deflections - work-energy methods
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Deflections - Work-Energy Methods Virtual Work (Unit Load Method)
Introduction
The method of virtual work, or sometimes referred to as the unit-load method, is one of the several techniques available that can be used to solve for displacements and rotations at any point on a structure.
The following paragraphs briefly describe this concept. Please refer to an introductory text book on structural analysis for a complete description of this approach.
The following relationships are used to calculate:
Deflection (translation) at a point:
(1)
Rotation at a point:
(2)
Where Q is a virtual load applied at the point of interest in the direction of interest (i.e., in the direction of which a displacement needs to be calculated). This Q load is often taken to be unity and must be consistent with the units being used in the analysis (i.e., the load Q is a unit force or a unit moment in the case of calculating a translation and rotation, respectively). The moments M and m are the moments induced in a structure due to the applied "real" loads and the virtual load, Q, respectively. E and I are the Young's modulus and the moment of inertia for the member over which the integration is being evaluated.
The integration to solve for the displacement can be carried out using either direct integration or by utilizing a visual integration method. With direct integration, the equations of M and m for each segment of the structure must be developed for use in the equation,
(3)
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The determination of the moments M and m due to the applied "real" loads and the virtual load respectively can be quite difficult and is prone to error, especially with complex bending moment diagrams. An alternative to this approach is to construct the moment diagrams by using either the method of superposition or the cantilever method (examples for each method are given below).
The visual integration technique is a simplified process that completes the integration of equations (1) and (2) by utilizing the following relationship,
(4)
Where n is the number of segments in the M diagram. The segments are selected and numbered to simplify the integration of equation 4. A is the area of the moment diagram of each segment and h is the respective height of the m diagram at the centroid of each segment of the moment diagram, M.
By using equation (4), the calculation of deflections and rotations becomes a simple matter of addition rather than integration.
IMPORTANT NOTES:
In performing the integration in equation 4 using visual integration, the following rules must be observed.
1) Construct the moment diagram due to the applied loads on the structure.
2) Divide the moment diagram, M, to segments that you can easily be able to calculate the area and locate the center of each segment (see note 5 below). Calculate the area and locate the center of each segment on the M-diagram. Project the location of the center of each area on the m-diagram.
3) Draw the m-diagram due to a virtual load Q. The virtual load Q, has an arbitrary value, most of the time a value of one is used. This load is applied at the point of interest and in the direction of which a displacement is to be calculated. Measure the height, hi, on the moment diagram of the virtual load. Note: Q is a unit force when calculating horizontal or vertical displacement and is a unit moment when calculating rotation.
4) Both moment diagrams must be continuous over the length over which the integration being performed.
5) If the moment diagram due the applied loads or the moment diagram due to the virtual load is not continuous, one MUST divide the integration into segments, each of which is continuous over the integration length. See the following example:
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6) Another alternative to perform the integration is illustrated below:
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Examples
Beam Deflection - Determine the deflection at a point on a beam. - Cantilever Method - Superposition Method
Beam Rotation - Determine the rotation at a point on a beam. - Cantilever Method - Superposition Method
Frame Deflection - Determine the deflection at a point on a frame. - Cantilever Method - Superposition Method
Truss Deflection - Determine the deflection at a point on a truss due to applied loads, temperature changes, and fabrication errors.
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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,
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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
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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.
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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
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§ 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-ft2/EI) Location of centroid from support B (ft)
1. -56x20/EI=-1120/EI X1 = 1/2x20 = 10 2. 1/3x20x-400/EI=-2666.67/EI X2 = 1/4x20 = 5 3. 1/2x20x420/EI=4200/EI X3 = 1/3x20 = 6.67 4. 1/2x6x-36/EI=-108/EI X4 = 1/3x6 = 2
§ determine heights of virtual moment diagram at centroids
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Determine the values - heights (hi) - 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 (hi) on the moment diagram (m). For example, using similar triangles from the shared angle (location of X1, X2, X3 & X4 were determined previously)
Figure 14 - Heights on virtual load diagram
The heights (hi) 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-ft2/EI)
Height (h) from m diagram (k-ft) Ai*hi (k2-ft3/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 k2-ft3/EI.
