Submission Date: 11/15/2003 TAMUK Truss Company Trusses Made Simple
Table of Contents Introduction..3 Proposal.3 Solution..5 Hand Calculations 5 TRUSS2D 7 NENastran 7 Comparison of Results... 8 Data Analysis.10 Removal of 0 Force Members 10 Modification Structural Elements..11 Conclusion..12 Appendices.13 Appendix A: Tables..13 Appendix B: Figures.....15 Appendix C: Data. 20 2
Introduction A simple truss system will be evaluated using hand calculations and finite element software, TRUSS2D and NENastran. Nodal displacements, elements forces, element stresses, and factors of safety for each element will be calculated. The truss setup is explained in the proposal. Methods of calculation and use of programs are exhibited in the solution section. The results of the calculations are compiled and compared in the comparison of results section. There are some alterations proposed for the current design of the truss. These alterations are discussed in the data analysis section. Proposal An eleven-element truss system has been proposed to support a 150,000-pound force. A two-dimensional representation of the model is shown in Figure 1a and Figure 1b: Figure 1: Two-Dimensional Truss Layout 7ft. 9ft. 9ft. 9ft. 150000 lbs. 9ft. y x AISI 4340 Steel E=30,000,000 psi 3
Figure 1b: Node and Element Identification 6 e11 7 e5 e6 e7 e8 e9 e10 1 e1 2 e2 3 e3 4 e4 5 Figure 1a gives the dimensions of the truss system, the load placement, and the support layout. Figure 1b shows the node identification numbers (1-7) and element identification numbers (e1-e11). Elements e1, e2, e3, e4, e6, and e9 are twelve-inch area rods, and bolded elements e5, e7, e8, e10, and e11 are eighteen-inch area rods. The rods are made from AISI 4340 steel with a modified Young s modulus of 30,000,000 pounds per square inch. A support at node 3 prevents movement in the vertical (y) direction and a support at node 5 prevents movement in both the horizontal and vertical (x and y) directions. The 150,000-pound force will be set at node 1 in the negative y direction. The truss is evaluated for translation at each node, stresses in each member, and reactions at the supports. Failure analysis and a factor of safety are calculated using the Max Stress Failure criteria. Truss analysis is done by hand and by computer programs TRUSS2D, a two dimensional truss analysis program in Excel Visual Basic, and NENastran, a finite element analysis package. 4
Solution The truss is initially evaluated by hand. The results are then checked with TRUSS2D and NENastran. Figure 2 shows the forces in each element. Hand calculations, TRUSS2D input and output, and NENastran output are included in the appendix. Table 2 and Table 3 give a comparison of the hand calculations, the TRUSS2D output, and the NENastran output. Hand Calculations: The forces in each member and the reaction forces are calculated by balancing the forces in each node. An example of balancing forces at a node is shown in Figure 3. First, the forces at the reaction are determined by balancing the forces in the y direction. Then starting at node 1, the force in each member can be found by balancing the forces in the x and y directions. Figure 3 shows the forces for each element. Figure 4 gives the element forces and support reactions in kilo pounds (kip equals 1000 pounds). The maximum failure theory is used to determine the factor of safety for each element. The factor of safety for each element can be calculated by dividing the yield point stress by the beam s axial stress. A yield point stress of 60,000 pounds per square inch in tension and 60,000 pounds per square inch in compression is used for the calculations. If the factor of safety for a beam is calculated to be less than 1, the beam has failed. The factor of safety for each beam is shown in Table 1. 5
