Force Method for One Degree of Indeterminacy
Contents
Introduction
The Force Method allows for the exact analysis of externally and internally indeterminate structures. The following sections will explain the analysis of an indeterminate structure to the first degree, meaning one unknown reaction force will be considered as the redundant.^{[1]}
The Value of the Force Method
The Force Method, or Method of Consistent Deformations as it is sometimes called, is a valuable tool that can be used to solve systems (frames, beams, trusses) that are indeterminate.^{[2]} With this notion in mind, we must realize that at the end of the day, this method is just a tool. We need to have prior structural analysis knowledge in order to use this tool effectively. In the first lecture of CIVE 3203 Structural Analysis, Professor Erochko reiterated the statement from Professor M.P. Collins of the University of Toronto, that: "To find the answer, you must know the answer."^{[3]} This statement holds incredible value in the case of The Force Method because in order to solve a force method problem (and get the correct answer), you must be proficient with the skills that you have learned beforehand to get a general sense of what the solution to the problem will be. These skills include: Understanding what makes a structure indeterminate or determinate (Determinacy, Indeterminacy and Stability), knowing how to properly draw shear and bending moment diagrams, and the principle of Virtual Work (Truss Virtual Work, Beam Virtual Work, Frame Virtual Work).^{[2]}
Important Terms
Terms | Definitions |
---|---|
Indeterminate | A structure is called indeterminate when it has more than 3 unknowns(or more than 6 unknowns in 3-D)and it is not possible to solve for the unknowns.^{[2]} (See Determinacy, Indeterminacy and Stability) |
Redundant | When having an indeterminate structures, the extra number of unknowns is called Redundant.^{[2]}(See Determinacy, Indeterminacy and Stability) |
Compatibility | Compatibility provides extra equations to solve for unknowns in an indeterminate structure.^{[2]} (See Equilibrium and Compatibility) |
Beam Deflection Equation | Deflection of the beams can be calculated using Beam Deflection Equations.^{[2]} More information can be found at: http://www.clag.org.uk/beam.html |
Flexibility Coefficient | A deflection at a point (e.g. i ) due to unit load of another point is called Flexibility Coefficient and it is denoted as $ Fij $.^{[2]} |
General Steps
This method was adapted from the Kassimali Textbook^{[4]} and Prof. Erochko's 14th Lecture^{[5]}
Step 1: Given an indeterminate beam choose one of the unknown reaction forces or reaction moments to be redundant. If removed this redundant force must leave the structure stable. For the beam below Cy was selected as the redundant
Step 2: Remove this redundant force and apply external loads. There should now only be three unknown forces which can be found using the equations of equilibrium.
Step 3: Now set up the compatibility equation. For the redundant Cy the compatibility equation is: $ \Delta_C= \Delta_{CO} + f_{cc} C_y= 0 $^{[6]} If a reaction moment, such as Ma, is selected as the redundant the equation would be: $ \Theta_A= \Theta_{AO} + f_{aa} M_a= 0 $^{[7]} These equations are equal to zero because at a support reaction there is no deformation or rotation.
Step 4: After solving for the 3 unknown forces, draw the shear and bending moment diagrams for the beam.
Step 5: Next return to the primary beam and solve for its virtual system.
Step 6: Draw the shear and bending moment diagrams for this virtual system.
Step 7: Next return to the primary beam again and remove all external loads. Then add the redundant force (selected in step 1) as a unit load (1kN or 1kNm). **NOTE: this beam is identical to the virtual system and therefore has the same bending moment diagram.
Step 8: Using the three bending moment diagrams unknowns in the compatibility condition can be found.
Step 8a: $ \Delta_{CO} $ (the deflection at C) is found by integrating the product of the primary structure's moment diagram (without the redundant) by the virtual system's moment diagram. This can be done using the integration tables or by direct integration. Note that if the reaction moment Ma was chosen as the redundant then the product of the two moment diagrams would give $ \Theta_a $ instead of $ \Delta_{CO} $
Step 8b: $ \ f_{cc} $ (the flexibility coefficient) is found by integrating the product of the virtual system's moment diagram by the primary structure with a unit redundant load moment diagram.
