Hi there
I get some results from the balance moments
Now I need to calculate the carry over moments by taking one-half of the balanced moment at a joint and transferring it to the far end of the member connected at that joint. The carry-over moment has the same sign as the balanced moment.
It is it should look like this table
I have to repeat these steps, first calculating the balance moment and then the carry over moment unitl I get zeros in all rows ... and then calculate the final moment by adding all the moments in each joint.
Like this table :
Can you show me how to do that in the attached sheet ?
Thanks
Yusra
Solved! Go to Solution.
Hi Yusra
It is now morning and I have checked my sheet. Two small alterations are required.
Terry
Hi,
Can you show the topography of the beams in your table example?
It does not appear to be one long continuous beam?
Cheers
Terry
hi there
Thank you for replying
the beam looks like this
all I need is to make the FEM moment in the left of the joint to be the right moment of the next step
for example I need to use half of FSM AB as the carry_over_ BA ..
carry_over AB= FSM A / 2
carry_over BA=(FSM AB)/2
carry_over BC= FSM BA / 2
carry over CB = FSM FSM CD /2
Hi,
Enclosed is a file that does moment distribution on the beam.
It is set up with pinned ends on the beam and needs little update for fixed ends.
Checked against other software and it works.
Cheers
Terry
Hi Terry
Thank you for your reply . I will check it right away
Yusra
Hi Yusra
It is now morning and I have checked my sheet. Two small alterations are required.
Terry
Hi Terry ,
Thank you for checking that again . Can I ask you something ?
Is this swiping the data of it is taking every scond number in Bal
Thanks
Yusra
Hi Yusra
Matrix "Bal" is where the distribution is assembled. Each column of "Bal" is either a balance or a carry over so "j" steps every second column one a balance and one a carry over. The carry over of 0.5 happens every second column and balance happens every second column. j*2+2 carry overs the previous j*2+1 column. Column one of "Bal" is the distribution factors, Rest of columns are moments as the distribution happens
Rows of "Bal" are along the beam. Two values for each support including the end supports. So with seven spans there are eight supports. Each sopport has a left hand side and right hand side for each support so 16 values. This means each span has a left hand moment being the right hand moment of left support and a left hand moment of right support. In carry over the rows representing the left and right hand ends of the seven spans are carried over. So there are two carry over statements in the carry over function for each span. Every second row of "Bal" not counting the first row is a left hand end of span.
Every second row of "Bal" not counting the first two rows is a right hand end of a span. (i-1)*2+2 is the left hand end of spans, (i-1)*2+3 is the right hand end of spans. i is from one to seven for the seven spans.
Cheers
Terry