This document discusses unsymmetrical bending of beams. It begins by defining pure bending and symmetrical bending. It then discusses assumptions and stress calculations for unsymmetrical bending. It defines product second moment of area and how to determine the principal axes of a section. It provides equations for direct stress distribution, deflection, and solving problems involving cantilever beams under unsymmetrical bending loads. The key points are that unsymmetrical bending involves calculating principal axes and moments of inertia, and that deflection and stresses are determined relative to the principal axes rather than the original axes.
1. The document discusses unsymmetrical bending of beams. When a beam bends about an axis that is not perpendicular to a plane of symmetry, it is undergoing unsymmetrical bending.
2. Key aspects discussed include determining the principal axes, direct stress distribution, and deflection of beams under unsymmetrical bending. Equations are provided to calculate stresses and deflections.
3. An example problem is given involving finding the stresses at two points on a cantilever beam subjected to an unsymmetrical loading. The principal moments of inertia and neutral axis orientation are calculated.
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 ...
Centroid and Moment of Inertia from mechanics of material by hibbler related ...FREE LAUNCER
Centroids and moment of inertia are important concepts in mechanics. The centroid of a plane figure is the point where the entire area is considered to be concentrated. It is found by suspending the figure from different corners and finding the intersection point of vertical lines. The center of gravity is where the entire mass is considered concentrated. For uniform plane figures with no weight, the centroid and center of gravity coincide. The moment of inertia is a measure of an object's resistance to changes in rotation or bending and depends on the object's mass distribution and axis of rotation. It is calculated based on the object's area or mass distances from the axis of rotation.
The document discusses key concepts related to the section properties of structural members including:
- The center of gravity is the point where the total weight of a system can be considered to be concentrated.
- The center of mass is calculated similarly using the total mass of a system rather than total weight.
- The centroid is the geometric center of an object, independent of forces or weights, and depends only on the object's shape.
- Moments of inertia measure the resistance of an area to bending and twisting forces, and are calculated based on the area properties and distance from specific axes.
1) A point charge moving through an electric field is shown to follow a parabolic trajectory given by z = -1.5×1010t2 m. At t = 3 μs, its position is found to be P(0.90, 0, -0.135) m.
2) A point charge moving through a uniform magnetic field follows a circular path. The equations of motion are derived and solved, giving the position, velocity, and kinetic energy at t = 3 μs.
3) Forces on current loops and filaments in various magnetic field configurations are calculated.
Prof. V. V. Nalawade, UNIT-3 Centroid, Centre off Gravity and Moment of InertiaVrushali Nalawade
The document discusses concepts related to center of gravity and moment of inertia. It defines center of gravity as the point where the entire weight of a body acts and centroid as the point where the entire area of a plane figure acts. It provides formulas for calculating the centroid of composite figures and discusses the parallel axis theorem and perpendicular axis theorem for calculating moment of inertia about different axes. The document also defines radius of gyration and provides formulas for calculating moment of inertia and radius of gyration of common plane figures.
This document discusses finding the centroid of solids of revolution. It explains that the centroid of a solid generated by revolving a plane area about an axis will lie on that axis. To find the coordinates of the centroid, one takes the moment of an elementary disc about the coordinate axes and sums these moments by treating the discs as an integral. Two examples are worked through to demonstrate finding the coordinates of the centroid. Five exercises are then posed asking to find the centroid coordinates for solids generated by revolving given bounded regions.
- The document describes the conjugate beam method for analyzing beams with varying rigidities.
- The method involves drawing an imaginary conjugate beam that is loaded based on the bending moments of the real beam.
- The slope and deflection of points on the real beam can then be determined from the shear and bending moment of the corresponding points on the conjugate beam.
- An example problem is worked out in detail to demonstrate calculating the slope and deflection at the end of a cantilever beam with varying moment of inertia along its length using the conjugate beam method.
1. The document discusses unsymmetrical bending of beams. When a beam bends about an axis that is not perpendicular to a plane of symmetry, it is undergoing unsymmetrical bending.
2. Key aspects discussed include determining the principal axes, direct stress distribution, and deflection of beams under unsymmetrical bending. Equations are provided to calculate stresses and deflections.
3. An example problem is given involving finding the stresses at two points on a cantilever beam subjected to an unsymmetrical loading. The principal moments of inertia and neutral axis orientation are calculated.
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 ...
Centroid and Moment of Inertia from mechanics of material by hibbler related ...FREE LAUNCER
Centroids and moment of inertia are important concepts in mechanics. The centroid of a plane figure is the point where the entire area is considered to be concentrated. It is found by suspending the figure from different corners and finding the intersection point of vertical lines. The center of gravity is where the entire mass is considered concentrated. For uniform plane figures with no weight, the centroid and center of gravity coincide. The moment of inertia is a measure of an object's resistance to changes in rotation or bending and depends on the object's mass distribution and axis of rotation. It is calculated based on the object's area or mass distances from the axis of rotation.
The document discusses key concepts related to the section properties of structural members including:
- The center of gravity is the point where the total weight of a system can be considered to be concentrated.
