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## Statically indeterminate beam (fixed ends) and the moments at the supports

I'm working through my professor's solution for this problem, and I don't understand how he comes up with the reaction force at B without taking into account the moment at B.

Any help would be greatly appreciated.

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I think I've been able to reason it out:  The moment at the right support is a reaction to the bending moment at that support, and the two are equal and opposite, therefore having no affect at the left end. Compare this to the moments caused by the forces acting between the supports. These forces are supported by both ends of the beam, therefore their moments affect both ends of the beam.  I'm sure I could have worded that better, but I think I got the gist of it.

Well I make MA = 8820kN-m and MB  = 9420kN-m  So the end moments are not equal.  Can you tell us more about these notes?

## Statically indeterminate beam (fixed ends) and the moments at the supports

When it comes to statically indeterminate, its harder. Solving the differential equations in Mathematica, I get: $$R_a=6095/72=84.6528...$$ $$R_b=6145/72=85.3472...$$ $$M_a=2165/12=180.417...$$ $$M_b=-2215/12=-184.583...$$ The force equation is seen to be solved since $Ra+Rb-20-30-10\times12=0$ and the torque equation is seen to be solved because $Ma+Mb-3\times 20-8\times 30+Rb\times 12-10\times 12\times 6=0$. Here I use counterclockwise torques as positive, and upward forces as positive, and I calculate the torque about x=0.

Since the problem is statically indeterminate, you cannot use the force and torque equations to solve for the four unknowns. Looking at the solutions in the notes, the Ra and Rb given (85) are close to the actual solutions, so I tend to think that some approximations have been made in the solution, but I don't know where. I agree with the OP, the exact solution cannot be obtained without taking into account the moment at point B.

Since this is important and not homework I have written it out at length in the attachments.  I think that there was more to the professor's presentation which is why I asked about this.  First Fig1 shows a fully restrained general beam with loads P1 and P2 and length L.  As a result of the loads it is subject to shears VA and moments MA at end A and similarly at end B.  In general MA ≠ MB and VA ≠ VB  It is vital to realise that these shears are different from the reactions attributable to a simply supported beam.   Fig 2  breaks the loading and support down to two simply suported beams which are added by superposition to yield the original conditions.  The first beam in Fig 2 offers two simple reactions R1 and R2 that resist loads P1 and P2.  They can thus be calculated by the ordinary rules of mechanical equilibrium as done at the end of Fig6 in the second attachment.  The second beam in Fig 2 offers end moments MA and MB. The difference is balanced by a couple formed from the extra reactions R and the lever arm of the length of the beam.  Fig3 shows the equations that result.  Now there are standard tables of 'fixed end moments' and Figs 4 and 5 show two extracts relevent to the OP beam loadings.  In Fig 6 I have addressed the OP loading case and used the standard fixed end moments to calculated MA and MB  Using the difference I can then calculate R.  The final calculation is to calculate R1 and R2 and add and subtract R as appropriate to obtain the end shears.  It is interesting to note that R1 = R2 = 85, the figure from the professor's calculation, so I suspect that which was presented in post#1 was this calculation only and the rest is missing. 		 		  		 		 			  			 				 					Attached Thumbnails

Oh dear this is what happens when you rush somehing to post without proper editorial checks.  Some arithmetical errors crept in.    The method is sound, however.  I have uploaded a corrected Fig6 and calculations. These confirm The results by Rap, as did working the dif equations by hand.  I have also just re-enroled in kindegarten arithmetic class - sums. 		 		  		 		 			  			 				 					Attached Thumbnails

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