Transcript Document
CHAPTER 5 THE ANALYSIS OF BEAMS & FRAMES 5.1 Introduction For a two-dimensional frame element each node has the capability of translating in two directions and rotating about one axis. Thus each node of plane frame has three degrees of freedom. Similarly three structure forces (vertical force, shear force and bending moment) act at a node. For a two-dimensional beam element each node has two degrees of freedom (one rotation and one translation). Similarly two structure forces (vertical force and a bending moment) act at a node. However in some structures a node has one degree of freedom either rotation or translation. Therefore they are subjected to a moment or a force as the case may be. The structure stiffness matrices for all these cases have been developed in the previous chapter which can be summarized as under:i) ii) iii) Beams and frames subjected to bending moment Beams and frames subjected to shear force and bending moment Beams and frames subjected to shear force, bending moment and axial forces Following steps provide a procedure for the determination of unknown deformation, support reactions and element forces (axial forces, shear forces and bending moment) using the force displacement relationship (W=K). The same procedure applies both to determinate and indeterminate structures. 5.2 PROCEDURE TO ANALYSE BEAMS AND FRAMES USING DIRECT STIFFNESS METHOD 5.2.1. Identifying the components of the structural system or labeling the Structures & Elements. As a first step, divide the structure into some finite number of elements by defining nodes or joints. Nodes may be points of supports, points of concentrated loads, corners or bends or the points where the internal forces or displacements are to be determined. Each element extends between the nodes and is identified by arbitrary numbers (1,2,3). a) Structure Forces and Deformations At a node structure forces are assumed to act in their positive direction. The positive direction of the forces is to the right and upward and positive moments and rotations are clockwise. Start numbering the known forces first and then the unknown forces. Structures Forces not acting at the joints Stiffness method is applicable to structures with structure forces acting at nodes only. However if the structure is subjected to concentrated loads which are not acting at the joints or nodal points or if it is subjected to distributed loads then equivalent joint loads are calculated using the following procedure. i) All the joints are considered to be fixed. [Figure-5(b)] ii) Fixed End Moments (FEM’s) and Reactions are calculated using the formulae given in the table as annex-I iii) If more than one FEM and reactions are present then the net FEM and Reaction is calculated. This is done by algebric summation. [Figure-5(c)] Equivalent structure forces or loads at the joints/nodes are obtained by reversing the signs of net FEM’s & Reactions. [Figure-5(d)] or Reversing the signs of Net FEM’s or reaction gives the equivalent structure loads on the joints/nodes. v) Equivalent element forces are calculated from these equivalent structure loads using equation 5.2, 5.3 and 5.4 as explained in article number 5. vi) Final element forces are obtained by the following equation w = wE + wF where wE = Equivalent element forces wF = Element forces while considering the elements to be fixed. A B M FAB M FBA C M FBC (a) Actual Structure M FCB (b) Fixed End Moments & Reactions (wf) R1 R2 R3 ( M FBA+ M FBC M FAB R1 ( R2 + R3 ) ( M FBA+ M FBC M FAB R4 M FCB (c) Net Fixed End Moments & Reactions R4 M FCB (d) R1 ( R2 + R3 ) Fig.5.1 R4 Equivalent Joint Loads (WE) b) Element Forces Specify the near and far end of each element. Draw free body diagram of each member showing its local co-ordinate and element forces. Arbitrary numbers can identify all the element forces of the structure. 