- The center of mass is calculated similarly using the total mass of a system rather than total weight.
- The centroid is the geometric center of an object, independent of forces or weights, and depends only on the object's shape.
- Moments of inertia measure the resistance of an area to bending and twisting forces, and are calculated based on the area properties and distance from specific axes.
1) A point charge moving through an electric field is shown to follow a parabolic trajectory given by z = -1.5×1010t2 m. At t = 3 μs, its position is found to be P(0.90, 0, -0.135) m.
2) A point charge moving through a uniform magnetic field follows a circular path. The equations of motion are derived and solved, giving the position, velocity, and kinetic energy at t = 3 μs.
3) Forces on current loops and filaments in various magnetic field configurations are calculated.
Prof. V. V. Nalawade, UNIT-3 Centroid, Centre off Gravity and Moment of InertiaVrushali Nalawade
The document discusses concepts related to center of gravity and moment of inertia. It defines center of gravity as the point where the entire weight of a body acts and centroid as the point where the entire area of a plane figure acts. It provides formulas for calculating the centroid of composite figures and discusses the parallel axis theorem and perpendicular axis theorem for calculating moment of inertia about different axes. The document also defines radius of gyration and provides formulas for calculating moment of inertia and radius of gyration of common plane figures.
This document discusses finding the centroid of solids of revolution. It explains that the centroid of a solid generated by revolving a plane area about an axis will lie on that axis. To find the coordinates of the centroid, one takes the moment of an elementary disc about the coordinate axes and sums these moments by treating the discs as an integral. Two examples are worked through to demonstrate finding the coordinates of the centroid. Five exercises are then posed asking to find the centroid coordinates for solids generated by revolving given bounded regions.
- The document describes the conjugate beam method for analyzing beams with varying rigidities.
- The method involves drawing an imaginary conjugate beam that is loaded based on the bending moments of the real beam.
- The slope and deflection of points on the real beam can then be determined from the shear and bending moment of the corresponding points on the conjugate beam.
- An example problem is worked out in detail to demonstrate calculating the slope and deflection at the end of a cantilever beam with varying moment of inertia along its length using the conjugate beam method.
1) Moment of inertia is the rotational analog of mass for linear motion and appears in the relationships for rotational dynamics.
2) The moment of inertia must be specified with respect to a chosen axis of rotation and is calculated by summing the products of small elements of mass and their distances from the axis.
3) The perpendicular axis theorem states that for a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia about two perpendicular axes in the plane.
Happy Birthday to you dear sir please find the attachment of my past but Ivishalyadavreso1111
1. Mohr's circle is a graphical representation of the state of plane stress at a point using a circle constructed from the known normal and shear stresses.
2. Key points on Mohr's circle correspond to the principal stresses and maximum shear stress, with their values determined from the circle's construction.
3. Drawing Mohr's circle involves plotting the known stresses, constructing the circle, and determining critical points that represent maximum and minimum normal and shear stresses on planes through the point of interest.
Prof. V. V. Nalawade, Notes CGMI with practice numericalVrushali Nalawade
Centre of gravity is a point where the whole weight of the body is assumed to act. i.e., it is a point where entire distribution of gravitational force is supposed to be concentrated
It is generally denoted “G” for all three dimensional rigid bodies.
e.g. Sphere, table , vehicle, dam, human etc
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Vibration Midterm-THEORETICAL SOLUTION AND STATIC ANALYSES STUDY OF VIBRATION...Mehmet Bariskan
1) The document analyzes the transverse vibration of channel beams using the Euler-Bernoulli beam theory. Equations of motion and boundary conditions are derived.
2) Natural frequencies and mode shapes are determined for different beam boundary conditions, including a simply supported beam.
3) Theoretical analyses are conducted for a 3m long stainless steel channel beam, including determining natural frequencies, reaction forces, stresses, deflections, and more. Results are compared to those from Solidworks.
Formula Bank and Important tips for Mechanical Engineering Students for Compe...Vinoth Jebaraj A
This document summarizes key concepts in engineering mechanics and strength of materials for mechanical engineering students. It covers topics like force equilibrium, stress and strain analysis, material properties, and failure theories. Key equations are presented for areas including static equilibrium, centroids, moments of inertia, stress-strain relationships, transformation of stresses, and bending stresses in beams. Diagrams illustrate stress distributions and Mohr's circle analyses for various loading conditions.
Relation between load shear force and bending moment of beamssushma chinta
This document discusses the relationships between loads, shear forces, and bending moments in beams. It states that shear forces and bending moments are internal stress resultants that can be calculated from equations of equilibrium. Distributed loads cause shear forces to vary linearly or quadratically along the beam and bending moments to vary quadratically or cubically. Concentrated loads cause an abrupt change in shear force but no change in bending moment. Couples cause no change in shear force but an abrupt change in bending moment.
Chapter 9 tutorial exercises with solutions 19 02 2013TRL4EVER
1. The document describes a system consisting of two blocks placed at the ends of a horizontal massless board that rests on an axis of rotation. The moment of inertia of the system is given as 12 kg·m2.
