CORRECTION - At ⏰Timecode 07:18 Unknown reactions are four: two at the left pin support and one each at the middle and right roller supports. Hence, Static indeterminacy (SI) = U - E = 4 - 3 = 1.
I love your videos, I’ve been watching them in LinkedIn and testing myself. I have a query at around 8:32 into the video. Where did the extra unknown come from? I think there’s a possible miscalculation but I maybe wrong, could you check my working out below. There are 4 unknown reactions, 2 (1 vertical & 1 horizontal) at A and 1 vertical at both B & C. We have a pinned support and two rollers making 2 equations Sum of Horizontal and Sum of Vertical forces; both support types are free to rotate so there is no moment equation. Therefore, SI = U-E = 4-2 = 2, indeterminate. Considering the internal hinge (i) we get SI = U-(E+i) = 4-(2+2) = 0, determinate
I’m grateful to you for pointing out the mistake. Reactions will be 4 and equilibrium equations are always three. SI=4-3=1. I've issued following amendment, “CORRECTION - At ⏰Timecode 07:18 Unknown reactions are four: two at the left pin support and one each at the middle and right roller supports. Hence, Static indeterminacy (SI) = U - E = 4 - 3 = 1.”
@@DrJQureshi no problem, I was just working it out my self and then watched your solution, so wanted to check I wasn’t doing anything wrong. I take it that this system becomes statically indeterminate as the additional 2 internal hinges would change the SI = 4-3 = 1-2 = -1?
CORRECTION - At ⏰Timecode 07:18 Unknown reactions are four: two at the left pin support and one each at the middle and right roller supports. Hence, Static indeterminacy (SI) = U - E = 4 - 3 = 1.
Wonderful Sir! The way you teach has developed my interest in learning more. Keep up the great work, Sir.
Thanks for your kinds words. I'm here to help people and make complex engineering simple.
I love your videos, I’ve been watching them in LinkedIn and testing myself. I have a query at around 8:32 into the video. Where did the extra unknown come from? I think there’s a possible miscalculation but I maybe wrong, could you check my working out below.
There are 4 unknown reactions, 2 (1 vertical & 1 horizontal) at A and 1 vertical at both B & C.
We have a pinned support and two rollers making 2 equations Sum of Horizontal and Sum of Vertical forces; both support types are free to rotate so there is no moment equation.
Therefore, SI = U-E = 4-2 = 2, indeterminate.
Considering the internal hinge (i) we get SI = U-(E+i) = 4-(2+2) = 0, determinate
I’m grateful to you for pointing out the mistake.
Reactions will be 4 and equilibrium equations are always three. SI=4-3=1. I've issued following amendment,
“CORRECTION - At ⏰Timecode 07:18 Unknown reactions are four: two at the left pin support and one each at the middle and right roller supports. Hence, Static indeterminacy (SI) = U - E = 4 - 3 = 1.”
@@DrJQureshi no problem, I was just working it out my self and then watched your solution, so wanted to check I wasn’t doing anything wrong.
I take it that this system becomes statically indeterminate as the additional 2 internal hinges would change the SI = 4-3 = 1-2 = -1?
how unkowns become 5
Reactions are 4. I made a mistake at ⏰Timecode 07:18, apologies for that. I realised later and issued correction note above.
Very informative. It really helps to recall the analysis and calcs. Keep up sir.
I appreciate your kind comments.