Mechanics of Solids & Structural Mechanics

Video Lectures

Displaying all 67 video lectures.
Lecture 1
Mohr's Circle Example
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Mohr's Circle Example
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This video presents another comprehensive example of Mohr’s circle for a given element asking to calculate principle stress 1 & 2, maximum shear stress, orientation of planes experiencing principle stresses and to draw the Mohr’s circle. First part of the video illustrates the given element with the values of normal and shear stress and subsequently shows a Mohr’s circle. Next, the video identify the coordinate position of the center of Mohr’s circle & the circle radius using the earlier described formula and successively points out over the drawn Mohr’s circle.

Moving on, the video calculates the maximum shear stress at top of the circle and thereafter shows the calculation of principle stress using the Mohr’s circle. Later, the video briefly shows the rotation of planes that experiences principle stresses and subsequently talks about the degree of rotation in reality of the planes with the direction that experiences principle stresses. Overall, the video tries to give a brief inside to solve these kinds of problems using the advantages of Mohr’s circle.
Lecture 2
Von Mises & Trescas Yield Criterion Example
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Von Mises & Trescas Yield Criterion Example
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This video presents a comprehensive example of Von Mises’ and Tresca’s yield criterion under the combined stresses. Given the same schematic of exemplary element with normal and shear stresses described before, the video for a given value of yield stress asks to check if the given element under combined action yielded according to Tresca or Von Mises’ yield criterion theory. The video recalls the earlier described formulas of Tresca’s and Von Mises’ yield criterion for doing the calculation.

Moving on, the video first shows the calculation of yielding under Tresca’s theory and finds that the element is yielded under combined action. Next, the video shows the yielding calculation under Von Mises theory & finds that the element does not yielded under the combined action. Later, the video clears the fact that why this difference has happened by doing a brief comparison of Von Mises’ and Tresca’s yield criterion theory explaining all the facts and figures in greater details.
Lecture 3
Von Mises Yield Criterion
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Von Mises Yield Criterion
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The objective of this video is to explain Von Mises yield criterion followed by a brief comparison of Von Mises’ and Tresca’s yield criterion. Von Mises’ criterion is based on the energy associated with the distorted shape of an element. The yielding is assumed to occur when the shearing distortion of an element under a combined stress state is equal to shearing distortion energy of uniaxial tension yield. Moving on, the video expresses the Von Mises yielding criterion in terms of three dimensions using the principle stresses explaining all the facts in details.

The video, then, represents Von Mises yield criterion as geometric interpretation of ellipse showing that how yielding takes place outside the ellipse and no yielding within the ellipse. Later, the video does a brief comparison of Von Mises’ and Tresca’s yield criterion over a graphical representation & concludes the fact both yield surfaces do not differ significantly. In addition, geometrically both the criterion gives better combined yield stress calculation.
Lecture 4
Tresca's Yield Criterion
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Tresca's Yield Criterion
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This video discusses on Tresca’s yield criterion by going back to the original stress-strain curve for a metal such as steel experiencing uniaxial tension. At first, the video presents a stress-strain curve showing the elastic, plastic & steel hardening regions followed by discussion on yield point over the curve. A material experiencing both normal and shear stresses will require a stress combination to yield and the combination that produces yielding is known as yielding criterion.

Next, the video clears the base of Tresca’s yield criterion stating that Tresca’s criterion is based on the maximum shear stress reaching a critical level followed by graphical presentation of an element which is yielding under uniaxial tension action over it. Moving on, the video does a comprehensive analysis over the element and subsequently demonstrates the Tresca’s yield criterion explaining all the necessary facts & figures. Overall, the video tried to give a brief inside of Tresca’s yield criterion.
Lecture 5
Mohr's Circle Summary Example
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Mohr's Circle Summary Example
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This video shows a comprehensive workout on an example of Mohr’s circle representing that how much rotation required for an element to reach at principle stresses conditions. Very first, the video gives an overview on the schematic diagram of a rectangular block section that experiences normal & shear stresses of certain values. Subsequently asks to draw the Mohr’s circle in order to calculate the principle stresses, maximum shear stress & the rotation of the planes experiencing the principle stresses and maximum shear stress.

Moving next, the video determines the coordinates of Mohr’s circle with radius as well as finds out the angle of rotation using the formula derived earlier & successively uses all the obtained values to plot the Mohr’s circle. Later, the video does geometric analysis using the principle of Mohr’s circle & combined stresses & thereafter evaluates the points on the Mohr’s circle that experiences maximum shear stresses and principle stresses. Hopefully, good understanding of this example will clear the concept of Mohr’s circle mastering which is very important for the study of structural mechanics.
Lecture 6
Combined, Normal and Shear Stress Example
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Combined, Normal and Shear Stress Example
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This video represents an example of combined shear stresses asking to calculate the stresses on an element rotated by 10 degree anticlockwise followed by the drawing of Mohr’s circle to shows this transformation. At first, the video illustrates the given schematic rectangular section of an element that experiences shear stress of 40 MPa and normal stresses of 100 MPa and 80 MPa respectively at horizontal & vertical direction. Next, the video clears the fact that as the element rotated 10 degree anticlockwise in reality so it will have an effect of 20 degree rotation clockwise in Mohr’s circle.

Moving on, the video recalls the formulas of finding principle normal & shear stresses that has been derived in previous video 9.3 and subsequently determines the effects of combined stresses on the given element. Later, the video finds out the radius and coordinates of the Mohr’s circle and uses all the obtained values to draw the Mohr’s circle. Finally, the video elaborates the drawn Mohr’s circle explaining all the necessary facts and figures in greater details.
Lecture 7
Mohr's Circle Equations & Theory
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Mohr's Circle Equations & Theory
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This video explains Mohr’s circle of stresses followed by the construction of Mohr’s circle. Mohr’s circle is a geometrical interpretation that can be constructed to explain an element with principle stresses 1 and 2. First part of the video recalls the derived formula of principle stresses and asks to consider an element with the normal stresses & shear stresses rotated anticlockwise at an angle, . Next, the video clears the concept of sign convention stating that shear stresses are positive if they produce anticlockwise moment and vice versa.

Moving on, the video briefly shows the construction of Mohr’s circle setting horizontal axis always experiences normal stresses and vertical axis always experiences shear stresses followed by work out on the formula of finding the coordinates of the circle with radius. Later, the video draws Mohr’s circle and explains the different properties of Mohr’s circle. It is very important to understand the different properties of Mohr’s circle to use it effectively in the calculations of combined stresses.
Lecture 8
Combined Shear Stress & Mohr's Circle
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Combined Shear Stress & Mohr's Circle
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This video continues the analysis of combined stresses followed by a brief introduction to Mohr’s circle. At first, the video talks about the stresses on an element aligned along the member axis and introduce the formula of determining the direction of maximum stress on a plane. Next, the video shows the two solutions of the introduced formula. In fact, these two solutions for determining the direction of stress correspond to alternating maxima & minima on the adjacent faces of an element.

Moving on, the video does brief analysis of combined shear stresses over the orientation diagram of an element that experiences shear stresses and talks about several important points of the diagram to note. Later, the video gives introduction to Mohr’s circle which is a geometric interpretation of the principle stresses that an element experiences along different directions followed by a graphical illustration of the Mohr’s circle for the combined shear stresses formulas derived in this lesson.
Lecture 9
Analysis of Combined Stress
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Analysis of Combined Stress
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The objective of this video is to give an overview on combined stresses followed by the analysis of combined stresses. First part of the video asks to consider an element loaded in pure tension having no horizontal strains, only the vertical strains. Next, the video asks to consider a rhomboid element oriented at 45 degree angles, the corners which coincide with the midpoint of the element already being considered. Moving on, the video briefly talks about the shear strain & deformation followed by the discussion on combined stresses in greater details.

