Approximate Lateral Load Analysis By Portal Method
Approximate Lateral Load Analysis by Portal MethodPortal FramePortal frames, used in several Civil Engineering structures like buildings, factories, bridges have the primarypurpose of transferring horizontal loads applied at their tops to their foundations. Structural requirementsusually necessitate the use of statically indeterminate layout for portal frames, and approximate solutions areoften used in their analyses.(i)(ii)(iii)(iv)Portal Frame StructuresAssumptions for the Approximate SolutionIn order to analyze a structure using the equations of statics only, the number of independent forcecomponents must be equal to the number of independent equations of statics.If there are n more independent force components in the structure than there are independent equations ofstatics, the structure is statically indeterminate to the nth degree. Therefore to obtain an approximate solutionof the structure based on statics only, it will be necessary to make n additional independent assumptions. Asolution based on statics will not be possible by making fewer than n assumptions, while more than nassumptions will not in general be consistent.Thus, the first step in the approximate analysis of structures is to find its degree of statical indeterminacy(dosi) and then to make appropriate number of assumptions.For example, the dosi of portal frames shown in (i), (ii), (iii) and (iv) are 1, 3, 2 and 1 respectively. Based onthe type of frame, the following assumptions can be made for portal structures with a vertical axis ofsymmetry that are loaded horizontally at the top1. The horizontal support reactions are equal2. There is a point of inflection at the center of the unsupported height of each fixed based columnAssumption 1 is used if dosi is an odd number (i.e., 1 or 3) and Assumption 2 is used if dosi1.Some additional assumptions can be made in order to solve the structure approximately for different loadingand support conditions.3. Horizontal body forces not applied at the top of a column can be divided into two forces (i.e.,applied at the top and bottom of the column) based on simple supports4. For hinged and fixed supports, the horizontal reactions for fixed supports can be assumed to be fourtimes the horizontal reactions for hinged supports1
ExampleDraw the axial force, shear force and bending moment diagrams of the frames loaded as shown below.B10 kCB10 k10CEF5A15ADD15(i)(ii)Solution(i) For this frame, dosi 3 3 4 3 4 1; i.e., Assumption 1MA 0 10 10 VD 15 0 VD 6.67 kFy 0 VA VD 0 VA 6.67 kHA HD 10/2 5 k10 k505056.676.675k6.67555k6.67 k6.67 kReactionsAFD (k)SFD (k)BMD (k-ft)(ii) dosi 3 3 6 3 4 3Assumption 1 HA HD 10/2 5 k, Assumption 2 BME BMF 0BMF 0 HA 5 MA 0 MA 25 k-ft; Similarly BME 0 MD 25 k-ftMA 025 25 10 10 VD 15 0 VD 3.33 kFy 0 VA VD 0 VA 3.33 k10 k253.3353.335k553.335k25 k-ft3.33 kReactions252525 k-ft3.33 kAFD (k)SFD (k)2BMD (k-ft)25
(iii)GB10 kC5HAMAA7.5HDVAMD7.5VAHADHDCE10FEGB10 kFVD7.57.5VDdosi 3 4 6 3 5 1 2; Assumption 1 and 2 BME BMF 0BME 0 (bottom)HA 5 MA 0 MA 5HA; Similarly BMF 0Also BME 0 (free body of EBCF) 10 5 VD 15 0 VD 3.33 kFy 0 VA VD 0 VA VD 3.33 kBMG 0 (between E and G) VA 7.5 HA 5 0 HA 5 kFx 0 (entire structure) HA HD 10 05 HD 10 0MD 5HDMA 5HA 25 k-ftHD 5 k MD 5HD 25 k-ft2510 k3.3353.335k253.33555k25 k-ft252525 k-ft3.33 k3.33 kReactionsAFD (k)SFD (k)BMD (k-ft)(iv) dosi 3 5 9 3 6 6 6 Assumptions needed to solve the structureAssumption 1 and 2 HA: HB: HC 1: 2: 1 HA 10/4 2.5 k, HB 5 k, HC 2.5 kMA MC 2.5 5 12.5 k-ft, MB 5 5 25 k-ftThe other 4 assumptions are the assumed internal hinge locations at midpoints of columns and one beam12.512.510 k12.5252.5 k5k2.5 k25 k-ft12.5 k-ftYBYA5012.5 k-ft12.5YCReactions1.6712.525BMD (k-ft)1.672.57.52.52.551.67SFD (k)1.67AFD (k)3