Table 1: Element Factor of Safety TAMUK Truss Company Factor of Safety Element Factor of Safety E1 3.20 E2 3.20 E3 3.20 E4 3.20 E5 3.99 E6 N/A E7 3.99 E8 3.99 E9 N/A E10 3.99 E11 2.40 According to Table 3, there were no failures. Elements e6 and e9 have a N/A in the factor of safety field because they are zero force members, and have no axial stress. Table 3 shows that element e11 has the lowest factor of safety at 2.4. The required factor of safety usually depends on types of service and industry standards. A factor of safety of around 1.5 is usually acceptable. Figure 3: Balancing Forces at Nodes e5 Reactive Force θ Given Load 150000 pounds e1 Reactive Force 6
Figure 4: Element Forces and Support Reactions 450 kips 270.416 kips 270.416 kips 270.416 kips 405.624 kips 0 kips 0 kips 225 kips 225 kips 225 kips 225 kips 150 kips 300 kips 150 kips TRUSS2D: TRUSS2D is an Excel Visual Basic program that analyzes two dimensional truss systems only. The number of nodes, number of elements, number and location of degrees of freedom, node location, element orientation with respect to the nodes, area of elements, load and load placement, and material properties are all needed to analyze a truss. TRUSS2D will output nodal displacements, element stresses, and support reactive forces when given the numerical representation of the truss. The element forces can then be calculated by multiplying the element stress by the element area. TRUSS2D input, input legend, and output can be found in Appendix C, Data 1, Data 2, and Data 3 respectively. NENastran: NENastran is a finite element package that is capable of analyzing many different types structures. To analyze this truss, the modeler must be used to draw the structure. First a material is selected. AISI 4340 steel is used in this analysis, but the Young s modulus has been changed to 30,000,000 pounds per square inch. Then the properties for the two different rods are set up, using areas of 12 square inches and 18 square inches for this 7
problem. Next, elements are drawn from node to node. Constraints are then set at node 3 and node 5. Node three is restrained in the y and z directions, and retrained from rotation about the x, y, and z axis. Node 4 is fixed, so there is no translation or rotation. Then the load is set at node 1 in the negative y direction. Figure 6 of Appendix B shows the truss layout in the NENastran Modeler. Once all the information is entered, the analysis can be run. The output can be sent directly to the printer, displayed in a list file, and displayed in a window directly below the viewing window. The outputs include total translation, CBAR forces, and CBAR stresses. NENastran can also display a deformed model of the truss. Deformation of the truss is shown in Figure 7 of appendix B. Note that deformations are exaggerated and not to scale. Comparison of Results The results of the hand calculations, the TRUSS2D and NENastran outputs are compiled in Table 2 and Table 3. The tables show that the forces and stresses are the same from all three methods of calculation. Table 4 displays a comparison of nodal displacements given by TRUSS2D and NENastran. 8
Table 2: Element Force Comparison TAMUK Truss Company Force Comparison Element Hand TRUSS2D NENastran e1-225000.0000-225000.0000-225000.0000 e2-225000.0000-225000.0000-225000.0000 e3-225000.0000-225000.0000-225000.0000 e4-225000.0000-225000.0000-225000.0000 e5 270416.3457 270416.3457 270416.3000 e6 0.0000 0.0000 0.0000 e7-270416.3457-270416.3457-270416.3000 e8-270416.3457-270416.3457-270416.3000 e9 0.0000 0.0000 0.0000 e10 270416.3457 270416.3457 270416.3000 e11 450000.0000 450000.0000 450000.0000 Table 3: Element Stress Comparison TAMUK Truss Company Stress Comparison Element Hand TRUSS2D NENastran e1-18750.0000-18750.0000-18750.0000 e2-18750.0000-18750.0000-18750.0000 