5.2.2. Calculation of Structure Stiffness Matrices of the members Properties of each element like its length, cross-sectional area, moment of inertia, direction cosines, and numbers identifying the structure forces acting at its near and far ends can be systematically tabulated. Using values of these parameters in equation 4.53 structure stiffness matrix of each member can be formed by applying equation 4.54,4.55 and 4.56 depending upon the situation. 5.2.3. Formation of Structure Stiffness Matrix of the Entire Structure According to the procedure discussed in chapter 3 article 3.1.3 stiffness matrix [K] of the entire structure is formed. 5.2.4. Calculation of Unknown Structure Forces and Displacements Following relation expresses the force-displacement relationship of the structure in the global coordinate system: [W] = [K] [] Where [W] is the structure load vector [K] is the structure stiffness matrix [] is the displacement vector Partitioning the above equation into known and unknown portions as shown below: Where K K Wk K11 Wu K21 12 u 22 k Wk = known loads Wu = unknown loads u = unknown deformation k = known deformation Expansion of the above leads to the following equation. [Wk] = [K11] [u] + [K12] [k] --------------------- (A) [Wu] = [K21] [u] + K22 k --------------------- (B) As k = 0 So, unknown structure displacement [u] can be calculated by solving the relation (A), which takes the following form. [u] = [K11]-1 [Wk] ---------------------(C) Unknown structure force i.e. reactions can be calculated by solving equation B which takes the following form Wu = [K21] [u] ---------------------- (D) 5.2.5. Calculation of element forces: Finally element forces at the end of the member are computed using the following equation (E). w = k = T w = kT --------------- (E) where [w] is the element force vector [kT] is the product of [k] and [T] matrices of the element where [w] is the element force vector [kT] is the product of [k] and [T] matrices of the element [] is the structure displacement vector for the element. Following are the [kT] matrices for different elements used in the subsequent examples. Case-I Beam/frame subjected to bending moment only 4 EI L kT 2EI L 2 EI L 4 EI L Case-II Beams subjected to Shear Forces & Bending Moment 4 EI L 2 EI kT 6LEI L2 6 EI L2 2 EI L 4 EI L 6 EI L2 6 EI L2 6 EI L2 6 EI L2 12EI L3 12EI L3 6 EI L2 6 EI L2 12EI L3 12EI L3 Case-IIIFor frame element subjected to axial force, shear force and bending moment. kT m 4 EI L 2 EI L 6 EI L2 6 EI L2 0 0 2 EI L 4 EI L 6 EI L2 6 EI L2 0 0 6 EI.l L2 6 EI.l L2 12EI.l 3 L 12EI.l 3 L AE m L AE m L 6 EI.l L2 6 EI.l L2 12EI.l 3 L 12EI.l 3 L AE m L AE m L 6 EI.m L2 6 EI.m L2 12EI.m L3 12EI.m L3 AE.l L AE.l L 6 EI.m L2 6 EI.m L2 12EI.m L3 12EI.m L3 AE.l L AE.l L Plotting bending moment and shearing force diagrams: Bending moment and shearing force diagrams of the structure are plotted using the element forces calculated in step-5. Examples on the next pages have been solved using the above-mentioned procedure. 5.3 ILLUSTRATIVE EXAMPLES EXAMPLE 5.1 Solve the beam shown in the figure using stiffness method. Solution Numbering element and structure forces 10k 10k 5' 2kip/ft 5' W1 , 1 15' 6' 6' W3 , 3 W2 , 2 W4 , 4 Structure loads and deformations w1, 1 w2, 2 1 w4 ,4 w3 , 3 2 Element forces and deformation w5 , 5 w6 ,6 3 Calculating fixed end moments and equivalent joint loads 12.5 12.5 37.5 FEF' s 37 . 