Analysis of Multi-storied Structures by Portal MethodApproximate methods of analyzing multi-storied structures are important because such structures arestatically highly indeterminate. The number of assumptions that must be made to permit an analysis bystatics alone is equal to the degree of statical indeterminacy of the structure.AssumptionsThe assumptions used in the approximate analysis of portal frames can be extended for the lateral loadanalysis of multi-storied structures. The Portal Method thus formulated is based on three assumptions1. The shear force in an interior column is twice the shear force in an exterior column.2. There is a point of inflection at the center of each column.3. There is a point of inflection at the center of each beam.Assumption 1 is based on assuming the interior columns to be formed by columns of two adjacent bays orportals. Assumption 2 and 3 are based on observing the deflected shape of the structure.122@10 20ExampleUse the Portal Method to draw the axial force, shear force and bending moment diagrams of the three-storiedframe structure loaded as shown below.Column shear forces are at the ratio of 1:2:2:1.18kMNOPShear force in (V) columns IM, JN, KO, LP are[18 1/(1 2 2 1) ] 3k, [18 2/(1 2 2 1) ] 6k,12kIJKL6k, 3k respectively. Similarly,VEI 30 1/(6) 5k, VFJ 10k, VGK 10k, VHL 5k; and6kVAE 36 1/(6) 6k, VBF 12k, VCG 12k, VDH 6kEFGHBending moments areABCDMIM 3 10/2 15k , MJN 30k , MKO 30k , MLP 15k101515MEI 5 10/2 25k , MFJ 50k , MGK 50k , MHL 25kMAE 6 10/2 30k , MFJ 60k , MGK 60k , MHL 30kThe rest of the calculations follow from the free-body 7236-5Column SFD (k)1515Column BMD (k-ft)6140-6-2-3-1Beam AFD -315-240-9-8-5.3361-8.13Beam BMD (k-ft)-12.2-2-8.13Beam SFD (k)4-115.5Column AFD (k)
Problems on Lateral Load Analysis by Portal Method1. The figure below shows the shear forces (kips) in the interior columns of a two-storied frame. Use thePortal Method to calculate the corresponding(i) applied loads P1 and P2, (ii) column bending moments, (iii) beam axial forces.P2310P151220202. The figure below shows the applied loads (F1, F2) and shear force (VEF) in column EF of a two-storiedframe. If F2 10 k, and VEF 5 k, use the Portal Method to calculate the(i) applied load F1, (ii) maximum column bending moments.AF1D10F2BE12VEFCF2015153. For the structure shown in Question 2, use the Portal Method to calculate the lateral loads F 1, F2 if theaxial forces in beams AD and BE are 10 kips and 15 kips respectively.4. For the structure shown below, use the Portal Method to(i) draw the bending moment diagrams of the top floor beams AB and BC(i) calculate the applied load F1 if the maximum bending moment in column EH is 30 k-ft.BAF2 10 kC10F1DEF14GIH15155. The figure below shows the exterior column shear forces (kips) in a four-storied fame.Calculate (i) the applied loads, (ii) beam shear forces.4 @ 10 40510152020105
