Page 1 Expressions of similar form as above may be obtained for all members. The sum of all horizontal components of individual forces gives us the stiffness coefficient and sum of all vertical component of forces give us the required stiffness coefficient . 11 k 21 k 4 2 4 4 3 2 3 3 2 2 2 2 1 2 1 1 11 cos cos cos cos ? ? ? ? l EA l EA l EA l EA k + + + = i i i i l EA k ? 2 4 1 11 cos ? = = (23.12) i i i i l EA k ? ? sin cos 21 ? = (23.13) In the next step, give a unit displacement along holding displacement along equal to zero and calculate reactions at 2 u 1 u E corresponding to unknown displacements and in the kinematically determinate structure. The corresponding reactions are denoted by and as shown in Fig.23.4d. The joint 1 u 2 u 12 k 22 k E gets displaced to E ' when a unit vertical displacement is given to the joint as shown in the figure. Thus, the new length of the member E A ' is 1 1 l l ? + . From the geometry, the elongation 1 l ? is given by 1 1 sin ? = ?l (23.14a) Thus axial force in the member along its centroidal axis is 1 1 1 sin ? l EA (23.14b) Resolve the axial force in the member along and directions. Thus, horizontal component of force in the member 1 u 2 u E A ' is 1 1 1 1 cos sin ? ? l EA (23.14c) and vertical component of force in the member E A ' is 1 2 1 1 sin ? l EA (23.14d) In order to evaluate , we need to sum vertical components of forces in all the members meeting at joint 22 k E .Thus, Page 2 Expressions of similar form as above may be obtained for all members. The sum of all horizontal components of individual forces gives us the stiffness coefficient and sum of all vertical component of forces give us the required stiffness coefficient . 11 k 21 k 4 2 4 4 3 2 3 3 2 2 2 2 1 2 1 1 11 cos cos cos cos ? ? ? ? l EA l EA l EA l EA k + + + = i i i i l EA k ? 2 4 1 11 cos ? = = (23.12) i i i i l EA k ? ? sin cos 21 ? = (23.13) In the next step, give a unit displacement along holding displacement along equal to zero and calculate reactions at 2 u 1 u E corresponding to unknown displacements and in the kinematically determinate structure. The corresponding reactions are denoted by and as shown in Fig.23.4d. The joint 1 u 2 u 12 k 22 k E gets displaced to E ' when a unit vertical displacement is given to the joint as shown in the figure. Thus, the new length of the member E A ' is 1 1 l l ? + . From the geometry, the elongation 1 l ? is given by 1 1 sin ? = ?l (23.14a) Thus axial force in the member along its centroidal axis is 1 1 1 sin ? l EA (23.14b) Resolve the axial force in the member along and directions. Thus, horizontal component of force in the member 1 u 2 u E A ' is 1 1 1 1 cos sin ? ? l EA (23.14c) and vertical component of force in the member E A ' is 1 2 1 1 sin ? l EA (23.14d) In order to evaluate , we need to sum vertical components of forces in all the members meeting at joint 22 k E .Thus, i i i i l EA k ? 2 4 1 22 sin ? = = (23.15) Similarly, i i i i i l EA k ? ? cos sin 4 1 12 ? = = (23.16) 3. Joint forces in the original structure corresponding to unknown displacements and are 1 u 2 u ? ? ? ? ? ? = ? ? ? ? ? ? 