e3-18750.0000-18750.0000-18750.0000 e4-18750.0000-18750.0000-18750.0000 e5 15023.1303 15023.1303 15023.1300 e6 0.0000 0.0000 0.0000 e7-15023.1303-15023.1303-15023.1300 e8-15023.1303-15023.1303-15023.1300 e9 0.0000 0.0000 0.0000 e10 15023.1303 15023.1303 15023.1300 e11 25000.0000 25000.0000 25000.0000 Note that the forces and stresses for elements e5, e10, and e11 are all negative. The negative solution shows the members to be in compression, the remaining elements are in tension. Element e11 is the most heavily loaded element at 450,000 lbs. Elements e6 and e9 are both zero force members. 9
The nodal displacements given by TRUSS2D and NENastran differ. Table 4 lists the displacements in the x and y direction for each node given by both programs. Table 4: Nodal Displacements TAMUK Truss Company Translation Comparison TRUSS2D NENastran Node x-displacement y-displacement x-displacement y-displacement 1 2.70E-01-1.41E+00 2.70E-01-1.41E+00 2 2.03E-01-6.06E-01 2.03E-01-6.06E-01 3 1.35E-01-2.41E-06 1.35E-01 0.00E+00 4 6.75E-02 1.01E-01 6.75E-02 1.01E-01 5-5.39E-22 1.21E-06 0.00E+00 0.00E+00 6-1.91E-01-6.06E-01-1.91E-01-6.06E-01 7-1.06E-02 1.01E-01-1.06E-02 1.01E-01 TRUSS2D gives small displacements at node 3 in the y direction and at node 5 in the x and y directions. The Penalty method gives such answers. The displacements are so small, they are considered to be 0. Other than the differences due to the Penalty method, TRUSS2D and NENastran agree on nodal displacements. The maximum x and y displacements are at node 1 with 0.27 and 1.41 inches of movement respectively. Data Analysis Analysis of the data shows that a few design alterations that can be made that will save in material costs, without significantly changing the displacements of the nodes. Removal of 0 Force Members: First, the 0 force members e6 and e9 can be eliminated. These rods carry no force and therefore they are not needed. Figure 8 of Appendix B gives a model of the truss in NENastran without the 0 force members. Table 5 of Appendix A shows a comparison of 10
the displacements for the original structure, which will be referred to as Structure 1, and structure with out the zero force members, which will be referred to as Structure 2. Note that nodes 2 and 4 are no longer present in the modified structure. A model of Structure 2 deformed is shown in Figure 9 of Appendix B. Since the displacements are the same, the stresses in each member will be the same. Modifying Structural Elements: The cross sectional area of the rods may also be decreased in order to save on material costs. The diameter of the 18 square inch rod is 4.787 inches, and the diameter of the 12 square inch area rod is 3.9088 inches. If all of the members were constructed of 4 inch diameter, 12.5664 square inch area, rods, the nodal displacement at the load would only drop 0.3593 inches in the y direction. Which is minimal compared to the overall length of the truss members. The axial stress in the most heavily loaded bar is increased by 43%, but the factor of safety is still 1.68, well above the 1.5 standard. By removing the 0 force members and changing the dimensions of the materials of construction, 5,478 cubic inches of steel can be saved. AISI 4340 steel has a density of 0.283 pounds per cubic inch. Although the body force of the structure was not considered in the analysis, the weight of the structure is reduced by 7.6% of the total weight, or 1,550 pounds. A representation of Structure 3 deformed is given in Figure 10 of Appendix B. Table 6 of Appendix A gives a comparison of the displacements of the Structure 2 and the modified structure with no zero force members, which will now be called Structure 3. Figure 5 shows the new identification numbers for structures 2 and 3. Table 7 of Appendix A compares the stresses in each element, and Table 8 of Appendix A will show the differences in the factors of safety. 11
