sorry...sometimes, i use BISAYA if i want to emphasize something....but if you want to ask something about the presentation, feel free to comment...i will try my best to explain it to you...thank you...
This is just to get your confirmation. At 12:50 mark, is the anti-clockwise rotation of 1.0 at point C' a radian unit? If it's radian, why the vertical distance from point C down is 6.0? Is it not be 7.56? Thanks in adavance for the reply.
yes sir...it's one radian...for very small angles sir, tangent of theta is approximately equal to sine of theta and approximately equal to theta..and since the tangent of theta is equal to rise over run, then 1 = y/6...which gives the value y=6 (ordinate of the influence line).
@@besfrenjohny Sir, is this what you mean? tan theta = rise/run = y/6, and tan theta = 1 radian, thus, 1 radian = y/6, y = 6? I thought that to get the theta (or the angle) it's tan (1radian x π/180) = tan 57.2958°. Therefore, y = 6 tan 57.2958° = 7.56
@@gnidnoeled786 same din sa moment area method sir...like for example simply suppurted beam...when you want to determine the slope( theta) of the support, it's rise over run...or deviation over distance...hindi na mg inverse tangent para makuha ang theta..
sometimes, I speak in our local dialect here in Mindanao, Philippines if I want to emphasize something hoping that my students can understand better some of the technical terms if ever I would use the local dialect. Feel free to ask question(s) if a portion of the video is not clear to you.
I commend you for a very nice presentation. However, there seems to be a disagreement of the I.L. for the shear at C. Please note that as the unit load travels from C to D, the value of the shear at C vanishes to zero due to the fact that when the travelling unit load coincides with the support at D, all vertical forces will now be fully absorbed by the reaction at D. Hence, a rectangular I.L. is not possible. In fact, when the unit load is positioned between C and D, shear at C is zero since Rb=0. Rb must be zero because C is a hinge and cannot resist moment.
@@gnidnoeled786 Yes sir...there is no hinged support...only two roller supports and a fixed support...about rotation, both hinged and roller supports cannot resist moment therefore it will rotate...regarding displacements, only the roller support allows it in one direction only.
If there is no hinge at E, the influence line for the moment at D will look like this: For segment ABC, the line has a negative slope with ordinates 5 @A, 0 @B, and -8 @C; for segment CD, the line has a positive slope with ordinates -8 @C and 0 @D; and for segment DEF, the line is horizontal with zero ordinate for the three point.
if hindi fixed ang F, magiging unstable cya...but know that ir you are using Muller Breslau, only the support that you consider will be displaced by one unit...the other supports same position lang.
so far, kana ra ang naa sa video...pero during online class, naay other problems i-solve....need raa man sabton ang principle....makuha na dayon nimo maski unsa ang arrangement sa support....naa ka channel? pls help promote my channel..basin maka-heelp ni sa uban ce students...
Thank you so much po para dito. Kahit di ko naintindihan ung dialect na gets ko parin ung concept nitong ILD.
sorry...sometimes, i use BISAYA if i want to emphasize something....but if you want to ask something about the presentation, feel free to comment...i will try my best to explain it to you...thank you...
Thank u kaayo sir kay clear jud kaayo imong pagpasabot.🙌🏽
thank you...
salamat kaayo sir nasabtan kaayo ang priciples og techniques na gitudlo nimo.
you are welcome...
Thank you. Mas nakasabot ko sa influence lines.
thank you...
This is just to get your confirmation. At 12:50 mark, is the anti-clockwise rotation of 1.0 at point C' a radian unit? If it's radian, why the vertical distance from point C down is 6.0? Is it not be 7.56? Thanks in adavance for the reply.
yes sir...it's one radian...for very small angles sir, tangent of theta is approximately equal to sine of theta and approximately equal to theta..and since the tangent of theta is equal to rise over run, then 1 = y/6...which gives the value y=6 (ordinate of the influence line).
@@besfrenjohny Sir, is this what you mean? tan theta = rise/run = y/6, and tan theta = 1 radian,
thus, 1 radian = y/6, y = 6?
I thought that to get the theta (or the angle) it's tan (1radian x π/180) = tan 57.2958°. Therefore, y = 6 tan 57.2958° = 7.56
@@gnidnoeled786 no sir...hindi na mg tangent... its's simple 1 = y/6...
@@gnidnoeled786 same din sa moment area method sir...like for example simply suppurted beam...when you want to determine the slope( theta) of the support, it's rise over run...or deviation over distance...hindi na mg inverse tangent para makuha ang theta..
