: Statically Indeterminate Trusses and Frames Approximate Analysis - In this supplement, we consider an approximate method of solving statically indeterminate trusses and frames subjected to lateral loads known as the Portal Method. The approximate analysis provides a means to solve a statically indeterminate problem using a simple model of the structure that is statically determinate. The method is based on the way the structure deforms under loaded conditions. The accuracy of the approximate method compares favorably with the more exact methods of analysis. Statically Indeterminate Frames Assumptions for the analysis of statically indeterminate frames by the Portal Method include the following: Points of inflection are located at the mid-points of all the beams (horizontal members). Points of inflection are located at the mid-heights of all the columns (vertical members). The horizontal shear force in all interior columns is the same. The horizontal shear force in the exterior columns is one-half of the horizontal shear force in the interior columns. A point of inflection is where the bending moment changes from positive bending to negative bending. Bending moment is zero at this point. Page 1 of 9
Example Problem - Statically Indeterminate Frame Given: The 3-story frame loaded as shown. Find: Analyze the frame to determine the approximate axial forces, shear forces, and bending moments in each member using the. Comments If we take a cut at each floor, we expose three actions (axial force, shear force, and bending moment) in each 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 18 th degree: 27 unknowns and only 9 equations of equilibrium for 3 free-body diagrams. By introducing the hinges at the midpoint of each beam and at the midheight of each column, we make 15 assumptions. In addition, for each floor we make 2 assumptions regarding the shear force - an additional 6 assumptions. In total, we made 21 assumptions, which allow the frame to be analyzed using the equations of equilibrium. Page 2 of 9
Solution Based on the third assumption, V is the shear force in the exterior columns and 2V is the shear force in the interior column. F x = 0 = 12 - V 2V V 4 V = 12 V = 3.0 kips Using the top left corner as a FBD: F x = 0 = 12 3 + N 1 N 1 = - 9.0 N 1 = 9.0 kips M A = 0 = - 3 (6) + V 1 (6) V 1 (6) = 18 V 1 = 3.0 kips F y = 0 = 3 N 2 N 2 = 3.0 kips Using the top center portion as a FBD: F x = 0 = 9 6 + N 1 N 1 = - 3.0 N 1 = 3.0 kips M B = 0 = + 3 (6) - 6 (6) + V 1 (9) V 1 (9) = - 18 + 36 = 18 V 1 = 2.0 kips F y = 0 = - 3 + 2 - N 2 N 2 = - 1.0 N 2 = 1.0 kip Page 3 of 9
Using the top right corner as a FBD: F y = 0 = - 2 N 1 N 1 = - 2.0 N 1 = 2.0 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 and equilibrium equations are not developed for each successive portion of the frame. Instead, portions of the frame are drawn and the equations of equilibrium for each portion of the frame are performed mentally without writing them out. The summation diagrams are shown on the following pages. Page 4 of 9
FBD of All Floors (Axial Forces and Shear Forces) Page 5 of 9
FBD of All Floors (Bending Moment in the Columns) FBD of All Floors (Bending Moment in the Beams) Page 6 of 9
Statically Indeterminate Trusses A truss is statically determinate if the following equation is satisfied. b + r = 2 j where b = the total number of bars (members) r = the total number of external support reactions j = the total number of joints A truss is statically indeterminate when b + r > 2 j b = 16, r = 3, j = 8: b + r = 19 > 2 j = 16 The truss is statically indeterminate to the 3 rd degree. Three assumptions regarding the bar forces are required. A statically indeterminate truss may be analyzed by one of the following methods. Method 1: Diagonal tension counters Long and slender diagonals tend to be relatively flexible in comparison to the other truss members. These slender diagonal members are incapable of resisting compressive forces due to their buckling tendency. The compressive diagonals are assumed to be zero force members and all panel shear is resisted by the tensile diagonal (known as diagonal tension counter ). This method was presented in Chapter 3. Page 7 of 9
Method 2: The diagonals are designed to support both tensile and compressive forces. Each diagonal is assumed to carry half of the panel shear. Example Problem - Statically Indeterminate Truss Given: The truss loaded as shown. Find: Forces in all members of the truss. Comments b = 11, r = 3, j = 6 b + r = 14 > 2 j = 12 This truss is statically indeterminate to the 2 nd degree. Solution Using the entire truss as a FBD, find the reactions at the supports. F x = 0 = F x F x = 0 M F = 0 = 8 C y 20 (4) C y = 10.0 k M C = 0 = - 8 F y + 10 (8) + 20 (4) F y = 20.0 k FBD Cut 1 F y = 0 = 20 10 (3/5) BF (3/5) AE 0 = 10 0.60 BF - 0.60 BF 0 = 10 1.20 BF BF = 8.33 k (T) AE = 8.33 k (C) (Assume AE = BF) M A = 0 = - 3 EF (4/5) BF (3) 0 = - 3 EF (4/5) 8.33 (3) EF = - 6.66 k EF = 6.66 k (C) M F = 0 = + 3 AB (4/5) AE (3) 0 = + 3 AB (4/5) 8.33 (3) AB = + 6.66 k AB = 6.66 k (T) Page 8 of 9
FBD Cut 2 F y = 0 = 10 (3/5) BD (3/5) CE 0 = 10 0.60 BD - 0.60 BD 0 = 10 1.20 BD BD = 8.33 k (T) CE = 8.33 k (C) (Assume BD = CE) M C = 0 = + 3 DE + (4/5) BD (3) 0 = + 3 DE + (4/5) 8.33 (3) DE = - 6.66 k DE = 6.66 k (C) M D = 0 = - 3 BC + (4/5) CE (3) 0 = - 3 BC + (4/5) 8.33 (3) BC = + 6.66 k BC = 6.66 k (T) FBD - Joint A F y = 0 = - 10 + AF (3/5) 8.33 AF = + 15.0 k AF = 15.0 k (T) FBD - Joint B F y = 0 = - 20 + BE + (3/5) 8.33 + (3/5) 8.33 BE = + 10.0 k BE = 10.0 k (T) FBD - Joint C F y = 0 = 10 + CD (3/5) 8.33 CD = - 5.0 k CD = 5.0 k (C) Page 9 of 9