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The 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 in4, and Q = 1 k, then
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Deflections - Method of Virtual Work Vertical Deflection of a Beam - Superposition
The following example illustrates the steps to be followed to calculate deflections of statically determinate structures using superposition and the method of virtual work.
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. (This problem is identical to the Vertical Deflection of a Beam - Cantilever example, except that the moment diagrams are developed using the method of superposition.)
Figure 1 - Beam structure to analyze
Solution:
§ calculate the support reactions
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Calculate the support reactions (caused by the applied "real" loads) using the following relationships.
Check these reactions by summing the shear forces.
The resulting system,
Figure 2 - Beam structure with support reactions
§ draw moment diagram (M) for the structure under applied "real" loads
Using the method of superposition, draw a moment diagram for each separate load applied to the beam.
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 3 - Moment diagram due to 56 ft-k moment
ii) Moment diagram due to the 2 k/ft applied load,
Figure 4 - Moment diagram due to 2 k/ft applied load
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iii) Moment diagram due to the 6k applied load at end C,
Figure 5 - Moment diagram due to 6 k applied load
Notice that the resultant moment diagram is equal to the sum of these three diagrams.
Figure 6 - 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 7 - Beam structure with virtual unit load applied
§ solve the support reactions due to the virtual load, Q
Following the same procedure used previously, calculate the support reactions (caused by the virtual load).
Sum moments about A and B.
Check these reactions by summing the vertical forces.
The resulting system,
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Figure 8 - Support reactions due to virtual unit load
§ draw virtual moment diagram (m)
The moment diagram due to the virtual load.
Figure 9 - Moment diagram 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. (the locations of X1, X2, X3 & X4 were determined previously)
Area No. Area/EI (k-ft2) Location of centroid from support (ft)
1. 1/2x-56x20/EI=-560/EI X1 = 1/3x20 = 6.67 2. 2/3x20x100/EI=1333.33/EI X2 = 1/2x20 = 10 3. 1/2x20x-36/EI=-360/EI X3 = 1/3x20 = 13.33 4. 1/2x6x-36/EI=-108/EI X4 = 1/3x6 = 2
§ determine heights of virtual moment diagram at centroids
Determine the values - heights (hi) - 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,
.
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Figure 10 - Heights of virtual moment diagram
The heights (hi) 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 (ft2-k)
Height (h) from m diagram (ft-k)
Ai*hi (k2-ft3)
1. -560/EI -2 1120/EI 2. 1333.33/EI -3 -4000/EI 3. -360/EI -4 1440/EI 4. -108/EI -4 432/EI
Total -1008/EI
Since EI is constant throughout the structure, the total deflection at C equals -1008 k2-ft3/EI.
The 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 in4, and Q = 1 k, then
Contact Dr. Fouad Fanous for more information. Last Modified: 08/25/2003 23:44:14
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Deflections - Method of Virtual Work Rotation of a Beam - Cantilever
problem statement
Using the same structure as used in the Beam Deflection examples, determine the rotation at A of the beam shown in the figure below using the Cantilever Method. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire 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.
Figure 1 - Beam structure to analyze
§ calculate the support reactions
Calculate the support reactions (caused by the applied loads) using the following relationships:
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Check these reactions by summing the vertical forces.
The resulting system,
Figure 2 - Beam structure with support reactions
§ draw moment diagrams (M) for the structure under applied "real" loads
Using the cantilever method, fix the structure at joint B and draw the resulting moment diagram induced by the applied "real" loads.
