Engineering Mechanics: Statics & Dynamics

Video Lectures

Displaying all 83 video lectures.
Lecture 1
Scalars and Vectors
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Scalars and Vectors
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The objectives of this video are to review the scalar & vector concept and to do distinguish between scalars and vectors. First of all, the video gives definition of scalar & vector where scalar represents quantities that can be fully described by a magnitude (a numerical value) alone & vector represents quantities that require both the magnitude and direction to get described. Effectively what it means is that a scalar is simply a number or unit & a vector is something having magnitude along with the direction acting towards. Next, the video illustrates scalar & vector vividly by a schematic figure.

Moving on, the video introduces with few examples of scalars & vectors demonstrating the grounds for being those to be scalars or vectors. Doing so, the video highlights the difference between scalar and vector explaining all the facts & figures in greater details. Later, the video gives an overview on how to do adding operation of scalar & vector and finally refers to watch next video to get in touch with the parallelogram law and triangle method used to do operations on vectors.
Lecture 2
Parallelogram Law and Triangle Method
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Parallelogram Law and Triangle Method
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The objectives of this video are to go through the parallelogram law for addition and subtraction of vectors followed by a brief discussion on triangle method used to do vectors addition & subtraction. At first, the video demonstrates the parallelogram law clearing the concept that if two vectors can be represented by two adjacent sides of a parallelogram then the diagonal of parallelogram will be equal to the resultant of these vectors regardless of the addition or subtraction operation. The video also talks about the difference of addition & subtraction operation explaining all the facts in details.

Next, the video briefly talks about the key points of parallelogram law-always join vectors tail to tail & subtraction is the same as adding the negative. Moving on, the video discusses on triangle method where vectors are joined tip to tail for addition and this is independent of orders of summation. The subtraction using the triangle method is very similar to parallelogram law. Later, the video explains vividly both the addition & subtraction operations of vectors applying triangle method.
Lecture 3
Unit Vectors and Components
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Unit Vectors and Components
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The objectives of this video are to discuss the concept of unit vectors & resolves vectors into x and y components followed by a brief discussion on vectors as a magnitude and angle. A unit vector is the vector having magnitude one and are often parallel to the coordinate axes pointing toward positive value of the coordinate. The unit vectors are useful as they ease the way to turn magnitude to vector by giving directions. Using the concept of unit vectors, the video resolves the given multiple vectors in x and y components and subsequently shows how to add those together.

Moving on, the video introduces with vectors operations as a means of magnitude & angle over the Cartesian coordinates. Subsequently, the video briefly illustrates how to find the value of magnitude of resultant vector & angle of the resultant vector formed by the original given vectors. Overall, the video tries to give a brief inside on unit vectors & its’ uses on doing vectors operations to the depth- understanding which is must to move forward with the course of engineering mechanics.
Lecture 4
Vectors Example
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Vectors Example
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This video presents a comprehensive example of finding resultant vector as a mean of magnitude & direction resolving three given vectors. First of all, the video illustrates the example vividly where it has shown that three different forces are acting on a object at different angle with the coordinates & each of the forces are varied in magnitude. Next, the video talks about the procedures to obtain the resultant vector using unit vectors concepts and resolving the given vectors into x & y components.

Moving on, the video shows how to resolve the given vectors into x and y components explaining all the facts and figures in details. The video, then, sum up all x and y components vectors separately & applies parallelogram formula to determine the magnitude and angle of the resultant vector. Later, the video shows the resultant vector in terms of magnitude & angle in the given exemplary figure. If you don’t understand any part of the tutorial, don’t hesitate to contact with our dedicated tutors.
Lecture 5
Vector Tower Example
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Vector Tower Example
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The objective of this video is to resolve vector to solve an engineering relevant problem. First of all, the video briefly illustrates the given exemplary figure of tower that is supported by 2 cables from opposite directions- one cable acts at an angle of 45 degree with the ground having 10 KN tension from right side & another one acts at an angle of 70 degree from left side. The video asks to find out the force exerted in left side cable. The height of the tower is given as 20 meter.

As the tower is stable at its position, so for equilibrium forces at left side must have to be equal with the forces acting from right side of the tower. Using this principle together with applying of vectors resolving technique, the video step by step shows how to determine the tension force exerted in the cable used to give support the tower from left side.
Lecture 6
3D Vectors
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3D Vectors
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The objectives of this video are to discuss 3-D axes followed by a brief introduction to plotting of 3-D vectors. At first, the video clears the concept of 3D axes graphically illustrating a coordinate frame having three axes respectively at x, y & z direction. Each axis is located at right angle from the other two axes. The video shows the three planes e.g., xy, yz & zx formed because of 3D axes sketching the region over the coordinate frame. It is necessary to have clear concept on 3D axes to understand the workouts on 3D vectors in doing the study of engineering mechanics.

Moving on, the video shows how to plot a point in 3D coordinates system followed by the plotting of 3-D vectors. Doing so, the video shows how magnitude and direction of vectors impacts on plotting & subsequently shows how the resultant represents the final coordinates of the 3-D vectors in a 3-D plot. Overall, the video tries to give a brief inside to 3-D vectors plotting which is quite simple yet very important for doing different operations of vectors over the 3-D coordinates system.
Lecture 7
3D Vector Example (Part 1)
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3D Vector Example (Part 1)
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This video shows the consideration of direction cosine for 3-D vectors followed by a comprehensive 3-D vector example. First part of the video displays a 3-D reference frame containing a 3-D vector & subsequently elaborates its components value along x, y & z axes with the cosine angle representing the direction of 3-D vector in 3D space. Next, the video introduces with the formula to find the value of resultant vector magnitude from the component values acting x, y & z axes. The video, then move on to a problem, with the objective to find the resultant vector magnitude from the given data.

In the exemplary problem, there are two vectors given with the acting angle over the 3-D reference frame. The absolute values of these vectors are also given. Moving on, the video briefly shows how to obtain these vectors in terms of unit vector along the x, y & z axis explaining all the facts & figures in details. Later, the video talks about the adding operation required to do to get resultant vector & finally refers to watch next video to continue the lesson of 3-D vectors example.
Lecture 8
3D Vectors Example (Part 2)
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3D Vectors Example (Part 2)
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This video, continuation of the previous video 1.8, very first demonstrates the key objective to add the unit component vector to get the resultant vector. The video step by step shows how to add unit vectors discretely at x, y and z axes followed by the addition operation of all three unit vectors to obtain the resultant vector over the 3D reference frame. Doing so, the video explains the addition of vectors having negative magnitude in value.

Moving on, the video briefly shows how to obtain the resultant vector absolute magnitude applying the formula introduced in earlier video and subsequently elaborates how to do angle calculation to get the resultant vector direction in 3D reference space. Later, the video shows the procedure to get unit vector of the obtained resultant vector. Overall, the video tries to give a brief inside to clear the concept of 3-D vector operation in 3-D reference space. Don’t forget to do on hand practices to have a strong basic on vectors which is required to understand many areas of engineering mechanics.
Lecture 9
Introduction to Forces
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Introduction to Forces
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This video discusses about forces and their relevance to engineering followed by a brief explanation on different types of forces and units. Forces are some kind of influence which results in changes to the motion of objects & often cause accelerations to object having mass by changing in velocity with respect to time. Moving on, the video talks about the relevance of forces to engineering mechanics pointing out its application to deformation, velocity changes, deterioration of materials, deflections and bending. Through understanding of forces are crucial for physical aspects of engineering study.

Second part of the video discusses on different types of forces including forces due to gravity, point & distributed load, spring actions and internal forces. Later, the video introduces with the SI unit of forces which is symbolized by N (Newton). 1N is the force required to accelerate 1 kilogram of mass at the rate of 1 meter per second squared. Overall, the video tries to give a brief inside to forces and its types, understanding which is vital to have a strong basic for the study of engineering mechanics.
Lecture 10
Introduction to Moments
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Introduction to Moments
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The objective of this video is to clear the moment concept followed by a workout on simple moment calculation. First of all, the video gives the definition of moment stating that moment is equal to the product of forces time distances. According to SI unit, the force is calculated by unit N (Newton) and distance is calculated by unit m (Meter). This makes the SI unit for moment is Nm (Newton-meter). The term moment is also described as being torque in the mechanics of solids.

Moving on, the video displays a figure illustrating the given data where it has asked to calculate the moment developed in a point 3m away from the applied vertical force, 10N. The video shows the moment calculation both by taking moment at clockwise & anticlockwise direction. Later, the video briefly shows what happens in moment calculations when a force is applied parallel to the point at which moment is needed to be calculated. Finally, the video concludes the facts that moment is only related to these forces which create rotation and rest forces just create axial forces.
Lecture 11
Moment Example 1
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Moment Example 1
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The objective of this video is to do a comprehensive workout on more complex moment calculation. Very first, the video describes the problem showing a free body diagram of a beam subjected to two different forces - one acting perpendicularly at leftmost point of the beam and the second one acting at rightmost point of the beam with an angle 45 degree. The beam is 12 meter long and a point ‘B’ is located 4 meter away from the rightmost end of the beam. The video asks to determine the moment at point A (leftmost end of the beam) and point B.