Second part of the video shows an analysis of combined stresses with respect to a rectangular block rotating in anticlockwise direction. Moving next, the video shows the stresses distributions over the block & derives the equation of combining stresses using the principle of equilibrium under certain condition. Later, the video briefly talks about combined stresses at different conditions and thereby does a comprehensive analysis on combined stresses.
Lecture 10
Closed Pipe, Hoop and Longitdunal Stress Pressure Vessel Example
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Closed Pipe, Hoop and Longitdunal Stress Pressure Vessel Example
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This video presents another comprehensive example of pressure vessels. At first, the video explains the schematic of a closed pipe having diameter of 500 mm and thickness of 5 mm and subsequently asks to determine the longitudinal & hoop stress if the given sealed pipe is subjected to an internal pressure, 30 MPa. The video also asks to draw a diagram that shows the stresses a typical element experiences. Next, the video recalls the formulas of longitudinal and hoop stresses.

Moving on, the video shows the calculations procedures of longitudinal and hoop stresses under the given conditions and draws a schematic showing the stresses distribution at horizontal and vertical direction. Later, the video gives a summary overview on the example described in this lesson. It is highly suggested to do on hand practice with this video to have good understanding on the workout done over here. Don’t hesitate to contact us, if you feel complexity to understand any part of it.
Lecture 11
Open Pipe, Hoop and Longitdunal Stress Pressure Vessel Example
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Open Pipe, Hoop and Longitdunal Stress Pressure Vessel Example
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This video presents the behavior of open pipe and pressure vessel followed by a detailed workout on a comprehensive example. At first, the video illustrates the given exemplary schematic of a large pipe having diameter of 2 m subjected to a design pressure of 600 KPa & subsequently asks to find out the minimum pipe thickness that will ensure the hoop stresses are no more than 50 MPa. Next, the video recalls the formula of finding the value of hoop stresses derived in previous video 8.1.

The formula shows that the hoop stress is equal to ratio of ‘the product of pressure and radius’ and thickness. Moving next, the video puts the known value of pressure, hoop stress and radius into the formula & this yields a thickness of 12 mm. Later, the video talks about the importance on the unit conversion that is must to following while doing any example to avoid any unwanted results.
Lecture 12
Longitudinal Stresss in Pressure Vessels
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Longitudinal Stresss in Pressure Vessels
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This video demonstrates longitudinal stresses considering a closed pipe with thickness t and radius R. At first, the video graphically explains the consideration of closed pipe experiencing longitudinal stresses and clears the fact that since the pipe is closed at both ends, so unlike the previous one this one will experiences longitudinal stresses at vertical direction. There are still hoop stresses present but their nature is altered due to flat plates sealing off the cylinder. This is beyond the scope of the course and hoop stress still remains as the ratio of ‘the product of pressure & radius’ and thickness.

Moving on, this video presents a free body diagram of cut of the cylinder showing the vertical forces only and subsequently derives the governing equation of longitudinal stress using the principle of vertical equilibrium. The final form of the equation shows that longitudinal stress of a closed pipe is equal to the ratio of ‘the product of pressure and radius’ and two times of thickness.
Lecture 13
Hoop & Cylinder Stress in Pressure Vessels
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Hoop & Cylinder Stress in Pressure Vessels
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The objective of this video is to give an introductory overview on hoop stress- a normal stress on tangential direction. First part of the video asks to consider an open pipe with radius R & thickness t and subsequently illustrates the free body diagram clearing the fact that a pipe open at its end only experiences hoop stresses. Next, the video briefly talks about the assumptions made to describe the theory of hoop stresses. Hoop stresses essentially are caused by pressure acting equally outwards.

This result in tension around the pipe circumference that will make it expands outwards across its radius. Moving on, the video talks about the notation used for hoop stress & talks about the fact that in a thin pipe it can be assumed the hoop stresses are constant throughout the thickness. Later, the video illustrates a free body diagram of half a pipe filled up with a fluid subjected to hoop stresses & subsequently derives the equation of hoop stress considering vertical force equilibrium.
Lecture 14
Cantilever Beam with Point Load at Free Edge Deflection Example
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Cantilever Beam with Point Load at Free Edge Deflection Example
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This video represents a harder deflection example of a cantilever beam with point load at end using integration of bending moments. Very first, the video shows the schematic of given cantilever beam pointing out the load condition followed by the illustration of bending moment diagram (BMD). The moment equation for the cantilever beam undergoing point load at end is also given. Next, the video recalls the governing differential equation of deflection containing the component bending moment.

Moving on, the video does integration two times over the differential equation & gets the deflection equation contacting two integration constants. Later, the video applies boundary conditions to find out the values of integration constants and by substituting them the ultimate equation of deflection has been derived down. Finally, the video shows the derivation procedure of maximum deflection explaining all the necessary facts and figures in greater details. Never hesitate to contact us, if you don’t understand any part of the lessons described over here.
Lecture 15
Cantilever Beam with Moment at Free Edge Deflection Example
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Cantilever Beam with Moment at Free Edge Deflection Example
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This video presents an example of finding the equation of deflection of cantilever beam subjected to a moment on free edge. Unlike previous examples, the load function or the moment equation is not given in this example. Therefore, the moment’ equation has to be derived in this lesson establishing the principle of equilibrium. First part of the video shows the derivation of moment equation taking cuts and drawing the bending moment diagram over the schematic of given cantilever beam using the equilibrium method explaining all the necessary facts and figures in details.

Next, the video recalls the differential equation of deflection in terms of moment and substitutes the parameter moment with the obtained equation of bending moment. Moving on, the video does the integrations 2 times to get the ultimate equation of deflection & successively shows the procedures to get the values of integration constants applying different boundary conditions as well as derives the formula of maximum deflection for the given structural section of cantilever beam.
Lecture 16
Cantilever Beam Deflection Example
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Cantilever Beam Deflection Example
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This video shows a workout on a comprehensive example of deflection of a cantilever subjected to uniformly distributed loading. First part of the video shows the schematic diagram of the cantilever beam given and successively demonstrates the boundary conditions. The equation of UDL in terms of load function, w with respect to distance is also given. Next, the video introduces the differential equation of deflection in terms of given load function which required integration four times to give the ultimate equation of deflection.

Moving on, the video briefly shows the integrations procedures that results four unknown constant parameters in the ultimate equation of deflection. Later, the video shows how to get rid of constant parameters applying the given boundary conditions and subsequently gets the ultimate equation of deflection in terms of known parameters distance, flexural rigidity & given loading function. Doing so, the video explains bending moment diagram and finally derives the equation of max deflection.
Lecture 17
Simply Supported Beam Deflection Example
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Simply Supported Beam Deflection Example
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This video presents an example of simple deflection using integration of the given bending moment equation. At first, the video shows the schematic of given exemplary structural section having a load function, w followed by the equation of moments. Next, the video recalls the differential equation of deflection in terms of moment and flexural rigidity introduced in previous video and substitutes the given moment equation into recalled differential equation of deflection. The video, then, integrates the differential equation 2 times & gets the equation of deflection containing constants constituents.

