Ind ndet etermi minat ate A Anal nalys ysis For Force M e Met ethod 1 • The force (flexibility) method expresses the relationships between displacements and forces that exist in a structure. • Primary objective of the force method is to determine the chosen set of excess unknown forces and/or couples – redundants . • The number of redundants is equal to the degree of static indeterminacy of the structure. 1 Also see pages 115 – 141 in your class notes. 1
Description of the Force Method Procedure 1. Determine the degree of static indeterminacy. Number of releases* equal to the degree of static indeterminacy are applied to the structure. Released structure is referred to as the primary structure . Primary structure must be chosen such that it is geometrically stable and statically determinate. Redundant forces should be carefully chosen so that the primary structure is easy to analyze 2 * Details on releases are given later in these notes.
Force Method – con’t 2. Calculate “errors” (displacements) at the primary structure redundants. These displacements are calculated using the method of virtual forces. 3. Determine displacements in the primary structure due to unit values of redundants (method of virtual forces). These displacements are required at the same location and in the same direction as the displacement errors determined in step 2. 3
Force Method – con’t 4. Calculate redundant forces to eliminate displacement errors. Use superposition equations in which the effects of the separate redundants are added to the displacements of the released structure. Displacement superposition results in a set of n linear equations (n = number of releases) that express the fact that there is zero relative displacement at each release. 4
Force Method – con’t These compatibility equations guarantee a final displaced shape consistent with known support conditions, i.e., the structure fits together at the n releases with no relative displacements. 5. Hence, we find the forces on the original indeterminate structure. They are the sum of the correction forces (redundants) and forces on the released structure. 5
Flexibility Analysis (1) R 2 R 1 = (2) D 2 D 1 + f 11 (x R 1 ) f 21 (x R 1 ) (3) 1 (R 1 ) + f 22 (x R 2 ) (3) 6 f 12 (x R 2 ) 1 (R 2 )
+ + = f R f R D 0 11 1 12 2 1 (4 ) + + = f R f R D 0 21 1 22 2 2 Solve for R 1 and R 2 . Using matrix methods: [F] {R} = -{D} ⇒ {R} = -[F] -1 {D} 7
f f ≡ flexibility 11 12 [F] = f f matrix 21 22 [F] -1 ( ≡ inverse flexibility matrix) − f f 1 22 12 = − − f f f f f f 21 11 11 22 12 21 ≡ primary D 1 = {D} structure D 2 displacement vector 8
≡ redundant R 1 = force {R} R vector 2 − − R f D f D 1 1 22 1 12 2 = − + R f D f D det[F] 2 21 1 11 2 − det [F] = f f f f 11 22 12 21 (5) With R 1 and R 2 known, remaining structure is statically determinate. 9
Releases Release is a break in the continuity of the elastic (displacement) curve . One release only breaks a single type of continuity. Figure 1 shows several types of releases. Common release is the support reaction, particularly for continuous beams. 10
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Flexibility Equations Primary structure displacements at the releases are related to the unknown redundant forces via − = D f R (1) i ij j ≡ f displacement at release i due ij to a unit force in the direction of and at release j; flexibility coefficients. Equation 1 for the case of three redundant forces is expressed as 12
− = + + D f R f R f R 1 11 1 12 2 13 3 − = + + D f R f R f R (2a) 2 21 1 22 2 23 3 − = + + D f R f R f R 3 31 1 32 2 33 3 Matrix form of (2a) (2b) -{D} = [F] {R} D 1 {D} = = <D 1 D 2 D 3 > T D 2 D 3 = displacement vector at the redundant degrees of 13 freedom
R 1 R {R} = = <R 1 R 2 R 3 > T 2 R 3 = redundant force vector f f f 11 12 13 f f f [F] = 21 22 23 f f f 31 32 33 = flexibility matrix 14
Displacement Calculations – Method of Virtual Forces ∫ ∫ = + φ D F d M d i Vi Vi (3) ⇒ subscript i direction of R i at release i d = differential axial displacement d φ = differential rotational displ 15