Analysis of Multi-storied Structures by Cantilever MethodAlthough the results using the Portal Method are reasonable in most cases, the method suffers due to thelack of consideration given to the variation of structural response due to the difference between sectionalproperties of various members. The Cantilever Method attempts to rectify this limitation by considering thecross-sectional areas of columns in distributing the axial forces in various columns of a story.AssumptionsThe Cantilever Method is based on three assumptions1. The axial force in each column of a storey is proportional to its horizontal distance from thecentroidal axis of all the columns of the storey.2. There is a point of inflection at the center of each column.3. There is a point of inflection at the center of each beam.Assumption 1 is based on assuming that the axial stresses can be obtained by a method analogous to thatused for determining the distribution of normal stresses on a transverse section of a cantilever beam.Assumption 2 and 3 are based on observing the deflected shape of the structure.122@10 20ExampleUse the Cantilever Method to draw the axial force, shear force and bending moment diagrams of the three storied frame structure loaded as shown below.The dotted line is the column centerline (at all floors)18kOMNPColumn axial forces are at the ratio of 20: 5: 5: 20.kAxial force in (P) columns IM, JN, KO, LP are12IKJL[18 5 20/{202 52 ( 5)2 ( 20)2} ] 2.12k, [18 55/(202 52 ( 5)2 ( 20)2} ] 0.53k, 0.53k, 2.12k6krespectively.FGEHSimilarly, PEI 330 20/(850) 7.76k, PFJ 1.94k, PGK ABCD1.94k, PHL 7.76k; and15155 5PAE 696 20/(850) 16.38k, PBF 4.09k, PCG 4.09k,PDH 16.38kThe rest of the calculations follow from the free-body 5.65-8.6116.384.09-4.09-16.38Column AFD .215.942.435.342.464.653.83.185.3011.656.35Column SFD (k)64.6Beam BMD (k-ft)9.70615.9-2.12Beam SFD (k)29.1Column BMD 00-1.05Beam AFD (k)
Results from ‘Exact’ Structural Analysis12MONPAll members have equal cross-sectionskIJKLEFGH6k122@10 2018kABC1510D1519.4 -16.3 -7.14-4.80-5.89-9.48-13.00Column AFD 51.5-48.143.6-48.143.5-60.9-60.7Column BMD (k-ft)-8.63-4.78-10.17-5.86Beam SFD 10.4516.2 -19.3-2.3638.1 -33.9 43.233.8 -37.947.4 -40.9 50.840.8 -47.1Beam BMD 98-2.87-6.01-2.27-2.97-2.40-51.0Column SFD (k)Beam AFD (k)Interior columns have twice the area of exterior columns39.2 -38.1 55.1-5.7312.2612.26-12.95-11.57Column AFD (k)18.543.342.4-80.4-79.8Column BMD (k-ft)-5.15Beam SFD (k)Beam BMD (k-ft)18.45.73-38.8-12.3712.3738.2 -39.112.215.69-4.66-2.65-38.4Column SFD (k)7Beam AFD (k)-1.31
Problems on Lateral Load Analysis by Cantilever Method1. The figure below shows the axial forces (kips) in the exterior columns of a two-storied frame.If the cross-sectional area of column ABC is twice the area of the other columns, use the CantileverMethod to calculate the corresponding applied loads P1 and P2.AP23101212BP1C20202. For the structure shown below, use the Cantilever Method to calculate the lateral loads F1, F2 if the shearforces in beams AB and DE are 10 kips and 15 kips respectively. Assume all the columns have the samecross-sectional M2@10 2018kColumns with area 2A3. Use the Cantilever Method to draw the axial force, shear force and bending moment diagrams of thethree-storied structure loaded as shown below.15D15103 @ 10 303 @ 10 304. Figure (a) below shows the exterior column axial forces (kips) in a three-storied fame.Use the Cantilever Method to calculate (i) the applied loads, (ii) beam bending moments, (iii) columnbending moments. Assume all the columns to have equal cross-sectional areas.5101520572010Fig. (a)210Fig. (b)5. Figure (b) above shows the column shear forces (kips) in a three-storied fame.Calculate the column BM, beam BM, beam SF and column AF.Also check if they satisfy the conditions for Cantilever Method (for equal column areas).8