5 39 39 W F -12.5k' 10k 12.5k' -39 2k/ft -37.5k' 37.5k' Fixed end moments W1F 12.5 W 25 Net Fixed End Moments 2 F W3 F 1.5 W 39 4F -1.5 -25 -12.5 1 2 Net fixed end moment 39 3 10k 2k/ft 39 Equivalent joint loads are : W K W1E 12.5 W 25 2E W3 E 1.5 W 39 4E 1.5 25 12.5 1 2 -39 3 Equivalent joint Moments Calculating structure stiffness matrices of element Following table lists the properties needed to form structure stiffness matrices of elements. Member Length (ft) I J 1 10 1 2 2 15 2 3 3 12 3 4 Structure stiffness matrices are: 1 2 EI 4 2 1 K 1 10 2 4 2 2 K 2 3 EI 4 2 2 15 2 4 3 3 1 4 2 0.4 0.2 1 EI 0 . 2 0 . 4 2 2 3 0.267 0.13 2 EI 0 . 13 0 . 267 3 3 4 0.34 0.167 3 EI 4 2 3 K3 EI 12 2 4 4 0.167 0.34 4 Forming structure stiffness matrix of the entire structure Using relation [K] = [K]1 + [K]2 + [K]3 structure stiffness matrix of the entire structure is: 1 2 3 4 0 .2 0 0 1 0.4 0.2 0.4 0.267 2 0 . 13 0 K EI 0 0.13 0.267 0.34 0.167 3 0 0 0 . 167 0 . 33 4 1 2 3 4 0 0 1 0 .4 0 .2 0.2 0.667 0.13 2 0 K EI 0 0.13 0.597 0.167 3 0 0 0 . 167 0 . 33 4 Finding unknown deformations Unknown deformations are obtained by using the following equation ] = [K]-1 [W] 1 0.4 0.2 2 1 EI 3 0 0 4 0.2 0 0.667 0.13 0.13 0.597 0 0.167 2.97 1 0.94 2 1 EI 3 0.24 0.123 4 18.8 1 24.85 2 1 EI 3 35 132.736 4 0 0 0.167 0.33 0.94 0.24 1.88 0.49 0.49 2.08 0.25 1.05 1 12.5 25 1.5 39 0.123 0.25 1.05 3.56 12.5 25 1.5 39 Calculating element forces Using relation [w]=[kT][] we get w1E 0.4 w 0.2 2 E w3 E 0 EI w 4E 0 w 0 5E w6 E 0 0.2 0 0.4 0 0.267 0.13 0.13 0.267 0 0.34 0 0.167 0 12.5 13.7 0 18.8 11.325 0 24.85 1 EI 0 35 12 . 67 10.27 0.167 - 132.74 0.34 39.88 Actual forces on the structure are obtained by superimposing the fixed end reactions on above calculated forces. w1 w1F w1E 12.5 12.5 0 w w w 12.5 13.7 26.2 2 2F 2E w3 w3 F w3 E 37.5 11.325 26.2 w w w 37 . 5 12 . 67 50 . 1 4 4 F 4 E w5 w5 F w5 E 39 10.27 50.1 w6 w6 F w6 E 39 39.08 0 w 1 w 1F w 1E 12.5 12.5 0 w w w 12.5 13.7 26.2 2 2F 2E w 3 w 3F w 3E 37.5 11.325 26.2 w w w 37 . 5 12 . 67 50 . 1 4 4F 4E w 5 w 5 F w 5 E 39 10.27 50.1 w w w 39 39 . 08 0 6 6 F 6 E EXAMPLE 5.2 Analyse the frame shown in the figure using stiffness method. Solution 2kip/ft W2 ,2 50k' W1 , 1 Numbering element forces and deformations W3 , 3 10' 20' Structure forces and deformations w1 , 1 w2 ,2 w3 ,3 1 2 Element forces and deformations w4 ,4 Finding Fixed end moments and equivalent structure load w1F 66.67 w 66.67 FEM ' s 2 F w3 F 0 w 0 4F W F W1F 66.67 Net fixed end moments W2 F 66.67 W3 F 0 2kip/ft 66.67k' Net fixed end moment -16.67k' Equivalent Joint Loads W 1E 16.67 Equivalent joint moment Calculating structure stiffness matrices of elements. Following table shows the properties of the elements required to form structure stiffness matrices of elements. Members Length (ft) i j 1 20 2 1 2 10 1 3 Structure stiffness matrices of both elements are: 2 1 2 1 0.2 0.1 2 EI 4 2 2 K 1 EI 1 0 . 1 0 . 2 20 2 4 1 1 K 2 EI 4 10 2 3 1 2 1 0.4 EI 0.2 4 3 3 0.2 1 0.4 3 Forming structure stiffness matrix for the entire structure. Structure stiffness matrix for the entire frame is obtained using relation [K] = [K]1 + [K]2 1 2 0.6 K EI 0.1 0.2 0.1 0.2 0 3 0.2 1 0 2 0.4 3 finding unknown deformation Unknown deformation can be calculated by using equation []u = [K11]-1 [W]k This can be done by partitioning the structure stiffness matrix into known and unknown deformations and forces Wk K11 Wu K 21 K12 u K 22 k [u] = [K11]-1[Wu] 16.67 0.6 0.1 0.2 W EI 0 . 1 0 . 