You said earlier in this video that only moment can't be transferred across the hinge only axial and transverse forces can ! Then afterwards why you didn't consider 50kN transverse point load on other side of hinge for calculating Rc ? As that can be transferred across hinge as you said ? Explain me please 🙏🏻🙏🏻🙏🏻
Hi, when you consider the free body diagram of the portion BC, there is a force "BY". This BY captures all the load transfer in Y direction from portion AB. "BY" can be anything, that's why I am assuming that as an unknown. Since equilibrium equations can be easily solved for the BC portion, I went ahead to solve for BY after summing up forces in Y direction
@@conceptsinengineering but when you calculate Reaction at B in vertical from left side of hinge then you will get Bv = 25kN. And then by considering that 25kN of Bv if you calculate Rc then the answer would be 15kN instead of current ans 20kN !! ..... You have to consider that 50kN force... talking in other context, inspite of having infinite amount of force on left side of hinge (in place of 50kN) there will be no effect on Rc and will remain same (20kN) which is not possible !!!!! Please explain in detail 🙏🏻
@@darpanshah8841 I didn't understand your logic of arriving at the value of reaction force at B as 25 kN (CONSIDERING THE LEFT SIDE OF HINGE ALONE). If you have applied the logic of symmetry, then you are wrong. Say I am drawing the free body diagram of the left portion then sum up force in Y direction BY -50 kN+ (Force reaction from fixed end) =0 (Eqn 1) Now, you sum moments about fixed point, BY * 6 -50kN* 3 + (Moment reaction from fixed end) =0 (Eqn 2) Did you see three unknowns (BY, Force reaction from fixed end, Moment reaction from fixed end) , but I have only two equations. How can you confidently tell BY=25kN? To answer the latter part of the question, load always flows through the stiff portion of a structure.
We are computing the moment about the point B. The resultant of the uniformly distributed load is 40 kN (10 kN/m x 4 m ). This resultant force is acting at a distance of 2m from point B, so the moment due to this distributed load about point B is 40 kN x 2 m = 80 kNm. Hope this answers your query. Happy learning!
@@conceptsinengineering For the propped cantilever beam shown, compute the support reactions using the method of consistent deformations by taking the reaction at support B as redundant. Also, draw the shear and bending moment diagrams. Use deflection formulas to determine deflections at B. Assume that EI is constant along the beam. (there is diagram for it)
th-cam.com/video/4w1tOvEYD7A/w-d-xo.html I have tried to answer all your questions in this video. Please watch the full video, If you still have some questions post it here.
![EQUILIBRIUM: Reactions at the Internal Hinge of a Compound beam- Problem #3 [1/2]](/img/n.gif)
Great video, its hard to find simple explanations for hinge beams. Very helpful!
this kind of videos will help student like me for years.much love for you human
@5:31 where did you get -40kn x "2m"
Because of the 10kN/m times the 4m results in the middle of the load. That's why it's 2m (middle of 4m)
@@khanzeref4496 alright thank you
why Rc is not 65, from vertical force balance as By is coming 25
How can we solve a cantilever with roller support at the other end?
You said earlier in this video that only moment can't be transferred across the hinge only axial and transverse forces can ! Then afterwards why you didn't consider 50kN transverse point load on other side of hinge for calculating Rc ? As that can be transferred across hinge as you said ? Explain me please 🙏🏻🙏🏻🙏🏻
Hi, when you consider the free body diagram of the portion BC, there is a force "BY". This BY captures all the load transfer in Y direction from portion AB. "BY" can be anything, that's why I am assuming that as an unknown. Since equilibrium equations can be easily solved for the BC portion, I went ahead to solve for BY after summing up forces in Y direction
@@conceptsinengineering but when you calculate Reaction at B in vertical from left side of hinge then you will get Bv = 25kN. And then by considering that 25kN of Bv if you calculate Rc then the answer would be 15kN instead of current ans 20kN !! ..... You have to consider that 50kN force... talking in other context, inspite of having infinite amount of force on left side of hinge (in place of 50kN) there will be no effect on Rc and will remain same (20kN) which is not possible !!!!! Please explain in detail 🙏🏻
@@darpanshah8841 I didn't understand your logic of arriving at the value of reaction force at B as 25 kN (CONSIDERING THE LEFT SIDE OF HINGE ALONE). If you have applied the logic of symmetry, then you are wrong.
Say I am drawing the free body diagram of the left portion then sum up force in Y direction
BY -50 kN+ (Force reaction from fixed end) =0 (Eqn 1)
Now, you sum moments about fixed point,
BY * 6 -50kN* 3 + (Moment reaction from fixed end) =0 (Eqn 2)
Did you see three unknowns (BY, Force reaction from fixed end, Moment reaction from fixed end) , but I have only two equations. How can you confidently tell BY=25kN?
To answer the latter part of the question, load always flows through the stiff portion of a structure.
why is clockwise negative here?
how -40*2 has come please tell me you havent explained thatpart
We are computing the moment about the point B. The resultant of the uniformly distributed load is 40 kN (10 kN/m x 4 m ). This resultant force is acting at a distance of 2m from point B, so the moment due to this distributed load about point B is 40 kN x 2 m = 80 kNm.
Hope this answers your query. Happy learning!
@@conceptsinengineering how did you know resultant force is 2m from B?
When there's an internal hinge, are the loadings always equal to reactions?
Yup
what if both side have fixed support
this video has helped me a lot. but i have another question related to this. is it possible for you to show me?
Post the question here. I will reply
@@conceptsinengineering For the propped cantilever beam shown, compute the support reactions using the method of consistent deformations by taking the reaction at support B as redundant. Also, draw the shear and bending moment diagrams. Use deflection formulas to determine deflections at B. Assume that EI is constant along the beam. (there is diagram for it)
th-cam.com/video/4w1tOvEYD7A/w-d-xo.html
I have tried to answer all your questions in this video. Please watch the full video, If you still have some questions post it here.
Chutiya Hindi me batana
Thanks..... it was helpful
good
Thanks
so chaotic
Ay is 70kn