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ENCI 451 (5)
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Final

# Virtual Work Make Up Example

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School
University of Calgary
Department
Civil Engineering
Course
ENCI 451
Professor
T.G.Brown
Semester
Fall

Description
Random Virtual Work Example – November 15, 2012 Given the truss in the figure above, find the forces in all the members. This is a simple truss, so the Force Method is sufficient enough to solve this problem. Step 1: Find i, the degree of static indeterminacy or degrees of freedom Two ways to do this: a) Inspection – 2 cross braced sections means i = 2. b) Calculation: Hence two forces releases are needed, so remove the diagonal members HB and HD Step 2: Find the Displacements {D} due to the released structure The axial displacement of a truss member using virtual work is given in the textbook as: Where: N ujthe member forces of the truss due to a VIRTUAL unit load at coordinate j, N = the member forces of the truss due to the released structure (Hence, the two diagonal members we removed SHOULD NOT EXIST in the truss), l = length of each truss member E = Young’s or elastic modulus of the truss member material, a = cross sectional area of the truss member Thus, So you must be saying “OK…so how do we do this?!” Well, ladies and gentlemen, we do the following:  Set E = a = 1, and yes, this is unrealistic because for example, E for steel is 200 GPa and for ice it is as high as 10 GPa which is extremely greater than 1. But, to make our lives easier, just set E = 1.  Find the length of each member. I believe this is trivial enough but there are some who will find this difficult to do!  Now the hard part… o Finding N To find N, we solve for all the forces in the released structure. The released structure is shown below: As you can see, from force and moment equilibrium: We find that the reactions are R A 5.25P and R =B1.75 P. Now we can find the member forces. So do the following: Joint equilibrium at A: Joint equilibrium at E: Then, find the rest of the member forces by:  Joint equilibrium at F  Joint equilibrium at J  Method of sections at GH, GC, and BC  Method of sections at HI, IC, and CD  Joint equilibrium at I  Joint equilibrium at G So then, the forces in each member due to the released structure are: Member N AF -5.25P AB 0 FB 8.75P FG -7P GB -5.25P BC 7P CG -2.916666667P GH -4.666666667P HC 0 CI 2.916666667P HI -4.666666667P CD 2.333333333P ID -1.75P IJ -2.333333333P JD 2.916666667P DE 0 JE -1.75P HB 0 HD 0 o Finding u1 On the same released structure, instead of the real load (7P at joint G), a VIRTUAL load of 1 is applied at Release 1: Analyzing the truss above, the forces in each truss member are then: Member Nu1 AF 0 AB 0 FB 0 FG 0 GB -0.6 BC -0.8 CG 1 GH -0.8 HC -0.6 CI 0 HI 0 CD 0 ID 0 IJ 0 JD 0 DE 0 JE 0 HB 1 HD 0 o Finding u2 Again, instead of the real load (7P at joint G), a VIRTUAL load of 1 is applied at Release 2: Again, the member forces due to the virtual load above are shown below: Member Nu2 AF 0 AB 0 FB 0 FG 0 GB 0 BC 0 CG 0 GH 0 HC -0.8 CI 1 HI -0.6 CD -0.6 ID -0.8 IJ 0 JD 0 DE 0 JE 0 HB 0 HD 1 So knowing now N, Nu1 Nu2 as well as the length of each member, we can 1ind D 2nd D : Member Length N Nu1 Nu2 Nu1*N*L Nu2*N*L AF 0.75 -5.25 0 0 0 0 AB 1 0 0 0 0 0 FB 1.25 8.75 0 0 0 0 FG 1 -7 0 0 0 0 GB 0.75 2.333333333 -0.6 0 -1.05 0 BC 1 7 -0.8 0 -5.6 0 CG 1.25 -2.916666667 1 0 -3.64583 0 GH 1 -4.666666667 -0.8 0 3.733333 0 HC 0.75 0 -0.6 -0.8 0 0 CI 1.25 2.916666667 0 1 0 3.645833 HI 1 -4.666666667 0 -0.6 0 2.8 CD 1 2.333333333 0 -0.6 0 -1.4 ID 0.75 -1.75 0 -0.8
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