2 0 2E W3E 0.2 0 0.4 1 1 0 EI 0 1 1 27.78 1 16.667 EI 0.6 EI Finding Unknown Reactions Unknown reactions can be calculated using the following equation. W u K 21 u W2 E 0.1 EI W 0.21 / EI 27.78 3E W2 E 2.78 W 5.556 3E W=WF+WE W2 66.67 2.78 69.45 W 0 5.556 5.56 3 Calculating element forces(Moments) Using relation w kT w1E 0.2 w 0.1 2 E EI w3 E 0 0 w4 E 0 2.78 0 0.2 0 1 5.56 - 27.78 0.4 0.2 EI 11.11 0 0.2 0.4 5.56 0.1 OR w1E 0.2 0.1 0 2.78 EI 1 / EI w 0.1 0.2 27.78 5.56 2E w3 E O.4 0.2 27.78 11.11 EI 1 / EI w 0.2 0.4 0 5.56 4E Actual forces acting on the structure can be found by superimposing the fixed end reactions on the forces calculated above. w wF wE w1 2.78 66.667 69.45 w 5.56 66.667 61.11 2 w3 11.11 11.11 0 w 5 . 56 0 5 . 56 4 69.45 61.11 11.11 11.11 1 Final element forces 61.11 2 69.45 5.56 BMD 5.56 EXAMPLE 5.3 Analyse the shown beam using direct stiffness method. Beam subjected to shear and moment. STEP-1 Numbering the forces and deformations 5kip/ft (0.41667kip/inch) 2kips 10' 15' W1 W4 W2 W3 W5 W6 Structure forces and deformations w2 w1 1 w3 w5 w6 2 w4 w7 Element forces and deformations w8 Finding fixed end forces and equivalent joint loads w1F 30 w 30 2F w3 F 1 w 1 FEF" s 4 F w5 F 1125 w 1125 6F w 37.5 7F w8 F 37.5 30k" 30k" 2kips 1k 1k 5kip/ft (0.41667kip/inch) -1125k" 1125k" 37.5k Fixed end forces 37.5k 30k" 1095k" Net fixed end forces [W]F=Net fixed end moments and forces = 30k" 1095k" Equivalent joint loads W1F 30 W 1095 2F W3 F 1125 W 1 4 F W5 F 38.5 W6 F 37.5 W1E 30 W 1095 2E Equivalent joint loads = Calculating Structure Stiffness Matrices of Elements Following table shows the properties of the elements required to form structure stiffness matrices of elements L I J K L 1 120 1 2 4 5 2 180 2 3 5 6 M From structural elements Member-1 1 2 4 5 0.016667 0.00041667 0.00041667 1 0.0333 0.016667 0.033333 0.00041667 0.00041667 2 K 1 EI 0.00041667 0.00041667 0.00000694 0.00000694 4 0.00041667 0.00041667 0.00000694 0.00000694 5 Member-2 2 K 2 3 0.0111 0.0222 0.0111 0.0222 EI 0.000185 0.000185 0.000185 0.000185 5 6 0.000185 0.000185 2 0.000185 0.000185 3 0.00000206 0.00000206 5 0.00000206 0.00000206 6 Forming structure stiffness matrix for the entire structure Structure stiffness matrix for the entire beam is obtained using the relation [K] = [K]1 + [K]2 1 2 3 0.016667 0 0.0333 0.016667 0.0555 0.0111 0 0.0111 0.0222 K EI 0 0.00041667 0.00041667 0.00041667 0.00023148 0.000185 0 0.00018519 0.000185 4 5 6 0.00041667 0.00041667 0 0.00041667 0.00023148 0.00018519 0 0.000185 0.000185 0.00000694 0.000000694 0 0.00000694 0.000009 0.00000206 0 0.0000206 0.00000206 1 2 3 4 5 6 Finding unknown deformations Unknown deformation can be calculated using equation [u] = [K11]-1 [Wk] This can be done by partitioning the structure stiffness matrix into known and unknown deformations and forces. Wk K11 K12 u W K K 22 k u 21 1 2 3 4 0.016667 0 W1E 0.0333 W 0.016667 0.0555 0.0111 2 E W3 E 0 0.0111 0.0222 EI 0 W4 E 0.00041667 0.00041667 W5 E 0.00041667 0.00023148 0.000185 0 0.00018519 0.000185 W6 E 5 1 0.00041667 0.00023148 0.00018519 2 0 0.000185 0.000185 3 0.00000694 0.000000694 0 4 0.00000694 0.000009 0.00000206 5 0 0.0000206 0.00000206 6 0.00041667 0.00041667 Using Equation [u] = [K11]-1 [Wk] 1 1 0.0333 0.016667 EI 2 0.016667 0.0555 6 1 30 1 10535.2946 EI 1095 22870.5886 Finding Unknown Reactions Unknown reactions can be calculated using the following equation [W]u = [K]21 []u 0 0.0111 W3 E 253.86 W4 E 0.00041667 0.00041667 10535.2946 1 5.14 EI W5 E 0.00041667 0.00023148 22870.5886 EI 0.91 W6 E 4.235 0 0.00018519 0 W = WE + WF W3 253.86 1125 1378.86 W4 5.14 1 4.14 W5 0.91 38.5 39.41 W6 4.235 37.5 41.735 Since all the deformations are known to this point we can find the element forces in each member using the relation [w]m = [kT]m []m Member-1: 0.0333 0.016667 0.00041667 0.0001667 10535.2946 30 w1E w 0.016667 22870.58863 586.7647 0 . 