Moving on, the video briefly shows how to resolve force components in x & y direction of the given inclined force. Next, the video shows the moment calculation at point A considering anticlockwise rotation caused by the applied forces explaining all the facts in details. Later, the video shows the moment calculations at point B considering the rotation caused by resolved forces in anticlockwise direction. Overall, the video tries to clear how to do moment calculations for complex setbacks.
Lecture 12
Moment Example 2
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Moment Example 2
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This video represents another comprehensive example of complex moment calculations. First of all, the video illustrates the given problem showing a free body diagram of a vertical structural member subjected to three different forces. It has shown in the diagram that one force is acting at top side of the vertical member towards right direction, second one is acting at 1 meter away from top towards left direction & the last one is acting at bottom side of the vertical member towards right direction.

A point ‘X’ has been shown to be located at 3m away from the top of the member. The video asks to do moment calculations at point X for the forces shown to be subjected over the vertical member. Moving on, the video takes moment at point X considering anticlockwise rotation and subsequently determines the moment developed by the forces applied in Nm (Newton-meter). It is recommended to do on hand practices to have better understanding on the content described in this lesson.
Lecture 13
Moments and Couple Moments
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Moments and Couple Moments
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The objectives of this video are to discuss about moment & couple moments followed by a workout on couple example. First of all, the video clears the concept of couple illustrating the fact that when two parallel forces acting on an object in opposite direction separated by a perpendicular distance, it results a phenomenon known as couple. Next, the video briefly discusses how a couple produces moments on a circular body and subsequently shows the couple moments calculation for it.

Moving next, the video does a workout on couple example where it has asked to calculate the force required to turn a wheel when applied at the edge of wheel and at the end of the bar. The necessary dimension is given with the free body diagram of the exemplary problem. It has also given that a 10 Nm momentum is required to turn the wheel. The video, then, shows the entire calculation & finally obtains the values of required forces to turn the wheel from particular positions.
Lecture 14
Equivalent Systems Theory and Example
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Equivalent Systems Theory and Example
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This video briefly demonstrates equivalent system theory followed by solving an equivalent system example. First part of the video shows a free body diagram of a structure subjected to a number of forces at different direction. The video, then, talks about equivalent systems theory clearing the fact that in an equivalent system the complex problems have been simplified into equivalent forces and moments to make the calculations much easier and less time consuming.

Next, the video does a comprehensive workout on an equivalent system example where it has asked to resolve the given forces into a single resultant force and to find couple moments at bottom of the structure. Moving on, the video illustrates the given free body diagram of the exemplary structure & step by step shows how to resolve the given forces to obtain a single resultant. Later, the video uses the obtained resultant force to determine the couple moment at bottom part of the structure.
Lecture 15
Distrubuted Loads
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Distrubuted Loads
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The objectives of this video are to give an introductory overview to distributed loads followed by a brief discussion on how to resolve the given distributed loads into an equivalent resultant force. At first, the video shows a schematic diagram of a beam subjected to a uniformly distributed loads i.e., forces that act across the length of the structure. Next, the video tries to clear the concept of UDL by means of a person horizontal & vertical positioning over a structural element. The video, then, talks about the steps of moment calculations for distributed loads.

Moving on, the video shows how to resolve the given distributed into an equivalent resultant point load doing a workout on an exemplary problem. A point load can be easily calculated from UDL by multiplying the given value of UDL with the length of the structure across which the UDL is acting over. Next, the video shows how to calculate the lever arm position followed by the calculations of moment at left edge of the beam. Overall, the video tries to give a brief inside to distributed loads.
Lecture 16
Solving Distributed Loads and Triangular Loads
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Solving Distributed Loads and Triangular Loads
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The objectives of this tutorial video are to discuss about different distributed loads combinations & to examine triangular distributed load. At first, the video starts up by looking at an exemplary beam structure subjected to 2 different distributed loads i.e., one UDL of 5 KN/m across 4 m length from right edge & another of 10 KN/m across 2 m length from left edge. Next, the video converts the UDL to points load and does moment calculations at right edge of the given beam.

Moving on, the video introduces with the triangular distributed loads and briefly demonstrates how to convert a triangular distributed load into a point load. Unlike the UDL, in a triangular distributed load the centroid position is required to determine in order to find the acting point of the converted point load. Later, the video also introduces with both the formulas of finding the resultant value of converted point load and distance of centroid from the edge that subjected to lower amount of load.
Lecture 17
Resolving Forces Advanced Example
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Resolving Forces Advanced Example
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The objective of this video is to consider more complex examples involving resolving of forces. First part of the video introduces with a diagram of a structural element subjected to two different forces and the fact is that the value of one force with its acting angle is not given which needed to be found out from the given values of another force and the resultant value of these forces. The acting angles both for another force and resultant force have been given.

Moving on, the video shows how to resolve the given forces into its components in x and y direction & subsequently determines the unknown force value solving the equations of obtained components forces. Later, the video uses the value of known angle and magnitude of forces to find out the acting angle of the respective force which working direction has not been given. This is quite a complex problem, so do practices to have better understanding on the lesson overviewed in this tutorial.
Lecture 18
Introduction to Equilibrium
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Introduction to Equilibrium
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The objectives of this video are to briefly discuss about equilibrium and relate equilibrium concepts to finding reaction forces. Basically equilibrium refers to analysis of forces subjected to an object at rest conditions. In this tutorial, the video firstly talks about the equilibrium of static condition and wanted to explain the equilibrium of dynamic condition afterward. Moving on, the video introduces with a diagram of coffee table having a box over it & subsequently discusses about the equilibrium of this condition.

The video vividly shows how the gravity force of the box and the reaction of the table balance each other following the Newton’s 3rd law of motion. Later, the video introduces with another condition where it has shown that a car is travelling over an inclined surface. Subsequently, the video explains the equilibrium of the gravity force, friction force and reactions forces subjected to the car when it is travelling. Overall, the video tries to give a brief inside about static and dynamic equilibrium.
Lecture 19
Introduction to Free Body Diagrams (FBD)
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Introduction to Free Body Diagrams (FBD)
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The objectives of this video are to give an introductory overview to free body diagrams followed by a brief discussion on the importance of free body diagram in the study of engineering mechanics. A free body diagram is a diagram used to show the relative magnitude & direction of all forces acting upon on an object in a given situation. Each force arrow on the diagram is mostly labeled to indicate the exact type of force. Moving on, the video shows a problem where it has shown that a box is rest over an inclined surface & subsequently elaborates how to draw a free body diagram for that case.

Next, the video demonstrates another exemplary problem of a simply supported beam drawing the free body diagram for the given condition. Doing so, the video briefly talks about the magnitude and acting direction of different forces explaining all the facts & figures in details. Later, the video talks about the importance of FBD briefly which allows to identify the forces that act upon on an object & to express such an understanding through the construction of FBD. Never hesitate to contact with our expert tutors if you face any difficulties in understanding the lesson overview in this tutorial.
Lecture 20
Free Body Diagram Example
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Free Body Diagram Example
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This video represents a comprehensive example of solving an equilibrium problem using free body diagram. First of all, the video introduces with the given exemplary diagram that shows a block is at rest condition over a slope having 30 degree angle with the ground. The mass of the block has given as 1 kg whereas the gravitational acceleration is 10m/s2. The block has been hold back by a tension force and it has been assumed that there is no frication force acting over the block. Moving on, the video works out on external forces and subsequently determines the value of gravity force.

Next, the video draws the free body diagram showing the working direction of gravity force, tension force and reaction force over the given slope. Later, the video uses the drawn free body diagram to resolve the forces into its components in x and y direction and successively finds out the unknown values of tension and reaction forces from the resolved equations of forces components. Overall, the video tries to give a brief inside on using FBD to establish equilibrium to solve mechanics problem.
Lecture 21
Introduction to Supports: Roller, Pin, Fixed
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Introduction to Supports: Roller, Pin, Fixed
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In this tutorial, the video demonstrates different types of supports e.g., pin, roller and fixed support. When a number of forces are acting on a body and the body is supported on another body then the 2nd body exerts a force known as reactions on the first body at the points of contact so that the first body is in equilibrium. The second body is known as support and the force, exerted by 2nd body on the first body, is known as support reactions. Moving on, the video briefly discusses on three types of supports commonly found in the construction of different structural elements.

