Continuous Beams - Flexibility Method

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Jeffery George
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1 ontinuous eams - Flexibility Method Qu. Sketch the M diagram for the beam shown in Fig.. Take E = 200kN/mm 2. 50kN 60kN-m = = 0kN/m D I = x 0 6 mm 4 Fig D M diagram in kn-m units Qu.2 Sketch the M diagram for the beam shown in Fig.2. Take EI = constant. 80kN = = 5kN/m D 20kN 4m m Fig D mid-span M diagram in kn-m units

2 ontinuous eams - Flexibility Method Qu.3 Sketch the M diagram for the beam shown in Fig.3. lso calculate the support reactions. Take EI const. [R = -29.9, R = 236.8, R = 38.kN] 60kN 40kN/m D 5m 8m m Fig.3 8m D mid-span 79.8 Qu.4 M diagram in kn-m units Sketch the M diagram for the beam shown in Fig.4. Support undergoes a settlement of 5mm. Take E = 200kN/mm 2 and I = x 0-4 m 4. 80kN D 6m 40kN 8m Fig.4 8m 4m D M diagram in kn-m units

3 End rotations of simply supported beams Span = L Flexural rigidity = EI f f2 2 f = L 3EI L f2 = 6EI f2 f22 2 L f2 = 6EI f22 = L 3EI UDL of intensity θ φ θ = φ = L 3 24EI θ φ a α = a L θ = L 2( - α)(2 - α)α 6EI φ = L 2( - α2)α 6EI

4 Flexibility method applied to continuous beams Example. Two-span continuous beam with the flexural rigidity EI = constant P Q X Y R R L L R Using statics Res. vertically: R + R + R = P + Q () Taking mom. about : R x (L + L ) + R x L = P x X + Q x Y (2) The three reactions are the unknowns but only two equations are available. Therefore the problem is statically indeterminate. We must use compatibility of displacements to generate an additional equation in order to complete the solution. One procedure for doing this is illustrated in the figures below: P Q X Δ Y Step Remove prop and calculate the deflection Δ δ Step 2 Introduce unit force at and calculate δ For compatibility R = Δ / δ (3) We use these three equations to solve for R, R and R respectively. The bending moment at is M = R x L - P x X (4)

5 P Q M ending Moment Diagram The major disadvantage of this method arises when EI varies from span to span. The deflections cannot be calculated from standard tables which assume that the flexural rigidity EI is constant. The free-body diagram for the individual spans are given below: P M Q R R R R R =R + R P M R R M Reactant M P Free M Final M = Reactant M + Free M powerful method that does not suffer from this defect is now described. It uses displacement compatibility and leads to the Three Moment Theorem. The method breaks a continuous beam into a series of simply supported spans. The slopes are therefore discontinuous over the supports. Rotations are then introduced via the reactant moments to heal the cuts.

6 Three Moment Theorem onsider a continuous beam consisting of N spans. I J K (a) (b) (i) (j) M tangent to curve I J φ J K R I R J R K L IJ L JK M I M J J M K I K R IJ R JI R JK R KL R J =R JI + R Jk I J K θ JI θ JK Stage : End rotations due to applied loads M I I M J α J β K M K Stage : End rotations due to reactant moments In order to restore continuity at joint J, φ J = θ JI - α = β - θ JK

7 f f 2 f 2 f Flexibility oefficients: End rotations due to unit couples Making use of the flexibility coef ts shown above we may write the compatibility condition as follows: or θ JI - M I f i 2 - M J f i 22 = M J f j - M K f j 2 - θ JK M I f i 2 + M J ( f j + f i 22) + M K f j 2 = θ JI + θ JK Note f i 2 = (L/6EI) IJ, f i 22 = (L/3EI) IJ, f j = (L/3EI) JK, f j 2 = (L/6EI) JK This is the so-called Three Moment Theorem (due to lapeyron). Settlement of supports The theorem can be easily extended to deal with settlement of supports. Let Δ J be the settlement of support J. The rotations on each side of support J due to the differential settlement are (Δ J - Δ I )/L IJ and (Δ J - Δ K )/L JK respectively. We get M I f i 2 + M J ( f j + f i 22) + M K f j 2 = (θ JI - (Δ J - Δ I )/L IJ ) + (θ JK - (Δ J - Δ K )/L JK ) Fixed end (zero rotation) M M R R R fixed end, in the figure above, does not rotate. The rotation θ must therefore be balanced by the reactant rotations. We have M f a + M f a 2 = θ - (Δ - Δ )/L

8 Worked examples Example. alculate the reactions at the supports of the beam shown in Fig.. The flexural rigidity EI is the same for both spans. 0kN/m Fig. Step Remove prop 0kN/m Δ Step 2 pply unit load δ Δ = 5 x 0 x 64/(384EI) = 68.75/EI δ = x 63/(48EI) = 4.5/EI Let R denote the reaction at the support in Fig. then R δ = Δ which yields R = 37.5kN Resolving vertically we get for the beam in Fig., R + R + R = 60 lso Therefore R = R by symmetry R =.25kN

9 Worked examples Example 2. alculate the rotations at the supports of the beam shown in Fig.2. The flexural rigidity is EI. L Fig.2 M diagram x/l x Deflection diagram θ θ pply EI d2y/dx2 = -M = x/l Integrating Integrating again pplying b.c s : EI dy/dx = x2/(2l) +, is a constant EI y = x3/(6l) + x + D [] [2] t x = 0, y = 0 D = 0 from Eq.[2] t x = L, y = 0 = -L/6 Substituting in Eq.{] EIθ = = -L/6 or θ = -L/(6EI) EIθ = L/2 + or θ = L/(3EI) Note: The difference in signs of the rotations is due to the direction of the rotation. positive sign denotes an clockwise rotation whilst the negative sign denotes an anti-clockwise rotation.

10 Worked examples Example 3. Draw the M diagram for the beam shown in Fig. 3 below. The flexural rigidity EI is the same for both spans. 5kN/m 200kN Fig.3 We note that M = 0 and M = 0, simple supports. pplying the three-moment theorem at we get M = θ + θ 3EI 3EI = 85/(9EI) Μ Β = 54.33kN-m Reactant M (kn-m) Free M (kn-m) 24.7 Final M (kn-m) 5.2

11 Worked examples Example 4. alculate the bending moments at the supports of the beam in Fig.4. The flexural rigidity EI is the same for all spans. 5kN/m 200kN = 400kN = D Fig.4 We note that M = 0 and M D = 0, simple supports. pplying the three-moment we get: Joint Taking EI =, [2/3 + 3/3] M + 3/6 M = 5/ /9 0 Μ Β + 3 M = [] Joint 3/6 M + [3/3 + 2/3] M = 000/ M + 0 M = [2] Solving Eqs.[] and [2] we get M = 8.0kN-m M = 2.3kN-m

12 Worked examples Example 5. Rework the problem of example 3 taking the support as being fixed. 5kN/m 200kN Fig.3 We note that M = 0, simple support. pplying the three-moment theorem at we get: Joint Taking EI =, 2/3 M + 2/6 M = 5/3 4 M + 2 M = 0 [] Joint 2/6 M + [2/3 + 3/3] M = 85/9 4 M + 20 M = [2] Eqs.[] and [2] M = -27.4kN-m M = 59.8kN-m Reactant M (kn-m) Free M (kn-m) Final M (kn-m)

Structural Analysis III Compatibility of Displacements & Principle of Superposition

Structural Analysis III Compatibility of Displacements & Principle of Superposition 2007/8 Dr. Colin Caprani, Chartered Engineer 1 1. Introduction 1.1 Background In the case of 2-dimensional structures