CVE 312_Case Study

docx

School

Cleveland State University *

*We aren’t endorsed by this school

Course

312

Subject

Civil Engineering

Date

Dec 6, 2023

Type

docx

Pages

10

Report

Uploaded by HighnessWillpower5296

1 CVE 312 Structural Analysis I Case Study
2 I. Abstract In this essay, they contrasted the Pratt trust and the Howe trust, two well-known truss designs. A truss is a structure that provides support for things like roofs and bridges. A truss is a type of steel beam used to support a bridge. Engineers use tools like RISA to help collect all the data a truss can offer. RISA-3D is a thorough CAD-like drawing environment that makes even the most complex buildings easy to create. Information about axial forces, shear forces, and bending moments is collected with the use of RISA. Results from RISA-3D were quantified for the Pratt trust and the Howe trust. They chose the joint reactions, joint deflections, member forces, and member deflections as their outcomes. The output of those choices was a joint and member with the indicated loads from the Truss model. The objective is to determine which truss bridge design is more effective at dissipating the force of a load, the Howe truss, or the Pratt truss. II. Problem Statement To complete this task, a presupposed that the live load, or P, was 100 kips and the dead load, or P, was 15 kips. Additionally, a presumed that the members are constructed of steel with a yield stress of 45. (KSI). They choose to utilize a critical stress value of 24 for compression members (KSI). They then used RISA 3D to perform a structural analysis on each truss. They had to include a summary in their own words and writing. An illustration of the axial forces, a representation of the member forces and reactions, and the deflected form of each truss. The cross-sectional area of each member was then determined. A consideration of 1 (in 2 ) was the minimum value to be used. A displayed member forces, cross-sectional areas, length, and weight using the steel's 490 (lb./ft 3 ) density as a starting point was to be used. It was then necessary to
3 figure out how much each truss weighed in total. And they needed to determine which truss has the most effective arrangement. III. Methods One must launch the RISA program and open a new file to begin with this task. To start, you must give the Project design a name by clicking global, adding a model title, and entering the required data into each slot. The next step is to choose a unit; in this instance, standard imperial was chosen. Next, adjust the grid using the modify tab. In this instance, the necessary settings were 6@15 and 20@1. The next step was data entry. The length of each joint coordinate was defined and then shown in the design. Then, one roller and one pin were used to define each boundary condition. Then, this needed to be displayed in the drawing. The following stage involved adding the correct joint loads to the truss, which also needed to be made visible in the design. In order to acquire the truss model with joint labels, member labels, and loads shown, data tables for each member force, support responses, deflections, an axial force diagram, and a deflected shape, you must lastly permit the software to run a structural analysis.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
4 IV. Results and Discussion Pratt Truss Membe r Length (ft) Area (in^2) P(Natural force) (lbs.) Yield limit (KSI) Critical limit (KSI) P/A (lbs/in^2) Area (ft^2) Volume (ft^3) Density (lbs/ft^3) Weight (lbs) AB 20 18.5 445(C) 45 24 24 0.128 2.569 490 1259 AC 15 1 0 45 24 0 0.007 0.104 490 51 BC 25 12.5 556.25(T) 45 24 45 0.087 2.170 490 1063 CE 15 7.5 333.75(T) 45 24 45 0.052 0.781 490 383 BD 15 14 333.75(C) 45 24 24 0.097 1.458 490 714 DE 25 7.5 333.75(T) 45 24 45 0.052 1.302 490 638 CD 20 18.5 445(C) 45 24 24 0.128 2.569 490 1259 DF 15 22 534(C) 45 24 24 0.153 2.292 490 1123 EF 20 11 267(C) 45 24 24 0.076 1.528 490 749 EG 15 12 534(T) 45 24 45 0.083 1.250 490 613 FG 25 2.5 111.25(T) 45 24 45 0.017 0.434 490 213 FH 15 