Moving on, the video shows how to find the values of integrating constant using different boundary conditions and subsequently derives the ultimate equation of finding simple deflection in terms of known parameters of distance, load function & flexural rigidity. Later, the video derives the formula of maximum deflection for a simply supported beam describing all the facts and figures. Finally, the video gives a quick summary overview of the example worked out in this lesson.
Lecture 18
Beam Delfection Equations
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Beam Delfection Equations
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This video demonstrates the relationship between moments, m & a loading function, w in deflection applications followed by the explanation of deflection properties under boundary conditions. Very first the video clears the fact that capital ‘V’ represents shear force & small ‘v’ represents deflection. Next, the video recalls the differential formula of loading & moment; consequently shows in greater details that the deflection could be found either by integrating twice the bending moment equation or integrating thrice the shear force equation or integrating four times the loading equation.

Moving on, the video briefly illustrates the schematic diagram of deflection over the schematic of load, shear force & bending moment diagram. Later, the video talks about the deflection properties at different condition of supports e.g., simple, fixed, free end etc. used in the structure & derives the formulas of deflection incorporating different boundary conditions explaining all the facts & figures in details. Finally, the video refers to watch next videos for the workout examples of deflection.
Lecture 19
Beam Deflection Theory
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Beam Deflection Theory
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This video gives an introductory overview on deflection explaining necessary terminology followed by the derivation of differential equation of deflection. The term ‘deflection’ refers to displacement of a structure under loading. The two important terminologies used while dealing with the problem are stiffness and second moment of area. Stiffness refers to the deformation response under loading condition & is usually associates with the elastic modulus. And second moment of area refers to the distribution of area away from the neutral axis.

The combination of stiffness and second moment of area is widely known as flexural rigidity which is an extremely important parameter in the concept of deflection. In deflection, it is very important to consider both the serviceability load and ultimate load acts on a structure. Moving on, the video briefly talks about the assumptions made for the derivation of equation of deflection; subsequently shows the derivation of differential equation of deflection explaining all the facts and figures.
Lecture 20
Torque & Torsion Summary Example
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Torque & Torsion Summary Example
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This video shows a workout on another comprehensive example of torsion of a hollow cylindrical section having three different thicknesses varying from top to bottom section- understanding which will give a depth of knowledge on torsion of circular sections. At first, the video illustrates the given data e.g., outer diameter of the cylinder, different twisting moments acting over the cylinder, value of modulus of torsion & different thicknesses of the cylindrical sections. The example asks to draw the TMD and calculates the value of max shear stress, rate of twist followed by the drawing showing the rotation with respect to the fixed base at any height.

Moving on, the video draws the torsional moment diagram (TMD) illustrating the facts and figures in details; subsequently shows how to find the max shear stress assessing the values of stress of top & bottom section and the value of torsion constant. Later, the video calculates the rate of twist and uses all the obtained values to plot a diagram showing the rotation with respect to the fixed base at any height over the schematic of the given hollow cylindrical section.
Lecture 21
Torsion on a Non-Circular Shaft Example
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Torsion on a Non-Circular Shaft Example
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The objectives of this video are to explain the theory of torsion of non-circular section followed by a workout on a comprehensive example. The torsion of non-circular section follows the same formula described in previous video 6.2 for circular section; however, the challenge for non-circular section is to find out the value of torsion constant. The video, very first, shows an approximated formula of torsion constant & asks to bear in mind that the approximation is better for slender section but not applicable for stocky rectangular elements.

In addition, the equations for determining shear strain and stress described previously for circular sections are not applicable for non-circular sections. Moving on, the video presents an example of I-section showing the entire calculation of finding the torsion constant value using the approximated formula introduced in the beginning of this lesson. Overall, the video tries to give an introductory overview on torsion of non-circular cross-section in details- understanding which is mandatory to master the study of structural mechanics.
Lecture 22
Uniform Torque on a Cylindrical Shaft Example
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Uniform Torque on a Cylindrical Shaft Example
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This video presents an example of torsion of solid shaft undergoing a uniform twisting moment. At first, the video illustrates the given exemplary cylindrical section having diameter of 50 mm, height of 500 mm and an anticlockwise moment of 20 KNm. The shear modulus of the cylindrical shaft has also given. The example asks to calculate the torsion constant, rate of twist, maximum shear stress, and number of rotation of the top cylinder with respect to base as well as asks to draw the twisting moment diagram.

Moving on, the video shows how to determine the torsion constant from the given data and formula described in previous video followed by the drawing of twisting moment diagram. Later, the video demonstrates the entire procedures to calculate rate of twist and maximum shear stress in details. The video finds that maximum shear stress occurs at the maximum radius away from the center of the circular cross section. Finally, the video shows how to find out the number of rotation of the top cylinder with respect to base.
Lecture 23
Torsion and Torque Equations
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Torsion and Torque Equations
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This video gives an overview on torque equations explaining the actions of torques followed by the derivations of torque equations used in the design of structural mechanics. Torque is defined as the turning effect of force about the axis of rotation; in short it is a moment of force applied in an object. So, how could be determined the torque acting over an object? It is quite simple- just by integrating the stress with respect to the area subjected to the twisting forces. The video derives the governing torque equations at the beginning of the lesson.

Doing so, the video introduces with the terminology shear modulus and torsional constant followed by an explanation in greater in details. Moving next, the video briefly demonstrates the equations of finding the required constituents of the governing equations of torque e.g., torsion constant, twist rate, shear strain, shear stress, maximum shear stress, total twist etc. Later, the video graphically talks about the torsion constant for a hollow section & successively derives the required formula.
Lecture 24
Torsion Shear Strain and Stress Distributions
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Torsion Shear Strain and Stress Distributions
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The objectives of this video are to explain concepts of torsion of circular & square sections followed by the derivation of governing equation of shear strain and stress distributions. Very first, the video illustrates the term torsion explaining the concept that torsion is the twisting of an object when it is subjected to an applied torque- essentially a twisting force. Next, the video demonstrates torsion of circular and square sections explaining the nature of plane under the action of twisting forces. The video, then, talks about the assumptions usually made for uniform torsion of circular sections.

Moving on, the video graphically illustrates a solid cylindrical section under uniform torsion actions and subsequently derives the governing equation for it. Later, the video shows the stress and strain distributions over the free body diagram of the described solid cylindrical section in greater details. Overall, the video tries to give an introductory overview on torsional actions over different bodies used as structural elements widely in constructions from the subject point of view.
Lecture 25
Method of Transformed Sections (Beams of 2 Materials) Steel a
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Method of Transformed Sections (Beams of 2 Materials) Steel a
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This video presents the last example of transformed sections. First part of the video demonstrates a composite section where an aluminum I beam has been strengthened by two timber sections. There are others data given e.g., length of the flanges is 200 mm, depth of the section is 300 mm, thickness of the flanges of the I-beam is 20 mm & width of the connecting element is 20 mm. The modulus of elasticity of both the aluminum and timber has also given. The example asks to draw stress & strain distribution when a positive 30 KNm moments is applied to the x axis.