A pin support can resist both horizontal & vertical force but not a moment. It only allows a member to rotate but not to translate in any direction. Whereas a roller support is free to rotate & translate along the surface upon which the roller rests and a fixed support can resist both the horizontal and vertical forces as well as moment. Last part of the video illustrates an exemplary frame having pin, roller and fixed supports explaining all the facts and figures in greater details.
Lecture 22
Simply Supported Beams Free Body Diagram Example
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Simply Supported Beams Free Body Diagram Example
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The objectives of this video are to give an introductory overview on how to use free body diagrams to deduce support reactions followed by a comprehensive workout on support reactions example. At first, the video illustrates a given diagram of simply supported beam having a pin support at left end and a roller support at right end and consequently shows the point load conditions acting over the beam. Next, the video explains the reactions forces exerted by the supports and draws the free body diagram for it successively deriving the equilibrium equations.

Moving next, the video presents an exemplary beam structure supported by pin and roller supports respectively at left and right edge and subjected to a uniformly distributed load and point load. The necessary dimensions & the subjected loads values are given. Using the given information, the video shows how to draw FBD & subsequently establishes equilibrium condition to find out the reactions forces value exerted by the supported bodies to the beam structure. It is highly recommended to do on hand practices to have good grasp on the content overviewed in this lesson.
Lecture 23
Cantilever Free Body Diagram Example
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Cantilever Free Body Diagram Example
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The key objective of this video is to consider support reactions for a fixed support. At first, the video refers previous video where the support conditions for pin and roller supports are overviewed and thereafter wants to look at what happens in fixed supports in this tutorial. The video, then, displays a cantilever beam subjected to a point load of 20 N at free edge of the beam in downward direction. The length of the beam has given as 3 m. Next, using the given information, the video shows how to draw the FBD illustrating what reactions and momentum have been caused by the fixed support.

Moving on, the video calculates the moment at cantilever support point developed by the point load subjected to the cantilever beam. The fixed support must have to be strong enough to withstand the calculated momentum in order to ensure that the beam will not rotate when subjected to the given amount of point load. Later, the video shows how to determine the reactions forces exerted by fixed support using the principle of equilibrium. At end, the video explains the scenario of moment effect over the cantilever beam pointing out the rotating condition of the given beam.
Lecture 24
Advanced Free Body Diagram Beam Example
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Advanced Free Body Diagram Beam Example
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The objective of this video is to solve harder support reaction problem. First of all, the video shows the diagram of given exemplary problem which is nothing but a beam supported by a roller support at left end and another pin support at 4 m away from the left end. Both the supports are working on the beam from upward direction. The beam is subjected to two different loads i.e., a point load of 30 KN acting downward at 2 m away from right end and a uniformly distributed load of 5 KN/m acting downward and over 2 m length of the beam from right end.

All the necessary dimensions are also given. Moving on, the video draws the free body diagram for the problem at first step explaining all the facts and figures in greater details. Next, the video shows how to establishes equilibrium condition & thereafter applies equilibrium principle to resolve force components in x and y direction to get forces equations. The video also shows how to take moments for such complex conditions and finally determines the unknown values of forces and moment.
Lecture 25
Introduction to Axial & Shear Forces and Bending Moments
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Introduction to Axial & Shear Forces and Bending Moments
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The objectives of this video are to discuss about internal forces followed by a brief demonstration on axial, shear & bending. This tutorial is pretty much all about internal forces & the way we resolve internal forces when we have given external forces. Basically, internal forces represent and explain what is happening inside a member and external forces are generally transferred through members via internal forces. The video clears the concept of internal forces drawing a FBD and subsequently explains all the facts and figures in greater details.

Moving on, the video introduces with a simply supported structure subjected to a heavy load at its center which results the sagging means the structure will be pushed down. Next, the video explains how the load condition and supports reactions develops axial, shear and bending and describes the physical occurrences through schematic diagrams. Overall, the video tries to give a brief inside to axial forces, shear forces and bending moments developed in structural element when subjected to different load conditions. Don’t hesitate to contact, if you don’t understand any part of the tutorial.
Lecture 26
Axial, Shear and Bending Diagrams
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Axial, Shear and Bending Diagrams
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The objectives of this video are to clear the concept on axial, shear and bending moments followed by a brief introduction to axial, shear and bending diagram. The video starts up by quickly looking at an exemplary problem of simply supported beam subjected to a uniformly distributed load over the length of the beam. The beam is supported by a fixed support at left end and a roller support at right end as well as a force is exerted outward from the beam at right end parallel with the beam.

Moving on, the video at first step draws the free body diagram of the beam finding out the reactions force developed by the supports and converting the given UDL to a point load acting at center of the beam. Next, the video demonstrates how to draw axial force diagram from the drawn FBD and later elaborates briefly the drawing of shear force and bending moment diagrams highlighting the max shear and moment conditions. Finally, the video gives a quick summary of the lesson described in this video. Don’t hesitate to contact, if you don’t understand any part of this tutorial.
Lecture 27
Method of Sections
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Method of Sections
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The objective of this video is to introduce method of sections for finding internal forces in members. The method of sections is converting external forces into internal forces at arbitrary points. At first, the video shows a simply supported beam having a fixed support at left end and a roller support at right end. The video, then, asks to find out the internal forces at point x means the axial force, shear force and bending moment of that point. The point x is basically an arbitrary point located x meter away from the left end of the given exemplary beam.

The method of section allows cutting the beam at the selected arbitrary point & separated the beam into two different parts. Moving on, the video shows the two cuts of the beam and elaborates what internal forces are developed due to the cuts at respective section of the given beam. Next, the video completes the FBD diagram of the respective sections & later shows the equilibrium conditions for left cut of the beam. Finally, the video refers to watch next video for a workout example.
Lecture 28
Method of Sections Simple Example
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Method of Sections Simple Example
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The objective of this video is to find internal forces and moments in a simply supported beam. First of all, the video shows the schematic of the beam supported by fixed support at left end and roller support at right end. The beam is subjected to a uniformly distributed load, 5 N/m across the length of the beam and a point load, 10 N at right end parallel with the beam. The video asks to find out the axial, shear and bending at 1 m away from the left end of the beam. Using the given information, the video then shows how to draw free body diagram for this exemplary problem.

Moving on, the video takes cut at critical point followed by the drawing of the FBD for the cut. Three internal reactions e.g., vertical and horizontal reaction forces and an anticlockwise bending moment are developed due to the cut which have been briefly examined in the video. Later, the video solves the unknown values of internal forces establishing the principle of equilibrium. It is suggested to do on hand practices to have good understanding on method of sections to solve such kind of problem.
Lecture 29
Method of Sections Advanced Example Part 1
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Method of Sections Advanced Example Part 1
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The objectives of this video are to consider more complex loading cases & determine internal forces from both ends of the beam. Very first, the video presents the schematic diagram of the exemplary problem & asks to find internal force at midspan. The schematic diagram shows a simply supported beam having a fixed support at left end & a roller support at right end. The beam is 8 m in length & subjected to 2 different types of loads i.e., a uniformly distributed load of 5 KN/m across 4 m length of the beam from right end and a point load of 10 KN at 3 m away from right end of the beam.

Moving on, the video draws the FBD followed by a workout showing how to determine the support reactions using the principle of equilibrium. Next, the video applies cuts at critical points and briefly talks about the two options available to calculate the internal forces at midspan. Later, the video considers the left cut & illustrates how to find internal forces i.e., axial, shear and bending moments at midspan of the beam. Finally, the video refers to watch next video to continue the lesson.
Lecture 30
Method of Sections Advanced Example Part 2
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Method of Sections Advanced Example Part 2
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This video, continuation of the earlier video, very first clears it objective to find the internal forces i.e., axial, shear and bending at midspan of the given exemplary beam considering the section to the right of the cut. After showing off the schematic of exemplary beam, the video examines section of right cut followed by the drawing of FBD. Next, the video applies the equilibrium principle to obtain forces and moments equations. Doing so, the video explains all the facts and figures in details.

Moving on, the video computes the unknown values of internal forces and subsequently verifies the results with the previously obtained results from section to left cut. Later, the video talks about the importance of considering a cut that minimize the computation time to find out the unknown values especially in exam. The method of sections is quite tricky, so better understanding on this method is very significant in doing many complex workouts of engineering mechanics.
Lecture 31
Introduction to Hooke's Law
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Introduction to Hooke's Law
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The objectives of this video are to give an introduction to Hooke’s law followed by a comprehensive workout on finding stress and strain in a column under axial loading. First of all, the video explains the theory of Hooke’s law that is a principle of physics which states that the forces needed to extend or compress a structure by some distance either in length contraction or elongation is proportional to that distance. The proportional constant is known as elastic modulus & it often refers to the ratio of stress and strain which allows deducing other when one of them has been known.