2 1 2 1 P P F F (23.17) Now the equilibrium equations at joint E states that the forces in the original structure are equal to the superposition of (i) reactions in the kinematically restrained structure corresponding to unknown joint displacements and (ii) reactions in the restrained structure due to unknown displacements themselves. This may be expressed as, () 2 12 1 11 1 1 u k u k R F L + + = () 2 22 1 21 2 2 u k u k R F L + + = (23.18) This may be written compactly as {} { } [ ] { } u k R F i + = (23.19) where, {} ? ? ? ? ? ? = 2 1 F F F ; {} () () ? ? ? ? ? ? = 2 1 L L L R R R [] ? ? ? ? ? ? = 22 21 12 11 k k k k k {} ? ? ? ? ? ? = 2 1 u u u (23.20) Page 3 Expressions of similar form as above may be obtained for all members. The sum of all horizontal components of individual forces gives us the stiffness coefficient and sum of all vertical component of forces give us the required stiffness coefficient . 11 k 21 k 4 2 4 4 3 2 3 3 2 2 2 2 1 2 1 1 11 cos cos cos cos ? ? ? ? l EA l EA l EA l EA k + + + = i i i i l EA k ? 2 4 1 11 cos ? = = (23.12) i i i i l EA k ? ? sin cos 21 ? = (23.13) In the next step, give a unit displacement along holding displacement along equal to zero and calculate reactions at 2 u 1 u E corresponding to unknown displacements and in the kinematically determinate structure. The corresponding reactions are denoted by and as shown in Fig.23.4d. The joint 1 u 2 u 12 k 22 k E gets displaced to E ' when a unit vertical displacement is given to the joint as shown in the figure. Thus, the new length of the member E A ' is 1 1 l l ? + . From the geometry, the elongation 1 l ? is given by 1 1 sin ? = ?l (23.14a) Thus axial force in the member along its centroidal axis is 1 1 1 sin ? l EA (23.14b) Resolve the axial force in the member along and directions. Thus, horizontal component of force in the member 1 u 2 u E A ' is 1 1 1 1 cos sin ? ? l EA (23.14c) and vertical component of force in the member E A ' is 1 2 1 1 sin ? l EA (23.14d) In order to evaluate , we need to sum vertical components of forces in all the members meeting at joint 22 k E .Thus, i i i i l EA k ? 2 4 1 22 sin ? = = (23.15) Similarly, i i i i i l EA k ? ? cos sin 4 1 12 ? = = (23.16) 3. Joint forces in the original structure corresponding to unknown displacements and are 1 u 2 u ? ? ? ? ? ? = ? ? ? ? ? ? 2 1 2 1 P P F F (23.17) Now the equilibrium equations at joint E states that the forces in the original structure are equal to the superposition of (i) reactions in the kinematically restrained structure corresponding to unknown joint displacements and (ii) reactions in the restrained structure due to unknown displacements themselves. This may be expressed as, () 2 12 1 11 1 1 u k u k R F L + + = () 2 22 1 21 2 2 u k u k R F L + + = (23.18) This may be written compactly as {} { } [ ] { } u k R F i + = (23.19) where, {} ? ? ? ? ? ? = 2 1 F F F ; {} () () ? ? ? ? ? ? = 2 1 L L L R R R [] ? ? ? ? ? ? = 22 21 12 11 k k k k k {} ? ? ? ? ? ? = 2 1 u u u (23.20) For example take P P P = = 2 1 , i i L L ? sin = , A A A A A = = = = 4 3 2 1 and ° = 35 1 ? , ° = 70 2 ? , ° = 105 3 ? and ° = 140 4 ? Then. {} ? ? ? ? ? ? = P P F (23.21) {} ? ? ? ? ? ? = 0 0 L R L EA L EA k i i 9367 . 0 sin cos 2 11 = = ? ? ? L EA L EA k i i 0135 . 0 cos sin 2 12 = = ? ? ? L EA L EA k i i 0135 . 0 cos sin 2 21 = = ? ? ? L EA L EA k i 1853 . 2 sin 3 22 = = ? ? (23.22) 1 2 0.9367 0.0135 0.0135 2.1853 u P EA u P L ? ? ?? ? ? = ?? ? ? ?? ?? ? ??? Solving which, yields EA L u 0611 . 1 1 = EA L u 451 . 0 2 = Page 4 Expressions of similar form as above may be obtained for all members. The sum of all horizontal components of individual forces gives us the stiffness coefficient and sum of all vertical component of forces give us the required stiffness coefficient . 11 k 21 k 4 2 4 4 3 2 3 3 2 2 2 2 1 2 1 1 11 cos cos cos cos ? ? ? ? l EA l EA l EA l EA k + + + = i i i i l EA k ? 