033333 0 . 00041667 0 . 00041667 2 E EI 1 w3 E 0.000416667 0.00041667 0.00000694 0.00000694 EI 5.1397 0 w 0 . 00041667 0 . 00041667 0 . 00000694 0 . 00000694 0 5 . 1397 4E Superimposing the fixed end forces for member-1 on the above w’s we get w = wE + wF 0 w1 30 30 w 586.7647 30 616.764 2 w3 5.1397 1 4.1397 w 5 . 1397 1 6 . 1397 4 Member 2 : 0.0111 w5 E 0.0222 w 0.0222 6 E EI 0.0111 w7 E 0.000185 0.000185 0.000185 0.000185 w8 E 0.000185 0.000185 22870.5886 508.2353 254.1176 0.000185 0.000185 1 0 4.23529 0.00000206 0.00000206 EI 0 0.00000206 0.00000206 0 4.23529 Superimposing the fixed end forces w wE wF w5 508.2353 1125 616.764 w 254.1176 1125 1379.117 6 w7 4.23529 37.5 33.2647 w 4 . 23529 37 . 5 41 . 73529 8 OR wE 0.01667 0 w1E 0.0333 w 0.01667 0.0333 0 2E w3 E 0.00041667 0.00041667 0 w 0.00041667 0.00041667 0 4 E EI w5 E 0 0.0222 0.0111 0 0.0111 0.0222 w6 E w 0 0.000185 0.000185 7E 0 0.000185 0.000185 w8 E wE w1E 30 w 586.53 2E w3 E 5.14 w 5 . 14 4E w5 E 508.42 w 254 . 21 6E w 4.24 7E w 8 E 4.24 w = wE +wF 0.00041667 0.00041667 10535.2946 0.00041667 0.00041667 0 22870.5886 0.00000694 0.00000694 0 1 0.00000694 0.000000694 0 0 0 0.000185 0.000185 0 EI 0 0.000185 0.000185 0 0 0.00000206 0.00000206 0 0 0.00000206 0.00000206 0 w1 30 30 w 586.53 30 2 w3 5.14 1 w 5 . 14 1 w 4 w5 508.42 1125 w 6 254.21 1125 w 4.24 37.5 7 w8 4.24 37.5 w1 0 w 616.53 2 w3 4.14 w 6 . 14 w 4 w5 616.58 w6 1379.2 w 33.26 7 w8 41.74 0 616.7647k" 1 -4.139 1379.177kip" -616.7647k" 2 -6.139 41.735 33.26 711.432 33.2647 -4.139 - + - -6.139 Shear Force Diagram -248.34 -41.735 - + -616.7647 -1379.117 Bending Moment Diagram EXAMPLE 5.4 Analyse the frame by STIFFNESS METHOD 3k/ft B 5' C 4k 6' 5' D A 10' W2,2 W5,5 W1,1 W4,4 W6,6 W3,3 W9,9 W12,12 W10,10 W7,7 W8,8 W11,11 Structure Forces and Deformation w9,9 w10,10 2 w11,11 w12,12 w7,7 w8,8 w4,4 w17,17 w2,2 w6,6 w13,13 w15 ,15 1 3 w16,16 w5 ,5 w14,14 w1,1 w18,18 w3,3 Element Forces and Deformations w1F 60 w 60 2F w3 F 0 w 0 4F w5 F 2 w 2 6F w 300 7F w 300 8F w 15 FEM' s 9 F w10 F 15 w 0 11F w12 F 0 w 0 13 F w 0 14 F w 0 15 F w 0 16 F w17 F 0 w18 F 0 3k/ft 2 300k" 300k" 15k 60k" 15k 2k 4k 1 3 2k 60k" Fixed End Moments and reactions [W]F= W1F 240 W 15 2F W3F 2 W 4 F 300 W 5 F 15 W6 F 0 W 0 7F W8 F 0 W 0 9F W10 F 60 W 0 11F W12 F 2 15k 240 K" 2k 15k 300 K" Net fixed end moments 0 [W]E= W1E 240 W 15 2E W3 E 2 W 300 4E W5 E 15 W 0 6E W 0 7E 15k 15k 300 K" 240 K" 2k 0 0 Equivalent joint loads Finding the structure stiffness matrices of elements. Next we will calculate the structure stiffness matrices for each element using the properties of members tabulated in below where E = 29000 ksi I = 100 inch4 and A = 5 inch2 (same for all members): Member Length (in.) l m I J K L M N 1 120 0 1 10 1 11 2 12 3 2 120 1 0 1 4 2 5 3 6 3 72 0 -1 4 7 5 8 6 9 For member-1 we have 10 1 11 2 12 3 0 0 1208.333 1208.333 10 96666.667 48333.333 48333.333 96666.667 1 0 0 1208 . 333 1208 . 333 0 0 1208.333 1208.333 0 0 11 K 1 0 0 1208 . 333 1208 . 333 0 0 2 1208.333 1208.333 0 0 20.13889 20.13889 12 0 0 20.13889 20.13889 3 1208.333 1208.333 For member-2 1 K 2 4 2 5 3 6 1208.333 0 0 96666.667 48333.333 1208.333 48333.333 96666.667 1208.333 1208 . 333 0 0 1208.333 1208.333 20.13889 20.13889 0 0 1208 . 333 1208 . 333 20 . 13889 20 . 