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Structural Funda by:- KheNiSiR
India
เข้าร่วมเมื่อ 26 ก.ย. 2020
“ANYONE CAN LEARN ANYTHING” if proper tutor, motivation, and conceptual teaching will be delivered. Regards this point, Myself, Hitenkumar Kheni (M.Tech Structural Engineering) wants to deliver my all fundamental knowledge regarding Structural Engineering deals with Applied Mechanics and Civil Engineering to all my dearest friends on the Globe.
For such, All fundamental, as well as important topics, Concept behind the solution of examples, the Easiest way to remember the methods, will be tried to cover.
“WE, ENGINEERS, BUILD THE NATION”
For such, All fundamental, as well as important topics, Concept behind the solution of examples, the Easiest way to remember the methods, will be tried to cover.
“WE, ENGINEERS, BUILD THE NATION”
CONJUGATE BEAM METHOD FOR SLOPE & DEFLECTION - SIMPLY SUPPORTED BEAM WITH UDL & POINT LOADS
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Conjugate Beam Method". Conjugate beam is defined as the Fictitious or Imaginary beam with the same length as that of the given beam and support conditions may be changed or as it is as per rule but loading on conjugate beam is bending moment divided by EI.
มุมมอง: 7 860
วีดีโอ
CONJUGATE BEAM METHOD - MAXIMUM DEFLECTION IN BEAM - SIMPLY SUPPORTED BEAM LOADED WITH POINT LOAD
มุมมอง 9K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Conjugate Beam Method". Conjugate beam is defined as the Fictitious or Imaginary beam with the same length as that of the given beam and support conditions may be changed or as it is as per rule but loading on conjugate beam is bending moment divided by EI.
MACAULAY METHOD FOR MAXIMUM DEFLECTION SOLVED EXAMPLE 04
มุมมอง 2.4K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Macaulay's Method" through which Slope and Deflection equation for given beam is formulated using Elastic Curve Equation. It is also known as advanced version of Double Integration Method with some assumption for taking moments as well as selection of section.
CONJUGATE BEAM METHOD FOR SLOPE & DEFLECTION - CANTILEVER BEAM WITH PARTIALLY LOADED UDL
มุมมอง 10K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Conjugate Beam Method". Conjugate beam is defined as the Fictitious or Imaginary beam with the same length as that of the given beam and support conditions may be changed or as it is as per rule but loading on conjugate beam is bending moment divided by EI.
CONJUGATE BEAM METHOD FOR SLOPE & DEFLECTION - CANTILEVER BEAM WITH UDL & POINT LOADS
มุมมอง 23K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Conjugate Beam Method". Conjugate beam is defined as the Fictitious or Imaginary beam with the same length as that of the given beam and support conditions may be changed or as it is as per rule but loading on conjugate beam is bending moment divided by EI.
WHAT IS CONJUGATE BEAM, DIFFERENCE BETWEEN REAL AND CONJUGATE BEAM , SUPPORT CONDITIONS CHANGE
มุมมอง 4.1K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Conjugate Beam Method". Conjugate beam is defined as the Fictitious or Imaginary beam with the same length as that of the givenl beam and supoort conditions may be changed or as it is as per rule but loading on conjugate beam is bending moment divided by EI.
MACAULAY METHOD ALTERNATE METHOD FOR SOLVED EXAMPLE 06 WITHOUT ADD AND REMOVE UDL UP TO SPAN
มุมมอง 1.8K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Macaulay's Method" through which Slope and Deflection equation for given beam is formulated using Elastic Curve Equation. It is also known as advanced version of Double Integration Method with some assumption for taking moments as well as selection of section.
MACAULAY METHOD FOR SLOPE & DEFLECTION SOLVED EXAMPLE 06 BY ADD AND REMOVE UDL UP TO SPAN
มุมมอง 3.3K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Macaulay's Method" through which Slope and Deflection equation for given beam is formulated using Elastic Curve Equation. It is also known as advanced version of Double Integration Method with some assumption for taking moments as well as selection of section.
MACAULAY METHOD FOR SLOPE & DEFLECTION SOLVED EXAMPLE 05
มุมมอง 3.8K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Macaulay's Method" through which Slope and Deflection equation for given beam is formulated using Elastic Curve Equation. It is also known as advanced version of Double Integration Method with some assumption for taking moments as well as selection of section.
