Determine the internal normal force, shear force, and moment at point C

Get the book: amzn.to/2h3hcFq
The beam weighs 280 lb/ft. Determine the internal
normal force, shear force, and moment at point C.

[@thenatmann1]

Hey, you usually do very well on these problems FinalAnswer, but on this one you kinda messed up, can you redo this one with the correct methods and solutions? I do still really appreciate all of your hard work and time and effort you put into these videos, they are usually spot on, I noticed that very rarely you mess up but we are just human after all.

[@michaelbabcock420]

fundamental part of chapter 7. when you take a segment the distributed load must be changed to fit that segment. if you take segment c-b it is 280(7)= 1960 lb at 3.5 ft.

[@andyh2045]

Thank you! this was really stumping me as to why it was wrong, but this helped! Thank you!

[@dariodedonato9126]

Replace the w in his equations with 1.96 kip (280*7=1960lb =1.96kip) and in the last moment equation, replace the 2 with 3.5. @Michael Babcock for the help

[@rickyspanish7587]

Nc=2.20 kip Vc=0.336 kip Mc=1.76 kip*ft

[@mroh5052]

solution manual: Nc=2.20kip : Vc=0.336kip : Mc=1.76kip ft

[@michaelbabcock420]

yea I got 1764 ftlb for mc as well. -1050(3)(4/5)+2800(3)(3/5)-840(1.5)(3/5) + MC =0 I used segment a-c. Fx= NC +1050(3/5)+2800(4/5)-840(4/5)=0 Fy= -VC -840(3/5)+2800(3/5)-1050(4/5)=0 BX=1050 NC= 2198 lb AX =1050 VC= 336 lb

[@yasmine4071]

I really like your videos, could you please solve some problems in chapter 7 about shear and moment diagrams ?

[@p09notankumar42]

Bro you forgot to take the weight of segment AC that is (280x3)=840\ lb=0.84 kip acting at the mid point of segment AC

[@lapsmfc0548]

Genio

[@syedyasirjaved2232]

Sir do you have viedo on chapter 8 problem