Axial Deformation Example with Self-Weight - Mechanics of Materials

[@RDunya]

Good Example. Thank You.

[@MrLazygrasshopper]

Thank you! Appreciate it very much. (= I hope you checked your PM to find a link to a video, having some interesting problems and variations to common problems. For ex, a truss with a predefined amount of movement allowed.

[@MsRoughman]

Relatively nice lectures.Keep up the good work. Can u please derive the deflection formula in case of a Conical bar for the effect of only "Self Weight". Thanks

[@Beyazhan1]

i don"t get the "w" thing. Previous example it was a triangular function. When we were trying to solve forces about the y-direction we took as fy = -20 - n (x) + 0.5 wx . x over here, like this example its rectangular. So do we have know its rectangular or triangular like do we have to ee that ourselves depending on the problem or in the question they share us that information.

[@BrunoPontoTxT]

The crosssection A at x = 0 is holding only the P Force, while at x = 0.5 it's holding P + half the weight and at x = 1 the Area is holding P + the entire weight so, ignoring the P force we should have a triangular w i might be wrong but i can't see why, any help?

[@semynrameshwarnath4397]

hi all. firstly, very helpful set of vidz. i am also having difficulty in understanding the N/m of the 38.5 that you calculated. i did some analysis of the dimensions and i got that the units would be kg/s^2, because the area is in m^2 which would cancell out the m^3 of the density. and i understand that because the area is not changing, we have a uniformly distributed load ( UDL ). the whole concept of how the self weight may vary w.r.t.the length needs to be made just a lillte bit clearer please. it would help alot. many thanks and keep up the great work.

[@dustinbrennan5271]

density is mass over volume. weight is mass times 9.8 m/s2. so weight is density times volume times 9.8. He messed up when he wrote density times area times 9.8. but because the length is 1m it just wasn't bothered to be mentiioned i presume.

[@prashikbhagat]

it should NOT be rectangular function of w....it should be tringular. let me explain as x increase the weight-force will increase gradually as we go up. how can it be uniform?????? plz correct me if i am mistaken

[@structurefree]

prashik bhagat it sounds like you are describing lateral fluid/earth pressure.

[@prashikbhagat]

I thought enough on that now i get it U R absolutely correct here. Thanks for correcting me....

[@wouterroos2359]

Can this be correct? By considering a differential element dx a distance x away from the bottom of the rod, I believe we yield a weight equation dependent on x. Then it cannot be a uniformly distributed load. By deriving an equation for deformation by self weight, with constant area, I yield the following equation: deformation = (total weight of the body)*(Length of the body)/2*(Modulus of elasticity)(cross sectional area of body) The two in the denominator implies a linear function. Not a constant function.

[@KurtRonaldEsteban]

Another approach which is formula-based is: The length of deformation (&) & = PL/AE + (Density)(9.81)(Length)(Length)/(2E)

[@jigsawz3438]

it would be great if you showed us which book you were using and gave easy/medium example to medium/hard example.

[@LosFSU]

good example. thank you

[@sheetanshutyagi6729]

Why didnt you assume the weight acting as a body force at the centroid instead ? Any specific reason ? Great video BTW. Love your work

[@gabenunez]

While these videos are beyond helpful I think it would be beneficial if you attached a file with some more challenging problems and there solutions for each topic.

[@rudryllontsi6398]

Thanks that was very helpful

[@0Sebek0]

thanks for this channel!

[@jasbirsarmal4449]

bro why did you left the graphs axial force vs length & strain vs length? they would have extremely helpful.

[@paulhendrikcortez5445]

why did you use A in w= PgA? Instead of w=PgV? Remember: P=m/v and w=mg; So w= PVg?.

[@anesmerazi603]

V=A(cross section) × L(length) , and L=1 meter so V=A

[@Yoaedn]

Thank YOu!! STRUCTUREFREEEE !

[@Beyazhan1]

have to see*** that

[@qpeueudjdh]

wrong!!!! calculation!! who treat self weight like that? nonsense

[@russellmanuba9859]

it should be 38.5N not 38.5N/m

[@structurefree]

it's a distributed weight since we multiplied by cross-sectional area, therefore the units are N/m. If we multiplied by volume, than it would just be a weight with units of N.