what is the force Fp up at the pin?
Fp(4/5)L = q L(L/2)
Fp = q L (5/8)
so the force up at he roller on the left is
Fr = q L - (5/8) q L = (3/8) q L
in the left sector
shear= (3/8) q L - integral from 0 to x of q dx
= (3/8) q L - q x
moment = integral of shear * dx
= integral [ (3/8) q L dx - q x dx ]
= (3/8) q L x - q x^2/2 + constant
when x = 0 bending moment must be 0 so constant is 0
so look for max of that
where (3/8) q L - q x = 0
x = (3/8) L
and there the BM is
(3/8) q L x - q x^2/2
= (3/8) q L(3/8) L - q (9/64)L^2/2
= (9 /128) q L^2
check my arithmetic !
A hollow cylinder with outer radius r_0 and inner radius r_i is loaded by a uniformly distributed load q across its length L. The beam is supported by a roller at its left end and a pin L/5 from its right end.
What is the maximum bending moment in the beam? Express your answer in terms of q and L.
5 answers
Hey thanks for your guidance this is correct.
Can you help me with the step where you did:
moment = integral of shear * dx
= integral [ (3/8) q L dx - q x dx ]
why does the derivative of x just become x?
moment = integral of shear * dx
= integral [ (3/8) q L dx - q x dx ]
why does the derivative of x just become x?
My apologies for separate messages.. but I'm also trying to figure out how the second term has a 2 in the denominator now:
M = (3/8) q L x - q x^2/2 + constant
M = (3/8) q L x - q x^2/2 + constant
Now that it's 2 days later than when you first posted this, I doubt the tutor who answered you will see this again. Please post your newer questions by using Ask a New Question, so the math tutors will see what you need help with. You might also include the following partial URL in case they need to refer to your initial question:
www.jiskha.com/questions/1822768/a-hollow-cylinder-with-outer-radius-r-0-and-inner-radius-r-i-is-loaded-by-a-uniformly
www.jiskha.com/questions/1822768/a-hollow-cylinder-with-outer-radius-r-0-and-inner-radius-r-i-is-loaded-by-a-uniformly