Step 9: By rearranging the compatibility equation the unknown reaction force can now be solved. $ C_y= - \frac{\Delta_{CO}}{{f_{cc}} $^{[2]}
Helpful Hints when solving Force Method Problems
This is a list of some helpful hints and tips that might come in handy when solving a force method problem:
-Establish a coordinate system at the beginning of each problem in order to set your positive and negative directions. This will ensure that you do not get your directions mixed up in the middle of the analysis, which could lead to incorrect reaction forces and shear and bending moment diagrams.
-Remembering that the moment arrow ALWAYS points towards the compression side of the beam when doing moment diagrams.^{[8]}
-Knowing how to determine the equations of shear and moment diagrams for the use in your virtual work calculation.^{[2]}
-Remembering how to properly use the virtual work integration table. This table can be found at: [1]
-The virtual system that you initially solve for the primary beam will also become your Real system and your Virtual system for the unit load redundant analysis.^{[5]} This essentially means that you only have to solve two separate systems in your analysis instead of four. Remember the pairings for virtual work this way: Primary goes with Virtual, and Real Virtual goes with Virtual Virtual.
-Remember to breathe when solving these problems! Sometimes they can be stressful to solve on a test or exam. Take your time and do not rush through the problem. If you do not understand the topic right away (hopefully you will gain a firm understanding of the force method after reading our wiki page), then move on to a different question and then come back to the one you struggled with later on in the exam.
Example Problems
Listed below are two example problems. The first one is an easier problem in order to get acquainted with the material. The second problem is more advanced because it is a frame, and has a distributed load over the angled member of the frame.
Example One
For the beam below, determine all the reaction forces.
We initially have to confirm that the beam given has one degree of indeterminacy. We know that the number of reactions of this beam is r=4, and since we only have three equilibrium equations, the degree of indeterminacy of the beam is given by r-3=1 Therefore we can conclude that the beam is one degree indeterminate.
Now following the steps given previously, we can set up our equation of compatibility. In this example we will choose the reaction at point D as our redundant. Therefore, we can write the equation of compatibility as:
$ \Delta_{DO} + f_{DD} D_y= 0 $
Now that we have our equation of compatibility and our redundant force is chosen, we can initially determine the deflection at D due to external loading.
Therefore, by using the deflection formula, we get:
(note that in this example using the deflection formula is a much easier method to find the deflection at point D. In the next example, virtual work integration table will be used)
$ \Delta_{DO}= -\bigg(\frac{P_2\times\(L_2)^3}{3EI}+\frac{P_2\times\(L_2)^2}{2EI}(L_1)+\frac{P_1\times\(L_1)^3}{3EI}+\frac{P_1\times\(L_1)^2}{2EI}(L_2)\bigg) $
$ \Delta_{DO}= -\bigg(\frac{80\times\6^3}{3EI}+\frac{80\times\6^2}{2EI}(3)+\frac{40\times\3^3}{3EI}+\frac{40\times\3^2}{2EI}(6)\bigg) $
$ \Delta_{DO}= -\frac{11520}{EI}\ kN.m^2 $
Now that we have our deflection at point D, in order to find the flexibility at point D, virtual work will be used and the primary beam will be subjected to a unit load in the same position and direction as reaction D.
Therefore, by using the deflection formula, we get:
$ f_{DD}=\frac{1\times\ 9^3}{3EI} $
$ f_{DD}=\frac{243}{EI} kN.m^2/m $
Now that we have found the deflection and flexibility at point D, we can substitute them back into our compatibility equation and solve for the reaction at point D.
$ \Delta_{DO} + f_{DD} D_y= 0 $
$ -\frac{11520}{EI} + \frac{243}{EI} D_y= 0 $
$ D_y= \frac{\frac{11520}{EI}}{\frac{243}{EI}} = \frac{11520}{243} = 47.41 kN $
Now that we have the value of the reaction at point D, we can solve for all other reactions.
$ \sum {A_x} = 0 $ ; $ A_{x} = 0 $
$ \sum {A_y} = 0 $ ; $ A_{y} + 47.41 - 40 - 80 =0 $ ; $ A_{y} = 72.6 kN $
$ \sum {M_A} = 0 $ ; $ M_{A} -40(3) - 80(6) + 47.4(9) =0 $ ; $ M_{a} = $ $ 173.4 kN.m^2 $
Example Two
For the frame below, determine all the reaction forces.