2. It then presents a problem involving a wrecking ball supported by a boom. Forces acting on the system include the weight of the wrecking ball, the weight of the boom, and the tension in a support cable.
3. The document solves the problems by applying principles of equilibrium, including analyzing torque equations and using trigonometric relationships between forces and angles. Solutions are obtained for requested values like tension in the support cable and magnitude of the force on the lower end of the boom
2 work energy power to properties of liquidsAntony Jaison
1) Work is done when a force causes an object to be displaced. It is defined as the product of the force and displacement in the direction of the force. Work is a scalar quantity measured in joules.
2) Energy is the ability to do work and exists in kinetic and potential forms. Kinetic energy is the energy of motion and potential energy is stored energy due to an object's position or state.
3) According to the work-energy theorem, the work done on an object equals its change in kinetic energy. For a variable force, the work is calculated as the area under the force-displacement graph.
2 work energy power to properties of liquidsarunjyothi247
Work is done when a force causes an object to be displaced. Work is defined as the product of the force and displacement in the direction of the force. Kinetic energy is the energy an object possesses due to its motion. Potential energy is the energy an object possesses due to its position or state. The law of conservation of energy states that energy cannot be created or destroyed, only changed from one form to another. Elastic collisions are collisions where both momentum and kinetic energy are conserved, while inelastic collisions conserve momentum but not kinetic energy.
Is ellipse really a section of cone. The question intrigued me for 20 odd years after leaving high school. Finally got the proof on a cremation ground. Only thereafter I came to know of Dandelin spheres. But this proof uses only bare basics within the scope of high school course of Analytical geometry.
The document discusses various types of combined stresses:
1) Biaxial bending stresses are the sum of stresses from bending in two perpendicular directions.
2) Combined bending and axial stresses add the axial stress to the bending stress.
3) Eccentrically loaded axial members have extra bending stress due to eccentricity.
4) Normal and shear stresses on a stress element are defined.
5) Principal stresses are the maximum normal stresses on perpendicular planes.
6) Maximum shear stress and its orientation are also defined.
7) Mohr's circle is introduced as a graphical method to represent the stresses on a plane.
- A column subjected to an eccentric axial load experiences both direct compressive stress and bending stress. The maximum stress occurs on the edge of the column closest to the load and the minimum stress is on the opposite edge.
- For a dam, the total water pressure acts below the centroid and combined with the weight of the dam creates an eccentric load. This leads to maximum and minimum pressures at the base that depend on the eccentricity.
- For a retaining wall, it experiences an eccentric load due to the total earth pressure acting below the midpoint. This causes maximum and minimum pressures similar to a dam.
- A chimney or wall subjected to wind pressure on one side experiences direct stress due to its self weight and bending
1. The document provides calculations for determining work done by electric fields on charges moving through various paths. It includes determining incremental work from given electric field expressions and calculating potentials from different charge distributions.
2. Key results include the work done on a 20 μC charge moving in different directions in a given electric field, the potential and potential difference from a uniform spherical charge distribution, and expressing the potential field of an infinite line charge with different references.
3. The final problem calculates the potential at a point from the combined electric fields of a uniform sheet charge, uniform line charge, and point charge, with the potential set to zero at a reference point.
The use of Calculus is very important in every aspects of engineering.
The use of Differential equation is very much applied in the concept of Elastic beams.
The document discusses different types of loads acting on columns and how they induce stresses. It defines axial load, eccentric load, and eccentricity. It explains that eccentric loads produce both direct and bending stresses while axial loads only produce direct stresses. It provides equations to calculate the maximum and minimum stresses in a column under eccentric loading. An example problem is worked out calculating the maximum stresses in a T-section column loaded eccentrically. The document also discusses loads, eccentricity, and stresses on dams, retaining walls, and chimneys/walls loaded by wind pressure.
This document discusses complex stress and strain resulting from combined loading. It begins by introducing plane stress and defining stress components acting on an infinitesimal element, including normal and shear stresses. It then derives equations to calculate stresses on inclined planes from the original stress components. These transformation equations allow determining principal stresses and angles, where normal stresses are maximum and minimum. Maximum shear stress can also be found and principal stresses calculated using properties of a right triangle. Principal angles correlate the two principal stresses. Shear stresses are zero on principal planes.
Welcome to the fascinating world of Work, Energy, and Power! In the realm of physics, these concepts form the cornerstone of understanding how objects interact with each other and how energy is transformed within systems. From the motion of everyday objects to the dynamics of celestial bodies, the principles of work, energy, and power are ubiquitous, shaping the very fabric of the universe.
For more information, visit-www.vavaclasses.com
Healthy Eating Habits:
Understanding Nutrition Labels: Teaches how to read and interpret food labels, focusing on serving sizes, calorie intake, and nutrients to limit or include.
Tips for Healthy Eating: Offers practical advice such as incorporating a variety of foods, practicing moderation, staying hydrated, and eating mindfully.
Benefits of Regular Exercise:
Physical Benefits: Discusses how exercise aids in weight management, muscle and bone health, cardiovascular health, and flexibility.
