Supplement: Statically Indeterminate Frames
Supplement: Statically Indeterminate FramesApproximate Analysis - Cantilever MethodIn this supplement, we consider another approximate method of solving staticallyindeterminate frames subjected to lateral loads known as the “Cantilever Method.” Like the “Portal Method,” this approximate analysis provides a means to solve astatically indeterminate problem using a simple model of the structure that isstatically determinate. This method is more accurate than the “Portal Method” for tall and narrowbuildings.Statically Indeterminate FramesAssumptions for the analysis of statically indeterminate frames by the cantilevermethod include the following. A tall building is conceptualized as a cantilever beam as far as the axial stresses inthe columns are concerned. Axial stresses in the columns of a building frame vary linearly with distance fromthe center of gravity of the columns.- The linear variation is on stress, not the axial force. Points of inflection are located at the mid-points of all the beams (horizontalmembers).- A “point of inflection” is where the bending moment changes from positivebending to negative bending.-Bending moment is zero at this point. Points of inflection are located at the mid-heights of all the columns (verticalmembers). The axial stresses in the columns ateach story level vary as the distancesof the columns from the center ofgravity of the columns. It is usually assumed that all columnsin a story are of equal area, at leastfor preliminary analysis.SupplementCantilever MethodPage 1 of 15
Example Problem: Statically Indeterminate FrameGiven: The 3-story frame, loaded asshown.Find: Analyze the frame to determinethe approximate axial forces, shearforces, and bending moments in eachmember using the cantilever method.CommentsIf we take a cut at each floor, weexpose three actions (axial force,shear force, and bending moment) ineach of the 3 columns. For each cut, there are only 3 equations of equilibrium, but 9 unknowns. For each floor, there are 6 more unknowns than equations of equilibrium.For the three cuts (three floors), the frame is statically indeterminate (internally)to the 18th degree: 27 unknowns and only 9 equations of equilibrium for 3 freebody diagrams. By introducing the hinges at themid-point of each beam and at themid-height of each column, wemake 15 assumptions. In addition, for each floor we makeassumptionsregardingthevariation of the column stress(relating 3 unknowns to a singleunknown) - an additional 6assumptions. In total, we made 21 assumptions,which allow the frame to beanalyzed using the equations ofequilibrium.SupplementCantilever MethodPage 2 of 15
SolutionTop storyFind the “center of gravity (centroid)” of the columns in the top story. Assume in this problem that all columns in the top story have equal areas (and letthat area A).x ( xiAi )/( Ai ) 0 (A) 12(A) 30 (A) 14.03AFor this problem, because the columns are of equal area, the column axial forces willvary linearly as well as the axial stress.Let F be the force in the left column. Relative axial forces in the other columns are (2/14) F in the middle column and(16/14) F in the right column.Then, sum moments about the mid-point of the left column (i.e. point O). MO 0 - 6 (12) - (2/14) F (12) (16/14) F (30)0 - 72 - 1.7143 F 34.2857 F32.5714 F 72F 2.21 kipsSupplementCantilever MethodPage 3 of 15
A similar procedure is carried out for the other stories.2nd story MO 0 - 12 (18) - 6 (12) - (2/14) F (12) (16/14) F (30)0 - 216 - 72 - 1.7143 F 34.2857 F32.5714 F 288F 8.84 kipsSupplementCantilever MethodPage 4 of 15
1st story MO 0 - 12 (30) - 12 (18) - 6 (12) - (2/14) F (12) (16/14) F (30)0 - 360 - 216 - 72 - 1.7143 F 34.2857 F32.5714 F 648F 19.89 kipsSummary of the axial loadsTop storyMiddle storyBottom storyLeft - (F) 2.21 8.84 19.89Middle - (2/14)F 0.32 1.26 2.84SupplementCantilever MethodPage 5 of 15Right - (16/14)F- 2.53- 10.10- 22.74