MACAULAY METHOD FOR SLOPE & DEFLECTION SOLVED EXAMPLE 03
มุมมอง 34K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Macaulay's Method" through which Slope and Deflection equation for given beam is formulated using Elastic Curve Equation. It is also known as advanced version of Double Integration Method with some assumption for taking moments as well as selection of section.
MACAULAY METHOD FOR SLOPE & DEFLECTION SOLVED EXAMPLE 02
มุมมอง 6K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Macaulay's Method" through which Slope and Deflection equation for given beam is formulated using Elastic Curve Equation. It is also known as advanced version of Double Integration Method with some assumption for taking moments as well as selection of section.
MACAULAY METHOD FOR SLOPE & DEFLECTION SOLVED EXAMPLE 01
มุมมอง 7K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Macaulay's Method" through which Slope and Deflection equation for given beam is formulated using Elastic Curve Equation. It is also known as advanced version of Double Integration Method with some assumption for taking moments as well as selection of section.
MACAULAY METHOD FOR SLOPE & DEFLECTION SOLVED EXAMPLE 04
มุมมอง 7K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Macaulay's Method" through which Slope and Deflection equation for given beam is formulated using Elastic Curve Equation. It is also known as advanced version of Double Integration Method with some assumption for taking moments as well as selection of section.
ASSUMPTIONS MADE IN MACAULAY METHOD FOR SLOPE & DEFLECTION AS WELL AS BOUNDRY CONDITIONS
มุมมอง 2.3K2 ปีที่แล้ว
As Slope an Deflection is important for any loaded beam, there are so many different methods are used. out of which one method is "Macaulay's Method" through which Slope and Deflection equation for given beam is formulated using Elastic Curve Equation. It is also known as advanced version of Double Integration Method with some assumption for taking moments as well as selection of section.
ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 04
มุมมอง 6132 ปีที่แล้ว
for analysis of determinate frame, first supports reactions are to be found by equilibrium conditions and then spliting the frame into separate members with given loading as well as reactions. If load is given on joint then it is to be settle while using system in equilibrium for each splitting members.
ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 03
มุมมอง 6662 ปีที่แล้ว
ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 03
STABILITY OF TRUSSES USING STATIC INDETERMINACY EQUATION
มุมมอง 6052 ปีที่แล้ว
STABILITY OF TRUSSES USING STATIC INDETERMINACY EQUATION
ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 02
มุมมอง 6822 ปีที่แล้ว
ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 02
ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 01
มุมมอง 8822 ปีที่แล้ว
ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 01
RESULTANT OF NON-CONCURRENT FORCE SYSTEM | EXAMPLE 02
มุมมอง 6342 ปีที่แล้ว
RESULTANT OF NON-CONCURRENT FORCE SYSTEM | EXAMPLE 02
RESULTANT OF NON-CONCURRENT FORCE SYSTEM | EXAMPLE 01
มุมมอง 4952 ปีที่แล้ว
RESULTANT OF NON-CONCURRENT FORCE SYSTEM | EXAMPLE 01
UNKNOWN WEIGHT ON CHORD FOR GIVEN POSITION | EXAMPLE 02
มุมมอง 3182 ปีที่แล้ว
UNKNOWN WEIGHT ON CHORD FOR GIVEN POSITION | EXAMPLE 02
UNKNOWN WEIGHT ON CHORD FOR GIVEN POSITION | EXAMPLE 01
มุมมอง 6702 ปีที่แล้ว
UNKNOWN WEIGHT ON CHORD FOR GIVEN POSITION | EXAMPLE 01
RESULTANT OF CONCURRENT FORCE SYSTEM | EXAMPLE 02
มุมมอง 4272 ปีที่แล้ว
RESULTANT OF CONCURRENT FORCE SYSTEM | EXAMPLE 02
RESULTANT OF CONCURRENT FORCE SYSTEM | EXAMPLE 01
มุมมอง 1.6K2 ปีที่แล้ว
RESULTANT OF CONCURRENT FORCE SYSTEM | EXAMPLE 01
FORCE AND FORCE SYSTEMS | FREE BODY DIAGRAM | CHARACTERISTICS | RESULTANT & EQUILIBRANT
มุมมอง 2912 ปีที่แล้ว
FORCE AND FORCE SYSTEMS | FREE BODY DIAGRAM | CHARACTERISTICS | RESULTANT & EQUILIBRANT
RESOLUTION OF FORCE | FUNDAMENTAL OF STATICS
มุมมอง 4132 ปีที่แล้ว
RESOLUTION OF FORCE | FUNDAMENTAL OF STATICS
SIMPLY SUPPORTED BEAM LOADED WITH UDL, POINT LOAD AND EXTERNAL MMOMENT
มุมมอง 4152 ปีที่แล้ว
SIMPLY SUPPORTED BEAM LOADED WITH UDL, POINT LOAD AND EXTERNAL MMOMENT
SIMPLY SUPPORTED BEAM HAVING THREE POINT LOADS AS PER GIVEN DATA
มุมมอง 6122 ปีที่แล้ว
SIMPLY SUPPORTED BEAM HAVING THREE POINT LOADS AS PER GIVEN DATA
SF & BM OF CANTILEVER BEAM WITH MAXIMUM MOMENT CALCULATION & UPWARD LOADING ON IT
มุมมอง 4422 ปีที่แล้ว
SF & BM OF CANTILEVER BEAM WITH MAXIMUM MOMENT CALCULATION & UPWARD LOADING ON IT
Excellent work, thanks for your efforts!
