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

*Consistent Deformations - Force Method*

Indeterminate Beam with Vertical Reaction as Redundant

problem statement

Using the method of consistent deformations, determine the reactions, moments and shears under the loading conditions shown. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire beam.

Figure 1 - Beam 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 redundant and remove restraint

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 B is selected as a
redundant to remove in order to obtain a __primary determinate structure__.

Figure 2 - Primary structure

- determine reactions and draw moment diagram for the primary structure

Calculate the support reactions of the released structure.

The resulting system, (_{B0} indicates the resulting
deflection or deformation at the location of the removed redundant for
the primary structure).

Figure 3 - Support reactions

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

In this example, the cantilever method is used to develop the moment diagram. (See a Virtual Work Cantilver Example for a complete description of this step)

Figure 4 - M_{0} - Moment diagram due to applied loads

*note: The moment diagram due to the 6k load is segmented as shown due to the value of the
moment diagram m _{b}, which is found later in the problem.*

- calculate deformation at redundant

Using the virtual work method, calculate the deflection
at support B that correponds to the redundant Y_{B}.
Remove all loads and apply a unit load in the direction of the redundant, draw the moment
diagram, m_{b}, and sketch the deflected shape due to the unit moment.

The resulting system, (*f*_{bb} is the deformation caused by the unit load).

Figure 5 - Primary structure with unit load applied and resulting
deflected shape

Figure 6 - Moment diagram m_{b} with Y_{B} = 1 k

Calculate the vertical translation corresponding to the redundant Y_{B} at support B
using the following equation:

Calculate the deformation at the redundant, _{B0}.
Use the method of virtual work, calculate the 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_{b} diagram:

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

A_{01} |
1/3 x 20 x -400/EI = -2666.67/EI | 15 | -40000/EI |

A_{02} |
1/2 x 20 x -120/EI = -1200/EI | 13.33 | -16000/EI |

A_{03} |
20 x -36/EI = -720/EI | 10 | -7200/EI |

A_{04} |
1/2 x 6 x -36/EI = -108/EI | 0 | 0 |

Total = Q(_{B0}) = |
-63200/EI |

Therefore, with Q = 1 k ;

*Q*(_{B0}) = -63200 (k^{2}-ft^{3})/EI

_{B0} = -63200 (k-ft^{3})/EI

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

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

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

A_{11} |
1/2 x 20 x 20/EI = 200/EI | 2/3 x 20 = 13.33 | 2666.67/EI |

Total = Q(f_{bb}) = |
2666.67/EI |

Therefore, with Q = 1 k ;

*Q*(*f*_{bb}) = 2666.67 (k^{2}-ft^{3})/EI

*f*_{bb} = 2666.67 (k-ft^{3})/EI

- write consistent deformation equation

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

_{B0} + *f*_{bb} * Y_{B} = 0 (1)

This equation is set equal to zero since the pinned support at B does not allow any vertical translation.

- solve consistent deformation equation

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

-63200 (k-ft^{3})/EI + 2666.67 (k-ft^{3})/EI * Y_{B} = 0

Y_{B} = 23.7

Multiply the unit load, Q, at Y_{B} by 23.7 to get the final reaction.
The positive answer indicates that the reaction is in the direction of the applied unit force.

- determine support reactions

Impose the value of the calculated Y_{B} 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 - Beam with support reactions

- draw moment, shear, and axial load diagrams

Shear:

Figure 8 - Final shear diagram

Moment:

Figure 9 - Final moment diagram

Deflected Shape:

Figure 10 - Deflected shape

Contact Dr. Fouad Fanous for more information.