Moving on, the video presents a workout of stress and strain where the value of elastic modulus as well as the schematic diagram of the problem has been given. Next, the video illustrates the FBD & AFD for the example followed by an elaboration on how to find stress using the formula overviewed in previous video. Later, the video uses the obtained value of stress to find strain and subsequently shows how to find the change in length of the given exemplary column due to the load subjected.
Lecture 32
Hooke's Law and Stress vs Strain
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Hooke's Law and Stress vs Strain
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In this tutorial, the video shows another workout example of Hooke’s law and stress vs. strain. Very first, the video overviews the example where it has given that the column length is 3.5 m and it gets shortened by 0.03 m when subjected to a load of F. The video asks to find the magnitude of the load F. The modulus of elasticity of the column material has been given as 40 MPa & the diameter of the column is 200 mm. The video, moving next, briefly talks about the procedures to solve the problem starting with FBD to AFD to use the method of sections cut to calculate the unknown values.

Using the given value of column length and shortened length of the column when subjected to load, the video shows how to find the value of strain. Moving on, the video uses the obtained strain value and given modulus of elasticity to determine the value of stress developed on the column member. Later, the video uses the average normal stress formula to find out the value of axial load F that causes the length contraction. Do on hand practices with this tutorial to have better understanding on the workout example overviewed in this lesson.
Lecture 33
Stress vs Strain Diagram
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Stress vs Strain Diagram
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The objectives of this video are to discuss & interpret stress vs. strain diagram for steel. First of all, the video shows the stress vs. strain diagram of steel where the x axis represents strain & the y axis represents stress. Next, the video starts talking about the curve plotted over the diagram where the origin represents zero stress & strain. The first zone of the curve which is linear in nature is known as elastic zone. In elastic zone, the material follows Hooke’s law and the constructions of structures must have to fall under this zone while doing the design phase calculations.

Moving on, the video briefly explains the importance & necessity of doing designs within the elastic zone. Later, the video elaborates other points and regions of the stress-strain curve e.g., yield point, ultimate strength and fracture. Doing so, the video also talks about the natures & properties of steel material. Overall, the video tries to give a crystal clear overview on stress vs. strain diagram of steel material- understanding which is must for the study of engineering mechanics.
Lecture 34
Rectilinear Motion |
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Rectilinear Motion |
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The objectives of this video are to discuss about rectilinear motion followed by an introduction to useful formulas needed to solve problems about rectilinear motion. Basically, rectilinear motion is a motion acting in a single direction where acceleration could be both constant and variable. Moving on, the video briefly talks about the scalar and vector terms e.g., distances, displacement, speed and velocity that are related to understand to clear concept on rectilinear motion of a body.

Next, the video introduces the formulas used to describe rectilinear motions using the terms initial velocity, final velocity, displacement, acceleration and time. Later, the video explains an important note clearing the fact that if acceleration is given as a function it is reasonably possible to find out the function which defines velocity & displacement. Finally, the video refers to watch next video to get in touch with a couple of workouts on the subject topic of rectilinear motions.
Lecture 35
Rectilinear Motion Examples |
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Rectilinear Motion Examples |
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The objective of this video is to use rectilinear concepts overviewed in last video to solve exemplary problems. At first, the video presents a problem of a man dropping a ball off a cliff. The video asks to calculate how tall is the cliff if the ball is travelling 40 m/s just before it hits to the ground. Moving on, the video overviews the problem through a schematic drawing & subsequently shows the three formulas described in previous video. The video, then, explains how to choose the required formula wisely to solve the problem based on the data and information available.

Second part of the video presents another problem of rectilinear motions where it has asked to find how much time is required to reach a terminal velocity of 60.0 m/s when a skydiver jumps out of a plane. Moving on, the video briefly shows how to assess parameters values from the given data and successively determines the time required to reach the stated terminal velocity. Overall, the video is tried to give an overview on the use of formulas of rectilinear motion in solving different problems.
Lecture 36
Rectilinear Motion with Variable Acceleration |
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Rectilinear Motion with Variable Acceleration |
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The objective of this video is to consider rectilinear motion with variable acceleration followed by a comprehensive workout on an exemplary problem. In the introductory video of rectilinear motion, it has been described that sometime there are needed to solve problem of rectilinear motion having variable acceleration where the previously used formulas are no longer effective to use to solve the problems. In such cases, if anyone of the displacement, velocity and acceleration is given as function of time, then it is necessary to work out on remaining parameters.

Moving on, the video presents an example of rectilinear motion with variable acceleration where it asks to calculate the displacement, velocity and acceleration after 6 seconds from start up based on the given function of displacement with respect to time. The video shows step by step how to do the calculations and later presents another example of finding out the time required to reach a particle velocity at constant value based on the given function of displacement with reference to time.
Lecture 37
Curvilinear Motion |
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Curvilinear Motion |
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The objective of this video is to introduce and discuss curvilinear motions. Principally, a curvilinear motion describes a particle motions along a curved path. Generally, there are two cases included in curvilinear motion- projectile motion and circular motion. Moving on, the video gives overviews on position vector, velocity vector and acceleration vector which represent the conditions of a particle at any time. Next, the video briefly talks about the projectile motion which analysis has been carried out over the Cartesian coordinates breaking up horizontal and vertical components.

Later, the video talks about circular and other motions which analysis have been carried out in the polar coordinates system breaking down position, velocity and acceleration into r and θ. The other motions include the motions which paths are not totally circular. At end, the video shows how to do analysis of these motions in the polar coordinates system & finally refers to watch next video to get in touch with the workout lesson of projectile motions.
Lecture 38
Projectile Motion |
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Projectile Motion |
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The objectives of this video are to discuss about projectile motion & to complete a projectile motion problem. Basically, a projectile motion is a form of motion in which an object moves along a curved path under the action of gravity only. First part of the video discusses about the correct analysis of a projectile motion that is useful enough to find the time of flight, initial and final velocity, max height, displacement and launch angle. The key of the complete analysis is the resolution of forces into x & y components- constant velocity & no acceleration in x axis & constant gravity acceleration in y axis.

Moving on, the video presents a comprehensive exemplary problem of projectile motion where the time of flight and the range of the throw have been asked to determine based on the data given. The initial velocity, gravity acceleration, launch angle and necessary dimension are given. Using the data available, the video draws the projectile and resolves the velocity into y components to determine the flight time. The video considers the time taken by the stone to reach ground by imagining that the stone has been thrown directly upward in direction.
Lecture 39
Projectile Motion Formulae Derivations |
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Projectile Motion Formulae Derivations |
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The objective of this video is to derive expressions for Vx (velocity in x direction) and Vy (velocity in y direction) for a standard projectile. The video also wanted to derive expressions for displacement in x direction, Sx and displacement in y direction, Sy at any time. First of all, the video illustrates the given projectile motion of initial velocity, V & launch angle, θ. Next, the video derives the expression of horizontal velocity as a function of time and successively works out on the formula of calculating displacement in x direction using the obtained formula of velocity component in x direction.

Moving on, the video derives the expression of vertical velocity component as a function of time and takes the gravity acceleration in consideration. The integration of vertical gravity acceleration gives the expression of velocity component in y direction and further integration of the obtained formula yields the formula of displacement in y direction. Doing the derivation, the video briefly illustrates how to find the values of integral constant elaborating all the facts and figures in greater details.
Lecture 40
Circular Motion and Cylindrical Coordinates |
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Circular Motion and Cylindrical Coordinates |
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The objectives of this video are to discuss about circular motion followed by a brief introduction to cylindrical coordinates. First of all, the video gives introductory overview to motions in cylindrical coordinates successively pointing outs the diversity of Cartesian coordinates system and cylindrical coordinate system in defining position, velocity and acceleration with reference to time. Moving on, the video expresses a circular motion over the cylindrical coordinates system where two important components- angle and radius vectors are used to describe the motion of any body at any time.

Next, the video introduces with unit vectors used in cylindrical coordinate and successively derives the equations used to represent the expression of velocity at any time. Doing so, the video describes the radical components and transverse components with their expressions elaborating all the facts & figures in details. The video uses the differentiae concepts of calculus to express the formulae of circular motion over the cylindrical coordinate- understanding which is very important to work out on different problems related to the branch of kinematics of engineering mechanics.
Lecture 41
Polar Coordinates Example |
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Polar Coordinates Example |
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The objective of this video is to solve a problem based on the dynamics using the polar coordinates system. First of all, the video defines the given exemplary problem that has asked to calculate down the position, velocity and acceleration at time 8.0 seconds based on the given equations of angle and displacement with reference to time. Next, the video illustrates a schematic diagram of the example and vividly elaborates the problem over the polar coordinate system.

Moving on, the video shows first derivation and second derivation of the given expressions of angle and displacement with respect to time and successively finds the position, velocity and acceleration at time 8 seconds giving the available data inputs to the expanded equations. At end, the video gives a summary overview on the workout illustrated in this lesson. It is highly suggested to do practices to have better understanding on how to solve such types of problem in polar coordinates system.
Lecture 42
Newton's Laws and Kinetics |
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Newton's Laws and Kinetics |
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The objective of this video is to introduce kinetics and to discuss Newton’s law. Basically, kinetics is used to analyze unbalanced forces and the changes to motion that they produce. At first, the video shows how to analyze the problems related to kinetic using the Newton’s second law of motion that states accelerations have caused when an object is subjected to forces. Next, the video talks about how kinetics uses this statement to solve different types of problems involving motion i.e., velocity, time of flight, displacement etc elaborating all the facts in greater details.