Flexibility Coefficients – Method of Virtual Forces a b = + f f f (4) ij ij ij = ∫ F Vj a f F dx ij Vi EA(x) ≡ axial deformation influence coefficient = ∫ M Vj b f M dx ij Vi EI(x) ≡ bending deformation influence coefficient 16
Force Method Examples 1. Calculate the support reactions for the two-span continuous beam, EI = constant. w L L = w Primary Structure w/ Load + 1 (x R 1 ) 17 Primary Structure w/ Redundant
2. Calculate the support reactions for the two-span continuous beam, EI = constant. w L L = w Primary Structure w/ Load + R 1 R 2 Primary Structure w/ Redundant Forces 18
Prismatic Member Displacements 19
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3. Calculate the support reactions for the two-span continuous beam using the internal moment at B as the redundant force, I AB = 2I and I BC = I; E = constant. P L 2 Primary Structure w/ Loading 22
M B Primary Structure w/ Redundant D B = __________________ f BB = _________________ M B = _________________ 23
4. Calculate the bar forces for the statically indeterminate truss. Statically Statically Determinate Indeterminate Released Truss Truss (Redundant X) 24
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Nonmechanical Loading ∆ = − + [F]{R} ({D} {D }) (5) ∆ ∆ ∆ ∆ T = < > {D } D D D 2 n 1 = relative dimensional change displacements calculated using principle of virtual forces Displacements due to dimension changes are all relative displace- ments , as are all displacements corresponding to releases. They are positive when they are in the same vector direction as the corres- 26 ponding release (redundant) .
Structure Forces Once the redundant forces are calculated from Eq. (5), all other support reactions and internal member forces can be calculated using static equilibrium along with the appropriate free body diagrams. This is possible since the force method of analysis has been used to determine the redundant forces or the forces in excess of those required for static determinacy. 27
Mathematical Expressions Calculation of the non-redundant forces A i (support reactions, internal shears and moments, truss member forces) can be expressed using superposition as N + ∑ R p = A A (A ) R (6) i ui j j i = j 1 p A where = desired action A i on i the primary structure due to the (A ) applied loading ; = action ui j A i on the primary structure due to a unit virtual force at redundant 28 R j and N R = number of redundants.
Example Beam Problem – Nonmechanical Loading E = 30,000 ksi I = 288 in 4 (a) Given structure (b) Primary structure 29
The interesting point of this example is that the flexibility equation will have a nonzero right hand side since the redundant displacement is prescribed to equal 0.72 ” downward. Thus the flexibility equation is D ∆ f BB R B = d B - (7) B where d B = prescribed displacement at redundant B = -0.72 " since R B is positive upward D ∆ = -0.24 " B D ∆ − = relative displacement d B B 30 at redundant B
Truss Example – Nonmechanical Loading For the truss structure on the next page, compute the redundant bar EC member force if the temperature in bar EF is increased 50 o F and member BF is fabricated 0.3 in. too short. EA = constant = 60,000 kips and α = 6x10 -6 / o F. 31
Truss Example 3 @ 20 ’ = 60 ’ B C A D 15 ’ E F B C A 1 D E F Primary Structure Subjected to F CE = 1 32
Truss Example Calculations Mem L F V F V F V L 240" AB 0 0 300" AE 0 0 240" BC -4/5 153.6 180" BE -3/5 64.8 300" BF 1 300 240" CD 0 0 300" CE 1 300 180" CF -3/5 64.8 300" DF 0 0 240" EF -4/5 153.6 33
m 1 ∑ = f F F L CE,CE Vi Vi i EA = i 1 m ∆ ∑ ∆ = δ D F CE Vi i = i 1 ∆ δ = α∆ = T L 0.072" EF EF EF ∆ δ = ∆ = − 0.3" BF BF ∆ + = f F D 0 CE,CE CE CE 34
Displacement Calculations Displacements for the statically indeterminate structure can be calculated using the exact member deformations for a truss or exact shear and moment expressions along with the virtual force expres- sions on the primary structure. For a truss structure , calculation ∆ of a joint displacement using the principle of virtual forces results in 35
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