Moving next, the video concerts the timber to aluminum using the transformed sections theory and shows the calculation of second moment of area under x-axis. Later, the video does the stress-strain calculations at bottom and top of the composite sections. Due to the symmetry of composite section, the stresses at bottom side are basically the mirror image but have positive value as they are under tension. Finally, the video plots the obtained values and shows the stress-strain distributions over the schematic diagram of the given composite section.
Lecture 26
Method of Transformed Sections (Beams of 2 Materials) Reinforced
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Method of Transformed Sections (Beams of 2 Materials) Reinforced
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This video, continuation of the example started at previous video 5.16, very first shows that the y bar is located 103 mm away from the top of the given rectangular reinforced concrete section. Next, the video recalls the formula of calculating the second moment of area and subsequently shows the assessments of required values of the constituents of the formula. Moving on, the video calculates down the I-value and later using the obtained values it calculates down the values of stresses at top and bottom side of the concrete and steel materials under the light of transformed sections theory.

The video, then, determines the value of strains of steel and concrete materials at top & bottom side of the given section; subsequently plots all the obtained values of stresses and strains over the free body diagram of the given rectangular reinforced concrete section & thus shows the stress & strain distributions. This example is quite complex, so it is highly recommended to do on hand practices with this video to have better understanding on the workout procedures of the described example.
Lecture 27
Method of Transformed Sections (Beams of 2 Materials) Reinforced
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Method of Transformed Sections (Beams of 2 Materials) Reinforced
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This video shows a workout on an example of finding the bending moment of reinforced concrete bar under the light of transformed sections theory. At first, the video illustrates the data given e.g., 200 mm by 100 mm rectangular reinforced concrete section having 2 steel bars of 12 mm diameter each castoff inside the section, distance of steel bars from the bottom of the section and the elastic modulus of steel & concrete materials. The example asks to determine the value of bending moment required to cause the concrete to crack in tension when stresses at tension equals to 5 MPa.

Moving on, the video uses the transformed sections theory to convert the steel bar material into the concrete material & subsequently shows the transformation of steel bars to concrete bars over the given section. The video clears the fact that each steel bar is replaced by a concrete bar and add an extra (n-1) concrete bar to meet the requirements of transformed sections theory. Later, the video shows the calculation of y bar taking arbitrary reference axis at top left corner of the given section.
Lecture 28
Method of Transformed Sections (Beams of 2 Materials) Timber Beam Example
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Method of Transformed Sections (Beams of 2 Materials) Timber Beam Example
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This video presents an example of bending through the application of transformed sections theory. Very first, the video explains the data given of the exemplary structural section e.g., 300 mm by 200 mm timber rectangular cross section subjected to an axial compression of 300 KN, bending moment of 11 KNm about major principle axis and bending moment of 0.1 KNm about minor principle axis. The example asks to identify the point on the cross section that experience the max and min values of stresses and calculate the value of these stresses.

Moving on, the video shows the calculation of finding the principle axis determining the I-value of x and Y axis; subsequently recalls the formulas of stresses due to x and y axis bending to find out the value of stresses at top and bottom side of the given structural section. Later, the video shows the calculation of tensile & compressive stresses & thereby finds out the cross section that experiences the max and min values of stresses over the given exemplary structural section.
Lecture 29
Method of Transformed Sections (Beams of 2 Materials) Example
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Method of Transformed Sections (Beams of 2 Materials) Example
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This video shows a workout on stress distribution due to the action of bending moment on sections made of two different materials through the application of transformed section. Very first, the video illustrates the given exemplary composite sections of two materials having steel at top and bottom & timber at the middle. Subsequently it explains the data given e.g., modulus of elasticity of steel & timber, value of positive bending moment subjected to the composite & the dimension details. Next, the video converts the composite to only steel material using the theory of transformed sections.

Moving on, the video works out on the new section and calculates the second moment of area under the given positive bending moment; successively does the calculation for stress distribution of steel material. Later, the video uses the transformed sections theory to find out the stress distribution of timber and finally plots all the obtained values over the schematic of composites sections to get the stresses distribution graphically.
Lecture 30
Method of Transformed Sections (Beams of 2 Materials)
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Method of Transformed Sections (Beams of 2 Materials)
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The key objective of this video is to explain transformed sections and this is just going on from the previous theoretical video 5.12 of stress and strain in composite beam. At first, the video assumes a cross sectional (composite) made from materials having elastic modulus E1 and E2. Next, the video shows how material 1 can be converted to material 2 by multiplying the width by a factor of n. The factor ‘n’ represents the modular ratio of the composite element.

Moving next, the video converts the top material to bottom material using the conversion formula. Even though the conversion takes place, however, the neutral axis remains unchanged. To work out on stress distribution, the I-value for the transformed section must have to be calculated. To meet the necessity, the video introduces the formula of stress distribution for the materials that has been transformed. Later, the video shows the conversion of bottom material to top material and explains the transformation of the sections in greater details.
Lecture 31
Beam Composite Actions: Stress and Strains
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Beam Composite Actions: Stress and Strains
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This video gives an overview on composite action theory followed by the stress-strain distribution explanation in a composite beam. Very first, the video defines the composite action by stating that a beam is considered as composite when both of its materials have deformed as one body. Next, the video illustrates the non-composite action and composite action of beam graphically & conclude the fact that fusion and bonding must be present between both materials for having composite action.

Moving on, the video talks about stress and strain in a composite beam showing schematic of beam elevation; followed by the beam cross section & beam composite action. Next, the video introduces with the term ‘modular ratio’ of an element which is the ratio of elastic moduli of both materials. In composite actions, both materials are straining by the same amount so that liner distributions take place. Last part of the video demonstrates stress & strain distribution in composite actions over the free body diagram of exemplary beam section.
Lecture 32
I-beam Centroid, Second Moment of Area (I-Value), Stress & Strai
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I-beam Centroid, Second Moment of Area (I-Value), Stress & Strai
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This video shows a workout on a comprehensive example of finding centroid, 2nd moment of area and bending moment about X and Y axis for a given exemplary unsymmetrical I-section. At first, the video illustrates the data given e.g., bottom flange thickness is 5 mm, top flange thickness is 10 mm, width of the connecting element is 6 mm, depth of the section is 200 mm & bottom flange length is 100 mm. Next, the video explains what needs to be found out during the workout on the example.

Moving on, the video recalls the formula of finding centroid coordinates with respect to an arbitrary reference axis and shows the calculation to determine the centroid position for the given I-section. Next, the video recalls the formula of second moment of area and using the determined value of the centroid the video successively finds the I-values along X & Y axis. Last part of the video shows the calculation of finding the bending moment about X & Y axis needed to produce the given maximum amount of stresses distribution over the unsymmetrical I-section.
Lecture 33
Example 8 Centroid Hollow section, I Value + Stresses due to bending
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Example 8 Centroid Hollow section, I Value + Stresses due to bending
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This video presents a comprehensive example of finding centroid, 2nd moment of area and bending moment for a given exemplary hollow circular section. First part of the video essentially illustrates the data given e.g., outer diameter and thickness of the circular section etc. Consequently it asks to determine the centroid position, second moment of area and the value of bending moment required to produce a maximum stress distribution of 150 MPa about the horizontal & centroidal axis.

A circle is doubly symmetric about its vertical & horizontal axis which results the centroid position to be in the center of the circle. Moving on, the video introduces the formula of second moment of area for symmetric circular section & shows the step by step calculations to determine the effective I value by eliminating the I-value of inner hollow section from the total I-value of the given circular section. Later, the video shows how to calculate the bending moment value required to produce the given maximum stress distribution in greater details.
Lecture 34
C-Section Stress & Strain Distributions from Bending
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C-Section Stress & Strain Distributions from Bending
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This video shows a workout on an example of stress-strain distribution under Y-axis bending. Very first, the video explains the given exemplary C section which flanges thickness is 10 mm, connecting bar width is 6 mm, width of the flanges is 100 mm and depth of the section is 200 mm. The centroid coordinates have been given & the video asks to find out the stress-strain distribution if the section is subjected to a bending moment of 10 KNm. Next, the video sets the reference coordinates at left bottom corner of the C section in order to determine the second moment of area.