25 60.75(C) 45 24 2 0.174 2.604 490 1276 HG 20 7.5 178(C) 45 24 24 0.052 1.042 490 511 GI 15 12 534(T) 45 24 45 0.083 1.250 490 613 HJ 15 25 600.76(C) 45 24 24 0.174 2.604 490 1276 GJ 25 2.5 111.25(T) 45 24 45 0.017 0.434 490 213 IK 15 7.5 333.75(T) 45 24 45 0.052 0.781 490 383 JL 15 22 534(C) 45 24 24 0.153 2.292 490 1123 IJ 20 11 267(C) 45 24 24 0.076 1.528 490 749 IL 25 7.5 333.75(T) 45 24 45 0.052 1.302 490 638 KL 20 18.5 445(C) 45 24 24 0.128 2.569 490 1259 KM 15 1 0 45 24 0 0.007 0.104 490 51 LN 15 14 333.75(C) 45 24 24 0.097 1.358 490 665 KN 25 12.5 556.25(T) 45 24 45 0.087 2.170 490 1063 MN 20 18.5 445(C) 45 24 24 0.128 2.569 490 1259
5 Howe Truss
6 Member Length (ft) Area (in^2) P(Natural force) (lbs.) Yield limit (KSI) Critical limit (KSI) P/A (lbs/in^2) Area (ft^2) Volume (ft^3) Density (lbs/ft^3) Weight (lbs) AB 20 1 0 45 24 0 0.007 0.139 490 68 AC 15 7.417 333.75(T) 45 24 45 0.052 0.773 490 379 BC 25 11.868 534(T) 45 24 45 0.082 2.060 490 1010 CE 15 1 0 45 24 0 0.007 0.104 490 51 BD 15 5.933 267(T) 45 24 45 0.041 0.618 490 303 DE 25 13.906 333.75(C) 45 24 24 0.097 2.414 490 1183 CD 20 1.977 89(T) 45 24 45 0.014 0.275 490 135 DF 15 13.35 600.75(T) 45 24 45 0.093 1.391 490 681 EF 20 22.25 534(C) 45 24 24 0.155 3.090 490 1514 EG 15 1 0 45 24 0 0.007 0.104 490 51 FG 25 13.35 600.75(T) 45 24 45 0.093 2.318 490 1136 FH 15 22.25 534(C) 45 24 24 0.155 2.318 490 1136 HG 20 11.866 534(T) 45 24 45 0.082 1.648 490 808 GI 15 13.906 333.75(C) 45 24 24 0.097 1.449 490 710 HJ 15 1.977 89(T) 45 24 45 0.014 0.206 490 101 GJ 25 5.933 267(T) 45 24 45 0.041 1.030 490 505 IK 15 7.417 333.75(T) 45 24 45 0.052 0.773 490 379 JL 15 1 0 45 24 0 0.007 0.104 490 51 IJ 20 1 0 45 24 0 0.007 0.139 490 68 IL 25 23.177 556.25(C) 45 24 24 0.161 4.024 490 1972 KL 20 13.906 333.75(C) 45 24 24 0.097 1.931 490 946 KM 15 4.635 111.25(C) 45 24 24 0.032 0.483 490 237 LN 15 4.635 111.25(C) 45 24 24 0.032 0.483 490 237 KN 25 13.906 333.75(C) 45 24 24 0.097 2.414 490 1183 MN 20 23.177 556.25(C) 45 24 24 0.161 3.219 490 1577 For both trusses, the length, yield limit, critical limit, and density were all constant.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
7 The total weight of the Pratt truss was 19,192 (lbs.) and the Howe truss was 16,418 (lbs.). The diagonal members are in tension, whilst the vertical members are in compression. As a result, the design can be simplified and more effectively by using less steel in the diagonal elements that are under tension. This has several benefits, including lowering the self-weight, lowering the cost of the structure due to more effective members, and making the structure easier to construct. The Howe Truss, whose diagonal members face the other way, is the closest relative of Pratt Trusses. The members exhibit an inverted compression/tension characteristic as a result. The Howe truss has no force on the center and the tension members are vertical instead. V. Conclusion Although both truss bridges significantly diffused the force more effectively than the beam bridge, the Pratt truss dissipated the load more effectively than the Howe truss. While Pratt truss uses diagonal beams that slope outward from the center of the bridge, here the structural diagonal beams slope toward the bridge center. With this method, the vertical web members of a Howe truss bridge are in tension while the diagonal elements are compressed. Additionally, the beam bridge deflected the most and held the least, whereas the Pratt truss deflected the least and held the most generally. This leads one to the conclusion that a structure is stronger and can support a greater weight if it is more rigid. Because of its stiff construction, the Pratt truss was able to distribute the load's force the most efficiently.
8 VI. Appendix Pratt Truss Figure 1- joint labels, member labels, and loads Figure 2- Axial force diagram Figure 3- Deflected shape
9 Howe Truss Figure 4- joint labels, member labels, and loads Figure 5- Axial force diagram Figure 6- Deflected shape
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
10