We have 4 unknown reactions to solve for:$ A_y , A_x , C_y , C_x $
Therefore, we have to choose one of these forces as our redundant. However, the structure still has to be stable.
For this example we will choose $ Ax $ as our redundant force that we have to solve for,
Because we are choosing $ Ax $ as our redundant, we have to create a real system without $ Ax $. This essentially turns the pinned support at $ A $ into a roller support.
We first need to solve our reaction forces:
$ \sum{F_x}=0 $ ; $ C_x=0 $
$ \sum{M_c}=0 $ ; $ (18)C_y - 15(8)(4)=0 $ ; $ C_y = 26.7 kN $
$ \sum{F_y}=0 $ ; $ A_y + c_y - 15(kN/m)(8m) $ ; $ A_y = 93.3 kN $
Rough view of FBD of frame:
The next step in our analysis is to determine the shear and moment diagrams for our real system. Since the frame member $ AB $ is on an angle, we have to determine the forces on the member relative to a new horizontal axis.Therefore, this means that we have to split up our forces into parallel and perpendicular components that act on member $ AB $.
We will create a new member called $ A'B' $ that will show those forces.
$ L=\sqrt{ 8^2 + 7^2} =10.6m $
Symbols
$ \parallel_w $ : parallel part of w
$ \perp_w $: perpendicular part of w
To find the perpendicular component of the distributed load, we use the formula:
$ \perp = \frac{w h cos(\theta)}{L}\Longrightarrow $ $ \frac{-15(kN/m)(8m)(Cos41.2)}{10.6m}= -8.52kN/m $
Now, for the parallel component, we have :
$ \parallel = \frac{-15(kN/m)(8m)(Sin41.2)}{10.6m}= -7.46kN/m $
To determine the perpendicular component to our point load $ A_y $, we use the equation:
$ A_y Cos(\theta) = (93.3kN) Cos(41.2)= 70.2 kN = A'_y $
Finally, for our parallel component of $ A_y $ we use:
$ A_y Sin(\theta) = (93.3kN) Sin(41.2)= 61.5 kN = A'_x $
Now, we have all of our necessary forces to solve our beam $ A'B' $
Now, we need to solve for our unknown forces at B( $ B_x , B_y , M_B $)
$ \sum{F_x)=0\Longrightarrow $ $ A_x - \parallel_w + B_x=0 \Longrightarrow $ $ 61.5 kN - (7.46(kN/m) (10.6m)+ B_x=0\Longrightarrow $ $ B_x = 17.6 kN $
$ \sum{F_y}=0 \Longrightarrow $ $ A_y - \perp_w + B_y=0 \Longrightarrow $ $ 70.2kN- ((8.52)(10.6))+B_y=0 \Longrightarrow $ $ B_y= 20.1kN $
$ \sum{M_B}=0\Longrightarrow $ $ M_B- A_y(10.6) + \perp_w (5.3m)=0 \Longrightarrow $ $ M_B= 70.2(kN)(10.6m)-((8.52kN/m)(10.6m)(5.3m))= 265.5 kN.m $
Therefore, with all of our forces solved, we can now draw our shear and bending moment diagrams for $ A'B' $. Remember, $ A'B' $ is just a horizontal view of $ AB $, and that out shear and bending moment diagram for $ A'B' $ is just a tilted version of the diagrams for member $ AB $.
The diagrams below are the V & BM diagrms:
Our next step is to solve member $ BC $ to determine its shear and bending moment diagrams.
$ \sum{F_x}=0 \Longrightarrow $ $ B_x + C_x= 0 \Longrightarrow $ $ B_x=0 $
$ \sum{F_y}=0 \Longrightarrow $ $ B_y + C_y=0 \Longrightarrow $ $ B_y + 26.7(kN)=0 \Longrightarrow $ $ B_y = -26.7kN = 26.7kN \downarrow $
$ \sum{M_B}=0 \Longrightarrow $ $ M_B + 26.7 (kN) (10m)=0 \Longrightarrow $ $ M_B = -267kN.m = 267kN.m (Clockwise) $
The FBD, shear and moment diagrams for member $ BC $ are shown below:
Moment diagram of real system
Note: Because of rounding errors, the moment values at the point $ B $ will be very close in value ( less than 2kN.m apart) and not exactly the same.