Moving on, the video introduces the formula of second moment of area along Y-axis & successively determines the Iy value assessing all the required constituents of the formula. Later, the video uses the Iy value, given bending moment value & centroid coordinates values to find out the stress-strain distributions over the schematic of stress and strain of the given C section. It is highly suggested to do practices with this video to have good understanding on the content described in this lesson.
Lecture 35
I-Beam Stress & Strain Distributions from Bending
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I-Beam Stress & Strain Distributions from Bending
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This video presents a comprehensive example on stress-strain distribution under x-axis bending. At first, the video illustrates the given exemplary symmetric I section which has a depth of 612 mm & the Y-coordinate of centroid is 306 mm away from the bottom of the section. The second moment of area is also given for the I-section & the section is subjected to a positive bending moment (sagging) of 966 KNm. Next, the video talks about the symmetric nature of the given I-section.

Moving on, the video introduces the formula of stress and strain distribution under x-axis bending that has been described in previous video 5.6; subsequently assesses all the required values and by putting them into the formula the stress & strain distributions both at tension and compression side of the I-section have been determined. Later, the video plots the obtained values into the schematic diagram of stress and strain and thereby shows the distributions over the free body diagrams.
Lecture 36
Bending of Beams: Stress & Strain Distributions (about y-axis)
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Bending of Beams: Stress & Strain Distributions (about y-axis)
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This video presents typical stress-strain distribution under Y axis bending. At first, the video refers previous video 5.6 where typical stress-strain distribution under X axis bending has been explained in greater details. Next, the video introduces the required formula of stress & strain for a structural section subjected to a Y axis bending moment; successively explains the stress & strain distribution graphically over the schematic of I-beam shape structural element. The stress-strain distribution is very similar to the distributions described in earlier video under the x-axis bending theory.

Moving on, the video briefly talks about the critical sections which could be found at furthest away from the neutral axis. Later, the video talks about the yielding phenomenon both for symmetrical and unsymmetrical structural section. Overall, the video has tried to give clear concepts of the effect of Y-axis bending over the typical stress vs. strain diagram- understanding which is very important to do workouts on many examples of structural mechanics.
Lecture 37
Bending of Beams: Stress & Strain Distributions
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Bending of Beams: Stress & Strain Distributions
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The objectives of this video are to give an introductory overview on bending terminology followed by a demonstration on a typical stress-strain distribution in bending. At first, the video briefly talks about positive & negative bending moment, also known as sagging & hogging moment respectively. Next, the video introduces with the formula of stress and strain for a structural section subjected to a bending moment; subsequently illustrates the stress and strain distribution graphically.

Moving on, the video introduces with the terminology ‘critical section’ which principally refers the part of a structural element that undergoes the highest level of stresses and strains and the one that tends to yield first. Later, the video talks about the yielding phenomenon both for symmetrical and unsymmetrical structural section. Overall, the video has tried to give clear conceptions of the effect of bending over the typical stress vs. strain diagram- understanding which is very important to do workouts on many examples of structural mechanics.
Lecture 38
Channel Section Centroid & Second Moment of Area (I value) Example
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Channel Section Centroid & Second Moment of Area (I value) Example
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This video shows the workout on another comprehensive example of finding the centroid positions & second moment of area for a given exemplary C channel structural element. First part of the video briefly illustrates the given C channel- the flanges thickness is 10 mm at each side, connecting bar thickness is 6 mm and total depth of the channel is 200 mm. Next, the video takes the bottom left corner as the reference axis and subsequently calculates down the values of centroidal coordinates.

Moving on, the video shows the calculation procedures of finding the second moment of area in x direction assessing all the required values & demonstrating all the facts & figures in greater details. It is highly recommended to watch the previous videos and do on hand practices with these videos to have better understanding on the calculation procedures of finding centroid position and second moment of area for different structural elements used in the construction extensively.
Lecture 39
Second Moment of Area (Iy value) of Band Beam Example
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Second Moment of Area (Iy value) of Band Beam Example
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This video shows how to work out on finding Iy value continuing straight from the earlier video 5.3. At first, the video gives emphasize on knowing the values of centroidal coordinates with respect to the reference axis before doing any calculation for finding the value of second moment of area. Next, the video once again vividly illustrates the centroidal position of the given exemplary band beam & introduces the formula of finding Iy value as described before in previous video 5.4.

According to the formula, the Iy value of the structural element with respect to the reference axis is equal to the sum of Iy value with respect to the centroidal axis plus the products of areas times the square values of horizontal distance between the centroidal and considered reference axis. Moving on, the video briefly shows the step by step calculations assessing all the required values from the schematic diagram of band beam and thereby finally determine the Iy value.
Lecture 40
Second Moment of Area (Ix value) of Band Beam Example
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Second Moment of Area (Ix value) of Band Beam Example
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The objective of this video is to show a comprehensive example of how to work out the I-values also known as the second moment of area. Very first, the video gives a brief inside on I-values referring graphically the most seen rectangular structural element of construction & subsequently introduces with the formula of finding out the coordinates position of I-values considering the centroidal axis as reference axis. Next, the video illustrates what happens to the formulas when an arbitrary axis is taken as reference axis. Subsequently, it introduces the formula effective for circular cross section.

Moving on, the video shows the same band beam described in previous video 5.2 and starts to find out the second moment of area considering the arbitrary axis at top left corner of the band beam as reference axis. The video shows step by step calculations assessing all the required values from the schematic of band beam & thereby finally determine the Ix value. Finally, the video refers to watch next video to continue the lesson of finding the Iy value.
Lecture 41
Centroid of Band Beam Example
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Centroid of Band Beam Example
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This video presents another comprehensive example of finding the centroid coordinate for a given exemplary band beam. Very first, the video illustrates the given band beam which has a thickness of 200 mm and a length of 1700 mm. The band beam has two legs; each leg is 400 mm in length, 300 mm in depth and the distance between the legs is 600 mm. Moving next, the video set the reference coordinate at top left corner of the band beam and starts to calculate down the centroid position.

The video shows how to use the benefits of having symmetric element in the calculation of centroid position as the given exemplary beam is symmetric in nature in horizontal direction. Moving on, the video shows the entire calculation of finding the coordinates x bar and y bar utilizing the symmetric nature and the earlier described formula of finding centroid coordinates for any structural element in greater details. On hand practice is suggested to do with this video to avoid confusions.
Lecture 42
Centroid of I-Beam Example
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Centroid of I-Beam Example
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This video shows a workout on a comprehensive example of finding the coordinates of centroid for an I-section structural element. Very first, the video illustrates the given exemplary I section having two flanges at top and bottom side of the I-section and a vertical bar element connecting the two flanges followed by the demonstration on the given dimension of the I-section structural element. Next, the video introduces with the formula of finding the centroid coordinate position.