13889 0 0 0 0 0 0 1208.333 1208.333 0 0 0 0 1208.333 1208.333 1 4 2 5 3 6 For member-3 4 K 3 7 5 8 6 9 0 0 3356.481 3356.481 161111.111 80555.556 80555.556 161111.111 0 0 3356 . 481 3356 . 481 0 0 2013.889 2013.889 0 0 0 0 2013 . 889 2013 . 889 0 0 3356.481 3356.481 0 0 93.2356 93.2356 3356.481 0 0 93.2356 93.2356 3356.481 Finding structure stiffness matrix for the entire frame. Using relation [K] = [K]1+[K]2+[K]3 we get the following structure stiffness matrix. (See next slide) 4 7 5 8 6 9 1 2 3 4 5 1208.3 193333.3 1208.3 1208 .33 48333.3 1 1208.3 1228.47 0 1208.3 20.138 2 1208.3 0 1228 .47 0 0 3 48333.3 1208.3 0 257777.78 1208.3 4 20.139 0 1208.3 2034.03 5 1208.3 0 1208.3 3356.48 6 0 K 7 0 0 0 80555.56 0 8 0 0 0 0 2013.89 9 0 0 0 3356.48 0 10 48333.3 0 1208.33 0 0 11 1208 .3 0 0 0 12 0 1208.3 0 20.139 0 0 6 0 0 1208.3 3356.48 0 1301.568 3356.48 0 93.2356 0 0 0 7 8 9 10 11 12 1208.3 0 1208.3 0 0 0 0 1208.3 0 20.139 80555.56 0 3356.48 0 0 0 0 2013 .89 0 0 0 0 3356.48 0 93.2356 0 0 0 161111.1 0 3356.48 0 0 0 0 2013.89 01 0 0 0 3356.48 0 93.2356 0 0 0 0 0 0 96666.67 0 1208.3 0 0 0 0 1208.3 0 0 0 0 1208.3 0 20.139 0 0 0 0 0 0 48333.3 0 Finding unknown deformations Unknown deformation can be calculated using equation [u] = [K11]–1[Wk] This can be done by partitioning the structure stiffness matrix into known and unknown deformations and forces. Wk K 11 W K u 21 K 12 u K 22 k Unknown deformations can be calculated using the equation ; []u=[K11]-1[W]k Solving the above equation we get , 1 193333.333 2 1208.333 1208.333 3 4 48333.333 1208.333 5 0 6 0 7 1208.333 1208.333 48333.333 1208.333 1228.4718 0 1208.333 20.13889 0 1228.722 0 0 1208.333 0 257777.778 1208.333 20.13889 0 1208.333 2034.0278 0 1208.333 3356.4814 0 0 0 80555.556 0 0 0 0 0 1208.333 0 3356.481 80555.556 0 0 1301.568 3356.481 3356.481 161111.111 1 240 15 2 300 15 0 0 1 .00184 2 .01202 .039859 3 4 .001527 .007682 5 6 .037023 .001535 7 Finding Unknown reactions Unknown reactions can be calculated using the following equation: Wu K 21 u 0.00184 0 0 0 0 2013.889 0 0 W8 E 0.01202 W 0 0 0 3356.481 0 93.2356 3356.481 0.039859 9E W10 E 48333.333 0.001527 0 1208.333 0 0 0 0 0 1208.333 0 0 0 0 0 W11E 0.007682 W12 E 1208.333 0.037023 0 20.13889 0 0 0 0 0.001535 15.4707 3.42501 40.7704 14.5242 1.42062 Now we will superimpose the fixed end reactions on the above calculated structure forces. [W]=[W]E+[W]F W8 15.47 0 15.47 W 3.42 0 3.42 9 W10 40.77 60 19.23 W 14 . 52 0 14 . 52 11 W12 1.42 2 0.58 Finding the unknown element forces. Up to this point all the deformations are known to us, we can find the element forces in each element using relation [w]m = [kT]m[]m For member-1 0 0 1208.333 1208.333 0 w1 96666.667 48333.333 w 48333.333 96666.667 0 0 1208.333 1208.333 0.00184 2 w5 1208.333 1208.333 0 0 20.13889 20.13889 0 w 1208 . 333 1208 . 333 0 0 20 . 13889 20 . 13889 0 . 01202 6 w3 0 0 1208.3330 1208.333 0 0 0 0 0 1208.333 1208.333 0 0 0.039859 w4 Equation gives w1 = 40.77 kips-inch c.w. w2 = 129.704 kips-inch c.w. w5 = –1.421 kips Downward w6 = 1.421 kips Upward w3 =14.524 kips Rightward w4 = –14.524 kips Leftward Superimposing the fixed end reactions in their actual direction we get w1 = 41.0269 – 60 = -18.973 kips-inch c.c.w. w2 = 130.2173 + 60 = 190.2173 kips-inch c.w. w5 = 1.427036 – 2 = -3.1427036 kips Rightward w6 = -1.427036 – 2 =0.57296 kips Rightward w3 = 14.5289 kips Upward w4 = -14.5289 kips Downward For member-2 0 0 w7 96666.667 48333.333 1208.333 1208.333 0.001845 w 48333.333 96666.667 1208.333 1208.333 0.001528 0 0 8 w9 1208.333 1208.333 20.13889 20.13889 0.012024 0 0 0 0 w10 1208.333 1208.333 20.13889 20.13889 0.0076822 w11 0 0 0 0 