Supplement: Statically Indeterminate Frames

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Bernadette McLaughlin
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1 : Statically Indeterminate Frames Approximate Analysis - In this supplement, we consider another approximate method of solving statically indeterminate frames subjected to lateral loads known as the. Like the Portal Method, this approximate analysis provides a means to solve a statically indeterminate problem using a simple model of the structure that is statically determinate. This method is more accurate than the Portal Method for tall and narrow buildings. Statically Indeterminate Frames Assumptions for the analysis of statically indeterminate frames by the cantilever method include the following. A tall building is conceptualized as a cantilever beam as far as the axial stresses in the columns are concerned. Axial stresses in the columns of a building frame vary linearly with distance from the 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 (horizontal members). - A point of inflection is where the bending moment changes from positive bending to negative bending. - Bending moment is zero at this point. Points of inflection are located at the mid-heights of all the columns (vertical members). The axial stresses in the columns at each story level vary as the distances of the columns from the center of gravity of the columns. It is usually assumed that all columns in a story are of equal area, at least for preliminary analysis. Page 1 of 15

2 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 cantilever method. 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 freebody diagrams. By introducing the hinges at the mid-point of each beam and at the mid-height of each column, we make 15 assumptions. In addition, for each floor we make assumptions regarding the variation of the column stress (relating 3 unknowns to a single unknown) - 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 15

3 Solution Top story Find 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 let that area = A). x = ( x i A i )/( A i ) = 0 (A) + 12(A) + 30 (A) = A For this problem, because the columns are of equal area, the column axial forces will vary 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). M O = 0 = - 6 (12) - (2/14) F (12) + (16/14) F (30) 0 = F F F = 72 F = 2.21 kips Page 3 of 15

4 A similar procedure is carried out for the other stories. 2 nd story M O = 0 = - 12 (18) - 6 (12) - (2/14) F (12) + (16/14) F (30) 0 = F F F = 288 F = 8.84 kips Page 4 of 15

5 1 st story M O = 0 = - 12 (30) - 12 (18) - 6 (12) - (2/14) F (12) + (16/14) F (30) 0 = F F F = 648 F = kips Summary of the axial loads Left - (F) Middle - (2/14)F Right - (16/14)F Top story Middle story Bottom story Page 5 of 15

6 The final step is to separate the frame into its component free-body diagrams, making cuts at the hinge locations to determine the rest of the internal forces. Using the top left corner as a FBD: M O = 0 = 2.21 (6) V 2 (6) V 2 (6) = V 2 = 2.21 kips F x = 0 = N 1 N 1 = N 1 = 9.79 kips F y = 0 = V 1 V 1 = 2.21 kips Using the top center portion as a FBD: M O = 0 = (15) (9) V 2 (6) V 2 (6) = = V 2 = 6.00 kips F x = 0 = N 1 N 1 = N 1 = 3.79 kips F y = 0 = V 1 V 1 = 2.53 V 1 = 2.53 kip Page 6 of 15

7 Using the top right corner as a FBD: F x = 0 = 3.79 V 1 V 1 = 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 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 7 of 15

8 FBD of All Floors (Axial Forces and Shear Forces) Page 8 of 15

9 FBD of All Floors (Bending Moment in the Columns) FBD of All Floors (Bending Moment in the Beams) Page 9 of 15

10 Comparison with other methods If the frame is designed according to the forces determined from the Portal Method or, the frame will be safe but not necessarily efficient. To compare the Portal Method and with a structural analysis using computer software (STRUDL), three cases were investigated. Case 1: The EI (where E is Young s modulus and I is the moment of inertia for the member) values for each member is proportional to the maximum bending moment in that member according to the Portal Method. Case 2: The EI value in each member is proportional to the maximum bending moment in that member according to the. Case 3: The EI value is the same for all members (i.e. all areas are equal). Reference for Members Relative Moment of Inertia Values for Case 1 Relative Moment of Inertia Relative Moment of Inertia Values for Case 2 Values for Case 3 Page 10 of 15

11 Beams Columns Portal Method Cantilever Method Portal Method Cantilever Method Portal Method Cantilever Method Member Axial Forces Shear Forces Bending Moments STRUDL Analysis Case 1 Case 2 Case 3 STRUDL Analysis Case 1 Case 2 Case 3 STRUDL Analysis Case 1 Case 2 Case 3 Near Far Near Far Near Far Page 11 of 15

12 Example Problem - Statically Indeterminate Frame Given: The 2-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 cantilever method. Solution Find the center of gravity (centroid) of the columns in the top story. In this example, the columns have different cross-sectional areas. x = ( x i A i )/( A i ) = 0 (625) + 18(500) (375) + 54 (625) x = 0 + 9, , ,750 x = ,125 For this problem, because the columns are not of equal area, only the column axial stresses 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. Left column Second column Third column Right column Stress f (7.68/25.68)f -(5.82/25.68)f (28.32/25.68)f Axial Load 625f (7.68/25.68)(500)f -(5.82/25.68)(375)f (28.32/25.68)(625)f Axial Load 625f f f f Page 12 of 15

13 Then, sum moments about the mid-point of the left column. M O = 0 = - 36 (6) f (18) f (31.5) f (54) 0 = ,691.54f + 2,677.19f + 37,219.50f 37,205.15f = 216 f = ksi The corresponding axial loads are 3.63 k in the left column, 0.87 k in the second column, k in the third column, and k in the right column. A similar procedure is carried out for the first story. 1 st story M O = 0 = - 36 (20) - 45 (8) f (18) f (31.5) f (54) 0 = ,691.54f + 2,677.19f + 37,219.50f 37,205.15f = 1,080 f = ksi The corresponding axial loads are k in the left column, 4.34 k in the second column, k in the third column, and k in the right column. Page 13 of 15

14 Summary of the axial loads Left column Second column Third column Right column Top story Bottom story The final step is to separate the frame into its component free-body diagrams, making cuts 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) Page 14 of 15

15 FBD of All Floors (Bending Moment in the Columns) FBD of All Floors (Bending Moment in the Beams) Page 15 of 15

Supplement: Statically Indeterminate Trusses and Frames

Supplement: Statically Indeterminate Trusses and Frames : 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