2 4 1 11 cos ? = = (23.12) i i i i l EA k ? ? sin cos 21 ? = (23.13) In the next step, give a unit displacement along holding displacement along equal to zero and calculate reactions at 2 u 1 u E corresponding to unknown displacements and in the kinematically determinate structure. The corresponding reactions are denoted by and as shown in Fig.23.4d. The joint 1 u 2 u 12 k 22 k E gets displaced to E ' when a unit vertical displacement is given to the joint as shown in the figure. Thus, the new length of the member E A ' is 1 1 l l ? + . From the geometry, the elongation 1 l ? is given by 1 1 sin ? = ?l (23.14a) Thus axial force in the member along its centroidal axis is 1 1 1 sin ? l EA (23.14b) Resolve the axial force in the member along and directions. Thus, horizontal component of force in the member 1 u 2 u E A ' is 1 1 1 1 cos sin ? ? l EA (23.14c) and vertical component of force in the member E A ' is 1 2 1 1 sin ? l EA (23.14d) In order to evaluate , we need to sum vertical components of forces in all the members meeting at joint 22 k E .Thus, i i i i l EA k ? 2 4 1 22 sin ? = = (23.15) Similarly, i i i i i l EA k ? ? cos sin 4 1 12 ? = = (23.16) 3. Joint forces in the original structure corresponding to unknown displacements and are 1 u 2 u ? ? ? ? ? ? = ? ? ? ? ? ? 2 1 2 1 P P F F (23.17) Now the equilibrium equations at joint E states that the forces in the original structure are equal to the superposition of (i) reactions in the kinematically restrained structure corresponding to unknown joint displacements and (ii) reactions in the restrained structure due to unknown displacements themselves. This may be expressed as, () 2 12 1 11 1 1 u k u k R F L + + = () 2 22 1 21 2 2 u k u k R F L + + = (23.18) This may be written compactly as {} { } [ ] { } u k R F i + = (23.19) where, {} ? ? ? ? ? ? = 2 1 F F F ; {} () () ? ? ? ? ? ? = 2 1 L L L R R R [] ? ? ? ? ? ? = 22 21 12 11 k k k k k {} ? ? ? ? ? ? = 2 1 u u u (23.20) For example take P P P = = 2 1 , i i L L ? sin = , A A A A A = = = = 4 3 2 1 and ° = 35 1 ? , ° = 70 2 ? , ° = 105 3 ? and ° = 140 4 ? Then. {} ? ? ? ? ? ? = P P F (23.21) {} ? ? ? ? ? ? = 0 0 L R L EA L EA k i i 9367 . 0 sin cos 2 11 = = ? ? ? L EA L EA k i i 0135 . 0 cos sin 2 12 = = ? ? ? L EA L EA k i i 0135 . 0 cos sin 2 21 = = ? ? ? L EA L EA k i 1853 . 2 sin 3 22 = = ? ? (23.22) 1 2 0.9367 0.0135 0.0135 2.1853 u P EA u P L ? ? ?? ? ? = ?? ? ? ?? ?? ? ??? Solving which, yields EA L u 0611 . 1 1 = EA L u 451 . 0 2 = Example 23.1 Analyze the plane frame shown in Fig.23.5a by the direct stiffness method. Assume that the flexural rigidity for all members is the same .Neglect axial displacements. Solution In the first step identify the degrees of freedom of the frame .The given frame has three degrees of freedom (see Fig.23.5b): (i) Two rotations as indicated by and and 1 u 2 u (ii) One horizontal displacement of joint B and C as indicated by . 3 u In the next step make all the displacements equal to zero by fixing joints B and C as shown in Fig.23.5c. On this kinematically determinate structure apply all the external loads and calculate reactions corresponding to unknown joint displacements .Thus, Page 5 Expressions of similar form as above may be obtained for all members. The sum of all horizontal components of individual forces gives us the stiffness coefficient and sum of all vertical component of forces give us the required stiffness coefficient . 11 k 21 k 4 2 4 4 3 2 3 3 2 2 2 2 1 2 1 1 11 cos cos cos cos ? ? ? ? l EA l EA l EA l EA k + + + = i i i i l EA k ? 