Figure 5: Node and Element Identification for Structure 2 and Structure 3 6 e7 7 e3 e4 e5 e6 1 e1 3 e2 5 Conclusion: Three methods of calculations are used to analyze the truss system of Project Number 4.14. The hand calculations, TRUSS2D and NENastran outputs all prove to be similar. Analysis software like TRUSS2D and NENastran allow for quick modification of the structure or structural elements. Removing the 0 force members, e6 and e9, proved to have no effect on the structures displacements or stresses. Significantly reducing the cross sectional area of the larger rods and slightly increasing the area of the smaller rods showed to have little effect on the displacement at the load, while reducing the weight by 7.7%. The factors of safety for the modified structure remain high. Analysis of the structure reveals that the truss is safe under its current design conditions, but some minor design changes will save money while keeping the truss safe. 12
Appendix A Tables Table 5: Displacements in Structure 1 and Structure 2 TAMUK Truss Company Translation Comparison Original Truss Zero Force Members Removed Node x-displacement y-displacement x-displacement y-displacement 1 0.2700-1.4100 0.2700-1.4137 2 0.2030-0.6060 N/A N/A 3 0.1350 0.0000 0.1350 0.0000 4 0.0675 0.1010 N/A N/A 5 0.0000 0.0000 0.0000 0.0000 6-0.1910-0.6060-0.1906-0.6056 7-0.0106 0.1010-0.0106 0.1013 Table 6: Displacements in Structure 2 and Structure 3 TAMUK Truss Company Translation Comparison Structure 2 Structure 3 Node x-displacement y-displacement x-displacement y-displacement 1 0.2700-1.4137 0.2578-1.8316 3 0.1350 0.0000 0.1289 0.0000 5 0.0000 0.0000 0.0000 0.0000 6-0.1906-0.6056-0.3053-0.8191 7-0.0106 0.1013-0.0474 0.0967 13
Appendix A Tables Table 7: Stresses in Structure 2 and Structure 3 TAMUK Truss Company Stress Comparison Element Structure 2 Structure 3 e1-18750.00-17904.89 e2-18750.00-17904.89 e3 15023.13 21519.00 e4-15023.13-21519.00 e5-15023.13-21519.00 e6 15023.13 21519.00 e7 25000.00 35809.78 Table 8: Factor of Safety for Structure 2 and Structure 3 TAMUK Truss Company Factor of Safety Element Structure 2 Structure 3 e1 12.80 13.40 e2 12.80 13.40 e3 14.31 9.99 e4 15.98 11.15 e5 15.98 11.15 e6 14.31 9.99 e7 8.60 6.00 14
Appendix B Figures Figure 6: NENastran Model of Structure 1 15
Appendix B Figures Figure 7: NENastran Model of Structure 1 Deformed 16
Appendix B Figures Figure 8: NENastran Model of Structure 2 17
Appendix B Figures Figure 9: NENastran Model of Structure 2 Deformed 18
Appendix B Figures Figure 10: NENastran Model of Structure 3 Deformed 19
Data 1: TRUSS2D Input Appendix C Data << 2D TRUSS ANALYSIS USING BAND SOLVER>> Project Number 4.14 NN NE NM NDIM NEN NDN 7 11 1 2 2 2 ND NL NMPC 3 1 0 Node# X Y 1 0 0 2 108 0 3 216 0 4 324 0 5 432 0 6 108 72 7 324 72 Elem# N1 N2 Mat# Area TempRise 1 1 2 1 12 0 2 2 3 1 12 0 3 3 4 1 12 0 4 4 5 1 12 0 5 1 6 1 18 0 6 6 2 1 12 0 7 6 3 1 18 0 8 3 7 1 18 0 9 7 4 1 12 0 10 7 5 1 18 0 11 6 7 1 18 0 DOF# Displacement 6 0 9 0 10 0 DOF# Load Solve the Problem Delete Results Sheet 2-150000 MAT# E Alpha 1 3.00E+07 0.00E+00 B1 i B2 j B3(Multi-point constr. B1*Qi+B2*Qj=B3) 20
Data 2: TRUSS2D Input Legend Appendix C Data Input Legend: NN NE NM NDIM NEN NDN ND NL NMPC Node# Elem# Mat# Area TempRise DOF# Displacement Load E Alpha Number of nodes Number of elements Number of different materials Number of coordinates per node Number of nodes per element Number of degrees of freedom per node Number of boundary conditions Number of applied component loads Number of applied multipoint constraints Individual node - give x and y coordinate Individual elements - specify from node to node (N1 to N2) Specify which material Give cross sectional area of element Give change in temperature Give degree of freedom at which constraint or load will be applied Give displacement at that node (0 if constrained) Give load at degree of freedom Youngs modulus Coefficient of thermal expansion 21
