Study Guides
(238,528)

Canada
(115,195)

University of Calgary
(1,210)

Civil Engineering
(33)

ENCI 451
(5)

T.G.Brown
(1)

Final

# Virtual Work Make Up Example

Unlock Document

University of Calgary

Civil Engineering

ENCI 451

T.G.Brown

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

More
Less
Related notes for ENCI 451