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Fb force resolved kartana Fx=Fbcos10 and Fy=Fbsin10 pahije

Could you please mention the book from where you took this problem

Many many thanks sir 🎉what easily you described ❤

E = 200 Gpa why u multiplied by 10^3

In your I section video, q2= q1 + L/0.5W, yet in this video, q2=q1+L/W, which one is correct????

Sir paper madhi hei question bank madhle question yeta na? Kasa te awgad zata baher ch question ala tr aple reply madadgar rahil amchya sathi

Moment of inertia is mm^ square hai

Derivation also plz...😊

Nice concept Btw music name plz

why do u divide by 12

Excellent teaching

Nice work

😍 THANK YOU SIR🙏

But i though the crippling load formula for fixed and free end column is divided by 4l^2 instead of le^2 ?

150 mm spacing can we provide?

My One Doubt Is Sir Why You Not Taken The Rb term in the summation of moment at b=0 It Is Also A Force It Will Be Also Coming In The Equation Naa ?

Radius of hollow shaft is 75 mm 150÷2

In joint A horizontal force is missing

Sir for q add the width is 50 or 250 Can u give me an clarity about this

Two blocks A and B of weight 100 N and 50 N are tied together by a rope parallel to the plane as shown in Figure 1.18. The coefficient of friction between block A and the plane is 0.2, and that between block B and the plane is 0.5. Determine the value of the angle θ at which sliding will occur. What is the tension in the rope?

A horizontal bar AB of length 4 m and weight 400 N is lying on inclined planes as shown in Figure 1.20. Find how close to the ends A and B a load of 500 N can be placed safely, if coefficient of friction between the bar and supports is 0.2. Facing problem with this numerical...pls explain

Thank you so much sir

i think the value of b for qadd above the NA should be 50mm or not ?

Sir please Hindi me kuchh video chodia please 😅

Thank You Sir, it's just that the Y and X axis the sine and cosine are been misplaced. But well comprehended. Your efforts are priceless you explained every things in details.

Thank you sir for solution 🙏

Ry =450 kha se aya sir

very good

Good work❤

🙏thank you sir

muy bueno

Sir how to find depth of beam if width is given and required to limit the max compressive bending stress os given please a solve a problem in this question

Sir how came , stress at impact load additional of p/a???

Thank you sir

For the second part of the question, if we take theta1/theta2 = T1IP2/T2IP1, then the answer happens to be different. Why?

This is good

❤❤

Sir please Upload fluid mechanics lecture sppu civil

🤗🌹

Sir please book ka nam bolaay with author

Sir why we converted kn into newton in this video

Sir in textbook e=80 is used formula 80 x 10 to power 3 i cant understand why this taken

Book ka nam

Each of the rods as shown below are of 6000 mm2 area at a temperature of 15 oC. Find the temperature at which the stress in the steel rod will be zero. Take 𝛂 for steel as 12 x 10 - 6 / 0C and 𝛂 for bronze as 18 x 10 - 6 / 0C. Young’s Modulus for steel is 200 GPa and for bronze is 80 GPa.

thank you very much

thanks sir 😍

What will be the stress at midpoint of each sides?

You deserve a gold medal for your clear explanations🎉

Nice 👍

Sir how to find thickness in Hollow circular giving rankines for mula