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 3 - Moment diagram due to 56 ft-k moment
ii) Moment diagram due to the 2 k/ft applied load,
Figure 4 - Moment diagram due to 2 k/ft applied load
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iii) Moment diagram due to the 21 k support reaction at A,
Figure 5 - Moment diagram due to 21 k support reaction
iv) Moment diagram due to the 6k applied load at end C
Figure 6 - Moment diagram due to 6 k applied load
Notice that the resultant moment diagram (figure 3 above) is the sum of these four diagrams.
Figure 7 - Resultant moment diagram
§ apply virtual load
Apply the virtual load at the point of interest in the desired direction. In this case, we want to know the rotation at point A. Therefore, apply a unit moment at point A in the positive (clockwise) direction.
Figure 8 - Beam with virtual unit load applied
§ solve the support reactions due to the virtual load
Following the same procedure used previously, calculate the support reactions (caused by the virtual load) using the following relationships:
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Check these reactions by summing the vertical forces.
The resulting system,
Figure 9 - Support reactions due to virtual unit load
§ draw virtual moment diagram (m)
Determine the moment diagram due to the virtual load using the same method as used to find the moment diagrams for the applied loads.
Moment diagram due to the virtual load by using the cantilever method and fixing the structure at joint B.
Figure 10 - Moment diagram 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 (ft2-k/EI) Location of centroid from support (ft)
1. -56x20/EI=-1120/EI X1 = 1/2x20 = 10 2. 1/3x20x-400/EI=-2666.67/EI X2 = 3/4x20 = 15 3. 1/2x20x420/EI=4200/EI X3 = 2/3x20 = 13.33 4. 1/2x6x-36/EI=-108/EI X4 = 1/3x6 = 2
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§ determine heights of virtual moment diagram at centroids
Determine the values - heights (hi) - 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,
Heights (hi) and locations by the cantilever method.
Figure 11 - Heights on virtual moment diagram
§ integrate
Integrate the equation by using the visual integration approach.
Multiply the areas of the "real" moment diagram by the heights found in the virtual moment diagram and add them together.
Area No.
Area (a) from M diagram (ft2-k/EI)
Height (h) from m diagram (ft-k) Ai*hi (ft3-k2/EI)
1. -1120/EI 1 -1120/EI 1. -1120/EI -1/2 560/EI 2. -2666.67/EI 1 -2666.67/EI 2. -2666.67/EI -3/4 2000/EI 3. 4200/EI 1 4200/EI 3. 4200/EI -2/3 -2800/EI 4. -108/EI 0 0/EI
Total 173.33/EI
Since EI is constant throughout the structure, the total rotation at A equals +173.33 ft3-k2/EI.
The positive sign indicates that the rotation is in the same direction as the unit moment applied at A - therefore the rotation is in the clockwise direction.
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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 in4, and Q = 1 ft-k, then
Contact Dr. Fouad Fanous for more information. Last Modified: 08/25/2003 23:44:14
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Deflections - Method of Virtual Work Rotation of a Beam - Superposition
problem statement
Using the same structure as used in the Beam Deflection examples, determine the rotation at A of the beam shown in the figure below using the method of Superposition. The modulus of elasticity (E) and the moment of inertia (I) are constant for the entire 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.
Figure 1 - Beam structure to analyze
§ calculate the support reactions
Calculate the support reactions (caused by the applied loads) using the following relationships:
Check these reactions by summing the vertical forces.
The resulting system,
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Figure 2 - Beam structure with support reactions
§ draw moment diagrams (M) for the structure under applied "real" loads
Using the method of superposition, draw a moment diagram for each separate load applied to the beam.
The resulting moment diagram can then be calculated by solving for each applied load separately and adding the results.
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 3 - Moment diagram due to 56 ft-k moment
ii) Moment diagram due to the 2 k/ft applied load,
Figure 4 - Moment diagram due to 2 k/ft applied load
iii) Moment diagram due to the 6k applied load at end C,
Figure 5 - Moment diagram due to 6 k applied load
Notice that the resultant moment diagram is equal to the sum of these three diagrams.