Second part of the video shows a comprehensive example of kinetics where it has asked to calculate the weight of bags based on the data given. The problem involves of bags that are being loaded onto a plane via steep ramp. A hydraulic ram at the bottom of the ramp brings the bags to rest in 150mm. It exerts a constant force of 2500N over this length. Moving on, the video step by step shows how to assess different values required to calculate down the weight of bags. At end, the video refers to do lot of practices to have better understanding on the procedures to solve problems relating kinetics.
Lecture 43
Introduction to Work |
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Introduction to Work |
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The objectives of this video are to give an introduction to work through a comprehensive workout on calculating work done on an object. Fundamentally, a force is said to do work if, when acting on an object causes a displacement from the point of application to the direction of the force. Very first, the video presents an exemplary problem asking to calculate the work done on the given block due to applied force. The mass of the block is given as 5 Kg which lies on a slope of 30 degree. It has also given that a force of 200 N is exerted on the block which pushes it 1 m down the slope from rest.

Moving on, the video illustrates the given schematic diagram of the exemplary problem and shows how to draw the free body diagram demonstrating all the facts & figures in greater details. Next, the video explains what to consider and what to ignore to resolve force component to find out the work done due to the situation of applied forces. It is highly recommended to do practices to have better understanding on the content overviewed in this lesson.
Lecture 44
Work Example |
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Work Example |
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The objectives of this video tutorial are to discuss about work and energy followed by completing a problem involving the concepts of work & energy. First of all, the video discusses on work & energy introducing the formula of finding work done in terms of initial and final kinetic energy of a system. Basically, the energy of a system should remain constant; however, if it does change with reference to time then it means work is done to the system. Next, the video points out the facts that often final and initial energy is given and it has asked to find the work done or vice versa.

Moving on, the video presents an exemplary problem of racing car where the driver is speeding to a corner at 200 km/hr and applied the brakes. It has asked to calculate the work done by the brakes in slowing the car if the car has entered to the corner at 60 km/hr & the mass of the car is 1000 kg. The video, then, briefly shows how to assess required values to give input to the introduced formula to finally calculate down the value of work done by the brakes in slowing the car.
Lecture 45
Power and Efficiency |
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Power and Efficiency |
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The objectives of this video tutorial are to introduce the concept of power & efficiency followed by completing an exemplary problem using the knowledge of power concepts. At first, the video gives definition of power stating that power is the amount of work performed per unit of time which unit is watt as per international standard. Next, the video presents a problem asking to determine power outputs when it takes a person 8 minutes to push a car 300 m to a mechanic due to breakdown. The constant force exerted by the person during the pushing time is also given as 150 N.

Moving on, the video step by step shows how to assess required values to calculate down the work outputs of the person elaborating all the facts & figures in details. Later, the video briefly discusses on efficiency which is the ratio of useful output power to the input power & successively illustrates the formula of calculating mechanical efficiency. One important note should be keep in mind that efficiency is always less than one due to the influence of frictional forces which dissipates energy.
Lecture 46
Work and Energy Example |
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Work and Energy Example |
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This video presents a comprehensive example of solving a complex problem using work and energy concepts that have been overviewed in previous video. First part of the video jumps straight to the problem where a person named John hits a hockey puck across a frozen lake at an initial velocity of 10m/s. The video asks to calculate the distance the puck will travel before stopping if the coefficient of kinetic friction between the puck & ice is 0.05. The mass & gravitational acceleration of the puck have respectively given as 0.5 kg and 10 m/s2.

Moving on, the video explains the procedures to solve the problem starting with finding the friction force from the product of reaction force and coefficient of kinetic friction. Doing so, the video shows how to find the reaction force from the given value of mass and gravity acceleration. Next, the video uses the formula of work & energy to find the net work done by the puck to travel before stopping. Later, the video uses the theory of work done equals to the product of working force and distance to finally determine the distance the puck travels before stopping.
Lecture 47
Potential Energy, Kinetic Energy & Conservation |
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Potential Energy, Kinetic Energy & Conservation |
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The objectives of this video are to briefly discuss on potential energy, kinetic energy & conservation of mechanical energy. At first, the video looks at potential energy which expresses the capacity of an object to do work & subsequently gives introductory overview to different types of potential energy that include gravitational potential energy & elastic potential energy. Next, the video demonstrates conservative force and conservation of energy clearing the core concept that the mechanical energy is conserved if only conservative forces are applied to an object or a system.

The theory of conservation of energy states that the total energy before and after must be the same and the energy may be converted from kinetic to potential & vice versa. Moving on, the video shows a workout on example that asks to find out the speed of a stone travelling at 10 m above the ground using the conservation of mechanical energy. It has given that the stone is kicked off a building and fall toward the ground. At 40m from the ground it is travelling at 30m/s. The mass and gravitational acceleration of the stone have also given respectively as 0.1 kg and 9.8 m/s2.
Lecture 48
Conservation of Mechanical Energy Example |
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Conservation of Mechanical Energy Example |
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The objective of this video is to use the conservation of mechanical energy concept to solve motion problem. First part of the video jumps straight to the given exemplary problem where it has shown that a ball on a string is connected to a hook in a wall and asks to calculate the velocity of the ball & the tension in the string just before it hits the wall if the ball is dropped from its initial position. The mass of the ball and gravitational acceleration have given respectively as 3 kg and 10 m/s2. Aside, the video briefly illustrates the given schematic diagram of the problem.

Moving on, the video illustrates the procedure to solve the problem starting with the consideration of initial and final states & successively calculates down the velocities of respective states. Next, the video equates initial state to final state that yields the velocity of the ball just before it hits the wall. Later, the video demonstrates how to draw FBD to deduce tension in the string resolving the force components in vertical direction using the concept of equilibrium. Do more practices to have better understanding on the workout example overviewed in this tutorial.
Lecture 49
Introduction to Impulse and Momentum |
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Introduction to Impulse and Momentum |
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The objectives of this video are to introduce with the concept of impulse and momentum followed by a brief discussion on the relationship between impulse & momentum. First of all, the video clears the concept of impulse stating that impulse is the integral of forces with reference to time and then talks about the SI unit of impulse. Basically, in an impulse the relationship between force & time is inversely proportional. Next, the video discusses on the concept of momentum which is nothing but the product of mass and velocity & subsequently introduces the respective formula of momentum.

Moving on, the video correlates the momentum and impulse with each other successively deriving the equation of impulse in terms of the initial and final momentum. This yields the verity that linear impulse is equal to the change to an object linear momentum and this relationship allows to solve a problem related to impacts, forces and velocities. Overall, the video tries to give a brief inside to the impulse and momentum understanding which is must for the study of engineering mechanics.
Lecture 50
Impulse, Momentum, Velocity Example 1 |
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Impulse, Momentum, Velocity Example 1 |
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The objective of this video is to calculate impulse of an object to deduce final momentum & velocity. First part of the video jumps straight to the given exemplary problem where it has shown that a box of 50 kg mass is dragged across a smooth, frictionless surface by a 90 N force applied at 30 degree to the horizontal. The video asks to find out the final velocity of a wooden box if the force is applied for 10 seconds. The gravitational acceleration is given as 10 m/s2.

Moving on, the video explains the given schematic diagram of the problem and shows how to draw the free body diagram elaborating all the facts & figures in greater details. Next, the video resolves force components in y direction to find the value of normal force. Later, the video uses the obtained value as input to the integral formula of impulse & thus calculates the value of impulse. Finally, the video uses the impulse and momentum equation to find out the final velocity of the wooded box.
Lecture 51
Impulse, Momentum, Velocity Example 2 |
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Impulse, Momentum, Velocity Example 2 |
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The objectives of this video are to use momentum relationships to work out average impulse and to relate kinematic concepts to linear momentum. First of all, the video describes the given exemplary problem where it has asked to find the average impulse based on the given data- a baseball player hits a pitch travelling at a speed of 10 m/s. The ball velocity after the impact is measured to be 15 m/s at 45 degree to the horizontal. The mass of the pitch and gravity acceleration has respectively given as 250 g and 10 m/s2. The schematic diagram of the problem has also given.

Moving on, the video shows how to find out the initial & final momentum and subsequently relates the obtained values to calculate down the average impulse. Next, the video explains how to find out the average force if the ball is in contact with the bat for 0.2 seconds. Overall, the video tries to give a brief inside to the course of action required to work out on complex problems relating to impulse, momentum and velocity. Don’t hesitate to contact, if you don’t understand any part of the tutorial.
Lecture 52
Introduction to Impact |
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Introduction to Impact |
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The objectives of this video are to discuss impact and the mechanics behind the analysis of impacts followed by a brief introduction and discussion to the coefficient of restitution. First of all, the video defines the term impact as two or more bodies collide with each other over a very short period that creates large impulsive forces. Next, the video briefly talks about central impact which occurs when the direction of motion of the mass centers of the objects is in line with the impact.

