Force Method for Indeterminate Structures - Intro to Structural Analysis

Learn how to calculate the reaction forces for indeterminate structures using the Force Method (sometimes called the flexibility method).
The force method is an intuitive way of computing reactions for systems with low degrees of indeterminacy. Supports are removed until the resulting structure is determinate, and then forces are applied at the removed supports to ensure that the displacements at those locations are zero, thereby satisfying the original constraints. These external forces are equivalent to the reaction forces.
For the two examples conducted here, the necessary displacements are available from common deflection tables. However, for more complex problems, the setup for the force method lends itself nicely to using the principle of virtual work (PVW) to find all the displacements. See here for how to compute displacements using PVW:
• Deflection of Frames u...

[@abdulrasheedabdul-lateef7054]

This is the best explanation of the force method. Thank you

[@gianlucacastro5281]

Such a well done video, thank you for it!

[@ahmedsalamaali4137]

Best explanation ever

[@StructuresProfH]

I'm humbled. Thank you!

[@mahindagamage3135]

You made life easy, Thanks

[@ralfjoshua9203]

Hi there professor, why are the displacements in the real system always negative? Would it be positive if I were to direct my virtual unit load in the same direction as the loads in my real system? Thank you very much.

[@StructuresProfH]

Yes. The signs of the displacements are relative to the directions of the virtual unit loads. Negative displacement means it moves opposite the direction of the unit load, and positive means in the same direction as the unit load.
If you reverse your unit load and repeat the problem, that will switch the signs for the real system displacements. Regardless of which way you point the unit loads, you should always find the same reactions from the force method, so the unit load directions is merely a matter of preference.

[@aiqnvc8883]

Thank you for your video! It is really inspiring!!

[@eliasvusi3451]

Thank you. Can someone explain why one deflection has unit of pbft^3 & the other only ft^3? I personally believe the reaction we are calculating through this method is a unit-less fictitious (scale factor turned into load).

[@StructuresProfH]

I apologize, the units are a bit weird here.
For the force method, normally we would calculate the deflection under the applied load with units of length and one or more flexibilities, which represents the deflection for a load of 1 (unitless) at one of the "removed" supports, with units of length/force.
In this video, I did not use values of modulus E or moment of inertia I because they were constants and would cancel out when applying the constraint equations. When you don't divide by E (force/length^2) and I (length^4), the deflection under applied loads looks like it has units of force*length^3 and the flexibility looks like it has units of length^3 (no force here because of that unitless load of 1). Regardless, when you put these back into the constraint equations, you will end up with units of force when computing the actual reaction. This is a real reaction with units that match the units you used for the applied loads (pounds, Newtons, whatever), not just a scale factor.

[@aibibiabuova764]

Thank youuuu

[@patrikengas6479]

hey, (i'm just taking a peek into the next semester curriculum), but one thing i dont get is why you use this method to find support reactions, are they initially unknown?

[@StructuresProfH]

Yes, the reactions are unknown. For an indeterminate structure, equilibrium alone is not enough to solve for the reactions, so that's why we need the force method (or other techniques for indeterminate structural analysis).

[@akramfadulalla2653]

I have been looking for an example for a fixed fixed frame. Third-degree determinacy with uniform loading. If you could do a similar example that would be great.
Thanks

[@muhammadtaha2401]

Where can I located the tables for deflection and flexibility?

[@alviniwenofu506]

Please can you share where you got it

[@brianmtshali4615]

Thank you!
Please do for indeterminate trusses

[@StructuresProfH]

Good idea. I'll put that in the pipeline for this summer.

[@allandionisio671]

sorry, I don't understand at 8:55 why is -51WL4/1152. where can i find tha value or that formula

[@StructuresProfH]

That is the deflection at the middle point of a cantilever of length L under a distributed load w. Formulas for simply supported beams and cantilevers are tabulated in a bunch of places, even Wikipedia: en.m.wikipedia.org/wiki/Deflection_(engineering)
(This wiki link has the formula as a function of x, but a simple substitution of x = L/2 should get the formula I used). Of course, you can always calculate these deflections using some other method, like moment-area if you wish - it’s just a lot more work.

[@paulpolycap2112]

A clamped beam AB of constant flexural rigidity is shown in Fig. 2.9a. The beam is subjected
to a uniform distributed load of
wKN / m
and a central concentrated moment
.
Draw shear force and bending moment diagrams by force method.
help with that

[@AyaulymB040]

Thanks!

[@jasonRhawt]

Are you able to upload the tables you used?

[@StructuresProfH]

I don't have one that's copyright free (I usually teach structural analysis from Hibbeler's textbook, and that has a table on the front cover). However, you can find typical tables just about anywhere. This website has most of the common cases: mechanicalc.com/reference/beam-deflection-tables

[@paulpolycap2112]

A clamped beam AB of constant flexural rigidity is shown in Fig. 2.9a. The beam is subjected
to a uniform distributed load of
wKN / m
and a central concentrated moment
.
Draw shear force and bending moment diagrams by force method.
help with that