1208.333 1208.333 0.0398595 0 0 0 0 1208.333 1208.333 0.0370233 w12 Above equation gives w7 = 109.782 kips-inch w8 = -53.25353 kips-inch w9 = - 0.47048 kips w10 = 0.47048 kips w11 = 3.427 kips w12 = -3.427 kips c.w. c.c.w. Downward Upward Rightward Leftward Similarly for member-2 we have to superimpose the fixed end reactions w7 = 109.782 – 300 = -190.218 kips-inch c.c.w. w8 = –53.25353 + 300 = 246.746 kips-inch c.w. w9 = –0.471071+15 = 14.5289 kips Upward w10 = 0.471071 +15 = 15.471 kips Upward w11 = 3.427 kips Rightward w12 = –3.427 kips Leftward And finally for member-3 we get 0 0 3356.481 3356.481 0.0015278 w13 161111.111 80555.556 w 80555.556 161111.111 0 0 3356.481 3356.481 0.00153523 14 w15 3356.481 3356.481 0 0 93.2356 93.2356 0.00768219 w 3356 . 481 3356 . 481 0 0 93 . 2356 93 . 2356 0 16 w17 0.0370233 0 0 2013.889 2013.889 0 0 0 0 2013.889 2013.889 0 0 0 w18 Solving the above equation we get w13 = -246.74227 kips-inch w14 = 0 kips-inch w15 = 3.427kips w16 = -3.427 kips w17 = 15.4711 kips w18 = -15.4711kips c.c.w. Upward Downward Rightward Leftward PLOTTING THE BENDING MOMENT AND SHEARING FORCE DIAGRAM. According to the forces calculated above bending moment and shearing force diagrams are plotted below: 231.959k" 14.528k + -190.218k" + - - + -15.471k -246.747k" - + 15.407k" -3.427k - 3.427k 3.1427k Shearing Force Diagram -18.973 Bending moment diagram EXAMPLE 5.5 To analyse the frame shown in the figure using direct stiffness method. 3k/ft 10' 40' 10' W 1,1 W7,7 W3 ,3 W9 ,9 W6 ,6 W2 , 2 W4 ,4 W 8,8 W5 ,5 Structure Forces and Deformation w7, 7 w2,2 w6, 6 w 11,11 w 12,12 w9, 9 w1, 1 w5, 5 w8, w4, 4 w3, 3 Element Forces and Deformation w10, 10 Fixed end moments [W]F=Net fixed end forces= w1F 0 w 0 2F w3 F 0 w4 F 0 w5 F 0 w 0 6F w 4800 7F w8 F 4800 w 60 9F w10 F 60 w11F 0 w12 F 0 W1F 4800 W 60 2F W3 F 0 W 0 4 F W5 F 0 W6 F 0 W 4800 7F W8 F 60 W 0 9F 4800k" 0 60 4800k" 0 60 Equivalent joint loads= W1E 4800 W 60 2E W3 E 0 The properties of each member are shown in the table below. E = 29 x 103 ksi , I = 1000 inch4 and A = 10 inch2 are same for all members. Member Length l m I J K L M N 1 169.7056 0.707 0.707 inch 4 1 5 2 6 3 2 480 inch 1 7 2 8 3 9 1 0 Using the structure stiffness matrix for frame element in general form we get struc stiffness matrix for member-1. K 1 683536.666 341768.33 4272.064 4272.064 4272.064 341768.33 683536.666 4272.064 4272.064 4272.064 4272.064 4272.064 890.0045 890.0045 818..804 4272.064 890.0045 890.045 818.804 4272.064 4272.064 4272.064 818.804 818.804 890.0045 818.804 890.0045 4272.064 4272.064 818.804 4272.064 4 4272.064 1 818.804 5 818.804 2 890.0045 6 890.0045 3 Similarly for member-2 we get, K 2 0 0 241666.667 120833.33 755.21 755.21 1 120833.33 241666.667 755.21 755.21 7 0 0 755.21 2 755.21 3.15 3.15 0 0 755 . 21 755 . 21 3 . 15 3 . 15 0 0 8 0 0 0 0 604.167 604.167 3 0 0 0 0 604 . 167 604 . 167 9 Calculating the structure stiffness matrix for the entire frame. After getting the structure stiffness matrices for each element we can find the structure stiffness matrix for the whole structure using following relation: [K]1+[K]2 = [K] 1 2 1 925203.32 2 3516.854 3 4272.064 4 341768.33 K 5 4272.064 6 4272.064 7 120833.33 8 755.21 9 0 3516.854 893.195 818.804 4272.064 890.0045 818.804 755.21 3.15 0 3 4272.064 818.804 1494.1715 4272.064 818.804 890.0045 0 0 604.167 4 341678.33 4272.064 4272.064 683536.66 4272.064 4272.064 0 0 0 5 4272.06 890.0045 818.804 4272.064 890.0045 818.804 0 0 0 6 7 8 9 4272.064 120833.33 755.21 0 818.804 755.21 3.15 0 890.0045 0 0 604.167 4272.064 0 0 0 818.804 0 0 0 890.0045 0 0 0 0 