Moving on, the video vividly illustrates the physical occurrence of impact assuming there is a time when the particles have the same velocity & afterward uses the coefficient of restitution to deduce velocity of particles after impact. Next, the video talks about the value of coefficient of restitution which value basically found in between 0 and 1 due to loss of mechanical energy in the collision in terms of heat, sound, light etc. Finally, the video refers to watch next video for a workout example.
Lecture 53
Central Impact Example |
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Central Impact Example |
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The key objective of this video is to consider a problem to work out on velocities of object following an impact. First part of the video jumps straight to the brief on the given exemplary problem where it has said a b’ute is speeding down a highway & rear-ends a hatch-back. If the b’ute was travelling at 130 km/hr and the hatch-back at 90 km/hr just before the collision, calculate the velocities after impact. Given the coefficient of restitution is 0.5. Moving on, the video vividly explains the physical occurrence highlighting all the facts and figures in greater details.

Next, the video step by step shows how to find initial and final momentum and subsequently shows how to relate them to finally calculate the value of velocities after impacts using the given value of coefficient of restitution. Overall, the video illustrates a comprehensive workout on central impacts example understanding which is very essential for the study of engineering mechanics. Always feel free to contact if you face any difficulties in understanding anything described over here.
Lecture 54
Shear Force Diagram Example
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Shear Force Diagram Example
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The objectives of this video are to develop shear force diagram from free body diagram and method of section. First of all, the video shows an exemplary beam asking to develop a shear force diagram for the beam. The beam is supported by a roller support at left end and a fixed support at 3 m away from the right end. It has also given that the beam is 7 m long and subjected to two point loads- one of 30 N at 2 m away from left end and another of 15 N just at the right end. Both the loads are acting downward in direction. Moving on, the video finds the support reactions and draws the FBD.

At second step, the video talks about to make cuts at critical points & the key doctrine for that is to make a cut every time a new force is applied to the beam. Using this concept, the video shows how to take cuts gradually one after another and to use the obtained value of internal forces to draw the shear force diagram. At end, the video briefly discusses the different point of shear force diagram as well as their nature explaining all the facts and figures in details.
Lecture 55
Bending Moment Diagram Example
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Bending Moment Diagram Example
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The objective of this video is to develop bending moment diagram from the free body diagram and method of section. At first, the video shows the same exemplary beam overviewed in previous video to develop shear force diagram & asks to develop the bending moment diagram. Doing so, the video follows the same doctrine of shear force diagram (SFD) that is to make a cut every time a new force is applied. The video also takes help from SFD for the drawing of bending moment diagram.

Moving on, the video step by step shows how to take cut gradually one after another & successively finds the momentum for every cut to use afterwards to draw the bending moment diagram (BMD). Later, the video briefly talks about the different points of BMD as well as their nature in details. At end, the video gives a quick summary on the content overviewed in this lesson. Feel free to contact with our expert tutors, if you face any difficulties in understanding any part of the tutorial.
Lecture 56
Shear and Bending Diagrams
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Shear and Bending Diagrams
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Continuing straight from the previous video, this video talks about the similarity and differences of shear force diagram (SFD) & bending moment diagram (BMD). First of all, the video points out the fact that wherever there is zero value at SFD, it always turns a max or min value of moment in BMD. Next, the video refers the concept of calculus stating the fact that SFD is basically the derivative of BMD which is also reflected in the slope of the diagram of BMD to SFD.

Moving on, the video briefly talks about how the slopes of BMD correlates with the SFD explaining the different points of the obtained SFD and BMD. Later, the video refers to watch next videos for complex workouts of SFD and BMD; understanding which is very important to master the study of engineering mechanics. Don’t hesitate to contact with our expert tutors if you don’t yet understand the model of shear force diagram and bending moment diagram.
Lecture 57
Beam Analysis Example Part 1
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Beam Analysis Example Part 1
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The objectives of this video are to draw the axial force diagram (AFD), shear force diagram (SFD) & bending moment diagram (BMD) for a given exemplary beam. First of all, the video shows the given exemplary beam having a roller support at 2 m away from left end and a fixed support at right end. The beam is subjected to three different loads- a point load of 10 N acting downward in direction at left end, a point load of 5 N acting parallel with the beam at left end & a uniformly distributed load of 2 N/m acting across 6 m length of the beam from the right end.

Moving on, the video draws the free body diagram and establishes equilibrium principle to find out the unknown values of reactions forces. Next, the video at 2nd step makes cuts at critical points to develop equations representing internal forces to use afterwards to draw the SFD & BMD. Later, the video shows the conversion of UDL to point load and subsequently develops moment equations to find moment at arbitrary points. Finally, the video refers to watch next video to continue the lesson.
Lecture 58
Beam Analysis Example Part 2
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Beam Analysis Example Part 2
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This video, continuation of the previous video, very first displays a summary table containing the obtained values of axial forces, shear forces & bending moment across different length of the beam. Next, the video examines the summary table and evaluates some unknown values of internal forces. Moving on, the video uses the derived equations & obtained values to draw the axial force diagram. Subsequently, the video also illustrates how to draw the SFD and BMD for the given beam.

In this tutorial, the video briefly talks about the lines & curves of SFD & BMD as well as the nature of different points of these diagrams for the workouts of the given exemplary beam. Overall, the video tries to give a brief overview on how to do complete internal analysis to find the unknown values of axial, shear & bending- understanding which are very important to master the study of engineering mechanics. Don’t hesitate to contact, if you don’t understand any part of this analysis.
Lecture 59
Introduction to Trusses
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Introduction to Trusses
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The objectives of this video are to discuss about what is truss and why do trusses are used. First of all, the video looks into truss clearing the concept that trusses are basically converts external loads into axial forces. This is done because materials such as steel & concrete can resist much larger axial forces than shear & bending moments. Therefore, trusses are designed eliminating shear & bending as much as possible so that they can take just axial loads. The video, then, quickly look at an analogy to describe the facts and figures of trusses in greater details.

The analogy involves the axial loads translation in a beam and column followed by a presentation of simple truss where loads act axially over the structure. Moving on, the video briefly discusses how the external loads transmit axially explaining the facts of shears and bending moments. Overall, the video tries to give an introduction to trusses understanding which is very important for the study of engineering mechanics. At end, the video refers to watch next video for workouts on trusses.
Lecture 60
Types of Trusses and Design Assumptions
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Types of Trusses and Design Assumptions
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The objectives of this video are to define a simple truss followed by a concise discussion on design assumptions of trusses. First of all, the video gives the definition of simple truss stating that a truss which begins with a triangular element and which can be expanded by adding two members & one joint as well as which meets relationship formula between members and joints is basically known as simple truss. The video introduces the stated relationship formula while giving the definition.

Moving next, the video shows different types of trusses structures and examines whether they are simple truss or not. Later, the video talks about the design assumptions needed to consider in doing the design of simple trusses e.g., all loads are applied at joint, self-weight of members are negligible & members are connected by smooth pins. Overall, the video tries to give a brief overview to simple trusses & design assumptions understanding which is must for the study of engineering mechanics.
Lecture 61
Method of Joints Truss Example
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Method of Joints Truss Example
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The objectives of this video are to introduce the method of joints & to resolve axial loads in a simple truss. First of all, the video displays the given exemplary problem of super simple truss having three members connected like a triangle and subjected to an axial force at top joint of the truss. The truss has been supported by a fixed support at left-bottom joint & a roller support at right-bottom joint. The video asks to resolve axial forces in two adjacent members by considering left-bottom joint.

Moving on, the video shows how to find support reactions and subsequently draws the FBD of the given truss. Next, the video draws the FBD of left-bottom joint by showing the assumed axial forces acting through the joint and successively applies equilibrium equation to solve the unknown values of internal forces. Thus, the video finds out the axial forces developed adjacent to two members of left-bottom joint explaining the facts of compression and tension developed in those members.
Lecture 62
Advanced Method of Joints Truss Example
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Advanced Method of Joints Truss Example
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The objective of this video is to use the method of joints to find axial forces in members of a simple truss. At first, the video presents the given exemplary simple truss asking to find the axial forces in two arbitrary members using the method of joints. The given truss has 7 members & 5 joints. Each of the members is 3 m long & the angles between members are also given. The truss has subjected to two axial loads- one of 10 N at left-top joint and another of 15 N at right-top joint.

Moving on, the video shows how to compute the distance of the joints subjected to axial loads from left-bottom end of the truss. Next, the video draws the FBD of the given truss, determines supports reactions forces using the principle of equilibrium. Once all the unknown values are computed out, the video considers left-bottom joint to find the axial forces developed adjacent to that joint. Later, the video uses the obtained values to determine the axial forces developed adjacent to left-top joint.
Lecture 63
Introduction to Method of Sections
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Introduction to Method of Sections
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The objectives of this tutorial are to discuss about the method of joints vs. the method of sections followed by a brief explanation on when to use method of sections. The method of sections allow to take a cut anywhere in a truss structure. It is not need to start at a support like the method of joints. Moving on, the video presents a super simple triangular truss and illustrates how to apply method of sections to compute the axial forces at members using the equilibrium principle.

