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

*Consistent Deformations - Force Method*

Indeterminate Truss

problem statement

Using the method of consistent deformations, determine the vertical and horizontal reactions at A and E and the resulting member loads for the truss in the accompanying figure.

The member properties are A = 2 in^{2} and E = 29x10^{3} ksi.

Figure 1 - Truss structure to analyze

- determine the degree of indeterminacy

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

- select redundant and remove restraint

To solve for the 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 horizontal reaction at support E is selected as a redundant to remove
in order to obtain a __primary determinate structure__.

Figure 2 - Primary determinate structure

- determine reactions and member forces

Calculate the support reactions of the primary structure, then determine the individual member forces by using either the method of sections, or the method of joints.

Figure 3 - Support reactions and member forces

- calculate deformation at redundant

Use the virtual work method, calculate the horizontal
translation of support E that corresponds to the removed redundant X_{E}.
Remove all loads and apply a unit force in the direction of the redundant.

Figure 4 - Primary structure with unit load applied

Calculate the member forces due to the unit load.

Figure 5 - Member forces due to unit load

Since the horizontal translation at E is equal to zero in the original structure, the horizontal translation of the released structure at support E must be countered by a force at E which causes an equal translation in the opposite direction.

Since the member forces are known in the released structure, and the virtual structure
with a unit load applied at E, the deflection at support E (_{E0})
caused by the applied loads on the released structure is determined by the following equation.

where m is equal to the number of members, n is the force in the member due to the virtual load, N is the force in the member due to the applied loads, L is the length, A is the area, and E represents Young's Modulus of Elasticity.

The force required to counter this deflection is found by determining the deflection caused by a unit load in the positive direction at the removed support, and multiplying this answer by the unknown support reaction.

- write consistent deformation equation

The consistent deformation euqation that correspondes to the redundant X_{E}
(the horizontal reaction at support E) is:

_{E0} + *f*_{ee}*X_{E} = 0 (1)

where _{E0} is the deflection at E of the primary structure,
*f*_{ee} is the deflection of the released structure
at E caused by the unit load, and X_{E} is the value of the unknown redundant at E.

- solve consistent deformation equations

The deflection of the released structure under applied loads, _{E0}.

*note: for trusses, only the members which are subjected to the unit load need to be included*

Member | n(k) | N(k) | L(ft) | nNL/AE (k^{2}-ft)/AE |
---|---|---|---|---|

AB | 1 | -25.83 | 4 | -103.33/AE |

BC | 1 | -25.83 | 4 | -103.33/AE |

CD | 1 | -44.167 | 4 | -176.67/AE |

DE | 1 | -44.167 | 4 | -176.67/AE |

Total = _{E0} = |
-560/AE |

The deflection of the released structure at E caused by the unit load, *f*_{ee}.

Member | n (k) | N (k) | L(ft) | nNL/AE (k^{2}-ft)/AE |
---|---|---|---|---|

AB | 1 | 1 | 4 | 4/AE |

BC | 1 | 1 | 4 | 4/AE |

CD | 1 | 1 | 4 | 4/AE |

DE | 1 | 1 | 4 | 4/AE |

Total = f_{ee} = |
16/AE |

Using equation (1), solve for X_{E}:

-560 (ft^{2}-k)/AE + 16 (ft^{2}-k)/AE*X_{E} = 0

X_{E} = 35

Multiply the unit load by this value to get the final reaction. The positive answer indicates that the reaction is in the direction of the applied unit load.

- determine support reactions

Impose the value of the calculated X_{E} along with the other applied loads on the
original truss. Calculate the remaining reactions using the trhee static equilibrium equations,
(F_{x} = 0,
F_{y} = 0 and
M = 0).

Figure 6 - Truss with resultant support reactions

The resulting member forces are now determined by using the method of sections, or method of joints. However, a much easier method is to use superposition and add the effects caused by the redundant load on the released structure.

In this example, the 35 k load will cause only the top chord of the truss to experience a 35 k tensile load in each member. This result can be added to those found in figure 3 above. The final results,

Figure 8 - Truss with final member forces

Contact Dr. Fouad Fanous for more information.