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Figure 6 - Resultant moment diagram
§ apply virtual load
Apply the virtual load at the point of interest in the desired direction. In this case, we want to know the rotation at point A. Therefore, apply a unit moment at point A in the positive (clockwise) direction.
Figure 7 - Beam structure with virtual unit load applied
§ solve the support reactions due to the virtual load
Following the same procedure used previously, calculate the support reactions (caused by the virtual load) using the following relationships:
Check these reactions by summing the vertical forces.
The resulting system,
Figure 8 - Reactions due to virtual unit load
§ draw virtual moment diagram (m)
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Determine the moment diagram due to the virtual load using the same method as used to find the moment diagrams for the applied loads.
Moment diagram due to the virtual load using superposition.
Figure 9 - Moment diagram 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-ft2/EI) Location of centroid from support (ft)
1. 1/2x-56x20/EI=-560/EI X1 = 1/3x20 = 6.67 2. 2/3x20x100/EI=1333.33/EI X2 = 1/2x20 = 10 3. 1/2x20x-36/EI=-360/EI X3 = 1/3x20 = 13.33 4. 1/2x6x-36/EI=-108/EI X4 = 1/3x6 = 2
§ determine heights of virtual moment diagram at centroids
Determine the values - heights (hi) - 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,
Heights and locations by superposition.
Figure 10 - Heights of virtual moment diagram
§ integrate
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Integrate the equation by using the visual integration approach.
Multiply the areas of the "real" moment diagram by the heights found in the virtual moment diagram and add them together.
Area No.
Area (a) from M diagram (k-ft2/EI)
Height (h) from m diagram (ft-k) Ai*hi (k2-ft3/EI)
1. -560/EI 2/3 -373.33/EI 2. 1333.33/EI 1/2 666.67/EI 3. -360/EI 1/3 -120/EI 4. -108/EI 0 0/EI
Total 173.33/EI
Since EI is constant throughout the structure, the total rotation at A equals +173.33 k2-ft3/EI.
The positive sign indicates that the rotation is in the same direction as the unit moment applied at A - therefore the rotation is in the clockwise direction.
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 in4, and Q = 1 ft-k, then
Contact Dr. Fouad Fanous for more information. Last Modified: 08/25/2003 23:44:15
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Deflections - Method of Virtual Work Horizontal Deflection of a Frame - Cantilever
The following example utilizes the cantilever method to determine the deflection of a frame.
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 horizontal displacement at D of the frame shown (indicated by XD in Figure 1).
Figure 1 - Beam structure to analyze
Solution:
§ calculate the support reactions
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Calculate the support reactions using statics. Positive moment is in the clockwise direction
Check these reactions by summing the moments at any point on the structure.
The resulting system,
Figure 2 - Support reactions due to applied loads
§ determine moment diagrams (M) for the structure under applied loads
Construct the moment diagrams caused by the applied loads utilizing the cantilever method.
The final moment diagram under the applied loads is;
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Figure 3 - Moment diagram due to applied loads
This diagram can be determined by using statics, or the cantilever method can be used.
In the cantilever method, a point on the structure is selected where a single fixed support can be used to replace all the degrees of freedom in the structure. This method is only useful in statically determinate structures.
In this example, point B is selected and a fixed support is inserted at this location.
The loaded system with a fixed support at point B.
Figure 4 - Frame structure fixed at B
The resulting moment diagram can be calculated by solving for each applied load separately and adding the results. This can be done by using superposition of the loads on the members acting as cantilevers off of support B.