Mental Benefits: Explains the psychological advantages, including stress reduction, improved mood, and better sleep.
Tips for Staying Active:
Encourages consistency, variety in exercises, setting realistic goals, and finding enjoyable activities to maintain motivation.
Maintaining a Balanced Lifestyle:
Integrating Nutrition and Exercise: Suggests meal planning and incorporating physical activity into daily routines.
Monitoring Progress: Recommends tracking food intake and exercise, regular health check-ups, and provides tips for achieving balance, such as getting sufficient sleep, managing stress, and staying socially active.
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1) Moment of inertia is the rotational analog of mass for linear motion and appears in the relationships for rotational dynamics.
2) The moment of inertia must be specified with respect to a chosen axis of rotation and is calculated by summing the products of small elements of mass and their distances from the axis.
3) The perpendicular axis theorem states that for a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia about two perpendicular axes in the plane.
Happy Birthday to you dear sir please find the attachment of my past but Ivishalyadavreso1111
1. Mohr's circle is a graphical representation of the state of plane stress at a point using a circle constructed from the known normal and shear stresses.
2. Key points on Mohr's circle correspond to the principal stresses and maximum shear stress, with their values determined from the circle's construction.
3. Drawing Mohr's circle involves plotting the known stresses, constructing the circle, and determining critical points that represent maximum and minimum normal and shear stresses on planes through the point of interest.
Prof. V. V. Nalawade, Notes CGMI with practice numericalVrushali Nalawade
Centre of gravity is a point where the whole weight of the body is assumed to act. i.e., it is a point where entire distribution of gravitational force is supposed to be concentrated
It is generally denoted “G” for all three dimensional rigid bodies.
e.g. Sphere, table , vehicle, dam, human etc
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Vibration Midterm-THEORETICAL SOLUTION AND STATIC ANALYSES STUDY OF VIBRATION...Mehmet Bariskan
1) The document analyzes the transverse vibration of channel beams using the Euler-Bernoulli beam theory. Equations of motion and boundary conditions are derived.
2) Natural frequencies and mode shapes are determined for different beam boundary conditions, including a simply supported beam.
3) Theoretical analyses are conducted for a 3m long stainless steel channel beam, including determining natural frequencies, reaction forces, stresses, deflections, and more. Results are compared to those from Solidworks.
Formula Bank and Important tips for Mechanical Engineering Students for Compe...Vinoth Jebaraj A
This document summarizes key concepts in engineering mechanics and strength of materials for mechanical engineering students. It covers topics like force equilibrium, stress and strain analysis, material properties, and failure theories. Key equations are presented for areas including static equilibrium, centroids, moments of inertia, stress-strain relationships, transformation of stresses, and bending stresses in beams. Diagrams illustrate stress distributions and Mohr's circle analyses for various loading conditions.
Relation between load shear force and bending moment of beamssushma chinta
This document discusses the relationships between loads, shear forces, and bending moments in beams. It states that shear forces and bending moments are internal stress resultants that can be calculated from equations of equilibrium. Distributed loads cause shear forces to vary linearly or quadratically along the beam and bending moments to vary quadratically or cubically. Concentrated loads cause an abrupt change in shear force but no change in bending moment. Couples cause no change in shear force but an abrupt change in bending moment.
Chapter 9 tutorial exercises with solutions 19 02 2013TRL4EVER
1. The document describes a system consisting of two blocks placed at the ends of a horizontal massless board that rests on an axis of rotation. The moment of inertia of the system is given as 12 kg·m2.
2. It then presents a problem involving a wrecking ball supported by a boom. Forces acting on the system include the weight of the wrecking ball, the weight of the boom, and the tension in a support cable.
3. The document solves the problems by applying principles of equilibrium, including analyzing torque equations and using trigonometric relationships between forces and angles. Solutions are obtained for requested values like tension in the support cable and magnitude of the force on the lower end of the boom
2 work energy power to properties of liquidsAntony Jaison
1) Work is done when a force causes an object to be displaced. It is defined as the product of the force and displacement in the direction of the force. Work is a scalar quantity measured in joules.
2) Energy is the ability to do work and exists in kinetic and potential forms. Kinetic energy is the energy of motion and potential energy is stored energy due to an object's position or state.
3) According to the work-energy theorem, the work done on an object equals its change in kinetic energy. For a variable force, the work is calculated as the area under the force-displacement graph.
2 work energy power to properties of liquidsarunjyothi247
Work is done when a force causes an object to be displaced. Work is defined as the product of the force and displacement in the direction of the force. Kinetic energy is the energy an object possesses due to its motion. Potential energy is the energy an object possesses due to its position or state. The law of conservation of energy states that energy cannot be created or destroyed, only changed from one form to another. Elastic collisions are collisions where both momentum and kinetic energy are conserved, while inelastic collisions conserve momentum but not kinetic energy.
Is ellipse really a section of cone. The question intrigued me for 20 odd years after leaving high school. Finally got the proof on a cremation ground. Only thereafter I came to know of Dandelin spheres. But this proof uses only bare basics within the scope of high school course of Analytical geometry.
