Thanks!🙏 You did a philanthropist work by your lectures on UA-cam for free. I almost watched every lecture of yours whether it is Fluid Mechanics or be it Strength of Material which is still treated as a Tough subject and Indeed it is, but with your guidance it won't be any more as of now. You nicely explained not only related topic but also cleared most of the doubts and warned about silly mistakes which students observe while understanding the concept and solving problems.
You prolly dont care but if you are bored like me during the covid times then you can stream pretty much all the latest movies on instaflixxer. I've been binge watching with my girlfriend during the lockdown =)
Location of centroid for top 70mm Moment of inertia 19106666.67mm4 Shear stress 0.62 3.76 4.42 No need to calculate for bottom because symmetry occur. In I section
At 38:21,why did you take Wl/2, और question में यह भी नहीं बताया कि कोणसे point पर का shear force लेना है (We could take middle point shear force also i.e zero but you took at the end sf value)
Sir app som ki lecture deflection moi and cg is topic e video jaldi upload krrega to bohut bohut help hoga.......app bohut ache se concept clear kar dete ho....pls sir pla....
Diploma in Mechanical Engineering ke 3rd semester main strain energy and concepts of buckling per lecture daliyega Sir apka lecture fully conceptual hai sir
Good evening sir. I attempted the question given by you. N values are. Y bar =70. I=10322666.66. Shear force just before joint = 1.16, after joint =5.812, n at NA =8.185. Please tell me sir is it right. Or there is any need to calculate for below section.
5) A bar of mild steel carries an axil pull of 10kN and transverse shear force of 5kN. Taking electric limit in tension as 240MPa, a FOS 3 and poisson’s ratio 0.3, calculate the diameter of bar according to (1) Maximum principle stress theory, (2) Maximum strain energy
Sir, aapke lecture ka bhut he siddat se wait karte hai hamlog...
Thanks!🙏 You did a philanthropist work by your lectures on UA-cam for free. I almost watched every lecture of yours whether it is Fluid Mechanics or be it Strength of Material which is still treated as a Tough subject and Indeed it is, but with your guidance it won't be any more as of now. You nicely explained not only related topic but also cleared most of the doubts and warned about silly mistakes which students observe while understanding the concept and solving problems.
I'll have a regret about why I didn't see your lectures 4 years ago.
All lectures of module 4, are so amazing sir..
It's still very very useful even in 2024...
Thank u so much...
That explanation of the reason for bent up bar provision was awesome!
All the lectures are superb and concept builder.
During the semester I missed your lectures but now I am enjoying ❤️
Sir, isi tarah lecture upload karte rahye because I am fully depend on your lecture..
You prolly dont care but if you are bored like me during the covid times then you can stream pretty much all the latest movies on instaflixxer. I've been binge watching with my girlfriend during the lockdown =)
@Kameron Bodhi yup, I have been using Instaflixxer for since december myself :)
Location of centroid for top 70mm
Moment of inertia 19106666.67mm4
Shear stress 0.62
3.76
4.42
No need to calculate for bottom because symmetry occur. In I section
Correct
Same here
I am getting different values I = 5673939.427
As b= 120 h= 20
b'= 20 h'= 100
A1= 2400 A2= 2000
y1= 10 y2= 70
@@vishalsinha9803 check it properly
@@surajkandari4065 does the value that i have taken is correct
As always great video Sir aiseyi upload krte rahiye ga
Very Nice presentation.
Thank you sir so much for uploading. Thank you. Please keep supporting us.
Sir thank you so much your way of explanation is outstanding sir Love from Assam
Tqq so much sir aapke sv module ki trh aapki explaining bhut hi achhi hoti h aapki vjh se ab pdai achhi hone lgi h
Sir aap bohot a66a parate Hain, please asahi videos upload karte rahie, we will support you
Outstanding teaching sir🙏
Excellent sir
Very nice explanation of concept
sir plzz start deflection of beam...love u sir
Bhut aacha lagta hai sir Nomrical
I'm proud your teaching sir. You are glamorous teacher
nice vedio also usefull for Amie☺☺☺
At 38:21,why did you take Wl/2, और question में यह भी नहीं बताया कि कोणसे point पर का shear force लेना है (We could take middle point shear force also i.e zero but you took at the end sf value)
Maximum shear force lenge beta
@@venugopalsharma1945 Tysm Sir For Som Subject
Thnx a million dear sir from my bottom of heart❤
Sir at right most plane of cantilever (exposed to udl), bending stress = MY/I = (wl^2)x/2i . Where x is distance of points from neutral axis .