Approximate Vertical Load AnalysisApproximation based on the Location of HingesIf a beam AB is subjected to a uniformly distributed vertical load of w per unit length [Fig. (a)], both thejoints A and B will rotate as shown in Fig. (b), because although the joints A and B are partly restrainedagainst rotation, the restraint is not complete. Had the joints A and B been completely fixed against rotation[Fig. (c)] the points of inflection would be located at a distance 0.21L from each end. If, on the other hand,the joints A and B are hinged [Fig. (d)], the points of zero moment would be at the end of the beam. For theactual case of partial fixity, the points of inflection can be assumed to be somewhere between 0.21 L and 0from the end of the beam. For approximate analysis, they are often assumed to be located at one-tenth (0.1L) of the span length from each end ending on the support conditions (i.e., hinge ended, fixed ended or continuous), a beam in general can bestatically indeterminate up to a degree of three. Therefore, to make it statically determinate, the followingthree assumptions are often made in the vertical load analysis of a beam1. The axial force in the beam is zero2. Points of inflection occur at the distance 0.1 L measured along the span from the left and right support.Bending Moment and Shear Force from Approximate AnalysisBased on the approximations mentioned (i.e., points of inflection at a distance 0.1 L from the ends), themaximum positive bending moment in the beam is calculated to beM( ) w(0.8L)2/8 0.08 wL2, at the midspan of the beamThe maximum negative bending moment isM( ) wL2/8 0.08 wL2 0.045 wL2, at the joints A and B of the beamThe shear forces are maximum (positive or negative) at the joints A and B and are calculated to beVA wL/2, and VB wL/2Moment and Shear Values using ACI CoefficientsMaximum allowable LL/DL 3, maximum allowable adjacent span difference 20%1. Positive Moments(i) For End Spans(a) If discontinuous end is unrestrained, M( ) wL2/11(b) If discontinuous end is restrained, M( ) wL2/14(ii) For Interior Spans, M( ) wL2/162. Negative Moments(i) At the exterior face of first interior supports(a) Two spans, M( ) wL2/9(b) More than two spans, M( ) wL2/10(ii) At the other faces of interior supports, M( ) wL2/11(iii) For spans not exceeding 10 , of where columns are much stiffer than beams, M( ) wL2/12(iv) At the interior faces of exterior supports(a) If the support is a beam, M( ) wL2/24(b) If the support is a column, M( ) wL2/163. Shear Forces(i) In end members at first interior support, V 1.15wL/2(ii) At all other supports, V wL/2[where L clear span for M( ) and V, and average of two adjacent clear spans for M( )]9
ExampleAnalyze the three-storied frame structure loaded as shown below using the approximate location of hinges todraw the axial force, shear force and bending moment diagrams of the beams and columns.2@10 201 k/MNOPIJKLEFGHABCD1 k/121 k/151015The maximum positive and negative beam moments andshear forces are as follows.For the 15 beam, M( ) 0.08 1 152 18 k-ftM( ) 0.045 1 152 10.13 k-ftV( ) 1 15/2 7.5 kFor the 10 beam, M( ) 0.08 1 102 8 k-ftM( ) 0.045 1 102 4.5 k-ftV( ) 1 10/2 5 kAxial Force P in all the beams 0The rest of the calculations follow from the free-body diagrams-7.57.55.07.5-15.0-22.5-12.5Column AFD 13 -4.5-37.5Beam SFD (k)10.1318.0-25.0Beam BMD .640.480.150.48-0.580.35Column BMD (k-ft)Beam AFD (k)Column SFD (k)Using the ACI coefficients (for pattern loading)-7.55.0 8.637.5-8.63Beam SFD mn AFD (k)106.25-15.63 -14.19Beam BMD (k-ft)