I don't know what language this is but its very strange because he reads the text in English but speaks in something else
sometimes, I speak in our local dialect here in Mindanao, Philippines if I want to emphasize something hoping that my students can understand better some of the technical terms if ever I would use the local dialect. Feel free to ask question(s) if a portion of the video is not clear to you.
Salamat kaayo sir!!
thank you also ..
I stand corrected, Sir. I thought it was at C. Good job!
no problem sir...God bless...
Salamat kaayo sir, galibog ko unsaon pagkuha sa ordinate sa shear pero dali ra kaayo diay 17:00
welcome...
Sir kung magiging engineer ako soon malaki po tulong nyo. salamat sa lahat sir.💖
Claim it! By God's grace....
I commend you for a very nice presentation. However, there seems to be a disagreement of the I.L. for the shear at C. Please note that as the unit load travels from C to D, the value of the shear at C vanishes to zero due to the fact that when the travelling unit load coincides with the support at D, all vertical forces will now be fully absorbed by the reaction at D. Hence, a rectangular I.L. is not possible. In fact, when the unit load is positioned between C and D, shear at C is zero since Rb=0. Rb must be zero because C is a hinge and cannot resist moment.
thank you for your comment sir...but it's notthe shear at C that is involved in the presentation but " C' " which is 6m to the right of C....
Pinned and roller supports allow rotation only? Hinged connection allows up and down vertical displacements?
what do you mean? what part of the video are you referring to?
@@besfrenjohny This video is showing what I'm trying to get a confirmation. There's no hinged support in this video, though.
@@gnidnoeled786 Yes sir...there is no hinged support...only two roller supports and a fixed support...about rotation, both hinged and roller supports cannot resist moment therefore it will rotate...regarding displacements, only the roller support allows it in one direction only.
Thank you, sir.
you are welcome ..
ty sirrrrr❤❤❤❤
welcome...
In which language r u teaching
local dialect in the Philippines..
Bisaya ni bay di mo kasabay
hi sir ask lang, condition na sir kanang distance nga 6m to the right of C og 2m to the right of D?
oo...part sa illustration sa Muller-Breslau...
@besfrenjohny Thank you sir
If there is no hinge at E then how Bending moment at D ,ILD will be
If there is no hinge at E, the influence line for the moment at D will look like this: For segment ABC, the line has a negative slope with ordinates 5 @A, 0 @B, and -8 @C; for segment CD, the line has a positive slope with ordinates -8 @C and 0 @D; and for segment DEF, the line is horizontal with zero ordinate for the three point.
@@besfrenjohny thank you
you are welcome sir...
Sir, paano po ang influence line ng Reaction B kung hindi fixed yung point F? Sasama po ba yung CDEF sa pagtaas? thank u po sa pagsagot
if hindi fixed ang F, magiging unstable cya...but know that ir you are using Muller Breslau, only the support that you consider will be displaced by one unit...the other supports same position lang.
sir, maayong buntag. applicable ba ni siya sa tanan beams? or naa ni limitations?
pang stat. determinate beam lang
Muller-Breslaue Principle plus neri's technique ang title dapat sir hehehe
hahaha...ayaw ky mosikat unya ta....joke...
Hehehe mas dali man imo technique kesa books sir
@@kennquilang5307 eeehhhhh.....tinuod na?
@@besfrenjohny oo kaau sir
Naa pakay lain example sa influence line sir?
so far, kana ra ang naa sa video...pero during online class, naay other problems i-solve....need raa man sabton ang principle....makuha na dayon nimo maski unsa ang arrangement sa support....naa ka channel? pls help promote my channel..basin maka-heelp ni sa uban ce students...
@@besfrenjohny ouh sir e promote ni nako sa akoa mga classmate nga CE pud
tnx...if naa mo querry about influence lines, comment lng...maybe i can explain sa comment ra pod....
Language barrier 🚧
sorry...this video was originally intended for my students....anyway, you can click the caption/subtitle....hopefully it can help you....thank you...
thanks po!!!!
you are welcome....i hope you were able to understand the way i explained the topic...
What language is this?
a local dialect here in the philippines...
Sir kala ko po arabo to hahahaha nabigla ako hahahaha nag tagalog hahahahagaha
Maayong gabii sir. Giunsa pagkuha sa -5 @E sa pag moment I.L. nimo sa F sir. Salamat.
1 rad man gd ang rotation (can be considered a slope ana nga figure)..and since 5 ang adjacent side, 5 pod ang opposite side...
@@besfrenjohny Thank you kaayo sir! tabang kaayo imohang discussion. dali rajud masabtan. hehe
Is this not english? I am so confused
local dialect in some part....do you have any questions sir?
Lmao 3:00
wtf is this language