The document discusses various types of combined stresses:
1) Biaxial bending stresses are the sum of stresses from bending in two perpendicular directions.
2) Combined bending and axial stresses add the axial stress to the bending stress.
3) Eccentrically loaded axial members have extra bending stress due to eccentricity.
4) Normal and shear stresses on a stress element are defined.
5) Principal stresses are the maximum normal stresses on perpendicular planes.
6) Maximum shear stress and its orientation are also defined.
7) Mohr's circle is introduced as a graphical method to represent the stresses on a plane.
- A column subjected to an eccentric axial load experiences both direct compressive stress and bending stress. The maximum stress occurs on the edge of the column closest to the load and the minimum stress is on the opposite edge.
- For a dam, the total water pressure acts below the centroid and combined with the weight of the dam creates an eccentric load. This leads to maximum and minimum pressures at the base that depend on the eccentricity.
- For a retaining wall, it experiences an eccentric load due to the total earth pressure acting below the midpoint. This causes maximum and minimum pressures similar to a dam.
- A chimney or wall subjected to wind pressure on one side experiences direct stress due to its self weight and bending
1. The document provides calculations for determining work done by electric fields on charges moving through various paths. It includes determining incremental work from given electric field expressions and calculating potentials from different charge distributions.
2. Key results include the work done on a 20 μC charge moving in different directions in a given electric field, the potential and potential difference from a uniform spherical charge distribution, and expressing the potential field of an infinite line charge with different references.
3. The final problem calculates the potential at a point from the combined electric fields of a uniform sheet charge, uniform line charge, and point charge, with the potential set to zero at a reference point.
The use of Calculus is very important in every aspects of engineering.
The use of Differential equation is very much applied in the concept of Elastic beams.
The document discusses different types of loads acting on columns and how they induce stresses. It defines axial load, eccentric load, and eccentricity. It explains that eccentric loads produce both direct and bending stresses while axial loads only produce direct stresses. It provides equations to calculate the maximum and minimum stresses in a column under eccentric loading. An example problem is worked out calculating the maximum stresses in a T-section column loaded eccentrically. The document also discusses loads, eccentricity, and stresses on dams, retaining walls, and chimneys/walls loaded by wind pressure.
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1. UNSYMMETRICAL BENDING OF
BEAMS
Under the guidance
of
Dr. M. V. RENUKA DEVI
Associate Professor
Department of Civil Engineering,
RVCE, Bangalore
By
VENKATESHAA
(1RV13CSE15)
2. PURE BENDING
Bending is a very severe form of stressing a structure
The simple bending theory applies when bending takes place about an
axis which is perpendicular to a plane of symmetry.
Bending moments acts along the axis of the member.
Assumptions made in pure bending
1. The normal planes remain normal even after bending.
2. There is no net internal axial force.
3. Stress varies linearly over cross section.
4. Zero stress exists at the centroid and the line of centroid is the
Neutral Axis (N.A)
4. Symmetrical bending : The plane of loading or the plane of
bending is co-incident with or parallel to, a plane containing
principal centroidal axes of inertia of the cross-section of the
beam.
Bending stress is given by
σz =
Mx
Ix
y +
My
Iy
x
Bending stress along N.A is σz= 0
5. UNSYMMETRICAL BENDING
Assumptions
1. The plane sections of the beam remain plane after bending
2. The material of the beam is homogeneous and linearly elastic.
3. There is no net internal axial force.
Sign conventions and notation u, v and w are the
displacement components of
any point within beam parallel to
x, y, z axes.
P = axial load and T = torque
𝑤𝑥 (z) and 𝑤𝑦 (z)are
distributed loads
𝑀𝑥and 𝑀𝑦are applied bending
moments
6. Fig. Representation of positive internal and external force systems
•We assume 𝑀𝑥and 𝑀𝑦 as positive when they each induce tensile stresses in the
positive xy quadrant of the beam section.
7. PRODUCT SECOND MOMENT OF AREA
The second moments of area of the surface about the X and Y axes are defined
as Ixx = x2
dA and Iyy = x2
dA
Similarly, the product second moment of area of the section is defined as
follows
Ixy = xy dA
8. Since the cross-section of most structural members used in bending applications
consists of a combination of rectangles the value of the product second moment
of area for such sections is determined by the addition of the Ixx value for each
rectangle.
Ixx = Ahk
Where h and k are the distances of the centroid of each rectangle from the X and
Y axes respectively (taking account of the normal sign convention for x and y)
and A is the area of the rectangle.