Thank you so much sir...
Sir plz explain in hindi ......aapka English me samjh nhi aa rha h
Dear friend Will soon upload explanation in hindi or try to dubbed surely.... Thanks for suggestion
Thanks sir
I don't understand how you get ordinates on the maximum bending moment 3.6,4.8,4.2 and 3.4 and also maximum shear ordinates 0.13,.0.23,0.3
Dear friend I used Similar triangle rule... For Bending moment.... Peak value 4.8 or say section BM ordinate value is as per derivation which is (L-x) *x/L After that as per location For left riangle use.... At 8 m value is 4.8 than at 6 m how much value??? Similarly for all ordinate, we can find values using similar triangles... Hope your query will resolve now
Thank you sir
Always welcome..... Hope you like learning
Thank you so much 🙏🏾
Sir what's the steps for finding SHEAR STRENGTH OF BOLT ?? Please help
Dear Aman Vdsb and Vdsf both are the equation for the Shear strength But as per Bolt Value or say Bolt strength We have to taken in account Vdpb Value In short Shear strength is similar to bolt value
The My =Pey
Dear.... As per Moment defination It is multiplication of Effective Load and perpendicular distance from CG of load to the point of application of load While you find Moment w.r.to y axis Then perpendicular distance must be in X axis
Sir aapme 6 baad 5.81 , 4.31, kaise nikale ho please tell me
Dear Friend Its Similar Triangle Rule Man lo ke Jese kisime 10m per ordinate value 40 hai to 3.8m par kitana hoga?? Ye rule lagake sare ordinate find kiya hua hai I hope you got the answer
Thank you so much sir❤
Sir both delta v are negative how to knowing which load is taking at c
Dear Friend.... Its trial method We have to start from first or last load of given train load as trial When delta V value changes its sign i.e negative to positive or vice versa... We have to take that trial load on section Hope you understand...
thankyou so much sir
Most welcome dear
Nice explanation sr
Happy you like that.... Keep learning
.5*6*22=?
❤❤❤
Thank you for the video sir, One question, if I asked to find the radius when the tensile stress is zero in any place, how it should be?
Thank you so much! Really helps.
Thanks....happy you like this....keep learning dear.
The FEM for ba should be (3/16)*P*L, for bc should be wL^2/8
Dear Friend, if propped cantilever beam is given than final moment at fixed end (ie Ma) will be (3/16)*p*l and for simply supported beam loaded with UDL over entire span, Max moment at centre of the span is wl^2/8... For Moment Distribution, each span is assumed as fixed for analysis purpose and the end moment is asuumed as Fixed End Moment (FEM) which are as described in example first step... Hope your doubt is clear now... Keep Learning
Thank you sir ❤
Most welcome dear ❤️
Thank you sir ❤
Most welcome dear 💗
Thank you sir ❤
Most welcome dear 💓
Thank you sir ❤
Most welcome
Let's say we have a moving loads crossing that beam. The wheel loads are 45kN, 25kN and 20kN. 45 and 25 is 4 m apart while 25 and 20 is 1.5 m apart. 20kN is the front wheel load of the truck. How shall we do it to get the max shear and moment at the midpoint of the beam?
Dear friend... Kindly provide your email id so that I can drop solution of your mentioned example...also need span value which is missing...