The final step is to separate the frame into its component free-body diagrams, makingcuts at the hinge locations to determine the rest of the internal forces.Using the top left corner as a FBD: MO 0 2.21 (6) – V2 (6)V2 (6) 13.26V2 2.21 kips Fx 0 - 2.21 12 N1N1 - 9.79N1 9.79 kips Fy 0 - 2.21 V1V1 2.21 kips Using the top center portion as a FBD: MO 0 2.21 (15) 0.32 (9) – V2 (6)V2 (6) 33.15 2.88 36.03V2 6.00 kips Fx 0 9.79 – 6.00 N1N1 - 3.79N1 3.79 kips Fy 0 - 2.21 – 0.32 V1V1 2.53V1 2.53 kip SupplementCantilever MethodPage 6 of 15
Using the top right corner as a FBD: Fx 0 3.79 – V1V1 3.79 kip The summation diagram for the top floor is shown below.The other floors are solved in a similar manner.A “short cut” version of this method of analysis may be used. The principle of equilibrium is still followed; however, separate FBDs andequilibrium equations are not developed for each successive portion of the frame. Instead, portions of the frame are drawn and the equations of equilibrium for eachportion of the frame are performed mentally without writing them out.The summation diagrams are shown on the following pages.SupplementCantilever MethodPage 7 of 15
FBD of All Floors (Axial Forces and Shear Forces)SupplementCantilever MethodPage 8 of 15
FBD of All Floors(Bending Moment in the Columns)FBD of All Floors(Bending Moment in the Beams)SupplementCantilever MethodPage 9 of 15
Comparison with other methodsIf the frame is designed according to the forces determined from the Portal Methodor Cantilever Method, the frame will be safe but not necessarily efficient.To compare the Portal Method and Cantilever Method with a structural analysis usingcomputer software (STRUDL), three cases were investigated. Case 1: The EI (where E is Young’s modulus and I is the moment of inertia for themember) values for each member is proportional to the maximum bending momentin that member according to the Portal Method. Case 2: The EI value in each member is proportional to the maximum bendingmoment in that member according to the Cantilever Method. Case 3: The EI value is the same for all members (i.e. all areas are equal).Reference for MembersRelative Moment of InertiaValues for Case 1Relative Moment of InertiaValues for Case 2Relative Moment of InertiaValues for Case 3SupplementCantilever MethodPage 10 of 15
BeamsColumns6.6327.0 19.8929.019.597.242-9.0-14.53 -3.393-18.0 -22.73 -14.48 -18.87 -13.99 -4.01.26-6.90-1.686-8.0 -10.10 -6.75-8.83-7.046.0-15.01 18.0 18.01 18.03 18.0311.57CantileverMethod9.0Case Case Case12312.8422.26 29.00Bending MomentsSTRUDL AnalysisPortalMethodCase Case Case123CantileverMethodShear ForcesSTRUDL thodAxial Forces54.039.7854.068.165.197.3436.0-7.62 12.0 11.99 12.04 12.0411.2972.05.3736.07.595.196.78Case 1Near66.18FarCase 2NearFarCase 3NearFar48.90 49.92 36.93 86.34 52.4714.17 108.0 108.06 127.77 88.62 128.73 87.66 96.75 73.2610.276.0STRUDL Analysis61.3239.1878.57 50.22 81.1542.0626.5247.3139.9331.6871.9473.5370.92 73.65 70.77 64.95 70.5045.5431.7730.5741.5230.57 38.79 49.3239.81 25.71 6.006.026.025.9536.036.0036.1836.0336.1836.00 30.06 41.3415.33 7.78 15.0 11.05 15.3511.7514.3490.066.3090.2194.0568.6172.33 91.26 80.857.9221.3311-3.0-3.77-3.18-3.96-4.90 10.0 12.637.7210.046.9590.0 113.6768.1070.95 88.9591.74 57.36 3254.039.7860.2162.4348.27 64.65 54.068.1344.6446.26 58.68 60.24 18.013.2620.2521.0015.5116.26 27.99 8.022.7715.0315.5719.7720.31 17.34SupplementCantilever MethodPage 11 of 1546.1146.6821.33
Example Problem - Statically Indeterminate FrameGiven: The 2-story frameloaded as shown.Find: Analyze the frame todetermine the approximateaxial forces, shear forces,and bending moments in eachmember using the cantilevermethod.SolutionFind the “center of gravity (centroid)” of the columns in the top story. In this example, the columns have different cross-sectional areas.x ( xiAi )/( Ai ) 0 (625) 18(500) 31.5 (375) 54 (625)625 500 375 625x 0 9,000 11,812.5 33,7502,125x 25.68’For this problem, because the columns are not of equal area, only the column axialstresses will vary linearly. The column axial forces will not vary linearly.Following is a summary of the relative axial stresses and axial loads in the columns. Set “f” as the stress in the left column.StressLeft columnfSecond column(7.68/25.68)fThird column-(5.82/25.68)fRight column–(28.32/25.68)fAxial LoadAxial entCantilever MethodPage 12 of 15