Approximate Analysis of Bridge Portal and Mill BentBridge Portals and Mill BentsPortals for bridges or bents for mill buildings are often arranged in a manner to include a truss between twoflexural members. In such structures, the flexural members are continuous from the foundation to the top andare designed to carry bending moment, shear force as well as axial force. The other members that constitutethe truss at the top of the structure are considered pin-connected and to carry axial force only.Mill BentBridge PortalSuch a structure can be statically indeterminate to the first or third degree, depending on whether thesupports are assumed hinged of fixed. Therefore, the same three assumptions made earlier for portal framescan be made for the approximate analysis of these structures also; i.e., for a load applied at the top1. The horizontal support reactions are equal2. There is a point of inflection at the center of the unsupported height of each fixed based columnExampleIn the bridge portal loaded as shown below, draw the bending moment diagrams of columns AB and CD.BC101 k/ftFE15AMAHADMDHDVAVDAssuming the total load to be applied equally (i.e., 25/2 12.5 k and 12.5 k) at A and B, the horizontal reactions areHA 12.5 12.5/2 18.75 k, HB 12.5/2 6.25 kAlso, BM 0 at the midpoint of the free height; i.e., at 15/2 7.5 from the bottom.MA 18.75 7.5 7.5 7.5/2 0 MA 112.5 k-ftMD 6.25 7.5 0 MD 46.88 k-ftMA 0112.5 46.88 25 12.5 VD 20 0VD 7.66 kFy 0 VA VD 0 VA 7.66 k20RCRBRERF56.2518.75 k112.5 k-ft7.66 k46.886.25 k112.5BMD (k-ft) of AB46.88 k-ft7.66 k1146.88BMD (k-ft) of CD
Approximate Analysis of Statically Indeterminate TrussesTwo approximate methods are commonly used for the analysis of statically indeterminate trusses. Themethods are based on two basic assumptionsMethod 1: Diagonal members take equal share of the sectional shear forceMethod 2: Diagonal members can take tension only (i.e., they cannot take any compression)ExampleCalculate member forces GC, BH, GH, BC of the statically indeterminate truss shown below assuming(i) Diagonal members take equal share of the sectional shear force,(ii) Diagonal members can take tension only.5k20 k10 k10 k10 kHI5kxFGJ15A36.9EB23.75 kxCD4 @20 80Fx 0 Ex 20 0 Ex 20ME 0 20 15 (5 10 10 10 5) 40 Ay 80 0Fy 0 Ay Ey 5 10 10 10 5 0 Ey 16.25 k16.25 kAy 23.75 k(i) At section x-x,Fx 0 FGH FBC FBH cos 36.9 FGC cos 36.9 20 0FGH FBC 0.8 FBH 0.8 FGC 20 0Fy 0 FBH sin 36.9 FGC sin 36.9 5 10 23.75 00.6 FBH 0.6 FGC 8.75Assuming diagonal members to take equal share of the sectional shear force0.6 FBH 0.6 FGC 8.75/2 4.375 FBH 7.29 k, FGC 7.29 kMB 023.75 20 5 20 20 15 0.8 7.29 15 FGH 15 0Fx 0 FGH FBC 0.8 FBH 0.8 FGC 20 0 FBC 30.83 kFGH 10.83 k(ii) Assuming the diagonal member GC to fail in compression (i.e., to be non-existent)At section x-x,Fx 0 FGH FBC FBH cos 36.9 20 0 FGH FBC 0.8 FBH 20 0Fy 0 FBH sin 36.9 5 10 23.75 0 FBH 14.58 kMB 023.75 20 5 20 20 15 FGH 15 0 FGH 5 kFx 0 FGH FBC 0.8 FBH 20 0 FBC 36.67 kNote: The actual values from GRASP (assuming identical member sections) areFBH 4.88 k, FGC 9.71 k, FGH 12.77 k, FBC 28.90 k1220 k
Problems on Approximate Analysis of Bridge Portal, Mill Bent and Truss1. In the mill bent shown below, use the portal method to calculate the axial forces in members BG and EHand draw the shear force and bending moment diagrams of ABC and DEF.10 kCDGH45451020EB1 k/ft40AF4 @20 802. In the mill bent shown below,(i) Use the Portal Method to draw the bending moment diagram of the member KLM.(ii) Calculate the forces in EG and FH, assuming them to take equal share of the sectional shear.D10 kEHIN1010 kCFGMJ10LB10KA4 @10 403. In the bridge portal shown below, compression in member DG is 10 kips. Use the Portal Method to(i) calculate the load w per unit length, assuming the diagonal members to share the sectional shearforce equally.(ii) draw the BMD and SFD of the member FGH for the value of w calculated in (i).w kip per ftCDH25BGE50FA2 @50 10013