9. DETERMINATION OF PRINCIPAL AXIS OF SECTION
Let,
U-U, V-V be the Principal Centroidal Axes,
X-X, Y-Y be the pair of orthogonal axes,
α be the angle between both the axes system,
𝜕𝑎 Be the elementary area with co-ordinates (u, v) referred to the principal
axes i.e., U-V axes and (x, y) referred to the X-Y axes, Since, U-V axes are
principal axes the product of inertia = 0
10. The relationship between (x, y) and (u, v) co-ordinates are given by
Now substituting for u and v in the above equations we get,
Iuv = uv 𝜕A = 0
By definition
Ixx = y2. 𝜕A ; Iyy = x2. 𝜕A ; Ixy = xy. 𝜕A ;
Iuu = v2. 𝜕A ; Ivv = u2. 𝜕A ; Iuv = uv. 𝜕A ;
u = AB+DP = 𝑥 cos ∝ + 𝑦 sin ∝
v = GP-HG =GP-EF = 𝑦 cos ∝ − 𝑥 sin ∝
Iuu = {y cos ∝ − x sin ∝}2. 𝜕A
Iuu = cos2 ∝ y2. 𝜕A + sin2 ∝ x2. 𝜕A - 2 sin ∝ cos ∝ xy. 𝜕A
Iuu = cos2 ∝ Ixx + sin2 ∝ Iyy - sin 2 ∝ Ixy
11. Similarly,
Ivv = cos2 ∝ Iyy + sin2 ∝ Ixx + sin 2 ∝ Ixy
Iuv = cos2 ∝ Ixy - sin2 ∝ Ixy - sin 2 ∝ (Ixx - Iyy)
we have Iuv = 0, therefore we can write
tan 2 ∝ =
2 Ixy
Ixx− Iyy
cos2 ∝ =
(1+cos 2∝)
2
and sin2 ∝ =
(1−cos 2∝)
2
Iuu =
Ixx+ Iyy
2
+
Ixx− Iyy
2
cos 2 ∝ - Ixy sin 2 ∝
Ivv =
Ixx+ Iyy
2
-
Ixx− Iyy
2
cos 2 ∝ + Ixy sin 2 ∝
12. we can write for sin 2 ∝ and cos 2 ∝ as follows
Thus knowing the values of Ixx, Iyy and Ixy , the principal moments of inertia
Iuuand Ivv can be calculated from the above analytical expression.
Note: moment of inertia of a section about its principal axes have maximum and
minimum values.
sin 2 ∝ =
−Ixy
(
Ixx− Iyy
2
)2+Ixy
2
cos 2 ∝ =
Ixx−Ixy
(
Ixx− Iyy
2
)2+Ixy
2
Substituting the values of sin 2 ∝ and cos 2 ∝
Iuu =
Ixx+ Iyy
2
+ (
Ixx− Iyy
2
)2+Ixy
2
Ivv =
Ixx+ Iyy
2
- (
Ixx− Iyy
2
)2+Ixy
2
13. DIRECT STRESS DISTRIBUTION
We know that a beam bends about the neutral axis of its cross section so
that the radius of curvature, R, of the beam is perpendicular to the neutral
axis.
Fig. bending of an unsymmetrical beam section
σZ = E
p
R
14. The beam section is subjected to a pure bending moment so that the resultant direct
load on the section is zero. Hence
𝐴
𝜎𝑧 𝑑𝐴 = 0
𝐴
𝐸
𝑃
𝑅
𝑑𝐴 = 0
For a beam of a given material subjected to a given bending moment
A
P dA = 0
Above equation states that the first moment of area of the beam section about the
neutral axis is zero. It follows pure bending of beams in which the neutral axis
always passes through the centroid of the beam section.
p = x sin ∝ + y cos ∝
σz =
E
R
(x sin ∝ + y cos ∝)
The moment resultants of the direct stress distribution are
Mx = A
σz y dA , My = A
σz x dA
15. Substituting for σz
Mx =
E sin∝
R A
xy dA +
E cos∝
R A
y2
dA
My =
E sin∝
R A
x2 dA +
E cos∝
R A
xy dA
Mx =
E sin∝
R
Ixy +
E cos∝
R
Ix ; My =
E sin∝
R
Iy +
E cos∝
R
Ixy
Bending stress is written as
σz =
Mx
Ix
y +
My
Iy
x
Where,
Mx =
My −Mx Ixy/Iy
1− (Ixy)2/Ix Iy
; My =
My −Mx Ixy/Ix
1− (Ixy)2/Ix Iy
In the case where the beam section has either Ox or Oy (or both) as an axis of
symmetry, then Ixy is zero and Ox, Oy are principal axes, Mx = Mx , My =
My
σz =
Mx
Ix
y +
My
Iy
x
16. Position of the neutral axis
The direct stress at all points on the neutral axis of the beam section is zero.