Mja aa gya sir.
No doubt sir.
Hand's off sir🙇
Sir app som ki lecture deflection moi and cg is topic e video jaldi upload krrega to bohut bohut help hoga.......app bohut ache se concept clear kar dete ho....pls sir pla....
38:35 shouldn't you convert 10 kN x 2 m to 10 x 2 x 10³ x 10³ Nmm ?
So the answer of the Tau would be 791.855 N/mm² ?
Hellow bro at that place we take shear force value only then it is converted from KN to N (the solution is correct )
Zabardast concept sir❤️
awesome lecture ❤️❤️🙏
Thanku so much sir... 🙏
मै आपका बहुत बहुत आभार हू
Nice
I=19175414.66mm4
Shear stress just above joint=0.5822N/mm2
Shear stress just below joint=3.493N/mm2
Shear stress on neutral axis = 4.0169 N/mm2
Sir in this question if we take ybar from the bottom fibre then the value of neutral axis and moi is coming different
Diploma in Mechanical Engineering ke 3rd semester main strain energy and concepts of buckling per lecture daliyega Sir apka lecture fully conceptual hai sir
God bless you sir
Sir simply supported beam me mid pe jo bending stress a raha ha uska liya complimentary shear stresses ki value kaya hogi?
Sir 40:45sec. me aap y2= y1 kyu kie ho jab y1=35mm hai but y2 to sir 25ka half hona chahie na?
THANKS SIR 🙏🙏🙏
While calculating V ie WL/2 how is L equal to 2?
Sir yhi same lec Maine pd course purchase Kiya h usme h... Aur same chij yha free m aur pd course m v vhi chij 🙆
ty alot sir..aap na hote tosom mei fail hojate
Thanks a lot sir
Best for isro exam sir
Sir, bars are bent up for top reinforcement in slab or beam
I =19106666.67
SHEAR STRESS just above joint=0.6280 mm²
SHEAR STRESS just below the joint= 3.768mm²
SHEAR STRESS at neutral axis N/A =4.55mm²
Correct..👍
Mera bhi yahi araha hai
Sir aap ke mechanical ke sare leacture kaise milege ??? Bcz apke video ke liye bhot wait krna hota hai 😭😭😭
Good evening sir. I attempted the question given by you. N values are. Y bar =70. I=10322666.66. Shear force just before joint = 1.16, after joint =5.812, n at NA =8.185. Please tell me sir is it right. Or there is any need to calculate for below section.
wrong
Check my ans
😁😁😁fantastic
5) A bar of mild steel carries an axil pull of 10kN and transverse shear force of 5kN. Taking
electric limit in tension as 240MPa, a FOS 3 and poisson’s ratio 0.3, calculate the diameter
of bar according to (1) Maximum principle stress theory, (2) Maximum strain energy
Sir deflection ka class nahi hua he
Thanku sir
Sir parallel axis theorem nhi btaya. Moment of inertia of i section and t section ek baar fir samjha dijiye
Sir I think the value of I should be 6,040000mm
Sir MOI top or bottom se diffrent aayega kya ??
Ku ki "y bar" toh diffrent aayega
For rectengular secttion as well as circular section ,,moi is same for top nd bottom
👍
⭐⭐⭐⭐⭐
sir please lecture continue rkhie
sir deflection of beam kyu nhi padhaya aapne iss module mei??
Sir , agr hame mechanical ke sare subject ka lecture mil jaye then I will definitely crack the I.E.S exam..
Paid lectures lo
Sir, aapka lecture chahye hame, mil jayega?
@@vikasjha8547 bhai vahi bata raha hu pd course lele
Sir, in the T X-sc sum,the udl is given in N/m.Should we not convert it in N/mm?
No bcz length is in m so when we find load m will be cut by N/m.
@@yme2003 👍
🙏🙏🙏🙏🙏🙏🙏
❤️❤️❤️🙏
Thank u sir..
What abt U section
Structural analysis ki video series upload karyei Please.
Thanks sir
Sir please upload more lecture
sir when r u going to upload vibration.
Sir please bhes disiye mdule7 , sir bhut dino se wait kar rha hu...
I beam ka shear stress diagram 😳🍆
Gd after noon sir, Sir SSCJE MAINS ki Kuch class le lijiye because I do not understand, what I write and how much write. So please help me Sir
Please support sir. Exams in December first week.
Thank you very much sir
Thanks sir..
Thanku sir
Thanks sir
thank you sir
Thank u sir
Thank you sir