According to the formula, the x coordinate of the structural element is equal to the ratio of ‘the sum of product of individual area times individual centroid x coordinate’ and the sum of individual area. And the y coordinate is equal to the ratio of ‘the sum of product of individual area times individual centroid y coordinate’ and the sum of individual area. Moving on, the video shows the calculation of finding centroid coordinate for the given I-section explaining all the facts and figures in details.
Lecture 43
Axial Loading & Temperature Effects
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Axial Loading & Temperature Effects
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The objective of this video is to examine the temperature effects on structural elements. First part of the video clears the conception of temperature effects on materials stating that many materials expand & contract with changes of temperature. Subsequently, the video explains the phenomenon graphically considering an element having length L which increases length by the amount L with the application of certain amount of heat. Generally, the heating causes expansion whereas cooling causes contraction for most of the materials used as structural elements.

Moving on, the video introduces the theory of temperature effects on structural materials stating that an unrestrained length of a bar will expand by the amount equal to the product of coefficient of thermal expansion of a material times the temperature difference. Later, the video talks about the unit of coefficient of thermal expansion and displays the values of coefficient of thermal expansion of common materials e.g., steel, copper, aluminum and concrete etc. widely used in construction.
Lecture 44
Axial Loading of a Composite Structural Member
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Axial Loading of a Composite Structural Member
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The objective of this video is to explain composite actions followed by a workout on an example of concrete filled steel hollow section column. Very first, the video talks about the rules of actions that strain and stress cannot be added together, however, deformation can be added together. Next, the video explains the composite action of material considering an aluminum column filled with copper in the interior subjected to compression. Subsequently, it talks about different facts that the entire column when compressed by same amount, both copper and aluminum, this behaves compositely.

To guarantee composite actions, a bond or a fusion is needed between the materials. For example, if the surface contact is greased and the load was applied on copper; the copper would compress only. Moving on, the video presents an example asking to calculate the axial shortening of concrete filled steel hollow column based on the data available. The video step by step shows the entire calculation and calculate down the required value applying the rules of composite actions.
Lecture 45
Axial Loading Example
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Axial Loading Example
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The objective of this video is to do an analysis of a member under axial load followed by an example of steel hollow section column. First part of the video illustrates a 3D representation of a structural element of varying cross section & 2D elevation of the cross sectional area of that element in details. Next, the video briefly talks about the different properties of the given structural element e.g., cross sectional properties of any slice of the element is dependent of the distance from origin; axial force is the function of distance and strain is same over the slices as the modulus elasticity is constant etc.

Moving on, the video introduces the formula of extension of any slice to total extension of member & explains special case of a member of constant area and constant force. Next, the video works out on an example of steel hollow column where it asks to calculate stress, strain & axial shortening of steel based on the data given. The video step by step shows the entire calculation finding the cross sectional area first and then followed by the formula of stress, strain and deformation.
Lecture 46
Stress & Strain: Non Linear, Ductile and Brittle Behaviour
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Stress & Strain: Non Linear, Ductile and Brittle Behaviour
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This video gives a brief inside explanation on nonlinear behavior followed by the ductile and brittle behavior. At first, the video refers previous video 3.4 of stress-strain behavior of linear material and then point out the fact that there are some other materials which behave in a nonlinear fashion and don’t have a defined yield stress. This causes an imprecise elastic-plastic approximation. In order to determine an approximate yield stress, a phenomenon known as proof stress is integrated.

Moving on, the video displays a stress-strain diagram for nonlinear material & defined proof stress as the stress with same slope as the dashed line starting from a strain of 0.002; subsequently shows how to determine approximate yield stress for nonlinear material. Next, the video explains concrete behavior followed by the ductile vs. brittle behavior in details- ductile behaviors are much desired where large amount of deformation takes place before failure and brittle behaviors are less desired where failure occurs suddenly with little plastic deformation may not giving warning before failure.
Lecture 47
Stress & Strain Diagram
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Stress & Strain Diagram
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This video talks more in depth about the stress-strain curve of a typical steel material starting with the origin representing zero stress and strain. The first region of the curve which is linear in nature is known as elastic region. In elastic region, the material follows Hooke’s law; there is no permanent deformation that has taken place & each material has a constant modulus of elasticity in this region. Next, the video explains yielding region at which the material has appreciable elongation or yielding without any increase in load and also referred as plastic deformation.

Moving on, the video talks about strain hardening region- subsequently to yielding; at higher strain the stress begin to increase until it reaches the peak of the graph & this is known as ultimate stress- the maximum load a member can take not necessarily the mode of failure. Later, the video explains necking & fracture over the stress-strain curve in details. No exception- as the understanding of the stress-strain diagram is mandatory to master the field of structural mechanics.
Lecture 48
Strain & Poisson's Ratio
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Strain & Poisson's Ratio
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This video demonstrates normal & shear strains, Poisson’s ratio & stress-strain relationships. First part of the video talks about normal strains recalling that normal stress tends to deform an object into an elongated shape or squashed shape. Strain is defined as the change in length divided by the original length & is a dimensionless quantity. And shear strain is defined as the angle change of the element & is also a dimensionless quantity. The video explains both the terminology- normal strain and shear strain vividly with necessary schematic diagrams.

Moving on, the video briefly explains Poisson’s ratio which represents the ratio of transverse strain to normal strain when the stress-strain behavior is linear; successively talks about the assumptions made in Poisson’s ratio i.e., materials are linear elastic, homogeneous & isotropic etc. in details. The video, then, displays a graphical plotting of stress vs. strain diagram containing a stress-strain curve and illustrates the different points of the curve over the stress-strain diagram.
Lecture 49
Normal & Shear Stress
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Normal & Shear Stress
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The objectives of this video are to briefly talk about general load deformation response following a discussion on normal and shear stresses. Load- deformation patterns are dependent on the size and shape of a structure which can have effect on many parameters e.g., force, stress, deformation etc. Generally force represents the load applied & is affected by the cross sectional area whereas stress is the measure of intensity of a force. And the term deformation represents change in overall length whereas strain is the change in length relative to the original length.

Moving on, the video graphically represents general load- deformation response taking 2 different cylinders as specimen; subsequently demonstrates the effect of load- deformation response against the force applied to the cross sectional area of these specimen in greater details. Later, the video explains the terms normal stresses and shear stresses- normal stresses represents the stresses that are perpendicular to the plane and shear represents the stresses that are parallel with the plane.
Lecture 50
Stress, Strain, Strength, Stiffness & Defomation
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Stress, Strain, Strength, Stiffness & Defomation
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This video demonstrates a brief inside of the deformation terminology of structural mechanics with graphical representations. At first, the video clears the fact that upon loading of a structure, there is always some sort of deformation. This always induces some sort of load deformation response that varies depending on material properties. Strength and stiffness are two important properties of the materials where strength represents how much loads an element can carry and stiffness represents slopes of the response at any given point.

Moving on, the video illustrates the load- deformation graphical response showing initial stiffness by taking slope e.g., ratio of rise over run of the load-deformation curve. Next, the video explains the strength point that represents maximum loading of a structure. Later, the video briefly discusses on other terminologies used in structural mechanics over the schematic diagram of load- deformation. Never hesitate to contact with our experts; if you don’t understand any part of the lecture.
Lecture 51
Twisting Moment Diagram Example
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Twisting Moment Diagram Example
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This video presents a lecture on twisting moment diagram following a workout on a comprehensive example of shaft with applied torque. Twisting moment develops because of the actions of torsion developed in a shaft or beam under the actions of a torque applied. First part of this video explains the two anticlockwise twisting actions acting at top and halfway through bottom of the shaft vividly in details. Next, the video demonstrates the rules of thumbs & clear the concept on twisting moment that different cut could be taken furthest point way through from the fixed point.

