Final member forces by superposition - 1 cơ học kế- 123docz.net

We obtain the final member forces by superposition

# Member

0[kN]

N N1 N =N0+x N1 1

1 AB 167.7 -2.236 -45.44 2 BC 167.7 -2.236 -45.44 3 AD -141.4 1.414 -6.61 4 DC -212.1 1.414 -77.32 5 DB 250.0 -2.000 59.35

δ11

2.236 2.236

1.414 1.414

2.000

C C

T T

Fig. 3.11 Final member forces and deflected shape.

3.5 Example 3.2 (Alternative) 1. Statically determinate base structure

In this section, we present an alternative analysis of the previous example by considering the force in memberBDas the redundant. The redundant force is now an internal force. When we release the axial force in this member (internal release), the remaining members and the two pin supports at A and Care sufficient to ensure stability of the structure (we can view the base structure as two independent two-member trusses). We can view this release as a sleeve-like mecha- nism. MemberBDretains its position in the structure but is unable to transmit an axial force.

Fig. 3.12 Statically determinate base structure.

2. Member forces for base structure

As in the previous examples, we have to analyze the determinate base structure for two load cases: Load case 0 is the two external forces applied to the base structure and load case 1 is a unit value of the redundantx1applied to the base structure. Since the redundant force is the force in memberBD, load case 1 is a unit force in memberBD.

45.44

45.44 6.61

77.32

59.35 C T

C C

N Δ=0

3 m 3 m

6 m 6 m

50 kN

100 kN A

C D

1 1

x =

Fig. 3.13 Member forces of base structure for external load (“0”-structure) and unit redundant force (“1-structure”).

3. Compatibility equation

As before, we calculate the displacements of the base structure in a table:

# Member

0[kN]

N N1 L[m] N1⋅N0⋅L N1⋅N1⋅L

1 AB -111.8 1.118 6.708 -838.5 8.385 2 BC -111.8 1.118 6.708 -838.5 8.385 3 AD 35.35 -0.707 8.485 -212.1 4.243 4 DC -35.35 -0.707 8.485 212.1 4.243 5 DB 0 1.000 3.000 0 3.000

∑ -1677 28.25

The redundant force (the force in memberBD) is thus

10 1

1677 59.35 kN 28.25

x δ

Δ −

=− =− = (3.12)

As expected, the answer for the redundant force matches the final force in memberBC of the first solution strategy.

Fig. 3.14 Illustration of displacementsδ11andΔ10.

Figure 3.14 illustrates the two displacements for the base structure. The displacementδ11is the overlap in the axial force release in memberBD. This overlap is the relative displacement betweenBandD(the amount by which joints B and Dapproach each other due tox1 =1 which is sum of virtual work of members 1 through 4 in column 7) plus the axial deformation within memberBD (member 5 in column 7). Likewise, the displacementΔ10is the relative displacement between jointsBandDdue to external loading. Since the external load causes jointsBandDto move away from each other (in the direction opposite to that ofx1), Δ10is negative.

1.118 C

T T

1.118

0.707 0.707

1.000 T

C C

T C

111.8 111.8

35.35 35.35

0 δ11

B D

1 1

x = D

B B

Δ10

B D

4. Final member forces by superposition

We obtain the final member forces by superposition, the results match those obtained before.

# Member

0[kN]

N N1 N =N0+x N1 1

1 AB -111.8 1.118 -45.44 2 BC -111.8 1.118 -45.44 3 AD 35.35 -0.707 -6.61 4 DC -35.35 -0.707 -77.32 5 DB 0 1.000 59.35

3.6 Example 3.3

Fig. 3.15 Example 3.3.

Problem: Find the bending moment and shear force diagrams for the two-span continuous beam above. Select the moment at the mid-support as redundant (internal release).

Solution:

1. Statically determinate base structure

The two-span continuous beam is indeterminate to the first degree. Instead of selecting a reaction force as redundant (see Example 3.1) we now consider the internal moment atBthe redundant. Thus we obtain the statically determinate base structure by removing the resistance to the bending moment at the middle support, i.e. inserting a hinge atB. The resul- ting base structure consists of two simply supported beamsABandBC . Figure 3.16 shows the original structure (the two-span continuous beam) as a superposition of two load cases acting on the base structure.

Fig. 3.16 Statically indeterminate structure as superposition of two load cases for the statically determinate base structure.

L L

A B C

L L

2. Moment diagrams and displacements for base structure

As before (Examples 3.1 and 3.2), we need to analyze the base structure for two load cases: (0) the external loadwand (1) the unit redundant momentx1=1acting at the middle support. When the external load acts on the primary structure, a slope discontinuity (angle change)Δ10occurs at supportB. Again we use the subscript 1 to indicate that the displacement is at the location where redundant 1 acts and the subscript 0 to indicate that the cause of the displacement is the given external load. Clearly, the slope discontinuity atBviolates the internal compatibility requirement of continuity such that the redundant moment must restore compatibility atB. To restore compatibility we apply a unit value of the redundant moment to the base structure, which causes slope discontinuityδ11. Since the compatibility requirement is that of a rotation, the displacementsΔ10andδ11are rotations. These rotations are the sum of the rotations atBof the two simply supported beamsBAandBC(see Figs. 3.17 and 3.18).

base structure under external load (“0”-structure)

Figure 3.17 shows the deflected shape and the moment diagram for the base structure under given external loadw.

Fig. 3.17 Deflected shape and moment diagram of base structure for external load (“0”-structure).

base structure under unit value of redundant (“1”-structure)

Figure 3.18 shows the deflected shape and the moment diagram for the base structure under unit value of the redundant forcex1=1.

Fig. 3.18 Deflected shape and moment diagram of base structure for unit redundant (“1”-structure).