Next, the video shows more complex trusses and subsequently talks about how to apply the method of sections to determine the axial forces at different members pointing out the key difference with the method of joints. Overall, the video tries to give a brief inside to the method of sections used to analysis different types of trusses in engineering mechanics. Finally, the video refers to watch next video to continue the lesson of method of sections theory.
Lecture 64
Method of Sections Theory
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Method of Sections Theory
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The objectives of this video are to introduce method of sections theory & steps involved in solving for axial forces. First of all, the video presents a simple truss asking to find out the axial forces in 3 arbitrary members. The truss has 7 members and 5 joints and subjected to two axial forces at top joints of the structure. The angles between the members are also given. Moving on, the video shows how to take cuts at arbitrary section and draws the FBD for both the left and right cuts showing the actions of internal forces that result because of the cuts.

Next, the video uses the principle of equilibrium to resolve forces components into x and y direction as well as takes anticlockwise moment at top-left joint to determine the unknown values of the axial forces; thereby finds out the axial forces in three arbitrary members of the given truss. Doing so, the video briefly demonstrates the actions of internal forces in the members and finally refers to watch next video for more complex workouts on method of sections theory.
Lecture 65
Method of Sections Truss Example
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Method of Sections Truss Example
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This video presents a comprehensive example of solving the axial forces in a truss using the method of sections. Very first, the video shows the given truss structure and asks to find the axial forces in 3 arbitrary members. All the necessary dimensions, angles and values of forces subjected to the truss have been given. The video, then, explains the shortcomings of using method of joints to solve this problem and subsequently talks about the advantages of using method of section to solve such type of problem. Moving on, the video determines the support reactions and draw the FBD of the truss.

At second step, the video shows how to take cut through critical members and examines the easier side. The video uses the section right to the cut followed by the drawing of FBD for the considered section showing the actions of axial forces over the FBD. Later, the video establishes equilibrium to find out the unknown values of axial forces and thereby determines the axial forces developed in 3 arbitrary members of the given exemplary truss. This is quite tricky, so do practices to have better understanding on how to use method of sections to solve such type of complex problems.
Lecture 66
Simple Frame Example
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Simple Frame Example
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The objective of this video is to analyze the members in a frame. At first, the video shows the given exemplary frame having two members. The member AB has supported by a fixed support at point A and connected with the member BC at point B. And the member BC has supported by roller support at point C & is subjected to a uniformly distributed load of 2 KN/m across the length of the member. The length of the member BC has given as 3 m and the distance between point A & B is given as 2 m.

Moving on, the video finds out the values of support reaction establishing the equilibrium equations and analyzing the frame as a whole. Subsequently the video draws the FBD of the frame explaining all the facts and figures in greater details. Next, the video splits frame into individual members and applies equilibrium equations at pin joint B. Later, the video applies equilibrium principle to get the unknown values of axial forces developed in member AB and BC through joint B.
Lecture 67
Advanced Frames Example
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Advanced Frames Example
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The objective of this video is to analyze a complex frame to deduce forces on individual members. First of all, the video vividly illustrates the given complex frame and asks to find reactions at A & F. The frame has four members- ABC, CD, DEF & BE which is subjected to two point loads- one of 60 N at point C parallel with the member CD and another of 100 N at midpoint of the member CD acting downward in direction. The frame has been supported by two hinge supports at point A & F. All the necessary dimensions are given representing the locations of different points in the frame.

Moving on, the video draws the free body diagram of the given complex frame and applies the law of equilibrium to get moments and force components equations. Doing so, the video has to examine different members of the frame and to relate the obtained equations to finally determine the values of reactions at point A & F. It is highly recommended to do on hand practices with this tutorial video to have better understanding on the solution of harder frames example overviewed in this lesson.
Lecture 68
Introduction to Friction
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Introduction to Friction
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The objectives of this video are to discuss static and kinetic friction followed by the consideration of friction-force plot. Basically, the term friction refers to the resistance offered to the movement of a body past another body with which it is in contact. It is a force that occurs between the surfaces of two bodies. A friction force resists motion, therefore acts in opposite direction to the movement of the bodies. Moving on, the video talks about static friction with an example which is the friction that resists all type of movement and in such case the FBD is found to be in equilibrium.

Next, the video introduces with the equation of static friction & shows the relation of reaction force with friction force explaining all the facts and figures in details. Later, the video discusses on kinetic friction which is the friction that slows movement of a body subjected to forces that cause motion. The video, then, introduces with the equation of kinetic friction & explains the physical occurrence of the body not in equilibrium condition. Finally, the video illustrates the friction-force plot showing the effects of static and kinetic friction over the plotted diagram.
Lecture 69
Static Friction Example
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Static Friction Example
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In this tutorial, the video presents a comprehensive example of static friction where it asks to find the coefficient of static friction based on the data given. First part of the video vividly illustrates the given problem where it has shown that a block of 10 Kg mass is rest upon on an inclined surface & the slope of the surface is given as 30 degree. The gravitational acceleration is also given as 10m/s2. Moving on, the video draws the FBD showing the actions of gravity force, normal force and friction force. In addition, the required angles of these forces are also shown over the FBD.

Next, the video applies the equilibrium principle and resolves force components into x & y direction and successively computes out the unknown values from the established equations. Later, the video introduces with the equation of static friction overviewed in the previous video & finally determine the coefficient of static friction putting the required values into the equation. Do on hand practices with the tutorial to have better understanding on the workout completed in this lesson.
Lecture 70
Tipping vs Slipping Friction
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Tipping vs Slipping Friction
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The objective of this video is to use the knowledge of friction to deduce behavior of a loaded ladder against a wall. At first, the video describes the problem asking to find the minimum force required to move the ladder (tip or slip) assuming the wall is smooth means no friction in the wall. The mass of the ladder is given as 1 kg and the coefficient of static friction is given as 0.8 for the floor. All the required dimensions are given for the workout of the problem. Moving on, the video draws the FBD of the ladder showing the actions of all forces acting over the ladder.

Next, the video points out the fact to take in consideration to establish equilibrium conditions from the drawn FBD- one case either slipping or tipping must has to presume first. The video considers tipping condition & successively calculates the minimum force required to move the ladder. Later, the video shows how to cross check the obtained result against the given value of static friction and finally discusses what will happen if the ladder would slip instead of being tipped.
Lecture 71
Introduction to Hyrdostatic Forces | Hyd
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Introduction to Hyrdostatic Forces | Hyd
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The objectives of this video are to introduce the hydrostatic forces concept and to solve hydrostatic force on a submerged plate. At first, the video briefly talks about hydrostatic & hydrodynamics; two branches of fluid mechanics. In the study of hydrostatic, the effect of shear force is not considered. The only force that is considered is the force developed by the pressure of fluids. With that in mind, the video examines a pressure vessel that experiences hydrostatic forces thorough out its walls. The video, then, introduces with the equation of pressure force explaining all the facts in details.

Moving on, the video presents a workout on an exemplary problem where the resultant force on the metal plate at the bottom of the given tank has been asked to calculate down. The video step by step shows how to find the pressure developed in the bottom plate from the unit weight & height of the fluid and subsequently determines the resultant force acting on the metal plate at the bottom of the tank just by multiplying the obtained pressure with the cross-sectional area.
Lecture 72
Hydrostatic Forces Example | Hyd
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Hydrostatic Forces Example | Hyd
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The objective of this video is to work out the equivalent hydrostatic force on a vertical tank wall. At first, the video describes the exemplary problem of a tank containing fluids in a 4 m by 4 m by 3.5 m volume space and asks to calculate the equivalent hydrostatic force developed on the shaded wall of the given tank as a result of pressure force developed by the fluid inside the tank. Moving on, the video calculates the pressure at top and bottom surface of the tank explaining the fact that how the pressure forces work at different depth of the tank.

Next, the video shows the conversion of bottom pressure force into a distributed line load & draws the triangular distributed loads FBD showing how resultant forces act on the shaded wall gradually increasing in value to the depth of the tank. Later, the video converts the triangular load into a point load and shows how it acts through the center of the shaded wall. Overall, the video tries to give a brief inside to the calculation procedures of hydrostatic forces in liquid vessels.
Lecture 73
Centroids
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Centroids
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The objectives of this video are to introduce the centroid concept followed by a brief discussion on centroid formulas used in the study of engineering mechanics. In engineering, the term ‘centroid’ is used to relate geometric properties to internal forces and stresses in members. Basically, a centroid refers to the center of the mass of an object. Moving on, the video briefly talks about the centroid of simple object that basically refers to the center of average position of areas. Next, the video explains the centroid of complex bodies introducing the necessary formulas to find the coordinates position.