Thus,
O =
Mx
Ix
yN.A +
My
Iy
xN.A
WherexN.A and yN.A are the coordinates of any point on the neutral
axis. Thus
yN.A
xN.A
= −
My Ix
Mx Iy
tan ∝ =
My Ix
Mx Iy
Since ∝ is positive when yN.Ais negative and xN.A is positive
17. DEFLECTION OF BEAM UNDER UNSYMMETRICAL BENDING
Let the bending moment “M” inclined at an angle “θ” with one of principal
planes (Say VV-axis)
Along UU-axis M component will be Mvv= M cos θ
Along VV-axis M component will be Muu= M sin θ
18. From the application of principal of virtual work (unit load method) the
deflection (δ) of the beam in any direction, due to a bending moment M is
given by
δ = 0
l M m
EI
dx
Where,
M = moment due to applied moment (say M)
m = moment due to unit load applied at the point in the direction of the desired
deflections,
dx = elementary length of beam, measured along the span of the beam
Hence the deflection of the beam in the direction of VV- axis is given by
δv = 0
l M cos θ
E Iuu
mv dx
Hence the deflection of the beam in direction of UU-axis is given by
δu = 0
l M sin θ
E Ivv
mu dx
The resultant deflection δ is given by
δ = δu
2 + δv
2
19. Since mv,mu are the moments developed due to unit load applied, it can be
taken both equal to m i.e. ( mv= mu= m)
If β is the inclination of neutral axis (NN-axis) with respect to UU-axis we
can write as
tan β =
Iuu
Ivv
tan θ
Let γ be the inclination of resultant deflection in the direction N!N!-axis
makes with UU-axis
tan γ = -
δu
δv
tan γ = -
0
lM sin θ
E Ivv
mu dx
0
lM cos θ
E Iuu
mv dx
tan γ = -
Iuu
Ivv
tan θ
tan γ = - tan β = tan(90 + β)
Therefore we can write as
γ = 90 + β
20. Hence the resultant deflection occurs in the direction exactly perpendicular
to the neutral axis (N!
N!
- axis perpendicular to NN-axis)
Let us consider the case of simply supported beam (SSB) subjected to
UDL, then,
δv =
5
384
w cos θ l4
E Iuu
; δu =
5
384
w sin θ l4
E Ivv
δ = δu
2 + δv
2
δ =
5
384
wl4 cos θ sec β
EIuu
Multiplying and dividing by cos(β − θ)
δ =
5
384
l4
E
w cos(β−θ)
Inn
Thus from the above expression for a simply supported beam (SSB) we
can conclude that the term w cos(β−θ) is the resultant udl acting along
N!N!- axis which is perpendicular to neutral axis.
21. PROBLEMS ON UNSYMMETRICAL BENDING
A Cantilever Problem
1.A horizontal cantilever 2 m long is constructed from the Z-section shown
below. A load of 10 KN is applied to the end of the cantilever at an angle
of 60°to the horizontal as shown. Assuming that no twisting moment is
applied to the section, determine the stresses at points A and B.
2.Determine the principal second moments of area of the section and hence,
by applying the simple bending theory about each principal axis, check
the answers obtained in part1.
3. What will be the deflection of the end of the cantilever? E = 200 GPa.
(Ixx = 48.3 x 10−6m4, Iyy = 4.4 x 10−6m4)
22. In the given section Ixy for the web is zero since its centroid lies on both axes
and hence h and k are both zero. The contributions to Ixy of the other two
portions will be negative since in both cases either h or k is negative.
Therefore, Ixy = -2(80 x 18) (40 - 9) (120 - 9) 10−12
= -9.91 x 10−6m4
Mx = +10000 sin60° x 2 = +17320 Nm
My= -10000 cos60° x 2 = -10000 Nm
But we have,
σz =
Mx
Ix
y +
My
Iy
x
Let
My
Iy
= P,
Mx
Ix
= Q
Mx = P Ixy + Q Ix ; My = -P Iy - Q Ixy
17320 = P (-9.91) x10−6 + Q 48.3x10−6
-10000 = P (-4.4x10−6) + Q 9.91x10−6
Solving the above two equations for P and Q,
P = 5725x106; Q = 1533x106
23. The inclination of the N.A relative to the X axis is then given by
tan ∝ =
My Ix
Mx Iy
= -
P
Q
= -
5725 x 106
1533 x 106 = - 3.735
∝ = −75°
1′
Now
σz =
Mx
Ix
y +
My
Iy
x = P x + Q y
Stress at A = 5725x106x 9x10−3+ 1533x106x 120x10−3
= 235 N/mm2
Similarly stress at B = 235 MN/mm2
The points A and B are on both side of neutral axis and equidistant from it.
Stresses at A and B are therefore of equal magnitude but with opposite sign.
