# Moment Distribution Method - Frame

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Hello,

I have question is with applying Moment distribution method in calculating moments in a frame with two forces; one vertical & one horizonal. (Please see attached)

• No sway is assumed.
• It's a frame i.e; joints B & C are fixed.

My lecturer has included an example with moments created from both vertical & horizontal forces applied to frame.

Then he goes on to say (in an email) that the Moment distribution method can't be used to calculate horizontal forces, only vertical forces

I have spent time trying to email the Lecturer & Tutors about this problem and they won't even take the time to read my email(s) properly and look at attachments & answer my questions to give me any real clarity.

I would like to know (if anyone can tell me) if it's possible to solve the attached problem with two forces in a single Moment distribution method table?

Otherwise how can I solve it using this method?CCF_000013.pdf

I've created an excel spreadsheet and can't get the numbers to work.

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The lecturer has posted a solution to solving the problem using the above method with only a vertical force.

Problem solved.

But upon reflection, I looked at a textbook & the only problems (MDM) which include frames & horizonal forces took sway (or horizontal deflection) & principle of superposition into account to solve the problem.

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23 hours ago, psyclones said:

Hello,

I have question is with applying Moment distribution method in calculating moments in a frame with two forces; one vertical & one horizonal. (Please see attached)

• No sway is assumed.
• It's a frame i.e; joints B & C are fixed.

My lecturer has included an example with moments created from both vertical & horizontal forces applied to frame.

Then he goes on to say (in an email) that the Moment distribution method can't be used to calculate horizontal forces, only vertical forces

I have spent time trying to email the Lecturer & Tutors about this problem and they won't even take the time to read my email(s) properly and look at attachments & answer my questions to give me any real clarity.

I would like to know (if anyone can tell me) if it's possible to solve the attached problem with two forces in a single Moment distribution method table?

Otherwise how can I solve it using this method?CCF_000013.pdf

I've created an excel spreadsheet and can't get the numbers to work.

I only just saw this one, but +1 for coming back to report even though no one answered you.

Since you have solved the problem (?) I will confine my remarks to discussion about other methods.

First it is necessary to correct your error on the pdf.

You have interchanged reactions Ax and Ay on the FBD.

Putting this right, yes there are several other methods available.

First to recast the equations of equilibrium correctly

${A_x} + {D_x} = F............horizontal\;equilibrium.........1$

${A_y} + {D_y} = F............vertical\;equilibrium.............2$

$F\left( {\frac{L}{2} + \frac{h}{2}} \right) = {D_y}L....moments\;about\;A................3$

$F\left( {\frac{L}{2} - \frac{h}{2}} \right) = {A_y}L....moments\;about\;D................4$

${D_y} = \frac{{F\left( {L + h} \right)}}{{2L}}.....................................................5$

${A_y} = \frac{{F\left( {L - h} \right)}}{{2L}}.....................................................6$

Equations 5 and 6 are rearrangements of 3 and 4 to obtain expressions for the vertical reactions Ay and Dy

If you add 5 and 6 together you recover equation 1 (a good check) but this means that there are onlt 3 effective equations.

There are thus 4 unknowns (Ax, Ay, Dx, Dy) and only 3 equations so the frame is singly indeterminate.

As can be seen the lecturer's comment that you can calculate the vertical reactions from the given information, but not the horizontal ones is correct.

To calculate these you can use Castigliano's second theorem (also called the principle of least work) or use the slope-deflection method, integrating around the whole frame.

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