truss and indicating whether each member is in tension or compression.
6161103 6.4 the method of sections
The method of sections involves cutting imaginary sections through structural members to isolate forces. Equilibrium equations are applied to the cut section to determine member forces. For a truss example, a section was drawn through three unknown members. The free body diagram of the cut section was drawn and the equations of equilibrium were used to determine the forces in each member as 800 N in tension, 800 N in compression, and 500 N in tension.
6161103 6.3 zero force members
Zero-force members are truss members that do not experience any internal forces from applied loads. They are used to increase stability during construction and allow the truss to support changed loadings. By examining the force and moment equilibrium at joints, members that are determined to have zero force can be identified. If two members meet at a joint with no external loads or supports, those members must be zero-force members. Similarly, if three members meet collinearly at a joint with no external loads, the third member is a zero-force member.
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6161103 2.7 position vectors
- 1. 2.7 Position Vectors x,y,z Coordinates - Right-handed coordinate system - Positive z axis points upwards, measuring the height of an object or the altitude of a point - Points are measured relative to the origin, O.
- 2. 2.7 Position Vectors x,y,z Coordinates Eg: For Point A, xA = +4m along the x axis, yA = -6m along the y axis and zA = -6m along the z axis. Thus, A (4, 2, -6) Similarly, B (0, 2, 0) and C (6, -1, 4)
- 3. 2.7 Position Vectors Position Vector - Position vector r is defined as a fixed vector which locates a point in space relative to another point. Eg: If r extends from the origin, O to point P (x, y, z) then, in Cartesian vector form r = xi + yj + zk
- 4. 2.7 Position Vectors Position Vector Note the head to tail vector addition of the three components Start at origin O, one travels x in the +i direction, y in the +j direction and z in the +k direction, arriving at point P (x, y, z)
- 5. 2.7 Position Vectors Position Vector - Position vector maybe directed from point A to point B - Designated by r or rAB Vector addition gives rA + r = rB Solving r = rB – rA = (xB – xA)i + (yB – yA)j + (zB –zA)k or r = (xB – xA)i + (yB – yA)j + (zB –zA)k
- 6. 2.7 Position Vectors Position Vector - The i, j, k components of the positive vector r may be formed by taking the coordinates of the tail, A (xA, yA, zA) and subtract them from the head B (xB, yB, zB) Note the head to tail vector addition of the three components
- 7. 2.7 Position Vectors Length and direction of cable AB can be found by measuring A and B using the x, y, z axes Position vector r can be established Magnitude r represent the length of cable
- 8. 2.7 Position Vectors Angles, α, β and γ represent the direction of the cable Unit vector, u = r/r
- 9. 2.7 Position Vectors Example 2.12 An elastic rubber band is attached to points A and B. Determine its length and its direction measured from A towards B.
- 10. 2.7 Position Vectors Solution Position vector r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k = {-3i + 2j + 6k}m Magnitude = length of the rubber band r= (− 3)2 + (2)2 + (6)2 = 7m Unit vector in the director of r u = r /r = -3/7i + 2/7j + 6/7k
- 11. 2.7 Position Vectors Solution α = cos-1(-3/7) = 115° β = cos-1(2/7) = 73.4° γ = cos-1(6/7) = 31.0°