Then, sum moments about the mid-point of the left column. MO 0 - 36 (6) - 149.53f (18) 84.99f (31.5) 689.25f (54)0 - 216 - 2,691.54f 2,677.19f 37,219.50f37,205.15f 216f 0.00581 ksiThe corresponding axial loads are 3.63 k in the left column, 0.87 k in the secondcolumn, -0.49 k in the third column, and -4.00 k in the right column.A similar procedure is carried out for the first story.1st story MO 0 - 36 (20) - 45 (8) - 149.53f (18) 84.99f (31.5) 689.25f (54)0 - 720 - 360 - 2,691.54f 2,677.19f 37,219.50f37,205.15f 1,080f 0.0290 ksiThe corresponding axial loads are 18.14 k in the left column, 4.34 k in the secondcolumn, -2.47 k in the third column, and -20.01 k in the right column.SupplementCantilever MethodPage 13 of 15
Summary of the axial loadsTop storyBottom storyLeft column3.6318.14Second column0.874.34Third column-0.49-2.47Right column-4.00-20.01The final step is to separate the frame into its component free-body diagrams, makingcuts at the hinge locations to determine the rest of the internal forces.The summation diagrams follow.FBD of All Floors (Axial Forces and Shear Forces)SupplementCantilever MethodPage 14 of 15
FBD of All Floors(Bending Moment in the Columns)FBD of All Floors(Bending Moment in the Beams)SupplementCantilever MethodPage 15 of 15
Approximate Analysis - Cantilever Method In this supplement, we consider another approximate method of solving statically indeterminate frames subjected to lateral loads known as the “Cantilever Method.” Like the “Portal Method,” this approximate analysis provides a means to solve a
Influence Lines for Statically Indeterminate Beams Reaction at A. 1 Scale Factor 1 E DE EE EE Vf f f §· ¹ Influence Lines for Statically Indeterminate Beams Shear at E. Influence Lines for Statically Indeterminate Beams Moment at E 1 Scale Factor 1
Statically Indeterminate Problems Example: Consider a bar AB supported at both ends by fixed supports, with an axial force of 12 kN applied at C as illustrated. Find the reactions at the walls A C B 500 mm 400 mm 12 kN Solution: First, draw a free body diagram: Statically Indeterminate Problems and Problems Involving Two Materials – p. 5/30
2. Analyse indeterminate trusses by approximate methods. 3. Analyse industrial frames and portals by approximate methods. 35.1 Introduction In module 2, force method of analysis is applied to solve indeterminate beams, trusses and frames. In modules 3 and 4, displacement based methods are discussed for the analysis of indeterminate structures.
alone are known as statically indeterminate structures. These, then, are structures that have more than 3 unknowns to be solved for. Therefore, in order to solve statically indeterminate stru
Using approximate methods to analyse statically indeterminate trusses and frames The methods are based on the way the structure deforms under the load Trusses Portal frames with trusses Vertical loads on building frames Lateral loads on building frames –Portal method –Cantilever method
Steps in Solving an Indeterminate Structure using the Force Method Determine degree of Indeterminacy Let n degree of indeterminacy (i.e. the structure is indeterminate to the nth degree) Define Primary Structure and the nRedundants Define the Primary Problem Solve for the n Relevant Deflections in Primary Problem Define the n Redundant Problems
C. Application of linear superposition to analyze indeterminate axially loaded members. Statically indeterminate structural systems are those where the number of unknown support reactions exceeds the number of global equilibrium equations. In such systems it is not possible to solve for the support
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