Force method example #2: two degrees indeterminate (part 1/3)

[@rosechoso]

I'm getting 1750 kn*m for the Moment at A during 7:55, not sure if its an error on my part but wanted to mention it.

[@Engineer4Free]

Ma = +1,750 kNm because Ma is drawn on the diagram in a counterclockwise sense. In other words, the reaction moment at A is positive 1,750 kNm based on the way we drew the free body diagram. The internal bending moment infinitesimally close to A has a magnitude of 1,750 kNm, but a negative sign. This is because if you were to draw a new free body diagram with a virtual cut infinitesimally close to A, regarding the moments, we would have Ma acting in the counterclockwise sense which we would draw on the left side of this new FBD, and the internal bending moment which we would draw on the right side of the virtual cut, would need to be equal and opposite, so a clockwise sense and a magnitude of 1,750 kNm. If we draw an internal bending moment on the right of a virtual cut that has a clockwise sense, that is opposite to the positive sign convention, so the sign must be negative. That is why it has a negative sign. I think the confusion probably comes from me using Ma to denote the reaction moment at A, AND the internal bending moment at A. Sorry about that, sometimes it just happens in these problems that the same letter will get used. If you need a refresher on the positive sign convention for internal bending moments, check out this video: www.engineer4free.com/4/internal-force-sign-convention

[@rosechoso]

I just saw this notification, sorry @@Engineer4Free. Thank you so much for responding and clearing that up for me, I got a little a confused, but I see it now. Have a happy new year!!

[@nomanaslam450]

Thank you sir, I have request you sir , Is their is other Example , in which beam is like that but uniformly distribution force is on fix side and point point load is on other side . Plz if any one know , give me link or picture or something for help . Plz give me link or something for that problem

[@Engineer4Free]

Hey check out engineer4free.com/structural-analysis I have a bunch of different examples there, also solved with several methods

[@greatpharoh0303]

How did you calculate are 5 and Xac for Area 5? Thanks again!

[@hongsonchu7431]

area for parabolic triangel is b*h/3 and Xac is 0.75b

[@mr.noname6109]

@@hongsonchu7431 thanks

[@hafizabdullah1312]

Can you solve the same example with unit load method?

[@Engineer4Free]

Hey, I don't think I have a video on that, you can check out all the beam videos I have at engineer4free.com/structural-analysis . Also videos 40-49 here engineer4free.com/mechanics-of-materials might also be helpful

[@collindainphillip3116]

When determining system determinacy what is the rule for counting the number of elements? Cause i'm counting 2 elements; 5

[@Engineer4Free]

Count all the unknowns. They are Ax, Ay, Ma, By, Cy. If we had 3 or less, the problem would be statically determinate, as we have 3 equations to work with in 2D statics (sum of force in x, sum of force in y, and sum of moments about a point). Each unknown beyond 3 is a degree of indeterminacy. So because we have 5 unknowns, we’ve got 2 too many for statics, so the problem is 2nd degree statically indeterminate.

[@behnammobaraki5091]

Hi everybody. Does anyone have the Matlab code of 'Force Method'?

[@shorove9956]

how did you solve the area 5 ?

[@hongsonchu7431]

area for parabolic triangel is b*h/3 and Xac is 0.75b

[@dr.engineering201]

Do videos on buoyancy and more on manometer problems, control volumes. Do Engineering degree level exam questions?. Phase change.

[@Engineer4Free]

I plan to make videos for all of those topics soon :)

[@jd_2625]

The problem of this method is u don't get the exact answer

[@nagasrisai3098]

Poor in structural analysis sir