Consistent Deformations - Force Method | Index of Examples | CCE Homepage |

*Consistent Deformations - Force Method*

Once Statically Indeterminate Frame with Vertical Reaction as Redundant

problem statement

Using the method of consistent deformations, determine the reactions, draw moment, shear, and axial load diagrams for the frame in the accompanying figure. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire structure.

Figure 1 - Frame structure to analyze

- determine the degree of indeterminacy

The structure is statically indeterminate to the first degree(r=4, e=3, n = r-e = 4-3 = 1).

- select redundants and remove restraints

To solve for this single degree of indeterminacy, the structure has to be reduced to a statically determinate and stable structure. This is done by removing a redundant support.

In this example, the vertical reaction at support D is selected as a
redundant to remove in order to obtain a __primary determinate structure__.

Figure 2 - Primary determinate structure

- determine reactions and draw moment diagram for the primary structure

Calculate the support reactions of the primary structure.

Figure 3 - Support reactions

Determine the moment diagram M_{0} due to the applied loads on the primary structure.

Figure 4 - Mo - Moment diagram of primary structure

- sketch deflected shape

Sketch an approximate deflected shape for the primary structure, (see Fig. 4). Label the deflection that occurs in the direction of the redundant at the released support.

Figure 5 - Deflected shape of primary structure

- calculate deformation at redundant

Using the virtual work method, calculate the vertical
translation of support D that corresponds to the redundant Y_{D}.
Remove all loads and apply a unit force in the direction of the redundant, draw the
moment diagram, m_{d}, and sketch the deflected shape due to this unit load.

Figure 6(a) - Primary structure with unit load applied

Figure 6(b) - Moment diagram m_{d} with Y_{D} = 1 k

Figure 6(c) - Deflected shape with Y_{D} = 1 k

Calculate the translation, _{D0} (see Fig. 5),
at Support D using the following equation:

Using the method of virtual work, calculate areas on the M_{0} diagram and multiply
each area by the corresponding heights, h_{i}, measured at the centroid of this area
on the m_{d} diagram:

Area No. | Area/EI (A) (k-ft^{2})/EI |
Height (h) on m_{d} diagram (k-ft) |
A_{i}*h_{i} (k^{2}-ft^{3})/EI |

A_{01} |
15 x -300/EI = -4500/EI | 30 | -135000/EI |

A_{02} |
1/3 x 15 x -112.5/EI = -562.5/EI | 30 | -16875/EI |

A_{03} |
1/2 x 15 x -300/EI = -2250/EI | 5/6 x 30 = 25 | -56250/EI |

Total = Q(_{D0}) = |
-208125/EI |

Therefore, with Q = 1 k ;

*Q*(_{D0}) = -208125 (k^{2}-ft^{3})/EI

_{D0} = -208125 (k-ft^{3})/EI

Calculate the flexibility coefficient, *f*_{dd}, by determining the deformations of the primary
structure when subjected to the redundant load, Y_{D} = 1 k .

Again, using the method of virtual work, calculate areas on the m_{d} diagram and
multiply by the corresponding heights, h_{i}, measured at the centroid of the area:

Area No. | Area/EI (A) (k-ft^{2})/EI |
Height (h) on m_{d} diagram (k-ft) |
A_{i}*h_{i} (k^{2}-ft^{3})/EI |

A_{11} |
15 x 30/EI = 450/EI | 30 | 13500/EI |

A_{12} |
1/2 x 30 x 30/EI = 450/EI | 2/3 x 30 = 20 | 9000/EI |

Total = Q(f_{dd}) = |
22500/EI |

Therefore, with Q = 1 k ;

*Q*(*f*_{dd}) = 22500 (k^{2}-ft^{3})/EI

*f*_{dd} = 22500 (k-ft^{3})/EI

- write consistent deformation equations

The consistent deformation equation that corresponds to the redundant Y_{D} (the
vertical reaction at support D) is:

_{D0} + *f*_{dd}* Y_{D} = 0 (1)

This equation is set equal to zero since the pinned support at D does not allow any translation in the direction of the redundant, i.e., in the vertical direction.

- solve consistent deformation equations

Using equation (1), solve for Y_{D}:

-208,125 (k-ft^{3})/EI + 22,500 (k-ft^{3})/EI* Y_{D} = 0

Y_{D} = 9.25

Multiply the unit load, Q, at Y_{D} by 9.25 to get the final reaction.
The positive answer indicates that this reaction is in the direction of the applied unit load.

- determine support reactions

Impose the value of the calculated support reaction at Y_{D} along with the other applied loads on the
original structure. Calculate the remaining reactions using the three static equilibrium
equations, (F_{x} = 0,
F_{y} = 0 and
M = 0).

Figure 7 - Frame with support reactions

- draw moment, shear, and axial load diagrams

All reactions are now known and are used to draw the complete shear and moment diagrams for the structure.

Shear:

Figure 8 - Final shear diagram

Moment:

Figure 9 - Final moment diagram

Axial Load:

Figure 10 - Final axial load diagram

Deflected Shape:

Figure 11 - Deflected shape

Contact Dr. Fouad Fanous for more information.