Note: The centroid of each area is indicated by the numbered arrow and dot.
i) Moment diagram due to the 1 k/ft distributed load,
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Figure 5 - Moment diagram due to 1 k/ft load Area: 1/3x(Base)x(Height) = 1/3 x 15 ft x -112.5 ft-k = -562.5 ft2-k Location of centroid: 1/4x(Base) = 1/4 x 15 ft = 3.75 ft from support B
ii) Moment diagram due to the 15 k support reaction at A,
Figure 6 - Moment diagram due to 15 k support reaction Area: 1/2x(Base)x(Height) = 1/2 x 15 ft x 225 ft-k = 1687.5 ft2-k Location of centroid: 1/3x(Base) = 1/3 x 15 ft = 5 ft from support B
iii) Moment diagram due to the 20 k concentrated load at C,
Figure 7 - Moment diagram due to 20 k applied load Area: 1/2x(Base)x(Height) = 1/2 x 15 ft x -300 ft-k = -2250 ft2-k Location of centroid: 1/3x(Base) = 1/3 x 15 ft = 5 ft from support B
iv) Moment diagram due to the 13.75 k support reaction at D,
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Figure 8 - Moment diagram due to 13.75 k support reaction Area: 1/2x(Base)x(Height) = 1/2 x 30 ft x 412.5 ft-k = 6187.5 ft2-k Location of centroid: 1/3x(Base) = 1/3 x 30 ft = 10 ft from support B
The sum of these four diagrams equals the total resultant for the structure.
Figure 9 - Resultant moment diagram
Notice that the answer at both sides of joint B is equal, i.e.;
225-112.5 = 112.5 ft-k 412.5-300 = 112.5 ft-k
§ apply virtual load and solve support reactions
Apply the virtual load at the point of interest in the desired direction. In this case, we want to know the horizontal deflection at point D. Therefore, apply a unit load at point D in the horizontal direction. (see Figure 3 below)
Figure 10 - Frame structure with unit load applied
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Following the same procedure used for the loaded structure, calculate the support reactions (caused by the virutal load). Positive moment is in the clockwise direction
note: the value of XA is negative because the resulting force acts in the opposite direction to how it is drawn in Figure 10, above.
Check these reactions by summing the moments at any point on the structure.
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.
The cantilever structure;
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Figure 12 - Cantilever structure with loads and reactions applied
i) Moment diagram due to the reaction load at A,
Figure 13 - Moment diagram due to reaction loads at A
ii) Moment diagram due to the reaction load at D,
Figure 14 - Moment diagram due to reaction loads at D
Again, notice that the moments on both sides of joint B are equal.
§ determine heights of virtual moment diagram at centroids
Determine the values - heights (hi) - 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 to determine the deflection.
Using the locations of the centroids, determined previously, determine the heights (hi) on the virtual moment diagram (m) at these locations.
Area No. Location from B Height (hi) (ft-k) 1. 3.75 ft -11.25
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2. 5 ft -10 3. 5 ft -12.5 4. 10 ft -10
Locations of heights (hi);
Figure 15 - Heights on virtual moment diagram
§ 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-ft2)/EI
Height (hi) from m diagram (k-ft)
Ai*hi (k2-ft3)/EI
1. -562.5 -11.25 6328.125/EI 2. 1687.5 -10 -16875/EI 3. -2250 -12.5 28125/EI 4. 6187.5 -10 -61875/EI
Total -44296.875/EI
Since EI is constant throughout the structure, the total horizontal deflection at D equals -44296.875 k2-ft3/EI.
The negative sign indicates that the displacement is opposite to the direction of the unit load that was applied at D - therefore the deflection is to the right.
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 in4, and Q = 1 k, then;
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Deflections - Method of Virtual Work Horizontal Deflection of a Frame - Superposition
The following example utilizes the superposition method to determine the deflection of a frame.
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 horizontal displacement at D of the frame shown (indicated by XD in Figure 1).
Figure 1 - Beam structure to analyze
Solution:
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§ calculate the support reactions
Calculate the support reactions using statics. Positive moment is in the clockwise direction
Check these reactions by summing the moments at any point on the structure.
The resulting system,
Figure 2 - Support reactions due to applied loads
§ determine moment diagrams (M) for the structure under applied loads
Construct the moment diagrams caused by the applied loads utilizing the superposition method.
