*Deflections - Method of Virtual Work*

Vertical Deflection of a Beam - Cantilever

The following example utilizes the cantilever method to determine the "real" and virtual moment diagrams used in the calculation of deflections of a beam.

**Note: The colors of the loads and moments are used to help indicate the contribution of each force to the deflection or rotation being calculated. The moment diagrams show the moments induced by a load using the same color as the load.**

problem statement

Determine the vertical displacement at end

*C*of the beam shown in the figure below. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire beam.Figure 1 - Beam structure to analyze

Solution:

- calculate the support reactions

Calculate the support reactions (caused by the applied "real" loads) using the following relationships.

Check these reactions by summing the forces in the vertical direction.

The resulting system,

Figure 2 - Beam structure with reactions

- draw shear (V) and moment diagrams
*(M)*for the structure under applied "real" loads

The resultant shear and moment diagrams can be determined using statics (see figures below).

Figure 3 - Resultant shear and moment diagrams

In this example we will use the cantilever method find an equivalent moment diagram in order to carry out the required integration.

To construct the moment diagrams caused by the applied "real" loads utilizing the cantilever method, a point on the structure is selected and a fixed support is assumed at this location. In this example, point

*B*is selected and a fixed support is inserted (see figure below).Notice that all reaction forces are applied as loads on the structure with the assumed fixed support at

*B*.Figure 4 - Cantilever beam structure

Plot the moment diagram for each applied load separately, i.e., by parts. The final results can then be obtained by utilizing the method of superposition i.e., by summing the contribution of each individual load to the displacement being calculated. This method is applicable since the structure is assumed to be elastic and the deflections are small.

Note: The centroid of each area is indicated by the numbered arrow and dot.

i) Moment diagram due to the 56 ft-k concentrated moment at

*A*,Figure 5 - Moment diagram due to 56 ft-k moment

ii) Moment diagram due to the 2 k/ft applied load,

Figure 6 - Moment diagram due to 2 k/ft applied load

iii) Moment diagram due to the 21 k support reaction at

*A*,Figure 7 - Moment diagram due to 21k support reaction

iv) Moment diagram due to the 6k applied load at end

*C*Figure 8 - Moment diagram due to 6k applied load

Notice that the resultant moment diagram (figure 3 above) is the sum of these four diagrams.

Figure 9 - Resultant moment diagram

- apply virtual load, Q

Apply the virtual load at the point of interest in the desired direction. In this case, apply a unit load at point

*C*in the vertical direction. (see figure below)Figure 10 - Beam with virtual load applied

- solve the support reactions due to the virtual load, Q

Following the same procedure as used previously, calculate the support reactions (caused by the virtual load).

Sum the moments about A and B.

Check these reactions by summing the vertical forces.

The resulting system,

Figure 11 - Support reactions due to unit load

- draw virtual moment diagram
*(m)*

Determine the moment diagram due to the virtual load using the same procedure used to draw the "real" moment diagram i.e., with a fixed support assumed at point

*B*.Figure 12 - Virtual unit load on cantilever structure

The resulting moment diagram due to the virtual load.

Figure 13 - Moment diagram on cantilever structure due to virtual unit load

- calculate areas and centroids

Once the "real" moment diagrams are determined, calculate the area enclosed by each moment diagram and determine the location of the centroid of each of these areas.

Area No. | Area/EI (k-ft^{2}/EI) | Location of centroid from support B (ft) |

1. | -56x20/EI=-1120/EI | X_{1} = 1/2x20 = 10 |

2. | 1/3x20x-400/EI=-2666.67/EI | X_{2} = 1/4x20 = 5 |

3. | 1/2x20x420/EI=4200/EI | X_{3} = 1/3x20 = 6.67 |

4. | 1/2x6x-36/EI=-108/EI | X_{4} = 1/3x6 = 2 |

- determine heights of virtual moment diagram at centroids

Determine the values - heights (h

_{i}) - on the virtual moment diagram (*m*) at the centroids of the moments due to the real loads. This is needed to carry out the integration by using the equation given in the introduction,Proportions can be used to determine these heights (h

_{i}) on the moment diagram (*m*). For example, using similar triangles from the shared angle*(location of X*_{1}, X_{2}, X_{3}& X_{4}were determined previously)Figure 14 - Heights on virtual load diagram

The heights (h

_{i}) are shown in the figure above at the locations of the centroids of the corresponding areas from the moment diagrams (*M).*- integrate

Integrate the equation , by using the visual integration approach.

Multiply the areas of the "real" moment diagram by the heights of the virtual moment diagram and add them together.

Area No. | Area (a) from M diagram (k-ft ^{2}/EI) | Height (h) from m diagram (k-ft) | A_{i}*h_{i} (k^{2}-ft^{3}/EI) |

1. | -1120/EI | -3 | 3360/EI |

2. | -2666.67/EI | -4.5 | 12000/EI |

3. | 4200/EI | -4 | -16800/EI |

4. | -108/EI | -4 | 432/EI |

Total | -1008/EI |

Since EI is constant throughout the structure, the total deflection at

*C*equals**-1008 k**.^{2}-ft^{3}/EIThe negative sign indicates that the displacement is opposite to the direction of the unit load that was applied at

*C*- therefore the deflection is upward.If values of E and I are specified, the vertical deflection at C in inches can be determined. For example, let E = 29,000 ksi, I = 144 in

^{4}, and Q = 1 k, then
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