The next step in our analysis is to solve for our virtual system. we apply a unit load of 1kN at the position of $ A_x $ and solve our system by finding our unknown reaction forces, and then determining the shear and moment diagrams.
$ \sum {F_x}=0 \Longrightarrow $ $ C_x + 1(kN)=0 \Longrightarrow $ $ C_x = -1kN= 1kN \leftarrow $
$ \sum{M_c}=0 \Longrightarrow $ $ -A_y(18m) + 1(17m)=0 \Longrightarrow $ $ A_y= 0.388kN $
$ \sum{F_y}=0 \Longrightarrow $ $ A_y + C_y=0 \Longrightarrow $ $ C_y =-0.388kN= 0.388kN \downarrow $
Just like we did in our real system, we have to tilt the axis of AB to get beam A'B' and solve for our shear and moment diagrams.
$ \sum{F'_x}= A_y Sin(41.2)+ A_x Cos(41.2) = 0.388 Sin(41.2) + 1 Cos(41.2)= 1.0 kN $
$ \sum{F'_y}= A_y Cos(41.2) - A_x Sin(41.2) = 0.386 Cos(41.2)- 1 Sin(41.2) = -0.366 kN = 0.366 \downarrow $
$ \sum{F_x}=0 \Longrightarrow $ $ B_x = -1.0 kN $
$ \sum{F_y}=0 $ $ B_y= 0.366 kN $
$ \sum{M_B}=0 $ $ M_B = -0.366 (10.6m)= -3.88(kN/m) = 3.88 kN/m (Clockwise) $
Now we will do, our V & M diagrams for our virtual $ A'B' $ member
Finally, we must solve member BC for the V & M diagram:
$ \sum{F_x}=0 \Longrightarrow $ $ B_x = 1 kN \rightarrow $
$ \sum{F_y}=0 \Longrightarrow $ $ B_y= 0.388 kN \uparrow $
Full moment diagram of virtual system:
Our next phase is determining our initial deflection $ \Delta A_{0} $ for when the redundant $ A_x $ is removed.
It is also important to outline our compatibility equation for the system:
$ \Delta A_{x0} + F_{A_x A_x} A_x =0 $
$ \Delta A_{x0} $ is determined by multiplying the moments from our real system by the moment of our virtual system using the virtual work integration table.
$ F_{A_x A_x} $ is determined by turning our virtual system into our new real system and multiplying that moment by our new virtual system, which is the exact same as our new real system.
$ W_{vi}= (1). \Delta A_{x0} = (AB_1(real) * AB_1 (virtual))+ (AB_2(real) * AB_2 (virtual))+(AB_3(real) * AB_3 (virtual))+ (AB_4(real) * AB_4 (virtual)) \Longrightarrow $
$ \frac{1}{EI} ( \frac{5LMQ}{12} + \frac{LM}{12} (5Qa + 3Qb) + \frac{LM}{2} (Qa+Qb)+ \frac{LMQ}{3} ) \Longrightarrow $ $ \frac{1}{EI} ( \frac{(5)(8.24)(289.2)(-3.02)}{12} + \frac{(2.36)(23.7)}{12} ((5)(-3.02)+(3)(-3.88)) + \frac {(2.36)(265.5)}{2} (-3.02-3.88) + \frac{(10)(26.7)(-3.88)}{3} ) = \frac {-8739}{EI} $
For $ F_{A_x A_x} $ our equation is much more simple:
$ W_{vi} = \frac{LMQ}{3} + \frac{LMQ}{3} = \frac{(10.6)(-3.88)(-3.88)}{3} + \frac{(10)(-3.88)(-3.88)}{3}\Longrightarrow $ $ W_{vi}= \frac {103.4}{EI}= F_{A_x A_x} $
Solve for $ A_x $
$ A_x= \frac{-\Delta A_{xo}}{F_{A_x A_x}} \Longrightarrow $ $ A_x = 84.5 kN $
Now, we can plug $ A_x $ back into our original system and solve for the other unknowns:
$ \sum{F_x}=0 \Longrightarrow $ $ A_x + C_x = 0 \Longrightarrow $ $ C_x = -84.5 kN= 84.5 kN \leftarrow $
$ \sum{M_A} = 0\Longrightarrow $ $ C_y(18) + C_x (7) - 15(8)(4)=0 \Longrightarrow $ $ C_y = -6.2 kN = 6.2 kN \downarrow $
$ \sum{F_y}=0 \Longrightarrow $ $ A_y + C_y - 120 =0 \Longrightarrow $ $ A_y=126.2 kN \uparrow $
Example Three
Determine the reaction at the roller support at point C, given that EI is constant.
