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

*Consistent Deformations - Force Method*

Twice Statically Indeterminate Frame with Horizontal and Vertical Reactions as Redundants

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 second degree (r=5, e=3, n = r-e = 5-3 = 2).

- select redundants and remove restraints

To solve for two degrees of indeterminacy, the structure has to be reduced to a statically determinate and stable structure. This is done by removing two redundant supports.

In this example, the horizontal and vertical reactions at support D are selected as the
redundants 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 each redundant at the released support.

Figure 5 - Deflected shape of primary structure

- calculate deformations at redundants

Using the virtual work method, calculate the vertical
translation of support D that corresponds to the redundants X_{1} and X_{2}.
Remove all loads and apply a unit force in the direction of each redundant, draw the
moment diagram, m, and sketch the deflected shape due to the unit load.

**For the redundant X _{1} acting in the direction of the reaction X_{D}:**

Figure 6(a) - Primary structure with unit load applied at X_{1}

Figure 6(b) - Moment diagram m_{1} with X_{1} = 1 k

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

Calculate the horizontal translation corresponding to the redundant X_{1},
_{D1},
at support D using the virtual work 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_{1} diagram:

Area No. | Area/EI (A) (k-ft^{2})/EI |
Height (h) on m_{1} diagram (k-ft) |
A_{i}*h_{i} (k^{2}-ft^{3})/EI |

A_{01} |
15 x -300/EI = -4500/EI | 1/2 x 15 = 7.5 | -33750/EI |

A_{02} |
1/3 x 15 x -112.5/EI = -562.5/EI | 3/4 x 15 = 11.25 | -6328.125/EI |

A_{03} |
1/2 x 15 x -300/EI = -2250/EI | 0 | 0/EI |

Total = Q(_{D1}) = |
-40078.125/EI |

Therefore, with Q = 1 k;

*Q*(_{D1}) = -40078.125 (k^{2}-ft^{3})/EI

_{D1} = -40078.125 (k-ft^{3})/EI

**For the redundant X _{2} acting in the direction of the reaction Y_{D}:**

Figure 7(a) - Primary structure with unit load applied at X_{2}

Figure 7(b) - Moment diagram m_{2} with X_{2} = 1 k

Figure 7(c) - Deflected shape with X_{2} = 1 k

Calculate the horizontal translation corresponding to the redundant X_{2},
_{D2},
at support D.

Again, 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_{2} diagram:

Area No. | Area/EI (A) (k-ft^{2})/EI |
Height (h) on m_{2} 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(_{D2}) = |
-208125/EI |

Therefore, with Q = 1 k;

*Q*(_{D2}) = -208125 (k^{2}-ft^{3})/EI

_{D2} = -208125 (k-ft^{3})/EI

Calculate the flexibility coefficients, *f*_{dd}, by determining the deformations of the primary
structure when subjected to the redundant loads.

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

**For f_{11}, the coefficient at X_{1} due to the load applied at X_{1}:**

Area No. | Area/EI (A) (k-ft^{2})/EI |
Height (h) on m_{1} diagram (k-ft) |
A_{i}*h_{i} (k^{2}-ft^{3})/EI |

A_{11} |
1/2 x 15 x 15/EI = 112.5/EI | 2/3 x 15 = 10 | 1125/EI |

Total = Q(f_{11}) = |
1125/EI |

Therefore, with Q = 1 k;

*Q*(*f*_{11}) = 1125 (k^{2}-ft^{3})/EI

*f*_{11} = 1125 (k-ft^{3})/EI

**For f_{12}, the coefficient at X_{1} due to the load applied at X_{2}:**

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

A_{11} |
1/2 x 15 x 15/EI = 112.5/EI | 30 | 3375/EI |

Total = Q(f_{12}) = |
3375/EI |

Therefore, with Q = 1 k;

*Q*(*f*_{12}) = 3375 (k^{2}-ft^{3})/EI

*f*_{12} = 3375 (k-ft^{3})/EI

**For f_{22}, the coefficient at X_{2} due to the load applied at X_{2}:**

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

A_{21} |
15 x 30/EI = 450/EI | 30 | 13500/EI |

A_{22} |
1/2 x 30 x 30/EI = 450/EI | 2/3 x 30 = 20 | 9000/EI |

Total = Q(f_{22}) = |
122500/EI |

Therefore, with Q = 1 k;

*Q*(*f*_{22}) = 122500 (k^{2}-ft^{3})/EI

*f*_{22} = 122500 (k-ft^{3})/EI

**For f_{21}, the coefficient at X_{2} due to the load applied at X_{1}:**

Area No. | Area/EI (A) (k-ft^{2})/EI |
Height (h) on m_{1} diagram (k-ft) |
A_{i}*h_{i} (k^{2}-ft^{3})/EI |

A_{21} |
15 x 30/EI = 450/EI | 1/2 x 15 = 7.5 | 3375/EI |

A_{22} |
1/2 x 30 x 30/EI = 450/EI | 0 | 0/EI |

Total = Q(f_{21}) = |
3375/EI |

Therefore, with Q = 1 k;

*Q*(*f*_{21}) = 3375 (k^{2}-ft^{3})/EI

*f*_{21} = 3375 (k-ft^{3})/EI

*note: f_{12} and f_{21} will always have the same result*

- write consistent deformation equations

The consistent deformation equations are:

_{D1} + *f*_{11}* X_{1} + *f*_{21}* X_{2} = 0 (1)

_{D2} + *f*_{12}* X_{1} + *f*_{22}* X_{2} = 0 (2)

These equations are set equal to zero since the pinned support at D does not allow translation in the direction of the redundants.

- solve consistent deformation equations

Using equation (1) and (2), simultaneously solve for X_{1} and X_{2}:

-40,078.125 (k-ft^{3})/EI + 1125(k-ft^{3})/EI * X_{1} + 3375(k-ft^{3})/EI* X_{2} = 0 (1)

-208,125 (k-ft^{3})/EI + 3375 (k-ft^{3})/EI* X_{1} + 22,500 (k-ft^{3})/EI* X_{2} = 0 (2)

X_{1} = 14.32

X_{2} = 7.10

The answers imply that the applied unit loads need to be multiplied, respectively, by the corresponding result in order to obtain the final reactions.

The positive answers indicate that the reactions are in the same direction as the applied unit loads.

- determine support reactions

Impose the new, correct values of X_{D} and 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 8 - 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 9 - Final shear diagram

Moment:

Figure 10 - Final moment diagram

Axial Load:

Figure 11 - Final axial load diagram

Deflected Shape:

Figure 12 - Deflected shape

Contact Dr. Fouad Fanous for more information.