Problem on Principle of superposition |Simple Stresses & Strains | Strength of Materials | MOM | MOS

This video explains simple solution to "Problem on Principle of superposition".

[@kerenkanzemwarabu]

This is really amazing so easy to understand thumbs up Sir 👍

[@user-so2uc7vf5n]

How can i thank you sir!
Until now, even after watching thousands of videos, I didn't understand, but your video made me understand just in 17minutes, so thank you very much sir from Nepal🇳🇵

[@SampurnaEngg]

Glad to hear that

[@THANUSHR-mk6yb]

Excellent explanation 😮🎉🎉

[@siddumoosewala2160]

Thank you so much sir for clearing the concept

[@user-tl1gi5mn7j]

ua-cam.com/video/3K9SLiftDPE/v-deo.html

[@dramoyoedward9229]

Thanks alot for the wonderful explanation

[@user-tl1gi5mn7j]

ua-cam.com/video/3K9SLiftDPE/v-deo.html

[@kiran3118]

Thank you so much for your help sir ❤

[@enjoyyourtripwithnature8151]

Good explanation....

[@kingsleyscepter7558]

This really helped me

[@deepakdagar591]

Thanks for the nice explanation 😊

[@chandachouhan9560]

Thank you so much 🥰 sir

[@SampurnaEngg]

Most welcome

[@c-116ujjurilavanya6]

Chala bada explain chesaruu sirrr thank you so much ❤

[@SampurnaEngg]

Follow the Sampurna Engineering UA-cam channel on WhatsApp: whatsapp.com/channel/0029VaALnef7dmehZn80DV11

[@ferdousalam1366]

Sir, pls give Ans
A Steel bar long 1m ang 20mmx20mm is cross section is subjected to a tensile force 40Kn. Determine the elongation of the rod. If the modules of elasticity of the rod 200GPA

[@ranjan7089]

0.5 mm

[@nanasochaugule817]

0.5

[@Biranchiofficial-bp8sc]

Thank you so much sir ❤❤

[@babarrind6757]

Really good sir ❤️

[@sleazyR5573]

Will this process work in the case case of fixed ends??

[@SampurnaEngg]

No,it won't work.

[@shivadeore146]

Sir how area is pi/4 *25'2

[@bangalkids9596]

Portion BC has comprehensive force of 50 kn .. please clear it

[@bhagwaa__01]

No u r wrong it is not comprehensive kindly note the original question

[@BarelyEditing57]

Yes you r correct

[@dheerajdexterous8929]

Thank you Sir..🥰

[@ikennaagu568]

Please I'm not still clear on this. This particular question is from exercise 2.2 of Rajput strength of materials but in the text book the lengths are not evenly distributed like how it is in this video. Is it that we can make it evenly distributed ourselves when solving or what?

[@ikennaagu568]

*2.12

[@SampurnaEngg]

You can take the respective length for the respective section.
Like (PL/AE) where L is the length of the respective section.

[@sumanthnenavath]

Thank you sir 😚

[@ranjan7089]

Tnx sir

[@omegaarcex2527]

thank u sir

[@SampurnaEngg]

Follow the Sampurna Engineering UA-cam channel on WhatsApp: whatsapp.com/channel/0029VaALnef7dmehZn80DV11

[@collinsokello-2613]

How does that force 60 comes from

[@chuchuscast]

Equal and opposite force at the beginning

[@souravjena4982]

Sir compression stress me -ve hoga

[@SampurnaEngg]

Didn't got your point.

[@fieryscout2110]

Yes in compression your P becomes -ve but as in the given sum each part having tensile force acting on it thus the values are +ve

[@allinformation.3a854]

🙏

[@hardiksolanki831]

Giga means 10^9 ? Why sir take as 10^3 ?

[@SampurnaEngg]

Yes.
But if it's GPA then we have to convert it into N/mm2.
So, We multiply it by 10^3.

[@user-tl1gi5mn7j]

ua-cam.com/video/3K9SLiftDPE/v-deo.html

[@Yourfather781]

Sir my calculations is going always wrong plz guide

[@SampurnaEngg]

Where you are going wrong, kindly let us know.

[@ANKITMAHALA-q6d]

one gpa = 1000000000

[@oscarabdallah1883]

Bro unajua

[@user-sx2rk4bn7m]

Please I can't see the handwriting well oo 😞😢

[@anushav550]

Thank u sir

[@SampurnaEngg]

Welcome