The simplest way to solve this exmaple is by using the beam deflections and slopes. This beam is indeterminate to the first degree. so we just need one compatibility eqaution. If it was indeterminate to the second degree, then we will need 2 compatibility eqautions.
step 1 - choose a redundant, in this case the redundant will be $ C_y $, because it's the force that we have to find.
step 2 - draw the beam without the redundant and with the deflection caused by the 9 KN force, as shown in the figure below.
figure 2: the beam without the redundant
step 3 - redraw the beam only with the redundant force applied, and the deflection caused by the redundant. as shown in the figure below.
figure 3: the beam with only the redundant
step 4 - write down the compatibility equation: assume deflection upwards is positive.
$ - \Delta_C + \Delta_'C = 0 $ ; $ \Delta_'C = C_y*f_c $ ;
Therefore equation 1:
$ - \Delta_C + (C_y*f_c) = 0 $ ;^{[9]}
step 5 - find $ \Delta_C $ from figure 2 by using the beam deflections and slopes table, witch is given below
$ \Delta_C = \Delta_B + \theta*(10m) $ ;
$ \Delta_C = \frac{PL^3}{3EI} +[ \frac{PL^2}{2EI}*(10m)] = \frac{9kN*10m^3}{3EI} +[ \frac{9kN*10m^2}{2EI}*(10m)] = \frac{7500}{EI} kN m^3\downarrow $ ;
step 6 - find $ F_C $ using the beam deflections and slopes table, witch is given below
when finding $ F_C $ always use $ P = 1 $, this is because we are using the beam's deflections and slopes in order to find the value of $ C_y $. and the 9 kN force only cause the deflection of $ \Delta_C $ but $ C_y $ causes the deflection of $ \Delta_'c $
$ F_C = \frac{PL^3}{3EI} = \frac{1*20m^3}{3EI} = \frac{8000}{3EI}m^3 \uparrow $ ;
step 7 - using equation 1, find $ C_Y $ ;
equation 1: $ - \Delta_c + (C_y*f_cc) = 0 $ ; by rearranging the terms we get
$ C_y = \frac{\triangle_C}{f_c} = \frac{7500/EI}{8000/3EI} *\frac{kN*m^3}{m^3} = 2.812 kN = 2.8 kN \uparrow $
click on this link to view the beam deflections and slopes table:
http://enggcharts.org/product/thumbnail/CH%202441c.jpg
References
- ↑ Dr. Swamidas, A.S.J. ENGI 6705 Class Notes 5 (Lecture, ENGI 6705 Structural Analysis, Memorial University, St John's, NF.)
- ↑ ^{2.0} ^{2.1} ^{2.2} ^{2.3} ^{2.4} ^{2.5} ^{2.6} ^{2.7} ^{2.8} Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning.
- ↑ Prof. Erochko, J. CIVE 3203 Lecture to class, September 6, 2013.)
- ↑ Modified from: Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.) pg515. Stamford, CT: Cengage Learning.
- ↑ ^{5.0} ^{5.1} Prof. Erochko, J. CIVE 3203 Lecture 14 Presentation (Lecture, CIVE 3203 Structural Analysis, Ottawa, ON, October 25, 2013.)
- ↑ Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.) pg511. Stamford, CT: Cengage Learning.
- ↑ Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.) pg513. Stamford, CT: Cengage Learning.
- ↑ Prof. Erochko, J. CIVE 3203 Lecture 20 Presentation (Lecture, CIVE 3203 Structural Analysis, Ottawa, ON, November 22, 2013.)
- ↑ Hibbeler, R.C., 2009. structural analysis. 7th ed.