Sir explain how sigma bxx is negative and sigma byy is positive waiting for your reply thanks..
Dear Friend.... Eccentric Stress will be positive or negative as per eccentric distance and its quadrant.... For example, during calculation of Sigma Bxx, we have to consider ey value and if ey distance is in negative quadrant then we have to consider sigma bxx as negative.... hope I can explain your query....
Sir can you teach us how to draw ild for three hinged arch also.? Your explanation for ild for beam is very understandable..🙏🏻
Dear Friend Will put the topic of ILD for arch in pipeline Will definitely cover in future as soon as possible Thanks for the suggestions 🙂
Can you explain if load 20 and 15 cannot interchange (ie 20 kn load always lead)
Dear Friend.... If 20kN load is always lead it means load position is fixed.... For such case Calculation steps are same as explained in this example but we have to take right side first load as 20kN and succeed load or say second load as 15kN Hope you understand the concept..😀
@@KHENISIR thank you so much sir🙏🏻
Welcome Dear friend 😊
Great video! <3
Happy you like it...
Next level and easiest way
Happy you like the method.... Keep learning
Well done sir
Thank you for appreciating dear Parthiv
Really nice sir
Keep watching
Great sir
Sir 4 ki jagah pe 1/4 ayega
Dear Friend Its 4 not 1/4 As its 1/4 is iin denomination of 1 So actually its 4 in nominator
Sir isme mass kahi pr bhi nhi ayega kya ...............how unit will satisfy
Dear Anurag ... Its for a 2D element And Moment of Inertia means Second moment of area... hence mass is not calculated
And unit of Moment of Inertia is Unit^4
Really very nice explanation sir, Thanks for the video ❤
Thank you for your great appraisal 🙏
if no. of bolts required are 5 then we have to take it as 6 bolts and put 3 bolts on one plate and 3 bolts on another , by your detailing its shown that you put total of 10 bolts not 5 bolts. do clarify
Dear Friend During calculation, if anyone consider bolt in double shear or load of both plates and you get 5 bolts then have to use total 6 bolt (3 in each) But during calculation of this example, its taken as single shear as well only one plate load, so total 5 nos are required for only 1 plate hence total 10 nos of bolts ( 5 in each) is provided Hope you got an answer now.....
Will the second moment of inertia formula be same for inverted triangle?
Yes Its same as Centre of gravity never changed either you orient it in any direction
Thank you sir 😊
How to find the cg distance??
Dear friends.... CG distances for predefined shapes are known.... For that kindly visit CENTRE OF GRAVITY playlist
Where does -M come from? Why does M negate?
Dear friend.... As per Euler's theory The moment is taken as negative due to the sign convention assumed by Euler.... And M means the sectional Moment on the span as per the method
Easily explained. Thank you sir
You are welcome
Thannnkuu
It should be 3EI/4L
Dear Indraneel.... If you consider relative factors then its right but if you consider columns individual distribution factor than its as explained .. hope you got the point
sir in the question u have taken the load 15 but in diagram u have taken load as 30 the video is very good but I thought to tell u this
Dear Abhishek.... UDL load intensity is taken as 30kN/m and in moment its (30x^2)/2 so its 15x^2
Is possible for you to show an example with a continuous with an overhang cantilever and 2 pin supports please
Definitely Ms Ria.....lecture of the mentioned points will be covered soon in future plan
For Parabola c.g. distance = 3/4
Dear Heri.... Its symmetrical parabola... Hence CG is at axis of symmetry... If not symmetrical than its 3/4 or 5/8 as per parabolic curve
Sir your Zxx and Zyy values are wrong, Zxx= 4.5*10^6 and Zyy= 36*10^6 But my concept are clear by your video sir Thanks
Thank you Dear Kushal....it's already written in a description about value.... Keep learning
Exactly you have seen what I have seen
Dear Friends It has been mentioned that in description about that Depending upon value and its by mistake interchanged..... But happy for you all that you got the concept
thank you ✔✔✔✔
Welcome 😊
In second triangle...??? Why you take cg distance 1/3×2....it is 2/3×2 hoga....???
Dear Friend..... As per rules the CG distance for traingle from apex is 2/3 and its 1/3 of distance from side...... If you go through this, you will get 1/3 of distance for 2nd traingle...
Thank you sir 😊❤
Most welcome 😊
Really great sir