yes, however because the units are directly related you can convert before or after the algebra. the answer comes out the same on either side of the math. 60 ft lbs is equal to 720 in lbs.
If you had an added torque of 300 going the opposite direction as the 500 applied somewhere in the CD section, would the first equation then be Ta+Tb=200? Also, would the twist equation for section DC then be Td+300?
i dont think plus. because TA and TD need to both offset the 500. as 500 is the only external T on the diagram. so it would be TA = 500-TD i believe. I noticed you just commented and ima be around. if u wana talk about it lmk
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Wonderful solids lessons!! really taking the time to explain the problems along with a good sense of humor :))
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but the 500 is in ft and the rest is in 'in', right? don't you have to convert ft to in for that 500??
yes, however because the units are directly related you can convert before or after the algebra. the answer comes out the same on either side of the math. 60 ft lbs is equal to 720 in lbs.
5:15 why is T(BC) = T(A)?
Wouldnt it be 500 lb*in since the T(A)s cancel out on the left?
If you had an added torque of 300 going the opposite direction as the 500 applied somewhere in the CD section, would the first equation then be Ta+Tb=200? Also, would the twist equation for section DC then be Td+300?
ı cant understand how we can write the 12 in/ft on shear stress equation?
Nice
Can u solve problems from the hebbler book ?
professor, should not section CD's torque be ===> TA = TD+500, I am cutting with respect to A. I believe TD is 122.24TD +30559 + 7.35TA = 0
i dont think plus. because TA and TD need to both offset the 500. as 500 is the only external T on the diagram. so it would be TA = 500-TD i believe. I noticed you just commented and ima be around. if u wana talk about it lmk
Hehe