Data 3: TRUSS2D Output Appendix C Data Results from Program TRUSS2D Project Number 4.14 Node# X-Displ Y-Displ 1 2.70E-01-1.41E+00 2 2.03E-01-6.06E-01 3 1.35E-01-2.41E-06 4 6.75E-02 1.01E-01 5-5.39E-22 1.21E-06 6-1.91E-01-6.06E-01 7-1.06E-02 1.01E-01 Elem# Stress 1-18750 2-18750 3-18750 4-18750 5 15023.13 6 0 7-15023.13 8-15023.13 9 0 10 15023.13 11 25000 DOF# Reaction 6 300000 9 6.7E-11 10-150000 22
Appendix C Data Data 3: NENastran Output for Structure 1 Translation X Y Z Rotation POINT T1 T2 T3 R1 R2 R3 1 2.70E-01-1.41E+00 0 0 0 0 2 2.03E-01-6.06E-01 0 0 0 0 3 1.35E-01 0 0 0 0 0 4 6.75E-02 1.01E-01 0 0 0 0 5 0 0 0 0 0 0 6-1.91E-01-6.06E-01 0 0 0 0 7-1.06E-02 1.01E-01 0 0 0 0 Axial Force ELEMENT AXIAL ID. FORCE TORQUE 1-2.25E+05 0 2-2.25E+05 0 3-2.25E+05 0 4-2.25E+05 0 5 2.70E+05 0 6 0 0 7-2.70E+05 0 8-2.70E+05 0 9 0 0 10 2.70E+05 0 11 4.50E+05 0 Axial Stress ELEMENT AXIAL ID. STRESS 1-1.88E+04 2-1.88E+04 3-1.88E+04 4-1.88E+04 5 1.50E+04 6 0 7-1.50E+04 8-1.50E+04 9 0 10 1.50E+04 11 2.50E+04 23
Appendix C Data Data 4: NENastran Output for Structure 2 Translation X Y Z Rotation POINT T1 T2 T3 R1 R2 R3 1 2.70E-01-1.41E+00 0 0 0 0 3 1.35E-01 0 0 0 0 0 5 0 0 0 0 0 0 6-1.91E-01-6.06E-01 0 0 0 0 7-1.06E-02 1.01E-01 0 0 0 0 Axial Force ELEMENT AXIAL ID. FORCE TORQUE 1-2.25E+05 0 2-2.25E+05 0 3 2.70E+05 0 4-2.70E+05 0 5-2.70E+05 0 6 2.70E+05 0 7 4.50E+05 0 Axial Stress ELEMENT AXIAL ID. STRESS 1-1.88E+04 2-1.88E+04 5 1.50E+04 7-1.50E+04 8-1.50E+04 10 1.50E+04 11 2.50E+04 24
Appendix C Data Data 5: NENastran Output for Structure 3 Translation X Y Z Rotation POINT T1 T2 T3 R1 R2 R3 1 2.58E-01-1.83E+00 0 0 0 0 3 1.29E-01 0 0 0 0 0 5 0 0 0 0 0 0 6-3.05E-01-8.19E-01 0 0 0 0 7-4.74E-02 9.67E-02 0 0 0 0 Axial Stress ELEMENT AXIAL ID. STRESS 1-1.79E+04 2-1.79E+04 3 2.15E+04 4-2.15E+04 5-2.15E+04 6 2.15E+04 7 3.58E+04 25
Data 6: Hand Calculations Appendix C Data Balancing Forces at a Node The vertical component of the reactive force for e5 (see Figure 3) can be found by balancing the forces in the y direction. Equation 1 shows that the summation of the forces in the y direction is equal to 0, which is equal to the negative of the given load plus the y component of e5. This gives a y component value of 150,000 pounds. Theta (θ) is equal to the inverse tangent of the division of the y dimension of the element by x dimension of the element, which is shown in Equation 2. Using Equation 3, the reactive force of e5 can be found by dividing the y component by the sine of theta (θ), giving a force of 270,416 pounds. Then, using Equation 4, the x component of e5 can be found by multiplying the reactive force of e5 by the cosine of theta, which results in a force of 225,000 pounds. Finally, the reactive force of e1 is given by the summation of the forces in the x direction. According to Equation 5, the summation of the forces in the x direction is equal to 0, which is equal to the x component of e5 minus the reactive force of e1. This gives the reactive force of e1 to be 225,000 pounds. Reactive forces, which point away from the node, denote an element in tension, and reaction forces in the direction of the node denote an element in compression. Variations of equations 1 through 5 will solve for the forces in the rest of the elements Equation 1: ΣF y = 0 = -150000 + F e5y F e5y = 150000 pounds Equation 2: θ = tan -1 (y/x) = tan - (71/108) = 33.69 Equation 3: F e5 = F e5y /sin(θ) = 150000/sin(33.69 ) = 270416 pounds Equation 4: F e5x = cos(θ) F e5 = cos(33.69 ) 270416 = 225000 pounds Equation 5: ΣF x = 0 = F e5x - F e1x F e1x = -225000 pounds 26
Data 7: Hand Calculations Appendix C Data 27