The final moment diagram under the applied loads is;
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Figure 3 - Moment diagram due to applied loads
This diagram can be determined by using statics, or a simplified method using superposition can be used.
To get the final moment diagram by superposition, start by selecting one of the applied loads, determine the support reactions due to the load and drawing the resulting moment diagram.
i) for the 20 k applied load at C,
Figure 4 - Frame structure with 20 k load applied
The resulting moment diagram,
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Figure 5 - Moment diagram due to 20 k load
ii) for the 1 k/ft applied load between AB,
Figure 6 - Frame structure with 1 k/ft load applied
The resulting moment diagram,
Figure 7 - Moment diagram due to 1 k/ft load
This diagram can be simplified by drawing the effects of each load separately, i.e.;
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Figure 8 - Moment diagram due to 1 k/ft load (by parts)
The sum of these diagrams are equal the total resultant for the structure.
Figure 9 - Resultant moment diagram
§ apply virtual load and solve support reactions
Apply the virtual load at the point of interest in the desired direction. In this case, we want to know the horizontal deflection at point D. Therefore, apply a unit load at point D in the horizontal direction. (see figure 3 below)
Figure 10 - Frame stucture with unit load applied
Following the same procedure used for the loaded structure, calculate the support reactions (caused by the virutal load). Positive moment is in the clockwise direction
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Check these reactions by summing the moments at any point on the structure.
The resulting system,
Figure 11 - Support reactions due to virtual 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) Moment diagram due to the unit load at D,
Figure 12 - Moment diagram due to virtual unit load
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Notice that the moments on both sides of joint B are equal.
§ determine heights of virtual moment diagram at centroids
Determine the values - heights (hi) - 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 to determine the deflection.
Using this locations of the centroids, determined previously, determine the heights (hi) on the virtual moment diagram (m) at these locations.
Area No. Location from B Height (hi) (ft-k) 1. 7.5 ft -7.5 2. 5 ft -10 3. 15 ft -7.5 4. 10 ft -10
Locations of heights (hi);
Figure 13 - Heights on virtual moment diagram
§ integrate
Integrate the equation , by using the visual integration approach.
Multiply the areas of the "real" moment diagram by the heights (hi) of the virtual moment diagram and add them together.
Area No.
Area (A) from M diagram (k-ft2)
Height (hi) from m diagram (ft-k)
Ai*hi (k2-ft3)/EI
1. 281.25 -7.5 -2109.375 2. 843.75 -10 -8437.5
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3. 2250 -7.5 -16875 4. 1687.5 -10 -16875
Total -44296.875/EI
Since EI is constant throughout the structure, the total horizontal deflection at D equals -44296.875 k2-ft3/EI.
The negative sign indicates that the displacement is opposite to the direction of the unit load that was applied at D - therefore the deflection is to the right.
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 in4, and Q = 1 k then;
Contact Dr. Fouad Fanous for more information. Last Modified: 08/25/2003 23:44:16
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Deflections - Method of Virtual Work Deflection of a Truss
The virtual work method can be used to determine the deflection of trusses. We know from the principle of virtual work for trusses that the deflection can be calculated by
the equation with n equal to the virtual force in the member and equal to the change in length of the member. Therefore, the deflection of a truss due to any condition that causes a change in length of the members can be calculated. This change in length can be caused by the applied loads acting on each member, temperature changes, and by fabrication errors.
Axial Deformation:
From statics we know how to determine member forces in a truss by using either the method of joints or the method of sections. Once these forces are known we can determine the axial deformation of each member by using the equation:
The equation for the deflection can be modified with this value for .
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 load, L is the length, A is the area, and E represents Young's Modulus of Elasticity.
Temperature Changes:
The axial deformation of a truss member of length L due to a change in temperature of is given by:
where is the coefficient of thermal expansion.
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The equation for the deflection is then modified with this value for .
where j is the number of members experiencing temperature change and n is the force in the member due to the virtual load.