4. In the structure shown below,(i) Use the Portal Method to calculate the reactions at support A, G and draw the BMD of ABC.(ii) Calculate the forces in members CD, BE, CF, assuming diagonal members to take tension only.5kD5k10 k5kCE10I10BHF20AG2 @20 405. In the bridge portal shown below,(i) Use the Portal Method to calculate the reactions at support A and force in member BE.(ii) Calculate the forces in members GI and FH, assuming diagonal members to take tension only.GMH10 kD10CE1 k/ftFLI10 k10BK10JA4 @10 4014
Deflection Calculation by the Method of Virtual WorkMethod of Virtual WorkAnother way of representing the equilibrium equat
The assumptions used in the approximate analysis of portal frames can be extended for the lateral load analysis of multi-storied structures. The Portal Method thus formulated is based on three assumptions 1. The shear force in an interior column is twice the shear force in an exterior column. 2.
9 Approximate Analysis of Rigid Frames 436 9.1 Gravity Load Method 438 9.2 Portal Method for Lateral Loads 458 9.3 Cantilever Method for Lateral Loads 473 9.4 Combined Gravity and Lateral Loads 490 Homework Problems 497 10 Approximate Lateral Displacements 514 10.1 Braced Frames—Story Drift Method 516
a story, is affected by vertical distribution of lateral loads, i.e., there is a unique displaced profile for each type of lateral load distribution. Conse quently, the lateral stiffness of a story is not a stationary property, but an apparent one that depends on lateral load distribution. In the analysis of
Earthquake Lateral Force Analysis The design lateral force shall first be computed for the building as a whole. This design lateral force shall then be distributed to the various floor levels. There are two commonly used procedures for seismic design lateral forces: 1. Equivalent static force analysis 2. Dynamic analysis
and the lateral surface area. Solve for the height. 16:(5 10.0 cm The surface area of a cube is 294 square inches. Find the length of a lateral edge. 62/87,21 All of the lateral edges of a cube are equal. Let x be the length of a lateral edge. 16:(5 7 in. Find the lateral area and surface area
1. femur 2. medial condyle 3. medial meniscus 4. posterior cruciate ligament 5. medial collateral ligament 6. tibia 7. lateral condyle 8. anterior cruciate ligament 9. lateral meniscus 10. lateral collateral ligament 11. fibula 12. lateral condyle/epicondyle 13. lateral meniscus 14. lateral collateral ligament 15. medial collateral ligament 16 .
turning radius speed drawbar gradeability under mast with load center wheelbase load length minimum outside travel lifting lowering pull-max @ 1 mph (1.6 km/h) with load without load with load without load with load without load with load without load 11.8 in 12.6 in 347 in 201 in 16 mp
Floor Joist Spans 35 – 47 Floor Joist Bridging and Bracing Requirements 35 Joist Bridging Detail 35 10 psf Dead Load and 20 psf Live Load 36 – 37 10 psf Dead Load and 30 psf Live Load 38 – 39 10 psf Dead Load and 40 psf Live Load 40 – 41 10 psf Dead Load and 50 psf Live Load 42 – 43 15 psf Dead Load and 125 psf Live Load 44 – 45
300-a02 abp enterprise sdn bhd. 7th floor menara lien hee no, 8 jalan tangung, 47700 petaling jaya. selangor p. j john c.o.d. 03-7804448 03-7804444 300-c01 control manufacturing 400-2 (tingkat satu) batu 1/2, jalan pahang, 51000 kuala lumpur kl lal net 60 days 03-6632599 03-6632588