2 4 1 11 cos ? = = (23.12) i i i i l EA k ? ? sin cos 21 ? = (23.13) In the next step, give a unit displacement along holding displacement along equal to zero and calculate reactions at 2 u 1 u E corresponding to unknown displacements and in the kinematically determinate structure. The corresponding reactions are denoted by and as shown in Fig.23.4d. The joint 1 u 2 u 12 k 22 k E gets displaced to E ' when a unit vertical displacement is given to the joint as shown in the figure. Thus, the new length of the member E A ' is 1 1 l l ? + . From the geometry, the elongation 1 l ? is given by 1 1 sin ? = ?l (23.14a) Thus axial force in the member along its centroidal axis is 1 1 1 sin ? l EA (23.14b) Resolve the axial force in the member along and directions. Thus, horizontal component of force in the member 1 u 2 u E A ' is 1 1 1 1 cos sin ? ? l EA (23.14c) and vertical component of force in the member E A ' is 1 2 1 1 sin ? l EA (23.14d) In order to evaluate , we need to sum vertical components of forces in all the members meeting at joint 22 k E .Thus, i i i i l EA k ? 2 4 1 22 sin ? = = (23.15) Similarly, i i i i i l EA k ? ? cos sin 4 1 12 ? = = (23.16) 3. Joint forces in the original structure corresponding to unknown displacements and are 1 u 2 u ? ? ? ? ? ? = ? ? ? ? ? ? 2 1 2 1 P P F F (23.17) Now the equilibrium equations at joint E states that the forces in the original structure are equal to the superposition of (i) reactions in the kinematically restrained structure corresponding to unknown joint displacements and (ii) reactions in the restrained structure due to unknown displacements themselves. This may be expressed as, () 2 12 1 11 1 1 u k u k R F L + + = () 2 22 1 21 2 2 u k u k R F L + + = (23.18) This may be written compactly as {} { } [ ] { } u k R F i + = (23.19) where, {} ? ? ? ? ? ? = 2 1 F F F ; {} () () ? ? ? ? ? ? = 2 1 L L L R R R [] ? ? ? ? ? ? = 22 21 12 11 k k k k k {} ? ? ? ? ? ? = 2 1 u u u (23.20) For example take P P P = = 2 1 , i i L L ? sin = , A A A A A = = = = 4 3 2 1 and ° = 35 1 ? , ° = 70 2 ? , ° = 105 3 ? and ° = 140 4 ? Then. {} ? ? ? ? ? ? = P P F (23.21) {} ? ? ? ? ? ? = 0 0 L R L EA L EA k i i 9367 . 0 sin cos 2 11 = = ? ? ? L EA L EA k i i 0135 . 0 cos sin 2 12 = = ? ? ? L EA L EA k i i 0135 . 0 cos sin 2 21 = = ? ? ? L EA L EA k i 1853 . 2 sin 3 22 = = ? ? (23.22) 1 2 0.9367 0.0135 0.0135 2.1853 u P EA u P L ? ? ?? ? ? = ?? ? ? ?? ?? ? ??? Solving which, yields EA L u 0611 . 1 1 = EA L u 451 . 0 2 = Example 23.1 Analyze the plane frame shown in Fig.23.5a by the direct stiffness method. Assume that the flexural rigidity for all members is the same .Neglect axial displacements. Solution In the first step identify the degrees of freedom of the frame .The given frame has three degrees of freedom (see Fig.23.5b): (i) Two rotations as indicated by and and 1 u 2 u (ii) One horizontal displacement of joint B and C as indicated by . 3 u In the next step make all the displacements equal to zero by fixing joints B and C as shown in Fig.23.5c. On this kinematically determinate structure apply all the external loads and calculate reactions corresponding to unknown joint displacements .Thus, () ? ? ? ? ? ? × × - + × × = 36 9 3 24 16 4 2 48 1 F D R (1) 24 18 6 kN.m =- = () 2 24 kN.m F D R =- ( ) 3 12 kN.m F D R = (2) Thus, () () () 1 2 3 6 24 12 F D F D F D R R R ?? ?? ?? ??? =- ??? ??? ?? ?? ?? ? ? ? (3) Next calculate stiffness coefficients. Apply unit rotation along and calculate reactions corresponding to the unknown joint displacements in the kinematically determinate structure (vide Fig.23.5d) 1 uRead More

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