Moving on, the video demonstrates that the exemplary shaft has no twisting moment at bottom end as the end is fixed; subsequently explains that the given two twisting moment values are positive in sign as they are acting in anticlockwise direction. Later, the video draws twisting moment diagram very similar to the drawing of bending moment diagram and illustrates what happens both for the positive and negative values of twisting moments over the schematic of twisting moment diagrams.
Lecture 52
Example 5: Axial Force, Shear Force and Bending Momemt Diagram (2/2)
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Example 5: Axial Force, Shear Force and Bending Momemt Diagram (2/2)
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Continuing the example 5 from previous video 2.10, this video very first shows how to take second cut for the stated problem and does the internal reactions and bending moment calculations similar to procedures previously done. Next, the video takes third cut from the right end for simplicity and highlights the sign convention in doing so to avoid any confusion in taking signs for the reactions & moments developed. Moving on, the video does the same calculation of finding unknown variables from the established equations of equilibrium.

The video briefly explains the calculation of maximum moments at different cuts by taking the first derivative of the developed moment equations & emphasizes on the finding procedure required for drawing the curve in bending moment diagram. Last part of the video vividly shows how to plot all the obtained values to get the axial force diagram, shear force diagram & bending moment diagram.
Lecture 53
Example 5: Axial Force, Shear Force and Bending Momemt Diagram (1/2)
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Example 5: Axial Force, Shear Force and Bending Momemt Diagram (1/2)
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This video presents example 5 of axial force diagram (AFD), shear force diagram (SFD) and bending moment diagram (BMD) for a given overhanging structure having a clockwise moment at leftmost end, then a pin support after 2 m and then a roller support after 4m. The structure contains a UDL of 2 KN/m over a distance of 6 m from the right end. Very first, the video vividly demonstrates the example and points out the internal reactions developed because of having pin and roller support.

Next, the video shows how to use the equilibrium principle to develop forces & moments equations so that the unknown values of internal restraints can be determined. Doing so, the video shows the conversion of uniformly distributed load to a point load to meet the demand of solving the example. Later, the video takes first cut just up to before the pin support and does the reactions and moment calculations. Finally, the video refers to watch next video to continue the lesson.
Lecture 54
Example 4: Axial Force, Shear Force and Bending Momemt Diagram (2/2)
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Example 4: Axial Force, Shear Force and Bending Momemt Diagram (2/2)
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This video, continuation of the example 4 from video 2.8, focuses on the rest two cuts over the given structure. At first, the video explains the schematic diagram of second cuts & finds out the unknown values of axial reaction & bending moment establishing the principle of equilibrium. Next, the video shows the same calculations for third cut span of the structure demonstrating the facts of maximum bending moment taking the first derivative of the moment equation and making it equals to zero.

Moving on, the video draws the original diagram showing loading and supports over the structure; consequently plots the obtained values of axial loading and reactions to get the axial force diagram. Later, the video uses the relationship between bending, shear & axial forces to draw the shear force diagram and bending moment diagram. Doing so, the video demonstrates all the facts and figures as well as the nature of lines and curves of SFD and BMD in greater details.
Lecture 55
Example 4: Axial Force, Shear Force and Bending Momemt Diagram (1/2)
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Example 4: Axial Force, Shear Force and Bending Momemt Diagram (1/2)
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This video presents a comprehensive example of bending moment diagram, shear force diagram & axial force diagram for a given structure having point support at left end, roller support at right end, point load & distributed load at certain distances from both ends. Just to start with, the video shows the internal restraints developed because of different supports & uses equilibrium principle to find out the unknown value. The video explains in details how to develop moment equation while doing the equilibrium making it simple for the learners.

Moving on, the video illustrates where to take cuts in order to draw AFD, SFD & BMD; subsequently shows the calculation of internal reaction & moment for first cut over the given structure. Doing so, the video develops moment equation to meet the demands of the problem. Finally, the video refers to watch next video for rest cuts & plotting of the aimed diagrams as part of continuing the lesson.
Lecture 56
Example 3: Axial Force, Shear Force and Bending Momemt Diagram (2/2)
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Example 3: Axial Force, Shear Force and Bending Momemt Diagram (2/2)
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This video, continuation of the example 3, focuses on different cuts from the beginning of the lesson. At first, the video demonstrates how to find out internal reactions and moments developed because of the second cut illustrating all the facts and figures in details. Next, the video takes 3rd cut adding the hinge support & expands the equilibrium equations. Moving on, the video takes forth cut adding the point load acting vertically upward and does the same calculations for this cut as well.

Later, the video shows the free body diagram of the structure showing the load, moment & support conditions; subsequently explains how to draw axial force diagram, shear force diagram & bending moment diagram plotting all the obtained values in greater details. Finally, the video discusses on the zero (0) moment condition developed because of hinge support present in the given structure. On hand practice is required to do to have good grasp on the content described in this lesson.
Lecture 57
Example 3: Axial Force, Shear Force and Bending Momemt Diagram (1/2)
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Example 3: Axial Force, Shear Force and Bending Momemt Diagram (1/2)
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The objective of this video is to work out on bending moment diagram, shear force diagram & axial force diagram for a given structure having roller support at left end, fixed support at right end, a point load and clockwise moment acting over the structure at certain distances and a hinge support at the middle of the structure. At first, the video vividly explains the different restraints developed because of different supports and shows how to establish equilibrium in order to find the unknown values of the internal reaction forces in greater details.

Next, the video takes advantage of hinge support- moments from either side of the structure about a hinge is equal to zero. This helps to calculate the unknown reactions from the equation developed using the principle of equilibrium. Moving on, the video takes first cut from left end and shows the entire calculation of finding internal reactions & bending moment in details. Finally, the video refers to watch next video to continue the lesson.
Lecture 58
Example 2: Axial Force, Shear Force and Bending Momemt Diagram (2/2)
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Example 2: Axial Force, Shear Force and Bending Momemt Diagram (2/2)
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This video, continuation of the previous video 2.4, very first once again reminds the sign convention when cutting from the right end & subsequently goes after the third cuts. The axial load & moments condition is exactly opposite in nature when cutting from left & right end so that they balance each other. The third cuts from right end save a lot of time because if this has to be taken from left end, then the first & second cuts have to be considered while carrying out the calculation. Moving on, the video shows all the consideration needed to be made while taking the third cuts from right end.

Next, the video establishes equilibrium and obtains a bending moment equation which can give the value of moments using the known variable distance. Later, the video uses all the obtained values to plot an axial force diagram below the free body diagram of the given simply supported beam. Next, the video illustrates how to draw the shear force diagram and bending moment diagram using the relationships between bending, shear and axial forces as described in previous video.
Lecture 59
Axial Force, Shear Force and Bending Momemt Diagram Example 2 (1/2)
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Axial Force, Shear Force and Bending Momemt Diagram Example 2 (1/2)
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This video shows another comprehensive example of AFD, SFD & BMD where it asks to draw these diagrams for a simply supported beam having point support at left end & roller support at right end and have point load & distributed load over the beam at certain distances from the end of the beam. Very first, the video explains its main focuses which is nothing but to show how to takes cuts from both ends of the beam in same equation & links them up together to make the problem less complex saving a lot of time. Next, the video starts to establish equilibrium to find out the internal reactions.