24. 2. The principal second moments of area may be found from the following
equations
Iuu =
Ixx+ Iyy
2
+ (
Ixx− Iyy
2
)2+Ixy
2
=
48.3x10−6+4.4x10−6
2
+ (
48.3x10−6− 4.4x10−6
2
)2+(−9.91 x 10−6)2
= 50.43x10−6
Ivv =
Ixx+ Iyy
2
- (
Ixx− Iyy
2
)2+Ixy
2
=
48.3x10−6+4.4x10−6
2
- (
48.3x10−6− 4.4x10−6
2
)2+(−9.91 x 10−6)2
= 2.27x10−6
tan 2 ∝ =
2 Ixy
Iyy− Ixx
=
−2 x 9.91 x 10−6
(4.4− 48.3)x 10−6 = 0.451
2 ∝ = 24°18′ , ∝ = 12°9′
25. The required stresses can now obtained from the above equation
σ =
Mv
Iv
u +
Mu
Iu
v
Mu= 10000 sin(60°-12°9′) x 2
= 10000sin(47°
51′
)
= 14828 Nm and
Mv = 10000cos(47° 51′) x 2
= 13422 Nm
And, for A,
u = xcos ∝ + ysin ∝
u = (9 x 0.9776) + (120 x 0.2105) = 34.05 mm
v = ycos ∝ - xsin ∝
v = (120 x 0.9776) - (9 x 0.2105) = 115.5 mm
σ =
14828 x 115.5 x 10−3
50.43x10−6
+
13422 x 34.05 x 10−3
2.27x10−6
= 235 MN/m2 as obtained before
26. 3. The deflection at free end of a cantilever is given by
δ =
WL3
3EI
Therefore the component of deflection perpendicular to the V axis
δv =
WvL3
3EIv
=
10000 cos(47°51′) x 23
3 x 200 x 109 x 2.27 x 10−6 = 39.4 mm
And component of deflection perpendicular to the U axis
δu =
WuL3
3EIu
=
10000 sin(47°51′) x 23
3 x 200 x 109 x 50.43 x 10−6 = 1.96 mm
The total deflection is then given by
δ = δu
2 + δv
2 = 39.42 + 1.962 = 39.45 mm
Its direction is normal to the N.A.
27. UNSYMMETRICAL CANTILEVER UNIT [11]
To demonstrate unsymmetrical bending of
beams
Determines deflections along u and v
directions
Consist of
1. Main column (cantilever specimen
clamped at its bottom)
2. Loading head at upper end – can rotate
180°
with 15°
intervals about vertical
axis
3. Set of pulley, located at the loading head,
to apply a horizontal load.
4. 2 Dial gauges of 0-25 mm and 0.01
mm accuracy, to measure and deflections.
Poligona industrial san jose de valderas,
spain demonstrated this model.
Limitations
-Dimensions: 400 x 300 x 400 mm approx.
-Weight: 14 Kg. approx.
28. IMPORTANCE OF UNSYMMETRICAL BENDING
If the plane of bending or plane of loading does not lie in or parallel to the
plane that contains the principal centroidal axes of cross-section, the
bending is called as unsymmetrical bending.
Members that are not symmetrical about the vertical axes and which are
typically composed of thin unsymmetrical sections (e.g. ISA, Channel)
undergo phenomenon of twisting under the transverse loads.
A channel section carrying the transverse load would twist because the line
of action of the load does not pass through shear centre of the member.
whereas rectangular beam would not twist because the loading would pass
through the centre of gravity of the section and for such two axis
symmetrical section the shear centre would coincide with the cg of the
section.
If one is desired to use unsymmetrical sections to carry transverse loads
without twisting, it is possible to do so by locating the load so that it passes
through the shear centre of the beam.
29. CONCLUSION
The axis of about which the product moment of inertia is zero is called as
principal axis. Hence we can conclude by saying that the simple bending
theory is applicable for bending about principal axes only.
It should be noted that moment of inertia of a section about its principal axes
have maximum and minimum values respectively.
The resultant deflection for simply supported beam subjected unsymmetrical
bending is
δ =
5
384
l4
E
w cos(β−∝)
Inn
The resultant deflection for cantilever beam is
δ =
L3
3E
w cos(β−∝)
Inn
Taking Inn = Iuu cos2
β + Ivv sin2
β
In order to overcome the effect of twisting when the beam subjected to
unsymmetrical loading, the study of unsymmetrical bending is useful.
30. REFERENCES
1. Arthur P. Boresi, Richard J.Schindt, “Advanced Mechanics of materials”, Sixth edition
John Wiley &Sons. Inc., New Delhi, 2005
2. Thimoshenko & J N Goodier, “Mechanics Of Solids”, Tata McGraw-Hill publishing
Co.Ltd, New Delhi, 1997
3. Seely Fred B. and Smith James O., “ Advanced Mechanics of Materials”, 2nd edition,
John Wiley & Sons Inc, New York, 1952
4. Srinath L.S., “Advanced Mechanics of Solids”, Tata McGraw-Hill publishing Co.Ltd,
New Delhi, 1980
5. Thimoshenko S., “Strength of Materials”, Part-1, Elementary Theory and Problems, 3rd
Edition, D. Van Nostrand company Inc., New York, 1955
6. Boresi A.P and Chong K.P(2000), “Elasticity In Engineering Mechanics” 2nd edition
New York ; Wiley – Interscience.
7. N Krishna Raju & D R Gururaja, “Advance Mechanics Of Solids & Structures”, 1997
8. B C Punmia & A K Jain. “Strength of Materials and Theory of Structures”, Vol.2
Lakshmi publications (P) Ltd.2002
9. Jiang Furu, “Unsymmetrical Bending Of Cantilever Beams”, Appl. Math. & Mech.
(English Ed.), 1982
10. G. D. Williams, D. R. Bohnhoff, R. C. Moody, “Bending Properties of Four-Layer Nail-
Laminated Posts”, United States Department of Agriculture, Forest Service (Forest
Products Laboratory), Research Paper FPL–RP–528, 1994
11. Poligona industrial san jose de valderas, spain, “Unsymmetrical Cantilever Unit”,
ED01/12, 2012