According to the formulas, the x coordinate of the centroid of a complex object with respect to an arbitrary axis is equal to the ratio of ‘the summation of the product of individual section area times the individual section x coordinate’ and the summation of all areas. And similarly, the location of the y coordinate is equal to the ratio of ‘the summation of the product of individual section area times the individual section y coordinate’ and the summation of all areas. Overall, the video tries to give a quick introduction to the centroid concept of simple to complex objects.
Lecture 74
Finding Centroids by Integration
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Finding Centroids by Integration
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The objective of this video is to find the centroid of a triangle using integration. It has given that the triangle is 4 m in length and 2 m is in height. First of all, the video clears the key step of finding the coordinates of x bar and y bar by deriving equations for f(x) and f(y). Next, the video considers the bottom-left point of the triangle as the reference coordinates and subsequently derives the formula of f(x) in terms of the distance variable x. The video, then, introduces with the integral equation of x bar coordinate and successively expands the equation considering a slice of the triangle.

The equation yields the value of x coordinate and later the video derives the integral equation for y coordinates. At end, the video generalizes the result of triangle centroid concluding the fact that the x coordinate of a centroid of a triangle is located two-third away of its length from the lower corner & the y coordinate of a centroid of a triangle is located one-third away from x axis. Overall, the video tries to give a brief overview on finding centroid coordinates by integration.
Lecture 75
Centroids of Composite Shapes Example
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Centroids of Composite Shapes Example
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The objective of this video is to consider centroid of composite shapes. First of all, the video talks about the theory of splitting complex area into a number of simple shapes to determine the centroid location in easier way. The video, then, presents an L-beam shape & asks to find out the location of centroid of the given beam. Using the theory of splitting complex area, the video shows how to split the given L-beam shape into two simple rectangular shape and subsequently find out the location of centroid using the coordinates formulas overviewed in previous video.

Moving on, the video presents another comprehensive workout of composite shape example where the video asks to find the centroid position of a given exemplary I-beam. The video splits the I-beam into three simple rectangular shapes and subsequently determine the location of centroid similar to before. Overall, the video tries to give a brief overview on how to calculate the position of centroid of different composite shapes in solving many problems of engineering mechanics.
Lecture 76
Moment of Inertia
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Moment of Inertia
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The objectives of this video are to introduce moment of inertia concept & to look at standard shape. The moment of inertia, also known as the second moment of area, is basically a geometric property related to some object resistance to rotational movements. In civil engineering, it is often combined this geometric property with material properties to determine physical quantities such as stiffness. Moving on, the video briefly talks about the unit of moment of inertia.

Next, the video introduces with Ix and Iy- Ix is related to the rotations relative to x axis & Iy is related to the rotations relative to y axis. Later, the video introduces with the standard moment formulas of rectangular and circular shape objects explaining all the facts and figures in greater details. Overall, the video tries to clear the concept on moment of inertia- understanding which is very important in doing many workouts of engineering mechanics.
Lecture 77
Moment of Inertia Standard Shapes
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Moment of Inertia Standard Shapes
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The objective of this video is to work out on moment of inertia of simple shape. First of all, the video shows a rectangular shape of 400 mm by 200 mm & asks to find the moment of inertia with respect to x and y axis. Next, the video briefly illustrates the given rectangular structure discussing the facts of finding moments of inertia with respect to reference axes and consequently introduces with the required formula to calculate Ix & Iy- Moving on, the video shows how to determine the moment of inertia with respect to x and y axis from the dimension data available.

Second part of the video shows another workout of finding moment of inertias for circular shape. In this example, the diameter of the circle has been given as 50 mm. The video moving next introduces with the equations required to calculate Ix & Iy that have been overviewed in previous video & finds out the values of moment of inertias successively. Later, the video refers to watch next videos for composite bodies’ workouts and to get in touch with parallel axis theorem.
Lecture 78
Parallel Axis Theorem Part 1
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Parallel Axis Theorem Part 1
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The objectives of this video are to introduce the parallel axis theorem and to find moment of inertia with respect to axis x for a composite section. At first, the video talks about the shortcomings of the formula used to find moment of inertias for simple shape. These formulas are only valid when the centroid of the shape is at the centroid of the cross-section; however, this is not true for composite sections. This demands necessity of parallel axis theorem for the calculation of moment of inertias for composite bodies. Moving on, the video introduces with the formula of parallel axis theorem.

The formula shows that the moment of inertias of a composite body with reference to an arbitrary axis is equal to the sum of moment of inertias for the complex shape plus the product of ‘area of the shape times the square of the distance between the section centroid and shape centroid’. The video moving forward shows a workout of finding the moment of inertia with respect to axis x for a given exemplary section of T-beam. At end, the video refers to watch next video to continue the lesson.
Lecture 79
Parallel Axis Theorem Part 2
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Parallel Axis Theorem Part 2
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This video, continuing straight from the previous video, very first clears its objective to find out the moment of inertias with respect to x axis. First part of the video presents the parallel axis theorem formula once again and focuses on finding the moment of inertia with respect to individual section centroid axis as input to the equation of finding moment of inertia with respect to shape centroidal axis. The video splits the given composite bodies into two shapes and once after another shows how to find the moment of inertia for them explaining all the facts in greater details.

Moving next, the video sums the moment of inertias for two split shapes and finally determines the moment of inertia for the given exemplary composite body. Later, the video briefly talks about few notes regarding the obtained values of moment of inertia successively elaborating the bending case of the composite body. It is highly suggested to do on hand practice to have better understanding on the workout done in this tutorial. Make a contact, if you don’t understand any part of this tutorial.
Lecture 80
Average Normal Stress
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Average Normal Stress
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The objectives of this video are to discuss about average normal stress and to work out on average stress in members. At first, the video introduces with the equation of average normal stress clearing the concept that the average normal stresses are equal to the ratio of axial resultant forces & area of cross-section. Next, the video shows an exemplary column structure asking to draw the axial force diagram (AFD) & to work out on the average normal stress developed in the column. The diameter of the cross-section is given as 500 mm and the column is subjected to an axial load of 10 KN.

Moving on, the video finds the support reaction of the column using the principle of equilibrium and subsequently calculates the axial force by making a critical cut at arbitrary section. Next, the video uses the obtained values to plot an axial force diagram & successively calculates the average normal stress giving input the required values into the respective equation of stress. Overall, the video tries give a brief inside to average normal stress and how to do the calculations of average normal stress.
Lecture 81
Average Stress Example
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Average Stress Example
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The objectives of this video are to complete a difficult axial force diagram & to calculate the average normal stress in an exemplary member. At first, the video illustrates the given exemplary structure having a cross-sectional diameter of 100 mm from point A to B and a cross-sectional diameter of 75 mm from point B to C. The structure is subjected to four different loads that have been overviewed with the progress of the tutorial and it has asked to find the average normal stress throughout the circular section of the given exemplary structural member.

Moving on, the video makes cuts one after another every time a new force is applied to the member and resolves the internal forces using the principle of equilibrium. Later, the video uses the gotten values to draw the axial force diagram (AFD) and successively calculates down the average normal stresses through the 2 different sections of the given member. On hand practices should have done with this tutorial to have better understanding on the workout overviewed in this lesson.
Lecture 82
Shear Stress Example
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Shear Stress Example
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The objectives of this video are to discuss about shear stress & to work out on finding shear stress in a rivet joint. First of all, the video clears the concept of shear stresses which is the result of forces applied along the same plane as the cross-section. It can also be described as a rubbing force. Next, the video illustrates the shear stress vividly representing a schematic of a structural member that is subjected to a couple. Moving on, the video presents an example of finding shear stress developed in a rivet that connects two structural members.

Both the members are subjected to point loads of 5 KN acting at free end parallel with the members. Moving next, the video illustrates how shear stress is developed at rivet joint due to the actions of applied axial forces and successively draws the free body diagram for it. Later, the video gives input to the respective equation and determines the value of shear stress. It is highly recommended to do on hand practices to have a good grasp on the content described in this lesson.
Lecture 83
Strain
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Strain
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The objectives of this video are to discuss engineering strain followed by a brief workout on finding strain in a given exemplary column. Basically, the engineering strain can be defined as the ratio of change in length to the initial length of a member. Moving on, the video talks about the expression & notation used to represent strain. The video, then, elaborates the fact that why engineering strain is a dimensionless quantity. Moving on, the video presents an example where the strain & length of a column have been given and it has asked to find the length contraction when subjected to axial load.

The video uses the introduced formula of strain & successively calculates down the value of length contraction when the column is subjected to axial load using the data available. The video, in its last part, briefly talks about the necessity of strain calculations in engineering designs to best reflects in meeting the physical demands. Finally, the video refers to watch next video tutorials to get in touch with the Hooke’s law and ways of relating stress and strain.