Fabrication Errors:
In the case of fabrication errors, the deformation of each member is known. Therefore, the original equation for deflection of a truss can be modified.
where k is the number of members undergoing fabrication errors and n is the force in the member due to the virtual load and is the change in length of the member due to fabrication errors.
The total deflection of a truss is made up of the sum of all of these cases.
This equation is now used to find the deflection of a truss. Please refer to an introductory text book on structural analysis for a complete description of this approach.
problem statement
Using the method of virtual work, determine the vertical deflection at joint G in the truss below, under the loading conditions show in figures i), ii), and iii).
The member properties are A=2 in2 and E=29x103 ksi.
The truss is subjected to the following applied loads:
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i) Figure 1 - Truss structure to analyze
And the following fabrication errors are present:
ii) Figure 2 - Fabrication errors present
NOTE: TO PREVENT ERRORS, CALCULATE THE INFLUENCE OF EACH CASE INDEPENDENTLY AND THEN ADD THE RESULTS AT THE END
§ calculate the support reactions, due to the applied loads
Figure 3 - Frame structure with applied loads
Calculate the support reactions (caused by the applied loads) by summing the moments about A and E: (answers in Kips)
Check these reactions by summing vertical and horizontal forces:
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The resulting system,
Figure 4 - Support reactions due to applied loads
§ use the method of joints to determine the force in each member, due to the applied loads
For equilibrium at joint A,
Figure 5 - Joint equilibrium at A
Sum vertical and horizontal forces to determine the force in each member, (Kips)
Remember that in the method of joints, a joint reaction is in the opposite direction to how the force acts on the member. Therefore, member AB is in compression.
Continue this method for each joint in the structure.
Truss diagram with internal forces due to applied loads,
Figure 6 - Truss member reactions
§ apply virtual load at G
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Apply the virtual load at the point of interest in the desired direction. In this case, we want to know the deflection at point G. Therefore, apply a unit load at point G.
Figure 7 - Truss with virtual unit load applied
§ solve the support reactions due to the virtual load
Following the same procedure used previously, calculate the support reactions (caused by the virtual load).
The resulting system,
Figure 8 - Support reactions due to virtual unit load
§ use method of joints to determine virtual force in each member
Use the method of joints as illustrated in Step 2 to determine the member results due to the unit virtual load. Add the results to your existing table:
Truss diagram with internal forces due to virtual load,
Figure 9 - Internal forces due to virtual unit load
§ calculate the deflection
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The deflection of the truss can now be determined by completing the equation:
For the case of Axial Deformation
Member n(k) N(k) L(in) AE (in2-ksi) nNL/AE (in-k)
AB -0.67 -33.33 48 58000 0.0184 BC -0.67 -33.33 48 58000 0.0184 CD -0.67 -46.66 48 58000 0.0257 DE -0.67 -46.66 48 58000 0.0257 AF 0.83 41.67 60 58000 0.0359 BF 0 -10 36 58000 0 CF -0.83 -25 60 58000 0.0216 FG 1.33 53.33 48 58000 0.0589 CG 1 0 36 58000 0 CH -0.83 -8.33 60 58000 0.0072 GH 1.33 53.33 48 58000 0.0589 DH 0 -30 36 58000 0 HE 0.83 58.33 60 58000 0.0503
Total 0.3209
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For the case of Fabrication Error
Member n(k) Change in
Length ( )(in) n( )(k-in)
AB -0.67 + 0.4 -0.268 FG 1.33 + 0.6 0.798 HE 0.83 - 0.3 -0.249
Sum 0.281
Since there were no temperature effects included in this example, the total deflection at point G is the sum of these two results.
(1 k)( ) = 0.281 in-k + 0.321 in-k = 0.602 in-k = 0.602 in-k / 1 k = 0.602 in
The positive answer of 0.602 in indicates that the structure will deflect down in the direction of the virtual load