241666.66 755.21 0 0 755.21 3.15 0 0 0 0 604.167 Finding unknown deformations Unknown deformation can be calculated using equation [u] = [K11]–1[Wk] This can be done by partitioning the structure stiffness matrix into known and unknown deformations and forces. Wk K 11 K 12 u W K K 22 k u 21 Unknown deformations can be calculated using the equation ; [u] = [K11]-1 [Wk] 1 925203.333 3516.854 4272.064 3516.854 893.195 818.804 2 3 4272.064 818.804 1494.1715 1 4800 60 0 Solving above equation we get 1 = 0.00669 2 = - 0.223186 3 = 0.141442 c.w. downward Rightward Wu K 21 u W 4 E 341768.33 4272.064 W 4272.064 890.0045 5E W6 E 4272.064 818.804 W7 E 120833.33 755.21 W8 E 755.21 3.15 0 0 W9 E 728.716 54.2432 85.4417 976 . 927 5.75539 85.4546 4272.064 0.00669 818.804 0.223186 890.0045 0.141442 0 0 604.167 Now we will superimpose the fixed end forces on the above calculated equivalent forces. W=WE+WF W4 728.716 0 W 54.2432 0 5 W6 85.4417 0 W 976 . 927 4800 7 W8 5.75539 60 W9 85.4546 0 728.716 54.2432 85.4417 5776 . 927 65.75539 85 . 4546 Calculating the unknown element forces All the deformations are known up to this point, therefore we can calculate the forces in all elements using relation [w] =[kT][] For member-1 we get [w]1=[kT]1[]1 w1E w 2E w3 E w4 E w5 E w6 E 683536.665 341768.332 341768.332 683536.665 6041.668 6041.668 6041.668 6041.668 0 0 0 0 4272.063 4272.063 4272.063 4272.063 50.34676 50.34676 50.34676 50.34676 1208.321 1208.321 1208.321 1208.321 4272.063 4272.063 0 4272.063 4272.063 0.00669 50.34676 50.34676 0 50.34676 50.34676 0.22319 0 1208.321 1208.321 0.14144 1208.321 1208.321 Solving the above equation we get following values of w’s for the equivalent loading condition provided by member-1 w1E= 729.0578 kip-inch c.w. w2E= 3015.829 kip-inch c..w. w3E= –22.066 kip Downward w4E= 22.0669 kip Upward w5E= 98.773 kip Rightward w6E= –98.773 kip Leftward w = wE+wF As wF are zero therefore w’s will be having the same values as those of wE ‘s For member-2 we get [w]2=[kT]2[]2 w7 E 241666.667 120833.333 755.208 w 120833.333 241666.667 755.208 8E 3.1467 w9 E 755.208 755.208 w10 E 755.208 755.208 3.1467 0 0 0 w11E w12 E 0 0 0 0.00669 755.208 0 0 0 0.223186 3.1467 0 0 0 3.1467 0 0 0 604.16667 604.16667 0.141442 0 0 604.16667 604.16667 755.208 0 0 solving the above equation we get the structure forces for the equivalent structure loads provided by member-2 w7E= 1785.543 kip-inch c.w. w8E= 977.047 kip-inch c..w. w9E= –5.7553 kip Downward w10E= 5.7553 kip Upward w11E= 85.454 kip Rightward w12E= –85.454 kip Leftward Superimposing the fixed end forces in their actual direction we get the element forces for the actual loading condition. w7= 1785.543 – 4800 = -3014.698 kip-inch c.c.w. w8= 977.047 + 4800 = 5776.927 kip-inch c..w. w9= –5.7553 +60 = 54.244 kip Upward w10= 5.7553 + 60 = 65.754 kip Upward w11 = 85.454 kip Rightward w12= – 85.454 kip Leftward Plotting the bending moment and shearing force diagrams. Bending moment and the shearing force diagram can be drawn according to the element forces calculated above. 54.244k 65.754k -85.454k 85.454k -3014.698k" 5776.927k" 2870.376 kip-inch 54.244 kip + - 263.016 " - + 729.0578 kip-inch + - - -65.7541 kips -22.059kip -3015.829 kip-inch -5797.047 kip-inch Shearing force Diagram Bending Moment Diagram P b a MA A B MB MA P ab2 L2 MB P a2 b L2 L P L/2 MA L/2 A B MB MA P ab2 L2 P a2 b MB L2 L A B w MA MB L wL2 MA 12 wL2 MB 12 L/2 A w B MA MB MA - 5 wL2 192 MB 11 wL2 192 L w A B MA MB MA - L L/2 L/2 w A B MA MB wL2 30 5wL2 MA 96 wL2 MB 20 5wL2 MB 96 L a MA A MA - b M B MB L Mb 3a - 1 L L MB Ma 3b - 1 L L The End