Moving on, the video determines the internal horizontal and vertical reactions developed due to the point and roller support; subsequently explains the sign convention that are extensively used while taking cuts over the beam. Next, it takes first cut from left end of the beam and just before the point load following the calculations of internal reactions & moments. Later, the video shows second cuts & accordingly does calculations for internal reactions & bending moments. Finally, the video refers to watch next video to continue the lesson.
Lecture 60
Axial Force, Shear Force and Bending Momemt Diagram Example 1
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Axial Force, Shear Force and Bending Momemt Diagram Example 1
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This video shows a workout on a comprehensive example where it asks to draw axial force diagram (AFD), shear force diagram (SFD) and bending moment diagram (BMD) for a given cantilever beam having a moment acting clockwise in direction at certain distance from the fixed end of the beam. At first, the video shows how to establish equilibrium for the given set of conditions finding the values of horizontal & vertical reactions & the bending moment gets developed at the cantilever end.

Moving on, the video takes two cuts possible over the cantilever beam and subsequently calculates down the reaction forces and bending moments over the cut span of the cantilever beam. Later, the video uses the obtained values to plot AFD below the FBD of the given cantilever beam; successively draws SFD and BMD demonstrating all the facts & figures in greater details. Finally, the video gives emphasize to make sure so that the unit used in the calculation must be written over the diagrams.
Lecture 61
Relationships Between Bending, Shear and Axial Forces
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Relationships Between Bending, Shear and Axial Forces
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The objectives of this video are to give a brief explanation on two terminologies namely sagging and hogging & their consequences with the bending moment following a discussion on the relationships between bending, shear & axial forces. At first, the video vividly illustrates the physical occurrences of sagging & hogging moments over a horizontally simply supported beam clearing the concept that sagging & hogging happen respectively when compression takes place at top & bottom of the beam.

Next, the video clears what happens for a vertical member stating that in a vertical member it is not possible to determine sagging and hogging condition; it is only possible to determine what side is in tension and what side is in compression from the acting direction of load. Beside the video explains why bending moment has to be drawn in compression side always. Later, the video discusses on the relationships between bending, shear & axial forces where SFD is the derivative of the BMD and the negative derivative of SFD yields the distributed loading function.
Lecture 62
Axial Force Diagram, Shear Force Diagram & Bending Moment Diagram
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Axial Force Diagram, Shear Force Diagram & Bending Moment Diagram
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This video briefly demonstrates the actions of internal loading of a structure followed by indicating bending moment diagrams (BMD), shear force diagrams (SFD) & axial force diagrams (AFD). When any structure is exposed to a loading, the loads are transferred from the point of action through the structural elements down to the base and/or supports. This procedure induces internal reactions or stress resultants; these may vary in magnitude throughout the structure.

Establishing a distribution of these internal actions is a key process in civil engineering design. The video, then, briefly illustrates AFD, SFD & BMD and their drawing in greater details. Moving on, the video shows the summary of procedures – STEPS ARE: draw the FBD, establish equilibrium, cutting structure at artificial locations, FBD of the cutting portions, sign convention, establish the equations of equilibrium and solve for internal actions and finally draw the AFD, SFD & BMD for the workout problems of structural elements.
Lecture 63
Equilibrium & Free Body Diagram Example 2
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Equilibrium & Free Body Diagram Example 2
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This video does a comprehensive workout on solving another example using equilibrium principle and free body diagram. Very first, the video highlights the key difference of this example from the previous one which is that - in this video it has been shown how to convert a uniformly distributed loading into a point loading over the free body of given beam section. Initially, the video gives brief overview on the given example illustrating all the facts and figures and then calculated down the point load from the UDL just by multiplying the UDL with the span distance.

Moving on, the video takes sum of all components of forces in both horizontal and vertical direction following the principle of equilibrium; subsequently takes moment at left end of the given beam and finally the unknown values are determined doing mathematics from the written equations. This is pretty simple; however, it is suggested to do on hand practices to have better understanding on the contents described in this lesson. Never hesitate to contact, if you don’t understand any part of it.
Lecture 64
Equilibrium & Free Body Diagram Example 1
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Equilibrium & Free Body Diagram Example 1
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This video briefly works out on solving an example using the principle of equilibrium and free body diagram. First part of the video asks to imagine a simply supported beam having a point support at left end and a roller support at right end. Next, the video talks about the data given e.g., distance between two ends of the beam and a bending moment acting clockwise over the beam at certain distance from left end. The video asks to consider black colored diagram only ignoring blue colored reactions that are actually unknown and needed to be calculated down using equilibrium method.

Moving on, the video explains the first step of solving the problem which is the drawing of reactions forces on the FBD; subsequently in next step shows how to establish equilibrium using the different components of forces and moments acting over the given simply supported beam. In last step, the video shows the solution procedures of the established equilibrium condition & thereby found out the unknown value of different reaction forces.
Lecture 65
Equilibrium & Free Body Diagrams
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Equilibrium & Free Body Diagrams
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The objectives of this video are to look at equilibrium both in two and three dimensionally; then at free body diagram - successively workout on relevant example. Very first, the video vividly explains a 3D Cartesian coordinate’s scheme showing the actions of forces and moments along its three axes. It clears the perception on bending moments elaborating graphically that anticlockwise & clockwise moments are respectively taken as positive & negative commonly. Next, it introduces equilibrium- the state when sum of components of all forces and moments in any direction at any point is zero.

Moving on, the video talks about statically determined structures where the numbers of unknown & known are equal & therefore are solvable by equilibrium method. Next, the video looks equilibrium in 2D; consequently demonstrates facts & figures of free body diagram (FBD) of structural element. Last part of the video works out on an example of reactions on a flag and shows how to find out the unknown values using the method of equilibrium.
Lecture 66
Loads, Transfer of Forces & Supports
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Loads, Transfer of Forces & Supports
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The objectives of this video are to give a brief explanation on loads and transfer of forces following a discussion on types of supports acting on a structure. At first, the video clears the concept of point of a structure by stating that it’s the acting point to transfer load down to the foundation effectively without any kind of failure. Next, the video asks question to itself that how does the loads transfer down? Subsequently it elaborates step by step how different loads are transmitted starting with the loads acting on a floor to its adjacent secondary beam in terms of KN/m.

Moving on, the video talks about uniformly distributed load (UDL) - forces distribution along length and shows how loads get transmitted from secondary beam to primary beam to column in greater details. Later, the video explains the concept of tension, necking, compression and bending moment with necessary figures and facts. Last part of the video briefly talks about different types of support e.g., point, roller and fixed supports; consequently explains how they works in a structure in details.
Lecture 67
Simply Supported Beam Deflection from Loading Function Example
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Simply Supported Beam Deflection from Loading Function Example
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This video shows a comprehensive workout example of integrating the given loading function to obtain the governing equation of deflection. Very first, the video demonstrates the given exemplary simply supported beam subjected to non-uniformly distributed load and explains the load function with respect to boundary conditions. Next, the video shows the equation of deflection with respect to the load function which requires integration four times to yield the deflection equation.

Moving on, the video integrates loading function & assess the integration constants for the 4 times integration and successively derived the final workout equation of deflection. It is very important to learn sharply how to apply different boundary conditions to get the workout equation of deflection. On hand practice is highly recommended to do with the workouts of the examples shown under this video series of deflection. Never hesitate to contact with our expert tutors, if you don’t understand any part of the lessons described over here.