team homework assignment #5 due date: thursday – …
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Team Homework Assignment #5
Due Date: Thursday – February 23, 2017
1. For problem 2 in homework assignment #4, calculate the vertical displacement at the internal hinge.
2. For the frame problem (problem 3) in homework assignment 4, calculate the horizontal
displacement at A and the rotation of joint C. Verify your hand-calculated solutions using a structural analysis computer program. Submit a printed copy of the output file (one per team). NOTE: Your virtual force calculations ignore axial deformation. To simulate zero axial deformation in your program, input a very large area (1000 x the moment of inertia) for the members.
3. For the given truss structure to the right, calculate the horizontal and vertical displacements at joint D. Verify your hand- calculated solution for problem 3 using a structural analysis computer program. Write your hand-calculated results next to the computer-generated results at each node or joint (one computer output per team). THEY SHOULD MATCH!
SEE HOMEWORK ASSIGNMENT #2 FOR SUGGESTED MATERIAL PROPERTY DATA.
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March 5, 2016Designer : 3:51 PMSolution
CE 382 Homework Assignment #5 Problem 2 - Frame Analysis Checked By:_____
GlobalDisplay Sections for Member Calcs 2
Joint CoordinatesJoint Label X Coordinate Y Coordinate
(m) (m)A 0 0B 0 12C 10 12
Boundary ConditionsJoint Label X Translation Y Translation Rotation
(kN/mm) (kN/mm) (kN-m/rad)A ReactionC Reaction Reaction
Member Data Moment of Elastic End Releases
Member Label I Joint J Joint Area Inertia Modulus I-End J-End Lengthmm^2 mm^4 MPa m
AB A B 4e+11 4e+8 200000 12BC B C 4e+11 4e+8 200000 10
Member Point LoadsMember Label I Joint J Joint Direction Magnitude Location
(kN, kN-m) (m or %)BC B C Y -90 5
Member Distributed LoadsMember Label Direction Start Magnitude End Magnitude Start Location End Location (kN/m, F) (kN/m, F) (m or %) (m or %)
AB X 25 25 0 0
Joint DisplacementsJoint Label X Translation Y Translation Rotation
(mm) (mm) (radians)A 1625.625 0 1.58e-1B 0 0 6.797e-2C 0 0 -3.047e-2
ReactionsJoint Label X Force Y Force Moment
(kN) (kN) (kN-m)A 0 225 0C -300 -135 0
Totals: -300 90
Member Section ForcesMember Label Section Axial Shear Moment
(kN) (kN) (kN-m)AB 1 225 0 0
2 225 -300 -1800BC 1 300 225 -1800
2 300 135 0
RISA-2D Educational Version 1.0 Page 1 [C:\Users\lca232\Desktop\Frame.r2e]
March 5, 2016Designer : 3:59 PMSolution
CE 382 Homework Assignment #5 Problem 3 - Truss Analysis Checked By:_____
GlobalDisplay Sections for Member Calcs 2
Joint CoordinatesJoint Label X Coordinate Y Coordinate
(ft) (ft)A 0 16B 12 0C 24 16D 12 32E 12 16
Boundary ConditionsJoint Label X Translation Y Translation Rotation
(k/in) (k/in) (k-ft/rad)A Reaction ReactionB Reaction
Member Data Moment of Elastic End Releases
Member Label I Joint J Joint Area Inertia Modulus I-End J-End Lengthin^2 in^4 ksi ft
AB A B 10 500 30000 PIN PIN 20BC B C 10 500 30000 PIN PIN 20CD C D 10 500 30000 PIN PIN 20AE A E 10 500 30000 PIN PIN 12EC E C 10 500 30000 PIN PIN 12BE B E 10 500 30000 PIN PIN 16ED E D 10 500 30000 PIN PIN 16
Joint Loads/Enforced DisplacementsJoint Label [L]oad or Direction Magnitude
[D]isplacement (k, k-ft, in, rad)D L X 15
Joint DisplacementsJoint Label X Translation Y Translation Rotation
(in) (in) (radians)A 0 0 0B -.033 0 0C .029 -.072 0D .192 .026 0E .014 .013 0
ReactionsJoint Label X Force Y Force Moment
(k) (k) (k-ft)A -15 -20 0B 0 20 0
Totals: -15 0
Member Section ForcesMember Label Section Axial Shear Moment
(k) (k) (k-ft)AB 1 25 0 0
RISA-2D Educational Version 1.0 Page 1 [C:\Users\lca232\Desktop\Truss.r2e]
March 5, 2016Designer : 3:59 PMSolution
CE 382 Homework Assignment #5 Problem 3 - Truss Analysis Checked By:_____
Member Section ForcesMember Label Section Axial Shear Moment
(k) (k) (k-ft)2 25 0 0
BC 1 25 0 02 25 0 0
CD 1 25 0 02 25 0 0
AE 1 -30 0 02 -30 0 0
EC 1 -30 0 02 -30 0 0
BE 1 -20 0 02 -20 0 0
ED 1 -20 0 02 -20 0 0
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